Iranian University Students Mathematics Competitions, 1973-2007 (Texts and Readings in Mathematics)

Iranian University Students Mathematics Competitions, 1973-2007 Bamdad R. Yahaghi School of Mathematics, Institute for Studies in Theoretical Physics and Mathematics (IPM), P.O. Box: 19395-5746 Tehran, Iran E-mail address: [email protected], [email protected]

Contents Preface

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Chapter 1. Problems 1.1. First Competition, University of Tehran, March 1973 1.2. Second Competition, Shiraz (former Pahlavi) University, March 1974 1.3. Third Competition, (former Jondi Shapour) University of Ahwaz, March 1975 1.4. Fourth Competition, University of Tabriz, March 1976 1.5. Fifth Competition, Sharif (former Aryamehr) University of Technology, March 1977 1.6. Sixth Competition, The University of Isfahan, March 1978 1.7. Seventh Competition, Ferdowsi University of Mashhad, March 1980 1.8. Eighth Competition, Shiraz University, March 1984 1.9. Ninth Competition, Tehran Teacher Training (Tarbiat Moallem) University, September 1985 1.10. Tenth Competition, University of Sistan and Baluchestan, March 1986 1.11. Eleventh Competition, The University of Birjand, March 1987 1.12. Twelfth Competition, Guilan University , March 1988 1.13. Thirteenth Competition, University of Tehran, March 1989 1.14. Fourteenth Competition, The University of Isfahan, March 1990 1.15. Fifteenth Competition, Ferdowsi University of Mashhad, March 1991 1.16. Sixteenth Competition, Razi (Rhazes or Rasis) University of Kermanshah, March 1992 1.17. Seventeenth Competition, Shahid Beheshti (former National) University, March 1993 1.18. Eighteenth Competition, Sharif University of Technology, March 1994 1.19. Nineteenth Competition, University of Kerman, March 1995 1.20. Twentieth Competition, Sharif University of Technology, February 1996 1.21. Twenty First Competition, University of Tehran, March 1997 1.22. Twenty Second Competition, University of Ahwaz, March 1998 1.23. Twenty Third Competition, Sharif University of Technology, March 1999 1.24. Twenty Fourth Competition, Khajeh Nasir Toosi University of Technology, May 2000 1.25. Twenty Fifth Competition, Imam Khomeini International University of Qazvin, May 2001 v

1 1 3 6 8 9 10 12 13 14 16 17 18 20 21 23 24 25 26 27 28 28 29 36 37 39

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1.26. Twenty Sixth Competition, Institute for Advanced Studies in Basic Sciences (IASBS) of Zanjan, May 2002 1.27. Twenty Seventh Competition, Bu-Ali Sina (Avecina) University of Hamedan, May 2003 1.28. Twenty Eighth Competition, Sharif University of Technology, May 2004 1.29. Twenty Ninth Competition, University of Mazandaran in Babolsar, May 2005 1.30. Thirtieth Competition, Tafresh University, May 2006 1.31. Thirty First Competition, Ferdowsi University of Mashhad, May 2007

40 41 43 45 46 48

Chapter 2. Solutions 2.1. First Competition 2.2. Second Competition 2.3. Third Competition 2.4. Fourth Competition 2.5. Fifth Competition 2.6. Sixth Competition 2.7. Seventh Competition 2.8. Eighth Competition 2.9. Ninth Competition 2.10. Tenth Competition 2.11. Eleventh Competition 2.12. Twelfth Competition 2.13. Thirteenth Competition 2.14. Fourteenth Competition 2.15. Fifteenth Competition 2.16. Sixteenth Competition 2.17. Seventeenth Competition 2.18. Eighteenth Competition 2.19. Nineteenth Competition 2.20. Twentieth Competition 2.21. Twenty First Competition 2.22. Twenty Second Competition 2.23. Twenty Third Competition 2.24. Twenty Fourth Competition 2.25. Twenty Fifth Competition 2.26. Twenty Sixth Competition 2.27. Twenty Seventh Competition 2.28. Twenty Eighth Competition 2.29. Twenty Ninth Competition 2.30. Thirtieth Competition 2.31. Thirty First Competition

51 51 56 64 71 74 83 86 91 96 109 115 119 125 134 143 149 152 155 160 165 172 176 179 184 196 202 205 210 215 220 226

Chapter 3. Problem Index First Competition Second Competition Third Competition

235 235 235 236

CONTENTS

Fourth Competition Fifth Competition Sixth Competition Seventh Competition Eighth Competition Ninth Competition Tenth Competition Eleventh Competition Twelfth Competition Thirteenth Competition Fourteenth Competition Fifteenth Competition Sixteenth Competition Seventeenth Competition Eighteenth Competition Nineteenth Competition Twentieth Competition Twenty First Competition Twenty Second Competition Twenty Third Competition Twenty Fourth Competition Twenty Fifth Competition Twenty Sixth Competition Twenty Seventh Competition Twenty Eighth Competition Twenty Ninth Competition Thirtieth Competition Thirty First Competition Classification of problems Index

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236 236 237 237 237 238 238 239 239 240 240 240 241 241 241 241 242 242 242 242 243 243 243 244 244 244 245 245 246 247

Preface taa khod raa be tchizi nadaadi be kol-liyat, aan tchiz sa’b-o doshvaar minomaayad. tchoon khod raa be kol-liyat be tchizi daadi, digar doshvaari namaanad. [Until you do not devote yourself to a task totally, that task looks hard and unreachable. As you devote yourself to a task totally, there remains no difficulty.] –Shamsoddin Tabrizi International mathematical competitions have gained great popularity in recent years. They provide the first test for a young person’s mathematical prowess, and success in these competitions often translates into gaining admittance to the best research institutions for graduate work. For this reason some colleges and even high schools have organized special classes for training mathematically talented students for these competitions. The purpose of this text is to provide a selection of mathematical problems that are not only suitable for special college level (and occasionally even at high school level) courses designed for these competitions, but can significantly improve the level of mathematical sophistication of students who are intrigued by and envision a career in mathematics in general. I was asked by the Iranian Mathematical Society (IMS) to prepare a problem book on the basis of college level competitions in Iran. Since Iranian students have done quite well in international competitions and their success to some extent reflects the training that they received in special courses, I decided to make the book available to a wider audience. The problems (and their solutions) that are presented in this book are from national mathematical competitions in Iran at college level from 1973 to 2007. I provided my own solutions to most of the problems and also utilized solutions from the files of the IMS, which contained approximately about 40% of all the solutions. I should point out that I have not edited the problems of the competitions and they are direct translations from Persian into English. Unfortunately, there were some typos and mistakes in the original version. The errors have been corrected and except for trivial typographical ones, the corrections are so indicated as footnotes throughout Chapter One. Some comments on the problems also appear in footnotes. I am indebted, and hence express my deepest gratitude, to the colleagues who contributed in different ways to these competitions and to this book — either as members of the Scientific Committee for the competitions, or proposed problems or assisted in finding elegant solutions. Unfortunately, some of those colleagues have left us; may they rest in peace. I am thankful to my dear friend Dr. Hossein ix

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Hajiabolhassan who constantly encouraged me and helped me with the preparation of the book. I would like to thank Professor Alireza Jamali, Professor Ebad Mahmoodian, the outgoing President of the IMS, Dr. Rashid Zaare-Nahandi, and Messrs. M. Shokoohi and M. Abdi-Zadeh and the administrative assistants at the main office of the IMS for their assistance in gaining access to the existing solution files of the IMS. I am grateful to Ms. Anahita Samie for drawing the figures for this book... Bamdad R. Yahaghi

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Acknowledgement A number of friends and colleagues took interest in this work and their contributions are implicit in some elegant solutions to the problems. While inadvertently I may have left out the names of some, the contributions of Saeed Akbari (the second solution of Problem 7 of 1.9.3), Kasra Alishahi (the third solution of Problem 1.b of 1.24.2), Rajendra Bhatia (the first proof of the lemma presented in Solution 3 of 2.3.1), Hossein Hajiabolhassan (Problems: 2 of 1.21.2, 2 of 1.23.2, and 6 of 1.28.1), Hossein Hajiabolhassan and the late Mojtaba Mehrabadi (Problem 2 of 1.14.2), Ramin Mohammadalikhani (the second solution of Problem 2 of 1.13.1), Ali Mohammadian (Problems: 6 of 1.27.2 and 5 of 1.25.1), Omid Naghshineh Arjmand (Problem 3.b of 1.3.1), are duly acknowledged. Needless to say I am responsible for any shortcomings or errors.

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Iranian University Students Mathematics Competitions – a historical introduction The history of the Iranian Mathematics Competitions for university students is an integral part of the contemporary history of mathematics in Iran. On March 31st, 1972, the first general assembly of the Iranian Mathematical Society (IMS) was held at the former National University (now Shahid Beheshti University). Upon a motion proposed by Mehdi Behzad, then president of the society –which at that time was called the Society’s secretary– the need for a mathematics competition aimed at university students was approved. Herewith a committee, consisting of Houshang Attarchi, Mehdi Behzad, Fereidoun Ghahramani, the late Mohammad-Ali Gheyni, and Bahman Vahidi, was first appointed to prepare a guideline for the competitions and to make preparations for holding theses competitions, which were intended for university students. This committee developed a guideline and required that the competitions must be held annually and simultaneously with the Annual Iranian Mathematical Society Conference. The primary goals of the competitions were to discover, encourage, and nurture mathematical talents throughout the country and to create a friendly and scientific rivalry among students and the Iranian universities. In those years, the mathematics courses in most of the universities across the country were heavily influenced by the traditional and old curriculum as opposed to the modern one. Only a handful of universities, such as the University of Tehran, Sharif (then Aryamehr) University of Technology, (former Pahlavi) University of Shiraz, and Mossaheb Institute of Mathematics offered courses such as Set Theory, Modern Algebra, Modern Analysis, and Combinatorics. The secondary goal of the competitions was to motivate the students to view mathematics in new ways to maintain and sustain a more diverse perspective of mathematics, thereby putting pressure on universities to change their educational systems. The following year, on March 30th, 1973, simultaneous with the Fourth Annual IMS Conference held at the University of Tehran, the First National Mathematics Competition was held with the participation of 22 students from five universities across the country. Half of the questions were in the areas of Mathematical Analysis and Algebra and the rest were in other areas categorized as General, i.e., innovative questions, Probability and Statistics, Differential Equations, and Topology. The second competition took place in March 1974 at the, former Pahlavi, University of Shiraz with the participation of 51 students from eleven universities. The students’ participation grew steadily in the following years so that in the thirtieth competition more than 180 students from 40 universities took part. Ever since 1973, the competitions were held annually except for 1979 with closure of universities because of the revolution and 1981-1983 when universities were closed due to the cultural revolution. Students’ and universities’ reception of the competitions caused a great increase in the number of participating students. On the other hand, when the competitions were held on the same date and location as the annual IMS conferences, the organizational matters of the event, in particular grading the student’s papers, had become so overwhelming that it would take a number of months to announce the results. This would sometimes lead to discontent and objections, which occasionally resulted in changes to the final scores. However, it was only for the tireless and diligent efforts of the questions committees in those years that the competitions survived and were kept alive despite all difficulties and shortcomings. In view of all this, the IMS was persuaded to hold

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the competitions in a date and location different from those of the Annual IMS conferences. Consequently, from 1996-2003 the competitions took place during a three day period. Since 2004 they have been taking place in a four day period at one of the Iranian universities during the Spring of every year. In June 1999, the IMS’ executive committee and board of trustees gathered at Tafresh University to reconsider the bylaws and regulations of the IMS. At that time, a committee was established to substantially change the regulations concerning the National Mathematics Competitions. From 2000, the competitions are being held according to the new regulations. It must, however, be said that some minor changes and improvements were made at some stages due to the experiences obtained from previous competitions. According to the current guideline of the competitions, each university or higher educational institution, can send only one team consisting of at most five students and a leader, possibly along with a deputy leader, to the competitions. Each university is responsible for selecting its own team, presumably through holding preliminary tests among its students. The contests take place in two sessions, with each session lasting three and a half hours. The students would answer twelve questions of which four are in the area of analysis, four in the area of algebra, and another four in other areas such as discrete mathematics, probability, number theory, etc, categorized as innovative questions. Attempts are made to propose problems whose solutions require not only mathematical knowledge but innovation. The questions of the competitions are now designed and selected by a questions committee consisting of a chairperson, a supervisor, and three other people as the heads of the following sections: analysis, algebra, and innovative questions. The heads of the three sections can each choose a vice head to assist them. The questions committee is appointed by the Executive Committee of the IMS for a period of two up to three years. The committee is responsible for designing questions and holding the competitions. It is also responsible for appointing a grading team, whose members are mostly chosen from the winners of the previous competitions. In the course of a number of months, the committee selects 24 questions which are divided into two sets of questions, each of which has 12 questions. The contests are held in two sessions. For each sitting, 6 questions are selected out of the 12 questions by a jury consisting of the leaders of the participating teams. The participants are ranked based on their individual scores. The top 5 individual scorers win gold medals, the next 10 top contestants are awarded silver medals, and up to the next top 15 receive bronze medals. It is worth mentioning that since April 2004, i.e., from the 28th competition on, the problems together with their proposed solutions are posted online, on the IMS website currently only in Persian. This is done for two reasons. Firstly, the universities can simultaneously hold the competitions for their interested students. Secondly, this way the problems and their proposed solutions are preserved on the Internet permanently. Now after thirty one competitions, it seems that most of the goals of the competitions have been achieved. A glance at the list of the winners of competitions reveals that many of them have gone on to become distinguished mathematicians and are now working at prominent universities and institutions throughout the world. Also, a glance at the questions of these thirty one competitions shows that the competitions and the Iranian mathematical community, on the whole, have

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evolved to a greater level of sophistication in terms of content, quantity and quality. In 1995, the executive committee of the IMS approved the preparation of a book containing the problems and solutions to the problems of the first twenty competitions. It took the IMS several months to collect the problems of the first twenty competitions. About one third of the proposed solutions of all problems were available in the IMS files. Rashid Zaare-Nahandi informed Bamdad Yahaghi about the decision made by the IMS and asked whether Yahaghi was interested in preparing such book. Yahaghi showed interest and readiness and it took him a year of hard work to prepare a preliminary copy of the manuscript. Under the recommendation of the IMS, the analysis and the algebra part of the manuscript was then refereed by the late Karim Seddighi and Mohammad Reza Darafsheh, respectively. This was simultaneous with Yahaghi’s departure for Canada to continue his PhD studies at Dalhousie University. Unfortunately, the original plan by the IMS to publish this book within a year could not be realized. Two years ago, shortly after the author’s return to Iran, his interest in the project on which he had spent such considerable time and energy was rekindled. He began to revise, update, and rewrite the book in both Persian and English. It must be said that the IMS did not have the proposed solutions to the problems of Competitions 21, 23, 24, and 27. It took Yahaghi almost two years to prepare both the English and Persian versions of the book. We must point out that this book could not have come into existence without the efforts of the colleagues who have contributed to these competitions. This book perhaps sheds light on the time and effort that have been put in organizing these competitions. It is hoped that this book succeeds in introducing these competitions to an international audience. Following this introduction, the names of the members of the questions committees1 and the date, location, and the winners of the competitions, together with their educational and employment affiliations, are quoted. Bamdad R. Yahaghi and Rashid Zaare-Nahandi

1Up until 1998, the executive committee of the IMS used to appoint only one person as the chairperson of the competition. This person was charged with the task of inviting other colleagues to submit questions and to mark the student’s papers. As outlined before, as of 1999, the questions committee is appointed by the Executive Committee of the IMS. And it is this committee that is responsible for designing questions and holding the competitions.

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The First Competition, University of Tehran, March 1973 Questions Committee: Mehdi Behzad. 1. Elizabeth Ebrahimzadeh, Sharif (former Aryamehr) University of Technology. PhD (University of California, Berkeley, 1984), Professor (California State University, Sacramento). 2. Mohammdad Reza Darafsheh, University of Tehran. PhD (University of Birmingham, UK, 1978), Professor (University of Tehran, Iran). 3. Hashem Madadi-Almousavi, Ferdowsi University of Mashhad. Deceased. 4. Mirebrahim Hashemi Aghdam, Sharif (former Aryamehr) University of Technology. 5. Yousef Bahrampour, (former Pahlavi) University of Shiraz. PhD (University of Oregon, USA, 1983), Professor (University of Kerman, Iran). The Second Competition, Shiraz (former Pahlavi) University, March 1974 Questions Committee: Mehdi Behzad. 1. 2. 3. 4. 5.

Mohammad Ali Najafi, Sharif (former Aryamehr) University of Technology. Firouz Khosroyani, University of Tehran. Mehdi Zekavat, (former Pahlavi) University of Shiraz. Hamid Hamed Akbari-Tousi, University of Tehran. Shahram Arshad-Riyazi, former National University.

The Third Competition, (former Jondi Shapour) University of Ahwaz, March 1975 Questions Committee: Mohammad Ali Gheyni. 1. 2. 3. 4. 5.

Pirouz Vakili, Sharif (former Aryamehr) University of Technology. Moslem Nikfar, University of Tehran. Ali Asghar Babadi Margha, University of Tehran. Saeed Ghahramani, Sharif (former Aryamehr) University of Technology. Homayoun Moeen, Sharif (former Aryamehr) University of Technology.

The Fourth Competition, University of Tabriz, March 1976 Questions Committee: Vahab Davarpanah. 1. 2. 3. 4. 5.

Pirouz Vakili, Sharif (former Aryamehr) University of Technology. Homayoun Moeen, Sharif (former Aryamehr) University of Technology. Nasser Hosseini, (former Pahlavi) University of Shiraz. Shahla Marvizi, Sharif (former Aryamehr) University of Technology. Ali Karimi, Tarbiat Moallem (Teacher Training) University.

The Fifth Competition, Sharif (former Aryamehr) University of Technology, March 1977

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Questions Committee: Mohammad Reza Nouri-Moghadam. 1. 2. 3. 4. 5.

Hamid Kazemi, Sharif (former Aryamehr) University of Technology. Hossein Masoumi Fakhar, Sharif (former Aryamehr) University of Technology. Masoud Khalkhali, Sharif (former Aryamehr) University of Technology. Ebrahim Sa´ atchi, (former Azarabadegan) University of Tabriz. Safa Nourbakhash, former National University.

The Sixth Competition, The University of Isfahan, March 1978 Questions Committee: Magerdich Toumanian. 1. 2. 3. 4. 5.

Nasser Boroujerdian, University of Tehran. Hamid Kazemi, Sharif (former Aryamehr) University of Technology. Ali Asghar Alikhani Kouhpaee, University of Isfahan. Mehdi Salehi Nejad, Ferdowsi University of Mashhad. Ali Rejali, University of Isfahan.

The Seventh Competition, Ferdowsi University of Mashhad, March 1980 Questions Committee: Akbar Hassani. 1. 2. 3. 4. 5.

Fereydoun Rezakhanlou, University of Tehran. Mehdi Alavi Shoushtari, University of Ahwaz. Seyed Esmail Seyedabadi, Sharif University of Technology. Rajabali Kamyabigol, Ferdowsi University of Mashahd. Ali Asghar Jodeyri Akbarfam, University of Tabriz.

The Eighth Competition, Shiraz University, March 1984 Questions Committee: Asadollah Niknam. 1. 2. 3. 4. 5.

Mohammad Hassan Jahanbakht, University of Isfahan. Mojtaba Moniri, University of Tehran. Mohammad Taghi Jahandideh, Shiraz University. Samad Ahmadi, Tarbiat Moallem (Teacher Training) University. Kamal Atigh, University of Tabriz.

The Ninth Competition, Tehran Teacher Training (Tarbiat Moallem) University, September 1985 Questions Committee: Rahim Zaare-Nahandi. 1. 2. 3. 4. 5.

Nasser Boroujerdian, University of Tehran. Mojtaba Moniri, University of Tehran. Majid Ashrafi, Shiraz University. Jamal Rouin, Tarbiat Moallem (Teacher Training) University. Ali Parsian, University of Tehran.

The Tenth Competition, University of Sistan and Baluchestan, March 1986

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Questions Committee: Karim Seddighi. 1. 2. 3. 4. 5.

Reza Jahani Nejad, University of Kashan. Amir Akbary Majdabadno, University of Tehran. Shaahin Ajoodani Namini, University of Tehran. Hamid Reza Farhadi, Tarbiat Moallem (Teacher Training) University. Masoud Amini, Ferdowsi University of Mashhad.

The Eleventh Competition, The University of Birjand, March 1987 Questions Committee: Karim Seddighi. 1. 2. 3. 4.

Vahid Tarokh, Sharif University of Technology. Masoud Amini, Ferdowsi University of Mashhad. Shaahin Ajoodani Namini, University of Tehran. Reza Karami, Isfahan University of Technology. The Twelfth Competition, Guilan University , March 1988

Questions Committee: Mohammad Ali Shahabi. 1. 2. 3. 4. 5.

Shaahin Ajoodani Namini, University of Tehran. Shaahin Amiri Sharifi, Sharif University of Technology. Gholam Hossein Eslamzadeh, Shiraz University. Bamdad Yahaghi, Sharif University of Technology. Ali Iranmanesh, Shiraz University.

The Thirteenth Competition, University of Tehran, March 1989 Questions Committee: Mohammad Ali Shahabi. 1. 2. 3. 4. 5.

Kambiz Mahmoodian, University of Tehran. Saeed Zakeri, University of Tehran. Mohammad Sal Moslehian, Ferdowsi University of Mashhad. Bamdad Yahaghi, Sharif University of Technology. Behrooz Mashayekhifard, Ferdowsi University of Mashhad.

The Fourteenth Competition, The University of Isfahan, March 1990 Questions Committee: Heydar Zahed Zahedani. 1. 2. 3. 4. 5.

Shaahin Amiri Sharifi, Sharif University of Technology. Hessam Hamidi Tehrani, Sharif University of Technology. Saeed Zakeri, University of Tehran. Hamid Mousavi, Tarbiat Moallem (Teacher Training) University. Shahab Shahabi, University of Tehran.

The Fifteenth Competition, Ferdowsi University of Mashhad, March 1991

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Questions Committee: Mohammad Reza Darafsheh. 1. 2. 3. 4. 5.

Hessam Hamidi Tehrani, Sharif University of Technology. Ali Rajai, Sharif University of Technology. Shahriar Mokhtari Sharghi, Sharif University of Technology. Pedram Safari, Sharif University of Technology. Ataollah Togha, University of Kerman.

The Sixteenth Competition, Razi (Rhazes or Rasis) University of Kermanshah, March 1992 Questions Committee: Mohammad Reza Darafsheh. 1. 2. 3. 4. 5. 6.

Shahriar Mokhtari Sharghi, Sharif University of Technology. Ataollah Togha, University of Kerman. Pedram Safari, Sharif University of Technology. Ali Sabetian, Shiraz University. Ali Rajai, Sharif University of Technology. Mehdi Najafikhah, Iran University of Science and Technology.

The Seventeenth Competition, Shahid Beheshti (former National) University, March 1993 Questions Committee: Mohammad Reza Darafsheh. 1. 2. 3. 4. 5.

Hossein Hajiabolhassan, Sharif University of Technology. Payman L. Kassaei, Sharif University of Technology. Behrang Noohi, Sharif University of Technology. Arash Rastegar, Sharif University of Technology. Ali Dadban, University of Tehran.

The Eighteenth Competition, Sharif University of Technology, March 1994 Questions Committee: Jafar Zafarani. 1. 2. 3. 4. 5.

Ramin Takloo-Bighash, Sharif University of Technology. Kasra Rafi, Sharif University of Technology. Behrang Noohi, Sharif University of Technology. Payman L. Kassaei, Sharif University of Technology. Aminollah Zargarian, University of Tehran.

The Nineteenth Competition, University of Kerman, March 1995 Questions Committee: Ahmad Haghani. 1. 2. 3. 4. 5.

Amir Jafari, Sharif University of Technology. Ali Lashgari Faghani, Isfahan University of Technology. Fatemeh Ayatollahzadeh Shirazi, University of Tehran. Mohammad Reza Raoofi, Isfahan University of Technology. Reza Naserasr, Sharif University of Technology.

The Twentieth Competition, Sharif University of Technology, February 1996

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Questions Committee: Omid Ali Karamzadeh and Yahya Tabesh. 1. 2. 3. 4. 5.

Keivan Mallahi-Karai, Sharif University of Technology. Hossein Movasati, Sharif University of Technology. Omid Naghshineh Arjmand, Sharif University of Technology. Ali Reza Amini Harandi, Isfahan University of Technology. Ebrahim Samei, Shahid Beheshti (former National) University.

The Twenty First Competition, University of Tehran, March 1997 Questions Committee: Omid Ali Karamzadeh. 1. 2. 3. 4. 5.

Kia Dalili, Sharif University of Technology. Maryam Mirzakhani, Sharif University of Technology. Ebrahim Samei, Shahid Beheshti (former National) University. Hadi Jorati, Sharif University of Technology. Hossein Abedi Andani, Isfahan University of Technology.

The Twenty Second Competition, University of Ahwaz, March 1998 Questions Committee: Omid Ali Karamzadeh. 1. 2. 3. 4. 5.

Maryam Mirzakhhani, Sharif University of Technology. Eaman Eftekhari, Sharif University of Technology. Payam Nasser Tayoub, University of Tehran. Mohammad Ahmadvand, Bu-Ali Sina (Avecina) University of Hamedan. Abolghassem Karimi, Shahid Beheshti (former National) University.

The Twenty Third Competition, Sharif University of Technology, March 1999 Questions Committee: Omid Ali Karamzadeh and Yahya Tabesh. 1. 2. 3. 4. 5.

Hadi Salmasian, Sharif University of Technology. Mohsen Bahramgiri, Sharif University of Technology. Mohammad Javaheri, Sharif University of Technology. Bijan Ahmadi, Shahid Beheshti (former National) University. Kamal Azizi, University of Tabriz.

The Twenty Fourth Competition, Khajeh Nasir Toosi University of Technology, May 2000 Questions Committee: Saeed Azam, Rouzbeh Tusserkani, Ali Reza Jamali (Chair), and Hossein Mohebi. 1. 2. 3. 4. 5.

Omid Amini, Sharif University of Technology. Kasra Alishahi, Sharif University of Technology. Maziar Mirrahimi, Sharif University of Technology. Seyed Reza Moghaddasi, Sharif University of Technology. Masoud Aryapour, Sharif University of Technology.

The Twenty Fifth Competition, Imam Khomeini International University of Qazvin, May 2001

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Questions Committee: Saeed Azam, Rouzbeh Tusserkani, Ali Reza Jamali (Chair), and Hossein Mohebi. 1. 2. 3. 4. 5.

Amir Mohammadi, Sharif University of Technology. Salman Abolfath Beigi, Sharif University of Technology. Hamid Reza Darbidi, Sharif University of Technology. Babak Amini, Shiraz University. Afshin Amini, Shiraz University.

The Twenty Sixth Competition, Institute for Advanced Studies in Basic Sciences (IASBS) of Zanjan, May 2002 Questions Committee: Saeed Azam, Rouzbeh Tusserkani, Ali Reza Jamali (Chair), and Hossein Mohebi. 1. 2. 3. 4. 5.

Salman Abolfath Beigi, Sharif University of Technology. Ali Shourideh, Sharif University of Technology. Javad Ebrahimi Boroujeni, Sharif University of Technology. Amin Aminzadeh Gohari, Sharif University of Technology. Majid Hadian, Sharif University of Technology.

The Twenty Seventh Competition, Bu-Ali Sina (Avecina) University of Hamedan, May 2003 Questions Committee: Gholam-Hossein Eslamzadeh, Hossein Hajiabolhassan, Mohammad Reza Pournaki, and Mohammad Taghi Dibaei (Chair). 1. 2. 3. 4. 5. 6. 7. 8.

Mohsen Sharifi Tabar, Sharif University of Technology. Ali Shourideh, Sharif University of Technology. Mohammad Farajzadeh Tehrani, Sharif University of Technology. Payam Valadkhan, Sharif University of Technology. Hamid Hassanzadeh, Tarbiat Moallem (Teacher Training) University. Rahbar Rasooli, University of Tehran. Maryam Khosravi, Tarbiat Moallem (Teacher Training) University. Amir Moradifam, Iran University of Science and Technology.

The Twenty Eighth Competition, Sharif University of Technology, May 2004 Questions Committee: Gholam Hossein Eslamzadeh, Hossein Hajiabolhassan, Majid Mirza-Vaziri, Abdolrasool Pourabbas, Mohammad Reza Pournaki, and Rashid Zaare-Nahandi (Chair). 1. 2. 3. 4. 5.

Iman Setayesh, Sharif University of Technology. Omid Haji Mirsadeghi, Sharif University of Technology. Armin Morabi, Sharif University of Technology. Sajad Lakzian, Amir Kabir University of Technology. Mohammad Kazem Anvari, Ferdowsi University of Mashhad.

The Twenty Ninth Competition, University of Mazandaran in Babolsar, May 2005

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Questions Committee: Hossein Hajiabolhassan, Majid Mirza-Vaziri, Mojtaba Moniri, Mohammad Reza Pournaki, Mehdi Radjabalipour (Chair), Bamdad R. Yahaghi, Rashid Zaare-Nahandi (Supervisor), and Manouchehr Zaker. 1. 2. 3. 4. 5. 6.

Iman Setayesh, Sharif University of Technology. Mohammad Farajzadeh Tehrani, Sharif University of Technology. Mohammad Abbas Rezai, Sharif University of Technology. Mohammad Hossein Mousavi, Sharif University of Technology. Fatemeh Doroodian, Amir Kabir University of Technology. Mahmoud Hassanzadeh, University of Tehran.

The Thirtieth Competition, Tafresh University, May 2006 Questions Committee: Hossein Hajiabolhassan, Majid Mirza-Vaziri, Mojtaba Moniri, Mohammad Reza Pournaki, Mehdi Radjabalipour (Chair), Bamdad R. Yahaghi, Rashid Zaare-Nahandi (Supervisor), and Manouchehr Zaker. 1. 2. 3. 4. 5.

Ali Akbar Daemi, Sharif University of Technology. Mohammad Gharakhani, Sharif University of Technology. Omid Haji Mirsadeghi, Sharif University of Technology. Mostafa Einollahzadeh Samadi, Sharif University of Technology. Behzad Mehrdad, Sharif University of Technology.

The Thirty First Competition, Ferdowsi University of Mashhad, May 2007 Questions Committee: Mojtaba Gheerati, Hassan Shirdareh Haghighi, Omid Naghshineh Arjmand, and Fariborz Azarpanah (Chair). 1. 2. 3. 4. 5.

Ali Akbar Daemi, Sharif University of Technology. Nasser Talebizadeh, Sharif University of Technology. Erfan Salavati, Sharif University of Technology. Nima Ahmadipour Anari, Sharif University of Technology. Mohammad Gharakhani, Sharif University of Technology.

CHAPTER 1

Problems 1.1. First Competition, University of Tehran, March 1973 1.1.1. Analysis. 1. A function φ that is a pointwise limit of continuous real functions is called a Baire function. (a) If f is a real function of the real variable x whose derivative exists everywhere, prove that the function f 0 is a Baire function. (b) By giving an example show that every Baire function is not necessarily the derivative of a function. 2. Let’s call the set of all n × n real matrices M. The set M is a metric space if we view the elements of it as n2 -dimensional vectors equipped with the Euclidean norm . That is, for every matrix A = (aij ), we define the norm of A as follows X
1 a2ij 2 . ||A|| = 1≤i,j≤n

Prove that the set of invertible matrices is (a) open. (b) disconnected. 3. Suppose that the function f is defined on the half-line (0, +∞) by  0 x∈ / Q, f (x) = p 1 x = , p, q ∈ N, gcd(p, q) = 1. p+q q (a) Show that the limit of this function exists at any point of (0, +∞). (b) At what points of (0, +∞) is the function f continuous? Prove your claim.

1.1.2. Algebra. 1. Let R be commutative ring with identity element. (a) Prove that an ideal P is prime if and only if ∀a ∈ R, ∀b ∈ R : ab ∈ P =⇒ a ∈ P or b ∈ P. (b) Prove that every maximal ideal is a prime ideal. • Hint. In a ring R, the product of two ideals T1 and T2 is defined as X xy : x ∈ T1 , y ∈ T2 . finite

An ideal M 6= R is called maximal if for any ideal T , we have M ⊆ T =⇒ T = R. An ideal P is prime if and only if for any two ideals T1 and T2 , from T1 T2 ⊆ P , it follows that T1 ⊆ P or T2 ⊆ P . 1

2

1. PROBLEMS

2. Suppose that every element of a group G satisfies the equality x2 = e. Prove that the group G is abelian (i.e., commutative). 3. Let E and F be two isomorphic sets (that is, there is a one-to-one correspondence between them), and that f is a function from P(E) into P(F ) satisfying the following three conditions, where P stands for the power set operation. (a) f (∅) = ∅. (b) ∀X ∈ P(E), ∀Y ∈ P(E) : f (X ∪ Y ) = f (X) ∪ f (Y ). (c) ∀X ∈ P(E) : cardX ≤ cardf (X). (cardA ≤ cardB means that there exists a one-to-one mapping from A into B.) 3.1. Prove that E and f (E) are isomorphic. Moreover, if X and Y are two finite subsets of E such that X is isomorphic to f (X) and Y is isomorphic to f (Y ), show that X ∪ Y and f (X ∪ Y ), and X ∩ Y and f (X ∩ Y ) are isomorphic sets, respectively, and that for any two such sets we have f (X ∩ Y ) = f (X) ∩ f (Y ). 3.2. Prove that if E is finite and normal, then there exists a normal subset X0 6= ∅ of E such that for any other normal subset X of E, we have X ∩ X0 = ∅ or X0 ⊆ X. (A subset A of E which is isomorphic to f (A) is called a normal subset of E.)

1.1.3. General. 1. A triangle with sides a, b, c such that a < b < c is given. Set S = max

a b c a b c , , min , , . b c a b c a

Show that S ≥ 1. 2. Show that the decimal fraction 0.123456789101112131415 . . ., which is formed by putting consecutive positive integers next to one another, is not periodic. 3. Find two distinct real or complex numbers in such a way that each of which is the cube of the other. 4. Let x, y be two positive real numbers with x + y = 1. Prove that √ xx + y y ≥ 2.

1.1.4. Differential Equations. Find the general solution of the following differential equation d2 x + a2 x = f (t), dt2 where a is a constant and f is a continuous real function.

1.2. SECOND COMPETITION

3

1.1.5. Probability and Statistics. Let X1 , . . . , X10 be ten independent rani e−µi µX i dom variables with the probability density functions fi (Xi ) = , where Xi µi = i (i = 1, . . . , 10) and that the codomain of the variable Xi is {0, 1, 2, . . .}. Set 10 1 X X= Xi . 10 i=1 What is the probability that X equals one?

1.1.6. Topology. Let Q be the set of all rational numbers. For each q ∈ Q, the set ]q, +∞[ is denoted by Aq . If T is the set consisting of R and ∅ and all Aq ’s, show that T is not a topology on R. √ • Hint. Consider the sets of the form Aq with q > 2 and q ∈ Q.

1.2. Second Competition, Shiraz (former Pahlavi) University, March 1974 1.2.1. Analysis. 1. Let f be a real continuous function with nonnegative values on the closed interval [0, 1]. Set Z 1  n1 n un = (f (x)) dx , M = sup f (x). 0≤x≤1

0

(a) Assuming that 0 < ε < 1, prove that there exist numbers α, β subject to 0 ≤ α < β ≤ 1 such that M (1 − ε) ≤ f (x) ≤ M for all x belonging to the open interval ]α, β[. (b) Prove that limn→+∞ un = M . 2. Evaluate  D = inf

f ∈F

sup 1 − f (t) + 0≤t≤1

Z

1

 1 − f (t) dt ,

0

where F is the vector space of all continuous functions from [0, 1] into R whose elements take the value zero at zero. Describe the geometrical interpretation of the number D. 3. Assuming that C is the Cantor set, show that C + C = [0, 2]. • Hint. 1. If A and B are two subsets of real numbers, by definition,  A + B := a + b : a ∈ A, b ∈ B . 2. The Cantor set consists of all real numbers between zero and one whose ternary expansions do not have any one.

4

1. PROBLEMS

1.2.2. Algebra. 1. Let F be a field with n elements. Prove that for every integer m with m ≥ 1, we have  X n−1 n − 1 | m, xm = 0 n − 1 - m. x∈F

∗

• Hint. F = F \ {0} is a cyclic group. 2. Let A be a ring. • Definition 1. An element z ∈ A is called right quasi-regular if there is a z 0 ∈ A such that z + z 0 − zz 0 = 0. • Definition 2. An element z ∈ A is called left quasi-regular if there is a z 0 ∈ A such that z + z 0 − z 0 z = 0. • Definition 3. An element z ∈ A is called quasi-regular if there is a z 0 ∈ A such that z + z 0 − zz 0 = z + z 0 − z 0 z = 0. (a) An element z ∈ A is quasi regular if an only if it is right and left quasiregular. (b) If A has an identity element, then the identity element is not right quasiregular. And if x is a right quasi-regular element, then 1 − x is invertible1 in A. 3. Let E be a finite-dimensional vector space over a field K and φ a homomorphism from the ring K[x] into the ring L(E) with the hypothesis that φ(1) is the identity element of L(E). (a) If P and D belong to K[x] and D is a divisor of P , prove that

ker φ(P ) ⊃ ker φ(D) . (b) If R is another element of K[x] and D is the greatest common divisor of P and R, prove that


ker φ(D) = ker φ(P ) ∩ ker φ(R) . • Hint. By K[x], we mean the ring of all polynomials in one indeterminate x with coefficients coming from K and L(E) is the ring of all endomorphisms of E. For any u belonging to L(E), ker u means the kernel of the linear transformation u.

1.2.3. General. 1. Find the minimum and the maximum number of “Wednesdays occurring on the 13th of the month” which might happen in a solar year, according to the Persian calendar that is. Remark. Recall that there are 12 months in the Persian calendar, each of the first six months has 31 days, each of the sixth-eleventh months has 30 days, and the last month has 29 days except that in a leap year it has 30 days. 2. Prove that

xn + y n + z n + tn , 4 4 where x, y, z, t are positive real numbers and n is a natural number.  x + y + z + t n

≤

3. A particle, moving along a straight line, goes one unit of distance in one unit of time in such a way that the speed of it at the initial point as well as the final point 1“invertible” must read “right invertible”!

1.2. SECOND COMPETITION

5

is zero. Prove that [the absolute value of] the particle acceleration at some point on its path is greater than or equal to four. 4. We know that in Iran the plate number of every car is a five-digit number none of whose digits is zero. In big cities, like Tehran, in addition to a number a [Persian] letter is also used to characterize the plate, e.g., “alef”, “be”, “pe”, etc. By explaining your reasoning, find the total number of the car plates that contain the letter “dal”. Remark. The Persian alphabet has 32 letters.

1.2.4. Probability. Let X and Y be two random variables subject to the following conditions. var(X) = var(Y ) = σ 2 , cov(X, Y ) = λ, where σ and λ are given real numbers. Define the random variables U and V by U = X + Y and V = X − Y . (a) Find the covariance of U and V . (b) Are U and V independent? If so, prove it; if not, by an example justify your answer.

1.2.5. Topology. Consider a nonempty topological space (E, T ) and two nonempty subsets A and K of E with K ⊃ A. Suppose that B and C are two closed subsets of the topological space (K, TK ) such that A ⊂ B ∪ C, A ∩ C 6= ∅, A ∩ B 6= ∅. Prove that if A is connected, then A ∩ B ∩ C 6= ∅. • Hint. By TK we mean the induced topology which is induced by T on K. The problem can be proved by contradiction.

1.2.6. Differential Equations. Find the general solution of the following differential equation subject to the two cases |x| > 1 and |x| < 1. 3(x2 − 1)y 2 y 0 + xy 3 = x3 + x2 − x − 1.

6

1. PROBLEMS

1.3. Third Competition, (former Jondi Shapour) University of Ahwaz, March 1975 1.3.1. Analysis. 1. Let f be a real function whose domain is [a, b]. If f 00 exists on [a, b] and is positive, then for all ξ ∈ [a, b],2 there exists a point x0 ∈ [a, b] such that f (a) − f (x0 ) f (b) − f (x0 ) f 0 (ξ) = or f 0 (ξ) = . a − x0 b − x0 2. A sequence (an )+∞ n=1 of numbers in the closed interval [0, 1] is given such that the elements of the sequence are all distinct. The goal is to find a continuous function f from [0, 1] into [0, 1] such that f (an ) = an+1 . Prove that (a) in general, the problem has no solution. (b) if the sequence is increasing or decreasing, the problem has a solution. 3. In each part of this problem, you can use the preceding parts. (a) If the complex number z is a root of the following equation with complex coefficients xp + c1 xp−1 + · · · + cp−1 x + cp = 0, then p ∃ k ∈ {1, . . . , p} : |z| ≤ 2 k |ck |. (b) The following two equations with complex coefficients are given. xp + c1 xp−1 + · · · + cp−1 x + cp = 0, xp + c01 xp−1 + · · · + c0p−1 x + c0p = 0. If the positive numbers K and δ satisfy the following
∀i = 1, . . . , p : |ci | < K i , |ci − c0i | < K i δ , then for any root zj of the first equation there exists a root zk0 of the second equation such that √ zj − zk0 < 2K p δ. (c) Suppose that the real numbers K and α are such that K > 0 and 0 < α ≤ 1. Also suppose that the complex valued functions b1 , . . . , bp of a real variable t belonging to the compact interval [t1 , t2 ] are such that |bi (t)| < K i , |bi (t) − bi (t0 )| < K i |t − t0 |α , for all t, t0 ∈ [t1 , t2 ] and each i = 1, . . . , p. Prove that if the complex valued function f is continuous on the closed interval [t1 , t2 ] and that it satisfies the equation f p + b1 f p−1 + · · · + bp−1 f + bp = 0, then p f (t2 ) − f (t1 ) < 4pK p (t2 − t1 )α . • Hint. If F : I → U is a function from an interval into an open subset of the complex plane, the image of F lies in one of the connected components of U . 2 “ξ ∈ [a, b]” must read “ξ ∈ (a, b)”!

1.3. THIRD COMPETITION

7

1.3.2. Algebra. 1. Let M be a module on a ring K and M1 and M2 two submodules of M . Set M3 = M1 + M2 , M4 = M1 ∩ M2 . M3 M2 and are isomorphic. M1 M4 (b) If K is a commutative field and M1 and M2 are finite-dimensional, show that (a) Show that the quotient modules

dim M1 + dim M2 = dim M3 + dim M4 . 2. Prove that a necessary and sufficient condition for a group to be commutative is that the following function is a homomorphism of groups. f : G → G, f (x) = x2 .

3. Suppose that for any element y of a semigroup S with a right identity element (that is, xe = x for all x ∈ S), there exists a y with the property that yy = e. (a) By giving an example, show that S might not be a group. (b) If the right identity element is unique, prove that S is a group.

1.3.3. General. 1. The real function F (x, y) where x and y are two real variables is considered. Suppose that this function has the following two properties. 3

∀x, y ∈ R : F (x, y) = 0 ⇐⇒ x = y, ∀x, y, z ∈ R : F (y, x) ≤ F (x, z) + F (z, y). (a) Prove that F (x, y) ≥ 0 for all x, y ∈ R. (b) Prove that F (x, y) = F (y, x) for all x, y ∈ R. 2. Without using mathematical induction, prove that 2 n  X n! (2n)! = , m!(n − m)! (n!)2 m=0 for all n ∈ N. 3. Two flat mirrors ∆ and ∆0 which are perpendicular to a plane P and two points M, N ∈ P are given. From the point M , shine a beam of light on the mirror ∆, in the plane P , in such a way that the reflected light beam after hitting the mirror ∆0 passes through the point N .

3 “∀x, y ∈ R : F (x, y) = 0 =⇒ x = y” is redundant!

8

1. PROBLEMS

M

N Figure 1

1.4. Fourth Competition, University of Tabriz, March 1976 1.4.1. Analysis. 1. In this problem R denotes the real line and R2 the Euclidean plane. Prove that if f is a continuous and one-to-one function from R into R2 , its inverse is not necessarily continuous. 2. Let (αn )+∞ n=1 be a sequence of nonnegative real numbers such that lim αn = 0.

n→+∞

Prove that there are infinitely many number of indices n such that αm ≤ αn for all m greater than or equal to n.

1.4.2. Algebra. 1. Let A be a commutative ring and B an ideal of A. Set  R(B) = x ∈ A : ∃r ∈ N  xr ∈ B . We call R(B) the radical of B. We call the ideal B of a commutative ring A a prime ideal if ∀a, b ∈ A, ab ∈ B =⇒ a ∈ B or b ∈ B. (a) Prove that R(B) is an ideal of A. (b) Prove that if the ideal B is prime, then so is the ideal R(B). (c) If we set A = Z, and if a is an element of Z and4

 B = a = aZ = aA, prove that there are prime numbers p1 , . . . , pr such that

 R(B) = p1 p2 · · · pr . 2. Prove that in a vector space E, whose dimension is at least 2, for any two vectors x and y which are not linearly dependent, one can find an automorphism u : E → E such that u(x) = x and u(y) = x + y. From this, conclude that if an endomorphism f : E → E commutes with any automorphism v (i.e., f ◦ v = v ◦ f ), then for every vector x ∈ E, the two vectors x and f (x) are linearly dependent whence f is just the scalar product of the identity automorphism by a fixed scalar (i.e., f = λidE ). 4 Obviously, we must have a 6= 0, ±1.

1.5. FIFTH COMPETITION

9

3. Let R be an integral domain with identity. Suppose that every descending chain of the ideals of R terminates. (that is, for every chain of ideals of R as follows I1 ⊇ I2 ⊇ · · · ⊇ Ik ⊇ · · · , there exists a positive integer n such that Im = In for all m ≥ n.) Prove that R is a field. • Hint. Assume that x is a nonzero element in R, investigate the ideals xi R (i = 1, 2, . . .).

1.4.3. General. 1. Evaluate the following integral. Z π2 sinn x dx, n ∈ N. I= cosn x + sinn x 0 2. We have three boxes A , B, and C, of which two are empty and one has a prize in it. We choose one of these boxes at random (say, for example B); it is plain that the probability that the chosen box contains the prize is 13 because of the three boxes only one contains the prize. Now, suppose that of the two not-chosen boxes A and C, we open the one that does not contain the prize (say, for example A). Therefore, the prize is in one of the boxes B and C. (a) What is the probability that the chosen box contains the prize? (b) What is the probability that the other box, i.e., C, contains the prize?5

1.5. Fifth Competition, Sharif (former Aryamehr) University of Technology, March 1977 1.5.1. Analysis. 1. If two functions are integrable on the interval (0, 1) in the Riemann sense, is it true that the composition of the two functions is integrable on the interval (0, 1) in the Riemann sense? Explain your reasoning. 2. Let R be the set of all real numbers and f a continuous function from R into R that does not assume any value more than twice. Prove that f assumes at least a value exactly once. 3. Let A be a subset of the real numbers R. A point p of A is called a congestion point of A if every neighborhood of p of the form (p − ε, p + ε) contains uncountably many points of A. Prove that all but countably many points of A are congestion points of A (A is uncountable).

5 This problem is also known as the Monty Hall problem.

10

1. PROBLEMS

1.5.2. Algebra. 1. Let V be a finite-dimensional vector space over C (the complex numbers). If A : V → V is a linear transformation such that A 6= I, A2 6= I, A3 = I, find the eigenvalues of A. Extend this to the case where Ak = I. Do the eigenvalues of A form a group under multiplication? Find a necessary and sufficient condition for this set to form a group. 2. Let G be group and a and b two elements of it satisfying the following relations. a 6= 1, b 6= 1, aba−1 = b2 , a7 = 1, where 1 is the identity element of the group. Find the orders of a and b. 3. Let R be a ring with identity element such that every x ∈ R satisfies the following relation: x3 + 2x2 + x = 0. (a) Prove that 2x = 0 for all x ∈ R. (b) Prove that R is a commutative ring.

1.5.3. General. 1. Let m, n ∈ N. Prove that the number

(mn)! is an m!(n!)m

integer. 2. If S = {z1 , . . . , zk } is a subset of the complex numbers, define the set C(S) by  C(S) := z = α1 z1 + · · · + αk zk |αi ≥ 0, α1 + · · · + αk = 1 . If f is a polynomial of degree greater than or equal to two and A and A0 denote the sets of the roots of the equations f (z) = 0 and f 0 (z) = 0, respectively, where f 0 denotes the derivative of f , prove that C(A) ⊃ C(A0 ). 3. Consider the trajectory of a billiard ball B which moves on an ellipse-shaped billiard table ξ. Suppose that the angle of incidence is equal to the angle of reflection. Assuming that the initial ball trajectory does not intersect the line segment joining the the two foci, show that there exists another ellipse η which is confocal with the ellipse ξ such that the ball trajectory is always tangent to η. • Hint. Recall that if P L and P L0 are two tangents from a point P to an ellipse, then the angles F P L and F 0 P L0 are equal. (F and F 0 are the foci of the ellipse.) 1.6. Sixth Competition, The University of Isfahan, March 1978 1.6.1. Analysis. 1. In the xy-plane a point is called rational if both of its coordinates are rational. Prove that if the center of a given circle in the plane is not rational, then there are at most two rational points on the circle. 2. Suppose that in a metric space M a sequence (fn )+∞ n=1 of continuous functions is uniformly convergent to a function f . Prove that for every sequence (xn )+∞ n=1 converging to a point x ∈ M , we have lim fn (xn ) = f (x).

n→+∞

1.6. SIXTH COMPETITION

11

R +∞ 3. If f is a continuous function on R and moreover 0 f (x) dx < +∞, prove that Z +∞ f (x + 1 ) − f (x) dx = 0. lim n→+∞ 0 n

1.6.2. Algebra. 1. Let G be a group with |G| = pn a, where p is a prime and p and a are relatively prime. If G has subgroups A and B satisfying the following conditions |A| = pn , |B| = pm , 0 < m ≤ n, B * A, prove that AB cannot be a subgroup of G. 2. Let A1 , . . . , An be mutually commuting m × m matrices such that A2i = 0 for all 1 ≤ i ≤ n. If m < 2n , prove that A1 A2 · · · An = 0. 3. If R is a ring with identity (1 6= 0) and e and f are two commuting elements of R such that e2 = e and f 2 = f , prove that (e − f )n = 0 =⇒ e = f, for all positive integers n.

1.6.3. General. 1. Let n1 , n2 , . . . , nk be k integers of which m1 , m2 , . . . , mk is a permutation. Prove that |n1 − m1 | + |n2 − m2 | + · · · + |nk − mk | is an even integer. 2. An n × n matrix whose elements are nonnegative integers is given. If the sum of the elements on any row and any column corresponding to any nonzero element of the matrix is at least n, prove that the sum of all elements of this matrix is no 2 less than n2 . 3. Suppose 1000000 points inside a circle are given. Can one find a straight line not passing through any of the points that divides the circle into two sections each of which containing 500000 of the points?

12

1. PROBLEMS

1.7. Seventh Competition, Ferdowsi University of Mashhad, March 1980

1.7.1. Analysis. 1. Is there a closed set S $ R2 such that for each x ∈ R2 \ S there are exactly two points in S as the closest point of S to x? 2. For each positive rational number r (r ∈ Q+ ), suppose Ir S is an open interval such that r < s =⇒ I r ⊂ Is . The function f on the set A = r∈Q+ Ir is defined by  f (x) = inf r : x ∈ Ir . Show that f is continuous. 3. Suppose that the sequence (xn )+∞ n=1 is defined by n

n + 1 X 2k , n = 1, 2, . . . . xn = n+1 2 k k=1

Prove that limn→+∞ xn exists and find its value.

1.7.2. Algebra. 1. (a) If in a ring R the element e is a left identity element (i.e., ex = x, ∀x ∈ R) that is unique, show that e is the identity element of this ring (i.e., ex = xe = x, ∀x ∈ R). (b) If zero is the only nilpotent element of the ring R, show that for every idempotent element a of R we have ax = xa, ∀x ∈ R. • Note. The element b of a ring R is called nilpotent if there exists a positive integer n such that bn = 0; b is called idempotent if b2 = b. 2. Let G be a group and G1 a normal subgroup of G such that the groups GG1 and G1 are commutative. Prove that for any arbitrary subgroup H of G, there is a commutative subgroup H1 of H such that H1 is a normal subgroup of H and HH1 is commutative. 3. An n-dimensional vector space V and a linear transformation θ : V → V are given. Consider the powers of θ, i.e., θ0 = 1, θ, θ2 , . . .. Prove that there is a nonzero integer s such that V = im(θs ) ⊕ ker(θs ).

1.8. EIGHTH COMPETITION

13

1.8. Eighth Competition, Shiraz University, March 1984 1.8.1. Analysis. 1. Let F and G be two closed subsets of Rn , where R denotes the real numbers endowed with its ordinary topology. Suppose that f : F → R and g : G → R are two continuous functions which coincide on F ∩ G. (a) Prove that there is a continuous function h : F ∪ G → R such that h is an extension of both of the functions f and g. (b) By giving an example, show that if F and G are not closed, the assertion in (a) is not correct. P+∞ 2. Suppose that (an )+∞ a decreasing sequence of real numbers and that n=1 an n=1 isP +∞ is convergent. Prove that n=1 n(an − an+1 ) is convergent.6 3. Let the function g be continuously differentiable on [0, 1]. Prove that Z 1 lim xn dg(x) = 0. n→+∞

0

1.8.2. Algebra. 1. In the multiplicative group of nonsingular 2 × 2 matrices with real entries, let     2 0 1 1 a= , b= . 0 1 0 1 If H is the subgroup generated by b, show that aHa−1 is a proper subgroup of H. 2. Let R be an infinite integral domain (i.e., a commutative ring with identity without divisors of zero). If R has finitely many units, prove that R has infinitely many maximal ideals. (A unit is an element which has a multiplicative inverse.) 3. Let A =P(aij ) be an n × n matrix over the field of real numbers such that for all n i we have j=1 aij = a. If A2 = I (I is the identity matrix), find a.

1.8.3. General. 1. Without using derivative, find the minimum of the three variable real function f (x, y, z) = x2 + 4y 2 + z 2 − 6x + 4y. 2. Five people have 719 rials altogether. If each person has an integer amount of money, that the money of no two people is equal, and that the ratio of a person’s money with respect to that of any other person with less amount of money is an integer, determine how much money each person has. 3. A bus which must pass through the city A passes through the crossroad B of the city with the probability of 31 . If the traffic-light of the crossroad is, consecutively, 30 seconds red and 30 seconds green, find the average of the stop time of the bus at the crossroad. 6 For a counterpart of this problem for sequences of real functions, see Problem 2 of 1.12.1.

14

1. PROBLEMS

1.9. Ninth Competition, Tehran Teacher Training (Tarbiat Moallem) University, September 1985 1.9.1. Analysis. 1. On the set of real numbers R, define the equivalence relation ∼ as follows a ∼ b ⇐⇒ a − b ∈ Q, where Q is the set of rational numbers. Prove that every equivalence class of ∼ is dense in R. 2. Prove that if f is a continuous and one-to-one function from R into R, then the inverse of f (from f (R) onto R) is continuous as well. 3. Let f be a continuous and increasing function from [a, b] into [a, b] and f (a) = a.  Prove that if E = x|a ≤ x ≤ b, f (x) ≥ x , then f (E) = E. 4. Let the function f be defined on [0, 1] by  0 x∈ / Q, f (x) = 1 x ∈ Q, x = pq , p, q ∈ N, gcd(p, q) = 1. q R1 Prove that 0 f (x)dx exists. 5. Let f : [0, 1] → [0, 1] × [0, 1] be a function. Prove that the function f can have any two properties of the following three properties but cannot have more than two properties. Continuity, Injectivity, Surjectivity.

1.9.2. Algebra. 1. Let G be a finite group and p the smallest prime dividing the order of G. Prove that every subgroup of index p in G is a normal subgroup of G. 2. Let Z3 denote the field of integers mod 3. Define, explicitly, an isomorphism Z3 [x] Z3 [x] from the field 2 onto the field 2 . (x + 1) (x + x + 2) 3. Find the greatest common divisor of the two polynomials 4x4 − 2x2 + 1 and −3x3 + 4x2 + x + 1 in Z7 [x]. 4. A quadratic extension K ⊇ F such that ch(F ) 6= 2 is given (here, ch(F ) denotes the characteristic of the field F ). Prove that there exists an element y ∈ K such that y 2 ∈ F and that {1, y} is a basis for K over F . 5. Let F be a field with characteristic 2, V an n-dimensional vectorspace over F , 2 and T : V → V a linear transformation such that T = I. Set W = v ∈ V : T v = v . Prove that dim(W ) ≥ n/2. (Here, I denotes the identity transformation on V .)

1.9. NINTH COMPETITION

15

1.9.3. General. 1. Evaluate the determinant of an n × n matrix whose diagonal entries are all equal to r and whose off-diagonal entries are all equal to λ. 2. A husband and a wife, who are working at one of the poultry farms of the country, have posed a problem on the number of their chickens as follows. If they sell 75 of their chickens, their chicken food will finish twenty days after they run out of it. But if they buy another one hundred chicken, their chicken food will finish fifteen days before they run out of chicken food. Find the number of chickens in the chicken farm. 3. Let α be a real number. Prove that there is no positive continuous function f : [0, 1] → R such that Z 1 Z 1 Z 1 f (x)dx = 1, xf (x)dx = α, x2 f (x)dx = α2 . 0

0

0

4. Find the number of the solutions of the following equation in the set of positive integers. x1 + x2 + · · · + xm = n, where m and n are natural numbers with m < n. 5. On a square grid paper the following figure is drawn. Can you cut the paper along the lines and divide it into two pieces in such a way that putting the pieces next to one another forms a chessboard?

6. The start hour in a factory is 8 o’clock in the morning. A worker has estimated that, by car, s/he goes the distance from home to work within 10 to 20 minutes (with uniform distribution). (a) If this worker leaves his/her home at 7 34 , what is the probability that s/he does not get to work on time? (b) If 15 minutes is needed to have breakfast at the work place, determine the latest time that this worker can leave home to get to work on time and with the probability of 75% to have time to eat breakfast. 7. Let c1 , . . .
, cn be n real numbers. Consider the matrix (ci cj )n×n . Evaluate det I + (ci cj ) , where I is the identity matrix of size n.

16

1. PROBLEMS

1.10. Tenth Competition, University of Sistan and Baluchestan, March 1986

1.10.1. Analysis. 1. The real function f satisfying the following conditions is given. (a) f (1) = 1. 1 (b) f 0 (x) =
2 for all x ≥ 1. x2 + f (x) Prove that limx→+∞ f (x) exists and is less than 1 + π4 . 2. The continuous function f : [a, b] → R satisfying the following conditions is given. (a) f (a) = f (b) = 0. (b) The second derivative of f on the interval (a, b) exists [and is bounded]. Show that Z b 3 f (x) dx ≤ M (b − a) , 12 a 00 where M = sup{ f (x) : x ∈ (a, b)}. • Hint. Use the auxiliary function g(t) = (t − a)(t − b)f (x) − (x − a)(x − b)f (t). 3. Prove that Z lim

n→+∞

0

3

x2 (1 − x)xn dx = 0. 1 + x2n

1.10.2. Algebra. 1. Prove that the additive group of R has no maximal subgroup. 2. Let A, B, C, D be ideals in a unital ring R such that A + B = A + C = A + D = R.

(∗)

If M = B ∩ C ∩ D, prove that A + M = R. Give an example of a unital ring R having ideals A, B, C, D which satisfy (∗). 3. Let E be an extension field of F and α ∈ E algebraic on F of degree n. If m < n and gcd(n, m!) = 1, then show that F (α) = F (αm ). 4. Prove that no set of nilpotent matrices can span Mn (F ), where Mn (F ) is the vector space of all n × n matrices over the field F .

1.11. ELEVENTH COMPETITION

17

1.10.3. General. 1. Consider a test in which the probability of success (S) is equal to p and the probability of failure (F) is equal to q = 1 − p. Perform this test two times. If the results turn out to be F S or SF , set the random variable X to be 0 or 1, respectively. If not, do the test another two more times. Again if the results of the two tests are F S or SF , we take the random variable X to be 0 or 1, respectively; and if not, we perform the test another two more times... Prove that P {X = 0} = P {X = 1} = 12 . 2. Show that every closed set A in the plane with the property that A◦ = ∅ is the boundary of an open set in the plane. (Here, A◦ denotes the interior of the set A.) 3. Prove that every solution of the differential equation (∗) is a solution of the integral equation (∗∗) and vice versa. x00 = f (t, x), x(0) = x0 , x0 (0) = y0 ,

(∗)

where f (t, x) is a continuous function in a region D which contains the point (0, x0 ). Z t x(t) = x0 + y0 t + (t − s)f (s, x(s))ds. (∗∗) 0

4. Prove that the product of r consecutive natural numbers is divisible by r! and n! from there conclude that is always an integer. r!(n − r)!

1.11. Eleventh Competition, The University of Birjand, March 1987 1.11.1. Analysis. 1. The function f : [0, 1] → R is differentiable and f and f 0 have no common zero. Prove that the set of zeros of f in [0, 1] is finite. (Recall that a solution of the equation f (x) = 0 is called a zero of the function f .) ··

·x

2. (a) Show that the sequence of functions fn (x) = xx is pointwise convergent 1 e to a function f on [1, e ] which satisfies the functional equation f (x) = xf (x) .
+∞ 1 • Hint. Prove that for all a ∈ [1, e e ] the sequence fn (a) n=1 is increasing and bounded. Then, in view of the fact that the function g(x) = ax is continuous on R, show that f (a) = af (a) . (b) Show that the function f is one-to-one. Then using the inverse of f , conclude that f is continuous. And then, prove that the sequence (fn )+∞ n=1 converges 1 e uniformly on [1, e ]. 3. Suppose that for every and  00n ∈ N, ϕn : [−1, 1] → R has second order derivative P+∞ 0 that ϕn (0) = 1. If sup ϕn (x) : n ∈ N, x ∈ [−1, 1] < +∞, prove that n=1 an is R1 convergent, where an = n1 −1 ϕn (t) cos nπtdt .

18

1. PROBLEMS

1.11.2. Algebra. 1. Suppose that the ring R has exactly two two-sided ideals. Prove that if there is an element u in R for which ux = x for all x ∈ R, then R is a ring with identity and u = 1R . 2. Prove that A, the field of algebraic numbers, is not a finite extension of Q. 3. Let G be a finite group with the following property: for every two elements x and y of G with x 6= e and y 6= e, where e is the identity element of G, there exists an automorphism θ ∈ Aut(G) such that y = θ(x). Prove that there is a prime number p such that G∼ = Zp ⊕ Zp ⊕ · · · ⊕ Zp .

1.11.3. General. 1. The matrix A is chosen at random from the set of all 2 × 2 matrices with entries from Z. What is the probability that the determinant of A is an even number? 2. A real number c is given. Show that if one of the roots of the equation 3 x3 − x + c = 0 4 is in the closed interval [−1, 1], then all of the roots of this equation are in the interval [−1, 1]. 3. Let X be a set with n elements and C a family of the subsets of X which satisfies the following conditions. (i) If A, B ∈ C, then A ∪ B ∈ C. (ii) If A ∈ C, then Ac := X \ A ∈ C. (iii) ∅ ∈ C. Show that the number of the elements of C is equal to 2k where k ≤ n.

1.12. Twelfth Competition, Guilan University , March 1988 1.12.1. Analysis. 1. Let f : R → R be a continuous function. Moreover, suppose that there is an M > 0 such that for all x, y ∈ R f (x) − f (y) ≥ M x − y . Show that the function f is one-to-one and surjective. 2. Let (fn )+∞ n=1 be a decreasing sequence of nonnegative functions on a nonvoid set S. That is, ∀n ∈ N : fn : S −→ R, fn ≥ fn+1 , fn ≥ 0. P Prove that the series n≥1 fn converges uniformly on S if and only if the series P 7 n≥1 n(fn − fn+1 ) does. 7 This problem, as stated, is not correct! More precisely, for the “if part” of the problem, we need to assume further that the sequence (fn )+∞ n=1 converges uniformly to zero on S.

1.12. TWELFTH COMPETITION

19

3. Let g : [0, 1] → R be a Riemann integrable function. Prove that  Z 1 Z 1 Z 1 g(t)dt dx = tg(t)dt. 0

x

0

1.12.2. Algebra. 1. Let G be a group of order 2p, where p is an odd prime. Prove that G has one and only one subgroup of order p. Also prove that G has p subgroups of order p, or one subgroup of order p. In the latter case, show that G is a cyclic group. 2. Let G be an abelian group of odd order and φ a homomorphism of G of order 2. Show that every element g ∈ G can, uniquely, be written as g = xy where φ(x) = x and φ(y) = y −1 . 3. Let R be a unital ring with characteristic 2 such that for every x 6= 1 and y 6= 1, we have xy 2 = xy. Show that R is commutative. 4. Suppose that R is a unital ring, that every ideal of R is principal, and that f : R → R is a surjective homomorphism. Prove that f is an isomorphism. 5. Let A be a 3 × 3 invertible matrix over a field F such that det A = 1 and tr(A) = tr(A−1 ) = 0. Prove that A3 = I. 6. If α and β are two distinct roots of the equation 1+x+

x2 xp + ··· + = 0, 2! p!

where p is a prime with p > 2, show that α − β and α + β and αβ are irrational numbers.

1.12.3. General. 1. A circle C centered at O is given and the center of a square is on the circle. If the area of the square is not greater than half of the area of the circle, prove that one of the vertices of the square is inside the circle. 2. 1700 people have participated in a true-false test. Knowing that 15 questions have been given in the test, that none of the participants has answered two consecutive questions correctly, and that all of them have answered all questions, do there exist two equal answer sheets? 3. Evaluate 1 − Cn2 + Cn4 − Cn6 + · · · , where n ∈ N, Cnk :=

n! if k = 0, . . . , n, and Cnk := 0 otherwise. k!(n − k)!

20

1. PROBLEMS

1.13. Thirteenth Competition, University of Tehran, March 1989 1.13.1. Analysis. 1. Let f : R → R be (Riemann) integrable and f (x + y) = f (x) + f (y) for all x, y ∈ R. Show that there exists a number c such that f (x) = cx for all x ∈ R. 2. Show that if an , bn are in R for all n ∈ N, (an + bn )bn 6= 0, and that both series +∞ +∞ +∞ X X X an an an
2 are convergent, then so is the series and . b b a + bn n n=1 n n=1 n n=1 3. Let (fn )+∞ n=1 , with fn : [0, 1] −→ R, be a sequence of differentiable functions such that ||fn0 ||∞ ≤ 1 for all n ∈ N. Show that if for any continuous function g : [0, 1] −→ R, we have Z 1 lim fn g = 0, n→+∞

then the sequence

(fn )+∞ n=1

0

converges uniformly to zero on [0, 1].

1.13.2. Algebra. 1. Let R be a unital, commutative, and uncountable ring such that for every ideal 0 6= I C R, the quotient ring R/I is countable. Prove that R is an integral domain. 2. Let R be a subring of Mn (Q). If   α 0 ∈ R for all α ∈ Q, 0 α

(∗)

prove that every left (or right) ideal of R is finitely generated. By finding a left ideal in the ring    z q1 T = |z ∈ Z, q1 , q2 ∈ Q , 0 q2 which is not finitely generated, show that the condition (∗) cannot be dropped. 3. Give an example of a group G which contains two elements a and b such that a and b are of order two but ab is of order infinity. 4. Let G = GL2 (Z3 ) (the multiplicative group of all 2 × 2 invertible matrices on Z3 ), K = Z(G) denote the center of G, and    a b H= : a, b, c ∈ Z3 , ac 6= 0 . 0 c (a) Prove K ≤ H ≤ G and that the order of H is 12. (b) Prove \ K= x−1 Hx. x∈G

(c) Prove G ∼ = S4 , K where S4 denotes the group of all permutations on four letters.

1.14. FOURTHEENTH COMPETITION

5. A = (aij ) is an n × n matrix, where  δi,n aij = δi,j−1

21

j = 1, j > 1.

Suppose that ξ is an nth root of unity in C. Set   1  ξ    2   ν(ξ) =  ξ .  ..   .  ξ n−1 Prove that ν(ξ) is an eigenvector of A and find its corresponding eigenvalue.

1.13.3. General. 1. A regular n-gon is inscribed in a circle with radius one. Choose an arbitrary point on the circle and find the squares of the distances from the point M to the vertices of the regular n-gon. Prove that the sum of these values is 2n. 2. Find all polynomials f (x) with rational coefficients with the property that f (x) is irrational whenever x is irrational. 3. Let n parallel lines in the space that do not lie in a plane be given. Show that there are at least n distinct planes each of which passes through at least two lines of these given lines (n > 2).

1.14. Fourteenth Competition, The University of Isfahan, March 1990

1.14.1. Analysis. 1. Let f : R −→ R be twice differentiable on R \ {x0 } for some x0 ∈ R. If f 0 (x) < 0 < f 00 (x) on x < x0 , and f 0 (x) > 0 > f 00 (x) on x > x0 , then f is not differentiable at x0 . 2. Let g : R −→ R be a continuous function with the property that g(x) > 0 for x 6= 0 and g(0) = 0. Let f : R −→ R be a uniformly continuous and bounded function such that g ◦ f is integrable on R. Prove that limx→∞ f (x) = 0. 3. Let f : R −→ R be a continuous function and [a, b] and [c, d] two closed intervals
such that [a,
b] ⊂ f [c, d] . Prove that there exists a subinterval [r, s] of [c, d] such that f [r, s] = [a, b].

22

1. PROBLEMS

1.14.2. Algebra. 1. Prove that if G is a finite p-group (p a prime number), then G0 6= G, where G0 is the commutator (a.k.a. derived) subgroup of G. 2. Let G be a finite group and K a normal subgroup of G of order p, where p is the smallest prime dividing |G|. Prove that K is a subgroup of Z(G), where Z(G) denotes the center of the group. 3. Give an example of a ring R having two elements which have the greatest common divisor in R but do not have the least common multiple in R. 4. Prove that if F is a field and n an integer greater than one, then xn y + xn−1 + 1 is irreducible in F [x, y]. What can be said in case n = 1? 5. Let V be a finite-dimensional vector space over a field F and V1 and V2 two subspaces of V with dim V1 = dim V2 . Prove that there exists a subspace U of V such that U ⊕ V1 = U ⊕ V2 .

1.14.3. General. 1. Let S be a set with n elements, and  A = A1 , . . . , An a family containing n distinct subsets of S. Show that there exists an element x ∈ S such that the sets A1 ∪ {x}, . . . , An ∪ {x} are distinct. 2. A publisher is to exhibit 1369 titles of its published books. The books are to be exhibited as follows. Each day 100 titles are to be placed on the exhibition table and that no two titles are placed on the table more than once. Determine the maximum number of the days of the publisher exhibition. 3. Find all nonzero real numbers a1 , . . . , an with the following property: n X i=1

am i =

n X i=1

ai ,

m = 1, . . . , n + 1.

1.15. FIFTEENTH COMPETITION

23

1.15. Fifteenth Competition, Ferdowsi University of Mashhad, March 1991 1.15.1. Analysis. 1. The real function f is defined on [0, +∞) and is increasing. The function ϕ is defined on [0, +∞) by Z x f (t)dt. ϕ(x) = 0

(a) Prove that for all x, y ≥ 0  x + y
1 ϕ ϕ(x) + ϕ(y) . ≤ 2 2 (b) Conclude that ϕ is convex. 2. The real function g is continuous on [0, 1] and g(0) = 0. Define the sequence (fn )+∞ n=1 of functions on [0, 1] by g(x)(sin x)n . 1 + nx is uniformly convergent on [0, 1].

fn (x) = Prove that the sequence (fn )+∞ n=1

3. A function f from (0, +∞) into (0, +∞) is defined; it is a one-to-one correspondence and for all x, y ∈ (0, +∞), we have 2xy ≤ xf (x) + yf −1 (y). (f −1 is the inverse of f .) (a) Show that for all x, y ∈ (0, +∞), y−x y−x f (x) ≤ f (y) − f (x) ≤ f (y). y x (b) Conclude that there exists a c ∈ R such that f (x) = cx for all x ∈ (0, +∞). 1.15.2. Algebra. 1. Let G be a group of order pα m, where p is an odd prime, a ∈ N, p - m, and that G has exactly (1 + p) Sylow p-subgroups. Show that p+1 \ Pi = pα−1 , i=1

where Pi ’s are the Sylow p-subgroups of G. • Hint. [G : A ∩ B] ≤ [G : A][G : B]. 2. Let R be a unital ring such that every left ideal of it is also a right ideal of R. Prove that the intersection of all prime ideals of R is equal to the set of the nilpotent elements of R. 3. Show that for every n × n matrix A, there is an n × n matrix B such that AB is an idempotent. 8 8 This problem, as stated, is trivial! To make the problem nontrivial, prove that for every n × n matrix A with entries from a field F , there is a matrix B ∈ Mn (F ) such that AB is an idempotent whose rank is equal to that of A (see Solution 3 of 2.15.2).

24

1. PROBLEMS

1.15.3. General. 1. Find a necessary and sufficient condition for the following property: the product of two integers is divisible by the sum of them. 2. Let f be nonnegative on [0, 1] and Z

1



Z x−

0

R1 0

f (x)dx = 1. Prove that 2

1

uf (u)du

f (x)dx ≤

0

1 . 4

3. A train has n wagons. Any passenger who is boarding the train (independent of others), chooses a wagon at random to get on. (a) What is the probability that at least one passenger gets on every wagon? (b) Using (a), evaluate the following sum.         n p n p n p n−1 n 1 − 2 + 3 − · · · + (−1) np , 1 2 3 n where 1 ≤ p ≤ n.

1.16. Sixteenth Competition, Razi (Rhazes or Rasis) University of Kermanshah, March 1992

1.16.1. Analysis. 1. Let g : [0, 1] −→ R be a continuous function with g(1) = 0. If fn (x) = xn g(x), show that the sequence (fn )+∞ n=1 converges uniformly on [0, 1]. 2. The function f : R −→ {0, 1, ..., 9} is defined by  a2 x 6= [x], f (x) = 9 x = [x], where [x] denotes the integral part of x and a2 is the second digit of the decimal expansion of x − [x]. (a) Prove that f is periodic and determine its period. (b) If c is the period of f , evaluate the following integral Z c xdf (x). 0

(If the decimal expansion of a number ends in zero, reduce the last nonzero digit by one and change all other digits on the right of it to 9.) 3. Let f : R −→ R be a uniformly continuous function. Prove that there are positive numbers a and b such that f (x) ≤ a x + b for all x ∈ R.

1.17. SEVENTEENTH COMPETITION

25

1.16.2. Algebra. 1. Let G be a finite nonabelian group, distinct abelian subgroups of G such that

and A, B two

[G : A] = [G : B] = p, where [G : A] and [G : B] denote the indexes of A and B in G, respectively, and p is the smallest prime dividing |G|. Prove that Inn(G) = Zp × Zp , where Inn(G) stands for the set of all inner isomorphisms of G. 2. Let R be a ring and r ∈ R be such that r − r2 is nilpotent. Prove that R has a nonzero idempotent element whenever r is not nilpotent. 3. Let A = (aij ) ∈ Mn (Q), where aij = gcd(i, j). Is A invertible? Why?

1.17. Seventeenth Competition, Shahid Beheshti (former National) University, March 1993

1.17.1. Analysis. 1. Suppose that the function f is continuous on the interval [a, b] and that it is differentiable on (a, b). Also suppose that the graph of f is not a line segment. Prove that there exists a c ∈ (a, b) such that 0 f (b) − f (a) f (c) > . b−a

2. If p1 (x), p2 (x), p3 (x), p4 (x) are real polynomials in the indeterminate x, prove that Z x Z x Z x Z x p1 (t)p3 (t)dt p2 (t)p4 (t)dt − p1 (t)p4 (t)dt p2 (t)p3 (t)dt −1

−1

−1

−1

is divisible by (x + 1)4 . 3. Let X be a metric space and f : X −→ R a continuous function. Show that the set {x ∈ X : f (x) = 0} is an open subset of X if and only if there exists a continuous function g : X −→ R such that f = gf 2 . 4. Let f : [0, 1] −→ R be a continuous function such that f (0) = f (1). Prove that there are two points a, b satisfying the following: 0 < a ≤ b ≤ 1, b − a = 12 , and f (b) = f (a).

26

1. PROBLEMS

1.17.2. Algebra. 1. Let G be a finite group and H ≤ G be such that
∀ x ∈ G if x ∈ / H =⇒ H ∩ (x−1 Hx) = {eG } . Prove that [G : H] and |H| are relatively prime. 2. Let R be the ring of all n × n matrices over a field F , and R[x, y] the ring of all polynomials in two indeterminates x, y with coefficients from R. (note: ∀a ∈ R : ax = xa, ay = ya, xy = yx.) Suppose f, g ∈ R[x, y] are such that f g = 1. Determine gf . 3. Let V be a finite-dimensional vector space over a field F and T : V −→ V a linear transformation. Determine the dimension of ker T ∩ T (V ) in terms of the ranks of the powers of T . 1.18. Eighteenth Competition, Sharif University of Technology, March 1994 1.18.1. Analysis. 1. Let (an )+∞ n=0 be a sequence of positive reals and An = an = 0, show that the radius of a0 + · · · + an . If limn→+∞ An = +∞ and lim n→+∞ An P+∞ convergence of the power series n=0 an xn is one. 2. Let f, g be continuous and periodic functions whose periods are one. Show that Z 1  Z 1  Z 1 lim f (x)g(nx)dx = f (x)dx g(x)dx . n

0

0

0

3. For any real number x, write the binary expansion of x − [x] as follows x − [x] =

∞ X ak k=1

2k

. Pn

Show that the function g : R −→ R defined by g(x) = lim sup

j=1

aj

maps every n nonempty interval onto [0, 1] (i.e., for every nonempty interval I, we have g(I) = [0, 1]). Using this, construct an open map from R into R that is not continuous. n→∞

1.18.2. Algebra. 1. Let G be a finite group and K a normal subgroup of G such that K as a group is simple and the square of its order does not divide the order of G. Prove that if H is a subgroup of G such that H ∼ = K, then H = K. 2. The following ring is given.      Z Q a b R= = : a ∈ Z, b, c ∈ Q 0 Q 0 c Prove that every ascending chain of right ideals of R is necessarily stable; that is, if I1 ⊆ I2 ⊆ I3 ⊆ · · · , ∀j ∈ N : Ij Cr R, then there is an n ∈ N such that In = In+1 = In+2 = · · · .

1.19. NINETEENTH COMPETITION

27

3. Let V be a finite-dimensional vector space over a field F . If {Vi }i∈I is a set of proper subspaces of V such that dim Vi = dim Vj for all i, j ∈ I and if |I| < |F |, prove that there is a proper subspace U of V such that V = Vi ⊕ U, for all i ∈ I. 1.19. Nineteenth Competition, University of Kerman, March 1995 1.19.1. Analysis. 1. If f : [0, 1] −→ [0, 1] is a continuous function, show that the equation Z x

2x −

f (t)dt = 1 0

has only one root in [0, 1]. 2. Let f : (0, +∞) −→
R be a continuously differentiable function such that limx→+∞ f (x) + hf 0 (x) = 0 for some h ∈ R+ . Show that limx→+∞ f (x) = limx→+∞ f 0 (x) = 0. 3. Let A be the set of all continuous real functions on [0, 1] 9 and, for each n ∈ N, En denote the set consisting of all f ∈ A for which there is an a ∈ [0, 1] such that |f (x) − f (a)| ≤ n|x − a|, for all x ∈ [0, 1]. (a) Show that En ’s are closed sets with empty interior. (b) Conclude that there is a continuous function on [0, 1] which is nowhere differentiable.

1.19.2. Algebra. 1. Let a group G be given. Suppose that the set  G ∼ A= N CG: =G N is ordered by inclusion and that A has a maximal element. Prove that the trivial group {e} is the only maximal element of A. P 2. Let R be a ring with identity. Set S = J, where J can be any minimal left ideal of R (in case such ideals are non-existent, set S=0). (a) Prove that S is a two-sided ideal in R (i.e., S C R). (b) If S 6= 0 and the product of any T two nonzero two-sided ideals of R is a nonzero two-sided ideal of R, then S = I, where I can be any nonzero two-sided ideal of R. 3. A square matrix A = (aij )n×n over a field F is given. Suppose that there are scalars xi , yi ∈ F such that aij = xi + yj (1 ≤ i, j ≤ n). Prove that rank(A) ≤ 2. 4. Let F be a field, n ∈ N, and A ∈ Mn (F ) with rank(A) = 1. Prove det(I + A) = 1 + tr(A). 9 The set, in fact the algebra, A is assumed to be endowed with the uniform metric of A, which is induced by the uniform norm of A, usually denoted by ||.||∞

28

1. PROBLEMS

1.20. Twentieth Competition, Sharif University of Technology, February 1996 1. Let C be the field of complex numbers and  (x, y), (z, t), (x0 , y 0 ), (z 0 , t0 ) ⊆ C2 . Prove that there are scalars α, β ∈ C which are not simultaneously zero such that the vectors θ = α(x, y) + β(z, t) and θ0 = α(x0 , y 0 ) + β(z 0 , t0 ) are linearly dependent. 2. The sequence (an )+∞ n=1 of nonnegative real numbers satisfies the following property: 1 + am+n ≤ (1 + am )(1 + an ), √ n 1 + an is for all m, n ∈ N. Prove that the sequence (xn )+∞ n=1 defined by xn = convergent. 3. Let G be a group such that for all σ ∈ Aut(G) and for all x ∈ G we have σ(x) = x or σ(x) = x−1 . Prove that G is solvable. 4. The function f : ( 41 , 1) −→ R for all x ∈ ( 41 , 1) satisfies the following xf (x) = f (x). Prove that f is uniformly continuous on ( 14 , 1). 5. Let R be a commutative ring with identity and with the following properties: (a) The intersection of all of its nonzero ideals is nontrivial. (b) If x and y are zero divisors in R, then xy = 0. Prove that R has exactly one nontrivial ideal. 6. Let f , g : [0, +∞) −→ R be two functions. Suppose that Rf is decreasing and p limx→+∞ f (x) = 0 and that g is a periodic function such that 0 g = 0, where p is R +∞ the period of g. Show that 0 f g is convergent. 1.21. Twenty First Competition, University of Tehran, March 1997

1.21.1. Analysis. 1. A set S ⊂ R and a function f : S −→ R are given.  (a) Assume that α is a limit point of the set S and that for every ε > 0 the set x ∈ S : |f (x)| ≥ ε is finite. Show that limx→α f (x) = 0. (b) Assume that S is compact and point α of S, we have  that for every limit limx→α f (x) = 0. Show that the set x ∈ S : |f (x)| ≥ ε is finite for all ε > 0. 2. Let g : R −→ R be a continuous function satisfying the following
lim g(x + t) − g(x) = 0, x→+∞

for all t ∈ R. (a) Suppose that K is a compact subset of R. Show that ∀ε > 0 ∃M > 0  x > M ⇒ g(x + t) − g(x) < ε (∀ t ∈ K).

1.22. TWENTY SECOND COMPETITION

(b) Use (a) to show that  Z x+1 g(u)du − g(x) = 0, lim x→+∞

x

lim

x→+∞

29

g(x) = 0. x

3. Let f : [0, 1] −→ R be a bounded function whose limit from the left exists at any point in [0, 1]. Prove that f is Riemann integrable on [0, 1].

1.21.2. Algebra. 1. Let R be a ring and A and B ideals of R with A ⊆ B. s

By definition, A is said to be small in B and we write A ⊆ B if for every ideal C s

s

of R for which A + C = B, then C = B. Show that if A ⊆ B and C ⊆ D and s

A ∩ C = B ∩ D = {0}, then A + C ⊆ B + D.10 2. Let G be a group and H a subgroup of G such that H ≤ Z(G), where Z(G) denotes the center of G. Prove that if [G : H] = p2 , where p is prime, then the derived subgroup of G is cyclic. 3. Let A be a nonzero real n × n matrix such that A = (aij )1≤i,j≤n , aik ajk = akk aij , ∀i, j, k. Prove that (a) tr(A) 6= 0. (b) The matrix A is symmetric.
(c) The characteristic polynomial of A is equal to xn−1 x − tr(A) . 1.22. Twenty Second Competition, University of Ahwaz, March 1998

1.22.1. Analysis. 1. Let f : [0, 1] −→ R be a differentiable function with f (0) = f (1) − 1 = 0. Show that for each n ∈ N, there are x1 , . . . , xn ∈ [0, 1] such n X 1 that = n. 0 (x ) f i i=1 2. Let X be a metric space, f : X −→ R a continuous function, and (fn )+∞ n=1 a sequence ofPcontinuous nonnegative functions on X. +∞ (a) If n=1 fn = f 2 , then there exists a sequence (gn )+∞ n=1 of continuous real functions on X suchP that fn = gn f for all n ∈ N.P +∞ +∞ (b) Prove that n=1 gn = f provided that n=1 gn is uniformly convergent on X and that the interior of f −1 ({0}) is the empty set. 3. Let (fn )+∞ n=1 be a sequence of analytic functions from the region D ⊂ C into C such that fn uniformly converges to f on D. Prove that if γ is a simple closed curve inside D, then f is analytic inside and on the curve γ and that for all z0 inside γ, we have Z Z fn (z) f (z) lim dz = dz. n z − z z − z0 0 γ γ 10 The hypothesis “A ∩ C = B ∩ D = {0}” is redundant!

30

1. PROBLEMS

1.22.2. Algebra. 1. Let G be a group, H ≤ K ≤ G, [G : K] an odd number, and [K : H] = 2. Also suppose that there is a k ∈ K of order 2 which is not conjugate to any element of H in G. Prove that G has a subgroup of index 2. 2. In a commutative ring R (which is not necessarily unital), the ideal M is called maximal if R2 * M and M is not contained in any ideal other than itself and R . Now, we define a subring J of R as follows. If R has no maximal ideal, set J = R. If not, set J to be the intersection of all maximal ideals of R. Prove that J has no maximal ideal. 3. Let V be an n-dimensional vector space over a field F (n ∈ N) and S, T : V −→ V two linear transformations such that the characteristic polynomial of one of them is irreducible over F . If L = T S − ST 6= 0, then the rank of L is greater than one.

1.22.3. General. 1. If lim

x→+∞

 b + x x b−x

= 9, then b is equal to

(b) 3e . (c) ln 3. (d) ln 9. Z x 1 t2 √ 2. If lim dt exists and is nonzero, then which of the following x→0 2bx − sin x 0 t−a is admissible? (a) 3.

(a) a < 0, b = 21 . (b) a = 0, b = 12 . (c) a > 0, b = 21 . (d) a = −1, b = −1 2 . 3. Which of the following is the general solution of the differential equation ex (1 + x)dx − (xex − yey )dy = 0? (a) x2 ex−y + y 3 = c. (b) y 2 + 2xex−y = c. (c) y 2 ex−y + x2 = c. (d) xex−y + 12 y 2 = 0. 4. If A and B are two ends of the diagonal of a cube with side length 3, and an ant wants to travel from A to B, what is the smallest distance in meters that the ant needs to travel? √ √ √ √ (a) 3 3. (b) 3 + 3 3. (c) 3 6. (d) 3 5. 5. If f (x) =

+∞ X (x + 2)n , then which one is the domain of f ? 2n (n + 1) n=1

(a) [−4, 0). (b) [−4, 0]. (c) (−4, 0]. (d) None of (a), (b), and (c).

1.22. TWENTY SECOND COMPETITION

31

B

1cm

A

Figure 2

6. If x3 + y 3 = xy + 1 and in a neighborhood of (1, 1), y is a function of x, which one is y 0 (1)? (a) −1.

(b)

−2 3 .

(c) 1.

(d) 32 .

7. A particle, initially located in the origin, is discretely moving along the x-axis one unit to the right or one unit to the left with probability p and q = 1 − p, respectively. The probability that after 2k moves (k ≥ 5), the particle is 10 units far away from the origin is k+5 k k 10 p q (p + q 10 ). (a) C2k k+5 k+5 k−5 (b) C2k p q . k+5 k−5 k+5 q . (c) C2k p k+5 k−5 k−5 10 p q (p + q 10 ). (d) C2k

8. An arrow, as shown in the figure, is moving according to the following law. If the arrow is put at a point B, then it jumps as much as the length of the arc AB in the trigonometric direction. At how many point(s) one can put the arrow so that after 36 seconds it hits the point A? (a) Exactly at one point. (b) Exactly at 18 points. (c) Exactly at 35 points. (d) At no point.  9. Consider the function f (x) =

x2 0

is differentiable is/are (a) the two points x = 0, 1. (b) exactly one point. (c) rational points. (d) no point. 10. Consider the series

+∞ X cos(xn ) . 1 + n2 n=0

(a) The series converges only at zero.

x irrational . The point(s) at which f x rational

32

1. PROBLEMS

(b) The series converges everywhere. (c) The series converges at nonzero points. (d) The series converges absolutely at one point. 11. If we consider the sequence (xn )+∞ n=1 (x1 6= 0) of real numbers satisfying the xn following recursive relation xn+1 = , then 1 + xn (a) xn → 1 as n → +∞. (b) (xn )+∞ n=1 has a subsequence converging to one. (c) (xn )+∞ n=1 has a subsequence converging to zero. (d) (xn )+∞ n=1 is divergent. 12. Which of the following propositions is correct? R n dx exists and is positive. (a) limn 0 1+x R n dx 2 exists and is negative. (b) limn 0 1+x R n dx 2 (c) limn 0 1+x2 is divergent. 1 (d) f (x) = 1+x 2 is not integrable on the interval [0, n]. 13. If the function f : R −→ R is differentiable and its derivative is bounded, then (a) the function f (b) the function f (c) the function f (d) the function f

is bounded. is increasing. is uniformly continuous. is decreasing.

14. If the function f : [0, 1] −→ [0, 21 ] is continuous, then (a) the graph of f intersects the line y = 2x. (b) there exists only one point x0 for which f (x0 ) = x0 . (c) f (x) 6= 0 for all x ∈ [0, 1]. (d) the equation f (x) = x has at most two solutions. |x| for all x ∈ R, then 1 + |x|

15. If f (x) =

(a) the function f is increasing. (b) f (x + y) =
f (x) + f (y).

(c) f f (x + y) ≤ f f (x) + f f (y) . (d) the function f is continuous only at zero and f (x + y) ≤ f (x) + f (y). 16. Below is drawn the graph of the polar curve r = θ. Which one is the area of the shaded region? (a)

R

(b)

R

(c)

R

(d)

R

π 2 3 π 6 −π 1 6 −π 2 3 5π 6 1 2π 2 3 11π 1 6 5π 2 3

θ dθ. θ2 dθ.

θ2 dθ −

R

θ2 dθ −

π 3 π 6

R

π 3 π 6

1 2 2 θ dθ. 1 2 2 θ dθ.

1.22. TWENTY SECOND COMPETITION

33

Figure 3

17. The value of (a) −44e. Z

R1 0

x5 ex dx is

(b) 120 + 44e.

x2

18. If y =

(x + t)dt, then sin x

(a) 5.

(b) 0.

(c) 44e.

(d) e.

dy at x = 0 is equal to dx (d) −1.

(c) 2.

Z 19. For what values of p, does the integral 1

2

dx converge? x(ln x)p

(a) p = 1. (b) p > 1. (c) p < 1. (d) p = 2. Z +∞ −ax e − e−bx dx (0 < a < b) is equal to 20. x 0 (a) ln( ab ).

(b) ln( ab ).

(c) ln(ab).

(d) eab .

21. Which of the following statements is not correct? (a) A power series might converge at both of the end points of its interval of convergence. (b) A power series can converge conditionally or diverge at any of the end points of its interval of convergence. (c) A power series can converge absolutely at one of the end points of its interval of convergence and converge conditionally at the other end point. (d) A power series can converge conditionally at one of the end points of its interval of convergence and diverge at the other end point. 22. Using the spherical coordinates to compute the volume of the ellipsoid, the Jacobian of the transformation is equal to (a) ρ2 sin(ϕ).

(b) ρ sin(ϕ).

(c) abcρ2 sin(ϕ).

(d) abc.

→ − → − → − 23. Letting F (x, y, z) = (x − y) i + (y − z) j + (x − y) k (−1 ≤ x, y, z, ≤ 1) be a vector field, how much is the flux of the field F through the surface of a cube of side length two centimeters? (a) 8 cm3 .

(b) 16 cm3 .

(c) 1 cm3 .

(d) 14 cm3 .

34

1. PROBLEMS

→ − → − → − 24. For the vector field H F (x, y, z) = f (x, y, z) 2i + 2g(x, y, z) j + h(x, y, z) k , we have Curl(F ) = 0 but C F.dR 6= 0, where C : x + y = 1. Then (a) div(F ) < 0. (b) the domain of F is connected. (c) the domain of F is simply connected. (d) the domain of F is not connected. 25. The family of orthogonal curves to the curves r = c(1 + cos θ) is (a) r = c(1 − sin θ). (b) r = c sin θ. (c) r = c(1 + sin θ). (d) r = c(1 − cos θ). 26. The general solution of the differential equation y 00 − 2y 0 − 3y = 64xe−x is (a) y = c1 e3x + c2 e−x − e−x (8x2 + 4x + 1). (b) y = c1 e3x + c2 e−x . (c) y = c1 e3x + c2 e−x − 8x2 e−x . (d) y = c1 e3x + c2 e−x − e−x (8x2 + 1). 27. Using the Gamma function for the definition of the factorial of numbers, ( 21 )! is √ p √ a) π. (b) 2π . (c) √1π . (d) π2 . 28. The equation x4 + x2 − 1 = 0 has roots as follows. (a) Two positive roots, two negative roots. (b) One positive root, one negative root, two nonreal roots. (c) No real root. (d) Four real roots. 29. The equation f (x) = e−x − sin x = 0 has a real root in the interval [0.4, 0.8]. The appropriate fixed point iteration function to find this root is (a) g(x) = x + e−x − sin
x. (b) g(x) = − ln sin(x) . (c) g(x) = sin−1 (e−x ).
(a) g(x) = 21 2x + e−x − sin x . 30. In the bisection method, the number of times needed in order to obtain accuracy within ε is equal to log b−a ε . log 2 (b) log(b − a) − log(ε). (c) cannot be determined. log(b − a) (d) 1 + . log 2

(a) 1 +

1.22. TWENTY SECOND COMPETITION

31. In the iteration method, Pn+1 = g(Pn ) (n ≥ 0) where g(x) = x −

35

f (x) , as the f 0 (x)

multiplicity of the root x = P increases, (a) g 0 (P ) approaches zero. (b) the speed of the convergence does not change. (c) g 0 (P ) gets very close to one. (d) the method is not convergent. 32. Which of the following propositions is correct? (a) If (b) If (c) If (d) If

P+∞ an n=1 an . n=1 an+1 converges, then so does the series P+∞ nan = 0. n=1 an converges, then limnP +∞ limn nan = 0, then the series n=1 an is convergent. P+∞ P+∞ an+1 n=1 an . n=1 an converges, then so does the series P+∞

33. Which one is the equation  of the plane which includes the line 2y − 2 = 2x = z x−y =0 and is parallel to the line ? z=0 (a) y − x = 1. (b) y = 2x. (c) z − 2y − 2x = 2. (d) y − x = −1. 34. Which of the following statements is a more complete definition of “algorithm”?

(a) “Algorithm” is a method for solving a problem. (b) “Algorithm” is a logical method together with a terminating condition for solving a problem. (c) “Algorithm” is a logical procedure without ambiguity, which includes a finite set of stages that are related to one another, together with a terminating condition. (d) “Algorithm” just means method. 35. Every digit in octal expansion is equivalent to (a) one digit in decimal expansion. (b) five digits in binary expansion. (c) two digits in hexadecimal expansion. (d) three digits in binary expansion. 36. A is infinite if and only if (a) A is countable. (b) card(A \ B) = card(A) for all B ⊆ A. (c) A is equivalent to any of its infinite subsets. (d) for all natural numbers n, A includes a subset which is equivalent to an n-set.

36

1. PROBLEMS

37. Let A be an ordered set and B ⊆ C ⊆ A. Which of the following claims is true? (a) If sup(A) and sup(C) exist, then sup(B) ≤ sup(C). (b) If sup(B) and sup(C) exist, then sup(C) ≤ sup(B). (c) If sup(B) and sup(C) exist, then inf(B) = inf(C). (d) There is no relation between sup(B) and sup(C). 38. Which of the following is equivalent to Zorn’s Lemma? (a) Every subset of an ordered set having an upper bound has sup. (b) There is a choice function for every nonempty family of nonempty subsets of an arbitrary set X. (c) If card(A) ≤ card(B) ≤ card(P (A)), then card(A) = card(B) or card(B) = card(P (A)). (d) Every bounded subset of an ordered set has a maximal element. Z a sin xdx 39. The integral is equal to x + e−x e −a (a) 2. (b) 0.R a xdx (c) 2 0 esin x +e−x . (d) It does not exist. 40. If f (a + b − x) = f (x) for all x ∈ [a, b], then

Rb a

xf (x)dx is equal to

Rb f (x)dx. (a) a−b 2 Rb a (b) a f (x)dx. Rb f (x)dx. (c) a+b 2 a R b (c) (a + b) a f (x)dx. 1.23. Twenty Third Competition, Sharif University of Technology, March 1999

1.23.1. Analysis. 1. Let X = Cb (R) be the space of all continuous bounded functions g : R −→ R which is endowed with the norm ||g|| = supt∈R g(t) . For a given function f ∈ X , define fα : R −→ R by fα (t) = f (t + α), where α ∈ R. Prove or disprove the following propositions. (a) If fα : α ∈ R is compact, then f is uniformly continuous on R. (b) If f is uniformly continuous on R, then the set fα : α ∈ R is compact in X. 2. Let α ∈ R and α 6= 2kπ for all k ∈ Z. Find the sum of the series justify your answer. Ditto for

+∞ +∞ X X sin(nα) cos(nα) and . n n n=1 n=1

+∞ inα X e and n n=1

1.24. TWENTY FOURTH COMPETITION

37

3. Let f, f1 , f2 , . . . be real continuous functions on the interval [a, b] whose derivatives are also continuous on [a, b]. If the sequence (fn )+∞ n=1 converges to f pointwise and the sequence (fn0 )+∞ converges pointwise to a continuous function g, show that n=1 f 0 (x) = g(x) for all x ∈ [a, b]. (Note: at the end points a and b, only the right and left derivatives are to be considered.)

1.23.2. Algebra. 1. Let G be a nonabelian group. Prove that the group of the inner automorphisms of G cannot be nonabelian of order 8.11 2. Let R be a ring and H the intersection of all nonzero right ideals of R. Show that if H 6= 0, then H is a two-sided ideal in R and that we have H 2 = 0, or else R is a division ring. 3. Let A be an n × n matrix with entries from a field F . Prove that if for every matrix B with trace zero, we have tr(AB) = 0, then A = λI for some scalar λ in F. 1.24. Twenty Fourth Competition, Khajeh Nasir Toosi University of Technology, May 2000

0 1.24.1. First Day. 1. Let f,
g : [a, b] −→ R
be continuous and that g exists 0 0 on [a, b]. Show that if f (a) − g (a) g (b) − f (b) > 0, then there exists a c ∈ (a, b) such that f (c) = g 0 (c).

2. Let D be a domain in the complex plane and u a real valued harmonic function on D. Show that if the set  A = (x0 , y0 ) ∈ D : ∃r > 0  ∀(x, y) ∈ Br (x0 , y0 ) ∩ D ⇒ u(x, y) ≤ u(x0 , y0 ) is nonempty, then u is constant on D. Recall that a domain is a connected open set and that p  Br (x0 , y0 ) = (x, y) ∈ R2 : (x − x0 )2 + (y − y0 )2 < r . 3. Let t(n) be the smallest prime factor of n for all 1 < n ∈ N. Prove that t(n) < t(3n − 2n ). 4. Let G be an infinite group with the property that for every two infinite subsets X and Y of G, there exist x ∈ X and y ∈ Y such that xy = yx. Prove that if the center of G has finite index in G, then G is abelian. 5. Let M (n, d) denote the maximal number of n-tuples from {0, 1} such that every two of which differ at least in d components. (For instance, M (4, 3) = 2.) Prove that M (2d − 1, d) ≤ 2d. 6. A ruthless governor has made three mathematicians, who have been sentenced to death, play the following game. 11 This problem is wrong!

38

1. PROBLEMS

Two lookalike boxes, one of which contains two black marbles and a white marble, and the other contains two white marbles and a black marble, are provided. Each of the convicts picks a marble, which is not to be replaced, from one of the boxes at random. If the picked marble is black, the convict will be executed; if not, s/he will be set free. Assuming that every convict witnesses the choice(s) made by the convict(s) prior to oneself and that the convicts will do the most logical effort, what is the probability that the second person survives? The probability of survival for the third person is higher or for the second person?

1.24.2. Second Day. 1. Let B be a nonempty bounded set in Rn with the property that for each pair of points x, y in B, there exists an open ball U such that U ⊆ B and x, y ∈ U . (a) Prove that B is an open ball. (b) Show that if we replace Rn by a complete metric space, then the conclusion of (a) does not necessarily hold. 2. Let f be a continuous function on the interval [a, b] such that Show that for each k ∈ (0, 1), there exists a c ∈ (a, b) such that Z c Z b f (t)dt = k f (t)dt. a

Rb a

f (t)dt 6= 0.

a

3. Let matrices A, B, C be such that AB, BC, and ABC make sense. Denoting the rank of any matrix P by r(P ), prove that (a) r(BC) + r(AB) ≤ r(ABC) + r(B). (b) r(A) + r(B) ≤ r(AB) + n, where n is the number of columns of the matrix A. 4. Let R be a ring. Prove that if for any left ideal I of R, we have I 2 = I, then for any two-sided ideal K of R and any left ideal I of R, we have I ∩ K = KI. (Remark: Note that R is not necessarily unital.) 5. In an athletic tournament, n teams have participated and any two teams have played once against one another. Assuming that the game result is to be either win or lose, prove that at the end of the tournament there are two teams whose wins are equal if and only if there are three teams A, B, C such that A has won B, B has won C, and C has won A. 6. Two persons are playing the following game. First, the first person chooses a triple from {0, 1} (for instance, (0, 1, 0)). The second person, who knows the triple chosen by the first person, chooses a different triple. Then a machine which at random generates 0 with probability 21 and 1 with probability 12 is turned on. We write the numbers generated by the machine from left to right and the player whose chosen triple comes first is to win the game. Show that no matter what choice is made by the first person, there is a choice for the second person so that the probability of winning the game by the second person is greater than 21 .

1.25. TWENTY FIFTH COMPETITION

39

1.25. Twenty Fifth Competition, Imam Khomeini International University of Qazvin, May 2001

1.25.1. First Day. 1. Let G be a finite group of order n such that [G : Z(G)] = 4. Prove that 8|n. For any given natural number n such that 8|n, construct a group with the aforementioned property. (Here, Z(G) denotes the center of G.) 2. Let S, T be two linear transformations on a vector space V . Prove that if S 2 = S, T 2 = T , ker T ⊆ imS, and imT ⊆ ker S, then T + S is the identity transformation. 3. Let f be a twice differentiable function such that f 00 (t) < 0 for all t ∈ R. Prove that if for two real numbers x and y we have f 0 (y) + x < f (y + 1), then f (y) > x. 4. Let f : (a, b) −→ R be continuously differentiable and lim f 2 (x) = 0, lim− f 2 (x) = e − 1.

x→a+

x→b

Prove that if 2f (x)f 0 (x) − f 2 (x) ≥ 1 for all x ∈ (a, b), then 0 < b − a ≤ 1. Give an
2 example in which b − a = 1. (Note: f 2 (x) = f (x) .) 5. Seven boxes, on each of which a number from 1 to 7 is written, are at our disposal. Seven marbles numbered 1, seven marbles numbered 2, ..., and seven marbles numbered 7 are provided. We place the marbles in the boxes at random in such a way that each box has exactly seven marbles. Play the following game. In stage one, a marble is drawn at random from box number one (without replacement). In stage i (i ≥ 2), a marble is drawn at random from the box whose number is identical to that of the marble drawn in stage (i − 1) (without replacement). The game is stopped when it is needed to draw a marble from an empty box. Find the probability that all the marbles are drawn from all boxes. 6. Prove that the number of triangles, with integer sides, each of which having perimeter n is equal to the number of partitions of the number n into the summands 2, 3, and 4 in which the summand 3 appears at least once.

1.25.2. Second Day. 1. Let R be an integral domain and U (R) the multiplicative group of the units of R. Prove that every finite subgroup of U (R) is cyclic. 2. Let p be an odd prime. Prove that every prime factor of 2p + 1 which is different from 3 is of the form 2kp + 1. 3. Let (X, d) be a compact metric space and f : X −→ X a surjective function. Show that if
d f (x), f (y) ≥ d(x, y), for all x, y ∈ X, then f is continuous. 4. Let f : C −→ C be a function with f (0) = 0 and such that for all z ∈ C and w ∈ {0, 1, i}, we have f (z) − f (w) = z − w .

40

1. PROBLEMS

Find the function f explicitly. (i2 = −1.) 5. Let n be a an odd number and A an n × n matrix whose entries are from the set {−1, 1}. If the product of the entries of the ith row is shown by ai and the product of the entries of the jth column is shown by bj , prove that n X i=1

ai +

n X

bj 6= 0.

j=1

Does the above assertion hold for an even n? 6. A society is called ideal if for every two distinct members a and b of the society, there exists a member c of it such that c is acquainted with one and only one of a or b (We assume that acquaintance is a symmetric and nonreflexive relation.) Prove that every ideal society of n people (n ≥ 2) contains an ideal society of n − 1 people. 1.26. Twenty Sixth Competition, Institute for Advanced Studies in Basic Sciences (IASBS) of Zanjan, May 2002

1.26.1. First Day. 1. Let f : [a, b] → [a, b] be a continuous function which is differentiable on (a, b) and f (a) = a , f (b) = b. Prove that there exist two distinct points x1 and x2 in (a, b) such that f 0 (x1 )f 0 (x2 ) = 1.  2. Suppose that U = z ∈ C : |z| < 1 and D is an open set in C such that U ⊆ D, and that the function f is analytic on D. Also let f (0) = 1 and that |f (z)| > 2 for all z for which |z| = 1. Prove that there exists z0 ∈ U such that f (z0 ) = 0. 3. Letting p be an odd prime, prove that pp−1 + (2p − 2)p−1 ≡ 1


mod p(2p − 2) .

4. Let R be a unital commutative ring such that every ideal of it is principal. Prove that if R has only one maximal ideal, then Rx ⊆ Ry or Ry ⊆ Rx for all x, y ∈ R. 5. A k-element cover for a set S is a collection of k nonempty subsets of S such that the union of it is S. An n-element cover of the set S is called minimal if no n − 1 elements of it cover S. If the number of the minimal k-element covers of a set with n elements is denoted by M (n, k), prove that
n n M (n, n − 1) = 2 −n−1 . 2 6. Let X be a set with n elements and positive integers m, k be such that m ≤ k and m + k ≤ n. Denote the set of all subsets of X with m elements by X {m} . is the set of real numbers. Prove Suppose that f : X {m} → R is a function where RX {k} that if for every member S of X we have f (T ) = 0, then f ≡ 0. T ∈X {m} ,T ⊆S

1.27. TWENTY SEVENTH COMPETITION

41

 1.26.2. Second Day. 1. Let r1 ,r2 , . . . be the set of rational numbers in [0, 1]. For each x ∈ [0, 1], we let Ax = n ∈ N : rn ≤ x and define the function f : [0, 1] → R by X 1 f (x) = . 2n n∈Ax

Prove that f is continuous at any irrational point of the interval [0, 1].
2. Let f : R → R be a continuous function and that f ◦ f ◦ f (x) = x for all x ∈ R. Prove that f (x) = x for all x ∈ R. 3. Let G be a group which has two distinct maximal subgroups, say, H and K. Prove that if H and K are abelian and Z(G) = {e}, then H ∩ K = {e}. 4. Let A be an n × n matrix with entries in a field F . Let λ ∈ F be an eigenvalue of An corresponding to an eigenvector x such that for all a0 , a1 , . . . , an−1 if

n ∈ F , 12 Pn−1 i n a A x = 0, then a = a = · · · = a = 0. Prove that A − λI = 0. i 0 1 n−1 i=0 5. An extended die is a homogeneous cube on each of whose sides an arbitrary positive integer is written. (These numbers are not necessarily distinct.) Can one design two extended dice (which are not necessarily identical) in such a way that the probability that the sum of the numbers on the rolled dice is the same as those 1 , the of two ordinary dice? (That is, the probability of getting a sum of 2 is 36 2 probability of getting a sum of 3 is 36 , etc.) In case, you answer in the affirmative, find all possible answers; and if you answer in the negative, prove your claim. 6. A matrix Mm×n = (mij ) with real entries is said to be balanced, if whenever for all i, i0 , j, j 0 we have 1 ≤ i < i0 ≤ m and 1 ≤ j < j 0 ≤ n, then mij + mi0 j 0 ≤ mij 0 + mi0 j . Suppose that there are two rows i1 i2 such that by interchanging the rows, the matrix remains balanced. Prove that for all i01 and i02 , if i1 < i01 < i02 < i2 , then by interchanging the rows i01 and i02 the matrix remains balanced. 1.27. Twenty Seventh Competition, Bu-Ali Sina (Avecina) University of Hamedan, May 2003

1.27.1. First Day. 1. Let n be a natural number greater than 1. Denote the set of all n × n matrices with real entries by Mn (R). Define the following metric on Mn (R). For A = (Aij ) and B = (Bij ),  d(A, B) = max Aij − Bij : i, j = 1, . . . , n . Prove that GLn (R), the set of nonsingular matrices, is an open and disconnected subset of Mn (R). 2. Let f be a function that is analytic on C. Let L and M be two orthogonal lines intersecting at a point A such that f (L) = L and f (M ) = M . Prove that if z1 and z2 are two complex numbers that are symmetric with respect to A, then f (z1 ) is the image of f (z2 ) with respect to A. 12 Prove that An − λI = 0!

42

1. PROBLEMS

3. Let (an )+∞ n=1 be a sequence of real numbers defined as follows q p a0 = 0, a1 = b, an+1 = an 1 + a2n−1 + an−1 1 + a2n , n ≥ 1. Determine an in terms of b.  4. Let K be a nonempty set and ∅ = 6 I ⊆ K. Let Ai : i ∈ I be a family of the subsets of K. Prove that if    i:i∈ / Ai ∈ Ai : i ∈ I ∪ ∅ , then i ∈ Ai for all i ∈ I. 5. Let A be a 3 × 2 matrix and B a 2 × 3  8 AB =  2 −2

matrix with complex entries for which  2 −2 5 4 . 4 5

Determine the matrix BA. (Hint: Compute (AB)2 .) 6. Let G be a group and H a subgroup of it such that for all x ∈ G \ H and all y ∈ G, there is a u ∈ H such that y −1 xy = u−1 xu. Prove that H is normal in G and G/H is abelian.

1.27.2. Second Day. 1. Let f : R → R be a differentiable function, a, b ∈ R and a < b. If f (a) = f (b) = 0, f 0 (a) > 0, and f 0 (b) > 0, prove that f 0 has at least two roots in the interval (a, b). 2. Prove that the interval [0, 1] cannot be written as a union of pairwise disjoint closed intervals each of which having a positive length less than one. 3. Let D be a countable subset of the Euclidean plane, i.e., R × R. Prove that there is a partition for D into two subsets X and Y such that any line parallel to the x-axis intersects X in finitely many points and any line parallel to the y-axis intersects Y in finitely many points.  Sn 4. Let A1 , . . . , An be a family of finite sets and S = i=1 Ai . Suppose that a fixed number k, 1 ≤ k ≤ n, satisfies the following conditions.  (a) The union of any k elements of the family A1, . . . , An is equal to S. (b) The union of any k − 1 elements of the family A1 , . . . , An is not equal to S.
n (c) |S| = k−1 Find the number of the elements of all of Ai ’s. 5. Let a, b be two natural numbers such that gcd(a, b) = 1. Prove that ordab (a + b) = lcm[ordb (a), orda (b)] (Here, by ordn (m), we mean the order of m modulo n.) 6. Let R and R0 be two rings whose elements are all idempotents and consider a function f : R → R0 that is one-to-one and onto such that f (xy) = f (x)f (y) for all x, y ∈ R. Prove that R ∼ = R0 .

1.28. TWENTY EIGHTH COMPETITION

43

1.28. Twenty Eighth Competition, Sharif University of Technology, May 2004

1.28.1. First Day. 1. Prove that any one-to-one and entire function is of the form f (z) = az + b, where a and b are fixed complex numbers and a 6= 0. 2.  Let d1 and d2 be the Euclidean metrics on R and R × R, respectively, A = (x, y) ∈ R × R : x = 0 or y = 0 , and d = d2 |A the metric induced by the Euclidean metric of R × R on A. Prove that if f : (R, d1 ) → (A, d) is continuous and surjective, then f −1 {(0, 0)} has at least three elements.13 3. P Let f be an arithmetic P function with the property that for every natural number n, d|n f (d) = n2 . ( d|n means sum over all positive divisors of n.) If φ is Euler’s totient function, prove that  Y 1 f (n) =n 1+ , φ(n) p p|n,p prime

for all natural numbers n > 1. 4. Let G be a subgroup of Sn with the property that for all 1 ≤ i ≤ n and 1 ≤ j ≤ n, there exists σ ∈ G such that σ(i) = j. Prove that for all 1 ≤ k ≤ n, Gk ∩ Z(G) = {e} where Gk = τ ∈ G : τ (k) = k , Z(G) denotes the center of G, and e is the identity element of G. 5. Let A be the set of all vectors of n-tuples whose entries are zero or one in such a way that in each vector the numbers of one entries is odd. By explaining your reasoning, determine, the maximal number of vectors from the set A such that every two of which share an even number of one entries. 6. Let (P, ≤) be a poset (reflexive, antisymmetric, and  transitive)  with n elements. For every element i ∈ P , set Ui = j : j ∈ P, j > i and Li = j : j ∈ P, j < i . Suppose that relative to every element i of P , there corresponds a real number Xi subject to the following ( 1− j∈Li Xj if Li 6= ∅, n−|Ui | Xi = 1 if Li = ∅. n−|Ui |

P

Prove that 0 ≤ Xi ≤ 1 for all i ∈ P .

1.28.2. Second Day. 1. Let, in Figure 4, the function f be one-to-one and continuous and that for every point P on the curve y = 2x2 , the areas of the regions A and B are equal to one another. Determine the function f explicitly. 2. Let f : [a, b] → (a, b) be a continuous function. Prove that for every natural number n, there exist a positive number α and c ∈ (a, b) such that n f (c) + f (c + α) + · · · + f (c + nα) = (n + 1)(c + α). 2 13 Prove that f −1 {(0, 0)} has infinitely many elements!

44

1. PROBLEMS

P B

A 1.5 Figure 4

3. A matrix O is called orthogonal if OOt = Ot O = I, where Ot denotes the transpose of the matrix O . Let A and B be two orthogonal square matrices with entries in C and det(A) + det(B) = 0. Can one conclude that det(A + B) = 0? Why? 4. Suppose that R is a unital ring and that there is a natural number n with the property that for all k ∈ {n, n + 1, n + 2} and for all x, y ∈ R, (xy)k = xk y k . Prove that R is commutative. 5. Of three people who are suspected to have committed a murder, one is the murderer. The three people are to be interrogated by taking a test which has five questions. If the person being tested is innocent, the probability that the person responds positively to any question is 0.4; and if guilty, the probability of responding positively is 0.8. Of these three people, one who is chosen at random is being interrogated. This person responds positively to four questions and negatively to one question. What is the probability that the person is the murderer? 6. Let X be a set and r a natural number. Let Xr be the set of all subsets of X which have r elements. Suppose that F is a subset of Xr with the property that the intersection of every k elements of F is nonempty (k ≥ 2 is a fixed number). If we set  I(F ) = min |T | : T ⊆ X, T ∩ A 6= ∅, ∀A ∈ F , prove that I(F ) ≤

r−1 k−1

+ 1.

1.29. TWENTY NINTH COMPETITION

45

1.29. Twenty Ninth Competition, University of Mazandaran in Babolsar, May 2005

1.29.1. First Day. 1. Let f : [0, a] → R be a continuous and positive function. Prove that  Z a Z a  dx f (x)dx ≥ a2 . 0 f (x) 0 2. Let (ni )+∞ i=1 be an increasing sequence (not necessarily strictly) of natural num+∞ X 1 converges to a real number x. bers with n1 ≥ 2 such that the series n · · · ni i=1 1 Prove that x is rational if and only if there exists a natural number ` such that ni = n` for all i ≥ `. 3. Consider the sequence of all natural numbers whose digits consist of 1. 1, 11, 111, . . . Prove that if the natural number m is relatively prime with respect to 30, then infinitely many terms of the above sequence are divisible by m. 4. Let R be an arbitrary ring (not necessarily unital) that has no nonzero nilpotent two-sided ideal. Prove that every nonzero right ideal in R has an element whose square is not zero. 5. Let Z, E, and O be the set of integers, even integers, and odd integers, respectively. Set  X := A ∈ P(Z) : both A ∩ E, A ∩ O are infinite ,  Y := A ∈ P(Z) : A is infinite . We know that there exists a one-to-one correspondence from X into Y . Exhibit a law for a surjective function f : X → Y . 6. Let S be the k-dimensional vector space of all binary (zero and one) sequences of length n over the field Z2 . The distance between two elements X, Y of S are to be defined as the number of entries of X and Y that are different from one another. (More precisely, if X = (x1 , . . . , xn ) and Y = (y1 , . . . , yn ), then the distance between X and Y is equal to the number of i’s for which xi 6= yi .) Suppose that the minimum distance between two distinct elements of S is equal to d. Prove that n2k−1 . d≤ k 2 −1

46

1. PROBLEMS

 1.29.2. Second Day. 1. Let D = z ∈ C : |z| < 1 and f : D → C be an analytic function such that f ( n1 ) ∈ R for all natural numbers n ≥ 2. Prove that for all natural numbers n, f (n) (0) ∈ R, where f (n) denotes the nth derivative of the function f . 2. (i) Prove that if X is a connected metric space, then for all ε > 0 and every two points x, y ∈ X, there exist an n ∈ N and points x1 , . . . , xn ∈ X such that x1 = x , xn = y and that d(xi , xi+1 ) < ε for all i < n. (ii) Give an example showing that the converse of the assertion of (i) does not hold. (iii) Prove that the converse of the assertion of (i) holds provided that X is compact. 3. Let G be a group and K a subgroup of it. NG (K) is isomorphic to a subgroup of Aut(K). (i) Prove that CG (K) (ii) Prove that if K is cyclic and K E G = G0 , then K ≤ Z(G). 4. Let F be a field, Mn (F ) the set of all n × n matrices with entries in F , A ∈ Mn (F ), and the invertible matrix P ∈ Mn (F ) be such that P −1 AP is upper triangular. Prove that every two invariant subspaces of A are comparable with respect to inclusion if and only if there exist a λ ∈ F and a nilpotent matrix N ∈ Mn (F ) with N n−1 6= 0 such that A = λI + N . 5. Two people, named A and B, are playing a coin-flipping game as follows. They throw their coins, if the result of flipping the two coins turns out to be the same, A is going to win both coins; if not, B is going to take both coins. Suppose that A has m coins and B has n coins. On average, how many times should the game be played till one of the players runs out of the coins? 6. Let C be the Cantor  set. Prove that C − C = [−1, 1]. (Hint: it is worth mentioning that C − C = x − y : x, y ∈ C and that C is equal to the set of all numbers in [0, 1] whose ternary expansions have only 0 or 2.)

1.30. Thirtieth Competition, Tafresh University, May 2006

1.30.1. First Day. 1. Let f : [0, +∞) −→ R be a continuous function and limx→+∞ f (x) = 1. Evaluate the following limit. Z 2006 lim f (nx)dx. n→+∞

1385

2. Let m ∈ N, c ∈ C, aj ∈ C, and |aj | = 1 for all 1 ≤ j ≤ m. If lim

n→+∞

m X j=1

then c = m and aj = 1 for all 1 ≤ j ≤ m.

anj = c,

1.30. THIRTIETH COMPETITION

47

3. Let R be a commutative ring with identity and a an element of R with a3 −a−1 = 0. Prove that if J is an ideal of R such that the quotient ring R/J has at most four elements, then J = R. 4. Let p, q be prime numbers such that p = 2q + 1 and q ≡ 1 (mod 4). Prove that 2 is a primitive root modulo p. 5. Prove that

X  2m  ≥ 22m−1 , m+k √

|k|< m

for all m ∈ N with m ≥ 1 Hint. By the Chebyshev Inequality if X is a random variable whose mean and variance are µ and σ 2 , respectively, then P (|X − µ| ≥ λσ) ≤ λ12 for all λ > 0. 6. A financial group has n members each of whom has a number of coins. Let k ∈ N be a fixed positive integer. A business among four people, say, p1 , p2 , p3 , and p4 , of the n people, who are chosen arbitrarily, can be done subject to the following conditions:
(a) sum of the number
of coins owned by p3 and p4 − sum of the number of coins owned by p1 and p2 + 2k > 0, (b) each of p1 and p2 has at least k coins. If the conditions (a) and (b) are met, then a business is done as follows. Each of p3 and p4 earns exactly k coins which are lost by each of p1 and p2 . Prove that after a finite number of doing business, the condition (a) or (b) in the above does not hold for any four people of the n people.

1.30.2. Second Day. 1. Let X be a separable metric space, that is, X has a countable dense subset. Let f : X −→ R be a function for which limx→a f (x) exists for all a ∈ X. Prove that the set of points at which f is discontinuous is at most countable. P+∞ 2. Suppose that f (z) = n=0 an z n defines a nonconstant analytic function, where the radius of the convergence of the series is equal to R > 0. Prove that the closest 0| distance from a zero of f to the origin is at least MR|a +|a0 | , where M = M (R) = sup|z|=R |f (z)|. 3. Let G be a group with the property that the order of any element of G0 , the derived subgroup of G, is finite. Prove that the set of all elements of G whose orders are finite is a subgroup of G. 4. Let K be a field, F a subfield of K, n ∈ N, and A an F -algebraic n×n matrix with entries in K such that rank(A) = rank(A2 ). Prove, firstly, that K n = imA ⊕ ker A and, secondly, that there is a polynomial f ∈ F [x] such that E := f (A) is an idempotent matrix satisfying E(x + y) = x for all x ∈ imA and y ∈ ker A.  5. For an arbitrary subset C of the natural numbers, set C ⊕ C := x + y|x, y ∈ C, x 6= y . Prove that there is a unique partition of the set of natural numbers into two subsets A and B such that none of A ⊕ A and B ⊕ B contains a prime.

48

1. PROBLEMS

Hint. Recall that by Bertrand’s conjecture (a.k.a. Bertrand’s principle) for every natural number n there exists at least a prime number p such that n < p ≤ 2n. 6. Let C be a circle whose perimeter is equal to one and 0 < α < 21 . The distance between two points of the circle is defined to be the length of the shortest arc joining the two points. Let T ⊆ C and T = I1 ∪· · ·∪Im , where m ∈ N and Ij ’s are disjoints arcs of the circle (1 ≤ j ≤ m). Prove that if the distance between every two points of T is less than or equal to α, then m X `(Ij ) ≤ α, j=1

where `(Ij ) denotes the length of the arc Ij . 1.31. Thirty First Competition, Ferdowsi University of Mashhad, May 2007

1.31.1. First Day. 1. We have colored a circle by two colors. Does there necessarily exist a monocolored triangle inscribed in the circle which is (a) an equilateral triangle? (b) a right triangle? (c) an isosceles triangle? 2. The Minkowski sum of two sets A, B ⊆ Rd is defined as follows  A+B = a + b ∈ Rd |a ∈ A, b ∈ B . Prove that if A is bounded and B is closed, then (A + B)0 = (A0 + B) ∪ (A + B 0 ), where A0 stands for the set of the limit points of the set A. 3. Suppose that there is a group G which has exactly n subgroups of index 2 (n is a natural number.). Prove that there exists a finite abelian group which has exactly n subgroups of index 2. 4. Can one find two biased dice in such a way that the probability of getting a sum of j, for all 2 ≤
j ≤ 12, for the two simultaneously thrown dice is a number in the 4 2 , 33 ? interval 33 5. Show that R2 has a dense subset whose no three points are collinear. 6. Let A be an n × n matrix with real entries. Prove  tr(A) 1 0  tr(A2 ) tr(A) 2   3 2  tr(A ) 1 tr(A ) tr(A) det(A) = det   . . .. n! .. ..  .   tr(An−1 ) tr(An−2 ) · · · tr(An ) tr(An−1 ) · · ·

that ··· 0 3 .. . ··· ···

··· ··· .. .

0 0 .. .



    .  .. . 0   tr(A) n − 1  tr(A2 ) tr(A)

1.31. THIRTY FIRST COMPETITION

49

1.31.2. Second Day. 1. Let n be an odd natural number. Prove that there are n consecutive natural numbers whose sum is a complete square. In this spirit, do there exist twelve natural numbers whose sum is a complete square? 2. For what values of the nonzero reals α and β the following limit exists (and is finite)? x2α y 2β lim + 3α + y 3β x,y→0 x 3. Let A be a nonvoid set and An be the set of all ordered n-tuples of the elements of A. For α = (α1 , . . . , αn ) and β = (β1 , . . . , βn ) in An , define d(α, β) = the number of corresponding components of α and β that differ with one another. For two arbitrary elements x and y in An , prove that there is a one-to-one correspondence between the following two sets  C = z ∈ An : d(x, z) < d(y, z) ,  D = z ∈ An : d(y, z) < d(x, z) . 4. Let R be a commutative ring with identity. Prove that the ring R[x] has infinitely many maximal ideals. 5. Let x1 , . . . , x2n be real numbers such that removing any of them the remaining ones can be partitioned into two sets with equal sums (n ≥ 2). Prove that xi ’s are all zero. 6. Let T be the union of all line segments with one end at M = (0, 1) and the other end at a point on the x-axis having rational coordinates. In other words,  T = (tq, 1 − t) ∈ R2 |t ∈ [0, 1], q ∈ Q . (a) Assume that A, B ∈ T such that A, B, and M are not collinear. Show that any continuous path from the point A to the point B in the set T has to pass through the point M (by the aforementioned path, we mean a continuous function γ : [0, 1] → T with γ(0) = A and γ(1) = B.). (b) Prove that every continuous function from T into T has at least a fixed point.

CHAPTER 2

Solutions 2.1. First Competition

2.1.1. Analysis. 1. (a)
For the given function f , let fn : R → R be defined by fn (x) = n f (x + n1 ) − f (x) , where n ∈ N. It is plain that the sequence (fn )+∞ n=1 converges pointwise to f 0 on R, finishing the proof of this part. (b) First, we recall the following which is known as Darboux’s Theorem. Darboux’s Theorem. The derivative of any differentiable function on a closed interval has the intermediate value property. Proof. Suppose that f : [a, b] −→ R is a differentiable function, f 0 (a) 6= f 0 (b), and λ ∈ R is between f 0 (a) and f 0 (b). We need to show that there exists a c ∈ (a, b) such that f 0 (c) = λ. To this end, if necessary replacing f by −f , without loss of generality, we may assume that f 0 (a) < f 0 (b). There are two cases to consider. (a) (i) f 0 (a) < λ ≤ f (b)−f . b−a Define the function g : [a, b] −→ R by  f 0 (a) x = a, g(x) = f (x)−f (a) a < x ≤ b. x−a It is readily seen that g is continuous on [a, b]. We have g(a) < λ ≤ g(b). It thus follows from the Intermediate Value Theorem that there exists an x0 ∈ (a, b] such (a) that g(x0 ) = f (xx00)−f = λ. Therefore, by the Mean Value Theorem, λ = f 0 (c) for −a some c ∈ (a, b), which is what we want. (a) (i) f (b)−f < λ < f 0 (b). b−a In this case, defining the function g : [a, b] −→ R by  f (b)−f (x) a ≤ x < b, b−x g(x) = f 0 (b) x = b, the assertion follows in a similar fashion. Define the function f : R → R by  0 f (x) = 1 51



x ≤ 0, x > 0.

52

2. SOLUTIONS

The function f is a Baire function because the sequence (fn )+∞ n=1 of continuous functions, where fn : R → R is defined by  x ≤ 0,  0 nx 0 < x < n1 , fn (x) =  1 1 n ≤ x, converges to the function f pointwise. Yet, in view of Darboux’s Theorem , the function f is not the derivative of any function on R because f does not have the intermediate value property (a.k.a. Darboux’s property ).  2. (a) For X = (xij ) ∈ M, by definition, we have X sgn(σ)x1σ1 · · · xnσn , det(X) = σ∈Sn

where Sn denotes the symmetric group on {1, . . . , n} and sgn(σ) stands for the sign of the permutation σ ∈ Sn . Hence, the function det : M → R is continuous, for f is a polynomial function in n2 variables xij where 1 ≤ i, j ≤ n. We have
U (M) = det−1 R \ {0} , where U (M) denotes the set of invertible matrices in M. It follows that U (M) is an open subset of M because R \ {0} is an open subset of R. (b) To see that M is disconnected, just note that

U (M) = det−1 R− ∪ det−1 R+ + where R− = (−∞, 0)
and R −1= (0,
+∞). Thus, in view of the proof of (a), the −1 − open sets det R and det R+ form a disconnectedness for U (M), finishing the proof. 

3. (a) We need the following lemma. Lemma. Let N ∈ R+ and I a bounded interval of R. Then, in the interval I there are finitely many r ∈ Q whose denominators are less than or equal to N . Proof. Without loss of generality assume that N is a natural number. Just note that, by the Archimedean property of real numbers and the well-ordering SN principle of natural numbers, the set b=1 { ab ∈ I : a ∈ Z} is a finite set, proving the assertion.  We claim that limx→a f (x) = 0 for all a ∈ R+ . To see this, for a given ε > 0, first pick N ∈ N such that N1 < ε. It follows from the lemma above that there exists δ > 0 such that x = pq ∈ Q with gcd(p, q) = 1 and 0 < |x − a| < δ imply that q > N . So for this given ε > 0, find δ > 0 as in the above. If 0 < |x − a| < δ, there are two cases to consider: (i) x = pq ∈ Q with gcd(p, q) = 1. Noting that q > N , we can write |f (x) − 0| =

1 1 1 < < < ε. p+q q N

(ii) x ∈ / Q. In this case, we have |f (x) − 0| = 0 < ε. From these two cases, we see that limx→a f (x) = 0, which is what we want.

2.1. FIRST COMPETITION

53

(b) In view of the claim we made in (a), it follows that f is continuous at the irrational points of (0, +∞) and is discontinuous at the rational points of (0, +∞). 

2.1.2. Algebra. 1. (a) “if”: Suppose that I1 , I2 C R with I1 I2 ⊆ R and I1 * P . We show that I2 ⊆ P . Choosing an x0 ∈ I1 \ P and assuming that y ∈ I2 is arbitrary, we have x0 y ∈ I1 I2 ⊆ P . Hence, y ∈ P , for x0 ∈ I1 \ P . As y was arbitrary, we see that I2 ⊆ P , which is what we want. “only if”: Let hai and hbi denote the ideals generated by a and b in R, respectively. The ring R is commutative with identity element. So we have hai = aR, hbi = bR, and haihbi = habi. Now from ab ∈ P , it follows that haihbi ⊆ P . But P is prime. Therefore, hai ⊆ P or hbi ⊆ P , implying that a ∈ P or b ∈ P , which is what we want. (b) Let M be a maximal ideal in R. In view of (a), it suffices to show that M is prime. To this end, suppose that ab ∈ M , where a, b ∈ R. We need to show that a ∈ M or b ∈ M . If a ∈ / M , then M + hai = R because M is maximal and a ∈ / M. In particular, we see that there are m ∈ M and x ∈ R such that 1 = m + ax, implying that b = bm + abx ∈ M , for M is an ideal and ab ∈ M .  2. It follows from the hypothesis that g −1 = g for all g ∈ G. We can write ab = (ab)−1 = b−1 a−1 = ba, for all a, b ∈ G. That is, G is abelian, which is what we want.



3. (3.1) In view of (c), we can write cardE ≤ cardf (E) ≤ cardF = cardE. This, together with the Schroder-Bernstein Theorem, implies that cardE = cardf (E). Now, suppose that X and Y are two finite subsets of E such that cardX = cardf (X) and cardY = cardf (Y ). We can write
cardf (X ∪ Y ) = card f (X) ∪ f (Y )
= cardf (X) + cardf (Y ) − card f (X) ∩ f (Y ) . (∗) On the other hand, card(X ∪ Y ) = cardX + cardY − card(X ∩ Y ). (∗∗)
But card(X ∪ Y ) ≤ cardf (X
∪ Y ) = card f (X) ∪ f (Y ) . So it follows from (∗) and (∗∗) that card f (X) ∩ f (Y ) ≤ card(X ∩ Y ). As f (X ∩ Y ) ⊆ f (X) ∩ f (Y ), we can write
card(X ∩ Y ) ≤ cardf (X ∩ Y ) ≤ card f (X) ∩ f (Y ) ≤ card(X ∩ Y ). Therefore,
cardf (X ∩ Y ) = card f (X) ∩ f (Y ) = card(X ∩ Y ). That is, X ∩ Y and f (X) ∩ f (Y ) are isomorphic sets. Now, since f (X ∩ Y ) ⊆ f (X) ∩ f (Y ) and these two sets are finite, it follows that f (X ∩ Y ) = f (X) ∩ f (Y ). This equality together with (∗) and (∗∗) implies that card(X ∪ Y ) = cardf (X ∪ Y ), finishing the proof of assertion in this part.

54

2. SOLUTIONS

(3.2) Set  T = card(A) : ∅ 6= A ⊆ E and card(A) = cardf (A) . It is obvious that the set T is a nonempty subset of N. It thus follows from the well-ordering principle of N that the set T has an initial element, say, cardX0 , where X0 is a nonempty normal subset of E. We claim that for any other normal subset X of E, we have X ∩ X0 = ∅ or X0 ⊆ X. Note that X0 ∩ X ⊆ X0 . If X0 ∩ X = X0 , we obtain X0 ⊆ X; if not, we show that X0 ∩ X = ∅, completing the proof. Suppose that X0 ∩ X $ X0 . Since X0 is finite, it follows that card(X0 ∩ X) < card(X0 ). On the other hand, by (3.1), card(X0 ∩ X) = cardf (X0 ∩ X). We claim that X0 ∩ X = ∅. Otherwise, we see that card(X0 ∩ X) ∈ T , yielding card(X0 ) ≤ card(X0 ∩ X), for card(X0 ) is the initial element of T . This is a contradiction. Hence, X0 ∩ X = ∅ and we are done. 

b c a 2.1.3. General. 1. From a < b < c, we see that < 1 , < 1 and 1 < . b c a Thus, a b c
c a b c
c max , , = , min , , 6= . b c a a b c a a There are two cases to consider. a b c
a a b c
b (i) min , , = and (ii) min , , = . b c a b b c a c In both cases, in view of the above inequalities, it is easily seen that S > 1.  2. We show that the decimal fraction a = 0.123456789101112131415 . . . cannot be periodic after any digit. To prove this by contradiction, assume that the decimal fraction a is periodic after its first k digits and that its period is l. Obviously, the number 10k+l occurs somewhere after the kth digit of the decimal fraction a. As 10k+l has k + l zeros in its decimal expansion and that the period of a after the kth digit is assumed to be l, it follows that the digits of a after the kth digit are all zero, which is obviously impossible, settling the proof.  3. We need to solve the following system of equations in C  x = y3 , x 6= y. y = x3 It is plain that x, y 6= −1, 0, 1, for otherwise x = y ∈ {−1, 0, 1}. By substitution, we obtain x9 = x, x 6= 0 =⇒ x8 = 1 =⇒ x = ω j , 0 < j < 7, j 6= 4, where ω = cos π4 + i sin π4 . It is now easily seen that  x = ωj , 0 < j < 7, j 6= 4, y = ω 3j is the general solution of the above system of equations.



2.1. FIRST COMPETITION

55

4. Using the AM-GM Inequality, we can write √ xx + y y ≥ 2 xx y y = 2ef (x) , where f : (0, 1) → R is defined by f (x) = x ln x + (1 − x) ln(1 − x). We have f 0 (x) = − ln( x1 − 1). Thus, f 0 (x) < 0 on (0, 12 ) and f 0 (x) > 0 on ( 21 , 1), implying that f attains its only local minimum and hence its absolute minimum at 12 . Consequently, we have √ 1 xx + y y ≥ 2ef (x) ≥ 2ef ( 2 ) = 2, for all x, y ∈ R+ with x + y = 1 and that the equality happens if and only if x = y = 12 , which is what we want. 

2.1.4. Differential Equations. We know that the general solution of a nonhomogeneous linear equation is y = yc + yp , where yc is the general solution of the corresponding homogeneous equation and yp is a particular solution of the nonhomogeneous linear equation x00 + a2 x = f (t). In view of this, first of all, it is easily seen that yc (t) = c1 cos at + c2 sin at (t ∈ R), where c1 , c2 ∈ R, is the general solution of the homogeneous equation. Using the method of variation of parameters, a straightforward calculation shows that the function yp defined by Z 1 t yp (t) = f (x) sin a(t − x)dx (t ∈ R), a t0 where t0 ∈ R is a fixed arbitrary real number, is a particular solution of the nonhomogeneous linear equation x00 + a2 x = f (t). Therefore, Z 1 t f (x) sin a(t − x)dx (t ∈ R), y(t) = c1 cos at + c2 sin at + a t0 is the general solution of the equation x00 + a2 x = f (t).




P10 2.1.5. Probability and Statistics. We need to calculate P i=1 Xi = 10 . To do this, let, first of all, X be a Poisson random variable with parameter λ defined by its probability density function as follows  −λ i e λ i ∈ N ∪ {0}, i! fX (i) = p(X = i) = 0 otherwise. Next, we find the moment generating function of X. mX (t)

+∞ X
X tXi = E etX = e fX (Xi ) = eti fX (i) i=0

Xi

= e−λ

+∞ X (λet )i i=0

i!

= eλ(e

t

−1)

.

56

2. SOLUTIONS

Now, since each Xi is a Poisson random variable parameter µi and that Xi ’s Pwith 10 are independent random variables, letting Y = i=1 Xi , we can write = E et

mY (t)

=

10 Y

P

10 i=1

Xi




=E

P (

10 Y

10
Y
etXi = E etXi

i=1

i=1

10 i=1

mXi (t) = e

µi )(et −1)

.

i=1

Therefore, Y = whence P

P10

Xi is a Poisson random variable with parameter

10 X

e−

i=1


Xi = 10

= fY (10) =

P

10 i=1

i=1

P10

i=1

µi ,

P10 ( i=1 µi )10 . 10!

µi

But µi = i. So, we have
P X=1

= P

10 X


e−55 5510 Xi = 10 = . 10! i=1 

2.1.6. Topology. To prove the assertion by contradiction, suppose that T is a topology on R. We must have [ √ Aq = ( 2, +∞) ∈ T, q>

implying that

√

√

2

q∈Q

2 ∈ Q, which is a contradiction, settling the proof.



2.2. Second Competition 2.2.1. Analysis. 1. (a) Since f is continuous and the interval [0, 1] is compact, there exists an x0 ∈ [0, 1] such that M = sup f (x) = max f (x) = f (x0 ). 0≤x≤1

0≤x≤1

If M = 0, there is nothing to prove. If not, as f is continuous at x0 , we see that for given M ε > 0, there exists a neighborhood of x0 , say, (α, β), where 0 ≤ α < β ≤ 1, such that |f (x) − f (x0 )| < M ε whenever α < x < β. Noting that f (x0 ) = M ≥ f (x) for all x ∈ [0, 1], in view of the above inequality, we obtain M (1 − ε) < f (x) ≤ M, for all x ∈ (α, β), which is what we want. (b) For given 0 < ε < 1, find 0 ≤ α < β ≤ 1 from (a). We have 0 < M (1 − ε) ≤ f (x) ≤ M on [α, β], for f is continuous on [0, 1]. We can write Z β Z 1 n n n M (1 − ε) (β − α) ≤ f ≤ f n ≤ M n. α

0

2.2. SECOND COMPETITION

57

Taking nth root, we obtain M (1 − ε)(β − α)

1 n

1

Z ≤

f

n

 n1 ≤ M.

(∗)

0

Now taking limn and limn of both sides of the left and right inequalities above, we obtain Z 1  n1 Z 1  n1 n ≤ limn ≤ M. f fn M (1 − ε) ≤ limn 0

R

 n1

1 0

R

0

 n1

1 0

Thus, 0 ≤ limn fn − limn fn ≤ M − M (1 − ε) = M ε. As 0 < ε < 1 is arbitrary, we conclude that Z 1  n1 Z 1  n1 n f = limn fn . limn 0

That is, limn

R

1 0

fn

 n1

0

exists. Letting n → +∞ in (∗), we obtain 1

Z M (1 − ε) ≤ lim

f

n

n

 n1 ≤ M.

0

Again, 0 < ε < 1 being arbitrary, we see that Z 1  n1 = M = f (x0 ) = sup f (x), lim fn n

0≤x≤1

0

finishing the proof.



2. Note first that 1

Z

sup 1 − f (t) ≥ 1 − f (0) = 1, 0≤t≤1

1 − f (t) dt ≥ 0,

0

and hence sup 1 − f (t) + 0≤t≤1

Z

1

1 − f (t) dt ≥ 1,

0

for all f ∈ F. This implies that  Z D = inf sup 1 − f (t) + f ∈F

0≤t≤1

1

1 − f (t) dt

 ≥ 1.

0

We claim that D = 1. To see this, just note that for the sequence (fn )+∞ n=1 , where fn : [0, 1] → R (n ∈ N) is defined by  nt 0 ≤ t ≤ n1 , fn (t) = 1 1 n < t ≤ 1, we have  lim n

sup 1 − fn (t) + 0≤t≤1

Z 0

1

1 − fn (t) dt



  1 = lim 1 + = 1. n 2n

58

2. SOLUTIONS

As for the geometrical interpretation of D, we view F as a subspace of the normed space all continuous functions on [0, 1], denoted by C[0, 1], which is equipped with the following norm Z 1 f (t) dt. ||f || := sup f (t) + 0≤t≤1

0

If 1 : [0, 1] → R is defined by 1(x) = 1 for all x ∈ [0, 1], we would then have d({1}, F)

= =

inf ||1 − f ||  Z inf sup 1 − f (t) +

f ∈F

f ∈F

0≤t≤1

1

1 − f (t) dt

 = D = 1.

0

That is, the distance from the vector 1 ∈ C[0, 1] to the subspace F of C[0, 1] is equal to D = 1.  3. It suffices to show that [0, 2] ⊆ C + C. To this end, let x ∈ [0, 2] be arbitrary. If x = 0.x1 x2 . . . 2 denotes the ternary expansion of x2 where xi ∈ {0, 1, 2}, then we can write x = 0.a1 a2 . . . + 0.b1 b2 . . . , 2 where ai , bi ∈ {0, 1} and xi = ai + bi for all i ∈ N. So we have x =

2(0.a1 a2 . . .) + 2(0.b1 b2 . . .) = 0.a01 a02 . . . + 0.b01 b02 . . . := a + b,

where a0i = 2ai and b0i = 2bi , and hence a0i , b0i ∈ {0, 2} for all i ∈ N. Therefore, a, b ∈ C because there is no one in their ternary expansions. This finishes the proof. 

2.2.2. Algebra. 1. To prove the assertion we need the following general lemma. Lemma. Let G be a finite group with the property that the equation xn = e has at most n solutions in G for all n ∈ N, where e is the identity element of G. Then the group G is cyclic. Remark. In case the group G is abelian, the lemma is a quick consequence of the Fundamental Theorem of finite abelian groups . First proof. Let |G| = n and C be a cyclic group with n elements so that C = hai, where ord(a) = n. Suppose that for a divisor d of n, there is an element gd ∈ G such that ord(gd ) = d. It follows that gd , gd2 , . . . , gdd−1 , gdd = e are d solutions of the equation xd = e. As d n = |G| = |C| and C is cyclic, there is an ad ∈ C such that ord(ad ) = d. Consequently, hgd i ∼ = had i, and hence C has at least as many elements of order d as G has. But G and C both have n elements. Thus, G must have the same number of elements of order d as C, for all d dividing n. It follows that G is cyclic because C has an element of order n, namely the element a. So the proof is complete.

2.2. SECOND COMPETITION

59

Second proof. Define the relation ∼ on G as follows a ∼ b ⇐⇒ ord(a) = ord(b). It is obvious that ∼ is an equivalence relation on G. Hence, the relation ∼ partitions G into its equivalence classes. In other words, we can write G=

k [

[gi ],

i=1

where k ∈ N is less than |G|, {g1 , . . . , gk } is a maximal set of nonequivalent elements of G, and [gi ] denotes the equivalence class containing the element gi . Note that if G happens to be a cyclic group of order n so that G = hai with ord(a) = n, we can write k X [gi ] , n = G = i=1

where gi ’s (1 ≤ i ≤ k) are as in the above and gi = aji for some 1 ≤ ji ≤ n. Let di = gcd(ji , n). It is easily checked that ord(ak ) =

n ord(a)
= gcd(k, n) gcd k, ord(a)

(∗)

for all 1 ≤ k ≤ n. In view of this, we see that di 6= di0 whenever i 6= i0 and n that ak ∈ [gi ] if and only if gcd(k,n) = nd if and only if gcd(k, n) = d if and only if gcd( kd , nd ) = 1. Thus, [gi ] = φ( nd ), where φ denotes Euler’s totient function . Consequently, k X X n
X [gi ] = n= φ = φ(d). d i=1 d|n

d|n

finite group G, letting di = ord(gi ), we claim that Now back to a general [gi ] = φ(di ). That [gi ] ≥ φ(di ) follows from the hypothesis that di = ord(gi ) and that, in view of (∗), the cyclic group hgi i has exactly φ(di ) generators. On the other hand, [gi ] ≤ φ(di ), for otherwise there must exist a g ∈ G \ hgi i with ord(g) = di , which, in turn, implies that the equation xdi = e has at least di + 1 solutions, namely, g, gi , gi2 , . . . , gidi = e, which is a contradiction. Thus, [gi ] = φ(di ), and hence k X X n = G = φ(di ) = φ(d). i=1

d|n

This, in particular, implies that G has an element of order n. Therefore, G is a cyclic group of order n, which is what we want.  Corollary. Let D be a division ring and G a finite abelian subgroup of the multiplicative group of D, i.e., D∗ = D \ {0}. Then, G is a cyclic group. In particular, the multiplicative group of any finite field is cyclic.  By the corollary above, there exists an a ∈ F ∗ such that F ∗ = hai. Letting b = am , we can write S :=

X x∈F

m

x

=

n−2 X i=0

i m

(a )

=

n−2 X i=0

bi .

60

2. SOLUTIONS

If n − 1|m, then, as P an−1 = 1, we have b = 1. This together with the above equality implies that x∈F xm = n − 1. If n − 1 - m, then bm 6= 1. So we have S = bS = 1 + b + · · · bn−1 , yielding (b − 1)S = 0 which, in turn, implies that S = 0, proving the assertion.  2. (a) “⇐=” This is obvious. “=⇒” Assume that z is right and left quasi-regular. Thus, there exist z 0 , z 00 ∈ A such that z + z 0 − z 0 z = z + z 00 − zz 00 = 0. (∗) We can write (z + z 0 − z 0 z)z 00 = 0 =⇒ zz 00 + z 0 z 00 − z 0 (z + z 00 ) = 0 =⇒ zz 00 = z 0 z. This together with (∗) implies that z 0 = z 00 . That is, z is quasi-regular, which is what we want. (b) Proceed by contradiction. The identity element of A being right quasiregular means there exists 10 ∈ A such that 1 + 10 − 10 = 0, yielding 1 = 0, which is impossible. Likewise, the identity element is not left quasi-regular either. As pointed out in the footnote of this problem, from the element x being right quasi-regular one can only conclude that 1 − x is right invertible. First, let x be right quasi-regular. So there exists x0 ∈ A such that x + x0 − xx0 = 0. From this, we obtain (1−x)(1−x0 ) = 1. That is, 1−x is right invertible. In fact, as this argument is reversible, in a similar fashion, one can prove that the element x is right (resp. left) quasi-regular if and only if 1 − x is right (resp. left) invertible. We now show that in general from x being right quasi-regular one cannot conclude that 1 − x is invertible. To this end, in view of the aforementioned comment, it suffices to show that in general from x being right quasi-regular one cannot conclude that x is left quasi-regular as well. To see this, let F be a field of characteristic zero and F [x] the ring of all polynomials with coefficients from the field F . View F [x] as a vector space over F and use A to denote L(F [x]), the ring of all linear transformations from F [x] into F [x]. Let I denote the identity transformation, D the differentiation linear transformation, i.e. D(f0 + f1 x + · · · + fn xn ) = f1 + 2f2 x + · · · + nfn xn−1 , and I1 and I2 denote the linear transformations defined by f1 fn n+1 I1 (f0 + f1 x + · · · + fn xn ) = 1 + f0 x + x2 + · · · + x , 2 n+1 f1 fn n+1 I2 (f0 + f1 x + · · · + fn xn ) = 2 + f0 x + x2 + · · · + x . 2 n+1 Clearly, DI1 = DI2 = I. That is, D has two right inverses, and hence D is not invertible. This implies that I − D is right quasi-regular but it is not left quasi-regular. Because otherwise I − (I − D) = D will become invertible, which is impossible.  3. (a) Since D divides P , there exists Q ∈ K[x] such that P = QD. We conclude that φ(P ) = φ(Q)φ(D) because φ is a ring homomorphism. It is now obvious that ker φ(D) ⊆ ker φ(P ), which is what we want. (b) Suppose that D = gcd(P, R), where P, R ∈ K[x]. Since K[x] is a Euclidean ring, it follows that there exist M, N ∈ K[x] such that D = M P + N R. As φ

2.2. SECOND COMPETITION

61

is a ring homomorphism, we conclude that φ(D) = φ(M )φ(P ) + φ(N )φ(R). This implies ker φ(R) ∩ ker φ(P ) ⊆ ker φ(D). On the other hand, since D = gcd(P, R), in view of (a), we see that ker φ(D) ⊆ ker φ(P ) and ker φ(D) ⊆ ker φ(R), yielding ker φ(D) ⊆ ker φ(R)∩ker φ(P ). Therefore, ker φ(D) = ker φ(R)∩ker φ(P ), finishing the proof. 

2.2.3. General. 1. If the first day up to the seventh day of the first month of the Persian calendar, called “Farvardin”, corresponds to 1 up to 7, respectively, then the thirteenth day of the first month corresponds to 6. From this convention, we will obtain the 12-tuple (6, 2, 5, 1, 4, 7, 3, 5, 7, 2, 4, 6) whose ith component is corresponded to the thirteenth day of the month i of the Persian calendar (1 ≤ i ≤ 12). For instance, the number 1 being the fourth component of the above 12-tuple means that the thirteenth day of the fourth month of the Persian calendar occurs on the same day of the week as does the first day of the first month of the calendar. For 1 ≤ i ≤ 7, let n(i) denote the number of times that the number i occurs in the above 12-tuple. Obviously,  1 i ∈ {1, 3}, n(i) = 2 i∈ / {1, 3}. As the first day of “Farvardin” can occur on any day of the week, we see that “Wednesday” can correspond to any of the numbers 1, 2, . . . , 7. Therefore, in the Persian calendar, the minimum and the maximum number of “Wednesdays occurring on the 13th of the month” is 1 and 2, respectively.  2. Recall that a real function f on an interval I is called convex if
f λx + (1 − λ)y ≤ λf (x) + (1 − λ)f (y), for all x, y ∈ I and 0 ≤ λ ≤ 1. It is well-known that a function f : I → R which is twice differentiable is convex if and only if f 00 ≥ 0 on I. From this, it follows that the function gn : (0, +∞) → R defined by gn (x) = xn , where n ∈ R+ , is convex if and only n ≥ 1. In particular, gn is convex whenever n ∈ N. This yields
gifn (x)+g x+y n (y) for all n ∈ N and x, y ∈ R+ . So, we can write gn 2 ≤ 2  x + y + z + t n x + y + z + t
1 x + y
1 z + t
= gn ≤ gn + gn 4 4 2 2 2 2 gn (x) + gn (y) + gn (z) + gn (t) xn + y n + z n + tn ≤ = , 4 4 for all n ∈ N and x, y, z, t ∈ R+ , which is what we want.  3. We prove the following proposition of which the assertion is a quick consequence. Note that if the particle were moving along a straight line, then the tangential acceleration of the motion would be equal to the acceleration of the motion along the straight line. A particle moving along a curve goes s0 unit(s) of distance in t0 unit(s) of time in such a way that the instantaneous speed of it at the initial point as well

62

2. SOLUTIONS

as the final point is zero. Prove that the absolute value of the particle tangential 0 unit of distance acceleration at some point on its path is greater than or equal to 4s . t2 (unit of time)2 0

Assume that the initial and final moments of the motion are 0 and t0 , respectively. If the equation of the motion is given by a vector function x : [0, t0 ] → R3 , it follows that x is twice differentiable, x0 (0) = x0 (t0 ) = 0, and s(0) = 0, s(t0 ) = s0 , Rt 0 where x0 = dx dt denotes the instantaneous velocity vector and s(t) = 0 ||x (τ )||dτ the length of the path at time t. It is plain that s0 (0) = s0 (t0 ) = 0. Also, recall that the tangential acceleration of the motion, denoted by at , is, by definition, the d2 s 00 time derivative of the instantaneous speed, i.e., at = dv dt = dt2 = s (t). As s is continuous, we see from the Intermediate Value Theorem that there is a moment 0 < c < t0 at which we have s(c) = s(t0 )+s(0) = s(t20 ) . There are two cases to 2 consider. (i) 0 < c < t20 . In this case, using the extension of the Mean Value Theorem for second order derivatives, we obtain a moment t1 with 0 < t1 < c such that s(c) = s(0) + s0 (0)(c − 0) + s00 (t1 )

(c − 0)2 c2 = s00 (t1 ) , 2! 2

2

0) from which, we get s(t20 ) = s00 (t1 ) c2 , and hence s00 (t1 ) = s(t c2 . This together 4s(t t0 with the hypothesis that 0 < c < 2 implies that s00 (t1 ) ≥ t2 0 ) > 0, yielding 0

|s00 (t1 )| ≥

4s(t0 ) t20

4s0 , t20

=

which is what we want.

(ii) t20 ≤ c < t0 . Again using the extension of the Mean Value Theorem for second derivatives, we obtain a moment t1 with t20 < t1 < t0 such that s(c) = s(t0 ) + s0 (t0 )(c − t0 ) + s00 (t1 ) from which, we obtain This together with is what we want.

t0 2

s(t0 ) 2

(c − t0 )2 , 2!

2

−s(t0 ) (c−t0 )2 < 0. 4s(t0 ) 0 = 4s , which t20 t20

= s(t0 ) + s00 (t1 ) (c−t2 0 ) , and hence s00 (t1 ) =

≤ c < t0 implies that |s00 (t1 )| =

s(t0 ) (c−t0 )2

≥



4. Since the digits can be selected independent of the letter, using the product rule of combinatorics, it is obvious that the total number of the car plates that contain the letter “dal” is equal to 9 × 9 × 9 × 9 × 9 × 1 = 59049. 

2.2.4. Probability. (a) We can write cov(U, V )

= E(U V ) − E(U )E(V )
= E (X + Y )(X − Y ) − E(X + Y )E(X − Y )
= E(X 2 − Y 2 ) − (E(X))2 − (E(Y ))2
= E(X 2 ) − (E(X))2 − E(Y 2 ) − (E(Y ))2 =

var(X) − var(Y ) = σ 2 − σ 2 = 0.

2.2. SECOND COMPETITION

63

Therefore, cov(U, V ) = 0. Note: As is obvious now the condition cov(X, Y ) = λ was redundant. (b) Not necessarily. Suppose that U, V are two random variables with cov(U, V ) = 0, which are not independent. If we let X = 12 (U + V ) and Y = 12 (U − V ), we would have
1
1 var(U ) + var(V ) + 2cov(U, V ) = var(U ) + var(V ) , var(X) = 4 4
1
1 var(Y ) = var(U ) + var(V ) − 2cov(U, V ) = var(U ) + var(V ) . 4 4
1 Therefore, var(X) = var(Y ) = 4 var(U ) + var(V ) = σ 2 . Note that the random variables X + Y = U and X − Y = V are not independent by our hypothesis. To give a concrete example, let U and V be two random variables defined by their joint distribution table as follows. U V 1 1 0.1 2 0.1 3 0.1 0.3

2 0.1 0.1 0.1 0.3

3 0.1 0.1 0.1 0.4

0.3 0.4 0.3

It is easily verified that UV

1 0.1

2 3 0.2 0.2

4 0.1

6 0.3

9 0.1

From the above table, we obtain E(U V ) = 4.2. On the other hand, we have E(U ) = 2 and E(V ) = 2.1, implying that E(U V ) = E(U )E(V ), and hence cov(U, V ) = 0. But P (U = 1, V = 1) = 0.1 and P (U = 1) = P (V = 1) = 0.3, yielding P (U = 1, V = 1) 6= P (U = 1).P (V = 1). In other words, U and V are not independent, which is what we wanted. 

2.2.5. Topology. Proceed by contradiction. As A ⊆ B ∪C, we have A = (A∩ B)∪(A∩C). Also, by the contradiction hypothesis, (A∩B)∩(A∩C) = A∩B∩C = ∅. But A ∩ B 6= ∅ and A ∩ C 6= ∅. Hence, it suffices to show that A ∩ B and A ∩ C are closed sets in the induced topology of A, for A would then be disconnected, a contradiction. To this end, note that B and C are closed in (K, TK ). Thus, there exist closed sets B1 and C1 in (E, T ) such that B = B1 ∩ K and C = C1 ∩ K, from which, we obtain A ∩ B = A ∩ B1 , A ∩ C = A ∩ C1 . Now as B1 and C1 are closed in E, it follows that A ∩ B and A ∩ C are closed in A, which is what we want. 

64

2. SOLUTIONS

2.2.6. Differential Equations. Two cases to consider. (i) |x| > 1. Rewrite the equation as x (x2 − 1)(x + 1) y3 = = x + 1. −1 x2 − 1 Multiplying both sides of the equation by Z  p p xdx 2 − 1| = = |x x2 − 1, µ(x) = exp (x2 − 1) (y 3 )0 +

x2

we obtain  p d  3p 2 y x − 1 = (x + 1) x2 − 1, dx from which, we get y3

p

x2 − 1 =

Z

Z p p x x2 − 1dx + x2 − 1dx.

√ Using integration by parts and dividing both sides of the above equality by x2 − 1, we see that the general solution of the equation is p x 1 1 c y 3 = (x2 − 1) + + √ ln |x + x2 − 1| + √ , 2 2 3 2 2 x −1 x −1 where c is an arbitrary fixed constant. (ii) |x| < 1. √ In this case, note that |x2 − 1| = 1 − x2 yields µ(x) = 1 − x2 . From this, a similar argument as in (i) shows that the general solution of the equation is x 1 −1 2 c (x − 1) + + √ arcsin x + √ , 3 2 2 1 − x2 1 − x2 where c is an arbitrary fixed constant. y3 =



2.3. Third Competition 2.3.1. Analysis. 1. Remark. It is worth mentioning that the assertion holds under the weaker hypothesis that f 00 is nonnegative on (a, b) or f 0 is monotonic on [a, b]. Also in the statement of the problem “ξ ∈ [a, b]” must read “ξ ∈ (a, b)” and that “ξ ∈ (a, b)” cannot be weakened, because for the function f : [0, 1] → R with f (x) = x2 and ξ ∈ {0, 1}, there is no x0 ∈ (0, 1) such that f 0 (ξ) =

f (0) − f (x0 ) f (1) − f (x0 ) or f 0 (ξ) = . 0 − x0 1 − x0

Since f 00 > 0 on [a, b], it follows that f 0 is strictly increasing on [a, b]. By the Mean Value Theorem, there exists a < ξ0 < b such that f (b) − f (a) . b−a Now, if f 0 (ξ) = f 0 (ξ0 ), there is nothing to prove. If not, there are two cases to consider. f 0 (ξ0 ) =

2.3. THIRD COMPETITION

65

(i) f 0 (ξ) < f 0 (ξ0 ) = Define the function g : [a, b] → R by  f 0 (a) g(x) = f (a)−f (x) a−x

f (b) − f (a) . b−a x = a, a < x ≤ b.

It is plain that f (a) − f (x) = f 0 (a) = g(a). a−x Hence, g is continuous at a from the right, and therefore it is continuous on [a, b]. As f 0 is strictly increasing on [a, b], we obtain f 0 (a) < f 0 (ξ). So we can write lim g(x) = lim+

x→a+

x→a

g(a) = f 0 (a) < f 0 (ξ) < f 0 (ξ0 ) = g(b). Thus, it follows from the Intermediate Value Theorem that there exists an a < x0 < b such that f (a) − f (x0 ) f 0 (ξ) = g(x0 ) = , a − x0 which is what we want. (ii) f (b) − f (a) f 0 (ξ) > f 0 (ξ0 ) = . b−a The proof, which is omitted, is almost identical to that of (i) except that we need to define the function g : [a, b] → R by  g(x) =

f (b)−f (x) b−x 0

f (b)

a ≤ x < b, x = b.


1 2. (a) Set B = Q ∩ [0, 1] \ n : n ∈ N . Plainly, we can write B = {bn : n ∈ N}. Define the sequence (an )+∞ n=1 by  1 n odd, n an = bn n even. We claim that for this sequence (an )+∞ n=1 , there is no continuous function f satisfying f (an ) = an+1 for all n ∈ N. Suppose to the contrary that f : [0, 1] → [0, 1] is a continuous function such that f (an ) = an+1 for all n ∈ N. By proving that f = 0 on (0, 1], and hence on [0, 1], we obtain a contradiction, proving the claim. For a given x ∈ (0, 1], pick a subsequence (ani )+∞ x. As x 6= 0, i=1 such that limi→+∞ ani =  if necessary by passing to a subsequence, we may assume that ani ∈ / n1 : n ∈ N for all i ∈ N. This implies that ni ’s are all even numbers, from which, we see that (ni + 1)’s are all odd numbers and hence 1 f (x) = lim f (ani ) = lim ani +1 = lim = 0, i→+∞ i→+∞ i→+∞ ni + 1 as desired. (b) We sketch the proof for the case in which the sequence (an )+∞ n=1 is increasing. In case the sequence is decreasing, the proof can be carried out in a similar fashion.

66

2. SOLUTIONS

Without loss of generality, if necessary by redefining the sequence (an )+∞ n=1 , we may assume that a1 = 0. Let a∞ = limn→+∞ an . Define the function f on [0, 1] by  f (x) =

an+2 −an+1 an+1 −an (x

− an ) + an+1 x

an ≤ x < an+1 , n ∈ N, a∞ ≤ x ≤ 1.

It is not difficult to see that f : [0, 1] → [0, 1] is continuous and that f (an ) = an+1 for all n ∈ N, as desired.  3. (a) We need the following proposition. Proposition. Let the complex number z be a root of the following equation with complex coefficients xp + c1 xp−1 + · · · + cp−1 x + cp = 0, Pp where p ∈ N. If λk > 0 for all 1 ≤ k ≤ p and k=1 λ1k ≤ 1, then p |z| ≤ max k λk |ck |. 1≤k≤p

Under the hypothesis of the proposition, taking λk = 2k for all 1 ≤ k ≤ p, we obtain |z| ≤ 2 max

1≤k≤p

p k |ck |,

proving the assertion. In order to prove the proposition, we need the following lemma. Lemma. Let the complex numbers z1 , . . . , zp (p ∈ N) be the roots of the following equation with complex coefficients xp + c1 xp−1 + · · · + cp−1 x + cp = 0. Set r0 = max1≤i≤p |zi |. If r > 0 and rp > |c1 |rp−1 + · · · + |cp |, then r0 < r. First proof. Let z ∈ C with f (z) z =

|z| ≥ r be arbitrary. We can write c1 cp + ··· + p 1 + z z c c 1 p ≥ 1 − − ··· − p z z |c1 | |cp | − · · · − p > 0. ≥ 1− r r

Thus, f (z) 6= 0, and hence r0 < r, as desired.

2.3. THIRD COMPETITION

67

Second proof. Let k(x) := xp − |c1 |xp−1 − · · · − |cp |. By the hypothesis, Pp |ci | k(r) > 0, which is equivalent to k(r) i=1 r i > 0. Thus, for all x ≥ r, we rp = 1 − have p p X X |ci | |ci | ≥ 1 − > 0. 1− i x ri i=1 i=1 Consequently, k 0 (x)

k(x)
k(x)
0 + xp p x xp p p X |ci |
X i|ci | p = pxp−1 1 − > 0, + x i i+1 x x i=1 i=1 =

(xp )0

for all x ≥ r. Thus, k is strictly increasing on the interval [r, +∞). Now, let z ∈ C be such that |z| ≥ r. It follows that k(|z|) ≥ k(r) > 0. So, we can write |z p + c1 z p−1 + · · · + cp |

≥ |z|p − |c1 ||z|p−1 − · · · − |cp | = k(|z|) ≥ k(r) > 0.

That is, z p +c1 z p−1 +· · ·+cp 6= 0 whenever |z| ≥ r. This yields r0 = max1≤i≤p |zi | < r, which is what we want. Third proof. Define f (z) = z p + c1 z p−1 + · · · + cp−1 z + cp , g(z) = z p , and h(z) = c1 z p−1 + · · · + cp = f (z) − g(z) on C. Note that f (z) 6= 0 for all z ∈ C with |z| = r, for otherwise p−1 rp = z p = − c1 z p−1 − · · · − cp ≤ c1 z + · · · + cp , from which, we obtain rp ≤ |c1 |rp−1 + · · · + |cp |, which is a contradiction. For all z on the circle |z| = r, we have h(z) ≤ c1 rp−1 + · · · + cp < rp = g(z) . So it follows from Rouch´e’s Theorem that the entire functions g(z) = z p and f (z) = h(z) + g(z) have the same number of zeros inside the circle |z| = r. Consequently, f has p zeros inside the circle |z| = r, implying that r0 < r, which is what we want.  p ProofPof Proposition. Set r1 = max1≤k≤p k λk |ck |, where λk ’s in R+ are p such that k=1 λ1k ≤ 1. For all r > r1 and for each k = 1, . . . , p, we obviously have 1 λk

>

|ck | . rk

Thus, 1≥

p p X X 1 |ck | > , λk rk

k=1

k=1

which obtains rp > |c1 |rp−1 + · · · + |cp |, and hence r0 < r in view ofpthe above lemma. As r > r1 was arbitrary, we conclude  that r0 ≤ r1 = max1≤k≤p k λk |ck |, which is what we want. (b) Let f (z) := z p +c1 z p−1 +· · ·+cp−1 z+cp , g(z) := z p +c01 z p−1 +· · ·+c0p−1 z+c0p , and h(z) := g(z + zj ). First, we prove that the absolute value of the product of the

68

2. SOLUTIONS

roots of h, i.e., |h(0)|, is less than (2K)p δ. To this end, we can write h(0)

p X = g(zj ) = g(zj ) − f (zj ) = (c0i − ci )zjp−i i=1 p p X 0 p−i X p−i ≤ ci − ci zj ≤ K i δ zj i=1 p X

≤ δ

i=1 p X p K i (2 max k |ck |)p−i < δ K i (2K)p−i 1≤k≤p

i=1

= δK p

p X

i=1

2p−i = δK p (2p − 1) < (2K)p δ,

i=1

yielding h(0) < (2K)p δ, as desired. Now, let z10 , . . . , zp0 be the roots of g(z) = 0. It follows that z10 − zj , . . . , zp0 − zj are the roots of h(z) = g(z + zj ) = 0. Obviously, there is a 1 ≤ k ≤ p such that |zk0 − zj | = min1≤i≤p |zi0 − zj |. Consequently, Y zi0 − zj = h(0) < (2K)p δ, |zk0 − zj |p ≤ 1≤i≤p

√ p

implying that |zk0 − zj | < 2K δ, which is what we want. (c) We have not been able to prove the assertion. However, we prove the following which can be thought of as a counterpart of the assertion at the local level. Under the hypothesis of the (c) part of the problem, prove that for every t ∈ [t1 , t2 ], there exists a δ = δ(t) > 0 such that if s ∈ [t1 , t2 ] and |s − t| < δ, then p |f (s) − f (t)| < 2K p |s − t|α .

For a given t ∈ [t1 , t2 ], let z1 = f (t), z2 , . . . , zk (1 ≤ k ≤ p) be distinct roots of z p + b1 (t)z p−1 + · · · + bp−1 (t)z + bp (t) = 0. If k = 1, in fact f need not be continuous and yet the assertion is a quick consequence of (b). To see this, note that by (b), for the root f (s) of z p + b1 (s)z p−1 + · · · + bp−1 (s)z + bp (s) = 0, where s ∈ [t1 , t2 ], there is a root of z p + b1 (t)z p−1 + · · · + bp−1 (t)z + bp (t) = 0, which must be the only root of the equation, namely f (t), such that p f (s) − f (t) < 2K p |s − t|α , as desired. So, we may without loss of generality assume that k p > 1. Now, set d0 = 21 min1≤i<j≤k |zi − zj |. Choose a δ = δ(t) > 0 such that 2K p |s − t|α < d0 and |f (s) − f (t)| < d0 whenever |s − t| < δ and s ∈ [t1 , t2 ]. It follows from (b) that for every s ∈ [t1 , t2 ] with |s − t| < δ, there is an is ∈ {1, . . . , k} such that p f (s) − zis < 2K p |s − t|α < d0 . On the other hand, f (s) − f (t) < d0 . Consequently, z1 − zi = f (t) − zi ≤ f (t) − f (s) + f (s) − zi < 2d0 , s s s

2.3. THIRD COMPETITION

69

implying that z1 = f (t) = zis . Thus, p f (s) − f (t) < 2K p |s − t|α , whenever s ∈ [t1 , t2 ] and |s − t| < δ, which is what we want.



2.3.2. Algebra. 1. (a) Just apply the First Isomorphism Theorem for mod2 ules to the mapping ϕ : M2 → M1M+M defined by ϕ(x) = x + M1 , where x ∈ M2 . 1 m (b) Suppose that {αi }m i=1 is a basis for M1 ∩ M2 . Enlarge {αi }i=1 to bases p m n m {αi }i=1 ∪ {βi }i=1 and {αi }i=1 ∪ {γi }i=1 for M1 and M2 , respectively. It is easily p n seen that {αi }m i=1 ∪ {βi }i=1 ∪ {γi }i=1 is a basis for M1 + M2 . We can write

dim M1 + dim M2

= (m + n) + (m + p) = (m + n + p) + m = dim(M1 + M2 ) + dim(M1 ∩ M2 ) = dim M3 + dim M4 ,

which is what we want.



2. If G is abelian, we see that the function f : G → G defined by f (x) = x2 is a homomorphism. Conversely, if f , defined by f (x) = x2 for all x ∈ G, is a homomorphism, then (xy)2 = x2 y 2 for all x, y ∈ G. Multiplying both sides by x−1 and y −1 on the left and right, respectively, we obtain xy = yx for all x, y ∈ G. That is, G is abelian, which is what we want.  3. (a) It is easily verified that the semigroup S defined by  S := (aij ) ∈ M2 (R) : a11 = a21 = 0, a12 , a22 ∈ R \ {0}   0 1 has a right identity element, namely , with respect to which every element 0 1 of S has a left inverse. And yet S is not a group, for its right identity element is not a left identity element. (b) Let x ∈ S be an arbitrary element with a left inverse x ∈ S so that xx = e. To prove the assertion, it suffices to show that xx = e. To this end, let x be a left inverse of x. We have ex = (x.x)x = x(xx) = xe = x. By showing that xx is a right identity element of S, we conclude that xx = e, finishing the proof. Suppose that y ∈ S is arbitrary. We can write y.(xx) = (ye)(xx) = y(ex)x = y(x.x) = ye = y. Thus, xx = e, which is what we want.



70

2. SOLUTIONS

S

N’

R

M’

M N Figure 5

2.3.3. General. 1. Letting z = x in the second condition, we see that F (y, x) ≤ F (x, x) + F (x, y) = F (x, y), for all x, y ∈ R. Therefore, F (y, x) ≤ F (x, y) for all x, y ∈ R, implying that F (x, y) ≤ F (y, x) for all x, y ∈ R. That is, F (x, y) = F (y, x) for all x, y ∈ R. This proves (b). Letting y = x in the second condition, we get 0 = F (x, x) ≤ F (x, z) + F (z, x), from which, in view of (b), we obtain 2F (x, z) ≥ 0, implying that F (x, z) ≥ 0 for all x, z ∈ R. This proves (a), completing the proof.  2. The identity (1 + x)2n = (1 + x)n (1 + x)n in R[x] together with the Binomial Theorem implies that n n 2n   X X X k k Cnk xk , Cnk xk C2n x = k=0

k=0

k=0

n! for all n ∈ N, where Cnk = is the binomial coefficient. Using the formula k!(n − k)! for Cauchy product of two polynomials, noting that Cnk = Cnn−k , and finally equating the coefficients of xn of both sides of the above identity, we see that n

n

k=0

k=0

2 X X
(2n)! n! n k 2 , = C = C = 2n n (n!)2 k!(n − k)! proving the assertion.



3. It is plain that if from the point M , we shine a beam of light onto a point R of the mirror ∆ so that the reflected light beam hits the mirror ∆0 at a point S, then the three points S , R, and M 0 are collinear, where M 0 is the image of M in the mirror ∆. Hence the problem is solved as follows. Find the image of M and N in the mirrors ∆ and ∆0 , respectively, to obtain the points M 0 and N 0 of the plane P . The line joining M 0 and N 0 intersects ∆ and ∆0 at the points R and S, respectively. It is now obvious that if we shine a beam of light onto the point R on the mirror ∆, then the reflected light beam after hitting the mirror ∆0 at S would pass through the point N (see Figure 5). 

2.4. FOURTH COMPETITION

(-2,0) (-1,0)

71

(0,0) (1,0) (2,0) Figure 6

2.4. Fourth Competition 2.4.1. Analysis. 1. Define the function f : R → R2 by  t < −2,  g(t) (t, 0) −2 ≤ t < 2, f (t) =  h(t) 2 ≤ t,     2(t+2) 2(t−2) −2 2 where g(t) = 1+(t+2) and h(t) = 1+(t−2) . It is straight2 , 1+(t+2)2 2 , 1+(t−2)2 forward to see that f is continuous and one-to-one. In fact, f is a one-to-one parametrization of the curve consisting of the upper semicircle with radius one centered at (−1, 0) going from (0, 0) to (−2, 0), the line segment joining (−2, 0) and (2, 0), and the upper semicircle with radius one centered at (1, 0) going from (2, 0) to (0, 0) (see Figure 6). We claim that the function f −1 is discontinuous at (0, 0), proving the assertion. To see this, define the sequence (an )+∞ n=1 by
( 1 n even, n, 0   an = 2n 2 n odd. 1+n2 , 1+n2 It is obvious that limn an = (0, 0) = f (0) but  1 n f −1 (an ) = n+2

n even, n odd,

whose limit does not exist as n → +∞. Therefore, f −1 is not continuous at (0, 0), proving the claim.   2. Set S := n ∈ N : αn > 0 . If the set S is finite, there is nothing to prove. So assume that S is an infinite set. For each k ∈ S with k ≥ 2, define n o Tk := N ∈ S|N ≥ k  αn ≤ αk ∀n ≥ N . As limn αn = 0, letting ε = ak in the definition of the limit for limn αn = 0, we see that Tk 6= ∅. Let Mk be the initial element of Tk . If Mk = k, we will have αn ≤ αk for all n ≥ k, in which case set Nk := k. If Mk > k, set Nk to be the greatest integer subject to k ≤ Nk < Mk and Nk ∈ S. It follows that Nk ∈ / Tk and that αn = 0 for all Nk < n < Mk . Consequently, αNk > αk > 0, for Mk is the initial element of Tk . Now, if n ≥ Nk , then either n = Nk , in which case

72

2. SOLUTIONS

αn = αNk ≤ αNk , or Nk < n < Mk , which yields αn = 0 < αNk , or n ≥ Mk , in which case αn ≤ αk < αNk . In other words, αn ≤ αNk for all n ≥ Nk . On the other hand, S is an infinite set and Nk ≥ k for all k ∈ S. This shows that there are an infinite number of Nk ’s such that αn ≤ αNk for all n ≥ Nk . 

2.4.2. Algebra. 1. (a) Let x, y ∈ R(B) and a ∈ A be arbitrary. It follows that there are M, N ∈ N such that xM ∈ B and xN ∈ B. Noting that in a commutative ring the Binomial Theorem holds, (ab)n = an bn for all a, b ∈ A and n ∈ N, and that B is an ideal of A, we see that (x − y)M +N =

M +N X

i i M +N −i CM ∈ B, (ax)M = aM xM ∈ B, +N x (−y)

i=0

yielding x − y ∈ R(B) and ax ∈ R(B). That implies R(B) is an ideal of A, which is what we want. (b) First proof. Let B be a prime ideal of A. To prove that R(B) is prime, suppose for given x, y ∈ A, we have xy ∈ R(B). We need to show that x ∈ R(B) or y ∈ R(B). As xy ∈ R(B), there is an n ∈ N such that xn y n = (xy)n ∈ B. But B is prime. So we must have xn ∈ B or y n ∈ B. In other words, x ∈∈ R(B) or y ∈ R(B), as desired. Second proof. We prove the assertion by showing that R(B) = B. Evidently, B ⊆ R(B). To show that R(B) ⊆ B, let x ∈ R(B) be arbitrary. It follows that there is an n ∈ N such that xn ∈ B. Now, a straightforward induction on n together with the hypothesis that B is prime reveals that x ∈ B. This proves that R(B) ⊆ B, and hence R(B) = B, as desired. (c) Remark. We must have a 6= 0, ±1. mk 1 Without loss of generality, assume that a > 1. Write a = pm 1 · · · pk , where mi ∈ N and pi ’s are distinct prime numbers (1 ≤ i ≤ k). We show that R(B) = hp1 p2 · · · pk i. First, let x ∈ hp1 p2 · · · pk i be arbitrary. It follows that x = p1 p2 · · · pk r for some r ∈ Z. Obviously, we can write xm1 +···+mk = as ∈ B for some s ∈ Z, implying that x ∈ R(B). This yields hp1 p2 · · · pk i ⊆ R(B). Next, let x ∈ R(B) mk n 1 be arbitrary. It follows that xn ∈ B = hpm 1 · · · pk i, form which, we obtain x = mk m1 n p1 · · · pk r for some r ∈ Z. Thus, pi |x for all 1 ≤ i ≤ k, yielding pi |x because pi is prime. This implies p1 · · · pk |x, for pi ’s are distinct primes. That is, x ∈ hp1 p2 · · · pk i, and hence R(B) ⊆ hp1 p2 · · · pk i. Therefore, R(B) = hp1 p2 · · · pk i, finishing the proof.  2. Let {x, y} be a linearly independent set in E, which is a vector space over a field F . Use Zorn’s Lemma to enlarge the independent sets {x, y} and {x, x + y} to {x, y}∪{αi }i∈I and {x, x+y}∪{βi }i∈I , respectively, each of which is a Hamel basis for E. It is well-known that there exists a linear transformation u : E → E satisfying u(x) = x, u(y) = x + y, and u(αi ) = βi for all i ∈ I. The linear transformation u is invertible because it maps the elements of the Hamel basis {x, y} ∪ {αi }i∈I onto those of the Hamel basis {x, x + y} ∪ {βi }i∈I . This means u is an automorphism of the vector space E. Now, suppose that a linear transformation f : E → E, also called an endomorphism of the vector space E, commutes with any invertible linear

2.4. FOURTH COMPETITION

73

transformation v : E → E, which is also called an automorphism of E. We claim, first of all, that for any nonzero vector x, we have f (x) = λx x for some λx ∈ F . To prove this by contradiction, letting ux : E → E be an automorphism such that ux (x) = x and ux (f (x)) = x + f (x), we must have f ux = ux f . In particular, we have f (ux (x)) = ux (f (x)) which means f (x) = x + f (x), yielding x = 0, which is impossible. Therefore, for any x ∈ E there exists λx ∈ F such that f (x) = λx x. Next, fix a nonzero x0 ∈ E. By showing that λx = λx0 , we prove the assertion. Without loss of generality, assume that {x, x0 } is linearly independent. Therefore, there is an automorphism ux : E → E such that ux (x) = x and ux (x0 ) = x + x0 . But f commuting with ux implies that
λx x + λx0 x0 = f (x + x0 ) = f ux (x0 )
= ux f (x0 ) = λx0 ux (x0 ) = λx0 x + λx0 x0 , yielding λx x = λx0 x. From this, we obtain λx = λx0 because x 6= 0. So we have shown that f (x) = λx0 x for all x ∈ E, or equivalently f = λx0 IE , which is what we want, finishing the proof.  3. It suffices to show that any nonzero x ∈ R has a multiplicative inverse. To this end, let a nonzero x ∈ R be given. It follows from the hypothesis that the following descending chain of the ideals of R xR ⊇ x2 R ⊇ x3 R ⊇ · · · must terminate. That is, there exists N ∈ N such that xn R = xN R for all n ≥ N . In particular, xN +1 R = xN R, implying that xN ∈ xN +1 R. Hence, there exists y ∈ R such that xN = xN +1 y, yielding xN (1 − xy) = 0. This, in view of the fact that R is an integral domain, implies that 1 − xy = 0, yielding xy = yx = 1. In other words, x is invertible, which is what we want. 

2.4.3. General. 1. Doing the substitution x = π2 − t, we can write Z 0 Z π2 sinn ( π2 − t) sinn x dx = I := (−dt) n π π π cosn ( cosn x + sinn x 0 2 − t) + sin ( 2 − t) 2 Z π2 cosn t dt. = sinn t + cosn t 0 We have I

=


1 1 I +I = 2 2

Z 0

π 2

sinn x + cosn x  1 π π dx = × = , cosn x + sinn x 2 2 4

for all n ∈ R. 2. (a) Define the events a, b, c, a1 , b1 , c1 as follows: a, b, c: The event that the prize is in box A, B, or C, respectively. a1 , b1 , c1 : The event that the box A, B, or C is opened, respectively.



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It is obvious that {a, b, c} is a partition of the probability space and that the desired probability is P (b|a1 ). Using the Inverse Probability Theorem or Bayes’ Theorem, we can write P (b|a1 )

= =

P (b)P (a1 |b) P (b)P (a1 |b) = P (a1 ) P (a)P (a1 |a) + P (b)P (a1 |b) + P (c)P (a1 |c) 1 1 1 3 × 2 = . 1 1 1 1 3 3 ×0+ 3 × 2 + 3 ×1

So the probability that the chosen box, i.e. B, contains the prize is 31 . (b) Plainly, the desired probability is P (c|a1 ). Again, using Bayes’ Theorem, we can write P (c)P (a1 |c) P (c)P (a1 |c) P (c|a1 ) = = P (a1 ) P (a)P (a1 |a) + P (b)P (a1 |b) + P (c)P (a1 |c) 1 × 1 2 3 = . = 1 1 1 1 3 × 0 + × + × 1 3 3 2 3 That is, the probability that the other box, i.e., C, contains the prize is 32 .



2.5. Fifth Competition 2.5.1. Analysis. 1. No, it is not. We present a counterexample on the closed interval [0, 1] which will work on the open interval (0, 1) as well. Define the functions f, g : [0, 1] → [0, 1] as follows  1 x ∈ Q ∩ (0, 1], x = pq , p, q ∈ N, gcd(p, q) = 1, q f (x) = 0 x ∈ [0, 1] \ (Q ∩ (0, 1]).  1 1 x ∈ { q |q ∈ N}, g(x) = 0 x∈ / { 1q |q ∈ N}. Obviously,  g ◦ f (x)

=

1 0

x ∈ Q ∩ (0, 1], x ∈ [0, 1] \ (Q ∩ (0, 1]).

We claim that f, g are Riemann integrable on [0, 1] but g ◦ f is not integrable on [0, 1]. In fact, it turns out that g ◦ f is not integrable on any closed interval [ε, δ], where 0 < ε < δ < 1, from which it follows that g ◦ f in not integrable on the open interval (0, 1). In view of Solution 3 of 2.1.1, we see that ∀a ∈ (0, 1) : lim f (x) = 0, lim f (x) = 0, lim f (x) = 0. x→a

x→0+

x→1−

Thus, the function f is continuous at the irrational points of (0, 1) and the set of discontinuity points of f , i.e., Q ∩ (0, 1), is countable. It is easily seen that g is discontinuous at any point of the set { 1q |q ∈ N} ∪ {0} and it is continuous on the set (0, 1] \ { 1q |q ∈ N}. Finally, using the fact that Q ∩ (0, 1] is dense in [0, 1], we see that g ◦ f is discontinuous at any point of the interval [0, 1]. From this point on, we present two proofs for the claim we made in the above. The first one uses some standard theorem from analysis and the second proof is self-contained.

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First proof. We will make use of Lebesgue’s Integrability Criterion for integrals in the Riemann sense which asserts that a bounded function f : [a, b] → R is integrable in the Riemann sense if and only if the set consisting of the points at which f is discontinuous has measure zero. Recall that a set A ⊆ R is said to be a set of measure zero if for any ε > 0, there exists a sequence (In )+∞ n=1 of intervals S+∞ P+∞ such that A ⊆ n=1 In and n=1 `(In ) < ε, where `(In ) := βn − αn and αn and βn are the end-points of the interval In . It is easy to see that any finite or countable subset of reals has measure zero and that a countable union of sets of measure zero has measure zero. In view of all that, we see that the functions f and g are Riemann integrable on [0, 1] and hence on (0, 1) but g ◦ f is not integrable (0, 1) because it is not integrable on any interval [ε, δ], where 0 < ε < δ < 1, for g ◦ f is continuous nowhere. +∞ Second proof. Define the sequences (fn )+∞ n=1 and (gn )n=1 , with fn , gn : [0, 1] → [0, 1], as follows  1 x = pq , p, q ∈ N, gcd(p, q) = 1, q ≤ n, q fn (x) = 0 otherwise.  1 1 x = q , q ∈ N, q ≤ n, gn (x) = 0 otherwise. +∞ It is easily verified that the sequences (fn )+∞ n=1 and (gn )n=1 converge uniformly on [0, 1] to f and g, respectively. As fn and gn (n ∈ N) are discontinuous at only finitely many points of the interval [0, 1], it follows that f and g are integrable in the Riemann sense on [0, 1]. Now, we see from the following lemma that g ◦ f is not integrable on any interval [ε, δ], where 0 < ε < δ < 1, which is what we want.

Lemma. Let f : [a, b] → R be bounded and integrable on [a, b] in the Riemann sense. Then the set of points at which f is continuous is infinite. Proof. Since f integrable on any closed subinterval of [a, b], it suffices to prove that f is continuous at some point of the interval [a, b], for it would then follow that any subinterval of [a, b] contains a point at which f is continuous. And hence, the set of points at which f is continuous would be infinite, which is what we want. Taking ε = b − a in Riemann’s criterion for integrability on the interval I0 := [a, b], we obtain a partition P0 : a = t00 < · · · < t0n0 = b of I0 such that U (P0 , f ) − L(P0 , f ) < b − a, where U (P0 , f ) =

n0 X

M0i ∆t0i , L(P0 , f ) =

i=1

M0i =

n0 X

m0i ∆t0i , ∆t0i = t0i − t0(i−1) ,

i=1

sup

f (x), m0i =

t0(i−1) ≤x≤t0i

inf

t0(i−1) ≤x≤t0i

f (x).

Hence, there exists 1 ≤ i0 ≤ n0 such that M0i0 − m0i0 < 1, for otherwise it would follow that U (P0 , f ) − L(P0 , f ) ≥ b − a, which is impossible. 1 Set I1 := [a1 , b1 ] = [t0(i0 −1) , t0i0 ]. As f is integrable on I1 , taking ε = b1 −a 2 , a similar argument shows that there exists a subinterval I2 := [a2 , b2 ] of I1 such that supx∈I2 f (x) − inf x∈I2 f (x) < 21 . Continuing this way, we obtain a sequence of

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nested closed interval I1 ⊃ I2 ⊃ · · · such that 0 ≤ supx∈In f (x) − inf x∈In f (x) < n1 . T+∞ Since n=1 In 6= ∅ because I0 is compact, in view of the above inequality, it follows T+∞ that f is continuous at any point of n=1 In 6= ∅, finishing the proof.  2. Remark. Adjusting the proof below, it is not difficult to prove the following problem, which characterizes all continuous real functions having the property that they do not assume any value more than twice. Let f : R −→ R be a continuous function that is neither increasing nor decreasing on R and that it assumes any value at most twice. Prove that one of the following statements holds. (i) There exists an a ∈ R such that f or −f is strictly increasing on (−∞, a] and is strictly decreasing on [a, +∞). (ii) There exist a, b ∈ R with a < b such that f or −f is strictly increasing on (−∞, a], is strictly decreasing on [a, b], and is strictly increasing on [b, +∞). Moreover, limx→−∞ f (x) ≥ limx→+∞ f (x). If the function f is strictly increasing or strictly decreasing, there is nothing to prove. So we may assume that f is neither strictly increasing nor strictly decreasing. At this point, we need the following lemma. Lemma. Let R be a totally ordered set with at least three elements and f : R → R a function that is neither strictly increasing nor strictly decreasing. Then,
there exist x1 , x2 , x3 ∈ R with
x1 < x2 < x3 such that f (x2 ) ≥ max f (x1 ), f (x3 ) or f (x2 ) ≤ min f (x1 ), f (x3 ) . Proof. To prove the assertion by contradiction, suppose
that for all x1 , x2 , x3 ∈
R with x1 < x2 < x3 , we have
f (x2 ) < max f (x1 ), f (x3 ) and
f (x2 ) > min f (x1 ), f (x3 ) . Now, since max f (x1 ), f (x3 ) = f (x1 ) or max f (x1 ), f (x3 ) = f (x3 ), we see that for all x1 , x2 , x3 ∈ R with x1 < x2 < x3 , we have f (x1 ) < f (x2 ) < f (x3 ) or f (x1 ) > f (x2 ) > f (x3 ).

(∗)

Pick a, b, c ∈ R with a < b < c. It follows that either f (a) < f (b) < f (c) or f (a) > f (b) > f (c). Suppose f (a) < f (b) < f (c). We show that f is strictly increasing, a contradiction. To see this, let x, y ∈ R with x < y be arbitrary. There are five cases to consider. (i) x < y < a, (ii) x < y = a, (iii) x < a < y, (iv) x = a < y, and (v) a < x < y. In each case, in view of (∗), it is easy to see that f (x) < f (y), from which, we conclude that f is strictly increasing, which is a contradiction. Likewise, if f (a) > f (b) > f (c), one can see, in a similar fashion, that f is strictly decreasing which is again a contradiction. So the proof is complete by contradiction.  In view of the lemma, there are two cases to consider.
(i) There are x1 , x2 , x3 ∈ R with x1 < x2 < x3 such that f (x2 ) ≥ max f (x1 ), f (x3 ) . In this case, from f (x2 ) ≥ max(f (x1 ), f (x3 )) and the continuity of f , we see that the function f attains its absolute maximum on the closed interval [x1 , x3 ] at an internal point of the interval, say, at xM ∈ (x1 , x3 ). By proving that f attains its only absolute maximum on R at xM , we complete the proof in this case. To this end, we first show that f (xM ) > f (x) for all x ∈ [x1 , x3 ] with x 6= xM . Suppose to the contrary that there exists an x0 ∈ [x1 , x3 ] with x0 6= xM such that f (x0 ) = f (xM ). Again, there are two cases to consider. (a) x1 ≤ x0 < xM and (b) xM < x0 ≤ x3 . Let x1 ≤ x0 < xM and pick a x0 < t < xM . As f

2.5. FIFTH COMPETITION

77

assumes the value f (xM ) at most twice and f (xM ) is the absolute maximum
of f f (xM )+max f (x3 ),f (t) on [x1 , x3 ], we see that f (t) < f (x0 ) = f (xM ). Set λ = . It 2 is obvious that f (t) < λ < f (x0 ), f (t) < λ < f (xM ), and f (x3 ) < λ < f (xM ). Thus, it follows from the Intermediate Value Theorem that there exist ξ1 , ξ2 , ξ3 with x0 < ξ1 < t , t < ξ2 < xM , and xM < ξ3 < x3 such that f (ξi ) = λ for each i = 1, 2, 3, contradicting the hypothesis. Similarly, if xM < x0 ≤ x3 , we obtain a contradiction. That is, we have shown that f (xM ) > f (x) for all x ∈ [x1 , x3 ] with x 6= xM . Next, we prove that f (x) < f (xM ) for all x ∈ R with x 6= xM . Again suppose to the contrary that there is a t ∈ R \ {xM } such that f (t) ≥ f (xM ). From what we have proved so far, it follows that t < x1 or t > x3 . If t < x1 , noting that f (x1 ) < f (xM ) ≤ f (t), we see that there exists an x0 ∈ R with
t ≤ x0 < x1 f (xM )+max f (x1 ),f (x3 )

. Again, it is such that f (x0 ) = f (xM ). Now, set λ = 2 obvious that f (x1 ) < λ < f (x0 ), f (x1 ) < λ < f (xM ), and f (x3 ) < λ < f (xM ). Hence, by the Intermediate Value Theorem we obtain ξ1 , ξ2 , ξ3 with x0 < ξ1 < x1 , x1 < ξ2 < xM , and xM < ξ3 < x3 such that f (ξi ) = λ for each i = 1, 2, 3, contradicting the hypothesis. Similarly, if x3 < t, we obtain a contradiction. This finishes the proof in this case.
(ii) There are x1 , x2 , x3 ∈ R with x1 < x2 < x3 such that f (x2 ) ≤ min f (x1 ), f (x3 ) . Adjusting the proof of case (i) or replacing f by −f and following the line of argument presented in case (i), we can see that the absolute minimum of f on R occurs in the interval (x1 , x3 ) and that it is unique. We omit the proof for the sake of brevity.   3. Let B = (p, q) : p, q ∈ Q, p < q , where (p, q) denotes the open interval with end points p and q in R. The set B is a countable basis for the ordinary topology of R. Write B = {In }n∈N , where In = (pn , qn ) for some pn , qn ∈ Q, and set T := {n ∈ N : card(In ∩ A) ≤ ℵ0 }. Obviously, T & N because otherwise A would be countable which is impossible. Now, set [
[ A1 = In ∩ A = (In ∩ A). n∈T

n∈T

The set A1 is countable because so is In ∩ A for all n ∈ T . By showing that any point x ∈ A \ A1 is a congestion point of A, we finish the proof. To prove this by contradiction, suppose that there exists an ε0 > 0 such that (x − ε0 , x + ε0 ) ∩ A is countable. As B = {In }n∈N is a basis for the topology of R, we see that there exists n0 ∈ N such that x ∈ In0 ⊆ (x − ε0 , x + ε0 ), from which, we obtain In0 ∩ A ⊆ (x − ε0 , x + ε0 ) ∩ A, implying that In0 ∩ A is countable because so is (x − ε0 , x + S ε0 ) ∩ A. That is, card(In0 ∩ A) ≤ ℵ0 , yielding n0 ∈ T , and hence x ∈ In0 ∩ A ⊆ n∈T (In ∩ A) = A1 . Consequently, x ∈ A1 , which contradicts the hypothesis that x ∈ A \ A1 . Thus, x is a congestion point of A, which is what we want. 

78

2. SOLUTIONS

2.5.2. Algebra. 1. Let A : V → V be such that A 6= I , A2 6= I, A3 = I. If x 6= 0 is an eigenvector corresponding to the eigenvalue λ ∈ C, we can write Ax = λx =⇒ A2 x = λ2 x =⇒ A3 x = λ3 x, which, in view of A3 = I, implies that λ3 = 1. That is, the set of the eigenvalues of A is a subset of the third roots of unity. Likewise, if A 6= I, . . . , Ak−1 6= I, Ak = I, then the set of the eigenvalues of A is a subset of the kth roots of unity. The eigenvalues of A do not necessarily form a group under multiplication. To see this, 2π let ωk = cos 2π k + i sin k , where k − 1 ∈ N, and just note that ωk 6= 1 is the only eigenvalue of the linear transformation ωk In , where In : Cn → Cn is the identity transformation. Obviously, {ωk } does not form a group under multiplication. As for a necessary and sufficient condition for the set of eigenvalues of A to form a group, nothing interesting can be said. Having said that, it is easily seen that the set of eigenvalues of A forms a group under multiplication, in fact a cyclic subgroup of the multiplicative group of the kth roots of unity, if and only if the set of the eigenvalues of A is closed under multiplication.  2. We have ord(a) = 7 because a7 = 1, a 6= 1, and 7 is prime. Induction on k together with the relations ab = b2 a and a = b2 ab−1 yields k

ak = b2 ak b−1 , 7

for all k ∈ N. In particular, a7 = b2 a7 b−1 together with a7 = 1 implies that 7 b2 −1 = 1. That is, b127 = 1. But 127 is prime and b 6= 1. Hence, ord(b) = 127. Therefore, ord(a) = 7 and ord(b) = 127, which is what we want.  3. (a) We can write (−x)3 + 2(−x)2 + (−x) = 0 =⇒ −x3 + 2x2 − x = 0, for all x ∈ R. On the other hand, x3 + 2x2 + x = 0 for all x ∈ R. Adding up the two equalities, we obtain 4x2 = 0 for all x ∈ R. We can also write (2x)3 + 2(2x)2 + (2x) = 0 =⇒ 8x3 + 8x2 + 2x = 0, for all x ∈ R. But 8x3 = 8x2 = 0 because 4x2 = 0 for all x ∈ R. This together with the above yields 2x = 0 for all x ∈ R, which is what we want. (b) Using (a), we can write (x + 1)3 + 2(x + 1)2 + (x + 1) = 0 =⇒ x2 + x = 0, for all x ∈ R. Consequently, x2 = −x = x because 2x = 0. Hence, x2 = x for all x ∈ R. Therefore, R is a Boolean ring, which is commutative by the following argument. (x + y)2 = x + y

=⇒ x2 + y 2 + xy + yx = x2 + y 2 =⇒ xy = −yx = yx,

for all x, y ∈ R, which is what we want.



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2.5.3. General. 1. We need the following lemma Lemma. The product of k consecutive integers is divisible by k!. Proof. It suffices to show that the product of k natural numbers is divisible by k!. To this end, let k, n ∈ N with k ≤ n be given. To show that n(n−1) · · · (n−k+1) n! is divisible by k!, it is enough to prove that Cnk := is an integer because k!(n − k)! n(n − 1) · · · (n − k + 1) = k!Cnk . By proving the polynomial identity below, which is also known as the Binomial Theorem n X (1 + x)n = Cnk xk , (∗) k=0

we conclude that Cnk is an integer, finishing the proof. Recall that, by Taylor’s Formula, for two polynomial functions p, q : R → R of degree n, we have p = q if and only if p(k) (0) = q (k) (0) for each k = 0, 1, . . . , n, where p(k) , q (k) denote the kth derivative of p and q, respectively, and the zeroth derivative of a function, by definition, is just the function. Also, it is readily verified that  1 ≤ i < k,  k(k − 1) · · · (k − (i − 1))(a + x)k−i
(i) k k! i = k, (a + x) =  0 i > k, where a ∈ R is a constant. Use p and q to denote the left and right hand side of (∗). Obviously, p(0) = q(0). We can write
(k) p(k) (0) = (1 + x)n (0) = n(n − 1) · · · (n − k + 1)(1 + 0)n−k = k!Cnk , n X
(k) q (k) (0) = Cni xi = Cnk (xk )(k) = Cnk k!, i=0

for each k = 1, . . . , n. So we have shown that p(k) (0) = q (k) (0) for each k = 0, 1, . . . , n, implying that p = q, which is what we want.  Now to prove the assertion, note first that   m−1 n Y Y  (kn + j) . (mn)! = k=0

j=1

But n Y j=1

(kn + j)

=

n−1 Y

(kn + j) × n(k + 1).

j=1

Qn−1 It follows from the lemma that j=1 (kn + j) is divisible by (n − 1)!, whence Qn by (n − 1)! × n(k + 1) = n!(k + 1). This implies (mn)! = j=1 (kn + j) is divisible  Qm−1 Qm−1 Qn m k=0 j=1 (kn + j) is divisible by k=0 n!(k+1) = (n!) m!. Thus, the number (mn)! is an integer, which is what we want.  m!(n!)m 2. The assertion is known as the Gauss-Lucas Theorem . As is shown in what follows, it suffices to prove that if the roots of a polynomial p lie in a closed half

80

2. SOLUTIONS

H’

z

h x

x’

Figure 7

plane, then so do those of p0 . For a subset S of C, the closed convex hull of S, denoted by the symbol C(S), is said to be the intersection of all closed convex subsets of C which include S, i.e., \ C(S) := C : S ⊆ C, C is closed and convex . It is straightforward to see that if S = {zi }ni=1 , where n ∈ N, then C(S) = C(S). We claim that \ C(S) = C(S) = H : S ⊆ H, H is a closed half plane . T Set C1 := H : S ⊆ H, H is a closed half-plane . As C1 is closed and convex, it c follows that C(S) = C(S) ⊆ C1 . We establish our claim by showing that C(S) = C(S)c ⊆ C1c . To see this, let x ∈ / C(S) = C(S) be arbitrary. As C(S) is closed and {x} is compact, it follows that there exists an x0 ∈ C(S) such that |x − x0 | =

inf

|x − y|.

y∈C(S)

Now, consider the perpendicular bisector ∆ of the line segment joining x and x0 . Let H 0 be the closed half-plane containing the point x0 ∈ S. The half plane H 0 includes S because if, contrary to this, there exists a point z ∈ S in the other half plane, then in the triangle x0 xz the perpendicular foot h of the altitude passing through the vertex x belongs S, for x0 , z ∈ S, and moreover |x − h| < |x − x0 |, which is a contradiction. Therefore, the closed half plane H 0 includes S, implying that C1 ⊆ C(S), which is what we want.

To complete the proof we need the following lemma. Lemma. If the roots of a polynomial with complex coefficients lie in a closed half plane, then so do those of the derivative of the polynomial. Proof. Suppose z1 , . . . , zn are the roots of a polynomial p so that p(z) = c(z − z1 ) · · · (z − zn ), where c ∈ C. Also suppose that the closed half plane H = {z ∈ C : Im z−a b ≤ 0} includes the roots of p. We show that H includes the roots of p0 , the derivative of p. If z = zi , for some 1 ≤ i ≤ n, is a root of p of multiplicity

2.5. FIFTH COMPETITION

81

greater than one, then zi is a root of p0 as well, in which case there is nothing to prove. So, without loss of generality, we may assume that p0 (z) 1 1 = + ··· + . p(z) z − z1 z − zn Now, suppose z ∈ / H. We complete the proof by showing that z cannot be a root of p0 . To this end, as z ∈ / H and zi ∈ H, we obtain z−a zi − a z − zi = Im − Im > 0, Im b b b b for all 1 ≤ i ≤ n. It follows that Im z−z < 0 for all 1 ≤ i ≤ n. From this, we obtain i Im

n X bp0 (z) b = < 0, Im p(z) z − zi i=1

implying that p0 (z) 6= 0, proving the lemma.



Now, assume that f is a polynomial of degree greater than or equal to 2 and A and A0 denote the sets of the roots of the equations f (z) = 0 and f 0 (z) = 0, respectively. Suppose that H is an arbitrary closed half plane including A. It follows from the lemma that H includes A0 . This yields \ A0 ⊆ H : A ⊆ H, H is a closed half plane = C(A). But since C(A) is a closed convex set which includes A0 , we see that C(A0 ) = C(A0 ) ⊆ C(A), which is what we want.



3. Remark. The lemma below together with the proof following the lemma shows that if the initial ball trajectory intersects the line segment joining the two foci, then there exists a hyperbola confocal with the original ellipse to which all trajectory segments are tangent. We need the following lemma. Lemma. (i) Let F and F 0 be two points and l a line in a plane not intersecting the line segment F 0 F (resp. intersecting the line segment F F 0 between its end points except the perpendicular bisector of F 0 F ). Then, there exists a unique ellipse (resp. hyperbola) whose foci are F 0 and F and to which the line l is tangent. (ii) Let F and F 0 be two points and P m and P n two half-lines through a point P in the plane such that the half-lines do not intersect the line segment F 0 F (resp. intersect F F 0 between its end points, i.e., between F and F 0 ). Then, there exists a unique ellipse (resp. hyperbola) whose foci are F 0 and F and to which the half-lines P m and P n are tangent if and only if ∠F P m = ∠F 0 P n. Proof. (i) Since the foci, i.e., F and F 0 , of the desired ellipse (resp. hyperbola) are fixed, it suffices to show that the sum (resp. the difference) of the focal radii of any point of the ellipse (resp. hyperbola) is uniquely determined by the given line l. To this end, suppose that the line l is tangent to an ellipse (resp. a hyperbola) whose foci are F and F 0 . By the optical property of ellipses (resp. hyperbolas) , the tangent point is obtained as follows. Reflect the focus F in the line l to get F1 . Then, join F1 and F 0 to intersect the line l at the tangent point, say, A ∈ l. It is plain that AF +AF 0 = F1 F 0 = F2 F (resp. |AF −AF 0 | = F1 F 0 = F2 F ), where F2 is

82

2. SOLUTIONS

F1

l

l F1

A

A

F’

F

F

F’ Figure 8

Figure 9

the reflection of F 0 across the line l. In other words, the sum (resp. the difference) of the focal radii of the point A, and hence any point, of the desired ellipse (resp. hyperbola) is F1 F 0 = F2 F , which is uniquely determined by the given line l. This proves (i). (ii) The uniqueness follows from (i). Let F1 and F2 be the reflection of F and F 0 across the half-lines P m and P n, respectively. First, suppose there exists an ellipse (resp. a hyperbola) whose foci are F 0 and F and to which the half-lines P m and P n are tangent. It is easily seen that the sum (resp. the difference) of the focal radii of any point of the ellipse (resp. hyperbola) is equal to F2 F = F1 F 0 . Now, the triangles 4F2 P F and 4F 0 P F1 are congruent because F2 F = F1 F 0 , F2 P = F 0 P , and P F = P F1 . This, in particular, implies ∠F2 P F = ∠F 0 P F1 , yielding ∠F 0 P F + 2∠F 0 P n = ∠F 0 P F + 2∠F P m

2.6. SIXTH COMPETITION

83

(resp. ∠F 0 P F − 2∠F 0 P n = ∠F 0 P F − 2∠F P m). Thus, ∠F 0 P n = ∠F P m, which, in turn, yields ∠F P n = ∠F 0 P F + ∠F 0 P n = ∠F 0 P F + ∠F P m = ∠F 0 P m, (resp. ∠F P n = ∠F 0 P F − ∠F 0 P n = ∠F 0 P F − ∠F P m = ∠F 0 P m, ) as desired. Next, suppose ∠F P n = ∠F 0 P m. We can write ∠F2 P F = ∠F 0 P F + 2∠F P n = ∠F 0 P F + 2∠F 0 P m = ∠F1 P F 0 , (resp. ∠F2 P F = ∠F 0 P F − 2∠F P n = ∠F 0 P F − 2∠F 0 P m = ∠F1 P F 0 ). That is, ∠F2 P F = ∠F1 P F 0 . It thus follows that the triangles 4F2 P F and 4F 0 P F1 are congruent because F2 P = F 0 P , ∠F2 P F = ∠F1 P F 0 , and P F = P F1 . In particular, we must have F2 F = F1 F 0 . In other words, the two ellipses (resp. hyperbolas) whose foci are F and F 0 and are tangent to the half-lines P m and P n coincide, proving the “if part” of the assertion.  Let F and F 0 be the foci of the ellipse ξ and P Q, QR, RS be three consecutive segments of the ball trajectory. It is well known that in any ellipse the focal radii of any point on the ellipse form equal angles with the tangent at the point. In other words, the normal and tangent lines at any point on an ellipse bisect the angles between the focal radii of the point. From this, it follows that if a billiard segment intersects the interior of the line segment F 0 F , so does all other billiard segments; if a billiard segment passes through one of the foci, then so does all other billiard segments; and finally, if a billiard segment does not intersect the line segment F 0 F , then neither does any other billiard segment. Now, with P, Q, R, S as in the above, it follows that ∠F 0 QP = ∠F QR and ∠F 0 RQ = ∠F RS. This together with the lemma implies that there are two ellipses ηQ and ηR confocal with ξ each of which is tangent to QR. We see from part (i) of the lemma that ηQ = ηR for any billiard segments QR. That is, there is an ellipse η confocal with the ellipse ξ such that the ball trajectory is always tangent to η, which is what we want.  2.6. Sixth Competition 2.6.1. Analysis. 1. To prove the assertion by contradiction, suppose that there are three rational points A = (x1 , y1 ) , B = (x2 , y2 ), and C = (x3 , y3 ) on the circle. Since A, B, C ∈ Q2 , it follows that the equations of the perpendicular bisectors of the line segments AB and BC are of the form a1 x + b1 y = c1 and a2 x + b2 y = c2 , respectively, where ai , bi , ci ∈ Q (i = 1, 2) and a1 b2 − b1 a2 6= 0. It is obvious that the center (x, y) of the circle is the solution of  a1 x + b1 y = c1 , a2 x + b2 y = c2 which must be rational because ai , bi , ci ∈ Q (i = 1, 2). This is a contradiction, proving the assertion, which is what we want. 

84

2. SOLUTIONS

2. As the sequence (fn )+∞ n=1 is uniformly convergent to f on M , for a given ε > 0, there exists a natural number N1 such that for all n ≥ N1 and x ∈ M we have
ε d fn (x), f (x) < . 6 Now, since the function fN1 is continuous at the point x ∈ M , for the given ε > 0, there exists a δ > 0 such that for all y ∈ M with d(x, y) < δ we have
ε d fN1 (y), fN1 (x) < . 3 From limn xn = x, we see that for δ > 0 obtained from the above, there exists a natural number N2 such that for all n ≥ N2 d(xn , x) < δ. Let N = max(N1 , N2 ). For the given ε > 0 and for all n ≥ N , we can write


d fn (xn ), f (x) ≤ d fn (xn ), f (xn ) + d f (xn ), fN1 (xn )

+d fN1 (xn ), fN1 (x) + d fN1 (x), f (x) ε ε ε ε + + + = ε. < 6 6 3 3 That is, limn→+∞ fn (xn ) = f (x), which is what we want.



R +∞ 3. Since 0 f (x) dx < +∞, we see that for given ε > 0, there exists a natural number N1 such that for all M ≥ N1 Z +∞ f (x) dx < ε . 3 M The function f is continuous on the compact interval [0, N1 + 1]. Hence, it is uniformly continuous on [0, N1 + 1]. Therefore, for given ε > 0, there exists a δ > 0 such that for all x, y ∈ [0, N1 + 1] with |x − y| < δ, we have f (x) − f (y) < ε . 3N1 Now, for a given ε > 0, find N1 > 0 and δ > 0 as in the above and pick N ∈ N such that N1 < δ. For all n ≥ N , we can write Z +∞ Z N1 Z +∞ f (x + 1 ) − f (x) dx = f (x + 1 ) − f (x) dx + f (x + 1 ) − f (x) dx n n n 0 0 N1 Z +∞ Z +∞ ε f (x + 1 ) dx + f (x) dx (N1 − 0) + < 3N1 n N1 N1 ε ε ε < + + = ε. 3 3 3 That is, Z +∞ f (x + 1 ) − f (x) dx − 0 < ε, n 0 R +∞ for al n ≥ N . Therefore, limn→+∞ 0 f (x + n1 ) − f (x) dx = 0, which is what we want. 

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85

2.6.2. Algebra. 1. Proceed by contradiction. Suppose that AB is a subgroup of G. Using Lagrange’s Theorem , we see that |AB| = pn b, where b ∈ N and b|a. Since gcd(p, a) = 1 and b|a, gcd(p, b) = 1. On the other hand, we can write |AB| =

|A| × |B| . |A ∩ B|

But A ∩ B $ B because B * A. Hence, |A ∩ B| < |B|. Now, in view of Lagrange’s Theorem, let |A ∩ B| = pi for some 0 ≤ i < m. We can write pn .pm =⇒ b = pm−i , pn b = pi implying that gcd(p, b) = p, which is a contradiction. Therefore, AB cannot be a subgroup of G, finishing the proof.  2. We prove the assertion for matrices over a general division ring ∆. Induct on m. If m = 1, the assertion is easily seen to hold. Assuming that the assertion holds for matrices of size less than m, we prove the assertion for matrices of size m. To this end, note first that r := rank(A1 ) ≤ dim ker A1 because A21 = 0. Now, from the Rank-Nullity Theorem, we see that r ≤ m 2 . Let K := ker A1 . Since Ai ’s commute, K is invariant under all Ai ’s (1 ≤ i ≤ n). Therefore, after a similarity, one can write   Bi Ci Ai = , 0 Di where Bi ∈ Mm−r (∆) with B1 = 0, Ci ∈ M(m−r)×r (∆), and Di ∈ Mr (∆) for all n−1 . As Ai ’s commute, so do 1 ≤ i ≤ n. We have m < 2n , yielding r ≤ m 2 < 2 Di ’s in Mr (∆) (1 ≤ i ≤ n). Hence, from the induction hypothesis, we see that D2 · · · Dn = 0, which, in turn, implies that   B C A2 · · · An = , (∗) 0 0 for some appropriate matrices B ∈ Mm−r (∆) and C ∈ M(m−r)×r (∆). On the other hand,   0 C1 A1 = . (∗∗) 0 D1 It now easily follows from (∗) and (∗∗) that A1 A2 · · · An = A1 (A2 · · · An−1 An ) = 0, which is what we want.



3. Since e2 = e, f 2 = f , and ef = f e, using the Binomial Theorem, we see that for all n ∈ N, there exists a Kn ∈ N such that (e − f )n = e + (−1)n f + Kn ef . Suppose (e − f )n = 0. Without loss of generality, we may assume that n is even. It follows that (e − f )n = e + f + Kn ef = 0. Multiplying both sides of the above by e and f , respectively, and noting that e2 = e, f 2 = f , and ef = f e, we obtain e + ef + Kn ef = 0 = f + ef + Kn ef, yielding e = f = −(1 + Kn )ef , which proves the assertion.



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2. SOLUTIONS

2.6.3. General. 1. Noting that (−1)|n| = (−1)n for all n ∈ N, we can write (−1)|n1 −m1 |+···+|nk −mk |

= (−1)|n1 −m1 | × · · · × (−1)|nk −mk | = (−1)n1 −m1 × · · · × (−1)nk −mk =

(−1)(n1 +···+nk )−(m1 +···+mk ) = (−1)0 = 1,

implying that |n1 − m1 | + · · · + |nk − mk | is an even number.



2. Proceed by induction on n. If n = 1, the assertion is easy. Assuming that the assertion holds for n×n matrices, we prove it for (n+1)×(n+1) matrices. Suppose that A is a (n + 1) × (n + 1) matrix such that the sum of the elements on any row and any column corresponding to any nonzero element of A is at least n + 1. As the entries of A are all nonnegative integers, if the entries are all nonzero, then the 2 sum of the entries of the matrix A is at least (n + 1)2 which is no less than (n+1) . 2 Now, assume that for some 1 ≤ i, j ≤ n + 1, aij = 0. Let Aˆij be the n × n matrix which is obtained from eliminating row i and column j of the matrix A. Obviously, the sum of the elements on any row and any column corresponding to any nonzero element of the matrix Aˆij is at least n. So it follows from the induction hypothesis 2 that the sum of the entries of Aˆij is no less than n2 . Noting that the sum of the entries of row i and column j of A is at least n + 1, we see that the sum of the entries of the matrix A is at least (n + 1)2 + 1 (n + 1)2 n2 +n+1 = > , 2 2 2 which is what we want, finishing the proof.  3. The assertion is a special case of the following: Suppose k + m points inside a circle are given. Then, there exists a straight line not passing through any of the points that divides the circle into two sections one of which containing k points and the other the remaining m points. To see this, note that the number of lines passing through a pair
of points from these k + m points is finite; in fact it is less than or equal to k+m . As the number of directions determined by these lines is 2 finite, there is a line d whose direction is not parallel to any of the lines determined by any pair of points from the given points. Let AA0 be a diameter of the circle whose direction is perpendicular to the line d. Project the given points on the line segment AA0 to get k + m distinct points. Going from A to A0 , if P is the kth point and Q the (k + 1)st point, then any line perpendicular to AA0 at any point between P and Q does not pass through any of the given points inside the circle and yet it divides the circle into two sections one of which contains k points, including the point whose image is P , and the other contains the remaining m points.  2.7. Seventh Competition 2.7.1. Analysis. 1. First of all, the hypothesis that S is closed is redundant because it follows from the other hypothesis of the problem that such an S is closed. The assertion is a consequence of the following. Let n ∈ N. Then there is no proper subset S of Rn with the property that for all x ∈ Rn \ S there exists nx ∈ N with nx > 1 such that there are exactly nx points in S as the closest point of S to x. To

2.7. SEVENTH COMPETITION

87

prove this by contradiction, pick a point x ∈ Rn \ S. It follows from the hypothesis that there exists an nx ∈ N with nx > 1 and nx points s1 , . . . , snx in S such that inf ||x − s|| = ||x − si || = δx ,

s∈S

for all 1 ≤ i ≤ nx . It is obvious that there is no point of S in Bδx (x), the open ball 1 centered at x with radius δx . Let y = x+s and δ = δ2x . Plainly, the only point of 2 S in B δ (y), the closed ball centered at y with radius δ, is s1 . As Bδ (y) ⊆ Bδx (x), it follows that for the point y ∈ Rn \ S there exists exactly one point s1 ∈ S such that inf ||x − s|| = ||x − s1 ||, s∈S

which is a contradiction, finishing the proof.



2. We need the following lemma. Lemma. Let (Ir )r∈Q+ and A and f be as in the statement of the problem. Then, (i) for all x ∈ A with x ∈ I r , we have f (x) ≤ r. (ii) for all x ∈ A with x ∈ / Ir , we have f (x) ≥ r. Proof. Set Sx = {r : x ∈ Ir }. We have f (x) = inf Sx . To prove (i), note that if x ∈ I r , then I r ⊆ Is whenever r < s where s ∈ Q, from which, we obtain x ∈ Is for all s ∈ Q with r < s. Consequently, Sx includes all rational numbers greater than r, whence f (x) = inf Sx ≤ r. To prove (ii), note that if x ∈ / Ir , then I s ⊆ Ir whenever s < r where s ∈ Q. This implies x ∈ / Is for all s ∈ Q with s < r. Therefore, Sx contains no rational number less than r, whence f (x) = inf Sx ≥ r, proving the lemma.  We now prove that f is continuous. Let x0 be an arbitrary point of A. It suffices to prove that for any open interval (c, d) with f (x0 ) ∈ (c, d) there exists an open neighborhood N around x0 such that f (N ) ⊆ (c, d). To this end, as c < f (x0 ) < d, pick rational numbers p and q such that c < p < f (x0 ) < q < d. We claim that the open set N = Iq \ I p contains x0 and moreover f (N ) ⊆ [p, q] ⊆ (c, d), finishing the proof. Firstly, x0 ∈ N because, in view of the lemma, x0 ∈ Iq and x0 ∈ / I p. Secondly, f (N ) ⊆ [p, q] ⊆ (c, d). To see this, let x ∈ N = Iq \ I p be arbitrary. It follows that x ∈ Iq ⊆ I q , which, in view of the lemma, implies f (x) ≤ q. On the other hand, since x ∈ / I p , we conclude from the lemma that f (x) ≥ p. That is, we have proved f (x) ∈ [p, q] for all x ∈ N . Consequently, f (N ) ⊆ [p, q] ⊆ (c, d), which is what we want.  3. First solution: To find the limit, we need the following lemma whose first and second parts are known as Stolz’s First and Second Theorems on limits. Lemma. (i) Let (pn )+∞ n=1 be a sequence of reals whose elements are eventually nonnegative, sn = p1 + · · · + pn , and limn sn = +∞. If (an )+∞ n=1 is a sequence of real numbers such that limn an ∈ R ∪ {±∞}, then lim n

p1 a1 + · · · + pn an = lim an . n p1 + · · · + pn

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2. SOLUTIONS

+∞ (ii) Let (xn )+∞ n=1 and (yn )n=1 be sequences of real numbers with limn yn = +∞ +∞ n−1 and such that (yn )n=1 is eventually increasing. If limn xynn −x −yn−1 ∈ R ∪ {±∞}, then

lim n

xn xn − xn−1 = lim . n yn − yn−1 yn

Proof. (i) There are two cases to consider. (a) limn an = ±∞. n (−an ) n an = − p1 ap11+···+p As p1 (−ap11)+···+p +···+pn +···+pn , it suffices to prove the assertion for the case when limn an = +∞. To this end, it follows from the hypothesis that for given M > 0, there exists an N1 > 0 such that an > 2M whenever n > N1 . Let AN1 = p1 a1 + · · · + pN1 aN1 . We can write p1 a1 + · · · + pn an p 1 + · · · + pn

= >

AN1 sn = 0 and sn −sN1 −M 3 and sn

But limn AN1 sn

> n > N , we have

limn >

2 3

sn −sN1 sn

AN1 pN +1 aN1 +1 + · · · + pn an + 1 sn sn AN1 sn − sN1 + 2M sn sn = 1. Thus, there exists N2 > 0 such that

whenever n > N2 . Letting N = max(N1 , N2 ), for all

p1 a1 + · · · + pn an 4M −M + = M. > p1 + · · · + pn 3 3 n an As M > 0 is arbitrary, this means limn p1 ap11+···+p = +∞, proving the assertion +···+pn in this case. (b) limn an = a ∈ R. First, let limn an = 0. It follows that for given ε > 0 there exists N1 > 0 such that |an | < 2ε whenever n > N1 . Let AN1 be as in (a). For all n > N1 , we can write p a + · · · + p a AN1 ε sn − sN AN1 ε 1 1 n n 1 + × ≤ + . ≤ p 1 + · · · + pn sn 2 sn sn 2 AN AN1 1 Now, as limn sn = 0, we see that there exists an N2 > 0 such that sn < 2ε whenever n > N2 . Letting N = max(N1 , N2 ), for all n > N , we have p a + · · · + p a ε ε 1 1 n n + = ε. < p1 + · · · + pn 2 2 n an That is, limn p1 ap11+···+p = 0, which is what we want. +···+pn Next, let limn an = a 6= 0. As limn (an − a) = 0, from what we just proved, it follows that   p1 a1 + · · · + pn an p1 (a1 − a) + · · · + pn (an − a) lim −a = lim n n p 1 + · · · + pn p 1 + · · · + pn = lim(an − a) = 0.

n

n an limn p1 ap11+···+p +···+pn

This implies that = a, finishing the proof. +∞ (ii) Define the two sequences (pn )+∞ n=1 and (an )n=1 , inductively, as follows pn = yn − yn−1 , p1 = y1 , (n ≥ 2), xn − xn−1 = pn an , x1 = p1 a1 , (n ≥ 2).

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89

Firstly, pn ’s are eventually nonnegative. Secondly, yn = p1 + · · · + pn → +∞ as n−1 n → +∞. Thirdly, an = xynn −x −yn−1 and p1 a1 +· · ·+pn an = xn , and p1 +· · ·+pn = yn . Therefore, (i) applies, proving the assertion.  Now, note that 21 1

xn =

+ ··· +

2n n

2n+1 n+1

=

Xn . Yn

First of all, it is easily seen that the sequence (Yn )+∞ n=1 , where Yn = n ∈ N, is increasing. Secondly, limn Yn = +∞. Thirdly, we can write lim n

Xn − Xn−1 = lim n Yn − Yn−1

2n n 2n+1 n+1

−

2n n

= lim n

n

n

for all

1 = 1. −1

2n n+1

Therefore, it follows from the lemma that limn xn = limn lim xn = lim

2n+1 n+1

Xn Yn

exists, and moreover

Xn Xn − Xn−1 = lim = 1, n Yn − Yn−1 Yn

which is what we want. Second solution: It is readily checked that xn+1 =

n+1
n+2 n + 2 X 2k = xn + 1 , n+2 2 k 2(n + 1) k=1

from which, we obtain xn+2 − xn+1 =

(n + 2)2 (xn+1 − xn ) − xn+1 − 1 , 2(n + 1)(n + 2)

for all n ∈ N. As xn > 0 for all n ∈ N, in view of the above equality, we see that xn+2 −xn+1 < 0 whenever xn+1 −xn ≤ 0. But a straightforward calculation reveals that x3 = x4 = 53 . Thus, by an easy induction on n ≥ 3, we see that the sequence (xn )+∞ n=1 is decreasing for n ≥ 3. From this, together with the fact that xn > 0 for all n ∈ N, we conclude that limn→+∞ xn exists, and hence limn→+∞ xn = L for some L ∈ R. Consequently,
1 n+2 L = lim xn+1 = lim xn + 1 = (L + 1) =⇒ L = 1. n→+∞ n→+∞ 2(n + 1) 2 Therefore, limn→+∞ xn = 1, which is what we want.



2.7.2. Algebra. 1. (a) Let x0 ∈ R be arbitrary. We show that x0 e = x0 . To this end, note that for all x ∈ R, we can write (x0 + e − x0 e)x = x0 x + ex − x0 (ex) = x0 x + x − x0 x = x. That is, (x0 + e − x0 e)x = x for all x ∈ R. As the left identity element is unique, we obtain x0 + e − x0 e = e, implying x0 e = x0 . This means e a right identity element as well. Consequently, e is the identity element of R. (b) Let a be an idempotent of R, i.e. a2 = a, and x ∈ R be arbitrary. It is straightforward to see that (axa − ax)2 = 0. That is, axa − ax is a nilpotent element of R, and hence axa−ax = 0, by the hypothesis. Likewise, (axa−xa)2 = 0,

90

2. SOLUTIONS

implying axa − xa = 0 for all x ∈ R. So we have shown that ax = xa = axa for all x ∈ R, which is what we want.  2. First solution: Let H be an arbitrary subgroup of G. Set H1 = H ∩ G1 . We show that H1 is commutative, H1 E H, and that HH1 is commutative. First, H1 is commutative because H1 ⊆ G1 and G1 is commutative. Next, H1 ≤ H because H1 = H ∩ G1 and H and G1 are subgroups of G. That H1 E H follows from G1 E G and H1 = H ∩ G1 . Finally, to show that HH1 is commutative, using the Second Isomorphism Theorem for groups, we can write H H HG1 ∼ = . = H1 H ∩ G1 G1 HG1 G 1 But HG1 ≤ G. Hence, HG G1 ≤ G1 . Consequently, G1 is commutative because so is GG1 . This, in view of the above isomorphism, completes the proof.

Second solution: Use G0 to denote the derived subgroup of G. Since abelian, it follows that G0 ≤ G1 . Now for a given subgroup H of G, set

  −1 H1 = H 0 := h1 h2 h−1 . 1 h2 : h1 , h 2 ∈ H

G G1

is

It is plain that H1 is a normal subgroup of H and that HH1 is abelian because H1 is the derived subgroup of H. To see that H1 is abelian, just note that H1 ≤ G0 ≤ G1 and that G1 is abelian by the hypothesis. This completes the proof.  3. For the given θ : V → V , it is obvious that ker θ ⊆ ker θ2 ⊆ · · · is an ascending chain of the subspaces of V . As V is finite-dimensional, it follows that there exists an s ∈ N such that ker θs = ker θs+1 . We claim that if ker θs = ker θs+1 , then ker θs = ker θs+k for all k ∈ N. We prove this by induction on k. If k = 1, there is nothing to prove. Suppose the assertion holds for k. To prove the assertion for k + 1, note first that ker θs ⊆ ker θs+k+1 . Next, let x ∈ ker θs+k+1 be arbitrary. We obtain θs+k (θx) = 0. That is, θx ∈ ker θs+k , which, in view of the induction hypothesis, yields θx ∈ ker θs . This implies θs+1 x = 0 which means x ∈ ker θs+1 . But ker θs+1 = ker θs . Thus, x ∈ ker θs . As x ∈ ker θs+k+1 was arbitrary, we see that ker θs+k+1 ⊆ ker θs . Therefore, ker θs+k+1 = ker θs , proving the claim by induction on k. We now show that V = im(θs ) ⊕ ker(θs ), where s is as in the above. Note that by the Rank-Nullity Theorem dim im(θs ) + dim ker(θs ) = dim V. On the other hand, by Problem 1(a) of 1.3.2, we can write dim(im(θs ) + ker(θs ))


= dim im(θs ) + dim ker(θs ) − dim im(θs ) ∩ ker(θs ) ,

which, in view of the preceding equality, implies V = im(θs ) ⊕ ker(θs ) as soon as we show that im(θs ) ∩ ker(θs ) = {0}. To see this, let x ∈ im(θs ) ∩ ker(θs ) be arbitrary. It follows that there exists x1 ∈ V such that x = θs x1 , from which, we obtain θs+s x1 = 0 because x ∈ ker θs . On the other hand, by the claim we made in the above, we have ker θs+s = ker θs , yielding x1 ∈ ker θs . In other words, x = θs x1 = 0. Since x ∈ im(θs ) ∩ ker(θs ) was arbitrary, we conclude that im(θs ) ∩ ker(θs ) = {0}, which is what we want. Therefore, V = im(θs ) ⊕ ker(θs ), completing the proof. 

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91

2.8. Eighth Competition 2.8.1. Analysis. 1. (a) Define the function h : F ∪ G → R by  f (x) x ∈ F, h(x) = g(x) x ∈ G. As f (x) = g(x) on F ∩ G, the function h is well-defined. It is obvious that h is an extension of f and g to F ∪ G. To prove that h is continuous, it suffices to show that h−1 (C) is a closed subset of F ∪ G whenever C is a closed subset of R. To this end, note that for any closed subset C of R, we can write h−1 (C) = f −1 (C) ∪ g −1 (C). Now, as f and g are continuous and C is a closed subset of R, we see that f −1 (C) and g −1 (C) are closed subsets of F and G, respectively. On the other hand, F and G are closed in F ∪ G. Hence, f −1 (C) and g −1 (C) are closed subsets of F ∪ G. This together with the above equality implies that h−1 (C) is a closed subset of F ∪ G, which is what we want.   (b) For given n ∈ N, let F = B1 (0) := x ∈ Rn : ||x|| < 1 and G = F c = x ∈ Rn : ||x|| ≥ 1 . Obviously, F ∩ G = ∅ and F ∪ G = Rn . Now, define the functions f : F → R and g : G → R by f (x) = 0 and g(x) = 1 whenever x ∈ F and x ∈ G, respectively. For this f and g, we show that the conclusion of (a) does not hold. Suppose to the contrary that h : Rn = F ∪ G → R is an extension of f and g to Rn = F ∪ G. We must have  0 x ∈ F, h(x) = 1 x ∈ G, which is not continuous on Rn = F ∪ G, for otherwise h(Rn ) = {0, 1}, which is disconnected whereas Rn is connected. This leads to a contradiction, proving the assertion.  2. Remark. It is worth mentioning that using a proof almost identical to the proof below one can show that if (an )+∞ of real numbers n=1 is a decreasing sequence P+∞ P+∞ with limn an = 0, then n=1 an is convergent if and only if n=1 n(an − an+1 ) is convergent. Also, see Problem 2 of 1.12.1. P+∞ We have limn an = 0 because n=1 an is convergent. Since (an )+∞ n=1 is decreasing, it follows that an ≥ an+k for all n, k ∈ N. Letting k → +∞, we obtain an ≥ 0 for P+∞ all n ∈ N. Now, as n=1 an is convergent, it follows that it is Cauchy. Thus, for given ε > 0, there exists an N > 0 such that n X ε ak < , 2 k=m

[ n2 ]

for all n ≥ m ≥ N . Let m = > N . We obtain n > m ≥ N . This together with the fact that an ’s are nonnegative implies n X k=[ n 2]

ak <

ε . 2

92

2. SOLUTIONS

As an ’s are decreasing, we have (n − [ n2 ] + 1)an ≤ nan 2

≤ =

Pn

k=[ n 2]

ak , yielding


nan n n + − [ ] + 1 an 2 2 2 n X
n ak n − [ ] + 1 an ≤ 2 n k=[ 2 ]

<

ε , 2

for all n > 2N + 2. Consequently, nan < ε for all n > 2(N + 1). That is, limn nan = 0. We can write +∞ X

n(an − an+1 ) =

n=1

+∞  X


nan − (n + 1)an+1 + an+1 .

n=1


P+∞ Since limn nan = 0, we see that the telescopic series n=1 nan − (n + 1)an+1 P+∞ converges to a1 . On the other hand, n=1 an+1 is convergent. Therefore, the P+∞ series n=1 n(an − an+1 ) converges and we have +∞ X

n(an − an+1 )

=

n=1

=

+∞  X n=1 +∞ X


nan − (n + 1)an+1 + an+1

+∞
X nan − (n + 1)an+1 + an+1

n=1

= a1 +

n=1 +∞ X

an+1 ,

n=1

finishing the proof.



3. First solution: As the function g is continuously differentiable on [0, 1], it follows that g is of bounded variation on [0, 1]. We show that g can be written as the difference of two increasing functions g1 and g2 each of which is continuously differentiable on [0, 1]. To see this, let g 0 = f , where f is continuous on [0, 1]. We can write f = f + − f − , where f + = max(f, 0) = f2 + |f2 | and f − = max(−f, 0) = |f | −f + and f − are nonnegative continuous functions on [0, 1]. 2 + 2 . Obviously, f Using the Second Fundamental Theorem of Calculus, we can write Z x Z x Z x
g(x) = g(0) + f = g(0) + f+ − f − = g1 (x) − g2 (x), 0

0

Rx

where g1 (x) = g(0) + 0 f + and g2 (x) = Theorem of Calculus, we obtain

0

Rx 0

f − . Applying the First Fundamental

g10 (x) = f + (x) ≥ 0, g20 (x) = f − (x) ≥ 0, for all x ∈ [0, 1]. Therefore, g = g1 − g2 , where g1 and g2 are continuously differentiable and increasing on [0, 1]. We can write Z 1 Z 1 Z 1 Z 1 n n n x dg = x d(g1 − g2 ) = x dg1 − xn dg2 . 0

0

0

0

2.8. EIGHTH COMPETITION

93

Now, as g1 and g2 are continuously differentiable on [0, 1], we have Z 1 Z 1 Z 1 Z 1
xn dg = xn g10 (x)dx − xn g20 (x)dx = xn g10 (x) − g20 (x) dx. 0

0

0

0

Hence, 1

Z

Z

n

x dg =

1

xn g 0 (x)dx.

0

0

Now, as g 0 is continuous on [0, 1], there is an M > 0 such that |g 0 | ≤ M on [0, 1], from which, we see that Z 1 Z 1 Z 1 n 0 n x g (x)dx ≤ xn g 0 (x) dx x dg = 0 0 0 Z 1 M xn dx = ≤ M , n+1 0 R1 M implying that limn 0 xn dg = 0 because limn n+1 = 0. Second solution: As g is continuously differentiable on [0, 1], it is of bounded variation on [0, 1]. Hence, there exists increasing functions g1 and g2 on [0, 1] such that g = g1 − g2 . By the definition of the Riemann-Stieltjes integrals with respect to integrands of bounded variations, we have Z 1 Z 1 Z 1 xn dg = xn dg1 − xn dg2 . 0

0

0

Using integration by parts, we obtain Z 1 Z 1 Z 1 n n n x dgi = x gi (x) − gi (x)dx = gi (1) − 0

0

for each i = 1, 2. So we have Z 1 Z xn dg = g(1) − 0

0

0

1

ng(x)xn−1 dx =

1

ngi (x)xn−1 dx,

0

Z

1


n g(1) − g(x) xn−1 dx.

0

Now, for given ε > 0, as g is continuous at 1, there exists a 0 < δ < 1 such that |g(x) − g(1)| < 2ε whenever 1 − δ < x < 1. We can write Z 1 Z 1 Z 1
n−1 n = ≤ x dg n g(1) − g(x) x dx n g(1) − g(x) xn−1 dx 0 0 0 Z 1−δ Z 1 n−1 = n g(1) − g(x) x dx + n g(1) − g(x) xn−1 dx. 0

1−δ

By the continuity of g on [0, 1], there exists M > 0 such that |g| ≤ M on [0, 1]. With this in mind, we have Z 1 Z Z 1−δ nε 1 n−1 n n−1 ≤ 2M n x dx x dg x dx + 2 1−δ 0 0
ε ε = 2M (1 − δ)n + 1 − (1 − δ)n < 2M (1 − δ)n + . 2 2 Consequently, Z 1 ε n x dg < 2M (1 − δ)n + . 2 0

94

2. SOLUTIONS

Now, limn (1 − δ)n = 0 because 0 < 1 − δ < 1. Hence, there exists N > 0 such that ε (1 − δ)n < 4M for all n ≥ N . Therefore, Z 1 n < 2M × ε + ε = ε, x dg 4M 2 0 R1 n for all n ≥ N . This means limn 0 x dg(x) = 0, which is what we want. 

2.8.2. Algebra. 1. Note that b = I + N , where I is the identity matrix, N = b − I, and N 2 = 0. From this, using the Binomial Theorem and noting that b−1 = I − N , it is easily verified that   1 n bn = , 0 1 for all n ∈ Z. Therefore, H = {bn : n ∈ Z} = −1

To find aHa

−1

, note that a

abn a−1 =



2 0

0 1

1 2

 =

0





0 1

1 n 0 1

1 n 0 1



 :n∈Z .

 . We can write 

1 2

0

0 1



 =

1 0

2n 1

 .

Therefore, aHa−1 =



1 0

2n 1



 :n∈Z .

This implies that aHa−1  H. That is, aHa−1 is a proper subgroup of H, which is what we want.  2. We need the “only if part” of the following standard lemma. ring with identity. Then, x ∈ J(R) := T  Lemma. Let R be a commutative m C R : m is maximal in R if and only if 1 − xy is a unit in R for all y ∈ R. Proof. “⇐=” Let x ∈ J(R). Suppose to the contrary that there exists a y0 ∈ R such that 1 − xy0 is not a unit. As 1 − xy0 is not invertible, a standard argument using Zorn’s Lemma shows that there exists a maximal ideal m0 such that 1 − xy0 ∈ m0 . On the other hand, since x ∈ J(R), x ∈ m0 , and hence xy0 ∈ m0 . So, we obtain 1 = (1 − xy0 ) + xy0 ∈ m0 , yielding m0 = R, which is a contradiction. Thus the assertion follows by contradiction. “=⇒” Again, we proceed by contradiction. Suppose that there exists a maximal ideal m0 such that x ∈ / m0 . As m0 $ m0 + xR and m0 is maximal, we see that m0 + xR = R. Hence, there exist a ∈ m0 and y0 ∈ R such that 1 = a + xy0 . This together with the hypothesis implies that a = 1 − xy0 ∈ m0 is a unit in R, which, in turn, implies m0 = R, a contradiction. Therefore, the assertion follows by contradiction.  To prove the assertion by contradiction, suppose that R has finitely many maximal ideals m1 , . . . , mn for some n ∈ N. Firstly, {0} $ mi for each i = 1, . . . , n because otherwise {0} and R are the only ideal of R, implying that R is a field,

2.8. EIGHTH COMPETITION

95

which is a contradiction, for R would then have infinitely many units. On the other hand, we have n \ mi ⊇ m1 · · · mn . i=1

We also have m1 · · · mn 6= {0} because {0} $ mi for each , n and R is Tni = 1, . . .T an integral domain. Now, pick 0 6= a ∈ m1 · · · mn ⊆ i=1 mi = {m C R : m is maximal in R}. It follows from the lemma that 1 − ax is a unit in R for x ∈ R. But, as R is an integral domain, 1 − ax 6= 1 − ay whenever x 6= y. This together with the fact that R is an infinite integral domain implies that R has infinitely many units, which is a contradiction. So the assertion follows by contradiction.  3. Let A = (aij ) ∈ Mn (R). Use the symbol (A2 )ij to denote the ij entry of the matrix A2 . Using A2 = I, we can write n n X n n X X X (A2 )ij = (I)ij = n. i=1 j=1

On the other hand, n n X X (A2 )ij

=

i=1 j=1

=

i=1 j=1

n X n n X X

aik akj =

n n X X

aik

n X

i=1 j=1 k=1 n X n X

i=1 k=1 n n X X

j=1 n X

i=1 k=1

i=1 k=1

i=1

aik a = a

aik = a

akj a = na2 .

2

Therefore, na = n, yielding a = ±1. Note that both cases a = 1 or a = −1 can happen because A = ±I yields a = ±1, respectively.  2.8.3. General. 1. Note that f (x, y, z) = (x − 3)2 + 4(y + 12 ) + z 2 − 10. This yields f (x, y, z) ≥ −10 for all x, y, z ∈ R and that f (x, y, z) = −10 if and only if x = 3 , y = − 12 , and z = 0. Therefore, the function f assumes it absolute minimum  at (3, − 12 , 0), which is what we want. 2. Let x1 < · · · < x5 be the amount owned by these five people in of money increasing order. We have x1 x2 , x2 x3 , x3 x4 , and x4 x5 . Hence, there exist natural numbers ki (1 ≤ i ≤ 4) such that xi+1 = ki xi . Using the hypothesis, we can write x1 (1 + k1 + k1 k2 + k1 k2 k3 + k1 k2 k3 k4 ) = 719. But 719 is prime. So, we obtain x1 = 1, 1 + k1 + k1 k2 + k1 k2 k3 + k1 k2 k3 k4 = 719. Consequently, k1 (1 + k2 + k2 k3 + k2 k3 k4 ) = 718 = 2 × 359. As 2 and 359 are primes, we must have k1 = 2, 1 + k2 + k2 k3 + k2 k3 k4 = 359, or, k1 = 1, 1 + k2 + k2 k3 + k2 k3 k4 = 718, or, k1 = 359, 1 + k2 + k2 k3 + k2 k3 k4 = 2, or, k1 = 718, 1 + k2 + k2 k3 + k2 k3 k4 = 1. The last three cases are easily seen to be refuted. Thus, k1 = 2, 1 + k2 + k2 k3 + k2 k3 k4 = 359.

96

2. SOLUTIONS

Simplifying, we obtain k2 (1 + k3 + k3 k4 ) = 358 = 2 × 179. Since 2 and 179 are primes, we see that k2 k2 k2 k2

= 2, 1 + k3 + k3 k4 = 179, or, = 1, 1 + k3 + k3 k4 = 358, or, = 179, 1 + k3 + k3 k4 = 2, or, = 358, 1 + k3 + k3 k4 = 1.

Once again, refuting the last three cases, we obtain k2 = 2, 1 + k3 + k3 k4 = 179, which yields k3 (1 + k4 ) = 178 = 2 × 89. Likewise, noting that 2 and 89 are primes, we get k3 = 2, 1 + k4 = 89, or, k3 = 1, 1 + k4 = 178, or, k3 = 89, 1 + k4 = 2, or, k3 = 178, 1 + k4 = 1. Again, refuting the last three cases, we obtain k3 = 2, 1 + k4 = 89, which yields k4 = 88. Therefore, the only solution of x1 + · · · + x5 = 719 subject to the imposed conditions is x1 = 1, x2 = k1 x1 = 2, x3 = k2 x2 = 4, x4 = k3 x3 = 8, and x5 = k4 x4 = 704. That is, these five people in increasing order have x1 = 1, x2 = 2, x3 = 4, x4 = 8, x5 = 704, amounts of money, which is what we want.



3. Without loss of generality, we may assume that starting with every minute, the traffic-light of the crossroad B is 30 seconds red and 30 seconds green, respectively. If t denotes the tth second of a particular minute at the crossroad B, then the stop time function of the bus at the crossroad, denoted by f , on that particular minute, is  30 − t 0 ≤ t ≤ 30, f (t) = 0 30 ≤ t ≤ 60. Let µf denote the average value of the function f , which means the average of the stop time at the crossroad B. By definition, we have Z 60 Z 30 1 1 f (t)dt = (30 − t)dt = 7.5. µf = 60 0 60 0 But the bus passes through the crossroad B of the city with the probability of 1 3 . Therefore, the average of the stop time of the bus at the crossroad is equal to 1 7.5  3 µf = 3 = 2.5, which is what we want. 2.9. Ninth Competition 2.9.1. Analysis. 1. Suppose that C is an equivalence class of the equivalence relation ∼. Pick an x0 ∈ C. It is plain that  C = x0 + r|r ∈ Q . It suffices to show that C ∩ (a, b) 6= ∅ for all open intervals (a, b) of R. To this end, for a given open interval (a, b), as Q is dense in R, we see that there exists an r ∈ Q ∩ (a − x0 , b − x0 ). This yields x0 + r ∈ C ∩ (a, b), implying C ∩ (a, b) 6= ∅, which is what we want. 

2.9. NINTH COMPETITION

97

2. First, we claim that the function f is strictly increasing or strictly decreasing. To prove this by contradiction, suppose that f is neither strictly increasing nor strictly decreasing. By the lemma presented in Solution 2 of 2.5.1, there
exist x1 , x2 , x3 ∈ R with x1
< x2 < x3 such that f (x2 ) ≥ max f (x1 ), f (x3 ) or f (x2 ) ≤ min f (x1 ), f (x3 )
. As f is one-to-one and x1
< x2 < x3 , we see that f (x2 ) > max f (x1 ), f (x3 ) or f (x2 ) < min f (x1 ), f (x
3 ) . First, suppose f (x2 ) >
f (x2 )+max f (x1 ),f (x3 ) max f (x1 ), f (x3 ) . Letting λ = , we see that f (x1 ) < λ < 2 f (x2 ) and f (x3 ) < λ < f (x2 ). Hence, by the Intermediate Value Theorem, there exist x1 < ξ1 < x2 and x2 < ξ2 < x3 such that f (ξ1 ) = λ = f (ξ2 ). As f is one-to-one, this yields ξ1 = ξ2 , which is a contradiction.
Likewise, if
f (x2 )+min f (x1 ),f (x3 ) , we obtain a f (x2 ) < min f (x1 ), f (x3 ) , considering λ = 2 contradiction, proving the claim. Thus, the function f is either strictly increasing or strictly decreasing. We prove the assertion in the case when f is strictly increasing. The case when f is strictly decreasing can be done similarly or by replacing f by −f and repeating the above argument. To prove that f −1 is continuous, it suffices to show that f is an open map. To this end, note first that f (−∞) = limx→−∞ f (x) and f (+∞) = limx→+∞ f (x) exist, where f (±∞) ∈ R ∪ {±∞},
and that, in view of the Intermediate Value Theorem, f (R) = f (−∞), f (+∞)
. Analogously, if (a, b) is an open interval in R, we obtain f (a, b) = f (a), f (b) . Now, suppose that G is an open subset of R. It follows that there exists a family {(ai , bi )}i∈I of open S intervals of R such that G = i∈I (ai , bi ). We can write [ [
[
f (G) = f (ai , bi ) = f (ai , bi ) = f (ai ), f (bi ) . i∈I

i∈I

i∈I



But f (ai ), f (bi ) is an open interval, hence an open subset, of f (R) = f (−∞), f (+∞) . This, in view of the above equality, implies that f (G) is an open subset of f (R). Therefore, f is an open map, which is what we want.  3. To show f (E) = E, assume first that x ∈ E is arbitrary. We need to show that x ∈ f (E). To see this, note that f (x) ≥ x. If f (x) = x, there is nothing to prove. If f (x) > x, we have f (a) = a ≤ x < f (x). It thus follows from the Intermediate Value Theorem that there exists a c ∈ [a, x) such that f (c) = x, implying x ∈ f [a, b]. By showing c ∈ E, we conclude that x ∈ f (E), which is what we / E, then f (c) < c, from which, as f is increasing, we see that
want. If c ∈ f f (c) < f (c). But f (c) = x, which obtains f (x) < x. This is a contradiction, yielding x ∈ f (E), implying E ⊆ f (E). Next, suppose that y = f (x) ∈ f (E), where x ∈ E, is arbitrary. As x ∈ E, we
have f (x) ≥ x, from which, as f is increasing, we obtain f f (x) ≥ f (x). That is, y ∈ [a, b] and f (y) ≥ y, which means y = f (x) ∈ E. In other words, f (E) ⊆ E, proving the assertion.  4. Remark. The function f , as defined, in not well-defined at 0. Since the value of f at any point has no affect on its integrability, we redefine f : [0, 1] → R by  1 x = pq , p, q ∈ N, gcd(a, b) = 1, q f (x) = 0 otherwise.

98

2. SOLUTIONS

First solution: Just as explained in Solution 1 of 2.5.1, the sequence (fn )+∞ n=1 , with fn : [0, 1] → [0, 1], defined by  1 x = pq , p, q ∈ N, gcd(p, q) = 1, q ≤ n, q fn (x) = 0 otherwise, converges uniformly to f on [0, 1]. As fn (n ∈ N) is zero everywhere but at finitely many points of the interval [0, 1], it follows that fn ’s are integrable on [0, 1]. Hence, f is Riemann integrable on [0, 1], and moreover Z 1 Z 1 fn (x)dx = 0. f (x)dx = lim 0

n

0

Second solution: As shown in Solution 3 of 2.1.1, we have limx→a f (x) = 0 for all a ∈ (0, 1), limx→0+ f (x) = 0, and
limx→1− f (x) = 0. Thus, the function f is continuous on the set [0, 1] \ Q ∩ (0, 1] and it is discontinuous on Q ∩ (0, 1], which is a countable set. From this point on, we present two proofs for the integrability of f on [0, 1]. First proof. Just note that the set of points at which f is discontinuous is countable and hence has measure zero. Therefore, using the Lebesgue’s Integrability Criterion for Riemann integrals, we see that f is Riemann integrable, which is what we want. Second proof. Using Lebesgue’s Number Lemma, we directly prove that f is Riemann integrable on [0, 1]. Recall that Lebesgue’s Number Lemma asserts that if an open cover of a compact metric space X is given, then there exists a number δ > 0, called a Lebesgue number for the cover, such that every subset of X of diameter less than δ is contained in some element of the cover. We use Riemann’s criterion for integrability to show that f is integrable. We need to show that for given ε > 0, there exists a partition P ∈ P[0, 1], the set of all partitions of [0, 1], such that n X (Mi − mi )∆xi < ε, U (P, f ) − L(P, f ) = i=1

where P : 0 = x0 < · · · < xn = 1, Mi = supxi−1 ≤x≤xi f (x), mi = inf xi−1 ≤x≤xi f (x), and ∆xi = xi − xi−1 . To this end, for given ε > 0, choose N ∈ N such that
+∞ 1 ε 1 1 < and let Q ∩ (0, 1] = {r } . Set N = r − , r + ∩ [0, 1]. i i=1 i i i 2 2N 2N +i 2N +i P+∞ 1 ε Obviously, i=1 l(Ni ) ≤ 2N < 2 , where l(Ni ) denotes the length of the interval S+∞ Ni . On the other hand, Q ∩ (0, 1] = {ri }+∞ ⊆ i=1 Ni . Therefore, for any i=1 S+∞ x ∈ C := [0, 1] \ i=1 Ni , the function f is continuous at x. Consequently, for given ε > 0, there exists an open interval Nx containing x such that ε MNx (f ) − mNx (f ) < , 2 where MNx (f ) = supt∈Nx f (t) and mNx (f ) = inf t∈Nx f (t). It is obvious that {Ni }+∞ i=1 ∪ {Nx }x∈C is an open cover for the compact interval [0, 1]. Let δ > 0 be a Lebesgue number for this open cover. Choose P ∈ P[0, 1] such that the length of its subintervals are all less than δ. That is, if P : 0 = x0 < · · · < xn = 1,

2.9. NINTH COMPETITION

99

then ∆xi = xi − xi−1 < δ for all 1 ≤ i ≤ n. Therefore, for all 1 ≤ i ≤ n, we have [xi−1 , xi ] ⊆ Nj for some j ∈ N or [xi−1 , xi ] ⊆ Nx for some x ∈ C. Letting  A = i ∈ N : 1 ≤ i ≤ n, ∃j ∈ N  [xi−1 , xi ] ⊆ Nj ,  B = i ∈ N : 1 ≤ i ≤ n, ∃x ∈ C  [xi−1 , xi ] ⊆ Nx , we can write U (P, f ) − L(P, f )

=

n X

(Mi − mi )∆xi

i=1

≤

X

(Mi − mi )∆xi +

i∈A

≤

X

(Mi − mi )∆xi

i∈B

1 × ∆xi +

i∈A

≤

X

+∞ n X Xε εX ∆xi ≤ l(Ni ) + ∆xi 2 2 i=1 i=1

i∈B

ε 1 + < ε. N 2 2

This means, U (P, f ) − L(P, f ) < ε, which is what we want. To find the value of the integral, it is obvious that L(P, f ) =

n X

mi ∆xi = 0,

i=1

for all P ∈ P[0, 1]. Since f is integrable, we have Z 1 f (x)dx = sup L(P, f ) = 0. 0



P ∈P[a,b]

5. First, we show that no function f : [0, 1] → [0, 1] × [0, 1] can have the three properties, namely, continuity, injectivity, and surjectivity. Suppose to the contrary that f is a continuous and one-to-one function from [0, 1] onto [0, 1] × [0, 1]. As f is invertible and [0, 1] is compact, it follows that f is a closed map, and hence f −1 : [0, 1] × [0, 1] → [0, 1] is a continuous function. Let f ( 12 ) = (a, b) for some a, b ∈ [a, b]. Obviously, f −1 (a, b) = 12 . Letting C = [0, 1] × [0, 1] \ {(a, b)}, we have 1 f −1 (C) = [0, 1] \ { }. 2 Since C is connected and f −1 is continuous, it follows that [0, 1] \ { 21 } is connected, which is a contradiction. Therefore, the assertion follows by contradiction. We now prove that f can have any two properties of the aforementioned properties. There are three cases to consider. (i) f : [0, 1] → [0, 1] × [0, 1] can be continuous and one-to-one. The function f : [0, 1] → [0, 1] × [0, 1] defined by f (t) = (t, 0) is continuous and one-to-one. (ii) f : [0, 1] → [0, 1] × [0, 1] can be continuous and onto.

100

2. SOLUTIONS

Such a function exists. We present the well-known example due to Schoenberg (1938). First, define the function φ on [0, 2] by  0 0 ≤ t ≤ 31 or 53 ≤ t ≤ 2,    1 2 3t − 1 3 ≤ t ≤ 3, φ(t) = 4 2 1  3 ≤ t ≤ 3,   4 5 −3t + 5 3 ≤ t ≤ 3. Then, extend φ to R via φ(t + 2) = φ(t). It is plain that φ : R → R is 2-periodic and that 0 ≤ φ(t) ≤ 1 for all t ∈ R. Now, define f1 , f2 : R → R by f1 (t) =

+∞ +∞ X X φ(32n−2 t) φ(32n−1 t) , f (t) = . 2 n 2 2n n=1 n=1

As 0 ≤ φ ≤ 1 on R, we see that the series defining f1 and f2 converge uniformly and absolutely on R, and hence f1 and f2 are continuous on R and moreover, 0 ≤ f1 , f2 ≤ 1 on R because 0 ≤ φ ≤ 1 on R. Thus, the function f : [0, 1] → [0, 1] × [0, 1] defined by f (t) = (f1 (t), f2 (t)) is well-defined and continuous. To prove that f is onto, let (a, b) ∈ [0, 1] × [0, 1] be arbitrary. We need to show that there exists a c ∈ [0, 1] such that f (c) = (a, b). To this end, write the binary expansions of a, b ∈ [0, 1] as follows +∞ +∞ X X an bn a = , b = , n 2 2n n=1 n=1 where an , bn = 0 or 1 for all n ∈ N. Let +∞ X cn c = 2 , n 3 n=1

where  cn =

a n+1 2 b n2

n odd, n even.

Thus, cn = 0 or 1 for all n ∈ N. Consequently, as 2 0 ≤ c ≤ 1. We claim that φ(3k c) = ck+1 ,

P+∞

1 n=1 3n

= 1, we obtain (∗)

for all k ∈ Z with k ≥ 0. Suppose that (∗) holds. We would then have φ(32n−2 c) = c2n−1 = an , φ(32n−1 c) = c2n = bn , for all n ∈ N. This easily yields f (c) = (f1 (c), f2 (c)) = (a, b), which is what we want. Now, to prove (∗), suppose k ∈ Z with k ≥ 0. We can write 3k c = 2

k +∞ X X cn cn + 2 = 2Nk + dk , n−k n−k 3 3 n=1 n=k+1

Pk

where Nk = n=1 3 period is 2, we have

k−n

cn ∈ N and dk = 2

P+∞

cn n=k+1 3n−k .

φ(3k c) = φ(dk ). If ck+1 = 0, then 0 ≤ dk ≤ 2

+∞ X 1 1 = , n 3 3 n=2

As φ is periodic and its

2.9. NINTH COMPETITION

101

in which case, φ(dk ) = 0, whence φ(3k c) = ck+1 = 0. If ck+1 = 1, then in which case, φ(dk ) = 1, whence φ(3k c) = ck+1 = 1. That is,

2 3

≤ dk ≤ 1,

φ(3k c) = ck+1 , for all k ∈ Z with k ≥ 0, proving the claim and hence finishing the proof. (iii) f : [0, 1] → [0, 1] × [0, 1] can be one-to-one and onto. Such a function exists. First, define f : [0, 1] → [0, 1] × [0, 1] by f (t) = (0.t1 t3 t5 . . . , 0.t2 t4 t6 . . .), where 0.t1 t2 t3 . . ., with ti = 0 or 1 for all i ∈ N, denotes the binary expansion of the number t. We set the convention that in the binary expansion of a number t the number 1 is not allowed to be repeated for all the digits after any digit except when t = 1, in which case its binary expansion, by definition, is set to be 0.111 . . .. It is obvious that f is onto. Next, as mentioned in (i), the function f : [0, 1] → [0, 1] × [0, 1] defined by f (t) = (t, 0) is (continuous and) one-to-one. It thus follows from the Schroder-Bernstein Theorem that there is a function f : [0, 1] → [0, 1] × [0, 1], which is one-to-one and onto, as desired. 

2.9.2. Algebra. 1. First solution: Proceed by contradiction. Suppose that there exists a nonnormal subgroup H of index p in G so that |G| = p|H|, where p is the smallest prime dividing the order of G. It follows that there is a g ∈ G such that K = H g := g −1 Hg = {g −1 hg : h ∈ H} 6= H. It is obvious that |K| = |H|. Also, we have |H ∩ K| < |H| = |K|, for otherwise |H ∩ K| = |H| = |K|, yielding H ∩ K = H = K, which is a contradiction. Now, we |K| |K| claim that |H∩K| ≥ p. To see this, let q be a prime dividing |H∩K| . It follows that q divides |K|, and hence |G|. This obtains p ≤ q ≤ prime dividing |G|. On the other hand, |G| ≥ |HK| = |H|

|K| |H∩K|

because p is the smallest

|K| ≥ |H|p = |G|, |H ∩ K|

implying that HK = G. Now, let g = hk for some h ∈ H, k ∈ K. We can write H g = H hk = (H h )k = H k . Thus, G = Gk = (HK)k = H k K k = H g K = KK = K, which is a contradiction. So, the assertion follows by way of contradiction. Second solution: Let H be a subgroup of G of index p in G, where p is the smallest prime dividing the order of G. First, we claim that if x ∈ / H, then xi ∈ /H for all i ∈ N with 1 ≤ i ≤ p − 1. By way of contradiction, suppose that there exists the smallest j ∈ N with 1 < j ≤ p − 1 such that xj ∈ H. It follows that xi ∈ / H for all 1 ≤ i < j. Set n = |G| and m = ord(x). We have m|n. As j < p and p is the smallest prime that divides n, it follows that m is not divisible by j. Hence,

102

2. SOLUTIONS

there are numbers r, q ∈ N with 0 < r < j such that m = jq + r. Noting that m = ord(x), we can write e = xm = xjq+r = (xj )q xr , yielding (xj )q xr ∈ H. But xj ∈ H, from which, we obtain xr ∈ H, a contradiction because 0 < r < j. Consequently, xi ∈ / H for all 1 ≤ i ≤ p − 1, as we claimed. Now, to prove the assertion, we proceed by contradiction again. Suppose that there exist g ∈ G and h ∈ H such that ghg −1 ∈ / H. This yields g ∈ / H, which, in turn together with the above claim, implies that g i ∈ / H for all i ∈ N with 1 ≤ i ≤ p − 1. On the other hand, g i H 6= g j H whenever 1 ≤ i < j ≤ p − 1 because otherwise g j = g i h0 for h0 ∈ H, implying g j−i = h0 ∈ H, which is a contradiction, for 1 ≤ j − i ≤ p − 1. Therefore, H, gH, . . . , g p−1 H are p distinct left cosets of H. Consequently,  G = H, gH, . . . , g p−1 H . H Let g1 = ghg −1 . Likewise, as g1 ∈ / H, we conclude that H, g1 H, . . . , g1p−1 H are p distinct left cosets of H, implying  G = H, g1 H, . . . , g1p−1 H . H Therefore, gH = g1r H for some 1 ≤ r ≤ p − 1, and hence g = g1r h1 for some h1 ∈ H. On the other hand, g1r = ghr g −1 , from which, we obtain g = ghr g −1 h1 . This yields g = h1 hr , implying g ∈ H, which is a contradiction. Therefore, H is a normal subgroup of G, finishing the proof. Third solution: Let H be a subgroup of G of index p in G, where p is the smallest prime dividing the order of G. Use L to denote the set of all left cosets of H in G, which has p elements by the hypothesis. For each g ∈ G define the mapping τg : L −→ L by τg (xH) = gxH. In other words, G acts on L by multiplication from the left. It is readily verified that τg ’s (g ∈ G) give rise to a group homomorphism φ : G −→ S(L) ∼ = Sp which is defined by φ(g) = τg , where S(L) denotes the group of all permutations of L which is isomorphic to Sp because |L| = p (here, as is usual, Sp denotes the symmetric group of degree p, whose order is p!). Obviously, K := ker φ ⊆ H. We prove the assertion by showing that H = K. By the First G is isomorphic to a subgroup of Sp implying Isomorphism Theorem for groups, K G G that | K | p!. On the other hand, every divisor of | K | divides |G|. Also, of the prime divisors of p! only p divides |G| because p is the smallest prime dividing the order G G G G of G. Thus, | K | = p or | K | = 1. The latter is impossible because | K | ≥ |H | = p. G G Therefore, | K | = p = | H |, implying that |H| = |K|. This obtains H = K because K ⊆ H, completing the proof. Fourth solution: By a standard result from Galois theory, e.g., see Proposition V.2.16 of “Algebra” by T.W. Hungerford, there are fields E, K such that K is an extension of E and that Gal(K/E) ∼ = G. In view of this, and two other standard results from Galois theory, e.g., see Theorem V.2.5 and Corollary V.3.15 of “Algebra” by T.W. Hungerford, it suffices to prove the following. Under the hypothesis of the problem, let K be an extension field of E such that Gal(K/E) ∼ = G. If F ⊆ K is an extension field of E such that [F : E] = p, then F is normal over E.

2.9. NINTH COMPETITION

103

To prove this, note first that every element of K is separable over E, i.e., the minimal polynomial of every element of K over E splits into distinct linear factors, for K is Galois over E. Therefore, F is a finite separable extension of E and hence, by the Primitive Element Theorem (see Proposition V.6.15 of “Algebra” by T.W. Hungerford, there is an α ∈ F such that F = E(α). Let g ∈ E[x] be an arbitrary irreducible polynomial that has a root, say h(α) for some h ∈ E[x], in F = E(α). It follows that the splitting field of g is contained in K, for h(α) ∈
F ⊆ K and K is Galois, and hence normal, over E. Note that deg(g) = [E h(α) : E] ≤ [E(α) : E] = p. We need to show that g splits into linear factors in E(α). Let β 6= h(α) be a root of g in the splitting field of g, which, as we just saw, is contained in K. It suffices to prove that β ∈ F = E(α). To this end, we note that [F (β) : F ] < deg(g) ≤ p
because g(β) = 0, g h(α) = 0, and h(α) ∈ F . This yields [F (β) : F ] = 1 because p is the smallest prime dividing |G| and [F (β) : F ] |G|, for β ∈ K and [K : E] = |G| since Gal(K/E) ∼ = G. Consequently, β ∈ F , and hence F is normal over E, as desired.  Z3 [x] Z3 [x] onto the field 2 . (x2 + 1) (x + x + 2) [x] 3 [x] Since f : hxZ23+1i → hx2Z+x+2i is an isomorphism of fields, we see that f (a+hx2 +1i) = a + hx2 + x + 2i for all a ∈ Z3 . Thus, to determine f , we only need to determine f (x + hx2 + 1i). To this end, suppose that f (x + hx2 + 1i) = ax + b + hx2 + x + 2i for some a, b ∈ Z3 . We must have 
2

f x + hx2 + 1i = f x2 + hx2 + 1i = f − 1 + hx2 + 1i ,

2. We determine all isomorphisms from the field

implying that (ax + b)2 + 1 ∈ hx2 + x + 2i. But the polynomial (ax + b)2 + 1 = a2 x2 + 2abx + b2 + 1 is divisible by x2 + x + 2 in Z3 [x] if and only if so is the polynomial (2ab − a2 )x + (a2 + b2 + 1), which is obtained from a2 x2 + 2abx + b2 + 1 by letting x2 = −x − 2 = −x + 1, if and only if a and b satisfy the equations a(2b − a) = 0 and a2 + b2 + 1 = 0. It is readily seen that a = 2, b = 1 and a = 1, b = 2 are the only solutions of the above equations in Z3 . Therefore, an [x] 3 [x] isomorphism f : hxZ23+1i → hx2Z+x+2i must be given by
f ax + b + hx2 + 1i = a(2x + 1) + b + hx2 + x + 2i, or by
f ax + b + hx2 + 1i = a(x + 2) + b + hx2 + x + 2i. That is, either
f ax + b + hx2 + 1i = 2ax + (a + b) + hx2 + x + 2i, or
f ax + b + hx2 + 1i = ax + (2a + b) + hx2 + x + 2i, which are both easily seen to be monomorphisms of rings. It follows that such an f Z3 [x] Z3 [x] is an isomorphism from the field 2 onto the field 2 because they (x + 1) (x + x + 2) both have 9 elements.  3. Note that the ring Z7 [x] is a Euclidean ring because Z7 is a field. With that in mind, we can use the Euclidean algorithm to find the greatest common divisor of

104

2. SOLUTIONS

the two polynomials 4x4 − 2x2 + 1 and −3x3 + 4x2 + x + 1 in Z7 [x]. We have 4x4 − 2x2 + 1 3

2

−3x + 4x + x + 1

=

(−3x3 + 4x2 + x + 1)(x − 1) + x2 + 2,

=

(x2 + 2)(−3x + 4).

Therefore, gcd(4x4 − 2x2 + 1, −3x3 + 4x2 + x + 1) = x2 + 2, which is what we want.



4. Let {1, y1 }, where y1 ∈ K, be a basis for the vector space K over F . If y12 ∈ F , there is nothing to prove. Suppose y12 ∈ / F . As y12 ∈ K, it follows that there exist 2 scalars a, b ∈ F with b 6= 0 such that y1 = a + by1 . We have y12 − by1 = b. On the other hand, since ch(F ) 6= 2, we have 2−1 =

1 2

∈ F , and hence we can write

b b y12 − by1 = (y1 − )2 − ( )2 , 2 2 whence b b (y1 − )2 = a + ( )2 ∈ F. 2 2 Let y = y1 − 2b . It is easily verified that {1, y} is a basis for K over F and y 2 = a + ( 2b )2 ∈ F , which is what we want.  5. Let S = T − I. Obviously, W = ker S. On the other hand, by the Rank-Nullity Theorem, we have dim ker S + dim SV = dim V = n. We claim that SV ⊆ ker S. To see this, let y = Sx ∈ SV , where x ∈ V , be arbitrary. We have Sy = S 2 x = (T − I)2 x = (T 2 − 2T + I)x = (2I − 2T )x = 0, because T 2 = I and ch(F ) = 2. This means, y ∈ ker S, proving the claim. Now, since SV ⊆ ker S, we can write n = dim ker S + dim SV ≤ 2 dim ker S, yielding dim W = dim ker S ≥

2.9.3. General. 1. Let      A =     

n 2,

r λ .. . .. . λ λ

which is what we want.

λ r .. . .. . λ λ

λ λ .. .

··· ··· .. .

λ λ .. .

..

..

..

. ··· ···

. λ ···

. r λ

λ λ .. . .. . λ r



     .    

2.9. NINTH COMPETITION

105

Subtracting the first row from all the other rows and then adding the jth column (2 ≤ j ≤ n) to the first column, we obtain an upper triangular matrix whose determinant is the product of the entries on its main diagonal. Thus, we can write   r λ λ ··· λ λ  λ−r r−λ 0 ··· 0 0     λ−r 0 r−λ 0 ··· 0    det A = det   .. .. .. .. .. ..   . . . . . .    λ−r 0 ··· 0 r−λ 0  λ−r 0 ··· ··· 0 r−λ   r + (n − 1)λ λ λ ··· λ λ  0 r−λ 0 ··· 0 0     0 0 r−λ 0 ··· 0    = det   .. .. .. .. .. ..   . . . . . .    0 0 ··· 0 r−λ 0  0 0 ··· ··· 0 r−λ

n−1 = r + (n − 1)λ r − λ .

n−1 That is, det A = r + (n − 1)λ r − λ , which is what we want.  2. Let x denote the number of chickens, a the amount of food a chicken consumes per day, and t the number of the days before the farmers run out of chicken food. The amount of food in the chicken farm is equal to (t + 20)(x − 75)a = (t − 15)(x + 100)a = txa. Simplifying, we obtain   (t + 20)(x − 75) = (t − 15)(x + 100) x = 5t. =⇒ (t − 15)(x + 100) = tx 20t − 3x − 300 = 0. Substituting x = 5t into the second equation yields x = 300, which implies t = 60. Therefore, there are x = 300 chickens in the farm.  3. To prove the assertion by contradiction, suppose that such a function f exists. We can write Z 1 Z 1 Z 1 Z 1 2 2 2 (x − α) f (x)dx = x f (x)dx − 2α xf (x)dx + α f (x)dx 0

0

0

0

= α2 − 2α2 + α2 = 0. R1 That is, 0 (x − α)2 f (x) = 0. As (x − α)2 f (x) is continuous and nonnegative on [0, 1], it follows that (x − α)2 f (x) = 0 for all x ∈ [0, 1]. This implies f = 0 on [0, 1] except possibly at x = α. Now, from the continuity of f , we see that f = 0 on [0, 1], a contradiction. Therefore, no such function f exists, finishing the proof.  4. First solution: Consider the points with integer coordinates in the closed interval [0, n] of the real line. To any solution (x1 , . . . , xm ) of the equation x1 + x2 + · · · + xm = n in N, there corresponds m − 1 points p1 < · · · < pm−1 from the set {i}n−1 i=1 as follows p1 = x1 , p2 = x1 + x2 , . . . , pm−1 = x1 + · · · + xm−1 .

106

2. SOLUTIONS

Conversely, for any m − 1 points p1 < · · · < pm−1 from the set {i}n−1 i=1 , a solution (x1 , . . . , xm ) of the equation x1 + x2 + · · · + xm = n is determined as follows x1 = p1 , x2 = p2 − p1 , . . . , xm−1 = pm−1 − pm−2 , xm = n − pm−1 . Therefore, the number of the solutions of the equation x1 + x2 + · · · + xm = n in N is equal to the number of ways of choosing m − 1
points {pi }m−1 i=1 from n − 1 points n−1 {i}n−1 , which, as is well-known, is equal to . i=1 m−1 Second solution: We only briefly elaborate on this solution. It is not difficult to see that the number of the solutions of the equation x1 + x2 + · · · + xm = n in N is equal to the coefficient of xn in the expansion of (x + x2 + · · · + xn−m )m or in that of (x + x2 + x3 + · · · )m = ( which is easily seen to be

n−1 m−1




 +∞  X x m m+n−1 n ) = xm x , 1−x m−1 n=0

.



5. Yes, just cut the paper along the heavy lines.

 6. Since the time needed to get to work from home obeys the uniform distribution, it follows that the density probability function is equal to  x < 10  0 1 10 ≤ x ≤ 20 . f (x) =  10 0 x > 20 (a) As the worker needs 15 minutes to get to work on time, the desired probability is Z 20 dx 1 P (x > 15) = = . 10 2 15

2.9. NINTH COMPETITION

107

(b) First solution. Noting that it takes the worker at least 10 minutes to get to work, let x0 > 10 denote the time needed for the worker to get to work on time and with the probability of 75% to have time to eat breakfast. We must have

Z

min(x0 ,20)

P (10 < x < x0 ) = 10

dx = 0.75 ⇐⇒ 10 ⇐⇒

min(x0 , 20) = 17.5 x0 = 17.5.

Therefore, the latest time that this worker can leave home to get to work on time and with the probability of 75% to have time to eat breakfast is equal to 7, 450 − 17.50 = 7, 27.50 = 7, 270 , 3000 . Second solution. Suppose the worker heads off for work m minutes before 8 o’clock in the morning. In order for the worker to get to work on time and have time to eat breakfast, we must have m − 15 ≥ 10, implying m ≥ 25. Assuming that it takes the worker x minutes to get to work, the worker has m − x minutes to have breakfast. Thus, to make sure that the m − x minutes, which is the remaining time before the start hour of the factory, is enough for the worker to have breakfast with the probability of 75%, we must have

P (m − x ≥ 15) = 0.75 ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒

P (x ≤ m − 15) = 0.75 Z min(m−15,20) dx = 0.75 10 10 min(m − 15, 20) = 17.5 m = 32.5.

Therefore, if the worker heads off for work 32.5 minutes before 8 a.m., i.e., at 8, 000 − 32.50 = 7, 27.50 = 7, 270 , 3000 , s/he will get to work on time and with the probability of 75% have time to eat breakfast. 

7. First solution: We solve the problem on a general field F . Let n ∈ N, A = (ci cj ) ∈ Mn (F ), and In the identity matrix of size n. To evaluate det(In + A), we may, without loss of generality, assume that ci 6= 0 for all 1 ≤ i ≤ n because if ci0 = 0 for some i0 ∈ {1, . . . , n}, then det(In + A) = det(In−1 + A0 ) where A0 = (c0i c0j ) ∈ Mn−1 (F ) with c0i ∈ {c1 , . . . , cn } \ {ci0 }, which has the same general form of A = (ci cj ) ∈ Mn (F ). Use the basic properties of the determinant function and perform the following operations: factor out ci of row i and of column i for each i = 1, . . . , n; then subtract the first row from all the other rows; then multiply column i by c2i (1 ≤ i ≤ n); and finally add row i to row 1 for each i = 2, . . . , n, to obtain an upper triangular whose determinant is equal to that of In + A. We can

108

2. SOLUTIONS

write 

det(In + A)

   2 2 = c1 · · · cn det     

1 + c12 1 1 .. . 1

1 1 + c12 2 .. . 1

··· ··· .. . ···

1

1

···

1 + c12 1 − c12 1 .. . − c12

    = c21 · · · c2n det    

.. . 0

··· ··· .. . ···

0

···

1 1 c22

=

   det    

=

=

   det   

1 + c21 −1 .. .

c22 1 .. .

··· ··· .. .

c2n−1 0 .. .

−1 −1

0 0

··· ···

1 0

1 0 .. . 0

c2n 0 .. .

1 c2n

0 

1 1 .. . 1

1 c2n−1

1

1 c2n−1

1



1+ 1 0 .. .

1

− c12

1 1 .. .

1+



1 c2n

      

        

     0  1

1 + c21 + · · · + c2n 0 .. .

c22 1 .. .

··· ··· .. .

c2n−1 0 .. .

0 0

0 0

··· ···

1 0

c2n 0 .. .



     0  1

(1 + c21 + · · · + c2n ) × 1 × · · · × 1 = 1 + c21 + · · · + c2n .

Therefore, det(In + A) = 1 + c21 + · · · + c2n , which is what we want. Second solution: Recall that if A ∈ Mn (F ), λi ’s (1 ≤ i ≤ n) are the eigenvalues of A in the algebraic closure of F , and f (x) = det(xIn − A) is the characteristic polynomial of A so that f (x) = xn + fn−1 xn−1 + · · · + f0 , Pn Pn then, fn−1 = − i=1 λi , f0 = (−1)n λ1 · · · λn , i=1 λi = tr(A), and λ1 · · · λn =
det(A). Also, if α ∈ F and g(x) = det xIn − (A + αIn ) is the characteristic polynomial of A + αIn , then
g(x) = det (x − α)In − A = f (x − α). With all that in mind, note first that the rank of A is 1 because row i is ci (c1 , . . . , cn ) which is a multiple of the fixed vector (c1 , . . . , cn ). It follows from the Rank-Nullity Theorem that dim ker A = n − 1, and hence 0 is an eigenvalue of A of multiplicity n − 1. Let λ be the only other eigenvalue of A. We must have f (x) = det(xIn − A) =

n Y i=1

(x − λi ) = xn−1 (x − λ) = xn − λxn−1 .

2.10. TENTH COMPETITION

109

On the other hand, n X

λi = λ = tr(A) =

i=1

implying λ = have

n X

c2i ,

i=1

Pn

2 i=1 ci . If g denotes the characteristic polynomial of In + A, we

g(x)

= f (x − 1) (x − 1)n − λ(x − 1)n−1

=

= xn + gn−1 xn−1 + · · · + g1 x + (−1)n (1 + λ). Recall that (−1)n (1 + λ) = (−1)n det(In + A), whence det(In + A) = 1 + λ = 1 + tr(A) = 1 +

n X

c2i ,

i=1

which is what we want.

 2.10. Tenth Competition

2.10.1. Analysis. 1. From (b), we see that f 0 (x) > 0 for all x ≥ 1, implying that f is strictly increasing on [1, +∞) and hence f (t) > f (1) = 1 for all t > 1. From this, we obtain 1 1 < , f 0 (t) = 2 t + (f (t))2 1 + t2 R +∞ 0 R +∞ dt π 0 for all t > 1. It thus follows that 1 f (t)dt < 1 1+t2 = 4 because f is 1 0 continuous and f (t) < 1+t2 on [1, +∞). Using the Second Fundamental Theorem of Calculus , we can write Z x Z x Z +∞ dt dt π 0 f (x) = 1 + f (t)dt < 1 + <1+ =1+ . 2 2 1+t 4 1 1 1+t 1 Therefore, f is bounded from the above. This together with the hypothesis that f is increasing implies that limx→+∞ f (x) exists and that Z +∞ Z +∞ 1 π lim f (x) = 1 + f 0 (t)dt < 1 + =1+ , 2 x→+∞ 1 + t 4 1 1 implying limx→+∞ f (x) < 1 + π4 , which is what we want.



2. For an x ∈ (a, b), define g : [a, b] → R by g(t) = (t − a)(t − b)f (x) − (x − a)(x − b)f (t). Plainly, g(a) = g(x) = g(b) = 0. From this, applying Rolle’s Theorem twice to g and g 0 , respectively, we obtain a c ∈ (a, b) such that g 00 (c) = 0. We can write g 00 (c) = 2f (x) − (x − a)(x − b)f 00 (c) = 0, which implies f (x) ≤ M (x − a)(b − x) , 2

110

2. SOLUTIONS

for all x ∈ [a, b]. Taking integrals of both sides of the above yields Z b Z b
(b − a)3 f (x) dx ≤ M − x2 + (a + b)x − ab dx = M . 2 a 12 a Rb 3 That is, a f (x) dx ≤ M (b−a) 12 , which is what we want.



3. First solution: We can write Z 3 2 Z 3 2 x (1 − x)xn x |1 − x|xn ≤ dx dx 2n 1+x 1 + x2n 0 0 Z 1 2 Z 3 2 x |1 − x|xn x |1 − x|xn = dx + dx 1 + x2n 1 + x2n 1 0 Z 1 Z 3 n x ≤ xn dx + 18 dx 2n x 0 1 1 18 18 = + − . n + 1 n − 1 3n−1 (n − 1) It thus follows from the Squeeze Lemma that Z 3 2 x (1 − x)xn dx = 0, lim n→+∞ 0 1 + x2n which is what we want. Second solution: In view of a standard theorem from classical analysis, it suffices 2 (1−x)xn
+∞ to show that the sequence x 1+x uniformly converges to the zero function 2n n=1 on [0, 3]. Note first that 2 x (1 − x)xn (∗) 1 + x2n ≤ |1 − x|, for all x ∈ [0, 3]. To see this, we have xn+2 ≤ 1 on [0, 1], implying xn+2 ≤ 1 + x2n on [0, 1]. Also if n ≥ 2, then xn+2 ≤ x2n on [1, 3], which implies xn+2 ≤ 1 + x2n on [1, 3]. Thus, xn+2 ≤ 1 + x2n on [0, 3] of which the above inequality is a quick consequence. Now, let 0 < ε < 1 be given. It follows from (∗) that 2 x (1 − x)xn 1 + x2n < ε, for all n ≥ 2 and x ∈ (1 − ε, 1 + ε). On the interval [0, 1 − ε], for all n > 2 we can write 2 x (1 − x)xn xn+2 n+2 ≤ (1 − ε)n+2 . 1 + x2n ≤ 1 + x2n ≤ x As 0 < 1 − ε < 1, we have limn (1 − ε)n+2 = 0. Hence, there exists an N1 > 1 such that (1 − ε)n+2 < ε for all n ≥ N1 . On the interval [1 + ε, 3], we have 2 x (1 − x)xn 2x2 2 2 2xn+2 1 + x2n < 1 + x2n = x−n + xn < xn−2 ≤ (1 + ε)n−2 , 2 2 for all n ≥ 2. As limn (1+ε) n−2 = 0, there exists an N2 > 1 such that (1+ε)n−2 < ε for all n ≥ N2 . Letting N = max(N1 , N2 ), for all x ∈ [0, 3] and n ≥ N we have 2 x (1 − x)xn 1 + x2n < ε.

2.10. TENTH COMPETITION

111

2 (1−x)xn
+∞ That is, the sequence x 1+x uniformly converges to the zero function on 2n n=1 [0, 3], yielding Z 3 2 Z 3 x (1 − x)xn x2 (1 − x)xn lim dx = lim dx = 0, 2n n→+∞ 0 1+x 1 + x2n 0 n→+∞ which is what we want. 

2.10.2. Algebra. 1. We prove the following lemma of which the assertion is a quick consequence. Lemma. Let V be a left (resp. right) vector space over a division ring D whose characteristic is zero. Then the additive group of V has no maximal subgroup. Proof. To prove the assertion by contradiction, suppose that M is a maximal subgroup of the additive group of V . As M ≤ V , V is abelian, and M is maximal, it V V is a simple abelian group. Consequently, M is a finite cyclic group, follows that M V implying that there exists a prime number p such that M = Zp , and hence M is of finite index in V . By proving that (V, +) has no proper subgroup of finite index, we obtain a contradiction, proving the assertion. Suppose to the contrary that there exists a proper subgroup H of the additive group of V such that [V : H] = n ∈ N. V As | H | = n, it follows from Lagrange’s Theorem that nx ∈ H for all x ∈ V . On the other hand, x = n( nx ) and nx ∈ V for all x ∈ V , implying x ∈ H for all x ∈ V . This yields V = H, a contradiction. Thus, V has no proper subgroup of finite index, which is what we want, finishing the proof.  To prove the assertion, just let V = D = R in the lemma.



2. Since the ring R is unital, it follows from (∗) that there exist elements a, a0 , a00 ∈ A, b ∈ B, c ∈ C, and d ∈ D such that a + b = 1, a0 + c = 1, a00 + d = 1. Multiplying the above equalities and simplifying, we see that there is a u ∈ A such that 1 = (a + b)(a0 + c)(a00 + d) = u + bcd, implying 1 = u + bcd ∈ A + M because bcd ∈ M . This obviously yields A + M = R, which is what we want. Let R = Z, A = 2Z, B = 3Z, C = 5Z, and D = 7Z. It is easily verified that M = B ∩ C ∩ D = 105Z and that R and its ideals A, B, C, D satisfy (∗).  3. We solve the problem under the weaker hypothesis that gcd(n, c) = 1, where c = lcm(1, 2, . . . , m − 1). It is obvious that F ⊆ F (αm ) ⊆ F (α). So we can write [F (α) : F ] = [F (α) : F (αm )][F (αm ) : F ].

(∗)

As [F (α) : F ] = n, it follows that there exists a polynomial P ∈ F [x] of degree n such that P (α) = p0 + p1 α + · · · + pn αn = 0. (∗0 ) Using the division algorithm, dividing k by m for each k = 0, . . . , n, and collecting terms appropriately in (∗0 ), we see that there exists a polynomial Q ∈ F (αm )[x] with deg(Q) ≤ m − 1 such that P (α) = Q(α) = q0 + q1 α + · · · + qm−1 αm−1 = 0.

112

2. SOLUTIONS

This clearly shows that [F (α) : F (αm )] ≤ m − 1, implying that [F (α) : F (αm )] divides lcm(1, 2, . . . , m − 1) = c. But as, in view of (∗), [F (α) : F (αm )] divides n and gcd(n, c) = 1, we conclude that [F (α) : F (αm )] = 1, whence F (α) = F (αm ), yielding [F (αm ) : F ] = [F (α) : F ]. This, in turn, implies F (α) = F (αm ), which is what we want.  4. Recall that the trace of any nilpotent matrix N ∈ Mn (F ) is zero. With that in mind, suppose by way of contradiction that there exist nilpotent matrices N1 , . . . , Nk ∈ Mn (F ) (k ∈ N) that span Mn (F ). In particular, for the matrix E11 ∈ Mn (F ), where E11 denote the matrix with 1 in the 11 place and zero elsewhere, there are scalars c1 , . . . , ck ∈ F such that E11 = c1 N1 + · · · + ck Nk , which implies 1 = tr(E11 ) = c1 tr(N1 ) + · · · + ck tr(Nk ) = 0, a contradiction, proving the assertion.



2.10.3. General. 1. As the tests are independent, we can write
P X = 0 = P (F S) + P (SSF S) + P (F F F S) + P (SSSSF S) +P (SSF F F S) + P (F F SSF S) + P (F F F F F S) + · · · = qp + p2 qp + q 2 qp + p4 qp + p2 q 2 qp + p2 p2 qp + q 4 qp + · · ·
= qp 1 + (p2 + q 2 ) + (p2 + q 2 )2 + · · · =

qp = 1 − p2 − q 2


That is, P X = 0 =

1 2.

1 4 1 2

=

1 . 2

Similarly, one can show that P X = 1




=

1 2.



2. We prove the assertion for the more general case where A is a subset of a topological space. With that in mind, let A be a closed subset of a topological space X such that A◦ = ∅. We prove the assertion by showing that A = ∂A = ∂Ac , where ∂A denotes the boundary of A. Recall that by definition ∂A = A ∩ Ac . So, we can write ∂Ac = Ac ∩ Acc = Ac ∩ A = ∂A. That is, ∂A = ∂Ac . On the other hand, ∂A = A \ A◦ = A \ ∅ = A = A. So we have A = ∂A = ∂Ac , where Ac is an open set because A is a closed set, which is what we want.  3. Let x(t) be a solution of (∗). We have x00 = f (t, x), x(0) = x0 , x0 (0) = y0 .

2.10. TENTH COMPETITION

113

As f is continuous, using the First Fundamental Theorem of Calculus and changing the order of integration, we can write 0

Z

0

x (t) = x (0) +

t


f s, x(s) ds

0

Z t Z

τ


f s, x(s) ds dτ =⇒ x(t) = x(0) + y0 t + 0 0 Z t Z t 
dτ f s, x(s) ds =⇒ x(t) = x0 + y0 t + s

0

Z =⇒ x(t) = x0 + y0 t +

t


(t − s)f s, x(s) ds,

0

which means x(t) is a solution of (∗∗). Now, let x(t) be a solution of (∗∗). We can write t

Z


(t − s)f s, x(s) ds,

x(t) = x0 + y0 t + 0

implying that Z x(t) = x0 + y0 t + t

t


f s, x(s) ds −

Z

0

t


sf s, x(s) ds.

0

Again, using the First Fundamental Theorem of Calculus, we can write 0

t

Z




f s, x(s) ds + tf t, x(t) − tf t, x(t) ,

x (t) = y0 + 0

which obtains 0

Z

x (t) = y0 +

t


f s, x(s) ds.

0

Taking derivative of both sides of the above yields
x00 (t) = f t, x(t) . It is obvious that x(0) = x0 and x0 (0) = y0 . Therefore, x(t) satisfies (∗), which is what we want. 

4. First Solution: See Solution 1 of 2.5.3.

Second solution: Just as in the first solution, we settle the problem by proving the polynomial identity (1 + x)n =

n X k=0

Cnk xk ,

(∗)

114

2. SOLUTIONS

n! . Proceed by induction on n. If n = 1, the assertion is easy. k!(n − k)! Suppose that (∗) holds for n. To prove that (∗) holds for n + 1, we can write

where Cnk =

(1 + x)n+1

=

(1 + x)n (1 + x) = (1 + x)

n X

Cnk xk

k=0

=

n X

Cnk xk +

k=0

= =

1+ 1+

n X

Cnk xk+1

k=0 n X k=1 n X

Cnk xk

+

n X

Cnk−1 xk + xn+1

k=1

(Cnk + Cnk−1 )xk + xn+1 .

k=1 k , from which, we easily A straightforward calculation shows that Cnk + Cnk−1 = Cn+1 obtain n X k (1 + x)n+1 = Cn+1 xk , k=0

proving the induction assertion, which is what we want. Third Solution: For n, k ∈ N with k ≤ n, use Pnk to denote n(n−1) · · · (n−k +1). n! . Thus, We have Pnk = k!Cnk for each n, k ∈ N with k ≤ n, where Cnk = k!(n − k)! to prove the assertion, we need to show that Pnk is divisible by k! for all n, k ∈ N with k ≤ n. We proceed by induction on n. If n = 1, there is nothing to prove. k is Suppose that Pnk is divisible by k! for all k ≤ n. We need to show that Pn+1 k divisible by k! for all k ≤ n + 1. If k = n + 1, then Pn+1 = k!, in which case, the assertion is trivial. If k ≤ n, we can write k Pn+1

= (n + 1)n(n − 1) · · · (n + 1 − k + 1) = (n − k + 1 + k)n(n − 1) · · · (n − k + 2) = kPnk−1 + Pnk .

By the induction hypothesis, Pnk−1 and Pnk are divisible by (k − 1)! and k!, respeck = kPnk−1 + Pnk is divisible by k!, which is what we want. tively. Thus, Pn+1 Fourth solution: As we know, it suffices to prove that Cnk is an integer. To this end, we need, first of all, the following lemma, which is due to Legendre (1808). Lemma. Let p a prime number and n ∈ N. Then, the largest power of p dividing n! is equal to p

P h i +∞ i=1

n pi

,

where [.] denotes the integer part function. Proof. As n! = 1 × 2 × · · · × n, without loss of generality, we may assume that p ≤ n. Among the consecutive integers 1, . . . , n those that are divisible by p are p, 2p, . . . , m1 p,

2.11. ELEVENTH COMPETITION

where m1 =

h i n p

. So, there are

h i n p

115

such integers. Likewise, there are i

h i n pi

consec-

i

utive integers from 1 up to n that are divisible by p (note that if p > n, then the number of integers between and including 1 and n that are divisible by pi is zero). Note that any multiple of pk which is not a multiple of pk+1 contributes as much as k to the largest power of p that divides n!. Therefore, the largest power of p that divides n! is equal to p

P h i +∞ i=1

n pi

,

proving the lemma.



Now to prove the assertion, note first that [a + b] ≥ [a] + [b], for all a, b ∈ R. Thus, we can write         n−k+k n−k k n = ≥ + i , pi pi pi p for all i ∈ N. Adding up these inequalities, we obtain   X  +∞  +∞  +∞  X X n n−k k ≥ + . pi pi pi i=1 i=1 i=1 This means that the largest power of any prime p that divides n! is divisible by the largest power of p that divides k!(n − k)!, proving the assertion. 

2.11. Eleventh Competition 2.11.1. Analysis. 1. To prove the assertion by contradiction, suppose that the set of zeros of f , i.e.,  Z := x ∈ [0, 1]|f (x) = 0 = f −1 ({0}) is infinite. As the interval [0, 1] is compact, it follows that there exists a sequence (xn )+∞ n=1 of distinct points in Z such that limn xn = x0 for some x0 ∈ [0, 1]. This implies f (x0 ) = 0 because f is continuous. But the function f is differentiable at x0 . So we can write f 0 (x0 ) = lim n

f (xn ) − f (x0 ) 0 = lim = 0. n xn − x0 xn − x0

That is, f 0 (x0 ) = 0. On the other hand, f (x0 ) = 0, implying that x0 is a common zero of f and f 0 , which is in contradiction with the hypothesis of the problem. Thus, Z is finite, which is what we want.  2. (a) To show that the sequence (fn )+∞ convergent to a function f on n=1 is pointwise
+∞ 1 e [1, e ], it suffices to prove that the sequence fn (a) n=1 is increasing and bounded 1

1

from the above for all a ∈ [1, e e ]. To this end, letting a ∈ [1, e e ] be arbitrary, we prove these assertions by induction on n. We have f1 (a) = a, fn+1 (a) = afn (a) ∀n ∈ N.

116

2. SOLUTIONS

First, by induction on n, we show that 1 ≤ fn (a) < e for all n ∈ N. If n = 1, then 1 1 ≤ f1 (a) = a ≤ e e < e. Assuming that 1 ≤ fn (a) < e, we can write 1

1 ≤ fn+1 (a) = afn (a) < (e e )e = e. This proves 1 ≤ fn (a) < e for all n ∈ N by induction. We note that the function g : R → R defined by g(x) = ax is increasing whenever a > 1. To prove that the
+∞ sequence fn (a) n=1 is increasing, again we use induction on n. If n = 1, we can write f2 (a) = g(a) > g(1) = f1 (a). Now assuming that fn+1 (a) > fn (a), we have

fn+2 (a) = g fn+1 (a) > g fn (a) = fn+1 (a).
+∞ This proves that the sequence fn (a) n=1 is increasing. Thus, the sequence (fn )+∞ n=1 1

is pointwise convergent to a function f : [1, e e ] → R. In view of the continuity of g, we can write


f (a) = lim fn+1 (a) = lim g fn (a) = g lim fn (a) = g f (a) = af (a) , n

n

f (a)

implying that f (a) = a

n

1

and 1 ≤ f (a) ≤ e for all a ∈ [1, e e ], as desired. 1

ln x

(b) Define the function g : [1, e] → [1, e e ] by g(x) = e x . By inspecting the 1 derivative of g, one can easily verify that the function g : [1, e] → [1, e e ] is one-to1 one and onto. Moreover, g is continuous and so is its inverse g −1 : [1, e e ] → [1, e]; in fact g and, hence its inverse, g −1 are differentiable. From f (x) = xf (x) , we 1 easily obtain g(f (x)) = x for all x ∈ [1, e e ]. This shows that f = g −1 . Thus, f is one-to-one and continuous because so is g −1 . Now, the functional sequence (fn )+∞ n=1 of continuous functions converges point1 wise to the continuous function f on the compact interval [1, e e ] and moreover 1 fn (x) < fn+1 (x) for all x ∈ [1, e e ]. It thus follows from Dini’s Theorem that the 1 e sequence (fn )+∞ n=1 converges uniformly to the function f on [1, e ], which is what we want.  3. Set  M = sup |φ00n (x)| : n ∈ N, x ∈ [−1, 1] . It follows from the Mean Value Theorem that there exists a c between 0 and x such that φ0n (x) − φ0n (0) = xφ00n (c), from which, we obtain 0 φn (x) = 1 + xφ00n (c) ≤ M + 1, for all n ∈ N and x ∈ [−1, 1]. Using integration by parts, we get Z 1 Z 1 0 φn (t) sin nπt dt ≤ 1 φ (t) cos nπtdt n nπ −1 −1 Z M + 1 1 M +1 ≤ sin nπt dt = , nπ nπ 2 −1

2.11. ELEVENTH COMPETITION

117

implying that Z M +1 1 1 φn (t) cos nπtdt ≤ 2 2 , an = n −1 n π for all n ∈ N. This, in view of the Limit Comparison Test for series, implies that P+∞  n=1 an is convergent, which is what we want.

2.11.2. Algebra. 1. Set I =



xu − x : x ∈ R .

It is obvious that I is a two-sided ideal of R. It thus follows from the hypothesis that I = {0} or I = R. If I = {0}, we see that xu = x, yielding xu = ux = x for all x ∈ R, which means u = 1R . By showing that the case I = R is impossible, we finish the proof. Suppose to the contrary that I = R. Thus, there exists an x0 ∈ R such that u = x0 u − x0 . As u2 = u, we can write u = u2 = (x0 u − x0 )u = x0 u − x0 u = 0, and hence x = ux = 0x = 0, for all x ∈ R. This means R = {0}, implying that R has only one ideal, which is a contradiction. Therefore, I = {0}, finishing the proof.  2. To prove the assertion by contradiction, suppose that A is a finite extension of Q so that [A : Q] = n for some n ∈ N. Pick a prime number p, e.g., p = 2, and note that, by Eisenstein’s Criterion, the monic polynomial xn+1 − p is irreducible over Z and hence over Q. If α is one of the roots of the equation xn+1 − p = 0, we have Q ⊆ Q(α) ⊆ A. On the other hand, [A : Q] = n < n + 1 = [Q(α) : Q], which is obviously impossible. Thus, A is not a finite extension of Q, which is what we want.  3. First we need to recall that if G is a finite p-group, where p is a prime number, then the center of G, denoted by Z(G), is nontrivial, i.e., |Z(G)| > 1. To prove this, first recall that the class equation of G is |G| = |Z(G)| +

k X

[G : CG (xi )],

i=1

where {x1 , . . . , xk } is a maximal set of nonconjugate elements of G \ Z(G) and CG (xi ) is the centralizer of xi in G (1 ≤ i ≤ k). Note that since xi ’s are in G\Z(G), CG (xi )’s are proper subgroups of G, implying that whose orders are all powers of p, and hence so are [G : CG (xi )]’s because G is a finite p-group. Consequently, from the class equation of G, in view of |G| = pn for some n ∈ N, we conclude that p divides |Z(G)|, yielding |Z(G)| > 1, as desired. We now prove the assertion. It follows from the hypothesis that any two elements of G which are different from the identity share the same orders. Now suppose that p is a prime dividing |G|. It follows from Cauchy’s Theorem that there exists an x ∈ G whose order is p. Hence, all of the elements of G but e have order p, whence the order of G is a power of p. That is, G is a p-group. Conse
quently, the center of G is nontrivial, i.e., Z(G) 6= {e}. Since α Z(G) = Z(G) for

118

2. SOLUTIONS

all α ∈ Aut(G), from the hypothesis, we see that Z(G) = G. That is, G is abelian. Therefore, by the Fundamental Theorem of finite abelian groups, we obtain G ∼ = Zp ⊕ · · · ⊕ Zp , which is what we want.



2.11.3. General. 1. Note that for an A ∈ M2 (Z), the evenness of det(A) is equivalent to saying that det(A0 ) = 0, where A0 ∈ M2 (Z2 ) is the matrix whose entries are obtained from those of A viewed as elements of Z2 . With that in mind, the desired probability is equal to the probability that a randomly chosen matrix A0 ∈ M2 (Z2 ) has determinant zero. Obviously, |M2 (Z2 )| = 24 . On the other hand, the matrices in M2 (Z2 ) with zero determinants are as follows           0 0 0 1 1 0 0 0 1 1 , , , , , 0 0 0 1 1 0 1 1 0 0 

1 1

1 1

  0 , 1

0 0

  1 , 0

0 0

  0 , 0

Therefore, the desired probability is equal to

10 16

1 0

  0 , 0

0 1

 .

= 58 , which is what we want.



2. Plainly, the equation is equivalent to the following 4x3 − 3x = p, where p = −4c. Now, letting x = cos z, where z ∈ C, the above equation becomes equivalent to the following equation cos 3z = p.

(∗)

As the original equation has a root x0 in the closed interval [−1, 1], it follows that p ∈ [−1, 1], whence there exists z0 ∈ R such that x0 = cos z0 . It is now obvious 4π that z0 , z0 + 2π 3 , and z0 + 3 satisfy (∗). Thus, the corresponding x’s which are 2π 4π cos z0 , cos(z0 + 3 ), cos(z0 + 3 ) ∈ [−1, 1] are all of the roots of the original equation, proving the assertion.  3. In view of the hypotheses of the problem, we see that C, endowed with the symmetric difference of sets, denoted by ∆, which is defined by A∆B = (A \ B) ∪ (B \ A) forms a subgroup of P(X), where P(X) denotes the power set of X which obviously forms a group under ∆ itself. As is well-known, |P(X)| = 2n , where n = |X|. Thus, it follows from Lagrange’s Theorem that |C| divides |P(X)| = 2n , yielding |C| = 2k for some k ≤ n, which is what we want. 

2.12. TWELFTH COMPETITION

119

2.12. Twelfth Competition 2.12.1. Analysis. 1. First Solution: Suppose that f (x) = f (y) for some x, y ∈ R. It follows that M |x − y| ≤ 0, implying x = y. That is, f is one-to-one. Now, as f is continuous and one-to-one, we see from Problem 2 of 1.9.1 that f is strictly monotonic. Without loss of generality, we may assume that f is strictly increasing. It now easily follows from the hypothesis that limx→−∞ f (x) = −∞ and limx→+∞ f (x) = +∞. This, in view of the Intermediate Value Theorem, implies that f is onto, which is what we want. Second Solution: Just as we saw in the first solution, the function f is one-toone and strictly monotonic, and hence the inverse of f , i.e., f −1 : f (R) → R is continuous. This implies that f is both an open and closed map. Now, since R is both open and closed in R, we see that so is f (R), yielding f (R) = R because R is connected. This proves the assertion.  P+∞ 2. “=⇒” Since n=1 fn converges uniformly on S, it is uniformly Cauchy on S, and hence for given ε > 0, there exists an N > 0 such that n X ε fk < , 2 k=m

  for all m, n ≥ N . Now, letting m = n2 > N , it follows that n > m ≥ N , from which, in view of the fact that (fn )+∞ n=1 is a decreasing sequence of nonnegative functions on S, we see that n X ε fk < 2 k=[ n 2] =⇒

n hni X ε nfn fk < , ≤ (n − + 1)fn ≤ 2 2 2 k=[ n 2]

for all n ≥ 2N + 1. This obviously implies nfn < ε for all n ≥ 2N + 1. That is, the sequence (nfn )+∞ n=1 converges uniformly to zero on S. On the other hand, +∞ X n=1

n(fn − fn+1 ) =

+∞  X


nfn − (n + 1)fn+1 + fn+1 .

(∗)

n=1

But (n + 1)fn+1 converges
uniformly to zero on S. Hence, the telescopic series P+∞ nf − (n + 1)f n n+1 converges uniformly to f1 on S. This together with the n=1 P+∞ P+∞ hypothesis and (∗) implies that n=1 n(fn −fn+1 ) converges uniformly to n=1 fn on S, which is what we want. “⇐=” Note that as pointed out in the footnote of the problem in Section 1.12.1, for this implication we need to assume further that the sequence (fn )+∞ n=1 converges uniformly to zero on the set S. That is because for the numerical sequence (fn )+∞ n=1 , P+∞ P+∞ where fn = 2+ n12 , the series n=1 n(fn −fn+1 ) converges but n=1 fn is divergent.

120

2. SOLUTIONS

x x=1 t

t=1 Figure 10

Now, since (fn )+∞ converges uniformly to zero on S, it follows that the teleP+∞ n=1 scopic series n=1 (fn − fn+1 ) converges uniformly to f1 on S. We can write nfn = n

+∞ X

+∞ X

(fk − fk+1 ) =

k=n

n(fk − fk+1 ) ≤

+∞ X

k(fk − fk+1 ),

(∗∗)

k=n

k=n

P+∞ for all n ∈ N. By the hypothesis, the series n=1 n(fn − fn+1 ) is uniformly Cauchy on S. This, in view of (∗∗), easily implies that the sequence (nfn )+∞ n=1 converges
P+∞ uniformly to zero on S. Thus, the telescopic series n=1 nfn − (n + 1)fn+1 converges uniformly to f1 on S. We can write +∞ X

fn+1 =

n=1

+∞  X

n(fn − fn+1 ) − nfn − (n + 1)fn+1




,

n=1

P+∞ P+∞ implying that n=1 fn+1 converges uniformly on S, for so do the series n=1 n(fn −
P+∞ P+∞ fn+1 ) and n=1 nfn − (n + 1)fn+1 on S. Therefore, n=1 fn+1 , and hence P+∞ f , converges uniformly on S. Moreover, n n=1 +∞ X

fn =

n=1

+∞ X

n(fn − fn+1 ).

n=1

So the proof is complete.



3. First solution: Using integration by parts, we can write 1  Z 1 Z 1 Z 1 Z 1 Z g(t)dt dx = x g(t)dt − xd 0

x

x

x=0

Z =

0+

1

xg(x)dx 0

Z 0

which is what we want.

1

xg(x)dx,

=

0

1

x

 g(t)dt

2.12. TWELFTH COMPETITION

121

Second solution: Changing the order of integration, we can write   Z 1 Z t Z 1 Z 1 g(t)dx dt g(t)dt dx = x=0

t=0 1

t=x

x=0

Z

Z =

t

g(t) t=0 Z 1

=

 dx dt

x=0

tg(t)dt, 0

which is what we want.



2.12.2. Algebra. 1. By Sylow’s Third Theorem, the number of subgroups of order p is equal to 1 + kp for some k ∈ N ∪ {0} and that 1 + kp divides |G|. That is, 1 + kp|2p. This yields 1 + kp = 1, implying k = 0. Thus, G has only one Sylow p-subgroup. Therefore, G has only one subgroup of order p. Using Sylow’s Third Theorem again, we see that the number of subgroups of order 2 is equal to 1 + 2k for some k ∈ N ∪ {0} and that 1 + 2k divides |G|. That is, 1 + 2k|2p, from which we obtain 1 + 2k = 1 or 1 + 2k = p. Therefore, G has either one and only one subgroup of order 2 or it has p subgroups of order 2. Now, suppose that G has only one subgroup of order 2. Let H and K be the only Sylow 2-subgroup and Sylow p-subgroup, respectively. We have |H| = 2 and |K| = p. This implies |H ∩ K| = 1. In other words, H ∩ K = e. Pick h ∈ H and k ∈ K such that ord(h) = 2 and ord(k) = p. As H ∩ K = e, we see that hk = kh because hkh−1 k −1 ∈ H ∩ K, and hence ord(hk) = 2p, for ord(h) = 2, ord(k) = p, and gcd(2, p) = 1 (for a more detailed proof of ord(hk) = 2p, see the lemma presented in Solution 1 of 2.25.2). This implies that the cyclic subgroup generated by hk is G. Therefore, G is a cyclic group, which is what we want.  2. Note first that G has no elements of even order, in particular of order 2, because G is of odd order. We shall show that the map ψ : G → G defined by ψ(g) = g 2 is an automorphism of G. To see that ψ is one-to-one, suppose ψ(a) = ψ(b) for some a, b ∈ G. It easily follows that (ab−1 )2 = e, where e is the identity element of G. But ord(ab−1 ) 6= 2. So ab−1 = e, yielding a = b. That is, ψ is one-to-one. It now follows that ψ is an automorphism of G because G is finite and abelian. Thus, for an arbitrary g ∈ G, there exists an a ∈ G such that g = a2 . Set x = aφ(a) and y = aφ(a−1 ). Noting that G is abelian, we can write

xy = aφ(a) aφ(a−1 ) = a2 φ(aa−1 ) = a2 = g. On the other hand, φ(x) φ(y)


= φ aφ(a) = φ(a)φ2 (a) = aφ(a) = x,
= φ aφ(a−1 ) = φ(a)φ2 (a−1 ) = a−1 φ(a) = y −1 ,

as desired. To show that such x, y are unique, suppose that for g ∈ G, there are x, y, z, t ∈ G such that g = xy = zt, φ(x) = x, φ(z) = z, φ(y) = y −1 , and

122

2. SOLUTIONS

φ(t) = t−1 . We have φ(g) = φ(xy) = φ(zt)

=⇒ φ(x)φ(y) = φ(z)φ(t) =⇒ xy −1 = zt−1 =⇒ xyy −2 = zt−1 =⇒ zty −2 = zt−1 =⇒ y 2 = t2 =⇒ ψ(y) = ψ(t) =⇒ y = t,

which, in view of xy = zt, yields x = z. Therefore, any element g ∈ G can, uniquely, be written as g = xy, where φ(x) = x and φ(y) = y −1 , finishing the proof.  3. As every Boolean ring is commutative, it suffices to show that R is a Boolean ring. To this end, pick x ∈ R such that x 6= 0, 1. As ch(R) = 2, there exists x1 ∈ R \ {0, 1} such that x = 1 + x1 . It follows from the hypothesis that xy = xy 2 for all y ∈ R. In other words, (1 + x1 )y = (1 + x1 )y 2 , from which, in view of x1 y = x1 y 2 , we obtain y = y 2 for all y ∈ R. That is, R is a Boolean ring, and hence commutative, which is what we want.  4. First, we claim that every ascending chain of ideals of R terminates. To this end, suppose I1 ⊆ I2 ⊆ · · · S+∞ is an ascending chain of ideals of R. Set I = n=1 In . It is easily seen that I is an ideal of R, and hence there exists an x0 ∈ R such that I = hx0 i = x0 R. As x ∈ I, it follows that there exists an N ∈ N such that x0 ∈ IN . We can write IN ⊆ I =

+∞ [

Ik = x0 R ⊆ IN

=⇒ IN =

+∞ [

Ik

k=1

k=1

=⇒

∀n ∈ N : In ⊆ IN .

On the other hand, IN ⊆ In for all n ≥ N , from which, in view of the above inclusion, we obtain IN = In for all n ≥ N . This proves the claim. Now, suppose that f : R → R is a surjective homomorphism. To see that f is one-to-one, we note that ker f ⊆ ker f 2 ⊆ ker f 3 ⊆ · · · is an ascending chain of ideals of R. It follows from the claim that there exists an N ∈ N such that ker f n = ker f N for all n ≥ N . In particular, ker f N +1 = ker f N . Suppose x ∈ ker f is arbitrary so that f (x) = 0. As f is onto, so is f n for all n ∈ N, and hence there exists an x0 ∈ R such that x = f N (x0 ). Consequently, f N +1 (x0 ) = 0, yielding x0 ∈ ker f N +1 = ker f N . Therefore, x = f N (x0 ) = 0, implying ker f = {0}. That is, f is one-to-one, which is what we want.  5. First solution: Let λi ’s (1 ≤ i ≤ 3) be the eigenvalues of the matrix A in the algebraic closure of F . As A is invertible, λ−1 i ’s (1 ≤ i ≤ 3) are the eigenvalues of A−1 . If f is the characteristic polynomial of A, we have f

=

(x − λ1 )(x − λ2 )(x − λ3 )

= x3 − (λ1 + λ2 + λ3 )x2 + (λ1 λ2 + λ1 λ3 + λ2 λ3 )x − λ1 λ2 λ3 .

2.12. TWELFTH COMPETITION

123

By the hypothesis, tr(A)

= λ1 + λ2 + λ3 = 0,

−1 −1 tr(A ) = λ−1 1 + λ2 + λ3 = 0, det(A) = λ1 λ2 λ3 = 1. −1

−1 −1 On the other hand, λ1 λ2 + λ1 λ3 + λ2 λ3 = λ−1 1 + λ2 + λ3 = 0. So, we must have 3 f (x) = x − 1. Now, by the Cayley-Hamilton Theorem, A3 = I, which is what we want.

Second solution: Let λi ’s (1 ≤ i ≤ 3) be as in the first solution. We see that −1 −1 λi ’s (1 ≤ i ≤ 3) are distinct because λ1 + λ2 + λ3 = 0 = λ−1 and 1 + λ2 + λ3 3 2 λ1 λ2 λ3 = 1. With that in mind, suppose f (x) = x +ax +bx+c is the characteristic polynomial of A. We have a = tr(A) = 0 and c = det(A) = −1. By the CayleyHamilton Theorem, A3 + bA − I = 0, yielding (A−1 )3 − b(A−1 )2 − I = 0. As λi ’s (1 ≤ i ≤ 3) are distinct, the characteristic polynomial of A−1 , denoted by g, is equal to g(x) = x3 − bx2 − 1. But b = tr(A−1 ) = 0. Therefore, (A−1 )3 − I = 0, implying A3 = I, which is what we want.  6. It is obvious that α and β are roots of the polynomial p! f (x) = p! + p!x + x2 + · · · + xp , 2! which is irreducible over Q by Eisenstein’s Criterion. Thus, f is the minimal polynomial of α and β over Q. We prove the assertion by way of contradiction. First suppose that α − β = r ∈ Q. We have f (β + r) = f (α) = 0. That is, β is a root of the polynomial f (x + r) whose coefficients are in Q. But f (x) is the minimal polynomial of β over Q. It follows that f (x) f (x+r). As f (x) and f (x+r) are both monic polynomials of the same degree, we obtain f (x + r) = f (x). Consequently, if α1 , . . . , αp are all of the roots of f (x) = 0, then so are α1 + r, . . . , αp + r. In particular, we must have (α1 + r) + · · · + (αp + r) = α1 + · · · + αp

=⇒ pr = 0 =⇒ r = 0 =⇒ α = β,

which is a contradiction. Thus, α − β = r ∈ / Q, as desired. Now , assume that α + β = r ∈ Q. We have −f (r − β) = f (α) = 0. That is, β is a root of the polynomial −f (r − x) whose coefficients are in Q. But f (x) is the minimal polynomial of β over Q. Thus, f (x) − f (r − x). As f (x) and −f (r − x) are both monic polynomials of the same degree, we obtain −f (r − x) = f (x). Consequently, letting x = 2r , we obtain −f ( 2r ) = f ( 2r ), which yields f ( 2r ) = 0. Hence, f (x) is divisible by x − 2r , which is in contradiction with the fact that f (x) is irreducible over Q. Therefore, α + β = r ∈ / Q, as desired. Next, suppose that αβ = r ∈ Q. Obviously, α, β 6= 0. We can write f ( βr ) = p f (α). It thus follows that β is a root of the polynomial xp! f ( xr ) whose coefficients xp r p are in Q. Again, we obtain f (x) p! f ( x ), from which, as f (x) and xp! f ( xr ) are p both monic polynomials of the same degree, we see that xp! f ( xr ) = f (x). There √ are two cases to consider: (i) r > 0 and (ii) r < 0. If r > 0, substituting x = r √ √ p into xp! f ( xr ) = f (x), we obtain f ( r) = 0. If r ∈ Q, then f (x) is divisible by

124

2. SOLUTIONS

A R

r M

O B Figure 11

√ √ of f over Q. And if r ∈ / Q, x− r, which is in contradiction with the irreducibility √ then √ x2 − r would be the minimal polynomial of r over Q. From this, in view of f ( r) = 0, we see that x2 − r f (x), implying that f is not irreducible over Q, √ p a contradiction again. If r < 0, substituting x = i −r into xp! f ( xr ) = f (x), we √ √ that x2 + r is the minimal polynomial of r over Q. obtain f (i −r) = 0. It follows √ From this, in view of f (i −r) = 0, we see that x2 + r f (x), implying that f is not irreducible over Q, which is again a contradiction. Therefore, αβ = r ∈ / Q, which is what we want. It might be worth mentioning that one can show, in a similar fashion, that α / Q.  β ∈

2.12.3. General. 1. Let R and a, respectively, be the radius of the circle C centered at O and the side length of the square whose√center M is on the circle. √ 2 From the hypothesis, we see that a2 < πR 2 , whence a 2 ≤ R π < 2R. That is, the diameter of the square is smaller than that of the circle. It follows that the vertices √of the square, with center M , are on the circle centered at M with radius r = a 2 2 < R. Let A and B be the points at which these two circle intersect one another. Consider two isosceles triangles OAM and OBM . We have AM = M B < OA = OB = OM . It is plain that ∠AM O > 60◦ and ∠BM O > 60◦ , and hence ∠AM B > 120◦ . Therefore, the arc AB from the circle centered at M is greater than 120◦ , and hence some vertex of the square must be on this arc which itself lies insides the circle with radius R centered at O. This proves the assertion.  2. The answer is: yes, it does. To see this, we count the number of the desired answer sheets in a test in which n questions are given. Suppose that there are n questions in the test. Use 1 and 0 to denote a false and a true answer, respectively, and An to denote the number of desired answer sheets. Then to any answer sheet, there corresponds an n-tuple of 0’s and 1’s. Let’s call an answer sheet or n-tuple admissible if there are no consecutive 1’s in it. First, for a given k (0 ≤ k ≤ n), we count the number of admissible n-tuples each of which contains k ones and use Nk to denote this number. To do so, note that in any such admissible n-tuple, there are k ones and n − k zeros and that we want to insert a zero between any two ones unless the zero occurs in the 1st position or the nth position. With that in mind, there are n − k − 1 + 2 = n − k + 1 positions and of those we need to choose k positions to each of which assign a one. Thus, the number Nk is equal to the

2.13. THIRTEENTH COMPETITION

125

number
of ways of choosing k positions out of n − k + 1 positions which is equal to  n−k+1 n+1 . Obviously, k ≤ p := and that we have k 2 An

=

p X

Nk =

 p  X n−k+1

k k=0      n+1 n n−p+1 + + ··· + . 0 1 p

k=0

 =

It thus follows that the number of desired answer sheets in the test in which 15 questions are given is equal to       16 15 8 A15 = + + ··· + = 1597. 0 1 8 As 1700 people have participated in the test, there are necessarily two equal answer sheets.  3. We have (1 + i)n =

n X

Cnk in−k ,

k=0

√
where i ∈ C with i = −1. On the other hand, 1 + i = 2 cos π4 + i sin π4 , and hence n nπ nπ
(1 + i)n = 2 2 cos + i sin . 4 4 So we have n X n nπ nπ
Cnk in−k = 2 2 cos + i sin , 4 4 2

k=0

whence

n n nπ nπ Cn0 − Cn2 + Cn4 − · · · + i Cn1 − Cn3 + Cn5 − · · · = 2 2 cos + i2 2 sin , 4 4 form which, we obtain Cn0 − Cn2 + Cn4 − · · · = Cn1 − Cn3 + Cn5 − · · · =

nπ , 4 n nπ 2 2 sin , 4 n

2 2 cos

which is what we want.



2.13. Thirteenth Competition 2.13.1. Analysis. 1. First solution: As f : R → R is integrable on any closed interval, for all x, y ∈ R, we can write Z x Z x Z x f (t + y)dt = f (t)dt + f (y)dt, 0

0

0

126

2. SOLUTIONS

which easily yields Z xf (y)

x

Z

f (t + y)dt − 0 Z Z x+y f (t)dt − = =

f (t)dt 0 x

f (t)dt

0

y

Z

x

x+y

Z

x

Z f (t)dt −

f (t)dt −

=

0

0

y

f (t)dt, 0

and hence xf (y) = yf (x) for all x, y ∈ R. Therefore, f (x) = f (1) = f (1) for all x 1 x ∈ R with x 6= 0. This implies f (x) = f (1)x for all x ∈ R, as desired. Second solution: As f : R → R is integrable, it follows from the lemma presented in Solution 1 of 2.5.1 that the function f is continuous at infinitely many points of any closed interval of R. On the other hand, f (x + y) = f (x) + f (y) for all x, y ∈ R and hence f (r) = rf (1) for all r ∈ Q. From this, it is easily seen that f is continuous on R and that f (x) = xf (1) for all x ∈ R, which is what we want.  2. First solution: We need the following well-known lemma which is known as Dedekind’s Extension of Abel’s Theorem. Lemma. Let (an )+∞ and (bn )+∞ If the series n=1 be sequences of real numbers. P+∞ P+∞ n=1 P+∞ a and |b − b | converge, then so does the series n n n+1 n=1 n=1 n=1 an bn . Proof. It is easily verified that an+1 bn+1 + · · · + an+p bn+p

(sn+1 − sn )(bn+1 − bn+2 ) + · · · + (sn+p−1 − sn )(bn+p−1 − bn+p ) + (sn+p − sn )bn+p , Pn P+∞ where sn = k=1 ak and n, p ∈ N. Note that as n=1 |bn − bn+1 | converges, so P+∞ does the telescopic series n=1 (bn − bn+1 ), from which, we see that limn bn exists, and hence the sequence (bn )+∞ the above equality, together with n=1 is bounded. Now,
+∞ Pn +∞ the facts that (sn )+∞ and |b − b | k+1 n=1 are Cauchy and that (bn )n=1 is n=1 k=1 k Pn P +∞ bounded, implies that ( k=1 ak bk )+∞ n=1 is Cauchy, and hence the series n=1 an bn is convergent, which is what we want.  =

To prove the assertion, letting cn =

an bn ,

we can write


an cn − c2n c2n
2 = = c − c 1 + . n n an + bn 1 − c2n 1 − c2n Let An = cn − c2n and Bn =

c2n 1−c2n .

We have

an = An + An Bn . an + bn P+∞ As n=1 c2n is convergent, limn c2n = 0, and hence there exists an N ∈ N such that c2n < 12 for all n ≥ N . It follows that for all n ≥ N we have 0 ≤ Bn =

c2n < 2c2n , 1 − c2n

2.13. THIRTEENTH COMPETITION

127

P+∞ from which, in view of the Comparison Test, we see that n=1 Bn is convergent. Note that Bn − Bn+1 ≤ Bn + Bn+1 = Bn + Bn+1 , for all n ≥ N . Again the Comparison Test together with the above inequality, in P+∞ P+∞ view of the convergence of n=1 Bn , implies that n=1 |Bn − Bn+1 | is convergent. P+∞ On the other hand, it follows from the hypothesis that n=1 An is convergent. Now, P+∞ using Dedekind’s Extension of Abel’s Theorem, we see that n=1 An Bn converges. P+∞ P+∞ This together with the convergence of n=1 An implies that n=1 (An + An Bn ) P+∞ n converges. Thus, so does the series n=1 ana+b , which is what we want. n Second solution: We can write an cn + c2n − c2n c2n = = cn − , an + bn 1 + cn 1 + cn where cn = abnn . As limn cn = 0, it follows that there exists an N ∈ N such that P+∞ 2 c2n n=1 cn , in view 1+cn > 0 for all n ≥ N . This together with the convergence of P+∞ c2n of the Limit Comparison Test, implies that the series n=1 1+cn converges. From P+∞ the above equality and the hypothesis that n=1 cn is convergent, we see that the P+∞ an series n=1 an +bn is convergent and that we have +∞ X

+∞ +∞ X X an c2n = cn − , a + bn 1 + cn n=1 n n=1 n=1

finishing the proof.



3. First solution: To prove the assertion by contradiction, suppose that the sequence (fn )+∞ n=1 does not converge uniformly to zero on [0, 1]. It follows that there are an  > 0, a sequence (nk )+∞ k=1 , with nk ≥ k, of natural numbers, and a +∞ sequence (xk )k=1 in [0, 1] such that |fnk (xk )| ≥  for all k ∈ N. If necessary, by passing to a subsequence of (xk )+∞ k=1 and replacing fnk by −fnk , without loss of generality, we may assume that fnk (xk ) > 0 for all k ∈ N and that there is an x0 ∈ [0, 1] such that limk xk = x0 . We can write Z xk fnk (xk ) − fnk (x0 ) = fn0 k (t)dt, x0

from which, in view of ||fn0 ||∞ ≤ 1 for all n ∈ N, we obtain fn (xk ) − fn (x0 ) ≤ xk − x0 , k

k

for all k ∈ N. Let K1 ∈ N be such that |xk − x0 | < 2 for all k ≥ K1 . We must have fn (x0 ) ≥ fn (xk ) − fn (xk ) − fn (x0 ) ≥  −  =  , k k k k 2 2 for all k ≥ K1 . Now, from this and Z x fnk (x) − fnk (x0 ) = fn0 k (t)dt, x0

in a similar fashion, we see that fn (x) ≥  , k 4

128

2. SOLUTIONS

for all x ∈ [0, 1] with |x − x0 | < 4 and k ≥ K2 , where K2 is such that |xk − x0 | < 4 for all k ≥ K2 . Since fnk ’s (k ≥ K2 ) are continuous functions, we conclude that fnk (x) > 0, and hence,  fnk (x) ≥ , 4 for all x ∈ [0, 1] with |x − x0 | < 4 and k ≥ K2 , for otherwise, by the Intermediate Value Theorem, there must exist zk ∈ [0, 1] with |zk − x0 | < 4 and k ≥ K2 such that fnk (zk ) = 0, which is impossible. Let [a, b] = [0, 1] ∩ [x0 − 4 , x0 + 4 ]. Now define g : [0, 1] → R by  0 0 ≤ x ≤ a,    3  b−a (x − a) a ≤ x ≤ a + b−a  3 , g(x) = ≤ x ≤ a + 2 b−a 1 a + b−a 3 3 ,  −3 b−a  (x − b) a + 2 ≤ x ≤ b,   3  b−a 0 b ≤ x ≤ 1. The function g is continuous and nonnegative on [0, 1]. For all k ≥ K2 , we can write Z a+2 b−a Z b Z 1 Z a+ b−a 3 3 fnk g + fnk g + fnk g fnk g = a+ b−a 3

a

0

Z

a+2 b−a 3

≥ a+ b−a 3

fnk ≥

contradicting the hypothesis that limn by contradiction.

R1 0

a+2 b−a 3

(b − a) , 12 fnk g = 0. Therefore, the assertion follows

+∞ Second solution: It suffices to show that every subsequence (gk )+∞ k=1 of (fn )n=1 , +∞ where gk = fnk (k ∈ N), in turn, has a subsequence (hj )j=1 , where hj = gkj (j ∈ N), which converges uniformly to zero on [0, 1]. We first show that the sequence (fn )+∞ n=1 is uniformly bounded on [0, 1]. To this end, noting that fn : [0, 1] → R and ||fn0 ||∞ ≤ 1 for all n ∈ N, from the Mean Value Theorem, we see that fn (x) ≤ 1 + fn (0) ,
+∞ for all x ∈ [0, 1] and n ∈ N. We claim that the sequence fn (0) n=1 is bounded.
+∞
+∞ Suppose to the contrary that there exists a subsequence fnk (0) n=1 of fn (0) n=1 such that limk fnk (0) = +∞ or limk fnk (0) = −∞. If necessary, replacing fn by −fn , we may assume that limk fnk (0) = +∞. Thus, there exists a K ∈ N such that fnk (0) ≥ 2 for all k ≥ K. The sequence (fn )+∞ n=1 is uniformly equicontinuous on [0, 1] because ||fn0 ||∞ ≤ 1 for all n ∈ N. It follows that the sequence (fnk )+∞ n=1 is uniformly equicontinuous on [0, 1] as well. So choosing ε = 1, we obtain 0 < 2δ < 1 such that fn (x) − fn (0) < 1, k k whenever 0 < x < 2δ and k ∈ N. This easily yields

fnk (x) ≥ 2 − 1 = 1, for all 0 < x < 2δ and k ∈ N with k ≥ K. Now, define g : [0, 1] → R by  1 0 ≤ x ≤ δ,  − xδ + 2 δ ≤ x ≤ 2δ, g(x) =  0 2δ < x ≤ 1.

2.13. THIRTEENTH COMPETITION

129

Obviously, g is continuous and nonnegative on [0, 1]. So it follows from the hypothR1 esis that limn 0 fn g = 0, from which, we obtain Z 1 lim fnk g = 0. k

0

Note that as g ≥ 0 on [δ, 2δ], for all k ≥ K, we have Z 1 Z 2δ fnk g = fnk g 0

0

Z =

δ

2δ

Z fnk +

0

fnk g δ

≥ 1 × (δ − 0) = δ. R1 That is, 0 fnk g ≥ δ for all k ≥ K, which is in contradiction with limk 0 fnk g = 0
+∞ and 0 < δ. Therefore, fn (0) n=1 is bounded and hence so is (fn )+∞ n=1 because |fn (x)| ≤ 1 + |fn (0)| for all x ∈ [0, 1] and n ∈ N. Now suppose (gk )+∞ k=1 , with gk = fnk (k ∈ N), is an arbitrary subsequence of +∞ (fn )+∞ n=1 . The sequence (gk )k=1 is uniformly bounded and equicontinuous on [0, 1] +∞ because so is (fn )n=1 on [0, 1]. It thus follows from Arzela’s Theorem that the +∞ sequence (gk )+∞ k=1 has a subsequence (hj )j=1 , with hj = gkj (j ∈ N), such that +∞ (hj )j=1 converges uniformly to a function h : [0, 1] → R on [0, 1]. The function h is continuous because it is a uniform limit of continuous functions. This together R1 with the hypothesis yields limj 0 hj h = 0. But it is plain that (hj h)+∞ j=1 converges uniformly to h2 on [0, 1]. So, we can write Z 1 Z 1 Z 1 2 h = lim hj h = lim hj h = 0, R1

0

R1

0

j

j

0

2

implying 0 h = 0, which, in turn, implies h = 0 because h is continuous on [0, 1]. That is, we have shown that every subsequence of (fn )+∞ n=1 , in turn, has a subsequence converging uniformly to zero on [0, 1]. This proves the assertion. 

2.13.2. Algebra. 1. Note first that every nonzero ideal I of R is uncountable. To see this, as R I is countable, we have  R = an + I : an ∈ R . I S+∞ It follows that R = n=1 (an + I), implying that I is uncountable because otherwise R would be countable, which is impossible. Now, to prove the assertion by contradiction, suppose that 0 6= a ∈ R is a divisor of zero in R. Define  I := r ∈ R : ar = 0 . It is plain that I is a nonzero ideal of R and hence it is uncountable. Also note that aR is a nonzero ideal of R. Define the map f : aR → R I by f (ax) = x + I. The map f is obviously onto. Suppose that f (ax1 ) = f (ax2 ). We have x1 + I = x2 + I, implying that x1 − x2 ∈ I, whence a(x1 − x2 ) = 0. That is, ax1 = ax2 which means f is one-to-one. Therefore, f is a one-to-one correspondence between the uncountable set aR and the countable set R I , a contradiction. Thus, R has no divisor of zero, and hence R is an integral domain, which is what we want. 

130

2. SOLUTIONS

2. Let S = M2 (Q). We have QI2 ≤ R ≤ S and dimQ R ≤ dimQ S ≤ 4, where I2 denotes the 2 × 2 identity matrix. Suppose that I is a nonzero left ideal of R. As R includes QI2 , the left ideal I can be viewed as a vector space over Q and we have dimQ I ≤ 4. Thus, there are xi ∈ I (1 ≤ i ≤ k ≤ 4) such that I = Qx1 + · · · + Qxk . This together with QI2 ≤ R implies that I = Rx1 + · · · + Rxk . In other words, I is finitely generated. Likewise, one can see that every right ideal of R is finitely generated as well. To show that the condition (∗) cannot be dropped, define   0 G J = , 0 0 1 where G = h 21i i+∞ i=1 ≤ Q is the additive group generated by 2i ’s in Q. It is easily verified that J is a left ideal of the ring T which is not finitely generated, which is what we want. 

3. First solution: Let G = GL2 (Q) be the multiplicative group of all 2 × 2 invertible matrices over the field Q and     1 1 −1 1 a= , b= . 0 −1 0 1 It is easily seen that a, b ∈ G, a2 = b2 = I2 , and ab = −I2 + 2N , where     1 0 0 1 I2 = , N= . 0 1 0 0 We have (ab)n = (−1)n I2 + (−1)n−1 2nN 6= I2 , for all n ∈ N. That is, ab has order infinity and yet both a and b have order two, which is what we want. Second solution: Let G be the group of isometries of the Euclidean plane under composition of isometries. Let A, B ∈ R2 be two distinct points in the plane and a, b the half turns around the points A and B, respectively. If e denotes the identity isometry, we have a2 = b2 = e. On the other hand, it is easy to see that ab is a −−→ translation isometry along the vector 2AB, and hence ab has order infinity because it is a translation. This is what we want, finishing the proof.  4. (a) It is well-known that |GLn (Zp )| = (pn − 1)(pn − p) · · · (pn − pn−1 ). To see this, note that a matrix A ∈ GLn (Zp ) if and only if the columns of A are linearly independent. In view of this, there are pn − 1 choices for the first column of an arbitrary element, say A, of GLn (Zp ) (the zero column vector being excluded). For an 1 ≤ i < n, assuming that the first i columns of an arbitrary element, say A, of GLn (Zp ) are chosen, there are exactly pn − pi choices for the (i + 1)st column of A, because the (i + 1)st column of A cannot be a linear combination of the first i columns of A. Therefore, by the product rule of combinatorics, GLn (Zp ) has exactly (pn −1)(pn −p) · · · (pn −pn−1 ) elements, as desired. Thus, |G| = (32 −1)(32 −3) = 48. As K = Z(G) includes only scalar matrices, we see that K = {I2 , 2I2 }, where I2 is the identity matrix, and hence |K| = 2. It is now obvious that K ≤ H ≤ G. To calculate |H|, we note that a, b, c ∈ Z3 can be chosen independent of one another

2.13. THIRTEENTH COMPETITION

131

and that a, c can be 1 or 2 and b can be 0, 1, 2. Thus, using the product rule of combinatorics, we have |H| = 2 × 2 × 3 = 12. T (b) Let N = x∈G x−1 Hx. First, we claim that N is the largest subgroup of H that is normal in G. We present two proofs for the claim. First proof. Firstly, N ⊆ e−1 He = H. Secondly, N is a subgroup of G because it is an intersection of subgroups of G, namely, x−1 Hx’s where x ∈ G. To see that N is aTnormal subgroup of G, suppose that a ∈ N and g ∈ G are arbitrary. As a ∈ N = x∈G x−1 Hx and xg −1 ∈ G, we see that a ∈ gx−1 Hxg −1 , yielding T g −1 ag ∈ x−1 Hx. It follows that g −1 ag ∈ x∈G x−1 Hx = N . That is, N is a normal subgroup of G. Now, suppose that N1 ⊆ H is a normal subgroup of G. We can write N1 = x−1 N1 x ⊆ x−1 Hx, T for all x ∈ G, implying that N1 ⊆ x∈G x−1 Hx = N , as desired. Second proof. We see from (a) that [G : H] = 4. Consequently, if X is used to denote the set of left cosets of H in G, then there exist g1 = e, g2 , g3 , g4 ∈ H such that  X = g1 H, . . . , g4 H . Observe that G acts on X via left multiplication. That is, g(gi H) = ggi H ∈ X, where g ∈ G. Now, for a fixed g ∈ G, define the map `g : X → X by `g (gi H) = ggi H. It is easily verified that `g is a one-to-one map which is onto because |X| = 4 < ∞. In other words, `g defines a permutation on X, i.e., `g ∈ S(X) ∼ = S4 , where S4 denotes the symmetric group of degree four. Now, define the map φ : G → S(X) ∼ = S4 by φ(g) = `g . As G acts on X via left multiplication, we see that `g1 g2 = `g1 `g2 for all g1 , g2 ∈ G. This means φ(g1 g2 ) = φ(g1 )φ(g2 ), for all g1 , g2 ∈ G. Thus, φ is a homomorphism of groups. We claim that \ ker φ = N = x−1 Hx ⊆ e−1 He = H, x∈G

from which, we see that N is a normal subgroup of G which is contained in H. To prove this last claim, suppose g ∈ ker φ. We have φ(g) = `g = `e , where e is the identity element of G. It follows that `g (x−1 H) = `e (x−1 H) for all x ∈ G. That is, gx−1 H =Tx−1 H, yielding xgx−1 ∈ H, and henceTg ∈ x−1 Hx for all x ∈ G. This implies g ∈ x∈G x−1 Hx = N . Conversely, if g ∈ x∈G x−1 Hx = N , we conclude T that g ∈ ker φ. Therefore, ker φ = x∈G x−1 Hx = N , proving the claim. We now show  that N= K. To this end, let n ∈ N be arbitrary. As N ⊆ H, a1 b1 we have n = , where a1 , b1 , c1 ∈ Z3 and a1 c1 6= 0. Suppose g = 0 c1   a b ∈ G, where a, b, c, d ∈ Z3 with ad − bc 6= 0, is arbitrary. As N is c d normal in G, we must have h = g −1 ng ∈ N ⊆ H. This implies that h21 = −aca1 −c2 b1 +acc1 = 0, where h21 denotes the 21 entry of the matrix h. Simplifying, ad−bc

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2. SOLUTIONS

we obtain ac(c1 − a1 ) + c2 (−b1 ) = 0 for all a, b, c, d ∈ Z3 with ad − bc 6= 0. This easily yields a1 = c1 and b1 = 0, proving that n ∈ K, which is what we want. (c) We have shown that the map φ : G → S(X) ∼ = S4 defined by φ(g) = `g is a homomorphism of groups and that \

ker φ =

x−1 Hx = N = K.

x∈G

In view of the First Isomorphism Theorem for groups, we can write G ∼ = im(φ) ≤ S4 . ker φ G ∼ That is, K = im(φ) ≤ S4 . On the other hand, |im(φ)| = [G : K] = 24 = |S4 |. G ∼ Thus, im(φ) ∼  = S4 , and hence K = S4 , which is what we want.


5. Let Aν(ξ) i1 (1 ≤ i ≤ n) denote the i1 entry of the column matrix Aν(ξ). We have
Aν(ξ) i1

=

n X


aik ν(ξ) k1 aik ν(ξ) k1 = ai1 ν(ξ) 11 +

n X k=1

k=2

= δi,n +

n X

δi,k−1 ξ k−1 =

k=2



1 ξi

i = n, i < n.

In other words, 

ξ ξ2 .. .



   Aν(ξ) =    ξ n−1 1



ξ ξ2 .. .

      =     ξ n−1 ξn





       = ξ    

1 ξ ξ2 .. . ξ n−1

     = ξν(ξ).  

That is, ν(ξ) is an eigenvector of A whose corresponding eigenvalue is ξ.



2.13.3. General. 1. Let M be an arbitrary point on the unit circle in the complex plane centered at the origin and A1 , . . . , An denote the vertices of a regular n-gon which is inscribed in the circle. We have M Aj

= =

exp iθ = cos θ + i sin θ, exp iθj = cos θj + i sin θj ,

2.13. THIRTEENTH COMPETITION

133

where θ, θ1 ∈ R, and θj = θ1 + (j − 1) 2π n (1 ≤ j ≤ n). We can write n X

M A2j

n X exp iθ − exp iθj 2

=

j=1

j=1 n X

=


2 − exp i(θ − θj ) − exp i(θj − θ)

j=1 n X 2π
j−1 2π
j−1 − exp i(θ1 − θ) exp i n n j=1 j=1

n n −1 −1 exp −i 2π exp i 2π n n = 2n − exp i(θ − θ1 ) − exp i(θ − θ) 1 2π 2π exp −i n − 1 exp i n − 1 = 2n − exp i(θ − θ1 ) × 0 − exp i(θ1 − θ) × 0 = 2n.

2n − exp i(θ − θ1 )

=

That is,

Pn

j=1

n X

exp −i

M A2j = 2n, which is what we want.



2. Let f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 , where ai ∈ Q (0 ≤ i ≤ n) and an > 0, be a polynomial with the property that f (x) is irrational whenever x is irrational. Let p be a prime and g(x)

=

lcm(a1 , . . . , an )f (x) − lcm(a1 , . . . , an )a0 − p

= gn xn + gn−1 xn−1 + · · · + g1 x − p, where gi = lcm(a1 , . . . , an )ai (1 ≤ i ≤ n). As the leading coefficient of g is positive and g(0) = −p < 0, it follows from the Intermediate Value Theorem that g has a positive root which must be rational because, in view of the above equality, the polynomial g has the property that g(x) is irrational whenever x is irrational. Now by the Rational Root Theorem, the positive root is of the form d1 or dp for some d dividing gn . If n > 1, it is obvious that by choosing p large enough none of the numbers d1 and dp , where d divides gn , can be a root of g(x) = 0. Therefore, n ≤ 1, yielding f (x) = ax + b for some a, b ∈ Q with a 6= 0.  3. To prove the assertion, we need the following theorem which is known as Sylvester’s problem. Theorem. Let S be a finite set of the points of a plane such that the line passing through any two points of S contains a third point of S. Then, all of the points of S lie on a straight line. Proof. To prove the assertion by contradiction, suppose that all of the points of S are not collinear. As S is finite, there exist A, B, C ∈ S such that the smallest height of the triangle ABC is minimal among all triangles with vertices from S. Without loss of generality, assume that the altitude AH drawn from the vertex A perpendicular to the base BC is the smallest height of the triangle ABC. If H = B or H = C, then the altitude drawn from H to the base opposite to it is smaller than AH, which is impossible. If not, then H is between B and C and there is a third point D of S on the line BC. Thus, if necessary by renaming the points B, C, and D, we may assume that B and C are both on the left or the right of AH such that BH < CH. Assume, with no loss of generality, that B and C are on the right of AH such that BH < CH. It is now easily seen that the altitude

134

2. SOLUTIONS

A B

C=H

A

B=H

A C

H

H’ B

B’ C

Figure 12

drawn from B to AC is smaller than AH, a contradiction. Thus, all of the points of S are collinear, finishing the proof.  Remark. Motivated by Sylvester’s problem, we pose the following problem. Let S be a finite set of the points of a plane containing no three collinear points such that the circle passing through any three points of S contains a fourth point of S. Then, all of the points of S lie on a circle. Hint. Perform an inversion with center at one of the given points and any radius, say, radius one. First, we show that there exists a plane which contains only two lines of the given n parallel lines. To this end, intersect the n parallel lines with a plane to obtain n points of intersection on the plane. It follows from the above theorem that in the plane there is a line which contains exactly two points of the n points. Now, this line and two lines of the given n parallel lines form a plane which passes through only two lines of the n parallel lines. With this in mind, we prove the assertion by induction on n. If n = 3, the assertion is obvious. Assuming that the assertion holds for n − 1 parallel lines, to prove the assertion for n parallel lines, as explained in the above find a plane that contains exactly two lines of the given n lines. Now, one of these two lines and the n − 2 remaining lines are not coplanar. It thus follows from the induction hypothesis that there exist at least n − 1 distinct planes each of which includes at least two lines of these n − 1 lines. These n − 1 distinct planes together with the plane that contains only two lines of the given n lines form n planes satisfying the desired property, which is what we want. 

2.14. Fourteenth Competition

2.14.1. Analysis. 1. To prove the assertion by contradiction, suppose that f 0 (x0 ) exists. As f 0 (x) < 0 on (−∞, x0 ), f is strictly decreasing on (−∞, x0 ], from which, we obtain f (x) > f (x0 ) for all x < x0 . Likewise, since f 0 (x) > 0 on (x0 , +∞), we see that f (x) > f (x0 ) for all x > x0 . Thus, f assumes it absolute minimum, and hence local minimum, at x0 , yielding f 0 (x0 ) = 0. Now, f 00 (x) < 0 on (x0 , +∞) implies f 0 is strictly decreasing on [x0 , +∞), which, in turn, implies f 0 (x) < f 0 (x0 ) = 0 for all x > x0 , which is in contradiction with the hypothesis. This shows that f 0 (x0 ) does not exist, which is what we want.  2. We need the following lemma.

2.14. FOURTEENTH COMPETITION

135

Lemma. Let g : R → R be a uniformly continuous and nonnegative function R +∞ on R such that −∞ g < ∞. Then lim g(x) = 0,

x→−∞

lim g(x) = 0.

x→+∞

Proof. We prove that limx→+∞ g(x) = 0. One can prove limx→−∞ g(x) = 0 in a similar fashion. Proceed by way of contradiction. Thus, there exists an ε0 > 0 such that for all n ∈ N, there is an xn > n for which g(xn ) = g(xn ) ≥ ε0 . From the hypothesis that g is uniformly continuous on R, we see that for ε20 > 0, there exists a 0 < δ0 < 1 such that g(x) − g(y) < ε0 , 2 for all x, y ∈ R with |x − y| < δ0 . In particular, for all i ∈ N and x ∈ R with |x − xi | < δ0 , we have g(x) − g(xi ) < ε0 , 2 from which, in view of |g(xi )| − |g(x)| ≤ |g(x) − g(xi )|, we obtain g(xi ) ε0 g(xi ) ε0 g(x) = g(x) ≥ g(xi ) − ≥ = ≥ . 2 2 2 2 Thus, g(x) ≥ ε20 on (xi − δ0 , xi + δ0 ) for all i ∈ N, yielding Z xi +δ0 ε0 g ≥ 2δ0 = ε0 δ0 . 2 xi −δ0 R xi +δ0 R xi −δ0 R xi +δ0 On the other hand, −∞ g − −∞ g = xi −δ0 g. Thus, letting i → +∞ in the above, in view of the hypothesis, we obtain 0 ≥ ε0 δ0 , which is a contradiction. Therefore, lim g(x) = 0, x→+∞

which is what we want.



First solution: Suppose by way of contradiction that there exists an ε0 > 0 such that for all n ∈ N there is an xn ∈ R with |xn | > n for which |f (xn )| ≥ ε0 . As f is uniformly continuous on R, for ε20 > 0 there exists a δ0 > 0 such that f (x) − f (y) < ε0 , 2 whenever x, y ∈ R and |x − y| < δ0 . In particular, for all i ∈ N and |x − xi | < δ0 , we have f (x) − f (xi ) < ε0 . 2 But f (xi ) − f (x) ≤ f (x) − f (xi ) from which, we see that f (x) ≥ f (xi ) − ε0 ≥ |f (xi )| ≥ ε0 . (∗) 2 2 2 On the other hand, f is bounded on R. So there exists an M > 0 such that |f (x)| ≤ M for all x ∈ R. By the continuity of g on R, g is continuous on the compact

136

2. SOLUTIONS

interval [−M, M ], and hence it is uniformly continuous on [−M, M ]. Consequently, for given ε > 0, there exists a δ1 > 0 such that g(x) − g(y) < ε, (∗0 ) whenever x, y ∈ [−M, M ] and |x − y| < δ1 . It follows from the uniform continuity of f on R that for the above δ1 > 0, which was obtained from the given ε > 0, there exists a δ > 0 such that f (x) − f (y) < δ1 , whenever x, y ∈ R and |x − y| < δ. Now, for all x, y ∈ R with |x − y| < δ, we have |f (x) − f (y)| < δ1 . This together with the fact that f (x), f (y) ∈ [−M, M ] for all x, y ∈ R with |x − y| < δ, in view of (∗0 ), implies g(f (x)) − g(f (y)) < ε. That is, g ◦ f is uniformly continuous on R. On the other hand, g ◦ f is integrable on R. It thus follows from the lemma that lim g ◦ f (x) = 0,

x→−∞

lim g ◦ f (x) = 0.

x→+∞

Now, set

 ε0 ε0 m = inf g(x)|x ∈ [−M, − ] ∪ [ , M ] . 2 2 As g(x) > 0 whenever x 6= 0 and that g is continuous on the compact set [−M, − ε20 ]∪ [ ε20 , M ], we see that there exists a c ∈ [−M, − ε20 ] ∪ [ ε20 , M ] such that m = g(c), and hence m = g(c) > 0. Now, from limx→∞ g ◦ f (x) = 0, it follows that for m > 0, there exists an N > 0 such that

g f (x) = g f (x) − 0 < m, whenever x ∈ R and |x| > N . On the other hand, we have limi |xi | = +∞, and hence for the above N > 0, there exists an I >
0 such that |xi | > N whenever i ≥ I. In particular, |xI | > N
, implying g f (xI ) < m. But, by (∗), we have ε20 < f (xI ) ≤ M , yielding g f (xI ) ≥ m, a contradiction. Therefore, limx→∞ f (x) = 0, which is what we want. Second solution: Again we proceed by contradiction and continue up to (∗) and set  ε0 ε0 m = inf g(x)|x ∈ [−M, − ] ∪ [ , M ] . 2 2 As we saw in the first solution, there exists a c ∈ [−M, − ε20 ] ∪ [ ε20 , M ] such that m = g(c), and hence m = g(c) > 0. It follows from (∗) that for all i ∈ N and x ∈ R such that x ∈ (xi − δ0 , xi + δ0 ),
we have M ≥ |f (x)| ≥ ε20 , implying f (x) ∈ [−M, − ε20 ] ∪ [ ε20 , M ], whence g f (x) ≥ m > 0. Thus, we can write Z xi +δ0 g ◦ f ≥ m(xi − δ0 − xi + δ0 ) = 2δ0 m (∗00 ) xi −δ0

On the other hand, Z

xi +δ0

Z

xi +δ0

g◦f = xi −δ0

Z

xi −δ0

g◦f − −∞

g ◦ f. −∞

Thus, letting i → +∞ in (∗00 ), in view of the hypothesis, we obtain 0 ≥ 2δ0 m, which is a contradiction. Therefore, limx→∞ f (x) = 0, which is what we want. 

2.14. FOURTEENTH COMPETITION

137

3. Define the sets A and B as follows 
A := x ∈ [c, d] : f (x) = a = g −1 {a} , 
B := x ∈ [c, d] : f (x) = b = g −1 {b} , where g = f |[c,d] , which is continuous on [c, d] because so is f on [a, b]. It follows that A = g −1 ({a}) and B = g −1 ({b}) are closed, and hence compact, subsets of the compact interval [c, d]. Thus, there exist r ∈ A and s ∈ B such that  s − r = inf x − y : x ∈ A, y ∈ B . It is plain that r, s ∈ [c, d] and r 6= s. Assume, without loss of generality, that r < s. By showing that f ([r, s]) = [a, b], we settle the proof. To this end, note first that f (r) = a and f (s) = b. Next, suppose that λ ∈ [a, b] is arbitrary. We have f (r) = a ≤ λ ≤ b = f (s). From this and the Intermediate Value Theorem, we see that there exists a c ∈ [r, s] such that f (c) = λ. That is, λ ∈ f ([r, s]), implying [a, b] ⊆ f ([r, s]). Now, suppose that λ = f (c) ∈ f ([r, s]), where c ∈ [r, s], is arbitrary. We need to show that λ ∈ [a, b]. Suppose to the contrary that λ ∈ / [a, b]. This implies λ < a or λ > b. If λ < a, we can write f (c) = λ < a < b = f (s). From this, in view of the Intermediate Value Theorem, we see that there exists r1 ∈ (c, s) such that f (r1 ) = a. On the other hand, r ≤ c < r1 < s, f (r) = f (r1 ) = a, f (s) = b, and |s − r1 | < |s − r|, which is in contradiction with our choice of r and s. This implies a ≤ λ. Likewise, one can see that λ ≤ b. Thus, a ≤ λ ≤ b, yielding f ([r, s]) ⊆ [a, b] because λ ∈ f ([r, s]) was arbitrary. Therefore, f ([r, s]) = [a, b], which is what we want. 

2.14.2. Algebra. 1. First solution: As G is a finite p-group, we have |G| = pn for some n ∈ N. It now follows from Sylow’s First Theorem that G has a normal |G| G subgroup H such that |H| = pn−1 . We have |H| = p. Thus, H is a cyclic, and G hence an ableian, group because p is a prime number. Since H is abelian, we obtain G0 ⊆ H $ G, implying that G0 6= G, as desired. Second solution: Recall that a group is solvable if and only if for some n ∈ N∪{0},
0 we have G(n) = {e}, where G(0) = G and G(n+1) = G(n) (n ∈ N ∪ {0}). It is well-known that every finite p-group is solvable. With all that in mind, proceed by contradiction. Suppose G0 = G. It follows that G(n) = G for all n ∈ N ∪ {0}. Consequently, G(n) 6= {e} for all n ∈ N ∪ {0}, and hence G cannot be solvable, which is in contradiction with the fact that every finite p-group is solvable. Thus, G0 6= G, which is what we want.  2. Let p be the smallest prime dividing |G| and K a normal subgroup of G with p elements. As p is prime, it follows that K is cyclic so that K = hai for some a ∈ G with ord(a) = p. Obviously, it suffices to prove that a ∈ Z(G). To this end, suppose that g ∈ G is arbitrary. It follows that g −1 ag ∈ K because K is a normal subgroup of G. Hence, there is an integer j with 0 ≤ j ≤ p − 1 such that g −1 ag = aj . Firstly, j 6= 0 because otherwise a = e, which is impossible, for ord(a) = p. So 0 < j ≤ p−1 and we can write aj = g −1 ag =⇒ (aj )j = g −1 aj g,

138

2. SOLUTIONS 2

n

whence aj = g −2 ag 2 . By an easy induction on n, we see that aj = g −n ag n for p−1 all n ∈ N. In particular, if n = p − 1, we obtain aj = g −(p−1) ag p−1 . Now, since 0 < j ≤ p − 1 and p is prime, it follows that gcd(j, p − 1) = 1. This together with p Fermat’s Little Theorem yields j p−1 ≡ 1. That is, there exists an integer k such that j p−1 = 1 + kp. As ord(a) = p, we have (g p−1 )−1 ag p−1 = aj

p−1

= a1+kp = a(ap )k = aek = a,

which yields (g p−1 )−1 ag p−1 = a. In other words, ag p−1 = g p−1 a for all g ∈ G. To finish the proof, we need to show that the map ϕ : G → G defined by ϕ(g) = g p−1 is onto. As |G| is finite, it suffices to show that the map ϕ is one-to-one. To this end, suppose ϕ(g1 ) = ϕ(g2 ) for some g1 , g2 ∈ G. We have g1p−1 = g2p−1 and that gcd(p − 1, |G|) = 1, for p is the smallest prime that divides |G|. It follows that there exist r, s ∈ Z such that r(p − 1) + s|G| = 1. Now, noting that g |G| = e for all g ∈ G, we can write r(p−1)+s|G| (p−1)
r |G|
s (p−1)
r |G|
s r(p−1)+s|G| g1 = g1 = g1 g1 = g2 g2 = g2 = g2 . That is, g1 = g2 . In other words, ϕ : G → G is one-to-one, and hence onto, finishing the proof.  3. Recall that if R is a commutative ring and a, b ∈ R are nonzero elements of it, then the greatest common divisor of a and b in R, denoted by gcd(a, b), is a nonzero element d ∈ R satisfying the following. (i) d|a and d|b, that is, there exist k1 , k2 ∈ R such that a = k1 d and b = k2 d. (ii) If d0 |a and d0 |b for some 0 6= d0 ∈ R, then d0 |d. Also, the least common multiple of a and b, denoted by lcm(a, b), is a nonzero element c ∈ R satisfying the following. (i) a|c and b|c. (ii) If a|c0 and b|c0 for some 0 6= c0 ∈ R, then c|c0 . We claim that in Z6 , the ring of integers mod 6, the greatest common divisor of 2 and 3 is 1, whereas 2 and 3 have no common multiple. First, it is obvious that 1|2 and 1|3. Next, suppose d0 |2 and d0 |3. It follows that d0 |3 − 2 = 1. Hence, 1 is a greatest common divisor of 2 and 3. Use contradiction to prove that 2 and 3 has no common multiple. Suppose to the contrary that for 0 6= c ∈ Z6 we have 2|c and 3|c. It follows that there exist k1 , k2 ∈ Z6 such that c = 2k1 and c = 3k2 . We can write 3c = 6k1 = 0, 2c = 6k2 = 0, whence c = 3c − 2c = 0 − 0 = 0. That is, c = 0, which is impossible. This proves the claim by contradiction.  4. First, if n ∈ N and n > 1, we show that xn y + xn−1 + 1 is irreducible in F [x, y]. Suppose to the contrary that there exist f, g ∈ F [x, y] such that xn y+xn−1 +1 = f g. As degy (xn y+xn−1 +1) = 1, in view of the above equality, without loss of generality, we may assume that degy (f ) = 0 and degy (g) = 1. Consequently, there exist a noninvertible f0 ∈ F [x] and g0 , g1 ∈ F [x] such that f = f0 and g = g0 + yg1 . We can write xn y + xn−1 + 1 = f0 g0 + f0 g1 y, yielding f0 g1 = xn and f0 g0 = xn−1 + 1. As f0 is not invertible, there exists 1 < j0 < n such that f0 = xj0 and g1 = xn−j0 . From this, we obtain g0 xj0 = xn−1 + 1,

2.14. FOURTEENTH COMPETITION

139

which is impossible because j0 > 1 and n > 1. Thus, the polynomial xn y + xn−1 + 1 is irreducible in F [x, y] whenever n > 1. Next, if n = 1, we show that xy + 1 + 1 = xy + 2 is irreducible in F [x, y] if and only if ch(F ) 6= 2. To this end, note first that if ch(F ) = 2, then xy + 2 = xy is reducible in F [x, y]. Now, suppose that xy + 2 is reducible in F [x, y]. We show that ch(F ) = 2. Suppose to the contrary that ch(F ) 6= 2. It follows that 2 has an inverse in F . Since xy + 2 is reducible in F [x, y], just as we saw in the above, there exist a non-invertible f0 ∈ F [x] and g0 , g1 ∈ F [x] such that f = f0 , g = g0 + yg1 , and xy + 2 = f g. We must have xy + 2 = f0 g0 + f0 g1 y, from which, we obtain f0 g0 = 2 and f0 g1 = x. Now, since 2 is invertible in F , it follows that so is f0 in F [x], a contradiction. Thus, ch(F ) = 2, which is what we want.  5. We may assume that V1 * V2 and V2 * V1 because otherwise, in view of dim V1 = dim V2 , we see that V1 = V2 , in which case the assertion is easy to prove. More precisely, pick a basis {αi }m i=1 for V1 = V2 , where m = dim V1 , and enlarge it n to a basis {αi }m ∪ {β } for V , where m + n = dim V . It is plain that i i=1 i=1 U ⊕ V1 = U ⊕ V2 = V, where U = hβ1 , . . . , βm i is the vector subspace spanned by βi ’s (1 ≤ i ≤ n). We prove the assertion by induction on dim V − dim V1 + 1. If dim V − dim V1 + 1 = 1, then dim V1 = dim V2 = dim V . Hence, for U = {0}, we obviously have U ⊕ V1 = U ⊕ V2 = V . Assuming that the assertion holds whenever dim V − dim V1 + 1 = n, we prove it whenever dim V − dim V1 + 1 = n + 1. As V1 * V2 and V2 * V1 , we see that V1 ∪ V2 $ V , and hence there exists a vector β ∈ V \ (V1 ∪ V2 ). Set V10 = V1 ⊕ hβi, V20 = V2 ⊕ hβi. We have dim V10 = dim V20 = dim V1 + 1, yielding dim V − dim V10 + 1 = dim V − dim V1 = n. It thus follows from the induction hypothesis that there exists a subspace U1 such that U1 ⊕ V10 = U1 ⊕ V20 = V. Letting U = U1 ⊕ hβi, we obviously obtain U ⊕ V1 = U ⊕ V2 = V, proving the induction assertion, which is what we want.



2.14.3. General. 1. First solution: Let A = {A1 , . . . , An } be a family containing n distinct subsets of the set S with n elements. Define the edge-labeled graph G with vertices in A as follows. For 1 ≤ i, j ≤ n with i 6= j, connect the distinct vertices Ai and Aj with an edge labeled s ∈ S whenever Ai = Aj ∪ {s} or Aj = Ai ∪ {s}. Now, define the relation ∼ on A as follows. For 1 ≤ i, j ≤ n, we write Ai ∼ Aj whenever i = j or there exists a path in the graph G joining the vertices Ai and Aj of the graph. It is easily verified that ∼ is an equivalence relation on A and that the equivalence classes of the relation ∼ are connected components of the graph G. Let {Gi }ki=1 (1 ≤ k ≤ n) denote the connected components of G. Also, let Ti (1 ≤ i ≤ k) be a maximal tree of Gi (recall that a tree is connected graph not having any cycle). That is, Ti is a subgraph of Gi which is a tree and that

140

2. SOLUTIONS

it is maximal as a tree in Gi (from this, we immediately see that Ti is a spanning tree of Gi , i.e., Ti a subgraph of Gi that contains all the vertices of Gi and is a tree). For a subgraph H of G, use L(H) to denote the set of labels of the edges of H. We claim that L(Ti ) = L(Gi ) for all 1 ≤ i ≤ k. Obviously, it suffices to show that L(Gi ) ⊆ L(Ti ). To this end, let s ∈ L(Gi ) be arbitrary. It follows that s ∈ S is the label of an edge Ai Aj of the subgraph Gi . If the edge Ai Aj is already in the tree Ti , there is nothing to prove. If not, as Ti is a maximal tree of Gi , we see that the graph obtain by adding the edge Ai Aj to the tree Ti is not a tree. It thus follows that the edge Ai Aj participates in a cycle whose all edges but the edge Ai Aj come from the tree Ti . Now, s ∈ S being the label of Ai Aj , we see that Ai = Aj ∪ {s} or Aj = Ai ∪ {s}. By symmetry, we may assume without loss of generality, that Ai = Aj ∪ {s}. Note that s ∈ / Aj , for i 6= j. Now, going from Aj to Ai along the other edges of the cycle, in which the edge Ai Aj participates, and noting that s ∈ / Aj but s ∈ Ai , we see that there must be an edge Ak Al ∈ Ti , where 1 ≤ k, l ≤ n, having s as its label, because otherwise s ∈ / Ai , which is impossible. Thus, s ∈ L(Ti ), whence L(Gi ) ⊆ L(Ti ), implying L(Gi ) = L(Ti ), as desired. It is plain that L(Ti ) ≤ ε(Ti ), where ε(Ti ) denotes the number of the edges of Ti . On the other hand, since Ti is a tree, by a standard theorem from the theory of graphs, we have ε(Ti ) = ν(Ti ) − 1, where ν(Ti ) denotes the number of vertices of Ti . The number k being the number of connected components of G, we can write L(G) =

k X

L(Gi ) ≤

i=1

k X

ε(Ti ) =

i=1

k X

ν(Ti ) − k,

i=1

implying L(G) ≤

k X

ν(Ti ) − k = ν(G) − k,

i=1

where ν(G) denotes the number of vertices of G. This implies L(G) ≤ ν(G) − k < ν(G). Thus, there exists an s0 ∈ S such that s0 ∈ / L(G). It is now plain that the sets A1 ∪ {s0 }, . . . , An ∪ {s0 } are distinct, for otherwise we must have Ai = Aj ∪ {s0 } or Aj = Ai ∪ {s0 } for some 1 ≤ i, j ≤ n with i 6= j, yielding s0 ∈ L(G), which is impossible. This completes the proof of the assertion. Second solution: This solution is taken from Linear Algebra Methods in Combinatorics by L. Babai and P. Frankl. Without loss of generality, assume that S = {1, . . . , n}. It suffices to show that there exists an element x ∈ S such that the sets A1 \ {x}, . . . , An \ {x} are all distinct because so will then be the sets A1 ∪ {x}, . . . , An ∪ {x}. To see this, just note that for some x ∈ S and 1 ≤ i, j ≤ n, Ai ∪ {x} = Aj ∪ {x} if and only if Ai \ {x} = Aj \ {x}. Define the n × n matrix M = (mij ) with

2.14. FOURTEENTH COMPETITION

141

mij ∈ {0, 1} (1 ≤ i, j ≤ n) as follows  mij =

1 0

j ∈ Ai , j∈ / Ai .

By the hypothesis, the rows of M are all distinct. We need to show that this remains to be the case after omitting an appropriate column from M . There are two cases to consider. (i) det M = 0; and (ii) det M 6= 0. If det M = 0, there is a column of M , say, column j for some 1 ≤ j ≤ n, which is linearly dependent on the other columns of M . We see that after deleting column j of M , the remaining rows are all distinct. Suppose on the contrary that this is not the case. That is, the matrix obtained from deleting column j of M has two equal rows, say rows i1 and i2 for some 1 ≤ i1 < i2 ≤ n. It follows that rows i1 and i2 of M are equal as well because column j of M depends linearly on the other columns of M . This is a contradiction, proving the assertion in this case. If det M 6= 0, if necessary by interchanging two rows of M , we may assume that the first row of M is a row with the minimum number of ones. Note that there might be several rows with the minimum number of ones. Expanding det M by the first row, we see that m1j det M1j 6= 0, where M1j is the matrix obtained by eliminating row 1 and column j from the matrix M . Consequently, m1j = 1 and no two rows of M1j are the same. It thus follows that deleting column j leaves no two equal rows. Suppose on the contrary that two rows of the matrix obtained from deleting column j of M are equal. Since no two rows of M1j are the same, we see that for some 1 < i ≤ n, row 1 and row i of the matrix obtained from deleting column j of M must be equal. This easily implies that mij = 0, for otherwise rows 1 and i of M would be equal, which is impossible. Thus, the number of ones of row i of M is less than that of row 1, which is a contradiction. Therefore, no two rows of the matrix obtained from deleting column j of M are equal, which is what we want.  2. We prove that the answer is 14 days. It is obvious that the maximum number of the exhibition days is attained provided that there is exactly one common book for any two days of the exhibition. With that in mind, first, we obtain the number of books needed to exhibit the books for k days in such a way that 100 books are to be exhibited in each day and that the exhibited books of any two days have exactly one book in common. Then, we find the maximum number of the days subject to the condition that the number of the books is less than or equal to 1369. In doing so, suppose that we would like to run the exhibition for k days. First, choose k − 1 books (b1i )ki=2 for the first day of the exhibition and set aside the book b1j to be exhibited for a second time in the jth day. Then, choose k − 2 books (b2i )ki=3 , from the books not already chosen, for the second day and set aside the book b2j (3 ≤ j ≤ k) to be exhibited for a second time in the jth day. Then, choose k − 3 books (b3i )ki=4 , from the books not already chosen, for the third day and set aside the book b3j (4 ≤ j ≤ k) to be exhibited for a second time in the jth day. Continue this way to finally choose one book bk−1 , from the books not already chosen, for k the (k − 1)st day and set aside the book bk−1 to be exhibited for a second time in k the kth day. So far, for each day we have k − 1 books to be exhibited and that the exhibited books of any two days, say day i and day j where i < j, have exactly one book in common, namely, the book bij . In order to exhibit 100 books per day, for each day we need another 100 − (k − 1) = 101 − k books. So the number of the

142

2. SOLUTIONS

books needed is (k − 1) + (k − 2) + · · · + 2 + 1 + k(101 − k)

k(k − 1) + k(101 − k) 2 k(201 − k) . 2

= =

Thus, for k = 14, the number of the books needed is 14(201−14) = 1309 and for 2 = 1395. Since we have only k = 15 the number of the books needed is 15(201−15) 2 1369 books, it follows that the maximum number of the days of the exhibition is 14, finishing the proof.  3. The assertion is a quick consequence of the following lemma. Lemma. Let F be a field and with ch(F ) = 0 or > n, K an extension of F , and {λ1 , . . . , λn } ⊆ K. Also, let m ∈ N ∪ {0} and c ∈ F . If m λk1 + · · · + λkn = ck−m (λm 1 + · · · + λn ),

for each k = m, m + 1, . . . , m + n, then λi = 0 or c for all 1 ≤ i ≤ n. Proof. We prove the assertion by induction on n. If n = 1, the proof is obvious. Suppose that the assertion holds for all k < n. Let {λ1 , . . . , λn } ⊆ K be a given subset as described in the lemma. There are two cases to consider. m (i) First, suppose c = 0 or λm 1 + · · · + λn = 0. Then λk1 + · · · + λkn = 0, for each k = m + 1, . . . , m + n. We claim that λi = 0 for all 1 ≤ i ≤ n. Suppose, by contradiction, that this is not the case. Let µi ∈ K (1 ≤ i ≤ l) be nonzero and distinct such that {µ1 , . . . , µl } ∪ {0} = {λ1 , . . . , λn } ∪ {0}. It follows that there exist n1 , . . . , nl ∈ N with n1 + · · · + nl ≤ n such that n1 µk1 + · · · + nl µkl = λk1 + · · · + λkn = 0, for each k = m + 1, . . . , m + n. In other words,  m+1    n1 µ1 · · · µm+1 l   ..    .. .. ..   .  =  . . . m+n nl µm+n · · · µ 1 l

 0 ..  , .  0

implying that   µm+1 · · · µm+1 det  1 l

1 .. . µn−1 1

··· .. . ···

1 .. . µn−1 l

   = 0.

This in turn, in view of Vandermonde’s determinant formula, implies µi = 0 for some 1 ≤ i ≤ l or µi = µj for some 1 ≤ i < j ≤ l, a contradiction in any event. Thus, λi = 0 for all 1 ≤ i ≤ n, which is what we want. m (ii) Next, suppose c 6= 0, and hence λm 1 + · · · + λn 6= 0. For each k = 1, . . . , n, let Sk denote the elementary symmetric polynomial in λ1 , . . . , λn of degree k, i.e., S1 = λ1 + · · · + λn , . . . , Sn = λ1 · · · λn . Obviously, we can write xn − S1 xn−1 + · · · + (−1)n Sn = (x − λ1 ) · · · (x − λn ), from which, we obtain λn+m − S1 λn+m−1 + · · · + (−1)n Sn λm = 0, i i i

(∗)

2.15. FIFTEENTH COMPETITION

143

for all 1 ≤ i ≤ n. Adding up these equations, we get n X

λn+m − S1 i

i=1

n X

λn+m−1 + · · · + (−1)n Sn i

i=1

n X

λm = 0. i

i=1

It now follows from the hypothesis that cn

n X

n−1 λm i − S1 c

i=1

=

n X

n X

n λm i + · · · + (−1) Sn

i=1

λn+m − S1 i

i=1

n X

n X

λm i

i=1 n X

λn+m−1 + · · · + (−1)n Sn i

i=1

λm i = 0.

i=1

Thus, n X


n
λm c − S1 cn−1 + · · · + (−1)n Sn = 0, i

i=1

yielding cn − S1 cn−1 + · · · + (−1)n Sn = 0. This together with (∗) implies that c = λi for some 1 ≤ i ≤ n. It is now plain that the induction hypothesis can be applied to the set {λ1 , . . . , λn } \ {c}, completing the proof.  Now, to prove the assertion, just let F = K = R and c = m = 1 in the lemma above. 

2.15. Fifteenth Competition 2.15.1. Analysis. 1. (a) Letting 0 ≤ x < y, we can write   Z x+y Z x Z x+y Z x+y 2 2 2 x+y = 2 f= f+ f. f+ 2ϕ 2 0 0 x 0 y−x is positive, we obtain 2 Z x+y Z y 2 y − x
≤ f t+ dt = f. x+y 2 x 2

As f is increasing on [0, +∞) and Z

x+y 2

f x

So, we have  2ϕ

x+y 2



Z

x

=

Z

x+y 2

f+ 0

Z

0 x

≤

Z

Z =

x+y 2

Z f+

0

y

f x+y 2

y

f 0

= ϕ(x) + ϕ(y). This proves (a).

Z f+

0 x

f x

f+ 0

x+y 2

Z f+

144

2. SOLUTIONS

(b) Note first that f is integrable on any closed and bounded subinterval of [0, +∞) because f is increasing. Consequently, ϕ is continuous on [0, +∞). On the other hand, it easily follows from (a) that
ϕ λx + (1 − λ)y ≤ λϕ(x) + (1 − λ)ϕ(y), (∗)  k k n where λ = 2k , 1 ≤ k ≤ 2 , and k, n ∈ N. Now, as the set 2k : k, n ∈ N, 1 ≤ k ≤ n 2 is dense in [0, 1], in view of the continuity of ϕ, we see that (∗) holds for all λ ∈ [0, 1]. In other words, ϕ is convex, which is what we want.  2. It is worth mentioning that the hypothesis g(0) = 0 is redundant. The function g is continuous on the closed interval [0, 1], so it is bounded on [0, 1]. Hence, there exists an M > 0 such that |g(x)| ≤ M for all x ∈ [0, 1]. Noting that 0 ≤ sin x ≤ 1 for all x ∈ [0, 1], we can write g(x)(sin x)n M sin x M ≤ |fn (x)| = ≤ . 1 + nx 1 + nx n From this, we conclude that the sequence (fn )+∞ n=1 uniformly converges to the zero function on [0, 1]. So the assertion follows.  3. (a) For all x, z ∈ (0, +∞), we have 2xz ≤ xf (x) + zf −1 (z). Letting z = f (y), we obtain 2xf (y) ≤ xf (x) + yf (y), for all x, y ∈ (0, +∞). We can write y−x f (y), x for all x, y ∈ (0, +∞). Interchanging x, y in the above, we obtain x−y y−x f (x) − f (y) ≤ f (x) =⇒ f (y) − f (x) ≥ f (x) y y xf (y) − xf (x) ≤ yf (y) − xf (y) =⇒ f (y) − f (x) ≤

for all x, y ∈ (0, +∞). This together with the preceding inequality proves (a). (b) First, we prove that f is increasing on (0, +∞). To this end, we have 2xf (y) ≤ xf (x) + yf (y), 2yf (x) ≤ yf (y) + xf (x), yielding

2 xf (y) + yf (x) ≤ 2 xf (x) + yf (y) , for all x, y ∈ (0, +∞). Consequently,
(y − x) f (y) − f (x) ≥ 0, for all x, y ∈ (0, +∞). This shows that f is increasing on (0, +∞). It thus follows that f has limit from the left and limit from the right at any point and that the two limits at any point coincide because f is increasing and onto. Thus, f is continuous on (0, +∞). From (a), we have y−x y−x f (x) ≤ f (y) − f (x) ≤ f (y), y x

2.15. FIFTEENTH COMPETITION

145

for all x, y ∈ (0, +∞). Dividing by y − x where x, y ∈ (0, +∞) with x < y, we can write f (x) f (y) − f (x) f (y) ≤ ≤ . y y−x x Letting y → x+ in the above, we obtain f (x) , x where f 0 (x+ ) denotes the right derivative of f at x. Likewise, dividing by y − x where x, y ∈ (0, +∞) with x > y, we can write f 0 (x+ ) =

f (x) f (y) − f (x) f (y) ≥ ≥ , y y−x x from which by letting y → x− , we obtain f (x) , x where f 0 (x− ) denotes the left derivative of f at x. That is, f 0 (x) exists for all f (x) 0 0 x ∈(0, +∞)  and moreover f (x) = x . This implies xf (x) − f (x) = 0 and hence f 0 (x− ) =

d dx

f (x) x

such that 

xf 0 (x)−f (x) = 0 for all x ∈ (0, +∞). Therefore, there x2 f (x) x = c, yielding f (x) = cx for all x ∈ (0, +∞), which is

=

exists a c ∈ R

what we want.

 2.15.2. Algebra. 1. Let Ω = P1 , . . . , Pp+1 denote the set of all (distinct) Sylow p-subgroups of G. The group G acts on Ω by conjugation. The kernel of the action is H :=



\ p+1 g ∈ G| ∀i = 1, . . . , p + 1 : g −1 Pi g = Pi = NG (Pi ), i=1

where NG (Pi ) denotes the normalizer of Pi in G. Recall that for a subset S of G, the normalizer of S in G, denoted by NG (S), is defined by  NG (S) = g ∈ G : g −1 Sg = S . We note that by Sylow’s Second Theorem H is a normal subgroup of G. From G the First Isomorphism Theorem for groups, we see that the group H is 2isomorphic to a subgroup of the symmetric group Sp+1 , and since p (p + 1)! but p - p + 1)!, it follows that pα |H| or pα−1 |H|. Note that the subgroup H ∩ Pi is a Sylow HPi H . Now, as H ∩ Pi p-subgroup of H for each i = 1, . . . , p + 1, for = Pi H∩Pi and H ∩ Pj are Sylow p-subgroups of H, by Sylow’s Second Theorem, they are conjugate in H, and hence there exists an h ∈ H such that h−1 (H ∩ Pi )h = H ∩ Pj . On the other hand, h ∈ H, implies h ∈ NG (Pi ) and hence H ∩ Pi = (h−1 Hh) ∩ (h−1 Pi h) = h−1 (H ∩ Pi )h = H ∩ Pj . That is, H ∩ Pi = H ∩ Pj ,

146

2. SOLUTIONS

for all 1 ≤ i, j ≤ p + 1. From this, we obtain H ∩ Pi ⊆ Pj for all 1 ≤ i, j ≤ p + 1, yielding p+1 \ H ∩ Pi ⊆ Pj , j=1

for all 1 ≤ i ≤ p + 1. On the other hand, p+1 \

Pi ⊆ H ∩ Pi ,

i=1

for all 1 ≤ i ≤ p + 1. Therefore, p+1 \

Pj = H ∩ Pi ,

j=1

Tp+1 for all 1 ≤ i ≤ p + 1. This implies that j=1 Pj is a Sylow p-subgroup of H. Now, Tp+1 Tp+1 if pα |H|, then j=1 Pj = pα = |Pi |, and hence j=1 Pj = Pi , implying that Pi = Pj , a contradiction. Thus, pα−1 |H|. Consequently, a Sylow p-subgroup of Tp+1 H must be of order pα−1 , and hence | i=1 Pi | = pα−1 , which is what we want.  2. First, we claim that if P is a prime ideal of R and ab ∈ P for some a, b ∈ R, then a ∈ P or b ∈ P . To see this, from ab ∈ P , we get Rab ⊆ P . Now, as the left ideal Ra is also a right ideal of R, we see that RaR ⊆ Ra, implying RaRb ⊆ Rab. This yields RaRb ⊆ P , from which, we obtain Ra ⊆ P or Rb ⊆ P , for P is a prime ideal of R. This, in turn, implies a ∈ P or b ∈ P because R is unital. Now to prove the assertion, first, suppose that x ∈ R is nilpotent. So there exists an n ∈ N such that xn = 0. We have xn = 0 ∈ P for all prime ideals of R. In viewTofthe above claim, we see that x ∈ P for all prime ideals of R. In other words, x∈ P : P C R, P is prime . Conversely, suppose that x ∈ R and that xn 6= 0 for all n ∈ N. In other words, for the multiplicative set S := {xn : n ∈ N}, we have S ∩ {0} = ∅. A standard argument using Zorn’s Lemma shows that there exists a prime ideal P such / P . In other words, we have T that P ∩ S = ∅. This yields x ∈ proved that if x ∈ P : P C R, P is prime , then x must be nilpotent. Therefore, the intersection of all prime ideals of R is equal to the set of the nilpotent elements of R, which is what we want.  3. If we do not require the matrix AB to be a nonzero idempotent, then the assertion is trivial. Just let B = 0. Then, AB = 0 is an idempotent, settling the proof. However, we state and prove the following nontrivial problem. Let D be a division ring, n ∈ N, and A ∈ Mn (D). Then, there exists a matrix B ∈ Mn (D) such that AB is an idempotent whose rank is equal to that of A. Let Dn denote the right vector space of all n × 1 column vectors with entries in D; that is, the addition x + y is defined componentwise and the multiplication of the scalar λ ∈ D into the vector x = (xi )ni=1 ∈ Dn is defined by xλ := (xi λ)ni=1 . The members of Mn (D) can be viewed as linear transformations acting on the left of Dn via the usual matrix multiplication; that is, we can write Mn (D) = L(Dn ), where L(Dn ) is the ring of all right linear transformations acting on the left of Dn . For x ∈ Dn and f ∈ (Dn )0 , where (Dn )0 = Dn denotes the dual of Dn which is the left vector space of all 1 × n row vectors with entries in D, define the rank-one linear transformation x ⊗ f ∈ L(Dn ) by (x ⊗ f )(y) := xf (y). Choose

2.15. FIFTEENTH COMPETITION

147

yi ∈ Dn (1 ≤ i ≤ r) such that {Ayi }1≤i≤r is a basis for the range of A. Set xi := Ai yi and enlarge {xi }1≤i≤r to a basis B ∪ {xi }1≤i≤r for Dn , where the set B is linearly independent. Now, let {fi }1≤i≤r be a dual subset with respect to B ∪ {xi }1≤i≤r so that hBi ⊆ ker fi and fi (xj ) = δij for each i, j = 1, . . . , r. Let B = y1 ⊗f1 +· · ·+yr ⊗fr . Since hBi ⊆ ker fi and fi (xj ) = δij for each i, j = 1, . . . , r, we easily see that AB = A(y1 ⊗ f1 + · · · + yr ⊗ fr ) = x1 ⊗ f1 + · · · + xr ⊗ fr is an idempotent whose rank is r = rank(A), proving the assertion. 

2.15.3. General. 1. We prove that a necessary and sufficient condition for the product of two integers a, b to be divisible by their sum is that there exist integers x, y, z such that x, y are relatively prime and that a = x(x + y)z, b = y(x + y)z. Sufficiency is easy. To prove necessity, from a + b ab, we see that ab = m(a + b), for some m ∈ Z. If one of a or b is zero, say, a = 0, then letting x = 0, y = 1, z = b proves the assertion. So without loss of generality, assume that a and b are nonzero. Set d = gcd(a, b) and b a , y= . d d It is obvious that x and y are relatively prime. From ab = m(a + b), we obtain dxy = m(x + y), implying x + y dxy. But xy and x + y are relatively prime because so are x and y. Thus, x + y d, and hence there exists an integer z such that d = (x + y)z. Form this, we obtain x=

a = x(x + y)z, b = y(x + y)z, which is what we want. R1 2. Set β = 0 xf (x)dx. We prove that Z 1 Z 1 2 (x − β) f (x)dx ≤ (x − α)2 f (x)dx, 0



(∗)

0

R1 R1 for all α ∈ R. To this end, using 0 f (x)dx = 1 and 0 xf (x)dx = β, we can write Z 1 Z 1
2 (x − α)2 f (x)dx = (x − β) + (β − α) f (x)dx 0

0

Z

1

(x − β)2 f (x)dx + (β − α)2 + 2(β − α)β − 2(β − α)β

= 0

Z

1

(x − β)2 f (x)dx + (β − α)2

= 0

Z ≥

1

(x − β)2 f (x)dx.

0

As (x − 12 )2 ≤

1 4

on [0, 1], we can write Z 1 Z 1 1
2 1 1 x− f (x)dx ≤ f (x)dx = . 2 4 4 0 0

148

2. SOLUTIONS

Now, in view of (∗), we have Z 1 Z (x − β)2 f (x)dx ≤ 0

0

1

x−

1
2 1 f (x)dx ≤ , 2 4

which is what we want.



3. (a) The desired probability is the ratio of the number of “favorable cases” to the number of “total cases”. The number of “total cases” is obviously np . To calculate the number of “favorable cases”, let Ai be the event that no one gets on the ith wagon. It follows that the number of “disfavorable cases” is equal to |A1 ∪ · · · ∪ An |. Using the inclusion-exclusion principle from combinatorics, we can write X X |A1 ∪ · · · ∪ An | = |Ai | − |Ai ∩ Aj | + · · · + (−1)n−1 |A1 ∩ · · · ∩ An | i

=

X i

i,j p

(n − 1) −

X

(n − 2)p + · · · + (−1)n−1 0p

i,j

      n n n 1p . (n − 1)p − (n − 2)p + · · · + (−1)n−2 = 2 n−1 1 Thus, the number of “favorable cases” is equal to       n n n p p p n−1 n − (n − 1) + (n − 2) + · · · + (−1) , 1 2 n−1 and hence the desired probability is        n n n 1 p p p n−1 n − (n − 1) + (n − 2) + · · · + (−1) , np 1 2 n−1 which is what we want. (b) First, if p < n, the probability calculated in (a) is obviously
zero,
implying n that the number of “favorable cases” is zero. This, in view of nr = n−r , yields         n p n p n p n−1 n 1 − 2 + 3 − · · · + (−1) np = 0. 1 2 3 n Next, if p = n, then the number of wagons is equal to that of the passengers. In this case, in order not to have any empty wagon, we must have one passenger in each wagon. Thus, the number of “favorable cases” is equal n!. On the other hand, if we let p = n in the above formula, which we obtained for the number of “favorable cases”, and equate these two values, we obtain         p p p p p p p p 1 − 2 + 3 − · · · + (−1)p−1 p = (−1)p−1 p!, 1 2 3 p finishing the proof.



2.16. SIXTEENTH COMPETITION

149

2.16. Sixteenth Competition 2.16.1. Analysis. 1. As g is continuous on the compact interval [0, 1], it follows that there exists an M > 0 such that g(x) ≤ M for all x ∈ [0, 1]. From the continuity of g at 1 and g(1) = 0, we see that for given ε > 0, there exists a 0 < δ < 1 such that g(x) < ε, whenever x ∈ [0, 1] and 0 < 1 − x < δ. On the other hand, for given ε > 0, there exists an N ∈ N such that (1 − δ)n M < ε for all n ≥ N . It thus follows that for a given ε > 0, for all x ∈ [0, 1] and n ≥ N , we have   fn (x) − 0 = xn g(x) ≤ max (1 − δ)n M, sup g(x) < ε, 1−δ≤x≤1

proving the assertion.



2. (a) Let x = [x] + 0.a1 a2 a3 . . .. We have x + 0.1 = [x] + a0 .a01 a2 a3 . . . , where a0 = 0 or 1 and a01 ∈ {0, 1, . . . , 9} and hence f (x) = f (x + 0.1) = a2 . Hence, 0.1 is a period for f . To prove that 0.1 is the period of f , suppose to the contrary that 0 < α < 0.1 is a period of f . If we let x1 = 0.1 − (0.1)α, we have 0.09 < x1 < 0.1. Consequently, x1 = 0.09α1 α2 . . ., where all of αi ’s are not equal to 9. It follows that f (x1 ) = 9. On the other hand, if we let x2 = α+x1 = 0.1+(0.9)α, we have 0.0999 . . . = 0.1 < x2 < 0.19 = 0.18999 . . . . Thus, x2 = 0.1β1 β2 . . ., where β1 ≤ 8. Hence, f (x2 ) = β2 ≤ 8, whence f (x1 ) 6= f (x2 ), which is a contradiction. This shows that α cannot be the period of f . (b) Using integration by parts, we can write Z c Z 0.1 Z xdf (x) = xdf (x) = 0.1f (0.1) − 0

0

=

(0.1).9 −

0.1

f (x)dx

0 9 X k=0

k

1 = 0.45, 100

which is what we want.



3. We prove the assertion in the normed linear space setting. Let X be a real or complex normed linear space and f : X → X a uniformly continuous function. Then, there exists a, b ∈ R+ such that ||f (x)|| ≤ a||x|| + b, for all x ∈ X, where ||.|| denotes the norm of X. As f is uniformly continuous on X, for ε = 1, there exists δ0 > 0 such that ||f (x) − f (y)|| < 1 whenever x, y ∈ Rn and ||x−y|| ≤ δ0 . We show that for a = δ10 and b = |f (0)|+1, we have ||f (x)|| ≤ a||x||+b

150

2. SOLUTIONS

for all x ∈ X. To this end, for x ∈ X, set N = integer part function. We can write

h

kxk δ0

i

, where bracket stands for the

||f (x)||

≤ ||f (0)|| + ||f (x) − f (0)||
≤ ||f 0 ||



N X


x
x
x




+

f kδ0 ||x|| − f (k − 1)δ0 ||x|| + f x − f N δ0 ||x|| . k=1 h i ||x|| From N = δ0 , we obtain N≤

kxk < N + 1, δ0

yielding



x − N δ0 x <

||x||



kδ0 x − (k − 1)δ0 x =

||x|| ||x||

kxk < δ0 , N +1

δ0

kxk x = δ0 ≤ δ0 ,

for all 1 ≤ k ≤ N . So, for all x ∈ X, we can write kf (x)k ≤ kf (0)k + N + 1 ≤

1 kxk + (kf (0)k + 1), δ0

whence ||f (x)|| ≤ a||x|| + b, for all x ∈ X, where a, b are as in the above, which is what we want.



G , where Z(G) denotes 2.16.2. Algebra. 1. First, recall that Inn(G) ∼ = Z(G) the center of the group G. To see this, it is easily checked that the map f : G −→ Aut(G) defined by fg (x) = gxg −1 is a homomorphism of groups and that ker f = Z(G). It thus follows from the First Isomorphism Theorem for groups G that Inn(G) ∼ . Next, we need to recall that if p is a prime, then every = Z(G) 2 group G of order p is abelian. To prove this by contradiction, suppose that G is not abelian. Note first that |Z(G)| > 1 because 3 GG is a p-group (see Solution = p, implying that G is of 2.11.2). Consequently, |Z(G)| = p. But then Z(G) Z(G) cyclic and hence G is abelian, a contradiction. Thus, G is abelian, as desired. We now prove the assertion. To this end, from the hypothesis that [G : A] = [G : B] = p, it follows that |A| = |B|. And since p is the smallest prime that divides |G|, from Problem 1 of 2.9.2, we see that A and B are both normal in G. It thus follows that AB ≤ G. By the Second Isomorphism Theorem for groups, we have B AB A = A∩B . So we can write

[G : AB] =

|G| [G : A] p = = . |AB| [B : A ∩ B] [B : A ∩ B]

If [B : A∩B] = 1, we obtain A = B, which is a contradiction. Thus, [B : A∩B] = p, and hence G = AB. We now prove that A ∩ B = Z(G). To this end, for x ∈ A ∩ B,

2.16. SIXTEENTH COMPETITION

151

as is usual, use CG (x) to denote the centralizer of the element x of G. As A and B are abelian, we have A, B ≤ CG (x), and hence p = [G : A] = [G : CG (x)][CG (x) : A], p = [G : B] = [G : CG (x)][CG (x) : B], from which, we see that CG (x) = G. That is, x ∈ Z(G), and hence A ∩ B ≤ Z(G). Now, as [G : A ∩ B] = [G : B][B : A ∩ B] = [G : B][AB : A] = [G : B][G : A] = p2 , we conclude that [G : Z(G)] = 1 or p or p2 because [G : Z(G)] [G : A ∩ B]. As G is nonabelian, the cases [G : Z(G)] = 1 or p are impossible. Thus, [G : Z(G)] = p2 , from which, in : A ∩ B] = p2 , we see that A ∩ B = Z(G). Consequently, Gview of [G 2 Inn(G) = Z(G) = p , and hence Inn(G) is abelian. This together with the G ∼ Fundamental Theorem of finite abelian groups implies Inn(G) ∼ = Z(G) = Zp2 G ∼ ∼ or Zp ⊕ Zp . From Inn(G) = Z(G) = Zp2 , it follows that G is abelian, which is impossible. Therefore, Inn(G) ∼  = Zp ⊕ Zp , which is what we want. 2. There exists an n ∈ N such that (r2 − r)n = 0, for r2 − r is nilpotent. It is plain that there exists a g ∈ Z[x] such that 0 = (r2 − r)n = rn − rn+1 g(r), whence rn = rn+1 g(r). Setting f (x) = xn g(x)n , we have f (r)2 = r2n g(r)2n , and hence we can write rn

= rn+1 g(r) = rg(r)rn = rg(r)rn+1 g(r) = rn+2 g(r)2 = r2 g(r)2 rn = r2 g(r)2 rn+1 g(r) = rn+3 g(r)3 = · · · = rn+n g(r)n = r2n g(r)n .

That is, rn = r2n g(r)n , yielding f (r)2 = r2n g(r)2n = r2n g(r)n g(r)n = rn g(r)n = f (r). If f (r) = 0, then 0 = rn f (r) = rn rn g(r)n = r2n g(r)n = rn , implying rn = 0, which contradicts the hypothesis that r is not nilpotent. Therefore, f (r) is a nonzero idempotent element of R, which is what we want.  3. Remark. Adjusting the proof below one can prove that
det lcm(i, j) = n!f (1) · · · f (n), P where f (n) = d|n dµ(d) and µ denotes the M¨obius function. Hint. Use lcm(i, j) = ij gcd(i,j) . We show that the matrix A is invertible. To this end, define the matrices B = (bij ) and C = (cij ) in Mn (Q) as follows   ϕ(i) i=j 1 j|i bij = , cij = , 0 i 6= j 0 j-i

152

2. SOLUTIONS

where ϕ is the Euler’s totient function. Let C t denote the transpose of C. We have n n n X X X (CBC t )ij = cik (BC t )kj = cik bkk0 cjk0 =

k=1 n X

k0 =1

k=1

X

cik ϕ(k)cjk =

k=1

X

ϕ(k)cik cjk =

k|i, k|j

ϕ(k)

k|gcd(i,j)

=

gcd(i, j) = aij = (A)ij , P for all 1 ≤ i, j ≤ n (for a proof of d|n φ(d) = n for all n ∈ N, see Solution 1 of 2.2.2). Consequently, A = CBC t . So we can write det A = =

det CBC t = det C det B det C t 1. det B.1 = det B = ϕ(1)ϕ(2) · · · ϕ(n),

implying that det A 6= 0. Thus, A is invertible, which is what we want.



2.17. Seventeenth Competition 2.17.1. 1. To prove the assertion by contradiction, suppose that Analysis. (a) for all x ∈ (a, b). There are two cases to consider. |f 0 (x)| ≤ f (b)−f b−a (i)

f (b)−f (a) b−a

≥ 0.

(a) Define the function g : [a, b] → R by g(x) = f (x) − f (b)−f (x − a). Now, from b−a the contradiction hypothesis, we see that

g 0 (x) = f 0 (x) −

f (b) − f (a) ≤ 0, b−a

for all x ∈ (a, b). Thus, g is non-increasing on [a, b]. On the other hand, we have g(a) = f (a) = g(b), and hence f (a) = g(b) ≤ g(x) ≤ g(a) = f (a) for all x ∈ [a, b]. Therefore, g(x) = f (a) for all x ∈ [a, b]. In other words, f (x) = (a) f (a) + f (b)−f (x − a) for all x ∈ [a, b]. That is, the graph of f is a line segment, b−a which is a contradiction. This proves the assertion in this case. (a) ≤ 0. (ii) f (b)−f b−a Adjusting the above proof or replacing f by −f and repeating the above argument, one can prove the assertion in this case as well.  2. Let Z

x

p(x) =

Z

x

Z

−1

x

p2 p 4 −

p1 p3 −1

Z

x

p1 p4 −1

p 2 p3 . −1

It is plain that p ∈ R[x] and p(−1) = 0. Form this, it follows that p is divisible by x + 1. We can write Z x Z x p0 (x) = p1 (x)p3 (x) p2 p4 + p2 (x)p4 (x) p1 p3 −1 −1 Z x Z x −p1 (x)p4 (x) p2 p3 − p2 (x)p3 (x) p1 p4 , −1

−1

2.17. SEVENTEENTH COMPETITION

153

from which, we obtain p0 (−1) = 0. Thus, p(x) is divisible by (x + 1)2 . Taking derivative of the both sides of the above equality yields Z x Z x p00 (x) = (p1 p3 )0 (x) p2 p4 + (p1 p3 p2 p4 )(x) + (p2 p4 )0 (x) p1 p3 −1 −1 Z x p2 p3 − (p1 p4 p2 p3 )(x) +(p2 p4 p1 p3 )(x) − (p1 p4 )0 (x) −1 Z x +(p2 p3 )0 (x) p1 p4 − (p2 p3 p1 p4 )(x), −1

implying p00 (−1) = 0, and hence p(x) is divisible by (x + 1)3 . Finally, we can write Z x Z x
000 00 0 00 p (x) = (p1 p3 ) (x) p2 p4 + (p1 p3 ) p2 p4 (x) + (p2 p4 ) (x) p 1 p3 −1 −1 Z x

p2 p3 − (p1 p4 )0 p2 p3 (x) + (p2 p4 )0 p1 p3 (x) − (p1 p4 )00 (x) −1 Z x
−(p2 p3 )00 (x) p1 p4 − (p2 p3 )0 p1 p4 (x), −1

which obtains p000 (−1)

= =



(p1 p3 )0 p2 p4 + (p2 p4 )0 p1 p3 (−1) − (p1 p4 )0 p2 p3 + (p2 p3 )0 p1 p4 (−1) (p1 p2 p3 p4 )0 (−1) − (p1 p2 p3 p4 )0 (−1) = 0.

Therefore, p(x) is divisible by (x + 1)4 , which is what we want.



3. “ =⇒” Suppose that Z(f ) := {x ∈ X : f (x) = 0} = f −1 ({0}) is an open subset of X. It follows that Z(f ) is both open and close, and hence so is Z(f )c = X \Z(f ). Now, since X = Z(f ) ∪ Z(f )c and Z(f ) and Z(f )c are both open sets, the function g : X → R defined by  1 x ∈ Z(f )c f (x) g(x) = 0 x ∈ Z(f ) is continuous on X and furthermore satisfies f = gf 2 on X. This is what we want. “ ⇐=” Suppose that there exists a continuous function g : X → R such that f = gf 2 . To show that Z(f ) := {x ∈ X : f (x) = 0} = f −1 ({0}) is an open subset of X, we prove that Z(f )c := X \ Z(f ) is closed. It suffices to show that for any c c sequence (xn )+∞ n=1 , with xn ∈ Z(f ) , we have x∞ = limn xn ∈ Z(f ) . To this end, 2 first note that f (xn ) = g(xn )f (xn ) for all n ∈ N. As f (xn ) 6= 0 for all n ∈ N, we see that f (xn )g(xn ) = 1 for all n ∈ N. Now, in view of the continuity of f and g, letting n → +∞, we obtain f (x∞ )g(x∞ ) = 1, yielding f (x∞ ) 6= 0. In other words, x∞ ∈ Z(f )c , which is what we want.  4. Define the function g : [0, 12 ] → R by g(x) = f (x + 21 ) − f (x). We have 1 1
1 g(0) = f ( ) − f (0) = − f (1) − f ( ) = −g( ), 2 2 2 yielding g(0)g( 12 ) = −g(0)2 ≤ 0. It thus follows from the Intermediate Value Theorem that there exists a c ∈ [0, 12 ] such that g(c) = 0. That is, f (c + 12 ) = f (c). Letting a = c and b = c + 12 , we have b − a = 12 and f (a) = f (a), which is what we want. 

154

2. SOLUTIONS

2.17.2. Algebra. 1. For a, b ∈ G, define a ∼ b if and only if h1 ah2 = b for some h1 , h2 ∈ H. It is readily verify that ∼ is indeed an equivalence relation on G. Also, for any x ∈ G, we have [x] = HxH, where [x] denotes the equivalence class of x. With all that in mind, let e denote the identity element of G and {x1 = e, x1 , . . . , xn } be a maximal set of nonequivalent elements of G. It follows from the hypothesis that n > 1. We have G =

n [

Hxi H.

i=1

This yields −1 n n H x Hxi X X |H|2 i G = , = H ∩ (x−1 Hxi ) H ∩ (x−1 Hxi ) i i i=1 i=1 from which, together with the hypothesis, we obtain n 2 X H = H + (n − 1) H 2 . G = H + i=2 |G| |H|

This implies [G : H] = = 1+(n−1)|H|. In other words, [G : H]−(n−1)|H| = 1, which easily yields gcd([G : H], |H|) = 1, which is what we want.  2. The assertion is a special case of the following. Let R be a unital ring with the property that ab = 1 implies ba = 1 whenever a, b ∈ R. Then, the ring R[x] has the same property, i.e., if a, b ∈ R[x] and ab = 1, then ba = 1. If D is a division ring and n ∈ N, it follows from the Rank-Nullity Theorem that for A, B ∈ Mn (D), we have AB = In if and only if BA = In , where In denotes the identity matrix. Thus, the ring R = Mn (F ) satisfies the above property. As a matter of fact, it can be shown that any left or right Noetherian ring satisfies the above property. To prove the above more general assertion, suppose that for f, g ∈ R[x] we have f g = 1. We prove that gf = 1. Suppose to the contrary that gf 6= 1. Note first that (gf )2 = g(f g)f = gf . Assuming that f = a0 + a1 x + · · · + an xn and g = b0 + b1 x + · · · + bm xm , we obtain a0 b0 = 1 because f g = 1. Thus, b0 a0 = 1. So we can write gf = 1 + cxk + · · · , where k is the least exponent of x such that c 6= 0. Now, from (gf )2 = gf , we see that (1 + cxk + · · · )2 = 1 + cxk + · · · , from which, we obtain 2c = c, yielding c = 0, which is a contradiction. Thus gf = 1, which is what we want.  3. We prove the counterpart of the problem for left (resp. right) finite-dimensional vector spaces over a division ring D. So assume that V is a left (resp. right) vector space over a division ring D and T : V −→ V a left (resp. right) linear transformation on V . As T 2 V ⊆ T V , the linear transformation S : T V → T V defined by S(T x) = T 2 x defines a left (resp. right) linear transformation on the left (resp. right) vector space T V . By the Rank-Nullity Theorem, we can write dim T V = dim S(T V ) + dim ker S. As S(T V ) = T 2 V and ker S = ker T ∩ T V , we see that dim(ker T ∩ T V ) = dim T V − dim T 2 V.

2.18. EIGHTEENTH COMPETITION

155

Therefore, dim(ker T ∩ T V ) = rank(T ) − rank(T 2 ), as desired.



2.18. Eighteenth Competition 2.18.1. Analysis. 1. We have an = An − An−1 . This together with the n hypothesis easily yields limn AAn−1 = 1, and hence limn AAn−1 = 1 because An > 0 n for all n ∈ Z P with n ≥ 0. From this, it follows that the radius of convergence of the +∞ power series n=0 An xn is 1. In particular, the series absolutely converges for all x ∈ (−1, 1). Now, as a0 = A0 and an < An for all n ∈ N, it follows that n n 0 ≤ an x ≤ An x , P+∞ n for all n ∈ N ∪ {0} and x ∈ (−1, 1). Thus, by the Comparison Test, n=0 an |x| P+∞ n converges for all x ∈ (−1, 1) because so does n=0 An |x| for all x ∈ (−1, 1). So, if P+∞ R is the radius of convergence of the series n=0 an xn , then R ≥ 1. But R > 1 is P+∞ impossible, for otherwise the series n=0 an must be convergent, which contradicts the hypothesis that limn An = +∞. Therefore, R = 1, which is what we want.  2. We prove the assertion under the weaker hypothesis that the functions f and g are Riemann integrable on [0, 1]. In the following integrals, perform the substitutions nx = t and s = t − k + 1, respectively, to obtain Z 1 Z n 1 t
f g(t)dt f (x)g(nx)dx = n n 0 0 n Z k X 1 t
= f g(t)dt n n k=1 k−1 n Z k X
1 s k − 1
= f + g s + (k − 1) ds. n n k−1 n k=1

But the period of g is one. So we can write Z 1 Z f (x)g(nx)dx = 0

1

fn (s)g(s)ds,

0


Pn . Suppose we have proved that the sequence where fn (s) = n1 k=1 f ns + k−1 n R1 +∞ (fn )n=1 uniformly converges to 0 f on [0, 1]. Since g is bounded on [0, 1], we see R1 that the sequence (fn g)+∞ n=1 uniformly converges to g 0 f on [0, 1]. Hence, we can write Z 1 Z 1 lim f (x)g(nx)dx = lim fn (s)g(s)ds n n 0 0 Z 1 Z 1 Z 1  = lim fn (s)g(s)ds = f g(s)ds n

0

Z =

1

0

 Z f

0

1

 g ,

0

0

156

2. SOLUTIONS

proving the assertion. It remains to show that the sequence (fn )+∞ n=1 uniformly R1 converges to 0 f on [0, 1]. To this end, as f is Riemann integrable, it follows from Riemann’s criterion for integrability that for given ε > 0, there exists a δ > 0 such that for any partition P ∈ P[0, 1], where P[0, 1] denotes the set of all partitions of the interval [0, 1], with ||P || < δ, we have n X

U (P, f ) − L(P, f ) =

(Mi − mi )∆xi < ε,

i=1

where P : x0 = 0 < x1 < · · · < xn = 1, ||P || = max1≤i≤n ∆xi U (P, f ) =

Pn

i=1

Mi ∆xi , L(P, f ) =

Pn

i=1

mi ∆xi ,

mi = inf x∈[xi−1 ,xi ] f (x), Mi = supx∈[xi−1 ,xi ] f (x). We also know that if U (P, f ) − L(P, f ) < ε for some P ∈ P[0, 1], then for any xi−1 ≤ ξi ≤ xi , we have Z 1 n X f (ξi )∆xi − f < ε. 0 i=1

Note that fn (s) corresponds to the Riemann sum with respect to the uniform partition of [0, 1] with n subintervals of the same length, i.e., ∆xi = n1 for all 1 ≤ i ≤ n, and the mid points ξi = ns + i−1 n . In view of this, for given ε > 0, find δ > 0 from the above and let N ∈ N such that N1 < δ. Now, for each n ≥ N , let Pn be the uniform partition of [0, 1] with n subintervals of the same length. We have ||Pn || = n1 < N1 < δ. It thus follows that n Z 1 Z 1 1 X s
i − 1 fn (s) − f + − f = f n n n 0 0 i=1 Z 1 n X = f (ξi )∆xi − f 0 i=1

≤ U (Pn , f ) − L(Pn , f ) < ε. That is, Z fn (s) −

0

1

f < ε,

for all s ∈ [0, 1] and n ≥ N . In other words, the sequence (fn )+∞ n=1 uniformly R1 converges to 0 f on [0, 1], which is what we want.  3. We need the following lemma. Lemma. If α ∈ [0,P 1], then there exists a sequence (ai )+∞ i=1 with ai ∈ {0, 1} n a i=1 i = α. (i ∈ N) such that lim n→+∞ n

2.18. EIGHTEENTH COMPETITION

157

Proof. If α = 0, let ai = 0 for all i ∈ N. If not, let  1 if i = b m α c for some m ∈ N, ai = 0 otherwise, where b.c stands for the integer part function. We have Pn o n jmk 1 i=1 ai ≤n = card m ∈ N : n n α n o bα(n + 1)c 1 = card m ∈ N : m < α(n + 1) = . n n This obtains Pn ai bα(n + 1)c lim i=1 = lim = α, n n n n as desired.



To prove assertion, note first that if a sequence (ai )+∞ i=1 has the property Pthe n a i=1 i that lim = α, then the property remains intact under changing a finite n→+∞ n number of the terms of (ai )+∞ i=1 . Now, let I be a nonempty open interval of R so that I = (x − r, x + r) for some x ∈ I and r > 0. Clearly, g(I) ⊆ [0, 1]. Let α ∈ [0, 1] 1 be arbitrary. Choose n ∈ N such that n < ε. And choose x0 ∈ R such that the 2 first n digits of its binary expansion are the same of those of x and from its (n+1)st digit onward its digits are equal to ai ’s. That is, x0 = 0.x1 . . . xn a1 a2 a3 . . . , where x = 0.x1 . . . xn xn+1 . . .. It is obvious that x0 ∈ I and that g(x0 ) = α. Consequently, [0, 1] ⊆ g(I), and hence g(I) = [0, 1], as desired. For the rest, note first that any open subset G of R includes an open interval I, implying that g(G) ⊇ g(I) = [0, 1]. This implies g(G) = [0, 1] because g(G) ⊆ [0, 1]. That is, for any open subset G of R, we have g(G) = [0, 1]. Now, define the function h : R → R by h = f ◦ g, where the function f : [0, 1] → R is defined by  x 0 < x < 1, f (x) = 1 x ∈ {0, 1}. 2 It is obvious that for any open subset G of R, we have

h(G) = f g(G) = f [0, 1] = (0, 1). In other words, h is an open map. By proving that h has no limit at any point of R, we show that h is not continuous on R. Suppose to the contrary that there exist x0 , ` ∈ R such that lim h(x) = `. x→x0

It follows that for every ε > 0, there exists a δ > 0 such that ` − ε < h(x) < ` + ε, whenever x ∈ (x0 − δ, x0 + δ). We have 0 ≤ l ≤ 1 because 0 < h < 1. Consequently, we can choose ε > 0 such that (` − ε, ` + ε) ∩ (0, 1) $ (0, 1). For this ε > 0, find δ > 0 from the above. Thus, for all x ∈ (x0 − δ, x0 + δ), we have h(x) ∈ (` − ε, ` + ε).

(∗)

158

2. SOLUTIONS

Letting G = (x0 − δ, x0 + δ), we obtain
h(G) = f g(G) = (0, 1), from which, in view of (∗), we see that (0, 1) ⊆ (` − ε, ` + ε), implying (` − ε, ` + ε) ∩ (0, 1) = (0, 1), contradicting our choice of ε > 0. Therefore, the function h has no limit at any point of R, which is what we want. 

2.18.2. Algebra. 1. The subgroup H ∩K is normal in K because K is normal in G. Since K is simple, we have H ∩ K = K or H ∩ K = {e}, where e is the identity element of G. If H ∩ K = K, then K ⊆ H, implying that K = H because |G| < ∞ and H ∼ = K. To finish the proof, we show that H ∩ K 6= {e}. Suppose to the contrary that H ∩ K = {e}. As K is normal in G, we have HK = KH, whence HK ≤ G, and hence |HK| |G|. But HK = H K = K 2 . H ∩ K In other words, |K|2 |G|, contradicting the hypothesis that the square of the order of K does not divide that of G. Thus, H ∩ K 6= {e}, which is what we want.  2. It suffices to show that any right ideal of R is finitely generated. To this end, let I be a right ideal in R. If I = 0, the assertion is trivial. Suppose I 6= 0. As is usual, use Eij to denote the 2 × 2 matrix whose ij entry is 1 and zero elsewhere. There are two cases to consider. (i) There exists A = a0 E11 + b0 E12 + c0 E22 ∈ I such that a0 6= 0. Using the well-ordering principle of natural numbers, let am be the least positive integer a for which there are b, c ∈ Q such that aE11 + bE12 + cE22 ∈ I. It is easily verified that if aE11 + bE12 + cE22 ∈ I for some a ∈ Z and b, c ∈ Q, then am divides a. We claim that I is generated by {am E11 , E12 , E22 }. Suppose am E11 + bE12 + cE21 ∈ I for some b, c ∈ Q. It follows that (am E11 + bE12 + cE22 )E12 = am E12 ∈ I, implying (am E12 )


1 E22 = E12 ∈ I. am

That is, E12 ∈ I. We can write (am E11 + bE12 + cE22 )E11 = am E11 ∈ I, yielding am E11 ∈ I. Now, as am E11 , E12 ∈ I, we see that if A = kam E11 + bE12 + cE22 ∈ I for some k ∈ Z and b, c ∈ Q, then cE22 ∈ I. If c = 0 for all A = kam E11 + bE12 + cE22 ∈ I, then I is generated by {am E11 , E12 }. Otherwise, just as we saw in the above, cE22 ∈ I for some nonzero c ∈ Q, yielding
1 (cE22 ) E22 = E22 ∈ I. c That is, E22 ∈ I, in which case the right ideal I is generated by {am E11 , E12 , E22 }. This proves the assertion in this case. (ii) Every A ∈ I is of the form A = bE12 + cE22 , where b, c ∈ Q.

2.18. EIGHTEENTH COMPETITION

159

Suppose A0 = b0 E12 +c0 E22 ∈ I, where b0 , c0 ∈ Q are such that (b0 , c0 ) 6= (0, 0). Set J = QA0 . It is easily verified that J is a right ideal of R. If I = J, then I is generated by A0 . If not, then there exist b1 , c1 ∈ Q such that b1 c0 − b0 c1 6= 0 and b1 E12 + c1 E22 ∈ I. It follows that (b0 c1 − b1 c0 )E12 = (b0 E12 + c0 E22 )(b1 E12 + c1 E22 ) − (b1 E12 + c1 E22 )(b0 E12 + c0 E22 ) ∈ I, and hence
1 (b0 c1 − b1 c0 )E12 E22 = E12 ∈ I. b0 c1 − b1 c0 That is, E12 ∈ I. So we have E12 (b0 E22 ) = b0 E12 ∈ I. From this, we obtain c0 E22 ∈ I because b0 E12 + c0 E22 ∈ I. If c0 = 0 whenever b0 E12 + c0 E22 ∈ I, then I is generated by E12 . If not, then c0 E22 ∈ I for some nonzero c0 ∈ Q. This implies E22 = (c0 E22 )( c10 E22 ) = E22 ∈ I. Thus, I is generated by {E12 , E22 } in this case. So in any event, I is finitely generated. We now prove that every ascending chain of right ideals of R necessarily terminates. To this end, let (In )+∞ n=1 be an ascending sequence of right ideals of R. Set +∞ I = ∪+∞ I . As (I ) is an ascending sequence of right ideals of R, it is easily n n n=1 n=1 verified that I is a right ideal of R. It follows from the above that there are at most three, not necessarily distinct, elements A1 , A2 , A3 ∈ I such that I = hA1 , A2 , A3 i. Since I = ∪+∞ n=1 In , there exist n1 , n2 , n3 ∈ N such that Ai ∈ Ini for each i = 1, 2, 3. Letting N = max(n1 , n2 , n3 ), we have Ai ∈ In for all n ≥ N and i = 1, 2, 3. This implies

 In ⊆ I = A1 , A2 , A3 ⊆ In , for all n ≥ N . Therefore, In = I for all n ≥ N . In other words, In = IN for all n ≥ N , which is what we want.  3. We need the following lemma. Lemma. Let V be a finite-dimensional vector space over a field F and {Vi }i∈I a family of proper subspaces of V such that |I| < |F |. Then, [ Vi $ V. i∈I

Proof. Without loss of generality, we may assume that {Vi }i∈I is a family of distinct proper subspaces of V . With that in mind, we prove the assertion by induction on dim V . If dim V = 1, we must have dim Vi = 0, implying Vi = {0} for all i ∈ I, in which case the assertion is trivial. Assuming that the assertion holds for any vector space V with dim V < k, we prove the assertion for any vector space V with dim V = k. Let Vi (i ∈ I) be as in the lemma. Set  S = dim Vi |i ∈ I . Plainly, S ⊆ N and S is nonempty and bounded from above by k = dim V . It follows that S has a terminal element. That is, there exists i0 ∈ I such that dim Vi ≤ dim Vi0 , for all i ∈ I. Set W = Vi0 , J = I \ {i0 }, and finally Wj = Vj ∩ Vi0 . Firstly, for all j ∈ J we have Wj $ Vi0 . Suppose to the contrary that Wj = Vj ∩Vi0 = Vi0 for some j ∈ J, then Vi0 ⊆ Vj , implying dim Vi0 ≤ dim Vj . On the other hand, dim Vj ≤ dim Vi0 , yielding Vi0 = Vj which is impossible because j 6= i0 . Thus, Wj $ W = Vi0

160

2. SOLUTIONS

for all j ∈ J. We have dim W = dim Vi0 < dim S V = k and |J| ≤ |I| < |F |. So it follows from the induction hypothesis that j∈J Wj $ W = Vi0 . Consequently, there exists a vector v0 ∈ Vi0 such that v0 ∈ / Wj = Vj ∩ Vi0 for all j ∈ J. That is, v0 ∈ / Vj for all j ∈ I with j 6= i0 . On the other hand, since Vi0 $ V , there exists a vector v1 ∈ V \ Vi0 . We claim that there exists an f0 ∈ F such that v1 + f0 v0 ∈ / Vi for all i ∈ I. To prove this by contradiction, suppose that for each f ∈ F there exists an if ∈ I such that v1 + f v0 ∈ Vif . Note that if f, f 0 ∈ F and f 6= f 0 , then v1 + f v0 6= v1 + f 0 v0 . Also note that the hypothesis |I| < |F | implies that there exist f, f 0 ∈ F with f 6= f 0 such that if = if 0 . Firstly, we observe that if 6= i0 , for otherwise v1 + f v0 ∈ Vi0 , yielding v1 ∈ Vi0 , because v0 ∈ Vi0 , which is impossible. Secondly, from v1 + f v0 , v1 + f 0 v0 ∈ Vif = Vif 0 , we obtain (f − f 0 )v0 ∈ Vif , yielding v0 ∈ Vif , which, in turn, implies if = i0 , contradicting if 6= i0 as observed in the above. That is, we obtain a contradiction in any event. Thus, there exists an S f0 ∈ F such that v + f v ∈ / V for all i ∈ I. In other words, v + f v ∈ / V 1 0 0 i 1 0 0 i∈I i , S and hence i∈I Vi $ V , which is what we want.  We now use the lemma to prove the assertion. From this point on, the proof is almost identical to that of Problem 5 of 1.14.2. Fix i0 ∈ I. We prove the assertion by induction on dim V − dim Vi0 . If dim V − S dim Vi0 = 1, then for U = {u}, where u ∈ V \ ( i∈I Vi ), we obviously have Vi ⊕ U = V for all i ∈ I. Assuming that the assertion holds whenever dim V −dim SVi0 = n, we prove it whenever dim V − dimSVi0 = n + 1. As, in view of the lemma, i∈I Vi $ V , there exists a vector β ∈ V \ ( i∈I Vi ). Set Vi0 = Vi ⊕ hβi. We have dim Vi0 = dim Vj0 = dim Vi0 + 1 for all i, j ∈ I, yielding dim V − dim Vi00 = dim V − dim Vi0 = n for all i ∈ I. It thus follows from the induction hypothesis that there exists a subspace U 0 such that Vi0 ⊕ U 0 = V, for all i ∈ I. Letting U = hβi ⊕ U 0 , we obviously obtain Vi ⊕ U = V, for all i ∈ I, proving the induction assertion, which is what we want.



2.19. Nineteenth Competition 2.19.1. Analysis. 1. Define the function g : [0, 1] → R by Z x g(x) = 2x − 1 − f. 0

It suffices to show that g has only one zero on the interval [0, 1]. As f is continuous on [0, 1], g is differentiable on (0, 1). Using the First Fundamental Theorem of Calculus, we can write g 0 (x) = 2 − f (x), for all x ∈ (0, 1). But 0 ≤ f (x) ≤ 1 for all x ∈ (0, 1). In particular, we must have g 0 (x) = 2 − f (x) ≥ 1 > 0 for all x ∈ (0, 1), implying that g is strictly increasing on [0, 1]. Therefore, g has at most one zero on [0, 1]. With that in mind, note that R1 g(0) = −1 < 0 ≤ 1 − 0 f = g(1). To see the second inequality, just note that

2.19. NINETEENTH COMPETITION

161

R1 R1 from f ≤ 1 on [0, 1], it follows that 0 f ≤ 1, yielding g(1) = 1 − 0 f ≥ 0. It thus follows from the Intermediate Value Theorem that there exists a c ∈ (0, 1] such that g(c) = 0. This completes the proof.  x

2. First Solution: Note that limx→+∞ e h = +∞ whenever h > 0. With that in mind, using L’Hopital’s rule, we can write
0 x x e h f (x) e h f (x) = lim lim f (x) = lim x x
0 x→+∞ x→+∞ x→+∞ eh eh =

x x 1 h 0 h h e f (x) + e f (x) lim x 1 x→+∞ h he

= lim

x→+∞


f (x) + hf 0 (x) = 0,

implying limx→+∞ f (x) = 0. This together with lim

x→+∞


f (x) + hf 0 (x) = 0,

where h ∈ R , yields limx→+∞ f 0 (x) = 0, which is what we want. (We note that the hypothesis that f is continuously differentiable is redundant. The differentiability of f is enough for the assertion to be true.) +

Second solution: It follows from the hypothesis that for given ε > 0, there exists an M = M (ε) > 0 such that f (x) + hf 0 (x) < ε , 2 d x
x ε h whenever x ≥ M . This easily implies dx e h f (x) < 2h e whenever x ≥ M . We can write x Z x d t
M h e h f (t) dt e f (x) − e h f (M ) = M dt Z x t M
ε x ε e h dt = e h − e h , ≤ 2h M 2 for all x ≥ M . Consequently, for all x ≥ M ,
f (x) − e Mh−x f (M ) < ε 1 − e Mh−x , 2 which obtains f (x) ≤ f (x) − e Mh−x f (M ) + e Mh−x f (M ) M −x
M −x ε ≤ 1 − e h + e h f (M ) 2 M −x ε < + e h f (M ) , 2 for all x ≥ M . Consequently, ε  ε M −x + e h f (M ) = , lim sup f (x) ≤ lim sup 2 2 x→+∞ x→+∞ for all ε > 0. Thus, lim supx→+∞ f (x) ≤ 0, and hence limx→+∞ f (x) = 0. As we pointed out in the
first solution, this together with the hypothesis that limx→+∞ f (x) + hf 0 (x) = 0, where h ∈ R+ , yields limx→+∞ f 0 (x) = 0, finishing the proof. 

162

2. SOLUTIONS

3. Recall that A equipped with the uniform metric, induced by the uniform norm of A, denoted by ||.||∞ , which is defined by d(f, g) = ||f − g||∞ = sup f (x) − g(x) , x∈[0,1]

is a complete normed space or a real Banach space. Also recall that by Problem 2 of 1.6.1 the following lemma holds. Lemma. Let X and Y be metric spaces and f, fn : X → Y continuous functions such that the sequence (fn )n∈N uniformly converges to f on X. Also let x0 , xn ∈ X be such that the sequence (xn )n∈N converges to x0 . Then, limn fn (xn ) = f (x0 ). With all that in mind, we now prove the assertion. (a) Let N ∈ N be arbitrary. To show that EN is closed, suppose that f : [0, 1] → R is a limit point of EN with respect to the topology induced by the uniform norm of A. We show that f ∈ EN . To this end, as f is a limit point of EN , we see that there exists a sequence (fn )n∈N in EN such that fn → f in A as n → +∞. In other words, fn ’s uniformly converge to f on [0, 1]. Now, from fn ∈ EN for each n ∈ N, it follows that there exists an an ∈ [0, 1] such that fn (x) − fn (an ) ≤ N x − an , for all x ∈ [0, 1]. It is plain that the sequence (an )n∈N in the compact interval [0, 1] has a subsequence (ank )k∈N converging to some a0 ∈ [0, 1]. Let bk = ank and gk = fnk for all k ∈ N. We can write gk (x) − gk (bk ) ≤ N x − bk , for all x ∈ [0, 1] and k ∈ N. Now, letting k → +∞ and using the above lemma, we see that f (x) − f (a) ≤ N x − a , for all x ∈ [0, 1]. That is, f ∈ EN , and hence EN is closed for all N ∈ N. To show that the interior of EN is empty for all N ∈ N, we prove that the set A\EN is dense in A. Suppose that f ∈ A is arbitrary and ε > 0 is given. From the continuity of f on the compact interval [0, 1], we see that f is uniformly continuous on [0, 1]. Thus, there exists a δ > 0 such that f (x) − f (y) < ε , 2 whenever x, y ∈ [0, 1] and |x−y| < δ. Choose n ∈ N such that n1 < δ and set xi = ni , where 1 ≤ i ≤ n. By constructing a function g satisfying ||f −g||∞ < ε and g ∈ / EN , we show that A \ EN is dense in A, finishing the proof. To construct g, it suffices to do so on any subinterval [xi−1 , xi ] (1 ≤ i ≤ n) in such a way that g is (piecewise) continuous, ||f − g||∞ < ε, and g ∈ / EN . Here is a description of the graph of g on the interval [xi−1 , xi ]. Let g be any function on   [xi−1 , xi ] (1 ≤ε i ≤ n) whose  graph ε , f (x ) + is a polygonal line in the rectangle x , x × f (x ) − i−1 i i−1 i 2 2 joining

the two points xi−1 , f (xi−1 ) and xi , f (xi ) satisfying the following properties. (i) The consecutive vertices of any segment of the polygonal line lie on the lines y = f (xi−1 ) − 2ε and y = f (xi ) + 2ε , respectively. (ii) The absolute value of the slope of any segment formed by any two consecutive vertices of the polygonal line is greater than N . It is plain that for any function g : [0, 1] → R as described in the above, we have ||f − g||∞ < ε and g ∈ / EN , which is what we want.

2.19. NINETEENTH COMPETITION

163

(b) As A is a complete metric space with respect to the metric induced by the uniform norm of A, it follows from the Baire Category Theorem that A cannot be written as a countable union of nowhere dense subsets of A. Recall that is subset of a topological space X is called nowhere dense if the complement of its closure is dense in X. It thus follows from (a) that S EN is nowhere dense for all N ∈ N, and hence, by the Baire Category Theorem, N ∈N EN $ A. Define the set E as follows [  E := f ∈ A|∃a ∈ [0, 1]  f ∈ Da = Da , a∈[0,1]

where Da , with a ∈ (0, 1), is the set of all functions that are differentiable at a ∈ (0, 1); the set D0 (resp. D1 ) is defined to be the set of functions that are right S (resp. left) differentiable at 0 (resp. 1). We claim that E ⊆ N ∈N EN . To prove this, suppose f ∈ E is given. It follows that there exists a ∈ [0, 1] such that f ∈ Da . Define the function g : [0, 1] → R by  f (x)−f (a) x ∈ [0, 1] \ {a}, x−a g(x) = 0 f (a) x = a. (a) = f 0 (a) = g(a) for all a ∈ (0, 1). Likewise, We have limx→a g(x) = limx→a f (x)−f x−a limx→0+ g(x) = g(0) and limx→1− g(x) = g(1). Therefore, the function g is continuous on the compact interval [0, 1], and hence it is bounded on [0, 1]. So there exists an M > 0 such that g(x) ≤ M, for all x ∈ [0, 1]. In other words, f (x) − f (a) ≤ M x − a ,

for all x ∈ [0, 1]. Consequently, if we let N ∈ N be such that N ≥ M , then f (x) − f (a) ≤ N x − a , S for all x ∈ [0, 1]. That is, f ∈ EN , yielding S f ∈ N ∈N EN , proving the claim. Now, to prove the assertion, recall that E ⊆ N ∈N EN $ A. This implies that there exists a function f ∈ A \ E. In other words, the function f : [0, 1] → R is continuous and yet it is nowhere differentiable because f ∈ / E. This proves (b). 

2.19.2. Algebra. 1. To prove the assertion by contradiction, suppose that N0 6= {e} is a maximal element of A. We have NG0 ∼ = G. It follows that there exists G an isomorphism φ : G → N0 . Since N0 6= {e} is a normal subgroup of G and φ is an isomorphism, we see that φ(N0 ) 6= {e} is a normal subgroup of NG0 . From 0 this, we obtain a normal subgroup M0 % N0 of G such that φ(N0 ) = M N0 . Thus, M0 M φ|N0 : N0 → N0 is an isomorphism of groups, yielding N0 ∼ = N00 . Now, the Third G ∼ 0 Isomorphism Theorem for groups together with N0 = G and N0 ∼ = M N0 implies that G G ∼ N0 ∼ G ∼ = M0 = = G. M0 N0 N0 In other words, G ∼ = G. But N0 $ M0 , contradicting the hypothesis that N0 is a M0

maximal element of A. Thus, the assertion follows by contradiction, which is what we want. 

164

2. SOLUTIONS

2. (a) Note first that for any nonzero a ∈ R and any minimal left ideal J of R, the left ideal Ja is also a minimal left ideal of R. To prove this by contradiction, suppose that Ja is not a minimal left ideal of R for some nonzero element a ∈ R and minimal left ideal J of R. It follows that there exists a nonzero left ideal I of R such that I $ Ja. Define  J1 := j ∈ J|ja ∈ I . It is plain that J1 is a left ideal of R, J1 $ J, because I $ Ja, and that J1 6= 0, because I 6= 0. This is in contradiction with J being a minimal left ideal of R. Thus, Ja is a minimal left ideal of R whenever J is a minimal left ideal of R and a ∈ R is nonzero. To prove the assertion, if R has no minimal P left ideal, the assertion is trivial because, by definition, S = 0. If not, as S = {J : J is a minimal left ideal of R}, we see S is a left ideal in R. To see that S is a right ideal of R as well, let r ∈ R and s ∈ S be arbitrary. We need to show that sr ∈ S. Since s ∈ S, it follows that there are minimal left ideals J1 , . . . , Jn , where n ∈ N, and jk ∈ Jk (1 ≤ k ≤ n) such that s = j1 + · · · + jn , yielding sr = j1 r + · · · + jn r. But jk r ∈ P Jk r and Jk r is a minimal left ideal of R for all 1 ≤ k ≤ n. It thus follows that sr ∈ {J : J is a minimal left ideal of R} = S. That is, S is a right ideal of R as well, and hence an ideal of R, as desired. (b) It follows from (a) that \

I ⊆ S,

ICR

for the two sided ideal S appears in the intersection. To prove the inclusion in the opposite direction, it suffices to show that for any minimal left ideal J of R, we have \ J ⊆ I. ICR

To this end, assuming that J and I are, respectively, an arbitrary minimal left ideal and an arbitrary two-sided ideal of R, it is enough to show that J ⊆ I. To see this, note first that IJ ⊆ J, for J is a left ideal of R. Secondly, IJ 6= 0 because otherwise I(JR) = (IJ)R = 0, where I 6= 0 and JR 6= 0 are two-sided ideals of R, contradicting the hypothesis. Thus, IJ 6= 0, and hence 0 6= IJ ⊆ J. That is, the nonzero left ideal IJ is contained in the minimal left ideal J. This implies IJ = J. Now, since I is a two-sided ideal of R, we see that IJ ⊆ I ∩ J, whence J ⊆ I ∩ J, yielding J ⊆ I, which is what we want. 

2.20. TWENTIETH COMPETITION

165

3. Recall that the rank of a matrix is equal to its column rank as well as its row rank. With that in mind, let Aj denote the jth column of the matrix A. We have       1 x1 x1 + yj  1   x2 + yj   x2        Aj =   =  ..  + yj  ..  = α + yj β, ..  .      . . 



xn + yj 

1

xn 

1  1       and β =  .. . That is, Aj ∈ hα, βi, yielding hA1 , . . . , An i ⊆  .   1 xn hα, βi. Thus, rank(A) = dimhA1 , . . . , An i ≤ dimhα, βi ≤ 2. Therefore, rank(A) ≤ 2, as desired.    where α =  

x1 x2 .. .

4. First solution: Let A ∈ Mn (F ) be a rank one n × n matrix over F . Since rank(A) = 1, it follows from the Rank-Nullity Theorem that the rank and nullity of the linear transformation T : Mn×1 (F ) → Mn×1 (F ) defined by T X = AX, are, n−1 respectively, equal to 1 and n−1. Let {αi }n−1 i=1 be a basis for ker T . Enlarge {αi }i=1 n−1 0 n to a basis B = {αi }i=1 for Mn×1 (F ), where αn is any vector in Mn×1 (F ) \ hαi ii=1 , where hαi in−1 i=1 denotes the vector subspace spanned by α1 , . . . , αn−1 . It is plain that the matrix of T with respect to the basis B 0 , denoted by B, has the form   0 · · · 0 b1  0 · · · 0 b2    B = (bij )n×n =  . . . .  .. · · · .. ..  0 · · · 0 bn On the other hand, the matrix of T with respect to the standard basis of Mn×1 (F ) is equal to A. It thus follows that there exists an invertible matrix P ∈ Mn (F ) such that B = P −1 AP , whence A = P BP −1 . We can write det(I + A)

=

=

det(P (I + B)P −1 ) = det(I + B)   1 ··· 0 b1  0 ··· 0 b2    det  .  . ..  .. · · · ..  . 0 ···

=

0

1 + bn

1 + bn = 1 + tr(B) = 1 + tr(A).

That is, det(I + A) = 1 + tr(A), which is what we want. Second solution: A proof identical to that of the second solution of Problem 7 of 1.9.3 shows that det(I + A) = 1 + tr(A), which is what we want.  2.20. Twentieth Competition 1. If {(x, y), (z, t)} or {(x0 , y 0 ), (z 0 , t0 )} is linearly dependent, then there exist α, β ∈ C, which are not simultaneously zero, such that θ = α(x, y) + β(z, t) = (0, 0) or θ0 = α(x0 , y 0 ) + β(z 0 , t0 ) = (0, 0).

166

2. SOLUTIONS

In this case, the assertion is trivial because θ = 0 or θ0 = 0. So suppose that {(x, y), (z, t)} and {(x0 , y 0 ), (z 0 , t0 )} are linearly independent. Set    0  x y x y0 A= , A0 = . z t z 0 t0 The matrices A and A0 are invertible because their rows are linearly independent. It follows that the matrix A−1 A0 ∈ M2 (C) is also invertible. Now, suppose that λ ∈ C is a nonzero eigenvalue of the invertible matrix A−1 A0 . We see that the matrix λI2 − A−1 A0 is not invertible. Hence, neither is λA − A0 = A(λI2 − A−1 A0 ). This implies that the rows of the noninvertible matrix    0  x y x y0 λA − A0 = λ − z t z 0 t0 are linearly dependent. Thus, there exist α, β ∈ C, which are not simultaneously zero, such that α(λx − x0 , λy − y 0 ) + β(λz − z 0 , λt − t0 ) = (0, 0). A straightforward calculation reveals that λθ − θ0 = α(λx − x0 , λy − y 0 ) + β(λz − z 0 , λt − t0 ) = (0, 0), where θ = α(x, y) + β(z, t) and θ0 = α(x0 , y 0 ) + β(z 0 , t0 ). In other words, the vectors θ and θ0 are linearly dependent, which is what we want.  2. Set xn = ln(1 + an ). It follows from the hypothesis that xm+n ≤ xm + xn , for all m, n ∈ N. We prove that limn xnn exists. To see this, letting α = inf n∈N xnn ∈ R, we show that limn xnn = α. To this end, let ε > 0 be given. Since α = inf n∈N xnn , there exists a positive integer N1 = N1 (ε) such that xN1 ε α ≤ ≤ α+ . N1 2 Now, for any n ≥ N1 , we can write n = N1 q + r, where q ∈ N and r ∈ N ∪ {0} with 0 ≤ r < N1 . In view of the hypothesis, we can write α−ε<α≤

xn n

≤ <

qxN1 + rx1 qxN1 rx1 ≤ + N1 q + r N1 q n N1 x1 ε N1 xN1 + ≤α+ + x1 . N1 n 2 n

Now, pick N2 ∈ N such that N2 > 2ε N1 x1 . Letting N = max(N1 , N2 ), for all n ≥ N , we have xn ε N1 α−ε< <α+ + x1 < α + ε. n 2 n That is, x n − α < ε, n √ xn for all n ≥ N . In other words, lim = α, implying that limn n 1 + an = eα , n n which is what we want. 

2.20. TWENTIETH COMPETITION

167

3. It follows from the hypothesis that σ 2 = id for all σ ∈ aut(G). Consequently, σ −1 = σ for all σ ∈ aut(G), from which we obtain σ1 σ2 = (σ1 σ2 )−1 = σ2−1 σ1−1 = σ2 σ1 , for all σ1 , σ2 ∈ aut(G). In other words, aut(G) is abelian. On the other hand, just as we saw in Solution 1 of 2.16.2 using the First Isomorphism Theorem for groups, G ∼ we have Z(G) = In(G), where In(G) denotes the set of all inner automorphisms of G, which is a normal subgroup of aut(G). It follows that In(G) is commutative and G hence so is Z(G) . This yields G0 ⊆ Z(G), where G0 denotes the derived subgroup (a.k.a. the commutator subgroup) of G. From this, we see that G00 = {e}, implying that G is solvable, which is what we want.  4. Note first that f (x) > 0 for all x ∈ ( 41 , 1) because f (x) = xf (x) for all such x’s. Next, since f (x) > 0, we see that 0 < ( 41 )f (x) < xf (x) < 1f (x) = 1, implying that 0 < f (x) < 1 for all x ∈ ( 14 , 1). Now, noting that xf (x) = f (x) and f (x) > 0 for all x ∈ ( 14 , 1), and taking ln, we obtain f (x) ln x = ln f (x) for all x ∈ ( 14 , 1). This implies ln f (x)

x = e f (x) , ln x for all x ∈ ( 14 , 1). Define g : (0, 1) → (0, 1) by g(x) = e x . The function g is x lnxx . Thus, g 0 (x) > 0 for all 0 < x < 1, differentiable and we have g 0 (x) = 1−ln x2 e and hence g is strictly increasing on (0, 1). The function g : (0, 1) → (0, 1) is surjective because limx→0+ g(x) = 0 and limx→1− g(x) = 1. From this, it follows that g has an inverse, say, g −1 : (0, 1) → (0, 1), and that g −1 is differentiable and strictly increasing on (0, 1). In particular, g −1 is continuous on (0, 1). Also, it is easily verified that g( 12 ) = 14 . With all that in mind, define the function ge : [0, 1] → [0, 1] by  x = 0,  0 g(x) 0 < x < 1, ge(x) =  1 x = 1. It is now plain that the function ge : [0, 1] → [0, 1] is surjective, one-to-one and continuous and that its inverse ge−1 : [0, 1] → [0, 1] is also continuous, and hence uniformly continuous, because [0, 1] is a compact interval. Noting 0 < f (x) < 1 for all x ∈ ( 41 , 1), we can write

ge f (x) = g f (x) = x, from which, we obtain f (x) = ge−1 (x) for all x ∈ ( 14 , 1). In other words, f = ge−1 |( 14 ,1) : ( 41 , 1) → ( 12 , 1). From this, it follows that f : ( 41 , 1) → ( 12 , 1) is uniformly continuous because so is ge−1 : [0, 1] → [0, 1], which is what we want.  5. Let I 6= 0 and Z, respectively, denote the intersection of all nonzero ideals of R and the set of all zero divisors of R including zero. By showing that Z is a maximal ideal of R and that Z = I, we settle the proof, for, from this, it follows that Z = I is the only nontrivial ideal of R. First, we show that I ⊆ Z. Suppose to the contrary there exists a nonzero x0 ∈ I such that x0 ∈ / Z. We see that x20 6= 0 because x0 is not a zero divisor of R. As R is unital and I is the intersection of all nonzero ideals of R, we obtain I ⊆ x0 R 6= 0 and I ⊆ x20 R 6= 0. On the other hand, x0 R ⊆ I and x20 R ⊆ I, for x0 ∈ I. It thus follows that I = x0 R = x20 R. In particular, we must have x0 ∈ x20 R, which yields x0 = x20 r for some r ∈ R.

168

2. SOLUTIONS

This obtains x0 (1 − x0 r) = 0, which, in turn, implies 1 = x0 r ∈ I, for x0 ∈ / Z, a contradiction. Thus, {0} 6= I ⊆ Z, and hence Z 6= {0}. Next, we show that Z is a maximal ideal of R. To this end, let x ∈ Z \ {0} and Ann(x) = {r ∈ R : rx = 0}. Note that Z ⊆ Ann(x) because Z 2 = 0. On the other hand, Ann(x) ⊆ Z whenever x 6= 0. It thus follows that Ann(x) = Z whenever x ∈ Z \ {0}. Now, let 0 6= x0 ∈ I ⊆ Z = Ann(x) be arbitrary. As we saw in the above Rx0 = I. It is plain that the map φ : R → Rx0 defined by φ(r) = rx0 is an epimorphism of the left modules R and Rx0 . From the First Isomorphism R ∼ ∼ R ∼ R Theorem for modules, we see that ker φ = Rx0 , implying that Rx0 = Ann(x) = Z because ker φ = Ann(x) = Z. We now show that Rx0 is a simple left module, i.e., it has no nontrivial submodules. To this end, suppose that M 6= 0 is a left submodule of Rx0 = I. Choose a nonzero y0 ∈ M . As M ≤ Rx0 = I, there exists an r0 ∈ R such that y0 = r0 x0 . We have y0 R ⊆ I because x0 ∈ I. On the other hand, I ⊆ y0 R, for R is unital and I is the intersection of all nonzero ideals of R. Therefore, I = y0 R. Consequently, we have M ⊆ I and I = y0 R ⊆ M because y0 ∈ M , yielding M = I = y0 R. This proves that Rx0 = I is simple as a left module. This together with Rx0 ∼ = R Z implies that Z is a maximal left ideal, and hence a maximal ideal of R because R is commutative. Finally, we show that Z ⊆ I. To this end, suppose that 0 6= x0 ∈ Z is arbitrary. We have x0 R 6= {0}, from which, we obtain {0} 6= I ⊆ x0 R. Now, choose an arbitrary nonzero element y0 ∈ I. It follows that there exist r0 ∈ R such that r0 x0 = y0 ∈ I ⊆ Z. Note that r0 ∈ / Z, for otherwise y0 = r0 x0 = 0 which is impossible. If r0 = 1, then x0 = y0 ∈ I yielding Z ⊆ I, as desired. If r0 6= 1, we obtain r0 R + Z = R because Z is a maximal ideal and r0 ∈ / Z. Consequently, there exist r1 ∈ R and x1 ∈ Z such that r0 r1 + x1 = 1, whence 1 − r1 r0 = x1 ∈ Z. On the other hand, x0 ∈ Z. Since Z 2 = {0}, we see that x0 (1 − r1 r0 ) = 0. This yields x0 = x0 r0 r1 = y0 r1 , which, in turn, implies x0 ∈ I because y0 ∈ I. That is, x0 ∈ I, and hence Z ⊆ I, for x0 was arbitrary. Therefore, Z = I and Z is a maximal ideal of R, proving the assertion, which is what we want.  6. First solution: We need the following lemmas. Lemma 1 (Cauchy’s criterion for the convergence of improper integrals). Let f, α : [a, +∞) → R be two functions such that α is increasing on [a, +∞) and f is integrable with respect to α in the sense of Riemann-Stieltjes, i.e., f ∈ R(α), on any closed interval [a, x], where x ∈ R with x > a. Then, a necessary R +∞ and sufficient condition for the convergence of a f dα is that for every ε > 0, there exists an x0 > a such that Z x2 < ε, f dα x1

whenever x2 > x1 > x0 .  As the lemma is standard, we omit its proof. The interested reader may consult standard books on analysis to see a proof of the lemma. The following lemma is also standard. We however include a proof for the reader’s convenience. Lemma 2 (Second Mean Value Theorem for integrals). [a, b] → R be a Riemann integrable function on [a, b].

Let g :

2.20. TWENTIETH COMPETITION

169

(a) If f : [a, b] → R is a nonnegative and decreasing function on [a, b], then there exists a number ξ ∈ [a, b] such that Z b Z ξ f g = f (a) g. a

a

(b) If f : [a, b] → R is a nonnegative and increasing function on [a, b], then there exists a number η ∈ [a, b] such that Z b Z b f g = f (b) g. a

η

Proof. (a) Set M = supx∈[a,b] |g(x)|. Obviously, g(x) + M ≥ 0 for all x ∈ [a, b]. With that in mind, suppose that P ∈ P[a, b] is a partition of the closed interval [a, b] which is given by P : x0 = a < x1 < · · · < xn = b. We can write Z b f (g + M )

=

a

=

n Z X i=1 n X

xi

f (g + M ) ≤

xi−1

n X

xi

f (xi−1 )

g+M xi−1

xi

n X

(g + M ) xi−1

i=1

Z

i=1

Z f (xi−1 )

f (xi−1 )(xi − xi−1 ).

i=1

Rx Now, let G : [a, b] → R be defined by G(x) = a g. As G(a) = 0 and f is nonnegative and decreasing on [a, b], we have Z xi n n X X
f (xi−1 ) g = f (xi−1 ) G(xi ) − G(xi−1 ) xi−1

i=1

i=1

= =

n X i=1 n X

f (xi−1 )G(xi ) −

n−1 X

f (xi )G(xi )

i=1


f (xi−1 ) − f (xi ) G(xi ) + f (b)G(b)

i=1

≤

n X


f (xi−1 ) − f (xi ) sup G(x) + f (b) sup G(x) x∈[a,b]

i=1

=

x∈[a,b]


f (a) − f (b) sup G(x) + f (b) sup G(x) x∈[a,b]

x∈[a,b]

= f (a) sup G(x). x∈[a,b]

So we can write Z b n X f (g + M ) ≤ f (a) sup G(x) + M f (xi−1 )(xi − xi−1 ). x∈[a,b]

a

i=1

On the other hand, by a result due to Darboux, we have Z b n X lim f (xi−1 )(xi − xi−1 ) = f, ||P ||→0

i=1

a

170

2. SOLUTIONS

where ||P || = max1≤i≤n (xi − xi−1 ). So letting ||P || → 0 in the above inequality, we obtain Z b Z b f, f (g + M ) ≤ f (a) sup G(x) + M x∈[a,b]

a

a

implying b

Z

f g ≤ f (a) sup G(x). x∈[a,b]

a

Replacing g by −g in the above inequality, we obtain Z b f (−g) ≤ f (a) sup −G(x), x∈[a,b]

a

yielding b

Z

f g ≥ f (a) inf G(x). x∈[a,b]

a

These inequalities together with the facts that f (a) ≥ 0 and that G is continuous, in view of the Intermediate Value Theorem, imply that there exists a number ξ ∈ [a, b] such that Z b Z ξ f g = f (a)G(ξ) = f (a) g, a

a

which is what we want. (b) A proof similar to that of (a) settles (b). However, here is a slick proof of (b) using (a). Define functions f1 , g1 : [0, b − a] → R as follows f1 (x) = f (b − x), g1 (x) = g(b − x). It is plain that f1 and g1 satisfy the hypotheses of (a). It thus follows from (a) that there exists ξ ∈ [0, b − a] such that Z b−a Z ξ f1 g1 = f1 (0) g1 . 0

0

R b−a

Rξ

That is, 0 f (b − x)g(b − x)dx = f (b) 0 g(b − x)dx. Performing the substitution b − x = t, we easily obtain Z b Z b f g = f (b) g, a

η

where η = b − ξ, as desired.

 Rp

Now, to prove the assertion, let M = 0 |g|. If M = 0, the assertion is trivial Rb because we would have a f g = 0 for all b > a, from which, the assertion trivially follows. So suppose M > 0. As limx→+∞ f (x) = 0 and f is decreasing on [0, +∞), we see that f (x) ≥ 0 on [0, +∞) and that for given ε > 0, there exists an x0 > 0 such that ε f (x) = f (x) < , 2M whenever x ≥ x0 . It now follows from Lemma 2 that for all x0 < x1 < x2 , there exists ξ ∈ [x1 , x2 ] such that Z Z x2 Z ξ ξ g . f g = f (x1 ) g = f (x1 ) x1 x1 x1

2.20. TWENTIETH COMPETITION

171

Since g is periodic with period p, so is |g|. And since Z p g = 0, 0

we see that Z

jp

Z

ip

g =

g = 0, ip

(i−1)p

Z

p

g ,

0

for all i, j ∈ N. Now, set T = {kp : k ∈ N} ∩ [x1 , ξ]. It is plain that T is a finite set. For x0 < x1 < x2 , there are two cases to consider. (i) T = ∅. (ii) T 6= ∅. First, if T = ∅, then we have [x1 , ξ] ⊆ [(i − 1)p, ip] for some i ∈ N. So we can write Z Z x2 Z ξ ξ g = f (x1 ) ≤ f (x ) g f g 1 x1 x1 x1 Z ip Z p ε g = ε g = ε M = ε < ε. ≤ 2M (i−1)p 2M 0 2M 2 That is, Z

x2

x1

f g < ε.

Next, if T 6= ∅, let ip and jp be the closest points of T to x1 and ξ, respectively. Note that i, j ∈ N and i ≤ j. We can write Z Z Z x2 Z jp Z ξ ξ ip g = f (x1 ) g+ g+ g f g = f (x1 ) x1 x1 ip jp x1 Z Z ξ Z ip Z ξ ! ip ε g + g . g+ = f (x1 ) g ≤ x1 2M jp x1 jp But [x1 , ip] ⊆ [(i − 1)p, ip] and [jp, ξ] ⊆ [jp, (j + 1)p]. So we can write Z x2 Z ip Z (j+1)p ! ε g + g f g ≤ 2M x1 (i−1)p jp Z p Z p  ε g + g = ε (M + M ) = ε. = 2M 2M 0 0 That is, Z

x2

x1

f g ≤ ε.

Thus, in any case, Z

x2

x1

f g < 2ε,

and hence from Lemma 1, it follows that

R +∞ 0

f g converges, which is what we want.

Second solution: Let G : [0, +∞) → R be defined by G(x) = R x+p from the hypothesis that x g = 0, and hence Z x Z x−np G(x) = g= g, np

0

Rx 0

g. It follows

172

2. SOLUTIONS

  where n = xp and b.c denotes the integer part function. Consequently, G(x) ≤ Rp g = M for all x ≥ 0. Thus, the function G is bounded. Now, we prove that the 0 assertion holds true under this hypothesis as well. That is, the assertionR will remain p true if in the problem one replaces “g is a p-periodic function such that 0 g = 0” by Rx “the function G : [0, +∞) → R, defined by G(x) = 0 g, is bounded”. It is worth mentioning that the idea of this proof is very much like that of Dirichlet’s Test for convergent series. Just as in the first it suffices to show that for given R y solution, ε > 0 there is an N > 0 such that x f g < ε whenever x, y ∈ R with N < x < y. By the First Fundamental Theorem of Calculus, G0 = g. So we can write Z y Z y Z y Z y y 0 = f G − = Gdf f dG = f G f g x x x x x Z y Z y Gd(−f ) Gd(−f ) ≤ 2M f (x) + ≤ |f (y)G(y) − f (x)G(x)| + x x Z y
≤ 2M f (x) + M d(−f ) = 2M f (x) + M f (x) − f (y) x

≤ 3M f (x). But limx→+∞ f (x) = 0. Thus, there is an N > 0 such that f (x) < x > N . Therefore, Z y f g < ε, x R +∞ whenever N < x < y, and hence 0 f g converges, as desired.

ε 3M

whenever



2.21. Twenty First Competition 2.21.1. Analysis. 1. (a) Suppose α is a limit point of the set S such that for every ε > 0 the set {x ∈ S : |f (x)| ≥ ε} is finite. For given ε > 0, let  δ = min α − x : f (x) ≥ ε . It is easily verified that |f (x)| < ε whenever 0 < |x − α| < δ. This proves the assertion. (b) To prove the assertion by contradiction, suppose there exist an ε > 0 and an infinite sequence (xi )+∞ i=1 of S such that |f (xi )| ≥ ε for all i ∈ N. Since the set S is compact, the subset {xi }+∞ i=1 has a limit point in S, say, α ∈ S. It is obvious that limx→α f (x) 6= 0, a contradiction. Thus, the assertion follows.  2. (a) To prove the assertion by contradiction, suppose that there exists an ε0 > 0 such that to each n ∈ N, there corresponds an xn ∈ R with xn > n and a tn ∈ K satisfying g(xn + tn ) − g(xn ) ≥ ε0 , for all n ∈ N. Since K is compact, if necessary, by passing to a subsequence of (tn )+∞ n=1 , we may assume that limn tn = t0 for some t0 ∈ K. Define f : [0, +∞) → R by f (x) = g(x + t0 ) − g(x). As limx→+∞ f (x) = 0 and g is continuous, we easily see that f is uniformly continuous on [0, +∞). Thus, for given ε0 > 0, there exists a δ0 > 0 such that g(x + t0 ) − g(y + t0 ) = f (x) − f (y) < ε0 , 2

2.21. TWENTY FIRST COMPETITION

173

whenever x, y ∈ [0, +∞) and |x − y| < δ0 . On the other hand, it follows from the hypothesis that for given ε0 > 0, there exists an M > 0 such that g(x + t0 ) − g(x) < ε0 , 2 whenever x > M . Now, since K is compact, and hence bounded, and limn tn = t0 , choose n ∈ N large enough so that xn + tn − t0 > 0, xn > M , and |tn − t0 | < δ0 , where M > 0 and δ0 are as in the above. We can write ε0 ≤ g(xn + tn ) − g(xn )
≤ g (xn + tn − t0 ) + t0 − g(xn + t0 ) + g(xn + t0 ) − g(xn ) ε0 ε0 < + = ε0 , 2 2 implying ε0 < ε0 . So the assertion follows by way of contradiction. R  x+1 (b) To prove limx→+∞ x g(u)du − g(x) = 0, note first that by the Mean Value Theorem for integrals, for all x ∈ R, there exists a tx ∈ [0, 1] such that Z x+1 g(u)du = g(x + tx ). x

Now, letting K = [0, 1] in (a), for given ε > 0, find M > 0 from (a). We have Z x+1 g(u)du − g(x) = g(x + tx ) − g(x) < ε, x R  x+1 whenever x > M . That is, limx→+∞ x g(u)du − g(x) = 0, as desired. To prove limx→+∞

g(x) x

= 0, it suffices to show that limx→+∞

g(x) [x]

= 0, where

limx→+∞ [x] x

= 1. To this end, note that [.] denotes the integer part function, for we can write


g [x] + tx − g [x] g [x] g(x) = + , [x] [x] [x] where tx = x −
[x] ∈ [0,
1) for all x ∈ R
with x > 1. We need to show that g [x]+tx −g [x]

g [x]

limx→+∞ = limx→+∞ [x] = 0. To this end, letting K = [0, 1] in [x] (a), for given ε > 0, there exists an M > 0 such that g(x + t) − g(x) < ε, whenever x > M and t ∈ [0, 1]. In particular, if x > M + 1, then [x] > M , implying that
g [x] + tx ) − g([x] < ε,
because tx ∈ [0, 1]. Thus, limx→+∞ g(x) − g([x]) = 0, and hence

g [x] + tx − g [x] = 0. lim x→+∞ [x]
g [x] To prove limx→+∞ [x] = 0, it suffices to show that limn→+∞ g(n) n = 0. To see this, note that part (ii) of the lemma presented in Solution 3 of 2.7.1, which is known as Stolz’s Second Theorem, together with the hypothesis yields
g(n) g(n) − g(n − 1) = lim = lim g(n) − g(n − 1) = 0, lim n→+∞ n n→+∞ n→+∞ n − (n − 1)

174

2. SOLUTIONS

as desired. Therefore, limx→+∞

g(x) x

= 0, which is what we want.



3. We need the following lemma which is (essentially) due to Leo M. Levine (1977). Lemma. Let f : (a, b) → R be a bounded function. Let D, L, and R denote the set of points at which f is discontinuous, has a left hand limit, and has a right hand limit, respectively. Then, D ∩ (L ∪ R) is countable. Proof. Define ω : (a, b) → [0, +∞) by 

 ω(x) = lim+ sup f [x − δ, x + δ] − inf f [x − δ, x + δ] . δ→0 S It is readily verified that x ∈ D if and only if ω(x) > 0. This implies D = n∈N Dn , where Dn = {x ∈ (a, b) : ω(x) > n1 }, and hence [
[ [
D ∩ (L ∪ R) = (Dn ∩ L) (Dn ∩ R) . n∈N

n∈N

Thus, to establish the lemma, it suffices to show that Dn ∩ L and Dn ∩ R are countable for all n ∈ N. To this end, suppose x0 ∈ Dn ∩ L (resp. x0 ∈ Dn ∩ R). As 1 x0 ∈ L (resp. x0 ∈ R), there exists a δ > 0 such that |f (x) − f (x− 0 )| < 2n (resp. + 1 |f (x) − f (x0 )| < 2n ) whenever x ∈ (x0 − δ, x0 ) (resp. x ∈ (x0 , x0 + δ)). From this, we obtain f (x1 ) − f (x2 ) < 1 , n whenever x1 , x2 ∈ (x0 − δ, x0 ) (resp. x1 , x2 ∈ (x0 , x0 + δ)). Therefore, ω(x) ≤ n1 for all x ∈ (x0 − δ, x0 ) (resp. x ∈ (x0 , x0 + δ)), implying x ∈ / Dn . It thus follows that any point of Dn ∩ L (resp. Dn ∩ R) is the right (resp. left) end point of an open interval which contains no point of Dn ∩ L (resp. Dn ∩ R). But these intervals are obviously disjoint and hence form a countable set. So Dn ∩ L (resp. Dn ∩ R) is countable for all n ∈ N, as desired.  It is now obvious that the lemma together with Lebesgue’s Integrability Criterion for Riemann integrals implies that a bounded function f : [a, b] → R is integrable in the Riemann sense if and only if [a, b] \ (L ∪ R) is a set of measure zero. This clearly proves the assertion. It is also worth mentioning that, in view of the lemma, one can mimic the second proof presented for Solution 4 of 2.9.1 to give a direct proof of the assertion. We omit the details for the sake of brevity. 

2.21.2. Algebra. 1. It is easily checked that if I, J, K are ideals of R with s

s

J ⊆ I, then I ∩ (J + K) = J + I ∩ K. Let A ⊆ B and C ⊆ D. We prove that s

A+C ⊆ B +D. To see this, suppose E is an ideal of R such that A+C +E = B +D. We need to show that E = B + D. To this end, we can write B ∩ (A + C + E) = B ∩ (B + D). But A ⊆ B and B ⊆ B + D. So, we have A + B ∩ (C + E) = B, s

implying that B ∩ (C + E) = B because A ⊆ B. Thus, B ⊆ C + E. Likewise, from D ∩ (A + C + E) = D ∩ (B + D), we obtain D ⊆ A + E. Note that A ⊆ B and B ⊆ C + E. Thus, A ⊆ C + E, and hence B + D = A + C + E = C + E.

2.21. TWENTY FIRST COMPETITION

175

Now, from C + E = B + D, we obtain D ∩ (C + E) = D ∩ (B + D), yielding s

C + (D ∩ E) = D, which, in turn in view of C ⊆ D, implies D ∩ E = D. Therefore, D ⊆ E. Analogously, since C ⊆ D ⊆ A + E, we have B + D = C + A + E = A + E, s

yielding B ∩(A+E) = B ∩(B +D). Hence, A+B ∩E = B, which in view of A ⊆ B, implies B ∩E = B. Therefore, B ⊆ E. It follows that B +D ⊆ E +E = E ⊆ B +D. In other words, E = B + D, as desired.  2. First we need to recall that if p is a prime, then every group G of order p2 is abelian (for a proof see Solution 1 of 2.16.2). We now prove the assertion. From H ≤ Z(G) ≤ G, we see that p2 = [G : H] = [G : Z(G)][Z(G) : H], implying that [G : Z(G)] = 1 or p or p2 . If [G : Z(G)] = 1, then G = Z(G). In other words, G is abelian and hence G0 = {e}, which is cyclic. If [G : Z(G)] = p, G then Z(G) is cyclic, implying that G is abelian. This again yields G0 = {e}, which G is cyclic. Finally, if [G : Z(G)] = p2 , then Z(G) is abelian because every group of 2 order p is abelian whenever p is a prime. It thus follows from the Fundamental G Theorem of finite abelian groups that Z(G) is either cyclic or there are x, y ∈ G G G such that Z(G) = hxZ(G), yZ(G)i. Just as we saw in the above, if Z(G) is cyclic, then G is abelian, in which case the assertion trivially holds. So assume that G Z(G) = hxZ(G), yZ(G)i for some x, y ∈ G. It follows that there is a z ∈ Z(G) such G that xy = yxz, for Z(G) is abelian. Note that z = x−1 y −1 xy ∈ Z(G). By proving that G0 = hx−1 y −1 xyi = {z k : k ∈ Z}, we complete the proof. A straightforward induction on i + j reveals that xi y j = y j xi z ij , for all i, j ∈ N. Also, as yx = xyz −1 and y −1 x = xy −1 z, inducting on i + j, we obtain y j xi = xi y j z −ij , y −j xi = xi y −j z ij , G for all i, j ∈ N. Again, as Z(G) = hxZ(G), yZ(G)i is abelian, for all g ∈ G there are i, j ∈ N and zg ∈ Z(G) such that g = xi y j zg . Therefore, for given g, g 0 ∈ G, there 0 0 are i, j, i0 , j 0 ∈ N and zg , zg0 ∈ Z(G) such that g = xi y j zg and g 0 = xi y j zg0 . We can write
0 0

0 0
g −1 g 0−1 gg 0 = y −j x−i y −j x−i xi y j xi y j

0 0 0 0
0 0
= z −ij x−i y −j y −j x−i xi y j y j xi z i j 0 0 0 0
0 0 = x−i y −(j+j ) x−i xi y j+j xi z i j −ij 0

0

0

0

= x−i y −(j+j ) x−i xi xi y j+j z −(j+j
0 0 0 = x−i y −(j+j ) xi y j+j z −j(i+i ) 0

0

0

0

)i0 i0 j 0 −ij

z

0

= x−i xi y −(j+j ) y j+j z i(j+j ) z −j(i+i ) = z ij −1 0−1

0

0

−i0 j

.

−1 −1

In other words, g g gg = (x y xy)k , where k = ij 0 − i0 j ∈ Z. Consequently, G0 is the cyclic group generated by z = x−1 y −1 xy ∈ Z(G), completing the proof. 

176

2. SOLUTIONS

3. (a) Let At ∈ Mn (R) denote the transpose of the matrix A ∈ Mn (R). For all 1 ≤ i, j ≤ n, use (AAt )ij to denote the ij entry of the matrix AAt . It follows from the hypothesis that (AAt )ij = tr(A)aij , for all 1 ≤ i, j ≤ n. It is now obvious that tr(A) 6= 0, for otherwise (AAt )ij = 0 for all 1 ≤ i, j ≤ n, yielding tr(AAt ) = 0, which obtains A = 0, contradicting the hypothesis. Thus, tr(A) 6= 0, which is what we want. (b) In view of (a), we have A=

1 AAt . tr(A)

Since AAt is symmetric, so is A, as desired. (c) In view of the second proof presented for Solution 7 of 2.9.3, it suffices to show that rank(A) = 1. To this end, suppose A ∈ Mn (R) satisfies aik ajk = akk aij for all 1 ≤ i, j, k ≤ n. For 1 ≤ j ≤ n, let us use Aj to denote the jth column of the matrix A. As tr(A) 6= 0, there exists a 1 ≤ k ≤ n such that akk 6= 0. It thus follows ajk . from the hypothesis that for all 1 ≤ j ≤ n, we have Aj = bjk Ak , where bjk = akk This obviously proves that rank(A) = 1, which is what we want.  2.22. Twenty Second Competition 2.22.1. Analysis. 1. By the Mean Value Theorem, there exists a c ∈ (0, 1) (0) = 1. If f 0 (x) ≥ 1 (resp. f 0 (x) ≤ 1) for all x ∈ (0, 1), we such that f 0 (c) = f (1)−f 1−0 see that f (x) = x for all x ∈ [0, 1], in which case the assertion is trivial. To see this, suppose to the contrary that there exists c ∈ (0, 1) such that f (c) > c or f (c) < c. Suppose f (c) > c. It follows from the Mean Value Theorem that there exists d ∈ (c) (0) (c, 1) (resp. d ∈ (0, c)) such that f 0 (d) = f (1)−f < 1 (resp. f 0 (d) = f (c)−f > 1), 1−c c−0 a contradiction in any event. Likewise, if f (c) < c, we obtain a contradiction. Thus, there exist a, b ∈ (0, 1) with a < c < b such that (f 0 (a)−1)(f 0 (b)−1) < 0. It follows from Darboux’s Theorem that the range of f
0 , denoted by R, includes the closed
interval [m, M ], where m = min f 0 (a), f 0 (b) < 1 < M = max f 0 (a), f 0 (b) . It thus follows that [1 − ε, 1 + ε] ⊆ R, where ε = min(1 − m, M − 1). It is obvious that for all y1 ∈ [1 − ε, 1 + ε], with y1 = f 0 (x1 ) < 1 for some x1 ∈ [0, 1], there exists y2 ∈ [1 − ε, 1 + ε], with y2 = f 0 (x2 ) > 1 for some x2 ∈ [0, 1], such that 1 f 0 (x1 )

+

1 f 0 (x2 )

= 2.

(∗)

This obviously proves the assertion when n ∈ N is even. If n ∈ N is odd, use (∗) and the fact that f 01(c) = 1, where c is as in the above, to prove the assertion.  P+∞ 2. (a) We have fn ≤ i=1 fi = f 2 for all n ∈ N because fn ’s are all nonnegative. Define gn : X −→ R by ( 0 x ∈ f −1 ({0}), gn (x) = fn (x) x ∈ X \ f −1 ({0}). f (x)

2.22. TWENTY SECOND COMPETITION

177

It is obvious that fn = gn f for all n ∈ N. It remains to show that gn is continuous for all n ∈ N. To this end, note that gn is continuous on the open set X \ f −1 ({0}) because so are f and fn on X \ f −1 ({0}) (n ∈ N). Suppose that x0 ∈ X \ f −1 ({0}) is arbitrary. It follows that for given ε > 0, there exists δ > 0 such that fn (x) fn (x0 ) < ε, − f (x) f (x0 ) whenever x ∈ X \ f −1 ({0}) and |x − x0 | < δ. Since X \ f −1 ({0}) is an open set and x0 ∈ X \ f −1 ({0}), choose δ small enough such that x ∈ X \ f −1 ({0}) whenever |x − x0 | < δ. So we have gn (x) − gn (x0 ) < ε, whenever |x−x0 | < δ. That is, gn is continuous at x0 . Next, suppose x0 ∈ f −1 ({0}) is arbitrary. As f is continuous, we see that for given ε > 0, there exists a δ > 0 such that f (x) = f (x) − f (x0 ) < ε, whenever |x − x0 | < δ. Now if |x − x0 | < δ, we can write ( 0<ε x ∈ f −1 ({0}) fn (x) gn (x) − g(x0 ) = gn (x) = . x ∈ X \ f −1 ({0}) f (x) ≤ f (x) < ε In other words, |gn (x) − g(x0 )| < ε whenever |x − x0 | < δ. That is, gn is continuous at x0 , as desired. P+∞ (b) Suppose that n=1 gn converges uniformly on X and that the interior of P+∞ f −1 ({0}) is empty. It follows that on X and n=1 gn is a continuous function P+∞ that X \ f −1 ({0}) is a dense subset of X. We see from (a) that n=1 gn f = f 2 , P+∞
P+∞ yielding f f − n=1 gn = 0. This easily implies f (x) − n=1 gn (x) = 0 whenever P+∞ x ∈ X \ f −1 ({0}). That is, the continuous function f − n=1 gn vanishes on
P +∞ the dense subset X \ f −1 {0} . Thus, f − n=1 gn vanishes on X, and hence P+∞ f = n=1 gn , which is what we want.  3. Note first that f is continuous on D because the sequence (fn )+∞ n=1 uniformly converges to f on D as n tends to infinity and that fn ’s are all analytic, and hence continuous. Thus, in view of Morera’s Theorem, it suffices to show that R f (z)dz = 0, where γ0 is any simple closed curve inside γ. We have γ0 Z Z f (z)dz = lim fn (z)dz = 0, γ0

n

γ0

(fn )+∞ n=1

because uniformly converges to f on γ0 and fn ’s are analytic. Assuming that z0 is inside γ, we see from Cauchy’s Integral Formula that Z Z fn (z) 1 f (z) 1 fn (z0 ) = dz, f (z0 ) = dz, 2πi γ z − z0 2πi γ z − z0 proving the assertion because fn (z0 ) → f (z0 ) as n → +∞.



178

2. SOLUTIONS

2.22.2. Algebra. 1. Suppose [G : K] = 2n − 1 for some n ∈ N. We have [G : H] = [G : K][K : H] = 2(2n − 1). The group G acts on Ω = {Hg : g ∈ G}, the set of all right cosets of H in G, by multiplication from the right. That is, the action is given by the map ϕ : G × Ω → Ω defined by ϕ(a, Hg) = Hga. Recall that ϕ gives rise to a group homomorphism ψ : G → S(Ω) defined by ψ(a) = ϕ(a, .) : Ω → Ω, where S(Ω) denotes the group of the permutations of Ω, i.e., the set of all one-toone maps from Ω onto Ω, which forms a group T under composition of maps. The kernel of the action, by definition, is ker ψ = g∈G g −1 Hg. It follows from the First Isomorphism Theorem for groups that kerGψ ∼ = im(ψ) ≤ S(Ω) ∼ = S2(2n−1) . That G is, ker ψ is isomorphic to a subgroup of the symmetric group of degree 2(2n − 1). Since k ∈ K has order 2, we see that for k ker ψ ∈ kerGψ , we have (k ker ψ)2 = k 2 ker ψ = ker ψ. Thus, the order of k ker ψ, as an element of kerGψ , is equal to 1 or 2. But k ∈ / ker ψ because otherwise, in particular, k ∈ H, which is impossible. Consequently, ord(k ker ψ) = 2. Note that kerGψ acts on Ω via (a ker ψ, Hg) → Hga and that the element (k ker ψ, .) ∈ S(Ω) ∼ = S2(2n−1) has no fixed point, for otherwise Hgk = Hg for some g ∈ G, yielding gkg −1 ∈ H, contradicting the hypothesis. Therefore, the element (k ker ψ, .) ∈ S(Ω) corresponds to an odd permutation of S(Ω) because, in view of (k ker ψ, .)2 = id ∈ S(Ω), it decomposes into the product of 2n − 1 transpositions or 2-cycles. Thus, the subgroup of kerGψ , say, kerJ ψ for some J ⊇ ker ψ, which consists of all elements that correspond to even permutations of S(Ω) forms a subgroup of kerGψ of index 2. Since [G : ker ψ] = [G : J][J : ker ψ], we can write h G J i [G : J] = : = 2, ker ψ ker ψ which is what we want.



2. If J = R or J = {0}, the assertion is trivial. So suppose that J is a nontrivial ideal of R and J has a maximal ideal, say, M . By obtaining a contradiction, J we settle the proof. Let f : J → M , defined by f (x) = x + M , be the natural J J homomorphism from J onto M . As M is a maximal ideal of J, it follows that M J has no nontrivial ideal. This implies that M is a field or it is a finite ring with zero multiplication. The latter is impossible because it would mean (x+M )(y+M ) = M J for all x, y ∈ J, yielding J 2 ⊆ M , which is impossible. Thus, M is a field. Let e+M J J denote the identity element of the field M , where e ∈ J \ M . Define f ∗ : R → M by ∗ ∗ f (x) = f (xe + M ). It is readily seen that f is a homomorphism of rings. So, we J see from the First Isomorphism Theorem for rings that kerRf ∗ ∼ . Consequently, =M J ∗ ∗ ker f is a maximal ideal of R, for M is field. Thus, J ⊆ ker f . This, in particular, implies f ∗ (e) = M . On the other hand, f ∗ (e) = f (e2 + M ) = (e + M )2 = e + M . So, we must have M = e + M , yielding e ∈ M , which is impossible. The assertion thus follows by contradiction.  3. To prove the assertion by contradiction, suppose that rank(L) = 1 so that LV = hαi for some α ∈ V . With no loss of generality, we may assume that the characteristic polynomial of T is irreducible over F . We claim that T k α ∈ ker L,

2.23. TWENTY THIRD COMPETITION

179

for each k = 0, 1, 2, . . .. As the rank of LT k is at most one and tr(LT k ) = tr(T ST k − ST k+1 ) = 0, for each k = 0, 1, 2, . . ., it follows that LT k is nilpotent. And hence its eigenvalues are all zero. But LT k α ∈ LV = hαi for all nonnegative integers k. That is, α is an eigenvector of LT k , and hence LT k α = 0 for all nonnegative integers k. Now, set

  W = T k α : k = 0, 1, 2, . . . . It is plain that W is invariant under T and that W is a nontrivial subspace of V because α is nonzero and W ⊆ ker L. It thus follows that the characteristic polynomial of T |W , the restriction of T to the invariant subspace W , divides the minimal polynomial of T , which is equal to the characteristic polynomial of T , for the characteristic polynomial of T is irreducible over F . This contradicts the hypothesis that the characteristic polynomial of T is irreducible. Thus, the assertion follows by contradiction. 

2.22.3. General. a b c 1. F 2. F 3. F 4. 5. F 6. F 7. 8. F 9. F 10. F 11. F 12. F 13. F 14. F

d

F

F

a 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.

b

c F F

d

F F F F F F F F F F F F

29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42.

a F F

b

c

d

F F F F F F F F F F

2.23. Twenty Third Competition 2.23.1. Analysis. 1. (a) We prove the assertion by way of contradiction. Suppose that the set {fα : α ∈ R} is compact with respect to the uniform norm of Cb (R) and yet f is not uniformly continuous on R. It follows that there are +∞  > 0 and sequences (xn )+∞ n=1 and (yn )n=1 in R such that limn xn = limn yn = ±∞, limn (xn − yn ) = 0, and |f (xn ) − f (yn )| ≥  for all n ∈ N. By the hypothesis, +∞ the sequence (gn )+∞ n=1 , where gn = fxn , has a subsequence, say (gnk )k=1 , which converges uniformly to fα on R for some α ∈ R. It follows that limk gnk (ynk −xnk ) = fα (0). On the
other hand, limk gnk (0) = fα (0). Consequently, limk gnk (ynk − xnk ) − gnk (0) = 0. But gn (yn − xn ) − gn (0) = f (yn ) − f (xn ) ≥ , k k k k k k

180

2. SOLUTIONS

for all k ∈ N, which is in contradiction with
lim gnk (ynk − xnk ) − gnk (0) = 0. k

Thus, f is uniformly continuous on R, as desired. (b) We disprove the proposition by showing that for the function f : R −→ R 2 defined by f (x) = e−x , the sequence {fα : α ∈ R} is not compact with respect to the uniform norm of Cb (R). Note first that f is bounded and uniformly continuous on R because limx→±∞ f (x) = 0. To disprove the proposition by contradiction, suppose that {fα : α ∈ R} is compact with respect to the uniform norm of Cb (R). It +∞ follows that the sequence (fn )+∞ n=1 has a subsequence, say (fnk )k=1 , which converges 2 2 uniformly on R to fα for some α ∈ R. Thus, 0 = limk e−(x+nk ) = e−(x+α) , 2 implying that e−(x+α) = 0 for all x ∈ R, which is obviously impossible. So the proposition is disproved by way of contradiction.  2. First solution: P+∞ First we need to recall Abel’s Continuity Theorem which asserts that if n=0 an is a convergent series of complex numbers, then lim

r→1−

+∞ X

an rn =

n=0

+∞ X

an .

n=0

Without loss of generality, if necessary by adding or subtracting multiples of 2π, we may assume that 0 < α < 2π. Set Sn (α) =

n X eikα k=1

k

.

We have Sn0 (α)

=i

n X k=1

eikα = i

ei(n+1)α − eiα . eiα − 1

Consequently, α

Z α ei(n+1)t 1
dt − i 1 + it dt it e −1 π e −1 π Z α Z α  ei(n+1)t  1 (cos t − 1) − i sin t = d − i(α − π) − i dt it − 1 e n + 1 2(1 − cos t) π π Z Z α 1 ei(n+1)t α i ei(n+2)t i α = + dt − i(α − π) + dt n + 1 eit − 1 π n + 1 π (eit − 1)2 2 π Z α 1  t − t d sin 2 π sin 2 Z α i(n+2)t (−1)n i i(α − π) 1 ei(n+1)α e = + + dt − iα it 2 n+1 e −1 2(n + 1) n + 1 π (e − 1) 2 α − ln sin . 2 Z

Sn (α) − Sn (π)

= i

2.23. TWENTY THIRD COMPETITION

181

So we can write α π − α
+i 2 2

Sn (α) − − ln 2 sin

=

1 ei(n+1)α n + 1 eiα − 1 Z α i(n+2)t n (−1) e i + dt. + 2(n + 1) n + 1 π (eit − 1)2
ln 2 + Sn (π) +

On the other hand, we have +∞ X (−1)k−1

lim Sn (π) = − n

k=1

k

= − ln 2.

n−1 P+∞ To see this, just apply Abel’s Continuity Theorem to the power series n=1 (−1)n xn = n P+∞ ln(1 + x) and note that the alternating series n=1 (−1) converges by Leibniz’s n Theorem. Also, it is plain that   1 ei(n+1)α (−1)n lim + = 0. n n + 1 eiα − 1 2(n + 1)

Finally, Z i n + 1

α

π

ei(n+2)t dt (eit − 1)2

≤

= for all n ∈ N and limn

Z 1 α n+1 π Z 1 α n+1 π

i(n+2)t e 2 dt eit − 1 1 dt , eit − 1 2

1 n+1

= 0, implying that Z α i(n+2)t i e lim dt = 0. n n + 1 π (eit − 1)2

Consequently,  π − α
 α lim Sn (α) − − ln 2 sin + i = 0. n 2 2 In other words, +∞ inα X e α π−α = − ln 2 sin + i . n 2 2 n=1

Now, equating the real and imaginary parts of the above equality obtains +∞ X cos(nα) n n=1

= − ln 2 sin

+∞ X sin(nα) n n=1

=

α , 2

π−α , 2

which is what we want. Second solution: For this solution, first we need to recall Dirichlet’s Theorem +∞ which asserts that if (an )+∞ n=1 and (bn )n=1 are sequences of complex and real numbers, respectively, such that the sequence of partial sums of (an )+∞ n=1 is bounded

182

2. SOLUTIONS

and that (bn )+∞ n=1 is monotonic and tends to zero as n → +∞, then converges. Note that, as we saw in the first solution, we have n X

eikα =

k=1

P+∞

n=1

an bn

ei(n+1)α − eiα , eiα − 1

implying that n X 2 , eikα ≤ iα e − 1 k=1

for all n ∈ N. On the other hand, the sequence ( n1 )+∞ n=1 is decreasing and tends to zero as n → +∞. Thus, by Dirichlet’s Theorem, the series +∞ inα X e n n=1

converges whenever α ∈ R \ {2kπ : k ∈ Z}. Again, without loss of generality, we may assume that 0 < α < 2π. Let z = reiα , where 0 ≤ r < 1. Note that 1 − z = (1 − r cos α) + i(−r sin α). We can write +∞ n +∞ inα X X z e − ln(1 − z) = = rn , n n n=1 n=1 where ln denotes the principal value of the natural logarithm, which is defined by π π ln z = ln |z| + i arg z, where arg z = arctan Im(z) Re(z) ∈ (− 2 , 2 ). On the other hand,
− ln(1 − z) = − ln |1 − z| + i arg(1 − z)   p −r sin α 2 . = − ln 1 − 2r cos α + r + i arctan 1 − r cos α Letting r → 1− , in view of Abel’s Continuity Theorem, we obtain +∞ inα X e n n=1

√ − sin α = − ln 2 − 2 cos α − i arctan 1 − cos α r α α = − ln 4 sin2 + i arctan cot( ) 2 2 α π−α = − ln 2 sin + i . 2 2

Thus, +∞ inα X e α π−α = − ln 2 sin + i . n 2 2 n=1

Equating the real and imaginary parts of the above equality, we obtain

which is what we want.

+∞ X cos(nα) n n=1

= − ln 2 sin

+∞ X sin(nα) n n=1

=

α , 2

π−α , 2 

2.23. TWENTY THIRD COMPETITION

183

3. It is worth mentioning that the hypothesis that f 0 and fn0 ’s are continuous is redundant. Let x ∈ [a, b) be arbitrary. In view of the definition of the derivative, the Mean Value Theorem, and the continuity of g, we can write
1 f 0 (x) = lim n f (x + ) − f (x) n→+∞ n
1 = lim lim n fk (x + ) − fk (x) = lim lim fk0 (xn ) n→+∞ k→+∞ n→+∞ k→+∞ n = lim g(xn ) = g(x), n→+∞

where x < xn < x + n1 . Thus, f 0 (x) =
g(x) for all x ∈ [a, b). If x = b, just write f 0 (b) = limn→+∞ −n f (b − n1 ) − f (b) and repeat the above argument to get f 0 (b) = g(b). So the proof is complete. 

2.23.2. Algebra. 1. This problem is wrong! For n ∈ N, use Dn to denote the dihedral group of order 2n. A presentation of Dn is

 a, b : an = e, b2 = e, ak 6= e (0 < k < n), ab = ba−1 ,  and hence Dn = ai bj : 0 ≤ i < n, j = 0, 1 . We show that D8 is a counterexample. G , where Z(G) denotes the center of the group G First, recall that Inn(G) ∼ = Z(G) ∼ D8 . To show that D8 is a (see Solution 1 of 2.16.2). Therefore, Inn(D8 ) = Z(D8 ) D2n ∼ counterexample, it suffices to prove that Z(D D = n for all n ∈ N. To this end, 2n )  n just note that Z(D2n ) = e, a , where a is the element of order 2n in the above presentation of D2n , and that  i j D2n = A B : A = aZ(D2n ), B = bZ(D2n ), 0 ≤ i < n, j = 0, 1 Z(D2n ) ∼ = Dn , for one can easily verify that An = Z(D2n ) = B 2 , AB = BA−1 , and that Ak 6= Z(D2n ) for all 0 < k < n.  2. First, we prove that H is a two-sided ideal of R by showing that for every r ∈ R and x ∈ H, there is a t ∈ R such that rx = xt. To this end, note first that rR is a right ideal of R. If rR = {0}, there is nothing to prove. If rR 6= {0}, then x ∈ rR, and hence there is a y ∈ R such that x = ry. Likewise, considering the right ideal yR, we obtain a t ∈ R such that x = yt. So we can write rx = (ry)t = xt. That is, for r ∈ R and x ∈ H, there is a t ∈ R such that rx = xt, as desired. So far, we have actually shown that the intersection of all nonzero right ideals of R is the same of those of the nonzero left ideals of R. Next, suppose that H 2 6= 0. We prove that R is a division ring. To this end, we first show that R has no zero divisors. Suppose to the contrary that there are nonzero elements x, y ∈ R such that xy = 0. As H 2 6= 0, there are a, b ∈ H such that ab 6= 0. For a, b ∈ R, just as we saw in the above, there are s, t ∈ R such that xs = a and yt = b because a, b ∈ H. Again, for s ∈ R and b ∈ H, there is a u ∈ R such that sb = bu. We can write xy = 0 =⇒ xyt = 0 =⇒ xb = 0 =⇒ xbu = 0 =⇒ xsb = 0 =⇒ ab = 0,

184

2. SOLUTIONS

a contradiction. Thus, R has no zero divisors. To show that R has a multiplicative identity, choose a nonzero a ∈ H and note that a ∈ aR, implying that a = ae for some e ∈ R. Since R has no zero divisors and a(ea − a) = 0, we obtain ea = a. Now for an arbitrary x ∈ R, we have (xe − x)a = a(ex − x) = 0, from which, we obtain xe = ex = x. That is, e is the multiplicative identity of R. For the rest, since H is an ideal of R, we need to prove that every nonzero element of H is invertible. To this end, for a nonzero a ∈ H, note that a2 6= 0, and hence a = a2 t for some t ∈ R. It follows that a(e − at) = 0, which, in turn, yields at = e. This implies that ta(e − ta) = 0, from which, we obtain e = ta. Thus e = ta = at. That is, a is invertible, which is what we want.  3. Let A = (aij ) ∈ Mn (F ). Use Eij and In to denote the matrix whose ij entry is 1 and zero elsewhere and the identity matrix, respectively. If 1 ≤ i, j ≤ n with i 6= j, then tr(Eij ) = 0. By the hypothesis, aji = tr(AEij ) = 0 for all 1 ≤ i, j ≤ n with i 6= j. Thus, A is a diagonal matrix. For 1 ≤ i ≤ n, let Bi = diag(1, 0, . . . , 0)−diag(δ1i , δ2i , . . . , δni ), where δji denotes the Kronecker delta. Since tr(Bi ) = 0, the hypothesis implies aii − a11 = tr(ABi ) = 0 for all 1 ≤ i ≤ n. Therefore, A = a11 In , as desired.  2.24. Twenty Fourth Competition Rx 2.24.1. First Day. 1. Define F : [a, b] −→ R by F (x) = a f (t)dt. By the First Fundamental Theorem of Calculus, F 0 (x) = f (x) for all x ∈ [a, b]. Let h = F − g. Obviously, h is differentiable on [a, b] and we have h0 = f − g 0 . It follows from the hypothesis that h0 (a)h0 (b) < 0. This together with Darboux’s Theorem implies that there exists a c ∈ (a, b) such that h0 (c) = 0. In other words, f (c) = g 0 (c), which is what we want.  2. Note first that all real harmonic functions satisfy the Mean Value Property. That is, if D is a domain such that Br0 (a0 ) ⊆ D for some a0 = (x0 , y0 ) ∈ D and r0 > 0 and u : D −→ R is a harmonic function, then Z 2π 1 u(a0 ) = u(a0 + r0 eiθ )dθ. 2π 0 To see this, let B be an open ball such that Br0 (a0 ) ⊆ B ⊆ D and f a holomorphic function on the simply connected domain B such that u|B = Ref . It easily follows from Cauchy’s Integral Formula that Z 2π 1 f (a0 ) = f (a0 + r0 eiθ )dθ. 2π 0 Taking the real part of the both sides of the above equation proves the counterpart of it with f replaced by u, as desired. Next, we need the following version of the Maximum Modulus Theorem for harmonic functions, which in fact holds true for all functions satisfying the Mean Value Property. Theorem. Let D be a domain and u : D −→ R be a harmonic function. If there is a point a0 = (x0 , y0 ) ∈ D such that u(z) ≤ u(a0 ) for all z ∈ D, then u(z) = u(a0 ) for all z ∈ D.

2.24. TWENTY FOURTH COMPETITION

185

Proof. Set L = {z ∈ D : u(z) = u(a0 )}. It is plain that L is a nonempty closed subset of D. By proving that L is an open subset of D as well, in view of the connectedness of D, we see that L = D, proving the assertion. To this end, suppose to the contrary that there is a z0 ∈ L which is not an interior point of L. Choose r > 0 such that Br (z0 ) ⊆ D. It follows that there is a b0 ∈ Br (z0 ) such that u(b0 ) < u(a0 ) = u(z0 ). And hence, from the continuity of u, we see that u(z) < u(z0 ) for all z in a small enough neighborhood of b0 . Letting r0 = |z0 − b0 | and b0 = z0 + r0 eiθ0 for some 0 ≤ θ0 ≤ 2π, there thus exists a proper subinterval I of [0, 2π] such that θ0 ∈ I and u(z0 + r0 eiθ ) < u(z0 ) for all θ ∈ I. This, in view of the Mean Value Property of harmonic functions, yields Z 2π 1 u(z0 + r0 eiθ )dθ < u(z0 ), u(z0 ) = 2π 0 which is a contradiction. Thus, L is open, completing the proof.  To prove the assertion, let the set A be as in the statement of the problem. By the Maximum Modulus Theorem for harmonic functions, the set A is open. The assertion follows as soon as we prove that A is closed because D is connected and A is nonempty by the hypothesis. To prove that A is closed, suppose that (an )+∞ n=1 is a sequence of the elements of A such that limn an = a for some a ∈ D. We need to show that a ∈ A. To this end, choose r > 0 such that Br (a) ⊆ D. It follows that there is an n ∈ N such that an ∈ Br (z0 ). As an ∈ A, there is an rn > 0 such that Brn (an ) ⊆ Br (a) ⊆ D and that u(z) ≤ u(an ) for all z ∈ Brn (an ). Since Br (a) is simply connected, there is a holomorphic function f on Br (a) such that u|Br (a) = Ref . By the Maximum Modulus Theorem for harmonic functions, Ref |Brn (an ) = Ref (an ), which, in view of the Cauchy-Riemann Equations, yields f |Brn (an ) = f (an ). Since f is holomorphic on Br (a), it follows that f |Br (a) = f (an ) = f (a), which, in turn, implies u(z) = u(a) for all z ∈ Br (a). Therefore, a ∈ A, which is what we want, completing the proof.  3. First solution: Let p = t(3n − 2n ). In particular, p 3n − 2n . In other words, p

3n ≡ 2n . Obviously, gcd(6, p) = 1. And hence, 2 has an inverse modulo p. That is, p there is an a ∈ N with 1 < a < p such that 2a ≡ 1. So we can write p

p

p

3n an ≡ 2n an ≡ (2a)n ≡ 1, p

implying (3a)n ≡ 1. Use m to denote the order of 3a modulo p, i.e., m is the p least positive integer satisfying (3a)m ≡ 1. It follows that m n. By Fermat’s Little p Theorem, (3a)p−1 ≡ 1, which obtains m p−1. Consequently, m gcd(n, p−1). Note p

p

that m > 1, for otherwise 3a ≡ 1, yielding 3 ≡ 2, which is impossible. This implies that any prime divisor of m, say q, is a divisor of n as well and that q ≤ p − 1 < p. Thus, t(n) ≤ q < p = t(3n − 2n ), as desired. p

Second solution: Let m be the least positive integer such that 3m ≡ 2m . Ob p p viously, 2m 3m ≡ 6 0. Note that if 3k ≡ 2k for some k ∈ N, then m k. To see this,

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2. SOLUTIONS

letting k = qm + r, where r + 1, q ∈ N and 0 ≤ r < k, we can write p

p

p

p

3r ≡ 3k−qm ≡ 3k (3m )−q ≡ 2k (2m )−q ≡ 2k−mq = 2r , p implying that 3r ≡ 2r . Thus, r = 0, yielding k = qm, i.e., m k. This together with p the hypothesis implies that m n. On the other hand, as 6 ≡ 6 0, in view of Fermat’s Little Theorem, we can write p

p

3p−1 ≡ 1 ≡ 2p−1 , yielding m p − 1. Consequently, m gcd(n, p − 1), which, as we saw in the first solution, entails t(n) < t(3n − 2n ), as desired.  4. Use Z to denote the center of G. It follows from the hypothesis that there exist n ∈ N and gi ∈ G (1 ≤ i ≤ n) such that G  = g1 Z, . . . , gn Z . Z It is obvious that gi Z’s (1 ≤ i ≤ n) are all infinite subsets of G. It suffices to show that gi gj = gj gi for all 1 ≤ i, j ≤ n. To this end, for given 1 ≤ i, j ≤ n with i 6= j, it follows from the hypothesis that there exist zi , zj ∈ Z such that (gi zi )(gj zj ) = (gj zj )(gi zi ). This easily yields (gi gj )(zi zj ) = (gj gi )(zi zj ), which, in turn, implies gi gj = gj gi for all 1 ≤ i, j ≤ n, completing the proof.  5. First solution: First, we show that if there are m n-tuples from the set {0, 1} in such a way that every two of which differ at least in d components, then mn ≥ (m − 1)d. 2 To this end, the given n-tuples form the following m × n matrix   a11 · · · a1n  .. ..  .  . ··· .  am1

···

amn

whose row i is the ith n-tuple (1 ≤ i ≤ m). Denote by A the total number of differP ences between any two rows of these m rows. In other words, A = 1≤i<j≤m dij , where dij is the number of components in
which row i and row j differ. As dij ≥ d for all 1 ≤ i < j ≤ m, we obtain A ≥ m 2 d. To find an upper bound for A, fix a column, say column j, and use cj to denote the number of differences that P these m vector can mutually have in their jth component. It is obvious that A = 1≤j≤n cj . 2

m m If there are xj zeros in column j, then cj = xj (m − xj ) ≤ m 2 (m − 2 ) = 4 because the polynomial function f (x) = x(m − x) assumes its maximum at x = m 2 . It P P 2 2 follows that A = 1≤j≤n cj ≤ 1≤j≤n m4 = m4 n . Consequently,   m2 n m m(m − 1) ≥A≥ d= d, 4 2 2

from which, we obtain mn ≥ (m − 1)d, 2 as desired.

2.24. TWENTY FOURTH COMPETITION

187

Now, to prove the assertion, let m = M (2d − 1, d) and note that n = 2d − 1. In view of the above inequality, we have m(2d − 1) ≥ (m − 1)d, 2 yielding 2md − m ≥ 2md − 2d =⇒ m = M (2d − 1, d) ≤ 2d, which is what we want. Second solution: First we need to recall Tur´an’s Theorem from Graph Theory. For the sake of completeness, we quote a simple proof of it which is due to William Staton. Theorem (Tur´ an, 1941). Graphs with n vertices containing no Kr have no 2 edges, for r ≥ 2. more than (r−2)n 2r−2 Proof (William Staton). Induct on r. If r = 2, the result is obvious. Now if the statement is true for Kr -free graphs it must be shown that Kr+1 -free graphs 2 have no more than (r−1)n edges. Let G be such a graph, and let x be the number 2r of vertices in a largest Kr -free induced subgraph of G. Since the neighbors of any vertex induce a Kr -free subgraph, no vertex of G has degree exceeding x. Let A be 2 a largest induced Kr -free subgraph of G. By induction, there are at most (r−2)x (2r−2) edges in A. Each edge of G not in A is incident with at least one of the n − x vertices not in A, so summing the degrees of these vertices counts each such edge at least once. Hence there are at most x(n − x) such edges and so G has at most (r−2)x2 (2r−2) + x(n − x) edges. Since  2 r−1 2 r (r − 1)n r−2 2 (x ) + x(n − x) = n − x− , 2r − 2 2r 2r − 2 r the result follows.



To prove the assertion by contradiction, suppose that M (2d−1, d) > 2d so that there are 2d+1 (2d−1)-tuples any two of which differ in at least d components. Let G be an edge-labeled graph as follows. The set of vertices is the 2d+1 (2d−1)-tuples. For any two vertices, if the two vertices differ in their k component (1 ≤ k ≤ 2d−1), then connect the two vertices with an edge with label k. Consequently, there are at least d labeled edges connecting any two vertices of the graph G and hence the
2d+1 graph G has at least × d = (2d + 1)d2 edges. Since 2d − 1 numbers are 2 assigned to the edges and there are at least (2d + h1)d2 edges, i we see that there is a 2 label 1 ≤ k0 ≤ 2d − 1 which is assigned to at least (2d+1)d + 1 edges of the graph, 2d−1 where [.] denotes the integer part function. Now consider the induced graph on these edges to each of which k0 ishassigned.i This induced graph is indeed a simple 2 graph having 2d + 1 vertices and (2d+1)d + 1 edges. Note that 2d−1     (2d + 1)d2 (2d + 1)d2 (2d + 1)2 +1> > d2 + d = . 2d − 1 2d − 1 4 It thus follows from Tur´ an’s Theorem that the induced graph has a K3 , i.e., a triangle. This means that there are three vertices any two of which differ in their

188

2. SOLUTIONS

k0 component. This is a contradiction because the components are either 0 or 1, completing the proof.  6. First we need the following simple lemma. Lemma. Let two lookalike boxes contain w1 white marbles and b1 black marbles, and w2 white marbles and b2 black marbles, respectively. The probability of
w2 1 + picking a white marble from one of the boxes is equal to 21 w1w+b . w2 +b2 1 Proof. Use A to denote the event that one picks a marble from the box that contains w1 white marbles and b1 black marbles, and use B to denote the event that one picks a marble from the other box. If W denotes the event of picking a white marble from one of the boxes, as {A, B} is a partition of the probability space, we can write
P (W ) = P (A ∩ W ) ∪ (B ∩ W ) = P (A)P (W |A) + P (B)P (W |B) w1 w2 w1 1 1 w2
1 × , + × = + = 2 w1 + b1 2 w2 + b2 2 w1 + b1 w2 + b2 proving the lemma.



Suppose that the first person has picked a marble from one of the boxes, say, box b. Use b0 to denote the other box. For each i = 1, 2, 3, define the following events Wi (resp. Wi0 ): the event that the ith person picks a white marble from b (resp. 0 b ). Bi (resp. Bi0 ): the event that the ith person picks a black marble from b (resp. 0 b ). By the lemma above, we have 1 1 2
1 P (W1 ) = + = . 2 1+2 1+2 2 Likewise, P (B1 ) = 12 . 0 0 Use B011 and B001 (resp. B011 and B001 ) to denote the events that box b (resp. 0 b ) contains one black marble and two white marbles and two black marbles and one white marble, respectively. First, suppose that the first person has picked a white marble from b. Using Bayes’ Theorem, we can write P (B011 |W1 )

=

P (B011 )P (W1 |B011 ) = P (W1 )

1 2

P (B001 |W1 )

=

P (B001 )P (W1 |B001 ) = P (W1 )

1 2

×

2 3

1 2

× 1 2

1 3

=

2 , 3

=

1 . 3

0 0 And hence, P (B011 |W1 ) = 1 − P (B011 |W1 ) = 1 − 23 = 13 and P (B001 |W1 ) = 1 2 0 |W1 , and 1 − P (B001 |W1 ) = 1 − 3 = 3 . Let U = B011 |W1 , V = B001 |W1 , U 0 = B011 0 V 0 = B001 |W1 . We have P (U ) = P (V 0 ) = 32 and P (V ) = P (U 0 ) = 13 . Noting that {U, V } is a partition of the probability space, we have

P (W2 ) = P (U )P (W2 |U ) + P (V )P (W2 |V ) =

2 1 1 1 × + ×0= . 3 2 3 3

2.24. TWENTY FOURTH COMPETITION

189

That is, assuming that the first person has picked a white marble from b, the probability that the second person picks a white marble from b is 13 . Analogously, as {U 0 , V 0 } is a partition of the probability space, we can write P (W20 ) = P (U 0 )P (W20 |U 0 ) + P (V 0 )P (W20 |V 0 ) =

4 1 2 2 1 × + × = . 3 3 3 3 9

In other words, assuming that the first person has picked a white marble from b, the probability that the second person picks a white marble from b0 is 49 , which is greater than 31 . So, in this case, the second person should pick a marble from the other box, i.e., b0 . Now, assuming that the first person has picked a black marble from b, again using Bayes’ Theorem, we will have P (B011 |B1 ) = 13 and P (B001 |B1 ) = 23 , and hence 0 0 P (B011 |B1 ) = 1 − P (B011 |B1 ) = 1 − 13 = 23 and P (B001 |B1 ) = 1 − P (B001 |B1 ) = 1 2 0 0 0 |B1 , and T 0 = B001 |B1 , 1− 3 = 3 . So, letting S = B011 |B1 , T = B001 |B1 , S = B011 we can write P (W2 ) P (W20 )

2 1 2 1 ×1+ × = , 3 3 2 3 2 2 1 1 5 = P (S 0 )P (W20 |S 0 ) + P (T 0 )P (W20 |T 0 ) = × + × = . 3 3 3 3 9 = P (S)P (W2 |S) + P (T )P (W2 |T ) =

Since P (W2 ) = 32 is greater than P (W20 ) = 59 , the second person should pick a marble from box b. Therefore, the probability of survival for the second person is 5 1 4 1 2 × + × = , 2 9 2 3 9 for, by the above lemma, the first person picks a white (resp. black) marble with the probability of 12 . To investigate the probability of survival for the third person, there are two cases to consider. (i) the first person survives. Without loss of generality, we may assume that the first person has picked a white marble from box b. In view of what we showed in the above, there are two subcases to consider. (a) The second person picks a white marble from box b0 ; and (b) The second person picks a black marble from box b0 . In case (a), since the first and second person have picked white marbles from b and b0 , respectively, the boxes do not make any difference for the third person to pick a marble from. Thus, by the lemma above, the third person picks a white marble from one of the boxes with the probability of 14 . So if we use S1 to denote the event that the third person survives in this case, we will have P (S1 ) =

1 4 1 1 × × = . 2 9 4 18

In case (b), letting C = W1 ∩ B20 , in view of Bayes’ Theorem, we can write P (B011 |C)

= =

P (C|B011 )P (B011 ) P (B011 )P (C|B011 ) + P (B001 )P (C|B001 ) 2 2 1 4 3 × 3 × 2 1 2 2 1 1 1 = 5. × × + × × 2 3 3 2 3 3

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2. SOLUTIONS

Thus, P (B001 |C) = 1 − P (B011 |C) = 15 . Consequently, the probability that the third person picks a white marble from b is equal to 4 1 2 1 P (B001 |C)P (W3 B001 |C) + P (B011 |C)P (W3 B011 |C) = × 0 + × = . 5 5 2 5 Likewise, the probability that the third person picks a white marble from b0 is equal to 1 4 1 3 P (B001 |C)P (W30 B001 |C) + P (B011 |C)P (W30 B011 |C) = × 1 + × = . 5 5 2 5 Thus, the third person should pick a marble from b0 , which is the box from which the second person picked a marble. So, the probability of survival for the third person in this case, denoted by P (S2 ), is equal to P (S2 ) =

1 1 5 3 × × = . 2 9 5 6

(ii) the first person does not survive. Again, we may assume that the first person has picked a black marble from box b. There are two subcases to consider. (a) The second person picks a white marble from box b; and (b) The second person picks a black marble from box b. In case (a), letting C = B1 ∩ W2 , in view of Bayes’ Theorem, we can write P (B001 |C)

= =

P (C|B001 )P (B001 ) P (B001 )P (C|B001 ) + P (B011 )P (C|B011 ) 2 1 1 1 3 × 2 × 2 = . 2 1 1 1 1 2 2 × 3 × 2 + 2 × 3 ×1

Thus, the boxes do not make any difference for the third person to pick a marble from. So the probability of picking a white marble becomes 42 = 12 . And hence, if S3 denotes the event that the third person survives in this case, we will have P (S3 ) =

1 1 2 1 × × = . 2 3 2 6

Finally, if the first and second person both pick black marbles from b, then the third person can survive by picking the remaining white marble from box b. So the probability of survival in this case, which is denoted by P (S4 ), is equal to P (S4 ) =

1 1 1 × ×1= . 2 3 6

Consequently, the probability of survival for this person, denoted by P (S), is equal to P (S) = P (S1 ) + · · · + P (S4 ) =

1 1 1 1 5 + + + = . 18 6 6 6 9

That is, the probability of survival for the third person is the same as that of the second person. 

2.24. TWENTY FOURTH COMPETITION

191

2.24.2. Second Day. 1. (a) Use B to denote the closure of B with respect to the Euclidean norm of Rn . As B is compact because it is bounded and closed, there is a positive real r such that  diam(B) := sup ||a − b|| : a, b ∈ B = 2r, . As diam(B) = diam(B), in view of the hypothesis, we see that for all n ∈ N there are xn , yn ∈ B and an open ball Brn (zn ), the ball centered at zn with radius rn , such that xn , yn ∈ Brn (zn ) ⊆ B and ||xn − yn || > 2r − n1 . Since zn ’s are in the compact set B, if necessary by passing to a subsequence of (zn )+∞ n=1 , we may assume that there is a z ∈ B such that limn zn = z. We claim that B = Br (z), the open ball with radius r centered at z, proving the assertion. First, we prove that Br (z) ⊆ B. To this end, let x ∈ Br (z) be arbitrary. We can write ||x − z|| = r −  for some  > 0. As limn n1 = 0, there is an N ∈ N such that N1 <  and ||z − zN || < 2 . We have  ||x − zN || ≤ ||x − z|| + ||z − zN || < r −  + 2  1 ||xN − yN || = r− r. Join the two points x, z to intersect the closed ball B r (z) at two antipodal points y, y 0 ∈ B r (z) so that y and y 0 are, respectively, on and off the line segment that joins x and z. It is obvious that ||x − y 0 || = ||x − z|| + ||z − y 0 || = ||x − z|| + r > 2r = diam(B) = diam(B), which is a contradiction because x, y 0 ∈ B. Therefore, B ⊆ Br (z), and hence B = Br (z), completing the proof. (b) First solution: Let X = (C[0, 1], ||.||∞ ) be the normed space of all real valued continuous functions on the closed interval [0, 1] equipped with the uniform norm. As is well-known, X is a Banach space, and hence in particular a complete metric space. Let Y = (C[0, 1], d) be the metric space equipped with the metric d which is defined on C[0, 1] by ||f − g||∞ d(f, g) := , 1 + ||f − g||∞ where ||f − g||∞ = supx∈[0,1] |f (x) − g(x)|. It is readily verified that Y is a complete metric space. We claim that the set B = {f ∈ C[0, 1] : f (x) > 0} is a bounded subset of Y with the property that for each pair of points x, y in B, there exists an open ball U such that U ⊆ B and x, y ∈ U and yet B is not an open ball in Y . Plainly, Y is bounded, and hence
so is B. Now, suppose f, g ∈ B are
arbitrary. Set M = supx∈[0,1] max f (x), g(x) and m = inf x∈[0,1] min f (x), g(x) . We have 0 < m, M ∈ R because f and g are continuous on the compact interval [0, 1]. x Define t : [0, +∞) → [0, +∞) by t(x) = 1+x . First, we show that f, g ∈ Br (h) ⊆ B, M− m

where r = t( 2 2 ) and h ∈ B is defined by h(x) = M +m for all x ∈ [0, 1]. To 2 see f ∈ Br (h), we need to show that d(f, h) < r. To this end, note first that m m ≤ f (x) ≤ M implying that 3m 4 < f (x) < M + 4 for all x ∈ [0, 1], from which,

192

2. SOLUTIONS

we easily obtain ||f − h||∞ < yields

M− m 2 2

. But t is strictly increasing on [0, +∞). This



M−m 2 d(f, h) = t ||f − h||∞ < t = r, 2 as desired. Likewise, we see that g ∈ Br (h). Next, we need to show that Br (h) ⊆ B. To this end, let k ∈ Br (h) be arbitrary. It follows that
M− d(k, h) = t ||k − h||∞ < r = t 2

m 2




,

which yields M−m M−m M +m 2 2 < k(x) − < , 2 2 2 for all x ∈ [0, 1] because t is strictly increasing on [0, +∞). This obtains −

3m m < k(x) < M + , 4 4 for all x ∈ [0, 1], implying that k ∈ B, which is what we want. It remains to prove that B is not a an open ball. Suppose to the contrary that B = Bt(s) (l), for some s > 0 and l ∈ B. Now, let k ∈ B = Bt(s) (l) be arbitrary. We see that
d(k, l) = t ||k − l||∞ < t(s), 0<

which obtains −s < k(x) − l(x) < s, for all x ∈ [0, 1]. Consequently, supx∈[0,1] k(x) < s+M0 , where M0 = supx∈[0,1] l(x). This is a contradiction because k ∈ B is arbitrary. Therefore, B is a bounded subset of the complete metric space Y = (C[0, 1], d) with the property that for each pair of points f, g in B there exists an open ball Br (h) such that f, g ∈ Br (h) ⊆ B and yet B is not an open ball in Y , proving the claim. Second solution: A simpler proof similar to that of first solution shows that the |x−y| open interval (0, +∞) of the complete metric space (R, d), where d(x, y) := 1+|x−y| and |.| denotes the absolute value function, is a counterexample to (a) when Rn is replaced by the complete metric space (R, d). We omit the details for the sake of brevity. Third solution: Let X = {A1 , . . . , A5 }, where A1 = (1, 0, 0), A2 = (0, 1, 0), A3 = −A1 , A4 = −A2 , and A5 = (0, 0, 34 ). Note that if we use d and s to, respectively, denote the diameter and side √ length of the square whose vertices are A1 , A2 , A3 , A4 , we have d = 2 and s = 2. Also note that s < A5 Ai = 53 < d for each i = 1, 2, 3, 4. It is readily checked that X with respect to the Euclidean metric of R3 is a complete metric space. Set B = {A1 , . . . , A4 }. It is now easily verified that B is a bounded subset of the complete metric space X with the property that for each pair of points x, y in B, there exists an open ball Br (z), where r = s+A25 A1 , such that x, y ∈ Br (z) ⊆ B and yet B is not an open ball in X, which is what we want.  2. First solution: It might be worth noting that the hypothesis that f is continuous is redundant. To prove the assertion, it is enough to assume that f is integrable

2.24. TWENTY FOURTH COMPETITION

193

Rb on [a, b]. Without loss of generality, we may assume that a f (t)dt > 0. For given k ∈ (0, 1), define the continuous function gk : [a, b] −→ R by Z b Z x f (t)dt. f (t)dt − k gk (x) = a

a

Rb

Rb We have gk (a) = −k a f (t)dt < 0 and gk (b) = (1 − k) a f (t)dt > 0. Thus, by the Intermediate Value Theorem, there is a ck ∈ (a, b) such that g(ck ) = 0, proving the assertion. Second solution: Define g : [a, b] −→ R by Rx f (t)dt g(x) = Rab . f (t)dt a Plainly, g is continuous on [a, b] and g(a) = 0 < 1 = g(b). Thus, by the Intermediate Rc Value Theorem, there exists a ck ∈ (a, b) such that g(ck ) = k, implying a k f (t)dt− Rb k a f (t)dt = 0, as desired.  3. (a) The inequality is known as the Frobenius Inequality. We prove the assertion over division rings. Let D be a division ring and A ∈ Mm×n (D), B ∈ Mn×p (D), C ∈ Mp×q (D), where m, n, p, q ∈ N. View A (resp. B, C) as a linear transformations acting on the left of Dn (resp. Dp , Dq ), the right vector space of all n × 1 (resp. p × 1, q × 1) column vectors over D. Let A1 = A|B(Dp ) . We can write
dim ker A1 |BC(Dq ) ≤ dim ker A1 = dim B(Dp ) − dim A1 B(Dp ) ,
which, in view of A1 B(Dp ) = AB(Dp ), yields dim ker A1 |BC(Dq ) ≤ r(B) − r(AB). On the other hand, dim ker A1 |BC(Dq )

=


dim BC(Dq ) − dim A1 BC(Dq )

= r(BC) − dim ABC(Dq ) = r(BC) − r(ABC). Therefore, r(BC) − r(ABC) ≤ r(B) − r(AB), implying r(BC) + r(AB) ≤ r(ABC) + r(B), as desired. (b) To prove the assertion which is known as the Sylvester Inequality, just set A = A, B = In , and C = B in (a).  4. Let I and K be a two-sided ideal and a left ideal of the ring R, respectively. It is plain that KI ⊆ I ∩ K. To prove I ∩ K ⊆ KI, choose an arbitrary element x ∈ I ∩ K. It suffices to show that x ∈ KI. To this end, by the hypothesis, we have (I ∩ K)2 = I ∩ K because I ∩ K is a left ideal of R. It follows that x ∈P (I ∩ K)2 . n And hence there are n ∈ N and xi , yi ∈ I ∩ K (1 ≤ i ≤ n) such that x = i=1 xi yi . But xi yi ∈ KI for all 1 ≤ i ≤ n. Therefore, x ∈ KI, which is what we want, completing the proof. 

194

2. SOLUTIONS

5. First solution: First, suppose that the number of wins of no two teams are equal. As there are n teams and any team has played against all of the other n − 1 teams, we see that for each i = 0, . . . , n − 1, there is exactly one team, which we call the ith team, whose wins is equal to i. It is obvious that the (n − 1)st team has won all the other n − 1 teams, the (n − 2)nd team losses to the (n − 1)st team and wins all the remaining n − 2 teams, and so on and so forth. For the ith team, where i ∈ {0, 1, . . . , n − 1}, let’s call i to be the label of the team. Thus, for any three teams, the team with the smallest label has lost to the other two teams. Therefore, there are no three teams A, B, C such that A wins B, B wins C, and C wins A. Next, suppose that there are no such three teams. Define the relation ≥ on the teams as follows. For two teams A and B, we write A ≥ B if and only if A = B or A wins B. Since there are no such three teams, in view of the hypothesis, it is easily verified that the relation ≥ is a linear order on the set of the participating teams. So if we put the teams in the decreasing order, say T1 > T2 > · · · > Tn , then Ti , the ith team, has won exactly n − i + 1 teams, namely, Ti+1 > · · · > Tn . Consequently, no two teams have scored the same number of wins, which is what we want. Second solution: First, suppose that there are two teams A and B that have scored k wins, where k ∈ {0, . . . , n − 1}. Without loss of generality, we may assume that A wins B. We claim that there is a team C such that A wins B, B wins C, and C wins A. To see this, as B has won k teams, one of them, say C, must have won A, for otherwise A must have won k + 1 teams, the k teams lost to B plus B itself, which is impossible. Thus, such team C exists, settling the implication. Next, suppose that there are three teams A , B, and C such that A has won B, B has won C, and C has won A. We prove the assertion by induction on n, the number of participating teams. If n = 3, the assertion is trivial. Assuming that the assertion holds for n − 1, to prove the assertion for n, argue by contradiction. So no two teams have scored the same number of wins. Consequently, there is exactly one team, say D, that has scored n − 1 wins, for n teams have participated in the game. Now, D is not one of A, B, or C because D has won them all. Exclude D and consider the game between the remaining n − 1 teams which include A, B, C. It follows from the induction hypothesis that there are two teams in the remaining n − 1 teams whose wins are equal. But the two teams have both lost to D. Thus, the two teams have scored the same number of wins in the original game with n teams, which is a contradiction. So the assertion follows.  6. We convert this problem into the following coin-flipping game of which the problem is a special case. We then present a proof of the counterpart of the assertion for the coin-flipping game, which we have taken from “Concrete Mathematics”, a book by Ronald L. Graham, Donald E. Knuth, and Oren Patashnik. Here is the counterpart of the problem. Two persons, A and B, are playing the following coin-flipping game. First, A chooses a pattern of length ` (` ≥ 3) of heads and tails (for instance, HT H, where H stands for heads and T for tails). Then B, who knows the pattern chosen by A, chooses a different pattern of the same length of heads and tails. Then a fair coin is flipped until one of the patterns is first obtained, in which case the the player whose pattern occurs first is to win the game. Show that no matter what

2.24. TWENTY FOURTH COMPETITION

195

choice is made by A, there is a choice for B so that the probability of winning the game by B is greater than 21 . Denote, respectively, by SA and SB the sum of A’s and B’s winning positions. Use N to denote the sum of the patterns each of which does not contain any occurrences of the patterns A and B chosen by A and B, respectively. For instance, if A chooses A = HT H and B chooses B = T T H, we have SA = HT H + HHT H + T HT H + HHHT H + T T HT H + T HHT H + · · · , SB = T T H + HT T H + T T T H + HHT T H + T T T T H + T HT T H + HT T T H + · · · , N = 1 + H + T + HH + T T + T H + HT + HHH + HHT + T HT + T T T + · · · . Obviously, if we set H = T = 21 , the resulting values for SA and SB , respectively, become the probability that A and B wins the game. We have 1 + N (H + T ) NA

= N + SA + SB , = SA

` X

A(`−k) δA(k) ,A(k) + SB

k=1

NB

= SA

` X

` X

A(`−k) δB(k) ,A(k) ,

k=1

B (`−k) δA(k) ,B(k) + SB

` X

B (k−`) δB(k) ,B(k) ,

k=1

k=1 (k)

where δα,β denotes the Kronecker delta, A and A(k) (resp. B (k) and B(k) ) denote the last and the first k characters of A (resp. B). To see the first equality, just note that every term on the left side of it either ends with A, or B, or it does not end with either of A and B meaning that the term belongs to N ; conversely, every term on the right of the first equality, is either empty or it belongs to N H or N T . The second equality holds because every term on the left either completes a term of SA in such a way that the last k characters of A coincides with its first k characters for some 1 ≤ k ≤ `, or a term of SB in such a way that the last k characters of B equals the first k characters of A for some 1 ≤ k ≤ `; and conversely because every term on the right belongs to the left. Analogously, the third equality holds. As noted in the above, by setting H = T = 12 , we obtain the wining probabilities for A and B, which we denote by P (A) and P (B), respectively. It follows from the first equality above that P (A) + P (B) = 1. Let A:A =

` X

2k−1 δA(k) ,A(k) , B:A =

k=1

A:B =

` X k=1

` X

2k−1 δB(k) ,A(k) ,

k=1

2k−1 δA(k) ,B(k) , B:B =

` X

2k−1 δB(k) ,B(k) .

k=1

Using the second and third equalities, we can write   N = 2 P (A)(A:A) + P (B)(B:A) ,   N = 2 P (A)(A:B) + P (B)(B:B) ,

196

2. SOLUTIONS

which obtains

P (A) B:B − B:A = . P (B) A:A − A:B We now claim that if A chooses the pattern A = τ1 τ2 · · · τ` , then B has a better chance of winning the game by choosing B = τ20 τ1 · · · τ`−1 , where τ20 is the heads/tails opposite of τ2 . It suffices to show that P (A) < P (B). Suppose the contrary, implying that B:B − B:A ≥ A:A − A:B. (∗) `−1 `−1 `−3 `−2 (`) Note that A:A ≥ 2 and B:B < 2 + 2 , and B:A ≥ 2 , for A = A(`) , B (`) = B(`) but B (`−2) 6= B(`−2) , and B (`) 6= A(`) but B (`−1) = B(`−1) . It follows that B:B − B:A < 2`−1 + 2`−3 − 2`−2 . Since A(`) 6= B(`) and A(`−1) 6= B(`−1) , we conclude A(`−2) = B(`−2) , for otherwise A:B ≤ 2`−3 , from which, we obtain A:A − A:B ≥ 2`−1 − 2`−3 ≥ 2`−1 + 2`−3 − 2`−2 > B:B − B:A, which is impossible. Consequently, A(`−2) = B(`−2) , yielding τ20 = τ3 , τ1 = τ4 , τ2 = τ5 , τ3 = τ6 , . . . , τ`−3 = τ` . But then, A:A ≥ 2`−1 + 2`−4 + · · · , A:B ≤ 2`−3 + 2`−6 + · · · , B:A ≥ 2`−2 + 2`−5 + · · · , and B:B < 2`−1 + 2`−4 + · · · , implying that A:A − A:B

≥ (2`−1 − 2`−3 ) + (2`−4 − 2`−6 ) + · · · > (2`−1 − 2`−2 ) + (2`−4 − 2`−5 ) + · · · > B:B − B:A.

In other words, A:A − A:B > B:B − B:A, which is in contradiction with (∗). Therefore, P (A) < P (B), proving the assertion.  2.25. Twenty Fifth Competition 2.25.1. First Day. 1. As |G| = n and [G : Z(G)] = 4, we have Z(G) $ CG (x) $ G, for all x ∈ G \ Z(G), where CG (x) := {g ∈ G : xg = gx} denotes the centralizer of the element x in G. We can write 4 = [G : Z(G)] = [G : CG (x)][CG (x) : Z(G)], which easily implies [G : CG (x)] = 2. It follows that the size of any conjugacy class G\Z(G) . On the other hand, of the elements of G\Z(G) is equal to 2, and hence 2 G \ Z(G) = 3n . Thus, 2 3n , implying 8 3n, which, in turn, yields 8 n. For a given 4 4 n ∈ N satisfying 8 n, set G := Q8 × Z n8 , where Q8 denotes the quaternionic group with 8 elements. It is obvious that Z(G) = {−1, 1} × Z n8 , yielding [G : Z(G)] = 4, as desired.  2. From T 2 = T , we see that V = ker T ⊕ imT . Now, let α ∈ V be arbitrary. It follows that there are β ∈ ker T and γ ∈ V such that α = β + T γ. In view of the hypotheses, we can write (T + S)α = (T + S)(β + T γ) = T β + T 2 γ + Sβ + ST γ = T γ + Sβ. Since β ∈ ker T , by the hypothesis, there is a δ ∈ V such that β = Sδ. So we can write (T + S)α = T γ + Sβ = T γ + S 2 δ = T γ + Sδ = β + T γ = α.

2.25. TWENTY FIFTH COMPETITION

197

Thus, T + S is the identity transformation, for α ∈ V was arbitrary, finishing the proof.  3. By the Mean Value Theorem, there is a c ∈ (y, y + 1) such that f (y + 1) − f (y) = f 0 (c). As f 00 (t) < 0 for all t ∈ R, f 0 is strictly decreasing on R. Thus, f 0 (y) > f 0 (c) because y < c. So we can write f (y + 1) − f (y) = f 0 (c) < f 0 (y) < f (y + 1) − x, yielding x < f (y), as desired.




4. Define g : (a, b) −→ R by g(x) = ln 1 + f 2 (x) − x. We have g 0 (x) =

2f (x)f 0 (x) 2f (x)f 0 (x) − f 2 (x) − 1 − 1 = ≥ 0, 1 + f 2 (x) 1 + f 2 (x)

for all x ∈ (a, b). Thus, g is nondecreasing on (a, b), implying that −a = limx→a+ g(x) ≤ limy→b− g(y) = 1 − b. Therefore, −a ≤ 1 − b, yielding b − a ≤ 1. As for an example √ for which b − a = 1, just let a = 0 = b − 1 and f (x) = ex − 1 on (0, 1).  5. Note first that the game stops exactly when the 7th numbered one marble is drawn from a box. To see this, it is obvious that if the 7th numbered one marble is drawn from a box, the the game stops because one needs to draw a marble from box one which is empty. Conversely, if the game stops when a marble numbered i is drawn from a box, then i = 1. Because otherwise the box numbered i must be empty for some i ≥ 2, which means seven marbles numbered i must have already been drawn from the ith box, implying that there are 8 marbles numbered i, a contradiction. Now consider the following extended game. Assume that, after a stop in the original game, the game is continued by choosing the box which is not empty yet and whose number is minimal among all nonempty boxes, and that the game is continued in this manner until all the marbles are drawn from all boxes. In this extended game, to each permutation of the 49 marbles, there corresponds a round of the game. Conversely, to each round of the game, there corresponds a permutation of the 49 marbles. Also, a round of the original game continues until all the marbles are drawn from all boxes only when the number on the last drawn marble, i.e., the 49th drawn marble, reads one, in which case the original game and the extended game are the same. It is now obvious that the probability of the event that all the marbles are drawn from all boxes in the original game is equal to that of that same event in the extended game which is equal to 48! 6!7!6 49! 7!7

which is what we want.

=

1 , 7 

6. Let 0 < c ≤ b ≤ a with a, b, c ∈ N and a + b + c = n for some n ∈ N, be the side lengths of a desired triangle. The following gives an algorithm for finding a partition of the number n into the summands 2, 3, and 4 in which the summand 3 appears at least once. As the following algorithm is reversible, the assertion follows. We have a < b + c, e.g., 7 < 6 + 3. We explain the algorithm in three stages.

198

2. SOLUTIONS

Stage (i): To create summands of 3, subtract one from the three side lengths, i.e., a, b, c, one at a time and continue this for c + b − a times so that the inequality a < b + c becomes an equality, e.g., 7 < 6 + 3 → 6 < 5 + 2 → 5 = 4 + 1. Stage (ii): To create summands of 4, in the equality obtained in stage (i), subtract two units and one unit, one at a time, from the left hand side and right hand side of the equality, respectively. Continue in this manner for a − b times till a zero appears in the equality , e.g., 5 = 4 + 1 → 3 = 3 + 0. Stage (iii): To create summands of 2, in the last equality obtained in stage (ii), subtract one unit at a time from each nonzero number of the equality and continue in this way for b − c times till all numbers become zero, e.g., 3 = 3 + 0 → 2 = 2 + 0 → 1 = 1 + 0 → 0 = 0 + 0. So, the corresponding partition for a triangle with side lengths 0 < a < b < c and a + b + c = n is given by n = 4(a − b) + 3(b + c − a) + 2(b − c), e.g., 16 = 4 + 3 + 3 + 2 + 2 + 2. 

2.25.2. Second Day. 1. The assertion is a quick consequence of the following proposition. Let R be a unital ring without zero divisors, i.e., ∀a, b ∈ R : ab = 0 ⇒ a = 0 or b = 0, and U (R) denote the multiplicative group of the units of R. Then, every finite abelian subgroup of U (R) is cyclic. We present two proofs for this proposition. First proof. Let G be an abelian subgroup of U (R) and Z denote the prime ring of R, i.e., Z = {k1 : k ∈ Z}. It follows from the hypothesis that Z[G] := {

n X

ki gi : n ∈ N, ki ∈ Z, gi ∈ G (1 ≤ i ≤ n)}

i=1

is an integral domain and that G is a subgroup of the multiplicative group of the units of Z[G]. So without loss of generality, we may assume that R is an integral domain. To prove the assertion, we need the following lemma. Lemma. (i) Let G be a group and g ∈ G with ord(g) = n1 · · · nk , where k ∈ N and nj ’s (1 ≤ j ≤ k) are pairwise relatively prime. Then, there exist unique g1 , . . . , gk ∈ G such that g = g1 · · · gk , gi gj = gj gi , and ord(gi ) = ni for each i, j = 1, . . . , k. Conversely, if for g ∈ G, there are g1 , . . . , gk ∈ G satisfying g = g1 · · · gk , gi gj = gj gi and ord(gi ) = ni for each i, j = 1, . . . , k such that nj ’s (1 ≤ j ≤ k) are pairwise relatively prime, then ord(g) = n1 · · · nk . (ii) Let G be a group and g1 , . . . , gk ∈ G be of finite order such that gi gj = gj gi
for each i, j = 1, . . . , k. Then there is a g ∈ G such that ord(g) = lcm ord(g1 ), . . . , ord(gk ) . Proof. (i) First, recall that if g ∈ G is such that ord(g) = n for some n ∈ N, Qk n for all i ∈ Z. Now, set mi = i6=j=1 nj . As nj ’s (1 ≤ j ≤ k) then ord(g i ) = gcd(i,n) are pairwise relatively prime, we see that gcd(m1 , . . . , mk ) = 1 and hence there are Pk cj ’s (1 ≤ j ≤ k) in Z such that j=1 cj mj = 1. We note that gcd(ci , ni ) = 1 for each i = 1, . . . , k. To see this, it suffices to show that no prime p divides

2.25. TWENTY FIFTH COMPETITION

199

gcd(ci , ni ). Suppose to the contrary that divides gcd(ci , ni ) there is a prime p that for some 1 ≤ i ≤ k. It follows that p ci and p ni and hence p ci and p mj for all Pk j ∈ {1, . . . , k} \ {i}. Consequently, p j=1 cj mj = 1, which is impossible. Thus, gcd(ci , ni ) = 1 for each i = 1, . . . , k. Now, letting gi = g ci mi , we can write ord(gi )

ord(g) mi ni = gcd(ci mi , ord(g)) gcd(ci mi , mi ni )

=

ord(g ci mi ) =

=

ni = ni . gcd(ci , ni )

Also, we can write g1 · · · gk = g c1 m1 · · · g ck mk = g

P

k j=1

cj mj

= g.

Finally, it is obvious that gj ’s (1 ≤ j ≤ k) commute. To see that gj ’s (1 ≤ j ≤ k) are unique, suppose that there are g10 , . . . , gk0 ∈ G such that g = g10 · · · gk0 , gi0 gj0 = gj0 gi0 , and ord(gi0 ) = ni for each i, j = 1, . . . , k. We can write c mj

g cj mj = (g10 · · · gk0 )cj mj = g 0 1j mj

But g 0 i

c mj

· · · g 0 kj

= e whenever i 6= j because ord(gi0 ) = ni and mj = c mj

g cj mj = g 0 jj

P

1−

= g0 j

k j6=i=1

ci mi

. Qk

j6=i=1

ni . Thus,

= gj0 ,

for all 1 ≤ j ≤ k, as desired. As for the converse, suppose that ord(g) = m. We need to show that m = n1 · · · nk . Note first that m n1 · · · nk because g n1 ···nk = (g1 · · · gk )n1 ···nk = e. It thus suffices to show that n1 · · · nk m. To this end, letting 1 ≤ i ≤ k be arbitrary but fixed, we can write k Y

g m = e =⇒ (g1 · · · gk )m = e =⇒ gim =

gj−m .

i6=j=1

As gini = e, we get (gim )ni = e, from which, we obtain ord(gim ) ni . On the other hand,

Q

(gim )

k i6=j=1

nj

=

k Y i6=l=1

gl−m




Q

k i6=j=1

nj

=

k Y

Q

gl

k i6=j=1

nj
−m

= e,

i6=l=1

Qk
Qk implying that ord(gim ) i6=j=1 nj . Thus, ord(gim ) gcd ni , i6=j=1 nj = 1, which yields ord(gim ) = 1. This obtains gim = e for all 1 ≤ i ≤ k. Hence, ni m for all 1 ≤ i ≤ k. It thus follows that n1 · · · nk m because nj ’s (1 ≤ j ≤ k) are pairwise relatively prime. Therefore, m = ord(g) = n1 · · · nk , which is what we want. (ii) It is plain that there exists an l ∈ N and primes p1 , . . . , pl such that il i1 ord(gi ) = pm · · · pm , 1 l

where mij ’s are nonnegative integers (1 ≤ i ≤ k, 1 ≤ j ≤ l). For 1 ≤ i ≤ k, we see from (i) that there are gi1 , . . . , gil ∈ G such that gi = gi1 · · · gil , that gij ’s (1 ≤ i ≤ m k, 1 ≤ j ≤ l) all commute, and that ord(gij ) = pj ij . It is obvious that for each

200

2. SOLUTIONS

 1 ≤ j ≤ l, there is an 1 ≤ ij ≤ k such that ord(gij j ) = max ord(g1j ), . . . , ord(gkj ) . Ql Let g 0 = j=1 gij j . It thus follows from (i) that  l Y
max m1j ,...,mkj 0 ord(g ) = pj = lcm ord(g1 ), . . . , ord(gk ) , j=1

which is what we want.



Note that U (R), and hence G, is commutative. Let a ∈ G be such that its order, say m, is maximal among the elements of G. Let b be an arbitrary element of G. Since G is commutative, by part (ii) of the above lemma, there is a c ∈ G whose order is equal to the least common multiple of ord(a) and ord(b). It follows that ord(c) = ord(a) because ord(a) is maximal. Thus, ord(b) ord(c) = ord(a) = m, yielding bm = 1. In other words, every element of G is a root of f = xm − 1 ∈ R[x]. As R is an integral domain, f has at most m roots, and hence |G| ≤ m. Therefore, |G| = m = ord(a). That is, G is cyclic, as desired. Second proof. Just as we saw in the first proof, we may assume that R is an integral domain. Consequently, the equation xn = 1 has at most n solutions in G for all n ∈ N. It thus follows from the lemma presented in Solution 1 of 2.2.2 that G is cyclic, as desired.  2. Let q be a prime factor of 2p + 1 which is different from 3. We have q 2p + 1, and q hence q 22p − 1, yielding 22p ≡ 1. Use t = ordq (2) to denote the order of 2 modulo q. It follows that t divides 2p. By showing that t 6= 1, 2, p, we conclude that t = 2p. q

q

Firstly, t 6= 1 because 21 6≡ 1. Secondly, t 6= 2 because 22 6≡ 1, for q 6= 3. Thirdly, q t 6= p because otherwise 2p ≡ 1, which, in view of q 2p + 1, obtains ∗ q 2, which is impossible, for q is odd. Thus, t = 2p. Now, since t = ordq (2) |Zq | = q − 1, where ∗ Zq = Zq \ {0}, we have 2p q − 1 or q − 1 = 2kp for some k ∈ N, which is what we want.  3. It is plain that f is a one-to-one function from X onto X. Thus, f is invertible. Use f −1 to denote the inverse of f . It follows from the hypothesis that
d f −1 (x), f −1 (y) ≤ d(x, y), for all x, y ∈ X. Thus, the function f −1 : X −→ X is a Lipschitz function, and hence it is (uniformly) continuous. Now, since f −1 is continuous and X is compact, we see that f −1 is a closed map, implying that f = (f −1 )−1 is continuous, which is what we want.  4. Let α = f (1) and β = f (i). By the hypothesis, |f (z)| = |z|, |f (z) − α| = |z − 1|, and |f (z) − β| = |z − i| for all z ∈ C. In particular, by substituting z = 1, i in the above equalities, we obtain √ |α| = |β| = 1, |α − β| = 2. We can write α2 + β 2

= α2 β β¯ + β 2 αα ¯ = αβ(αβ¯ + β α ¯)

¯ = αβ |α|2 + |β|2 − |α − β|2 = αβ αα ¯ + β β¯ − (α − β)(¯ α − β) = αβ(1 + 1 − 2) = 0,

2.25. TWENTY FIFTH COMPETITION

201

yielding β = εα, where ε = ±i. Simplify |f (z) − α| = |z − 1| and |f (z) − β| = |z − i| to, respectively, obtain α ¯ f (z) + αf (z) = z + z¯ and α ¯ f (z) − αf (z) = −εiz + εi¯ z for all z ∈ C. Adding up these two equalities, we obtain 2¯ αf (z) = (1 − εi)z + (1 + εi)¯ z, which, in view of α = f (1) and ε = ±i, yields f (z) = f (1)z for all z ∈ C or f (z) = f (1)¯ z for all z ∈ C, as desired.  5. Let n = 2k − 1 for some k ∈ N and Kn denote the n × n matrix whose elements are all 1. For a given n × n matrix A with entries in {−1, 1}, let’s call C(A) :=

n X i=1

ai +

n X

bj

j=1

the content of the matrix A, where ai ’s and bj ’s are as in the statement of the problem. We have C(Kn ) = n + n = 4k − 2. It is plain that any n × n matrix A, whose entries are in {−1, 1}, can be obtained from Kn within a finite number of steps so that at each step a fixed entry, say, aij = 1, of Kn is changed to −aij = −1. After each step, the number two will be added to or will be subtracted from each of the two sums in the definition of the content. Thus, the content remains intact or it will be increased or decreased by four at each step. Therefore, the content of any n × n matrix A will always be of the form 4k 0 − 2, for some k 0 ∈ Z, which is never zero. This proves the first part of the assertion. As for the case when n is even, the assertion does not hold because for the matrix A defined by   Km −Km A = , Km Km where m =

n 2,

we have C(A) = 0.



6. For a member a ∈ S, use A(a) to denote the set of all of the acquaintances of a. It follows from the hypothesis that A(a) 6= A(b) whenever a 6= b. Proceed by contradiction. Let x ∈ S be a member for which A(x) is maximal among all members of S. Since S is ideal but S \ {x} is not ideal, there are two distinct members p, q ∈ S \ {x} such that A(p) = A(q) \ {x}, x ∈ A(q).

(∗)

Likewise, since S is ideal but S \ {p} is not ideal, there are two distinct members r, s ∈ S \ {p} such that A(r) = A(s) \ {p}, p ∈ A(s).

(∗∗)

We claim that r = x. By (∗∗), p and s are acquainted. This, in view of (∗), implies that s 6= x and that s and q are acquainted. But since, by (∗∗), r is acquainted with all of the acquaintances of s but p, we see that r and q are acquainted. This implies r = x because r is not acquainted with p. Thus, A(x) = A(s) \ {p}. That is, the number of the acquaintances of s is greater than those of x by one, which is a contradiction, proving the assertion. 

202

2. SOLUTIONS

2.26. Twenty Sixth Competition 2.26.1. First Day. 1. First solution: Define g : [a, b] −→ R by g(x) = f (x) + x − (a + b). The function g is continuous on [a, b] and differentiable on (a, b). Moreover, g(a) = a − b < 0 and g(b) = b − a > 0. So, by the Intermediate Value Theorem, there is a c ∈ (a, b) such that g(c) = 0, which means f (c) = a + b − c. It now follows from the Mean Value Theorem that there are x1 , x2 ∈ (a, b) with x1 6= x2 such that f (a) − f (c) c−b a−c f (c) − f (b) = , f 0 (x2 ) = = , f 0 (x1 ) = a−c a−c c−b c−b yielding f 0 (x1 )f 0 (x2 ) = 1, proving the assertion. Second solution: Let g = f ◦ f : [a, b] −→ R. We have g(a) = a and g(b) = b. By the Mean Value Theorem, there is a c ∈ (a, b) such that g(b) − g(a) = f 0 (c)f 0 (f (c)), b−a yielding f 0 (c)f 0 (f (c)) = 1. There are two cases to consider. If f (c) 6= c, then the assertion is proved by letting x1 = c and x2 = f (c). If f (c) = c, then using the Mean Value Theorem for f on the intervals [a, c] and [c, b], respectively, we see that there are x1 ∈ (a, c) and x2 ∈ (c, b) such that 1=

f (c) − f (a) f (b) − f (c) = 1, f 0 (x2 ) = = 1, c−a b−c yielding f 0 (x1 ) = f 0 (x2 ) = 1, and hence f 0 (x1 )f 0 (x2 ) = 1, proving the assertion.  f 0 (x1 ) =

2. Suppose to the contrary that f has no zero in U . Define g : U −→ C by 1 g(z) = f (z) . It follows that g is analytic on U . Moreover, g(0) = 1 and |g(z)| ≤ 21 on |z| = 1. This contradicts the Maximum Modulus Principle, proving the assertion.  3. Since gcd(p, 2p−2) = 1, it suffices to prove the congruence modulo p and modulo p 2p − 2. Note that, by Fermat’s Little Theorem, (2p − 2)p−1 ≡ 1. This, together p p with pp−1 ≡ 0, yields pp−1 + (2p − 2)p−1 ≡ 1. Also, note that pp−1 − 1 = (p − 1)(pp−2 + · · · + p + 1) = 2k(p − 1), 2p−2

2p−2

which implies pp−1 ≡ 1. This, together with (2p − 2)p−1 ≡ 0, yields pp−1 + 2p−2

(2p − 2)p−1 ≡ 1. Therefore, pp−1 + (2p − 2)p−1

p(2p−2)

≡

1, as desired.



4. We need to show that for all x, y ∈ R, x divides y or y divides x. To prove this by contradiction, suppose that there are x, y ∈ R such that x - y and y - x. It follows from the hypothesis that there is a nonzero z ∈ R such that Rx + Ry = Rz. Thus, there are r, s ∈ R such that x = rz and y = sz. The elements r, s ∈ R are not units because otherwise x y or y x, which is not possible. Therefore, the elements r and s belong to M , the only maximal ideal of R. From Rx + Ry = Rz, we see that there are a, b ∈ R such that z = ax + by. This,
together with x = rz and y = sz, yields z = (ar + bs)z, implying 1 − (ar + bs) z = 0. On the other hand, 1 − (ar + bs) is a unit in R, for otherwise 1 − (ar + bs) ∈ M , yielding 1 ∈ M , which is a contradiction. Therefore, z = 0, a contradiction. So the assertion follows. 

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203

5. For the sake of brevity, we say a k-cover for the set S to mean a k-element cover for the set S. It is plain that any element of a minimal cover of S is nonempty and that it includes exactly one element of S by itself. Therefore, a minimal (n − 1)cover for the set S is either a partition of the set S, or there exists exactly one element x ∈ S which belongs to two or more elements of the (n − 1)-cover. In the first case, the number of such
(n − 1)-covers of S, i.e., such partitions of S into
n n (n − 1) subsets, is equal to . In the latter case, there are ways to choose 2 1
x ∈ S and 2n−1 − n1 = 2n−1 − n ways to put x in the elements of the (n − 1)-cover so that x is put in at least two elements of the cover. So, we must have    
n n n M (n, n − 1) = + 2n−1 − n = (2n − n − 1), 2 2 1 which is what we want.  P 6. Let g(R, S) = f (Y ) : Y ∈ X {m} , R ⊆ Y ⊆ S . We prove the following more general proposition. Under the hypotheses of the problem, g(R, S) = 0 whenever R ⊆ S and S \R ≥ k. The assertion is a consequence of the above proposition. To see this, note that for every T ∈ X {m} , there is an S ∈ X {m+k} such that T ⊆ S. This implies f (T ) = g(S, T ) = 0, as desired. We prove the above proposition by induction on |R|. If |R| = 0, assuming that |S| ≥ k, we have   X |S| g(∅, S) = g(∅, T ) = 0, k {k} T ∈X

,T ⊆S

for, by the hypothesis, g(∅, T ) = Y ∈X {m} ,Y ⊆T f (Y ) = 0 for all T ∈ X {k} such that T ⊆ S. On the other hand, the following recurrence equation holds.

g(R, S) = g R \ {α}, S − g R \ {α}, S \ {α} , P

for all α ∈ R. Therefore, the assertion follows by induction.



2.26.2. Second Day. 1. Let a ∈ [0, 1] \ Q be arbitrary. We show that f is P+∞ continuous at a. To this end, for a given ε > 0, choose k ∈ N such that i=k 21i < ε and let  δ = min a − r1 , . . . , a − rk . Note that δ > 0. If a − δ < x ≤ a, we can write X

f (x) − f (a) =

i∈{n∈N:x
+∞ X 1 1 ≤ < ε. 2i 2i i=k+1

Analogously, if a ≤ x < a + δ, then f (x) − f (a) =

X i∈{n∈N:a
That is, limx→a want.

+∞ X 1 1 ≤ < ε. 2i 2i i=k+1


f (x) − f (a) = 0. Thus, f is continuous at a, which is what we 

2. It easily follows from the hypothesis that f is one-to-one. Since f is continuous, we conclude that f is strictly increasing or decreasing on R. But f is not strictly decreasing, for otherwise f ◦ f would be strictly increasing and hence f ◦ f ◦ f = I

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2. SOLUTIONS

would be strictly decreasing, which is impossible. Thus, f is strictly increasing on R. Now, to prove the assertion by contradiction, suppose that there is an a ∈ R such that f (a) 6= a. Two cases If a < f (a), we see that f (a) < f (f (a)),  to consider.

 yielding f (a) < f f (a) < f f f (a) = a, a contradiction. Likewise, if a > f (a), we obtain a contradiction. So, the assertion follows by way of contradiction.  3. Suppose by way of contradiction that there is an x ∈ H ∩ K \ {e}, where e denotes the identity element of G. Since H is abelian and Z(G) = {e}, we have H ⊆ CG (x) & G, where CG (x) denotes the centralizer of the element x in G. Thus, H = CG (x) because H is maximal. Likewise, we see that K = CG (x). Therefore, H = K, which is a contradiction. So the assertion follows.  4. We prove the assertion by showing that An = λI, where I denotes the identity matrix of size n. Let F n denote the n-dimensional vector space of all n × 1 column vectors with entries in F . By the hypothesis, the vectors x, Ax, . . . , An−1 x are linearly independent and hence they form a basis for F n . We can write An (Ai x) = Ai (An x) = λAi x, implying (An − λI)(Ai x) = 0, for all 0 ≤ i ≤ n − 1. This yields An − λI = 0, and hence An = λI, because {x, Ax, . . . , An−1 x} forms a basis for F n .  5. Let (a1 , . . . , a6 ) and (b1 , . . . , b6 ) be the sequences of the numbers appearing on two extended dice. Let f (x) = xa1 + · · · + xa6 and g(x) = xb1 + · · · + xb6 . It follows from the hypothesis that f (x)g(x) = (x1 + · · · + x6 )2 = x2 (1 + x)2 (1 + x + x2 )2 (1 − x + x2 )2 . As f (0) = g(0) = 0, the factor x must appear in the factorizations of both of f and g. Since f (1) = g(1) = 6, the factors 1 + x and 1 + x + x2 must appear in the factorizations of both of f and g as well. Therefore, the only choices which do not lead to two ordinary dice are as follows f (x)

= x(1 + x)(1 + x + x2 ) = x + 2x2 + 2x3 + x4 ,

g(x)

= x(1 + x)(1 + x + x2 )(1 − x + x2 )2 = x + x3 + x4 + x5 + x6 + x8 .

That is, aside from the ordinary dice, the sequences of the numbers on the desired extended dice are (1, 2, 2, 3, 3, 4) and (1, 3, 4, 5, 6, 8).  6. In view of the definition of a balanced matrix and the hypothesis that the balanceness remains intact under interchanging the rows, it follows that for all j1 < j2 and i1 < i < i2 , we have   ci1 j1 + cij2 ≤ ci1 j2 + cij1 ⇐⇒ ci1 j1 + cij2 = cij2 + cij1 cij1 + ci1 j2 ≤ cij2 + ci1 j1 ⇐⇒ ci1 j1 + cij1 = ci1 j2 + cij2 . That is, the difference of two rows i and i0 is a constant vector whenever i1 ≤ i ≤ i0 ≤ i2 . Conversely, it follows from the above relations that if the difference of two rows is a constant vector, then the balanceness remains intact under interchanging the two rows. 

2.27. TWENTY SEVENTH COMPETITION

205

2.27. Twenty Seventh Competition 2.27.1. First Day. 1. First solution: See Problem 2 of 2.1.1. Second solution: Since all norms on the n2 -dimensional normed space Mn (R) are equivalent, it suffices to show that GLn (R) is open and disconnected with respect to the operator norm of Mn (R). To this end, let A ∈ GLn (R) be arbitrary. By showing that B ∈ GLn (R) whenever ||B − A|| < ||A−1 ||−1 , where ||.|| denotes the operator norm of Mn (R), we prove that GLn (R) is open. It is plain that B ∈ GLn (R) if and only if BA−1 ∈ GLn (R). But ||I − BA−1 || ≤ ||A − B|| ||A−1 || < 1. Thus, +∞
−1 X = (BA−1 )−1 = I − (I − BA−1 ) (I − BA−1 )n . i=1 −1

Consequently, BA ∈ GLn (R), implying B ∈ GLn (R), as desired. To show that GLn (R) is disconnected, suppose to the contrary that it is connected. Since GLn (R) is open and connected, it is path connected. Choose A, B ∈ GLn (R) with det(A) = − det(B) = 1. It follows that there is a continuous function φ : [0, 1] −→ GLn (R) with φ(0) = A and φ(1) = B. Since det : Mn (R) −→ R is
continuous, so is
det(φ) : [0, 1] −→ R. But det φ(0) = det(A) = 1 and det φ(1) = det(B) = −1. It thus follows from
the Intermediate Value Theorem that there is a c ∈ (0, 1) such that det φ(c) = 0. This implies φ(c) ∈ / GLn (R), which is a contradiction. Therefore, GLn (R) is disconnected, proving the assertion.  2. Let 0 ≤ θ < π be the angle of inclination of the line L, i.e., θ is the angle between
L and the x-axis. Define the function g : C −→ C by g(z) = f (eiθ z + A) − A e−iθ . It is plain that g is analytic on C, g takes the x-axis to the x-axis and the y-axis to the y-axis, and that g(0) = 0. We can write g(z) =

+∞ X

an z n ,

n=1

where an =

g (n) (0) n! .

Set ge(z) = g(¯ z) =

+∞ X

a ¯n z n ,

n=1

Since g maps the x-axis into the x-axis, we obtain (g − ge)(z) =

+∞ X

(an − a ¯n )z n = 0,

n=1

whenever z ∈ R. It thus follows from the Uniqueness Theorem for analytic functions that g = ge because the zeros of the analytic function g − ge has a limit point in C. Therefore, an = a ¯n which, in turn, implies an ∈ R for all n ∈ N. Define h : C −→ C by +∞ X h(z) = ig(iz) = an in+1 z n . n=1

Since g maps the x-axis into the x-axis and the y-axis into the y-axis, it follows that the analytic function h maps the x-axis into the x-axis. Hence, just as we saw

206

2. SOLUTIONS

in the above, we see that an in+1 ∈ R for all n ∈ N. This yields a2n = 0 for all n ∈ N. Therefore, g(z)

=

+∞ X

a2n−1 z 2n−1 ,

n=1

where a2n−1 ∈ R for all n ∈ N. Obviously, g(−z) = −g(z) for all z ∈ C. That is, if z1 , z2 ∈ C are symmetric with respect to 0, then so are g(z1 ) and g(z2 ) with respect to 0. Now, suppose z1 and z2 are symmetric with respect to A. So, z1 + z2 = 2A. We can write
f (z2 ) = eiθ g e−iθ (z2 − A) + A

= eiθ g − e−iθ (z1 − A) + A = −eiθ g e−iθ (z1 − A) + A = −f (z1 ) + 2A, implying f (z1 ) + f (z2 ) = 2A. Thus, f (z1 ) and f (z2 ) are symmetric with respect to A, finishing the proof.  x

−x

is one-to-one and 3. Note first that sinh : R −→ R defined by sinh(x) = e −e 2 onto. Thus, for each n ∈ N, there is a unique αn ∈ R such that an = sinh(αn ). We can write q q sinh(αn+1 ) = sinh(αn ) 1 + sinh2 (αn−1 ) + sinh(αn−1 ) 1 + sinh2 (αn ) = =

sinh(αn ) cosh(αn−1 ) + sinh(αn−1 ) cosh(αn ) sinh(αn−1 + αn ),

implying αn+1 = αn−1 + αn for each n ∈ N. The characteristic polynomial of √ 1± 5 2 this recurrence equation is x − x − 1 = 0, yielding x = 2 . Thus, αn = √
√
n n A 1+2 5 + B 1−2 5 for some A, B ∈ R, where n ∈ N ∪ {0}. To determine A and B, we have 0 sinh−1 (b) implying A = −B = −1

sinh

sinh−1 (b) √ . 5

sinh−1 (0) = α0 = A + B, √ √ 1 + 5
1 − 5
= A +B , 2 2

=

So we have

(an )

sinh−1 (b) √ = αn = 5

=

sinh−1 (b) √ 5

! √ √ 1 + 5
n 1 − 5
n − , 2 2

yielding an

sinh

for all n ∈ N, which is what we want.

!! √ √ 1 + 5
n 1 − 5
n − , 2 2 

4. To prove the assertion by contradiction, suppose that there is an i0 ∈ I such that i0 ∈ / Ai0 . It follows that {i : i ∈ / Ai } 6= ∅, from which, in view of the hypothesis, we see that {i : i ∈ / Ai } = Ai1 for some i1 ∈ I. Two cases to consider. If i1 ∈ Ai1 , then i1 ∈ / {i : i ∈ / Ai } = Ai1 , a contradiction. If i1 ∈ / Ai1 , then i1 ∈ {i : i ∈ / Ai } = Ai1 , a contradiction again. Therefore, i ∈ Ai for all i ∈ I, as desired. 

2.27. TWENTY SEVENTH COMPETITION

207

5. It is easily verified that (AB)2 = 9(AB). We can write (BA)3 = B(AB)2 A = 9B(AB)A = 9(BA)2 , yielding (BA)2 (BA − 9I2 ) = 0. If we use f to denote the minimal polynomial of BA, we see that f divides x2 (x − 9). As the degree of f is less than or equal to 2, there are four cases to consider. If f = x, then BA = 0, yielding AB = 0, for A(BA)B = (AB)2 = 9(AB), which is impossible. If f = x2 , then (BA)2 = 0, implying tr(AB) = tr(BA) = 0, which is again impossible. If f = x(x − 9), then tr(AB) = tr(BA) = 9, which is impossible. If f = x − 9, we obtain BA − 9I2 = 0. Thus, BA = 9I2 , which is what we want.  6. We first show that H is normal in G. To prove this by contradiction, suppose that there exist g ∈ G and h ∈ H such that g −1 hg ∈ / H. Since g −1 hg ∈ G \ H, it follows from the hypothesis that there is a u ∈ H such that g(g −1 hg)g −1 = u−1 (g −1 hg)u, from which, we obtain g −1 hg = uhu−1 ∈ H, a contradiction. Thus, H is normal in G. To prove that G/H is abelian, it suffices to show that g1 g2 H = g2 g1 H, or equivalently, g2−1 g1−1 g2 g1 ∈ H for all g1 , g2 ∈ G. If g2 ∈ H, we see that g2−1 (g1−1 g2 g1 ) ∈ H because H is normal in G. So we may, with no loss of generality, assume that g2 ∈ / H. Hence, for g1 ∈ G, there is a u ∈ H such that g1−1 g2 g1 = u−1 g2 u. This implies g2−1 g1−1 g2 g1 = (g2−1 u−1 g2 )u ∈ H, for H is normal in G. It thus follows that G/H is abelian, as desired.  2.27.2. Second Day. 1. First solution: Since f 0 (a) > 0, f 0 (b) > 0, and f (a) = f (b) = 0, there is a δ > 0 such that a + δ < b − δ, f (x) > 0 whenever a < x < a + δ, and that f (x) < 0 whenever b − δ < x < b. It follows that f (a + δ) ≥ 0 and f (b − δ) ≤ 0. So by the Intermediate Value Theorem, there is a c ∈ [a + δ, b − δ] such that f (c) = 0. We have f (a) = f (c) = f (b) = 0. Applying the Rolle’s Theorem, we obtain c1 ∈ (a, c) and c2 ∈ (c, b) such that f 0 (c1 ) = f 0 (c2 ) = 0, as desired. Second solution: Recall that by Darboux’s Theorem, the derivative function has the intermediate value property. If there is a c ∈ (a, b) such that f 0 (c) < 0, then f 0 (a)f (c) < 0 and f 0 (c)f 0 (b) < 0. Hence, by Darboux’s Theorem, there are c1 ∈ (a, c) and c2 ∈ (c, b) such that f 0 (c1 ) = f 0 (c2 ) = 0, proving the assertion in this case. If f 0 (x) ≥ 0 for all x ∈ (a, b), then f is increasing on [a, b]. This implies 0 = f (a) ≤ f (x) ≤ f (b) = 0, whence f (x) = 0 for all x ∈ [a, b], in which case the assertion is trivial. So the proof is complete.  2. First we need to recall the following lemma from theory of metric spaces. Also recall that a subset of a metric space is said to be perfect if it is closed and has no isolated points. Lemma. Every nonempty perfect subset of a complete metric space is uncountable. Proof. Let P be a nonempty perfect subset of a complete metric space X. As P is closed, P equipped with the metric X induces on P forms a complete metric space itself. Now to prove the assertion by contradiction, suppose that P is countable so that P = {xi }+∞ i=1 , where xi ∈ X for all i ∈ N. As P is perfect, each singleton {xi } (i ∈ N) is a nowhere dense subset of P , implying that P is a

208

2. SOLUTIONS

countable union of nowhere dense subsets, namely {xi }’s where i ∈ N. This is in contradiction with the Baire Category Theorem which asserts that every complete metric space is of second category in itself, i.e., the space cannot be written as a countable union of nowhere dense subsets of it. Thus, P is uncountable, which is what we want.  To prove the assertion by contradiction, suppose that [0, 1] is written as a union of mutually disjoint closed intervals each of which having a positive length less than one. There are a countable number of the intervals participating in the union because they are mutually disjoint. So we may assume that [0, 1] =

+∞ [

In ,

n=1

where In = [an , bn ] for some an , bn ∈ R with 0 < bn − an < 1 and Im ∩ In = ∅ whenever m, n ∈ N and m 6= n. Set  +∞  +∞ P = an n=1 ∪ bn n=1 . The set P is closed. Because if x ∈ [0, 1] \ P , then x ∈ (an , bn ) for some n ∈ N, and hence x is an interior point of [0, 1] \ P . We now show that any point x ∈ P \ {0, 1} is a limit point of P . Note that 0 and 1 are isolated points of P . If x = an > 0 for some n ∈ N, then for every 0 < δ < an , there exists an m ∈ N such that (an − δ, an ) ∩ Im 6= ∅. Since In ∩ Im = ∅, we see that bm ∈ (an − δ, an ). Thus, x = an is a limit point of P , as claimed. Likewise, if x = bn < 1 for n ∈ N, then x is a limit point of P . Therefore, P \ {0, 1} is a nonempty perfect subset of R, which is a complete metric space endowed with the ordinary metric of R. By the lemma above, the set P \ {0, 1} is uncountable, and hence so is P , which is obviously a contradiction. This completes the proof.  3. As D is countable, there are a countable number of lines which are all parallel x to the x-axis so that each of which interests D. Use {`xi }N i=1 to denote these lines, where `xi : x = xi and Nx ∈ N ∪ {+∞}. Likewise, there are a countable number Ny of lines, denoted by {`yj }j=1 , where `yj : y = yj and Ny ∈ N ∪ {+∞}, which are all parallel to the y-axis so that each of which intersects D. It is obvious that Ny Nx D ⊆ A × B, where A = {xi }i=1 and B = {yj }j=1 . Set  Γ = D ∩ (xi , yj ) : j ≤ i ,  ∆ = D ∩ (xi , yj ) : j > i . As D ⊆ A × B, {Γ, ∆} is a partition of D into two subsets. Now, any line parallel to the x-axis whose equation is given by x = a, where a ∈ R, intersects Γ at no point or at most at the i points (xi , y1 ), . . . , (xi , yi ) depending on whether a ∈ / A or a = xi for some 1 ≤ i ≤ Nx < +∞ or for some i ∈ N if Nx = +∞. Analogously, any line parallel the y-axis whose equation is given by y = b, where b ∈ R, intersects ∆ at no point or at most at the j − 1 points (x1 , yj ), . . . , (xj−1 , yj ) depending on whether b ∈ / B or b = yj for some 1 ≤ j ≤ Ny < +∞ or for some j ∈ N if Ny = +∞. So the assertion follows.  S S 4. For a subset I ⊆ Nn := {1, . . . , n}, set BI = ( i∈I Ai )c = S \( i∈I Ai ). Also, let K := {I ⊆ Nn : |I| = k − 1}. We claim that if I, J ∈ K, then BI ∩ BJ = ∅ whenever I 6= J. To see this, note first that BI ∩ BJ = BI∪J . Since I, J ∈ K and I 6= J,

2.27. TWENTY SEVENTH COMPETITION

209

S we see that |I ∪ J| ≥ k, which, in view of the hypothesis, yields i∈I∪J Ai = S, and hence BI ∩ BJ = BI∪J =
S \ S = ∅. It is now plain that the number of BI ’s,
n n where I ∈ K, is equal to k−1 and that such BI ’s form a partition of S into k−1 subsets. In other words, [ ˙  S= BI : I ∈ K , (∗) S where ˙ stands for the disjoint union.
Thus, |BI | = 1 for all I ∈ K, because the n number of such BI ’s is equal to k−1 = |S| and that they are all nonempty. Now, fix an arbitrary i0 ∈ I. Note that K is a disjoint union of
K0 := {I ∈ K : i0 ∈ I} and K00 := {I ∈ K : i0 ∈ / I}. It is plain that |K0 | = n−1 k−2 and that if x ∈ BI for some I ∈ K0 (resp. I ∈ K00 ), then x ∈ / Ai0 (resp. x ∈ Ai0 ). This together with (∗), in view of the fact that |BI | = 1 for all I ∈ K, implies that       n n−1 n−1 |Ai0 | = − = . k−1 k−2 k−1
In other words, Ai = n−1  k−1 for all i ∈ Nn , which is what we want. 5. Let k = ordab (a + b), m = ordb (a), n = orda (b), and [m, n] = lcm(m, n). We b

a

have am ≡ 1 and bn ≡ 1. Also, [m, n] = mq = nq 0 for some q, q 0 ∈ N. So we can write 0 a a a (a + b)[m,n] ≡ b[m,n] ≡ bnq ≡ 1. Analogously, b

b

b

(a + b)[m,n] ≡ a[m,n] ≡ amq ≡ 1. ab

Since gcd(a, b) = 1, it follows that (a + b)[m,n] ≡ 1, which implies ordab (a + b) ≤ ab

[m, n] = lcm[ordb (a), orda (b)]. On the other hand, (a + b)k ≡ 1. So, in particular, a

b

a

a

(a + b)k ≡ 1 and (a + b)k ≡ 1, which, in turn, yields bk ≡ (a + b)k ≡ 1 and b b ak ≡ (a+b)k ≡ 1. Thus, m k and n k, and hence lcm[ordb (a), orda (b)] = [m, n] k = ordab (a + b). Therefore, ordab (a + b) = lcm[ordb (a), orda (b)], as desired.



6. The rings R and R0 are both abelian and have characteristic 2 because they are Boolean rings. Since f is onto and 0 ∈ R0 , there is an x ∈ R such that f (x) = 0. We can write f (0) = f (0x) = f (0)f (x) = f (0)0 = 0. That is, f (0) = 0. Now, let a, b ∈ R be arbitrary. By the surjectivity of f , there is an x ∈ R such that f (a + b) − f (a) − f (b) = f (x). We need to show that x = 0. We can write
f (ax) f (a + b) − f (a) − f (b) = f (ax)f (x) = f (ax), implying that f (ax + abx) − f (ax) − f (abx) = f (ax), and hence f (ax + abx) = f (abx).

210

2. SOLUTIONS

Since f is one-to-one, we obtain ax + abx = abx, implying that ax = 0. Likewise, we see that bx = 0. Now, in view of ax = bx = 0, we can write
f (x) f (a + b) − f (a) − f (b) = f (x)f (x) = f (x), which yields f (ax + bx) − f (ax) − f (bx) = f (x). Hence, f (x) = 0 = f (0), from which, in view of the injectivity of f , we see that x = 0, which is what we want.  2.28. Twenty Eighth Competition P+∞ 2.28.1. First Day. 1. Let f (z) = n=0 cn z n . If f is a polynomial, then, by the Fundamental Theorem of Algebra, we have f (z) = c0 + c1 z because f is one-to-one. If not, then f ( z1 ) would have an essential singularity at zero. Thus, by Picard’s Great Theorem, f ( z1 ) is not one-to-one in any neighborhood of zero, and hence neither is f , a contradiction. This proves the assertion. 
2. We prove the assertion by showing that f −1 {(0, 0)} is not
bounded, and hence has infinitely many points. As f is
surjective, f −1 {(0, 0)} is nonempty.
Suppose to the contrary that f −1 {(0, 0)} is bounded. Thus, f −1 {(0, 0)} is a nonvoid closed bounded
subset of R. Consequently,
there exist a, b ∈ R such that a = min f −1 {(0, 0)} < b = max f −1 {(0, 0)} . Let A1 = {0} × (0, +∞), A2 = {0} × (−∞, 0), A2 = (0, +∞) × {0}, and A4 = (−∞, 0) ×
{0}. As f is continuous and [a, b] is a compact subset of R, we see that f [a, b] is a bounded subset of A. Now, since f is continuous and f (x) 6= 0 for all x ∈ (−∞, a) ∪ (b, +∞), there are i1 , i2 ∈ {1, 2, 3, 4}, not necessarily distinct, such that f (−∞, a) ⊆ Ai1 and f (b, +∞) ⊂ Ai2 . Therefore, f (R) ∩ Ai is a bounded subset Ai for each i ∈ {1, 2, 3, 4}\{i1 , i2 }, which is a contradiction because f is surjective. So the assertion follows.  3. By the hypothesis and the M¨obius Inversion Formula, we have X X µ(d) n f (n) = µ(d)( )2 = n2 = n2 g(n), d d2 d|n

for all n ∈ N, where g(n) =

d|n

µ(d) d|n d2 .

P

The arithmetic function g is multiplicative

µ(d) d2

because is a multiplicative function of d. If p is prime and α is a positive integer, we can write X µ(d) µ(1) µ(p) 1 g(pα ) = = + 2 = 1 − 2. 2 d 1 p p d|pα Q Thus, if n = p|n pαp , then  Y Y Y 1 αp αp g(n) = g( p ) = g(p ) = 1− 2 p p|n p|n p|n       Y 1 Y 1 φ(n) Y 1 1+ = 1+ . = 1− p p n p p|n

p|n

p|n

2.28. TWENTY EIGHTH COMPETITION

211

Consequently,    Y f (n) n2 g(n) n2 φ(n) Y 1 1 =n 1+ , = = 1+ φ(n) φ(n) φ(n) n p p p|n

p|n

for all n ∈ N with n > 1, as desired.



4. Let 1 ≤ i, k ≤ n be arbitrary and τ ∈ Gk ∩ Z(G). By the hypothesis, there exists a σ ∈ G such that σ(k) = i. We can write
λ ∈ Gi ⇐⇒ λ(i) = i ⇐⇒ λ σ(k) = σ(k) ⇐⇒

σ −1 λσ(k) = k ⇐⇒ σ −1 λσ ∈ Gk

⇐⇒

λ ∈ σGk σ −1 .

Thus, Gi = σGk σ −1 , and hence στ σ −1 ∈ Gi because τ ∈ Gk . On the other hand, τ ∈ Z(G) implies τ = στ σ −1 ∈ Gi . It follows that τ ∈ ∩ni=1 Gi = {e} because i was arbitrary. This proves Gk ∩ Z(G) = {e}, which is what we want.  5. We prove that the maximum number of such vectors is n. To this end, let the vectors v1 , . . . , vk ∈ A be such that any two of which share an even number of one entries. View the elements of A as vectors in the n-dimensional vector space Zn2 , where Z2 = {0, 1} is the field of integers modulo 2. We claim that the set {v1 , . . . , vk } is a linearly independent subset of Zn2 . To see this, suppose c1 v1 + · · · + ck vk = 0 for some c1 , . . . , ck ∈ Z2 . If “.” denotes the usual inner product on Zn2 , for all 1 ≤ i ≤ n, we have ci = vi .(c1 v1 + · · · + ck vk ) = 0, proving the claim. Thus, k ≤ n. On the other hand, if ei is the vector with 1 in the ith place and zero elsewhere, then any two elements of the set {e1 , . . . , en }, the standard basis of Zn2 , share an even number of one entries. Therefore, the maximum number of such vectors is n, as desired.  6. Let P = {p1 , . . . , pn }. As is usual, use Sym(P ) to denote the set of all permutations on (the poset) P . Let’s view Sym(P ) as a probability space. Choose a permutation σ on {1, . . . , n} at random which, in turn, gives rise to a permutation sσ ∈ Sym(P ) at random so that sσ (pi ) = pσi for all 1 ≤ i ≤ n. For any 1 ≤ i ≤ n, there exists a 1 ≤ j ≤ n such that σj = i, which is equivalent to pi = sσ (pj ) = pσj . With all that in mind, for an element pi ∈ P for which pi = sσ (pj ) = pσj for some 1 ≤ j ≤ n, let Fpi be the event that the permutation sσ ∈ Sym(P ) satisfies the following ∀p ∈ P \ Upi : p = sσ (pk ) = pσk =⇒ σj ≤ σk. In other words, if   p1 p2 · · · pn sσ = , pσ1 pσ2 · · · pσn then sσ ∈ Fpi if and only if the element pi precedes all of the elementsTofTP \Upi but pi in the second row of the above representation of sσ . Set Epi := Fpi ( p∈Lp Fpc ), i where Fpc := Sym(P ) \ Fp . We note that if pi and pi0 are distinct, then the events Epi and Epi0 cannot occur simultaneously, and hence they are disjoint. Let’s prove this. First, assume that for pi , pi0 ∈ P , Epi , and hence Fpi , occurs, and that pi and pi0 are not comparable. Then, in particular pi0 6> pi yielding pi0 ∈ P \ Upi , which, in turn, implies σj < σj 0 ,

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2. SOLUTIONS

where i = σj and i0 = σj 0 . Consequently, Fpi0 , and hence Epi0 , cannot occur, for otherwise, we must have σj 0 < σj because pi 6> pi0 , which is a contradiction. That is, Fpi and Fpi0 , and hence Epi and Epi0 , cannot occur simultaneously whenever pi and pi0 are not comparable. Next, if the elements pi , pi0 ∈ P are comparable, then it is plain that the events Epi and Epi0 cannot occur simultaneously. Thus, the events Epi and Epi0 are disjoint whenever pi , pi0 ∈ P are distinct. We now use induction on |Lpi | to prove Xpi = P (Epi ). If |Lpi | = 0, i.e., if pi is a minimal element of P , then Epi = Fpi and sσ ∈ Fpi with sσ (pj ) = pi if and p = s(pk ) = pσk ∈ P \ Upi for some 1 ≤ k ≤ n. As only if σj ≤ σk whenever P \ Up = n − Up , in this case, we can write i i
n (|Upi |)! n−|U (n − |Upi | − 1)! 1 pi | = . P (Epi ) = P (Fpi ) = n! n − |Upi | Now, assuming that Xp = P (Ep ) for all p ∈ Lpi , we prove Xpi = P (Epi ). We claim that \ \ Fpc = Epc , p∈Lpi

p∈Lpi

T for all pi ∈ P . To see this, first, let sσ ∈ p∈Lp Fpc be arbitrary. We show i T sσ ∈ p∈Lp Epc . Suppose the contrary, implying that there is a q ∈ Lpi such that i It follows that sσT∈ Eq , and hence sσ ∈ Fq . This is in contradiction sσ 6∈ Eqc . T T T with sσ ∈ p∈Lp Fpc . Thus, p∈Lp Fpc ⊆ p∈Lp Epc . Next, letting sσ ∈ p∈Lp Epc i i i i T be arbitrary, we show sσ ∈ p∈Lp Fpc . Again suppose the contrary, implying that i there is a minimal q ∈ Lpi so that sσ 6∈ Fqc but sσ ∈ Fpc for all p ∈ T Lq .TIt follows that sσ ∈ Fq but sσ 6∈ Fp for all p ∈ Lq . Consequently, sσ ∈ Eq = Fq ( p∈Lq Fpc ), T T which is a contradiction. Therefore, p∈Lp Epc ⊆ p∈Lp Fpc , proving the claim. i i Now, as Ep ’s (p ∈ Lpi ) are disjoint, in view of the claim, we can write \ [ X \


Fpc = P Epc = 1 − P Ep = 1 − Xp . P p∈Lpi

p∈Lpi

p∈Lpi

T

p∈Lpi

Fpc

are independent because the On the other hand, the events Fpi and p∈Lp i position of pi in the permutation has no affect on Fp ’s (p ∈ Lpi ). Thus, P 1 − p∈Lp Xp \ \ \

c c i . P (Epi ) = P ( Fp ) Fpi = P Fp P (Fpi ) = n − |Upi | p∈Lpi

p∈Lpi

Therefore, Xpi = P (Epi ) for all pi ∈ P , and hence 0 ≤ Xpi ≤ 1 for all pi ∈ P , which is what we want. 

2.28.2. Second Day. 1. Let g = f −1 . It follows from the hypothesis that for all x, y with y = 2x2 , we have √ Z x 2 3 1 A = (2t2 − t2 )dt = x3 = y2 = 3 12 0 √ Z y r Z y
t 2 3 2 B = − g(t) dt = y − g(t)dt, 2 3 0 0

2.28. TWENTY EIGHTH COMPETITION

213

P B

A 1.5 Figure 13

implying Z

√

y

g(t)dt = 0

2 3 y2. 4

Taking derivative of both sides yields √ 32 2 3 2 1 9 y 2 =⇒ x2 = y =⇒ y = x . x = g(y) = 8 32 9 2 Thus, y = f (x) = 32 9 x , which is what we want.



2. Set h(x) = f (x) − x. We have h(a) > 0 and h(b) < 0. Since h is continuous, there exists a δ > 0 such that h(a + x) > 0 and h(b − x) < 0 whenever x ∈ (0, δ). δ Set α = n+1 and g(x)

= h(x) + h(x + α) + · · · + h(x + nα)

= f (x) + · · · + f (x + nα) − x + · · · + (x + nα) .

It is readily seen that g(a) > 0 and g(b − nα) < 0. By the Intermediate Value Theorem, there exists a c ∈ (a, b − nα) such that g(c) = 0. This implies f (c) + f (c + α) + · · · + f (c + nα)

which is what we want.

= c + (c + α) + · · · + (c + nα) n(n + 1) n = (n + 1)c + α = (n + 1)(c + α), 2 2 

3. The answer is yes. To see this, noting that det(A + B) = det(At + B t ), we can write det(A) det(A + B)

= det(A) det(At + B t ) = det(I + AB t ) = det(I + BAt ) = det(B) det(At + B t ) = det(B) det(A + B).

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2. SOLUTIONS

Consequently, in view of det(A) + det(B) = 0, we have det(A) det(A + B) = det(B) det(A + B) = − det(A) det(A + B), yielding 2 det(A) det(A + B) = 0. But det(A) = ±1 because A is orthogonal. Thus, det(A + B) = 0, as desired.  4. Let x, y ∈ R be arbitrary. We can write xn+2 y n+2 = (xy)n+2 = (xy)n+1 xy = xn+1 y n+1 xy, yielding xn+1 (xy n+1 − y n+1 x)y = 0. As the above equality holds for all x, y ∈ R, changing x to x + 1 in the equality, we obtain (1 + x)n+1 (xy n+1 − y n+1 x)y = 0, or        n+1 n+1 n + 1 n+1 + x + ··· + x (xy n+1 − y n+1 x)y = 0. 0 1 n+1 Multiplying both sides of the equality by xn and using the aforementioned equality, we get xn (xy n+1 − y n+1 x)y = 0. Analogously, changing x to x + 1 in this equality, we see that xn−1 (xy n+1 − y n+1 x)y = 0. Continuing in this way, we finally obtain (xy n+1 − y n+1 x)y = 0.

(∗)

On the other hand, by the hypothesis, we have xn+1 y n+1 = (xy)n+1 = (xy)n xy = xn y n xy, yielding xn (xy n − y n x)y = 0. Likewise, from this, we obtain (xy n − y n x)y = 0. Multiplying both sides of the above by y from the left, we get yxy n+1 − y n+1 xy = 0, from which, in view of (∗), we see that (yx − xy)y n+1 = 0. Once again, preforming an argument involving changing y to y + 1 and multiplying both sides of the obtained equality by an appropriate power of y, we eventually conclude that yx − xy = 0. That is, xy = yx for all x, y ∈ R, which is what we want.  5. Suppose that a person is chosen at random. Define the following events. A (resp. B) := the event that the person is guilty (resp. innocent). + (resp. −) := the event that the person responds positively (resp. negatively). E := the event that the person responds positively to four questions and negatively to one question.

2.29. TWENTY NINTH COMPETITION

215

It follows from the hypothesis that P (A) = 31 , P (B) = 23 , P (+|A) = 0.8, P (+|B) = 0.4, P (−|A) = 0.2, and P (−|B) = 0.6. Using Bayes’ Theorem, we see that the desired probability is equal to P (A ∩ E) P (A)P (E|A) P (A|E) = = P (E) P (A)P (E|A) + P (B)P (E|B)
1 5 4 8 3 4 (0.8) (0.2)
= 1 5
. = 2 5 4 4 16 3 4 (0.8) (0.2) + 3 4 (0.4) (0.6) Thus, P (A|E) =

8 16

is the desired probability.



6. We claim that for each j = 1, . . . , k, there are subsets A1 , . . . , Aj satisfying the following inequalities. 1 ≤ |A1 ∩ · · · ∩ Aj | ≤ r − (j − 1)(I(F ) − 1)

(∗).

Obviously, the assertion follows from (∗) by letting j = k. We prove (∗) by induction on j. If j = 1, the assertion is trivial; in fact, any element A1 of F satisfies (∗) when j = 1. Suppose that (∗) holds for j, where j ≤ k − 1. We prove the assertion for j + 1. To this end, it follows from the hypothesis that A1 ∩ · · · ∩ Aj has nonempty intersection with any element of F . This implies |A1 ∩ · · · ∩ Aj | ≥ I(F ). Let S be a subset of A1 ∩ · · · ∩ Aj with I(F ) − 1 elements. Thus, there exists an Aj+1 ∈ F such that S ∩ Aj+1 = ∅. So we can write 1 ≤

|A1 ∩ · · · ∩ Aj ∩ Aj+1 | ≤ |(A1 ∩ · · · ∩ Aj ) ∩ S c |


= |(A1 ∩ · · · ∩ Aj ) \ S| = |A1 ∩ · · · ∩ Aj | − I(F ) − 1

≤ r − (j − 1) I(F ) − 1 − (I(F ) − 1) = r − j I(F ) − 1 , proving the induction assertion, finishing the proof.



2.29. Twenty Ninth Competition 2.29.1. First Day. 1. First solution: We can write Z a Z a dx I := f (x)dx 0 f (x) Z0 a Z a Z aZ a 1 1 = f (x) dxdy = f (y) dxdy, f (y) f (x) 0 0 0 0 from which, we obtain  Z aZ a f (x) f (y) 2I = I + I = + dxdy f (y) f (x) 0 0 Z aZ a f (x)2 + f (y)2 = dxdy. f (x)f (y) 0 0 On the other hand, obviously, we have Z aZ 2I ≥ 0

f (x)2 +f (y)2 f (x)f (y) a

2dxdy = 2a2 ,

0

yielding I ≥ a2 , as desired.

≥ 2 for all 0 ≤ x, y ≤ a. Thus,

216

2. SOLUTIONS

Second solution: By the Cauchy-Schwarz Inequality, we can write Z 2 Z  Z a  ap a dx 1 2 f (x)dx a = f (x) × p dx ≤ , 0 f (x) 0 f (x) 0 which is what we want.



2. The “if part” is trivial. We prove the “only if part” of the assertion by contradiction. To this end, suppose x = ab for some a, b ∈ N. It follows from the contradiction hypothesis that there exists a k ∈ N such that nk+1 − 1 > b. Set xk =

k X i=1

1 . n1 · · · ni

We can write a 1 1 − xk = x − xk = + + ··· b n1 · · · nk+1 n1 · · · nk+1 nk+2   1 1 1 1 1 1 + 2 + 3 + ··· = . < n1 · · · nk nk+1 nk+1 nk+1 n1 · · · nk nk+1 − 1 Consequently, 0 <

a 1 n1 · · · nk − xk n1 · · · nk < , b nk+1 − 1

whence b < 1, nk+1 − 1 which is a contradiction because an1 · · · nk − bxk n1 · · · nk ∈ N. Thus, the assertion follows by contradiction.  0 < an1 · · · nk − bxk n1 · · · nk <

3. Note that 10φ(m) ≡ 1 (mod m) for all positive integers m satisfying gcd(m, 10) = 1, where φ denotes Euler’s totient function. It follows that m 10kφ(m) − 1 whenever k, m ∈ N with gcd(m, 10) = 1. Moreover, if gcd(m, 3) = 1, then kφ(m) 10 −1 m = 1 + 10 + · · · + 10kφ(m)−1 . 10 − 1 kφ(m) times

Thus, if gcd(m, 30) = 1, then m

z }| { 11 . . . 1

for all k ∈ N, proving the assertion. 

4. Proceed by way of contradiction. Suppose I is a nonzero right ideal in R whose elements all have square zero. It follows that (x + y)2 = 0, and hence xy = −yx for all x, y ∈ R. Thus, (xI)(xI) = 0, from which, we see that (RxI)(RxI) ⊆ (Rx)I(xI) = R(xI)(xI) = 0, for all x ∈ R. In other words, the two-sided ideal RxI is nilpotent, which, in view of the hypothesis, yields RxI = 0. This shows that xI is contained in the left annihilator of R in R, denoted by l.ann(R). But l.ann(R) is nilpotent. Thus, xI ⊆ l.ann(R) = {0}, yielding xI = {0} for all x ∈ R. This easily implies I 2 = {0}, from which, we obtain (RI)(RI) ⊆ RI 2 = {0}. But RI is a two-sided ideal of R. So it follows from the hypothesis that RI = {0}. This implies that I is contained in the right annihilator of R in R, denoted by r.ann(R). But r.ann(R) is

2.29. TWENTY NINTH COMPETITION

217

obviously a nilpotent two-sided ideal. Thus, I ⊆ l.ann(R) = {0}, yielding I = {0}, a contradiction. This proves the assertion.  5. Define the function f : X −→ Y by  f (A) = k ∈ Z : 2k ∈ A . It suffices to show that for any B ∈ Y , there exists a A ∈ X such that f (A) = B. To this end, for given B ∈ Y , set  A := 2k : k ∈ B . It is obvious that A ∈ X and f (A) = B, as desired.



6. Note first that since S ≤ Zn2 is a vector space over Z2 , d is equal to the minimum of the number of ones occurring in the nonzero elements of S. It is plain that S has 2k elements. Use a matrix A ∈ M(2k −1)×n (Z2 ) to denote the nonzero elements of S, so that every nonzero element of S corresponds to a row of A. We may assume that the first k rows of A form a basis for S. Denote this k × n submatrix of A by B. It is plain that the number of one entries of the matrix A is at least d(2k − 1). By showing that this number is at most n2k−1 , we conclude that d(2k −1) ≤ n2k−1 , proving the assertion. We claim that the number of ones in any column of A is either 0 or 2k−1 . Since every row of A is a linear combination of the rows of B, the number of ones in any column of A is computed as follows. First, if there exists no “one” in column i (1 ≤ i ≤ n) of B, then neither does exist in the same column of A. Next, let t be the number of ones in column i (1 ≤ i ≤ n) of B. Since any linear combination of the rows of B gives rise to a row of A, and since the coefficients come from Z2 , it follows that the number of ones in column i of A, denoted by 1i , is equal to the number of all subsets of {1, 2, . . . , n} with k elements each of which contains an odd number of one entries from column i of B. Thus, 1i is equal to the number of all odd numbered subsets of a set with t elements times the number of all subsets of a set with k − t elements. In other words, 1i = 2t−1 2k−t = 2k−1 . Now, as A has k columns, the number of all ones is at most n2k−1 , finishing the proof. 

2.29.2. Second Day. 1. First solution: Let f (z) = f (¯ z ). It is plain that +∞ X g(z) = a ¯n z n ,

P+∞

n=0

an z n . Set g(z) =

n=0

and that g is analytic on D. It follows that f − g is analytic on D and moreover (f − g)( n1 ) = f ( n1 ) − g( n1 ) = 0 for all n ≥ 2. Hence, the set of zeros of the analytic function f − g has a limit point in D. It thus follows from the Uniqueness Theorem for analytic functions that f (z) = g(z) for all z ∈ D, whence an = a ¯n for all n ∈ N. Therefore, an ∈ R for all n ∈ N. This, in view of an = for all n ∈ N, as desired.

f (n) (0) n! ,

yields f (n) (0) ∈ R

(n) P+∞ Second solution: Let f (z) = n=0 an z n , where an = f n!(0) for all n ∈ N ∪ {0}. It suffices to show that an ∈ R for all n ∈ Z with n ≥ 0. We prove this by induction

218

2. SOLUTIONS

on n. If n = 0, the assertion is trivial because a0 = f (0) = limn f ( n1 ). Assuming that a0 , . . . , an−1 ∈ R, to prove an ∈ R, set  1
Pn−1 +∞ X ai z i z 6= 0, i n f (z) − i=0 z g(z) = ai+n z = an z = 0. i=0

It is readily verified that g is an analytic function satisfying g( n1 ) ∈ R for all natural  numbers n ≥ 2. Thus, an = g(0) = limn g( n1 ) ∈ R, which is what we want. 2. (i) The assertion can be stated as follows. For a given ε > 0, every point of the space X can be reached from any other point of the space within a finite number of steps each of which having a length less than ε. With this in mind, for given ε > 0, fix an x ∈ X and set  y ∈ X : ∃x1 = x, x2 , . . . , xn = y ∈ X  ∀i < n : d(xi , xi+1 ) < ε . Cε (x) := To prove the assertion, it suffices to show that Cε (x) = X. This, in view of the hypothesis that X is connected, follows as soon as we show that Cε (x) is a clopen subset, i.e., both a closed and open subset, of X. First, suppose z ∈ Cε (x). It follows that there exists a sequence (yn )+∞ n=1 in Cε (x) such that limn yn = z. Thus, there exists an N ∈ N such that d(yn , z) < ε whenever n ≥ N . On the other hand, since yn ∈ Cε (x), we have Bε (yn ) := {t ∈ X : d(t, yn ) < ε} ⊆ Cε (x). Consequently, z ∈ Cε (x) because z ∈ Bε (yn ). Thus, Cε (x) is a closed subset of X. Next, suppose y ∈ Cε (x) is arbitrary. It follows that Bε (y) ⊆ Cε (x). This implies that Cε (x) is an open subset of X. Therefore, Cε (x) = X because X is connected. (ii) Plainly, X = [0, 1) ∪ (1, 2] is an example showing the converse of (i) does not necessarily hold in general. (iii) To prove the assertion by contradiction, suppose X is disconnected. It follows that there is a clopen subset M of X. Since X is compact and both M and M c = X \ M are closed subsets of X, they both are compact subsets of X. But M ∩ M c = ∅. Hence, d(M, M c ) := inf{d(x, y) : x ∈ M, y ∈ M c } > 0. Pick x ∈ M and y ∈ M c . It is now obvious that x cannot be reached from y within a finite number of steps each of which having a length less than ε = d(M, M c ). Because, otherwise ε = d(M, M c ) ≤ d(xi , xi+1 ) < ε, where xi ∈ M and xi+1 ∈ M c are the consecutive foot steps in a walk from x to y with a finite number of steps each of which having a length less than ε. This is a contradiction. Thus, the assertion follows.  3. (i) For given g ∈ G, define ϕg : K −→ K by ϕg (x) = g −1 xg. Since g ∈ NG (K), the map ϕg is well-defined because g −1 xg ∈ K for all x ∈ K. This implies ϕg ∈ Aut(K). In other words, the map ϕ : NG (K) −→ Aut(K) defined by ϕ(g) = ϕg is well-defined. It is easily verified that ϕ is a homomorphism of groups and that  ker ϕ = g ∈ NG (K)| ϕ(g) = id  = g ∈ NG (K)| ∀x ∈ K : g −1 xg = x  = g ∈ NG (K)| ∀x ∈ K : gx = xg = CG (K). Now, the assertion is a quick consequence of the First Isomorphism Theorem for groups.

2.29. TWENTY NINTH COMPETITION

219

(ii) By (i), in view of the hypothesis that K E G, the group CGG(K) is isomorphic to a subgroup of Aut(K). Since K is cyclic, Aut(K) is abelian, and hence so is G 0 0 CG (K) , whence G ≤ CG (K). It follows that G ≤ CG (K) ≤ G because G = G. Thus, G = CG (K), implying K ≤ Z(G), which is what we want.  4. To prove the “if part”, it suffices to show that any two invariant subspaces of any nilpotent transformation N on F n with N n−1 6= 0 are comparable with respect to inclusion. Note first that N n = 0 and that there exists a vector α ∈ F n such that N n−1 α 6= 0 because N n−1 6= 0. It follows that the set {N n−1 α, . . . , N α, α} is a basis for F n . To see this, suppose c1 N n−1 α + · · · + cn−1 N α + cn α = 0 for some ci ∈ F (1 ≤ i ≤ n). Taking N n−1 of both sides of this equality yields cn N n−1 α = 0, implying cn = 0. Thus, c1 N n−1 α + · · · + cn−1 N α = 0. Now, taking N n−2 of both sides of the equality yields cn−1 = 0. Continuing in this manner, we obtain ci = 0 for all 1 ≤ i ≤ n. Thus, the set {N n−1 α, . . . , N α, α} is a basis for F n because dim F n = n. Set



 M0 = 0 = {0}, Mi = N n−1 α, . . . , N n−i α . We have

 M0 = {0} < M1 = N n−1 α < · · · < Mn = F n . It is plain that Mi is an invariant subspace of N for all 1 ≤ i ≤ n. This shows that any nilpotent transformation with N n−1 6= 0 is triangularizable. Now, suppose that M is an invariant subspace for N . We prove the assertion by showing that M = Mj for some 1 ≤ j ≤ n. To this end, let 1 ≤ j ≤ n be the greatest natural number for which there exists a nonzero vector x ∈ M such that x = c1 N n−1 α + · · · + cj N n−j α, j−1

(∗)

where cj 6= 0. We show that M = Mj . Taking N of both sides of (∗), we get cj N n−1 α = N j−1 x ∈ M, whence N n−1 α ∈ M because cj 6= 0. Now, using N n−1 α ∈ M and taking N j−2 of both sides of (∗), we see that N n−2 α ∈ M. Continuing this way, we conclude that N n−j α ∈ M. Thus, Mj = hN n−1 α, . . . , N n−j αi ⊆ M. On the other hand, M ⊆ Mj because N n−j α ∈ M. Therefore, M = Mj , as desired. To prove the “only if part”, note first that the matrix A can have at most one eigenvalue. Because, otherwise the corresponding eigenspaces of two distinct eigenvalues are not comparable with respect to inclusion, which is in contradiction with the hypothesis. So, with no loss of generality, we may assume that zero is the only eigenvalue of A, for A − λI and A share the same lattice of invariant subspaces. In other words, we may assume that A is nilpotent. We claim that there exists a nonzero vector α ∈ F n such that the set {An−1 α, . . . , Aα, α} is a basis for F n . Suppose to the contrary that the set {An−1 α, . . . , Aα, α} is not a basis for F n . Let M := hAn−1 α, . . . , Aα, αi. It follows that the subspace M is a nontrivial invariant subspace of A. Pick β ∈ F n \ M. It follows from the contradiction hypothesis that the subspace M0 := hAn−1 β, . . . , Aβ, βi is also a nontrivial invariant subspace of A. Obviously, the subspaces M and M0 are not comparable, a contradiction. Therefore, there exists a nonzero vector α ∈ F n such that the set {An−1 α, . . . , Aα, α} is a basis for F n . Since A is assumed to be nilpotent, we see that An = 0 but An−1 6= 0 because An−1 α 6= 0. This completes the proof. 

220

2. SOLUTIONS

5. Use A(m, n) to denote the average number of times that the players should play till one of them runs out of the coins. If the game is played once, A wins with the probability of 12 changing the number of the coins of A and B to m + 1 and n − 1, respectively; and, likewise, B wins with the probability of 21 changing the number of the coins of A and B to m − 1 and n + 1, respectively. This yields the following recurrence relation on m, n ∈ N. 1 1 A(m, n) = 1 + A(m − 1, n + 1) + A(m + 1, n − 1) 2 2 The boundary conditions are A(0, n) = A(m, 0) = 0. To solve this boundary recurrence equation, view A as a function of one variable, say n. It follows from the recurrence equation that on the line m + n = c, where c is a constant, we have A(m − 1, n + 1) + A(m + 1, n − 1) − 2A(m, n) = −2.

(∗)

Thus, A(m, n) must be a quadratic function. As A(0, n) = A(m, 0) = 0, we obtain A(m, n) = cmn. Substituting this into (∗), we obtain −2c = c(m − 1)(n + 1) + c(m + 1)(n − 1) − 2cmn = −2, implying c = 1. Consequently, A(m, n) = mn. Therefore, on average, the game is played as many as the product of the coins owned by A and B altogether.  6. It suffices to show that [0, 1] ⊆ C − C. To this end, let y ∈ [0, 1] be arbitrary. If 1−y 2 denotes the ternary expansion of

1−y 2 ,

=

0.y1 y2 . . .

where yi ∈ {0, 1, 2}, then we can write

1−y = 0.a1 a2 . . . + 0.b1 b2 . . . , 2 where ai , bi ∈ {0, 1} and yi = ai + bi for all i ∈ N. So we have 1−y

=

y

0.a01 a02 · · · + 0.b01 b02 · · · := a + b,

2(0.a1 a2 . . .) + 2(0.b1 b2 . . .),

implying =

where a0i = 2 − 2ai and b0i = 2bi , and hence a0i , b0i ∈ {0, 2} for all i ∈ N. Therefore, a, b ∈ C because there is no one in their ternary expansions. This finishes the proof. 

2.30. Thirtieth Competition 2.30.1. First Day. 1. First solution: It suffices to evaluate the following limit Z 2006 lim f (tx)dx. t→+∞

1385

Setting tx = u, we have Z

2006

lim

t→+∞

1385

1 f (tx)dx = lim t→+∞ t

Z

2006t

f (u)du. 1385t

2.30. THIRTIETH COMPETITION

221

As limt→+∞ t = +∞, using L’Hopital’s rule, we can write Z 2006   f (tx)dx = lim 2006f (2006t) − 1385f (1385t) = 621. lim t→+∞

t→+∞

1385

Thus, Z

2006

f (nx)dx = 621,

lim

n→+∞

1385

as desired. Second solution: We can write Z 2006 Z f (nx)dx = lim lim n→+∞

n→+∞

1385

2006


f (nx) − 1 dx + 621 = 621,

1385

R 2006


provided that limn→+∞ 1385 f (nx) − 1 dx = 0. It follows from the hypothesis ε that for given ε > 0, there is an N > 0 such that |f (x) − 1| < 621 whenever x > N . N Thus, for all n > 1385 and x ∈ [1385, 2006], we have f (nx) − 1 < ε. This yields 2006

Z 2006
f (nx) − 1 dx ≤ ε, f (nx) − 1 dx ≤ 1385 1385
R 2006 N for all n > 1385 . That is, limn→+∞ 1385 f (nx) − 1 dx = 0, settling the assertion.  Z

2. We need the following lemma. inα Lemma. Let α : n ∈ N} π 6∈ R \ Q, where π = 3.14159 . . .. Then, the set {e is dense in the unit circle. Proof. Use T to denote the unit circle. Set S = {einα : n ∈ N}. Plainly, the set S consists of distinct points, for α π 6∈ R \ Q, and it is a multiplicative semigroup. As T is compact, we see that S has a limit point in T. Thus, there is a strictly ink α increasing sequence (nk )+∞ exists. This k=1 of natural numbers such that limk e imk α implies that limk e = 1, where mk = nk+1 − nk . Now, let {eit : t ∈ (a, b)} be an arbitrary open arc of the unit circle, where a, b ∈ R and a < b. It follows that there is an ` ∈ N such that eimk α ∈ {eit : t ∈ (−δ, δ)} for all k ≥ `, where δ = b−a 2 . Consequently, {einm` α : n ∈ N} ∩ {eit : t ∈ (a, b)} 6= ∅. This shows that S is dense in T, which is what we want.  Remark. A proof similar to that of the above lemma shows that every additive subgroup of R is either dense or isolated, i.e., it has no limit points. A quick consequence of this is the following. Let α ∈ R \ Q. Then, the set m + nα : m, n ∈ Z is dense in R.

First, we claim that there is a sequence (kp )+∞ p=1 of natural numbers such that k

limp→+∞ aj p = 1 for all 1 ≤ j ≤ m. Proceed by induction on m. If m = 1, then limn→+∞ an1 = c. Firstly, c 6= 0 because |c| = limn→+∞ |a1 |n = 1. Secondly, by showing that a1 = 1, we prove the claim in this case. From c = limn→+∞ an1 = limn→+∞ an+1 = ca1 , we obtain c(1 − a1 ) = 0, which yields a1 = 1 because c 6= 0. 1 Assume that the assertion holds for m − 1. It follows from the induction hypothesis k` that there is a sequence (k` )+∞ `=1 such that lim`→+∞ aj = 1 for all 1 ≤ j ≤ m − 1.

222

2. SOLUTIONS

Without loss of generality, if necessary by passing to a subsequence of (k` )+∞ `=1 , we can assume that lim`→+∞ akj ` = 1 for all 1 ≤ j ≤ m − 1 and lim`→+∞ akm` = bm , where bm ∈ C and |bm | = 1. Set  A = (ak1 , . . . , akm−1 , akm ) : k ∈ N . Let A0 be the set of limit points of A. It is obvious that (1, . . . , 1, bkm ) ∈ A0 for all k ∈ N. As |bm | = 1, it is easily seen from the lemma above that there exists a subkr 0 0 sequence (kr )+∞ r=1 such that limr→+∞ bm = 1. Consequently, (1, . . . , 1, 1) ∈ (A ) ⊆ k k k p p p A0 . Thus, there is a subsequence (kp )+∞ p=1 such that limp→+∞ (a1 , . . . , am−1 , am ) = k

(1, . . . , 1, 1), and hence limp→+∞ aj p = 1 for all 1 ≤ j ≤ m, proving the claim. It thus follows that m m m X X X k +1 k lim aj p = c = lim aj p = 1 = m, p→+∞

yielding write

Pm

j=1

p→+∞

j=1

aj = m because m X j=1

k limp→+∞ aj p

Re(aj ) = Re

j=1

j=1

= 1 for all 1 ≤ j ≤ m. Now, we can

m X


aj = m,

j=1

Pm

yielding j=1 Re(aj ) = m. This implies Re(aj ) = 1, for Re(aj ) ≤ 1 for all 1 ≤ j ≤ m. As the aj ’s are on the unit circle, we see that aj = 1 for all 1 ≤ j ≤ m, which is what we want.  3. It is worth noting that the hypothesis that R is commutative is redundant. Consider the following five elements of R/J J, 1 + J, a + J, a2 + J, a3 + J. It follows from the hypothesis that
at least two of the above elements are equal. Thus, one of the following 10 = 52 cases will occur J = 1 + J, J = a + J, J = a2 + J, J = a3 + J, 1 + J = a + J, 1 + J = a2 + J, 1 + J = a3 + J, a + J = a2 + J, a + J = a3 + J, a2 + J = a3 + J. Consequently, at least one of the following elements of R belongs to J 1, a, a2 , a3 , a − 1, a2 − 1, a3 − 1 = a, a2 − a, a3 − a, a3 − a2 . In view of the identity a(a − 1)(a + 1) = 1, it is readily verified that the above elements are invertible elements of R, and hence J = R, as desired.  4. Since p is an odd prime, by Fermat’s Little Theorem, we have 2p−1 ≡ 1 (mod p). s Let s be the smallest natural number for which 2 ≡ 1 (mod p). It follows that s p − 1 = 2q, and hence s ∈ {1, 2, q, 2q}, for q is prime. To prove the assertion by contradiction, assume that s < 2q. If s = 1, then p 1, which is a contradiction. If s = 2, we obtain p = 3, yielding q = 1, which is a contradiction. Finally, if q
s = q, then the equation x2 ≡ 2p (mod p) is solvable for x in Z. Thus 2p = 1,
where ap , with a, p ∈ Z, denotes the Legendre symbol, which is defined to be ±1

2.30. THIRTIETH COMPETITION

223

depending on whether or not the equation x2 ≡ a (mod p) is solvable for x in Z. On the other hand,
q(p28−1)
q2 (q+1) 2q
2
q 2 = −1 = = −1 = −1, p p which is a contradiction. Note that the last equality in the above is implied by q ≡ 1 (mod 4). Therefore, we obtain a contradiction in any event. So the assertion follows.  5. Let X = X1 + · · · + X2m , where Xi ’s (1 ≤ i ≤ 2m) are independent random variables assuming the values zero and one with the probability of 21 . It is plain that m 2 µ = µX = m, σ = σX = . 2 √ Using the Chebyshev Inequality with λ = 2, we have √
1 P |X − m| ≥ m ≤ , 2 implying √
1 P |X − m| < m ≥ . 2 On the other hand, X  2m  1
2m √
1 ≤ P |X − m| < m = , 2 m+k 2 √ |k|< m

yielding X  2m  ≥ 22m−1 , m+k √

|k|< m

which is what we want.



6. Use p1 , . . . , pn to denote the members of the financial group. Suppose that the number of coins owned by person pi after doing business for j times is aij , where i = 1, . . . , n and j ∈ N. Without loss of generality, assume that the (j + 1)st business is done among p1 , p2 , p3 , and p4 . It follows that a3j + a4j − a1j − a2j + 2k > 0. Multiplying both sides of the above inequality by 2k, we easily conclude that (a3j + k)2 + (a4j + k)2 + (a1j − k)2 + (a2j − k)2 > a21j + a22j + a23j + a24j . Consequently, n X i=1

a2i(j+1) >

n X

a2ij .

i=1

On the other hand, the sum of the number of coins owned by p1 , . . . , pn is a fixed number. Therefore, a finite number of business can be done among these people, proving the assertion. 

224

2. SOLUTIONS

2.30.2. Second Day. 1. First solution: As X is separable, X has a countable base, say B = {Bi }+∞ i=1 , e.g., the family of all open balls with centers at the points of a countable dense subset of X and with positive rational radii is easily seen to be a countable base for X. Let D1 = {a ∈ X : limx→a f (x) < f (a)} and D2 = {a ∈ X : limx→a f (x) > f (a)}. It suffices to show that D1 ∪ D2 is at most countable. To this end, for a ∈ D1 (resp. a ∈ D2 ) choose an ra ∈ Q with limx→a f (x) < ra < f (a) (resp. limx→a f (x) > ra > f (a)) and Bia ∈ B with a ∈ Bia such that f (x) < ra (resp. f (x) > ra ) for all x ∈ Bia \ {a}. Define a function ι1 : D1 −→ Q × B (resp. ι2 : D2 −→ Q × B) by ι1 (a) = (ra , Bia ) (resp. ι2 (a) = (ra , Bia )). If (ra , Bia ) = (rb , Bib ) for some a, b ∈ X, then a = b, for otherwise f (x) < ra = rb (resp. f (x) > ra = rb ) whenever x ∈ Bia \ {a}. But Bia = Bib . Thus, f (b) < ra = rb < f (b) (resp. f (b) > ra = rb > f (b)), which is a contradiction. This proves that ι1 (resp. ι2 ) is one-to-one. Therefore, D1 and D2 are countable, and hence so is D1 ∪ D2 , which is what we want. Second solution: For a given f : X −→ R, use D and L to denote the the set of points at which f is discontinuous and has a limit, respectively. By showing that D ∩ L is at most countable, we prove the assertion. Define ω : X −→ [0, +∞) by 

 ω(x) = lim+ sup f Bδ (x) − inf f Bδ (x) , δ→0

where Bδ (x) = {y ∈ X : d(y, x) < δ} denotes the open ball with center at x S+∞ and radius δ > 0. Note that D = n=1 Dn , where Dn = {x ∈ X : ω(x) > n1 }. S+∞ Consequently, D ∩ L = n=1 (Dn ∩ L). It thus suffices to prove that Dn ∩ L is at most countable for all n ∈ N. To this end, for a given a ∈ Dn ∩ L, we have limx→a f (x) = la for some la ∈ R. It follows that there exists a δna > 0 such that 1 whenever x ∈ Bδna (a) \ {a}. This implies |f (x) − f (y)| < n1 for all |f (x) − la | < 2n x, y ∈ Bδna (a) \ {a}. Hence, ω(x) ≤ n1 for all x ∈ Bδna (a) \ {a}. In other words, x∈ / Dn for all x ∈ Bδna (a) \ {a}. This implies that Dn ∩ L is at most countable for all n ∈ N, for otherwise Dn ∩ L would have (uncountably many) limit points because X is separable. This obviously is in contradiction with x ∈ / Dn for all x ∈ Bδna (a) \ {a} and a ∈ Dn ∩ L, finishing the proof.  2. First solution: If M = +∞, the assertion is trivial. So, we may, without loss of generality, assume that M < +∞. Let D = {z ∈ C : |z| < 1} denote the unit disk (Rz)−a0 around zero. Define g : D −→ C by g(z) = fM +|a0 | . Plainly, g is analytic on D and g(0) = 0. By the Maximum Modulus Principle, |f (Rz)| ≤ sup|z|=R |f (z)| = M for all z ∈ D. It follows that g(z) ≤ f (Rz) + a0 ≤ M + a0 = 1, M + a0 M + a0 for all z ∈ D. Thus, by the Schwarz Lemma, |g(z)| ≤ |z| for all z ∈ D. Now, if f (z0 ) = 0 for some z0 ∈ C, then zR0 ∈ D because |z0 | < R. Consequently, |g( zR0 )| ≤ | zR0 |. But
a0 z z0
f R zR0 − a0 g = = ≤ 0 , R R M + a0 M + a0 yielding |z0 | ≥

R|a0 | M +|a0 | ,

finishing the proof.

2.30. THIRTIETH COMPETITION

225

Second solution: Again, we may, without loss of generality, assume that M < +∞. Let 0 < r < R be arbitrary. By the Cauchy Inequality, |an | ≤ Mr(r) for all n n ∈ N. In view of the Maximum Modulus Principle, we have M (r) ≤ M for all r < R, and hence we can write +∞ +∞ X X n M z z
n f (z) − a0 ≤ an z ≤ M (r) , = r r − z n=1

n=1

for all z ∈ C with |z| < R. On the other hand, M z r|a0 | < a0 ⇐⇒ |z| < . M + |a0 | r− z Consequently, f (z) 6= 0 whenever |z| < that f (z) 6= 0 whenever |z| <

R|a0 | M +|a0 | .

r|a0 | M +|a0 | .

Since r < R is arbitrary, we see

This proves the assertion.



3. Let H be the set of all elements of G whose orders are finite. Obviously, H is not empty because 1 ∈ H. Let x, y ∈ H be arbitrary. It follows that there is an n ∈ N such that xn = y n = 1. Consequently,
n (xy −1 )n G0 = (xy −1 G0 )n = (xG0 )(y −1 G0 ) (xG0 )n (y −1 G0 )n = (xn G0 )(y −n G0 ) = G0 ,
n where the equality (xG0 )(y −1 G0 ) = (xG0 )n (y −1 G0 )n follows from the hypothesis that GG0 is abelian. Thus, (xy −1 )n ∈ G0 , from which, we obtain an m ∈ N such that (xy −1 )nm = 1. That is, xy −1 ∈ H, and hence H ≤ G, as desired.  =

4. We prove the following, of which the assertion is a quick consequence. Let D be a division ring and F a subfield of its center. If A ∈ Mn (D) is algebraic over the subfield F and rank(A) = rank(A2 ), then there is an idempotent E in the F algebra generated by A which projects Dn onto the range of A along ker(A), i.e., Dn = im(A) ⊕ ker(A) and E(x + y) = x, where x ∈ im(A) and y ∈ ker(A). As is usual, the members of Mn (D) can be viewed as linear transformations acting on the left of Dn via the usual matrix multiplication. To show that Dn
= im(A) ⊕ ker(A), note that by the Rank-Nullity Theorem, we have dim im(A) +
dim ker(A) = dim(Dn ) = n. On the other hand,

dim im(A) + ker(A) = dim(im(A)) + dim(ker(A)) − dim im(A) ∩ ker(A) . Thus, to prove Dn = im(A) ⊕ ker(A), it suffices to show that im(A) ∩ ker(A) = {0}. To this end, let B := A|im(A) : im(A) −→ im(A). We can write
dim im(A2 ) = rank(A2 ) = rank(A)


= dim im(A) = dim ker(B) + dim im(B)

= dim im(A) ∩ ker(A) + dim im(A2 ) ,
yielding dim im(A) ∩ ker(A) = 0, whence im(A) ∩ ker(A) = {0}. For the rest, let f (A) = Am + am−1 Am−1 + · · · + a1 A + a0 I = 0, where m is minimal and a0 , a1 , ..., am−1 ∈ F . If a0 6= 0, then we have nothing to prove. If a0 = 0, then (Am−1 + am−1 Am−2 + · · · + a1 I)A = 0. We claim that a1 6= 0. If not, then (Am−2 + am−1 Am−3 + · · · + a2 I)A2 = 0. Now, since im(A) = im(A2 ) because rank(A) = rank(A2 ), it follows that (Am−2 + am−1 Am−3 + · · · + a2 I)A = 0,

226

2. SOLUTIONS

m−1 m−2 a contradiction. So a1 6= 0. Set G = a−1 + a−1 + · · · + a−1 1 A 1 am−1 A 1 a2 A + I n and note that D = ker(A) ⊕ im(A). Then E = I − G is the desired idempotent. Because for each x = Ax0 ∈ im(A) and y ∈ ker(A), we can write

E(x + y)

= Ex + Ey = =



−1 m−1 0 m−2 Ax0 − a−1 + a−1 + · · · + a−1 1 f (A)x − a1 A 1 am−1 A 1 a2 A y (Ax0 − 0) − 0 = Ax0 = x.

So the proof is complete.



5. It is plain that the sets of odd and even numbers, denoted by O and E, respectively, is a partition of the natural numbers such that neither O ⊕ O nor E ⊕ E contains any prime. To prove the uniqueness, suppose that N = A ∪ B and that neither A ⊕ A nor B ⊕ B contains any prime. Without loss of generality, assume that 1 ∈ A. By showing that A = O and B = E, we finish the proof. Proceed by induction. As 1 ∈ A and 1 + 2 = 3, 2 ∈ B. Assume that  1, 3, . . . , 2k − 1 ⊆ A, and 

2, 4, . . . , 2k ⊆ B.

It follows from Bertrand’s conjecture that for n = 2k + 1, there is a prime p such that 2k + 1 < p ≤ 4k + 2. We have 2k + 1 < p ≤ 4k + 2 because k ≥ 1. But p − (2k + 1) is an even number and 0 < p − (2k + 1) ≤ 2k. Hence, p − (2k + 1) ∈ B, whence 2k + 1 ∈ A. Likewise, from 0 < p − (2k + 2) ≤ 2k + 1, we see that 2k + 2 ∈ B. So the proof is complete by way of induction.  6. Let T denote the closure of T . It is plain that the distance between any two points of T is less than or equal to α. So, we may, with no loss of generality, assume that T is closed. Let x, y ∈ T be such that the distance between x and → to denote the shortest arc joining x and y in the clockwise y is maximal. Use − xy → then z ∈ T 0 . If z ∈ T direction. Construct T 0 from T as follows. If z ∈ T ∩ − xy, − → 0 0 0 and z ∈ / xy, then z ∈ T , where z denotes the antipodal point of z. In this case, we have z 0 ∈ / T , for otherwise the distance between z and z 0 is equal to one half → because otherwise the distance between z and x whereas 0 < α < 21 . Also, z 0 ∈ − xy, or z and y will be greater than the distance between x and y, which is impossible. → On the other hand, the sum of the lengths of the arcs whose union Thus, T 0 ⊆ − xy. is T is equal to the sum of the lengths of the arcs whose union is T 0 . This obviously proves the assertion.  2.31. Thirty First Competition

2.31.1. First Day. 1. (a) and (b) The answer is no. Color the circle by two colors, say black (b) and white (w), as shown in the figure. That is, (b) and (w) are, respectively, assigned to the upper and lower half-open semicircles. It is obvious that any equilateral triangle inscribed in the circle has a white vertex and two black

2.31. THIRTY FIRST COMPETITION

227

Figure 14

Figure 15

vertices or vise versa. Also, the hypotenuse of any right triangle inscribed in the circle is a diameter of the circle whose vertices are of the two colors. (c) The answer is yes. Proceed by contradiction. Assume that there is a coloring of the circle for which there is no monocolored isosceles triangle inscribed in the circle. First choose a monocolored chord, say with white ends, which is not a diameter. Use w1 w2 to denote the chord. The points on the circle that together with this chord form an isosceles triangle lie on the perpendicular bisector of w1 w2 , which is a diameter of the circle. By the contradiction hypothesis, the two ends of this diameter are both black. Use b1 b2 to denote this diameter. Likewise, the two ends of the diameter perpendicular to b1 b2 must be both white. Use w3 w4 to denote this diameter. Now consider the two chords w1 w3 and w2 w4 . It follows that the ends of the diameters perpendicular to w1 w3 and w2 w4 , respectively, must be all black. Use b3 b4 and b5 b6 to denote these diameters. It thus follows that the black-colored triangles b2 b4 b5 , b2 b3 b6 , b1 b4 b5 , and b1 b3 b6 are all isosceles. This is a contradiction, finishing the proof.  2. Using the Sequence Lemma, it is readily seen that A0 + B ⊆ (A + B)0 , A + B 0 ⊆ (A + B)0 , implying that (A0 +B)∪(A+B 0 ) ⊆ (A+B)0 . To see (A+B)0 ⊆ (A0 +B)∪(A+B 0 ), let x ∈ (A + B)0 be arbitrary. By the Sequence Lemma, there exists a sequence (an + bn )+∞ n=1 with distinct terms such that x = limn (an + bn ), where the sequences

228

2. SOLUTIONS

+∞ (an )+∞ n=1 and (bn )n=1 are in A and B, respectively. As A is bounded, if necessary, by passing to a subsequence, we may assume that there is an a ∈ A
:= A ∪ A0 such that a = limn an . It follows that limn bn = limn (an + bn ) − an = x − a, and hence b := x − a ∈ B because B is closed. There are two cases to consider. If a ∈ A0 , then x = a + b ∈ (A0 + B) ⊆ (A0 + B) ∪ (A + B 0 ), which is what we want. If a ∈ A \ A0 , then, if necessary by discarding a finite number of indices, we may assume that an = a for all n ∈ N. This implies that bn ’s (n ∈ N) are all distinct. Moreover, limn bn = x − a ∈ B 0 . Thus, x = a + (x − a) ∈ (A + B 0 ) ⊆ (A0 + B) ∪ (A + B 0 ), as desired. So, in any case, x ∈ (A0 + B) ∪ (A + B 0 ). Therefore, (A + B)0 ⊆ (A0 + B) ∪ (A + B 0 ). This completes the proof. 

3. Let H1 , . . . , Hn be all of the subgroups each of which has index 2 in G. It thus follows that each Hi (i = 1, . . . , n) is normal in G, and hence so is H1 ∩ · · · ∩ Hn in G. Also, H1 ∩ · · · ∩ Hn is of finite index in G because so is every Hi (i = 1, . . . , n) G . We claim that A is the desired group. Suppose that B is in G. Set A := H1 ∩···∩H n K a subgroup of A of index 2. It follows that B := H1 ∩···∩H , where H1 ∩ · · · ∩ Hn ⊆ n K E G. By the Third Isomorphism Theorem for groups , we have A = B

G H1 ∩···∩Hn K H1 ∩···∩Hn

G ∼ = . K

Thus, the index of K in G is 2. Therefore, K = Hi for some 1 ≤ i ≤ n. Consequently, A has exactly n subgroup of index 2 which are H1 H2 Hn , ,..., . H1 ∩ · · · ∩ Hn H1 ∩ · · · ∩ Hn H1 ∩ · · · ∩ Hn (Note that as Hi ’s are mutually disjoint, so are the aforementioned groups.) It remains to show that A is abelian. For each i = 1, . . . , n, the group HGi is abelian because Hi has index 2 in G. Thus, G0 ⊆ Hi for all 1 ≤ i ≤ n, from which we obtain G0 ⊆ H1 ∩ · · · ∩ Hn . Therefore, A is abelian, as desired.  4. No, one cannot find such two dice. To see this, proceed by contradiction. Consider two such dice A and B. For each i = 1, . . . , 6, let Ai and Bi , respectively, denote the probability of getting i for the two dice A and B. For each j = 2, . . . , 12, denote by Pj , the probability of getting a sum of j. ConPj sequently, Pj = k=1 Ak Bj−k . We have P2 = A1 B1 , P12 = A6 B6 , and P7 = A1 B6 + A2 B5 + A3 B4 + A4 B3 + A5 B2 + A6 B1 . So, we can write p p P7 ≥ A1 B6 + A6 B1 ≥ 2 A1 B6 A6 B1 = 2 A1 B1 A6 B6 r p 2 2 4 = 2 P2 P12 ≥ 2 . = , 33 33 33
2 4 which is impossible because Pi ∈ 33 , 33 for all 2 ≤ i ≤ 12.  5. Note that the set of all balls whose centers are rational points of R2 and whose radii are positive rationals is countable. Use {B1 , B2 , B3 , . . .} to denote this set. +∞ Choose a set {xi }+∞ i=1 such that xi ∈ Bi for all i ∈ N. As the set D := {xi }i=1 is evidently dense in the plane, it suffices to show that D can be so chosen that no three points of D are collinear. To this end, use induction on n to choose xn in such a way that for all i < j < n, the points xi , xj , xn are not collinear. This obviously can be done because no ball is a union of a finite number of lines. 

2.31. THIRTY FIRST COMPETITION

229

6. Let p(x) = xn + a1 xn−1 + · · · + an be the characteristic polynomial of A and let λ1 , . . . , λn be the eigenvalues of A, which are the roots of p(x) = 0. As, for each i = 1, . . . , n, the polynomial p(x) is divisible by x − λi , we can write p(x) = xn−1 + (λi + a1 )xn−2 + · · · + (λn−1 + a1 λn−2 + · · · + an−2 λi + an−1 ). i i x − λi We note that p0 (x) = p0 (x)

p(x) x−λ1

+ ··· +

p(x) x−λn .

It thus follows that

  = nxn−1 + tr(A) + na1 xn−2 + · · ·   + tr(An−1 ) + a1 tr(An−2 ) + · · · + an−2 tr(A) + nan−1 .

On the other hand, p0 (x) = nxn−1 + (n − 1)a1 xn−2 + · · · + an−1 . Thus, a1 a1 tr(A) + 2a2

= −tr(A) = −tr(A2 )

a1 tr(A2 ) + a2 tr(A) + 3a3 = −tr(A3 ) .. . . = .. a1 tr(An−2 ) + · · · + an−2 tr(A) + (n − 1)an−1 = −tr(An−1 ) a1 tr(An−1 ) + · · · + an−1 tr(A) + nan Note that the last equality in the above is obtained It now follows from Cramer’s Rule that  1 0 0 ···  tr(A) 2 0 ···   tr(A2 ) tr(A) 3 0  det  .. .. . . .. .  . . .  n−2 n−3  tr(A ) tr(A ) · · · tr(A) tr(An−1 ) tr(An−2 ) · · · ···  an = 1 0 0  tr(A) 2 0   tr(A2 ) tr(A) 3  det  .. .. . ..  . .   tr(An−2 ) tr(An−3 ) ··· tr(An−1 ) tr(An−2 ) tr(An−3 )

= −tr(An )

from p(λ1 ) + · · · + p(λn ) = 0. ··· ··· ··· .. .

−tr(A) −tr(A2 ) −tr(A3 ) .. .



       n − 1 −tr(An−1 )  tr(A) −tr(An )  . ··· ··· 0 ··· ··· 0   0 ··· 0   ..  .. .. . . .   tr(A) n − 1 0  ··· tr(A) n

Note that the denominator of the fraction above is equal to n!. Thus, in view of an = (−1)n det A, the assertion follows by substitution. 

2.31.2. Second Day. 1. For the first part, just note that


n−1 n−1 n−1 +1 + + 2 + ··· + + n = n2 . 2 2 2 For the rest, suppose that m, k are positive integers such that (m + 1) + (m + 2) + · · · + (m + 12) = k 2 . It follows that 12m + 78 = k 2 . Consequently, n is even, and hence k = 2t for some t ∈ N. By substitution, we obtain 2t2 − 6m = 39, which is impossible because the left hand side is an even integer. 

230

2. SOLUTIONS

2. Performing the substitution u = xα and y = y β and depending on the signs of α and β, we need to investigate the limit of u2 v 2 u3 + v 3 as u and v tend to 0 or +∞. There are four cases to consider. (a) α > 0, β > 0. 2 2 v In this case, we need to investigate limu,v→0+ uu3 +v But by the Sandwich 3. Theorem, this limit exists and equals zero because √  u
3 u2 v 2 v
32 −1 √ uv 2 = , 0≤ 3 + uv ≤ u + v3 v u 2 for all u, v > 0. (b) α > 0, β < 0. 2 2 v In this case, we need to investigate limu→0+ ,v→+∞ uu3 +v 3 . Again, by the Sandwich Theorem, the limit exists in this case and equals zero because 0≤

u2 v 2 = u3 + v 3

u2 u2 , ≤ v +v

u3 v2

for all u, v > 0. (c) α < 0, β > 0. 2 2 v Just as in (b), one can readily see that limu→+∞,v→0+ uu3 +v 3 = 0. (d) α < 0, β < 0. By showing that there is a path on which the limit is infinity, we conclude that the limit does not exist in this case. To this end, assuming that u = v, we have lim

u,v→+∞,u=v

u2 v 2 u = +∞, = lim 3 3 u→+∞ 2 u +v

as desired. Alternatively, the cases (a), (b), and (c) above could be proven using the following inequalities u2 v 2 u2 v 2
0≤ 3 ≤ min , ≤ max(u, v), u + v3 v u for all u, v > 0.  3. Let x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ). Define the function f : An → An by f (z) = z 0 , where z = (z1 , . . . , zn ), z 0 = (z10 , . . . , zn0 ), and   xj zj = xj 6= yj , yj zj = yj 6= xj , zj0 =  zj otherwise, for all 1 ≤ j ≤ n. It is readily seen that d(z, x) = d(f (z), y) = d(z 0 , y), d(z, y) = d(f (z), x) = d(z 0 , x). Therefore, f is a one-to-one correspondence between the sets C and D, which is what we want.  4. First, we prove that “if F is a field, then the ring F [x] has infinitely many maximal ideals.” To see this, noting that the ideal generated by an irreducible monic polynomial is a maximal ideal and that such ideals are distinct whenever

2.31. THIRTY FIRST COMPETITION

231

the generating polynomials are, it suffices to show that there are infinitely many irreducible monic polynomials in F [x]. To this end, motivated by Euclid’s proof of the infinitude of primes, suppose by contradiction that for some n ∈ N, there are only n irreducible monic polynomials in F [x], say f1 , . . . , fn . But the polynomial f1 · · · fn + 1 must have an irreducible monic divisor. This implies that some fi (1 ≤ i ≤ n) divides f1 · · · fn + 1. Consequently, fi must divide 1, which is a contradiction. Thus, there are infinitely many irreducible monic polynomials in F [x]. Now, to prove the assertion, let M be a maximal ideal of R. By the preceding R [x] has infinitely many maximal ideals. Since the mapping paragraph, the ring M R φ : R[x] → M [x], defined by f (an xn + · · · + a1 x + a0 ) = (an + M )xn + · · · + (a1 + M )x + (a0 + M ) is an epimorphism, i.e., a surjective homomorphism, of rings, the ring R[x] has has infinitely many maximal ideals as well. This is because for R [x], φ−1 (I) is also a maximal ideal of R[x] and that every maximal ideal I of M −1 −1 φ (I) 6= φ (J) whenever I 6= J. This completes the proof.  5. First solution: The hypothesis can be rephrased where    x1 0 ±1 · · · ±1    . . . . .. .. ..   ..  ±1   , X = A=   .  . .. ..  ..  .. . ±1  . ±1 · · · ±1 0 x2n 2n×2n

as follows. We have AX = 0, 



    

  ,0 =    2n×1

0 .. . .. . 0

     

.

2n×1

That is, the diagonal entries of A are all zero and the off diagonal entries of A are all either 1 or −1. It suffices to prove that A is invertible. Since the determinant 2

of A is an integer, we prove the assertion by showing that det A ≡ 1. We can write     0 1 ··· 1 0 ±1 · · · ±1   .  ..   1 . . . . . . ..   ±1 . . . . . . 2 .   ≡  det det A = det     . .  . .. .. .. ... 1   ..  .. . ±1  . 1 ··· 1 0 ±1 · · · ±1 0 = the number of derangements of 2n elements 1 1 1 1  = 2n! − + − · · · + (−1)2n 0! 1! 2! (2n)! 2

≡ 1. Recall that a derangement on n elements is a permutation on n elements having no fixed points. Also, a standard argument employing the inclusion-exclusion principle shows that the number of derangements on n elements is equal to n!

n X (−1)k k=0

k!

.

Second solution: We prove the following, of which the assertion is a quick consequence. Let G be an abelian torsion free group, i.e., the identity is the only element of the group having finite order. Let n, k ∈ N with n > k ≥ 2 and xi ’s (1 ≤ i ≤ n)

232

2. SOLUTIONS

be elements of G such that removing any of them the remaining ones can be partitioned into k subsets with equal sums. Prove that xi ’s are all equal. Therefore, either xi ’s are all zero or they are all equal to a nonzero element, in which case k

n ≡ 1. Since the subgroup generated by xi ’s (1 ≤ i ≤ n) is finitely generated, in view of the Fundamental Theorem of Finitely Generated Abelian Groups, we may, without loss of generality, assume that G = Z. Also, if necessary, by adding an appropriate positive integer to all xi ’s (1 ≤ i ≤ n), we may assume that xi ∈ N k Pn for each i = 1, . . . , n. First from the hypothesis, we easily obtain xj ≡ i=1 xi for each j = 1, . . . , n. Next, write xi ’s in the basis k. Plainly, there is an N ∈ N such that for all 1 ≤ j ≤ n, we can write xj =

N X

aij k i ,

i=0 k Pn where aij ∈ N with 0 ≤ aij < k and 0 ≤ i ≤ N . Note that xj ≡ i=1 xi for each j = 1, . . . , n obtains a0j = a01 for each j = 1, . . . , n. We use induction on N to prove that that xi ’s are all equal. If N = 0, the assertion is easy because in this case xj = a0j = a01 = x1 for all 1 ≤ j ≤ n, as desired. Assuming that the assertion holds for N , we prove it for N + 1. To this end, noting that a0j = a01 for each j = 1, . . . , n, define N +1 X xj − a01 = aij k i−1 , yj := k i=1

where 1 ≤ j ≤ n. It readily follows from the hypothesis that removing any of the yi ’s (1 ≤ i ≤ n) the remaining ones can be partitioned into k subsets with equal sums. Thus, the induction hypothesis applies to yi ’s (1 ≤ i ≤ n), and hence yi ’s are all equal. Therefore, so are xi ’s (1 ≤ i ≤ n), which is what we want.  6. (a) Let A = (t1 q1 , 1 − t1 ) and B = (t2 q2 , 1 − t2 ), where t1 , t2 ∈ (0, 1] and q1 , q2 ∈ Q are distinct. Let r be an irrational number between q1 , q2 ∈ Q. It follows that the line ` joining M and (r, 0) partitions the plane into the line ` and two open half-planes, say P1 and P2 . It is obvious that T \ {M } ⊂ P1 ∪ P2 and that {A, B} 6⊂ P1 whereas {A, B} ⊂ T \ {M } ⊂ P1 ∪ P2 . It thus follows that any continuous path from A to B in the set T must intersect the line ` at the point M because every other point of the line ` does not belong to the set T . This proves the assertion. (b) Let f : T → T be a continuous function and f (M ) = M 0 = (t0 q0 , 1 − t0 ) for some t0 ∈ [0, 1] and q0 ∈ Q. If t0 = 0, there is nothing to prove, for the point M would be a fixed point of f . If not, define the function g : [0, 1] → T by g(t) = f (tt0 q0 , 1 − tt0 ). Plainly, g is continuous. Define the set D as follows n o n o D = (tt0 q0 , 1 − tt0 ) ∈ T : t ∈ [0, 1] = t(t0 q0 , 1 − t0 ) + (1 − t)(0, 1) : t ∈ [0, 1] . It is plain that D is the line segment joining the points M = (0, 1) and M 0 = (t0 q0 , 1 − t0 ). Note that D is homeomorphic to the compact interval [0, 1]. We prove the assertion by showing that f has a fixed point in D. If f takes no point of D to the point M , the assertion is evident. That is because it would then follow from (a) that f (D) is contained in D, and hence f has a fixed point in D, for D

2.31. THIRTY FIRST COMPETITION

233

is homeomorphic to the compact interval [0, 1]. So assume that there is a t ∈ [0, 1] such that g(t) = M . Define n o tm = inf t ∈ [0, 1] : g(t) = M . Since g is continuous and g(0) = f (M ) 6= M , we see that tm > 0 and moreover g(tm ) = M . Also, it follows from (a) that the image of [0, tm ] under g is contained in D. Use π2 : R2 → R to denote the projection on the y-axis. We can write   π2 g(0) = 1 − t0 < 1 − 0t0 = 1 and

  π2 g(tm ) = π2 (M ) = 1 > 1 − tm t0 .   Consequently, there is a t1 ∈ (0, tm ) such that π2 g(t1 ) = 1−t1 t0 . Since g(t1 ) ∈ D, we obtain f (t1 t0 q0 , 1 − t1 t0 ) = (t1 t0 q0 , 1 − t1 t0 ), which means f has a fixed point, as desired. 

CHAPTER 3

Problem Index First Competition Analysis. Problem 1. (routine) Real analysis. Derivatives. Baire functions. Problem 2. Matrix analysis. Problem 3. Real analysis. Limits. Continuity. Algebra. Problem 1. (routine) Ring theory. Commutative rings with identity. Prime ideals. Maximal ideals. Problem 2. (routine) Group theory. Abelian groups. Problem 3. Baby set theory. Isomorphic sets. General. Problem 1. Problem 2. Problem 3. Problem 4.

(routine) (routine) (routine) (routine)

Triangles. Decimal expansion. Complex numbers. Real numbers. Inequalities.

Differential Equations. Problem 1. (routine) Second order differential equations. Probability and Statistics. Problem 1. (routine) Independent random variables. Topology. Problem 1. (routine) Point-set topology of real numbers. (Ir)rational numbers. Second Competition Analysis. Problem 1. Real analysis. Continuous functions. Limit of sequences. Problem 2. (routine) Real analysis. Integral norm of continuous functions. Problem 3. Point-set topology of real numbers. Cantor set. Algebra. Problem 1. (routine) Finite fields. Problem 2. (routine) Ring theory. (Right/Left) Quasi-regular elements. Problem 3. (routine) Homomorphism of rings. Polynomial rings in one variable. General. Problem 1. (routine) Arithmetic. Congruences. Problem 2. (routine) Real numbers. Inequalities. 235

236

3. PROBLEM INDEX

Problem 3. Real analysis. Second order derivatives. Mechanics. Particle acceleration. Problem 4. (routine) Counting. The product rule of combinatorics. Probability and Statistics. Problem 1. (routine) Covariance of random variables. Topology. Problem 1. (routine) Point-set topology. Connectedness. Differential Equations. Problem 1. (routine) First order differential equations. Third Competition Analysis. Problem 1. Real analysis. Derivatives. Problem 2. (routine) Real analysis. Continuous functions shifting forward sequences of real numbers. Problem 3(a). Roots of complex polynomials. Problem 3(b). Continuity of the roots of complex polynomials. Problem 3(c). Continuous algebraic functions. Holder functions. Algebra. Problem 1. (routine) Modules. Dimension of vector spaces. Problem 2. (routine) Group theory. Homomorphisms characterizing abelian groups. Problem 3. (routine) Semigroups. Groups. General. Problem 1. (routine) Real functions of two variables. Problem 2. (routine) Binomial coefficients. Problem 3. Plane geometry. Fourth Competition Analysis. Problem 1. Real analysis. Continuity. Injectivity. Problem 2. Sequences of nonnegative real numbers. Algebra. Problem 1. of ideal. Prime Problem 2. formations. Problem 3.

(routine) Ring theory. Commutative rings with identity. Radicals ideals. (routine) Linear algebra. Center of the algebra of all linear transRing theory. Artinian integral domains are fields.

General. Problem 1. Integral calculus. Problem 2. Conditional probability. Fifth Competition Analysis.

EIGHTH COMPETITION

237

Problem 1. Real analysis. Composition of Riemann integrable functions. Problem 2. Real analysis. Continuous functions not assuming any value more than twice. Problem 3. Point-set topology of real numbers. Congestions points. Algebra. Problem 1. (routine) Linear algebra. Eigenvalues of certain linear transformations. Problem 2. Group theory. Groups having two elements satisfying certain relations. Problem 3. Ring theory. Rings whose elements satisfy certain polynomial equations. General. Problem 1. Number theory. Divisibility. Problem 2. Roots of complex polynomials and those of their derivatives. Closed convex hulls. Problem 3. Plane geometry. Conic sections. Optics. Sixth Competition Analysis. Problem 1. (routine) Analytic geometry. Rational points. Problem 2. (routine) Metric spaces. Uniform convergence of sequences of continuous functions. Problem 3. Real analysis. Limits of improper integrals. Algebra. Problem 1. Group theory. Subgroups. Problem 2. Matrices. Problem 3. Ring theory. Rings. General. Problem 1. Integer numbers. Problem 2. Integer matrices. Problem 3. (routine) Plane geometry. Seventh Competition Analysis. Problem 1. Point set topology of real numbers. Problem 2. Real analysis. Continuity. Problem 3. Limits of sequences of of real numbers. Algebra. Problem 1. Ring theory. Nilpotent elements. Idempotent elements. Problem 2. Group theory. Normal subgroups. Problem 3. Linear algebra. Linear transformations. Eighth Competition Analysis. Problem 1. (routine) Real analysis. Continuous extension.

238

3. PROBLEM INDEX

Problem 2. Convergence of series of real numbers. Problem 3. (routine) Riemann-Stieltjes integral. Convergence of sequences of Riemann-Stieltjes integrals. Algebra. Problem 1. (routine) Real matrices. The general linear group of order 2. Problem 2. Ring Theory. Infinite integral domains. Problem 3. (routine) Real matrices. General. Problem 1. (routine) Minimum without using derivative. Problem 2. (routine) Arithmetic. Problem 3. Probability. Average function value. Ninth Competition Analysis. Problem 1. Problem 2. Problem 3. Problem 4. Problem 5. curves.

(routine) Point-set topology of real numbers. Equivalence relation. Real analysis. Continuity of inverse functions. (routine) Real analysis. Increasing continuous function. Real analysis. Riemann integrable function. Real analysis. Continuity. Injectivity. Surjectivity. Square filling

Algebra. Problem 1. Group theory. Finite groups. Subgroups of prime indexes. Normal subgroups. Problem 2. Field of integers mod 3, Z3 . Polynomial ring with coefficients in Z3 . Finite fields. Problem 3. (routine) Polynomial ring with coefficients in Z7 . Greatest common divisor. Problem 4. Field theory. Quadratic extensions. Problem 5. Linear algebra. Linear transformations. General. Problem 1. Problem 2. Problem 3. Problem 4. Problem 5. Problem 6. Problem 7.

Determinants. (routine) Algebraic equations. Real analysis. Continuous functions. Integrals. Diophantine equations. Number of solutions. Puzzles. Probability. Determinants. Tenth Competition

Analysis. Problem 1. Real analysis. Derivatives. Limits. Problem 2. Real analysis. Higher derivatives. Definite integrals. Problem 3. (routine) Real analysis. Limits of definite integrals. Convergence of function sequences. Algebra.

TWELFTH COMPETITION

Problem groups. Problem Problem Problem

239

1. Group theory. Additive group of real numbers. Maximal sub2. (routine) Ring theory. Ideals. 3. Field theory. Field extensions. 4. (routine) Matrix theory over general fields. Nilpotent matrices.

General. Problem 1. Probability. Random variables. Problem 2. Point-set topology of the plane. Problem 3. (routine) Differential and integral calculus. Differential equations. Integral equations. Problem 4. (routine) Arithmetic. Divisibility. Eleventh Competition Analysis. Problem 1. Real analysis. Derivatives. Problem 2. Real analysis. Convergence of function sequences. Functional equations. Uniform convergence. Problem 3. Real analysis. Higher order derivatives. Convergence of series. Algebra. Problem 1. Ring theory. Ideals. (Right) identity elements. Problem 2. Field theory. Algebraic numbers. Finite field extensions. Problem 3. Group theory. Finite groups. General. Problem 1. Probability. Determinants. Problem 2. Real numbers. Algebraic equations. Problem 3. Baby set theory. Group theory. Twelfth Competition Analysis. Problem 1. Real analysis. Continuity. Injectivity. Surjectivity. Problem 2. Real analysis. Uniform convergence of series. Problem 3. (routine) Integral calculus. Definite integrals. Algebra. Problem 1. Problem 2. Problem 3. Problem 4. Problem 5. Problem 6.

Group theory. Finite groups. Sylow theory. Group theory. Finite abelian groups. Homomorphisms. Ring theory. Ring theory. Principal ideals. Matrix theory over general fields. Determinant. Trace. Algebraic numbers.

General. Problem 1. Plane Geometry. Problem 2. Combinatorics. Pigeonhole principal. Problem 3. Binomial coefficients.

240

3. PROBLEM INDEX

Thirteenth Competition Analysis. Problem 1. Real analysis. Riemann integrable functions. Additive functions. Problem 2. Convergence of series of real numbers. Problem 3. Real analysis. Convergence of function sequences. Definite integrals. Algebra. Problem 1. Problem 2. Problem 3. Problem 4. Problem 5.

Ring theory. Integral domains. Ring theory. Right (Left) ideals. Group theory. Group theory. Permutations. (routine) Matrix theory. Eigenvectors. Eigenvalues.

General. Problem 1. (routine) Plane geometry. Problem 2. Rational polynomials. Problem 3. Space geometry. Fourteenth Competition Analysis. Problem 1. Real analysis. Differentiability. Problem 2. Real analysis. Uniform continuity. Limits. Problem 3. Real analysis. Continuous functions. Algebra. Problem 1. Problem 2. Problem 3. Problem 4. ducibility. Problem 5.

Group theory. Finite p-groups. Derived subgroups. Group theory. Finite groups. Normal subgroups. Center of groups. Ring theory. Greatest commons divisors. Least common multiples. (routine) Ring theory. Polynomial rings of two variables. IrreFinite-dimensional vector spaces. Direct sums.

General. Problem 1. Combinatorics. Problem 2. Combinatorics. Counting. Problem 3. Real numbers. Fifteenth Competition Analysis. Problem 1. Real analysis. Convexity. Problem 2. (routine) Real analysis. Continuous functions. Uniform convergence of function sequences. Problem 3. Real analysis. Real functions. Algebra. Problem 1. Group theory. Finite groups. Sylow p-subgroups. Problem 2. Ring theory. (Right/Left) ideals. Prime ideals. Nilpotent elements. Problem 3. (routine, as stated, but its nontrivial counterpart is nonroutine!)

NINETEENTH COMPETITION

241

General. Problem 1. Elementary number theory. Problem 2. Integral calculus. Problem 3. Combinatorics. Probability. Binomial coefficients. Sixteenth Competition Analysis. Problem 1. (routine) Real analysis. Uniform convergence. Problem 2. Real analysis. Problem 3. Real analysis. Uniform continuity. Algebra. Problem 1. Group theory. Finite abelian groups. Inner automorphisms. Problem 2. Ring theory. Nilpotent elements. Idempotent elements. Problem 3. Matrix theory. Number theory. Seventeenth Competition Analysis. Problem 1. Problem 2. visibility. Problem 3. Problem 4.

Real analysis. Derivatives. (routine) Differential and integral calculus. Real polynomials. DiReal continuous functions on metric spaces. Real analysis. Continuity.

Algebra. Problem 1. Group theory. Finite groups. Problem 2. Ring theory. Polynomial rings in two variables with coefficients from matrix rings over fields. Problem 3. Linear algebra. Linear transformations. Rank. Nullity. Eighteenth Competition Analysis. Problem 1. Real analysis. Power series. Problem 2. Real analysis. Periodic functions. Limits of definite integrals. Problem 3. Real analysis. Open maps. Algebra. Problem 1. Group theory. Finite groups. Normal subgroups. Problem 2. Ring theory. Ascending chains of right ideals. Problem 3. Vector spaces. Direct sums. Nineteenth Competition Analysis. Problem 1. (routine) Real analysis. Differential and integral calculus. Zeros of continuous functions. Problem 2. Real analysis. Limits. Problem 3. Real analysis. Continuous nowhere differentiable functions. Algebra. Problem 1. Group theory. Normal subgroups. Posets.

242

3. PROBLEM INDEX

Problem 2. Ring theory. Minimal left ideals. Ideals. Problem 3. (routine) Matrix theory. Rank. Problem 4. Matrix theory. Rank. Determinant. Trace. Twentieth Competition Problem Problem Problem Problem Problem Problem grals.

1. 2. 3. 4. 5. 6.

Vector spaces. Complex numbers. Linear independence. Real analysis. Sequences of real numbers. Limits. Group theory. Automorphism groups. Solvability. Real analysis. Functional equations. Uniform continuity. Ring theory. Commutative rings with identity. Zero divisors. Real analysis. Periodic functions. Convergence of improper inteTwenty First Competition

Analysis. Problem 1. Real analysis. Limits. Problem 2. Real analysis. Continuous functions. Limits. Integrals. Problem 3. Real analysis. One-sided limits. Riemann integrable functions. Algebra. Problem 1. Ring theory. Small ideals. Problem 2. Group theory. Center of groups. Derived subgroups. Problem 3. Matrix theory. Trace. Symmetric matrices. Characteristic polynomials. Twenty Second Competition Analysis. Problem 1. Real analysis. Derivatives. Problem 2. Real continuous functions on metric spaces. Series. Uniform convergence. Problem 3. Complex analysis. Sequences of analytic functions. Integrals. Algebra. Problem 1. Group theory. Group index. Problem 2. Ring theory. Commutative rings. Maximal ideals. Problem 3. Linear algebra. Characteristic polynomials. Rank. General. This section has forty routine multiple choice problems. Twenty Third Competition Analysis. Problem 1. Real analysis. Continuous bounded functions. Uniform norm. Compactness. Uniform continuity. Problem 2. Real analysis. Trigonometric series. Problem 3. Real analysis. Convergence of function sequences. Differentiability. Algebra. Problem 1. Group theory. Inner automorphism groups. Nonabelian groups. Problem 2. Ring theory. (Right) Ideals. Division rings.

TWENTY SIXTH COMPETITION

243

Problem 3. (routine) Matrix theory. Zero trace. Twenty Fourth Competition First Day. Problem 1. Real analysis. Derivatives. Problem 2. Real analysis. Real valued harmonic functions. Problem 3. Number theory. Primes. Divisibility. Congruences. Problem 4. (routine) Group theory. Infinite abelian groups. Problem 5. Combinatorics. Counting. Problem 6. Probability. Conditional probability. Second Day. Problem 1. Euclidean spaces. Metric spaces. Open balls. Problem 2. (routine) Real analysis. Differential and integral calculus. Continuous functions. Problem 3. Matrix theory. Rank inequalities. Problem 4. (routine) Ring theory. (Left) ideals. Problem 5. Combinatorics. Problem 6. Combinatorics. Game theory. Twenty Fifth Competition First Day. Problem 1. Group theory. Finite groups. Group centers. Problem 2. Linear algebra. Idempotents. Image. Kernel. Problem 3. Real analysis. Higher derivatives. Problem 4. Real analysis. Derivatives. Problem 5. Probability. Problem 6. Combinatorics. Triangles with integer sides. Partitions of numbers. Second Problem Problem Problem Problem Problem Problem

Day. 1. Ring theory. Integral domains. Multiplicative group of units. 2. Number theory. Prime numbers. Divisibility. Congruences. 3. Continuous functions on metric spaces. 4. Functions of one complex variable. 5. Combinatorics. Matrices with {−1, 1} entries. 6. Finite sets. Relations. Twenty Sixth Competition

First Day. Problem 1. Real analysis. Continuous functions. Derivatives. Problem 2. Complex analysis. Analytic functions. Problem 3. (routine) Number theory. Primes. Congruences. Problem 4. Ring theory. Commutative rings with identity. Maximal ideals. Problem 5. Combinatorics. Problem 6. Combinatorics. Finite sets. Second Day. Problem 1. Real analysis. Continuous functions. Problem 2. Real analysis. Continuous functions.

244

3. PROBLEM INDEX

Problem Problem Problem Problem

3. 4. 5. 6.

Group theory. Maximal subgroups. Matrix theory. Eigenvalues. Eigenvectors. Combinatorics. Probability. Discrete mathematics. Twenty Seventh Competition

First Day. Problem 1. Matrix analysis. Problem 2. Complex analysis. Analytic functions. Problem 3. Sequences of real numbers. Recurrence relations. Problem 4. Set theory. Problem 5. Matrices. Product of matrices. Problem 6. Group theory. Normal subgroups. Second Problem Problem Problem Problem Problem Problem

Day. 1. Real analysis. Derivatives. 2. Point set topology of real numbers. 3. Analytic geometry. 4. Combinatorics. Finite sets. 5. Number theory. Congruences. Divisibility. 6. Ring theory. Boolean rings. Isomorphisms. Twenty Eighth Competition

First Day. Problem 1. Complex analysis. Problem 2. Euclidean spaces. Continuous functions. Problem 3. Number theory. Arithmetic functions. Euler’s totient function. Problem 4. Group theory. Symmetric groups. Group center. Problem 5. Vector spaces over Z2 . Problem 6. Posets. Second Problem Problem Problem Problem Problem Problem

Day. 1. (routine) Integral calculus. 2. Real analysis. Continuous functions. 3. Complex matrices. Orthogonal matrices. Determinants. 4. Ring theory. Unital rings whose elements satisfy certain relations. 5. Probability. Conditional probability. 6. Combinatorics. Set theory. Twenty Ninth Competition

First Day. Problem 1. (routine) Integral calculus. Problem 2. Rational numbers. Series of rational numbers. Problem 3. (routine) Number theory. Divisibility. Problem 4. Ring theory. Nilpotent ideals. Right ideals. Problem 5. (routine) Baby set theory. Problem 6. Combinatorics. Second Day.

THIRTY FIRST COMPETITION

245

Problem 1. Complex analysis. Analytic functions. Problem 2. Connected metric spaces. Compact metric spaces. Problem 3. Group theory. Normalizer subgroup. Centralizer subgroup. Automorphism group. Cyclic groups. Derived subgroups. Group center. Problem 4. Matrix theory. Triangular matrices. Invariant subspaces. Nilpotent matrices. Problem 5. Probability. Expectation values. Problem 6. Point-set topology of real numbers. Cantor set. Thirtieth Competition First Day. Problem 1. (routine) Integral calculus. Problem 2. Limits of sequences of complex numbers. Problem 3. Ring theory. Problem 4. Number theory. Primes. Congruences. Divisibility. Problem 5. Binomial coefficients. Problem 6. Natural numbers. Second Problem Problem Problem Problem Problem Problem

Day. 1. Real functions on separable metric spaces. Points of discontinuity. 2. Complex analysis. Analytic functions. Power series. 3. Group theory. Derived subgroups. Elements of finite order. 4. Matrix theory. Rank. Image. Kernel. Direct sum. Idempotents. 5. Number theory. Primes. 6. Point-set topology. Circles. Arc lengths. Thirty First Competition

First Day. Problem 1. Plane geometry. Triangles. Coloring circles by two colors. Problem 2. Point-set topology of Rn . Minkowski sums. Limit points. Problem 3. Group theory. Subgroups of index two. Finite abelian groups. Problem 4. Probability. Dice throwing. Problem 5. Point-set topology of R2 . Collinearity. Problem 6. Matrix theory. Determinant. Trace. Second Day. Problem 1. Number theory. Complete squares. Problem 2. Calculus of two real variables. Limits. Problem 3. Set theory. One-to-one correspondence. Problem 4. Ring theory. Polynomial rings over commutative rings with identity. Maximal ideals. Problem 5. Real numbers. Partitions of sets with equal sums. Problem 6. Metric spaces. Fixed point theory.

246

3. PROBLEM INDEX

Classification of problems • Algebra, Ring theory,

Index Abel’s Continuity Theorem, 180, 182 Acceleration, 5 Algorithm, 35 AM-GM Inequality, 55 Algebraic, 16, 47 Algebraic numbers, 18 Archimedean property of real numbers, 52 Area, 19, 32, 43 Arzela’s Theorem, 129 Automorphism, 18

Connected, 5, 34, 37, 46 Connected component, 6 Convergent, 13, 17, 20, 28, 35 pointwise, 17 uniformly, 10, 23, 29 Darboux’s property, 52 Darboux’s Theorem, 51, 52, 169, 176, 184, 207 Dedekind’s Extension of Abel’s Theorem, 126, 127 Dense, 14, 47, 48 Derangement, 231 Derivative, 1, 10, 13, 16, 17, 32, 37, 46 Derived subgroup, 22, 29, 47, 90, 167 Determinant, 15, 18, 105, 107, 118, 231 Dihedral group, 183 Dini’s Theorem, 116 Dirichlet’s Theorem, 181 Disconnected, 1, 41 Division algorithm, 111 Division ring, 37, 59, 85, 111, 146, 154, 193, 225 Domain, 37

Baire Category Theorem, 163, 208 Baire function, 1, 52 Banach space, 162, 191 Bayes’ Theorem, 74, 188-190, 215 Bertrand’s conjecture, 48, 226 Bertrand’s principle, 48 Binomial Theorem, 70, 72, 79, 85, 94 Bisection method, 34 Boolean ring, 78, 122, 209 Cantor set, 3, 46 Cauchy Inequality, 225 Cauchy product of two polynomials, 70 Cauchy-Riemann Equations, 185 Cauchy-Schwarz Inequality, 216 Cauchy’s criterion for the convergence of improper integrals, 168 Cauchy’s Integral Formula, 177, 184 Cauchy’s Theorem, 117 Cayley-Hamilton Theorem, 123 Centralizer, 117, 151, 196, 204 Chebyshev Inequality, 47, 223 Clopen subset, 218 Chain ascending, 26, 90, 122, 122, 159 descending, 9, 73 Characteristic, 14, 19 Circle, 10, 11, 19, 21, 48 Class equation, 117 Closed convex hull, 80 Commutator subgroup, 22, 167 Compact, 6, 28, 36 Comparison Test, 127, 155 Complete metric space, 38 Complete square, 49 Congestion point, 9, 77

Eigenvalue, 10, 21, 41 Eigenvector, 21, 41 Eisenstein’s Criterion, 117, 123 Element idempotent, 12, 25, 42 identity, 4, 12 invertible, 4 left identity, 12 left quasi-regular, 4 maximal, 27 nilpotent, 12 right quasi-regular, 4 right identity, 7 quasi-regular, 4 Ellipse, 10, 81 Endomorphism, 4, 8 Equation differential, 2, 5, 17, 30, 34 functional, 17 integral, 17 Euclidean algorithm, 103 247

248

metric, 43, 192 norm, 1 plane, 8, 42, 130 ring, 60, 103 Euclid’s proof of the infinitude of primes, 231 Euler’s totient function, 43, 59, 152, 216 Expansion, binary, 26, 35 decimal, 24, 35 hexadecimal, 35 octal, 35 ternary, 3, 46 Extension, 13, 14, 16, 18 Fermat’s Little Theorem, 138, 185, 202, 222 Field, 4, 7, 9, 13, 14, 16, 18, 19, 22, 23, 26, 27, 28, 30, 37, 41, 45, 46, 47 First Isomorphism Theorem for groups, 102, 132, 145, 150, 167, 178, 218 for modules, 69, 168 for rings, 178 Flux, 33 Frobenius Inequality, 193 Function analytic, 29, 40, 41, 46, 47 arithmetic, 43 auxiliary, 16 bounded, 21, 29, 32 choice, 36 complex valued, 6 continuous, 1, 3, 6, 6, 8, 9, 10, 12, 13, 14, 15, 16, 17, 17, 18, 20, 21, 24, 25, 26, 27, 28, 29, 32, 37, 39, 40, 41, 43, 45, 46,49 continuous bounded, 36 continuously differentiable, 13, 27, 39 convex, 23 decreasing, 28, 32 differentiable, 17, 20, 21, 25, 29, 32, 39, 40, 42 entire, 43 harmonic, 37 increasing, 14, 23, 32 integral part, 24 nonnegative, 3, 24 nowhere differentiable, 27 one-to-one, 8, 14, 17, 18, 42, 43 periodic, 24, 26, 28 real, 1, 2, 6, 7, 16, 23 surjective, 18, 39, 43, 45 uniformly continuous, 21, 24, 28, 32, 36 Fundamental Theorem of Algebra, 210 Fundamental Theorems of Calculus, The First, 92, 113, 160, 184 The Second, 92, 109

INDEX

Fundamental Theorem of finite(ly generated) abelian groups, 58, 118, 151, 175, 232 Gamma function, 34 Gauss-Lucas Theorem, 79 Greatest common divisor, 14, 22 Group abelian, 2, 19, 37, 42, 48 commutative, 2, 19, 37 cyclic, 4, 19, 58, 59, 111, 121 finite, 14, 18, 22, 26, 26, 39 finite abelian, 48 finite nonabelian, 25 finite p-, 22 infinite, 37 nonabelian, 25, 37 of inner automorphisms, 25, 37, 150, 167 of permutations, 20, 102, 178, 211 of units, 198 p, 22, 117, 137, 150 symmetric, 52, 102, 131, 145, 178 quaternionic, 196 simple, 26 simple abelian, 111 solvable, 28, 137, 167 Hamel basis, 72 Hyperbola, 81 Ideal left, 20, 23, 38 maximal, 1, 13, 30, 40, 49 minimal left, 27 nilpotent, 45 prime, 1, 8, 23 principal, 19, 40 right, 20, 23, 26, 37, 45 small, 29 two-sided, 18, 27, 37, 38, 45 Inclusion, 27, 46 Inclusion-exclusion principle, 148, 231 Instantaneous speed, 61 velocity (vector), 62 Integral, 9, 24, 33, 36 Integral domain, 9, 13, 20, 39 Integration by parts, 64, 93, 116, 120, 149 Intermediate value property, 51, 207 Intermediate Value Theorem, 51, 62, 65, 77, 97, 119, 128, 133, 137, 137, 153, 161, 170, 193, 202, 205, 207, 213 Invariant subspace(s), 46 Inverse Probability Theorem, 74 Isomorphic sets, 2 Iteration method, 35 Jacobian, 33

INDEX

Labeled graph, 139, 187 Lagrange’s Theorem, 85, 111, 118 Largest prime power dividing a factorial, 114 Least common multiple, 22 Lebesgue number, 98, 98 Lebesgue’s Integrability Criterion for Riemann integrals, 75, 98, 174 Lebesgue’s Number Lemma, 98 Left annihilator, 216 Legendre symbol, 222 Leibniz’s Theorem, 181 L’Hopital’s rule, 161, 221 Limit, 1, 29, 46, 49 Limit Comparison Test, 117, 127 Limit point, 28, 48 Line, 4, 11, 21, 35, 41, 42 Linearly dependent, 8, 28 Linear transformation, 4, 10, 12, 14, 26, 30, 39 Matrix, 1, 11, 13, 15, 15, 18, 19, 21, 23, 27, 29, 37, 38, 40, 41, 42, 44, 46, 47, 48 Maximal subgroup, 16 Maximum Modulus Principle, 202, 224 Maximum Modulus Theorem for harmonic functions, 184, 185 Mean Value Property, 184 Mean Value Theorem, 51, 64, 116, 128, 176, 183, 197, 202 Mean Value Theorem for integrals, 173 Mean Value Theorem for second order derivatives, 62, 62 Metric space compact, 39, 46, complete, 38 connected, 46 separable, 47 Minkowski sum, 48 Module, 7 Morera’s Theorem, 177 M¨ obius function, 151 Inversion Formula, 210 Nilpotent matrix, 16, 46 Noetherian ring, 154 Normalizer, 145 Normal subgroup, 12, 14, 22, 26, 42 Normed linear space, 149 Nowhere dense subset, 163, 207 Odd prime, 19, 23, 39, 40 One-to-one correspondence, 2, 23, 45, 49 Open map, 26 Operator norm, 205 Optical property of ellipses, 81

249

hyperbolas, 81 Orthogonal matrix, 44 Partition, 39, 42, 47 Perfect subset, 207 Permutation, 11, 20 Persian alphabet, 5 Persian calendar, 4 Picard’s Great Theorem, 210 Plane, 7, 21, 35 Poisson random variable, 55, 56 Polynomial, 10, 14, 21, 25, 26, 47 characteristic, 29, 30, 108, 122, 178, 206, 229 minimal, 123, 179, 207 elementary symmetric, 142 Poset, 43 Prime ring, 198 Primitive root, 47 Probability, 3, 9, 13, 15, 17, 18, 24, 31, 38, 39, 41, 44, 48 Radical of an ideal, 8 Radius of convergence, 26, 47 Random variables, 3, 5, 17, 47 Rank, 23, 26, 27, 30, 38, 47 Rank-Nullity Theorem, 85, 90, 104, 108, 154, 165, 225 Rational Root Theorem, 133 Regular n-gon, 21 Riemann integrable, 9, 19, 20, 21, 29 Riemann’s criterion for integrability, 75, 98, 156 Right annihilator, 216 Ring commutative, 1, 8, 10, 19, 20, 28, 30, 40, 44, 47, 49 division, 37, 59, 85, 111, 146, 154, 193, 225 noetherian, 154 uncountable, unital, 16, 19, 20, 23, 40, 44 with identity, 1, 9, 10, 11, 12, 18, 27, 28, 47, 49 Rolle’s Theorem, 109, 207 Root, 6, 10, 18, 19, 21, 27, 34, 35, 42 Rouch´ e’s Theorem, 67 Sandwich Theorem, 230 Schroder-Bernstein Theorem, 53, 101 Schwarz Lemma, 224 Second Isomorphism Theorem for groups, 90, 150 Second Mean Value Theorem for integrals, 168 Semigroup, 7 Sequence, 6, 8, 10, 12, 17, 20, 23, 24, 26, 28, 29, 32, 37, 42, 45 decreasing, 6, 13, 18

250

increasing, 6, 17, 45 Series, 18, 20, 31, 35, 36, 45 power, 26, 33, 47 Set of measure zero, 75 partially ordered, a.k.a. poset, 43 power, 2 totally ordered, 76 Simple left module, 168 Speed, 4 Square, 19 Squeeze Lemma, 110 Stolz’s Theorems on limits the first theorem, 87 the second theorem, 87, 173 Sylow p-subgroup(s), 23, 121, 145 Sylow’s Theorems the first theorem, 137 the second theorem, 145 the third theorem, 121 Sylvester Inequality, 193 Sylvester’s problem, 133 Symmetric difference of sets, 118 Tangential acceleration, 62 Taylor’s Formula, 79 Third Isomorphism Theorem for groups, 163, 228 Tree, 139 Triangle, 2, 39, 48 Topological space, 5 Torsion free group, 231 Tur´ an’s Theorem, 187, 187 Uniqueness Theorem for analytic functions, 205, 217 Vandermonde’s determinant formula, 142 Variation of parameters, 55 Vector field, 33, 34 Vector space, 3, 4, 8, 10, 12, 14, 16, 22, 26, 30, 39, 45 Well-ordering principle of natural numbers, 52, 54, 158 Zero divisor(s), 28 Zorn’s Lemma, 36, 72, 94, 146

INDEX