College algebra

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College Algebra

The qualif1 of the materials used in the manufacture of this book is governed by cQnUnuedpostwar shortages.

College Alge bra BY

A. ADRIAN ALBERT Professor of Mathematics The University of Chicago

First Edition Second Impression

London

New York McGRAW-HILL

BOOK

1946

COMPANY,

INC.

COLLEGE ALGEBRA COPYRIGHT,

1946, BY TEE

MCGRAW-HILL BOOK COMPANY, INC. PRINTED IN TEE UNITED STATES OF AMERICA

All rights reseTlJea. This book, or parts thereof, may not be reproduced in any form 'Without permis8ion of the publishers.

TEE IUPLE PRESS COMPANY, YOBlt, PA••

PREFACE The mathematical material presented in standard texts on college algebra is a fundamental part of mathematics. The concepts involved are basic for an understanding of all more advanced mathematical topics, and the techniques are used in virtually all subsequent courses in algebra and analysis. Nevertheless, college algebra has been a most abused subje,ct. The time allotted to it is frequently inadequate for a genuinely good treatment, and indeed the entire course is sometimes omitted. This is due partly to a desire to bring students to a study of the calculus as early as possible. It is also due partly to the presentation of college algebra, in all texts thus far published, as a collection of seemingly unrelated topics. The desire to teach the calculus as early as possible tends to defeat its own ends. The building of a course in the calculus on what must be a weak foundation cannot result in a good student understanding of the subject. There is also no reason why the :ql8.terial of college algebra cannot be cohesively organized. ' About fifteen y~ars ago the author began to study the possibility of reorganizing the material of college algebra so as to present it as a sound and unified whole. The result of that study is the present text. He began with a formulation of what he felt should be standard minimum requirements for all college texts. These requirements are the following: 1. The material should be presented as a unified and compact body of mathematical theory. 2. The definitions and theorems should be stated accurately. Proofs of results should be given whenever it is reasonable to expect that the better students will be able to grasp them and provided that their inclusion will add to t4e understanding of the results. Whenever a proof is omitted for any reason it should be made clear that the v

-VI

proof is omitted. Proofs of special cases of theorems should not be presented as proofs of the theorems. 3. The important techniques and concepts of the text should be emphasized by the presentation of an adequat~· number of illustrative examples and exercises. Exercises should be given in sufficient numbers so that there are enough both for additional classroom illustration and for student home assignment. They should be solvable by" the use of the techniques of the text. 4. Where such material exists, the text should be rich in additional material on the same subject for the better student. It seems wasteful to use the trick problem as a device to test the student's native ability as a mathematician. His time can be employed far more profitably iu the learning of more advanced topics which are a part of the text subject and which are given in the text itself. The reader may judge for himself how well the present text lives up to these standards. It would seem that they should be met and that few college texts of today meet them. ' College algebra has a basic unity. It should consist of, a study of the number systems of elementary mathematics, polynomials and allied functions, algebraic identities, equations, and systems of equations. The unity of the present text is achieved by fitting the standard topics of college algebra into this pattern. Thus the text begins with three, chapters on number systems. The first chapter on Natural Numbers introduces the counting concept and so the fundamental idea of a sequence which appears so frequently in mathematics. As permutations and combinations involve nothing but <)ounting, this topic forms a part of the first chapter. The second chapter presents the theory of the factorization, of integers, the Euclidean greatest-common-divisor process, and a technique for listing the divisors of an integer. These concepts and techniques are a necessar.y . forerunner of the theory of the .integral roots of equations

PREFACE

vii

with rational coefficients. They are presupposed in most texts on opr, subject, arid there is no basis in fact for assuming that most students have ever seen them. The theory of rational, real, and complex numbers makes up the third chapter. It is here that the notion of a logari~, the summation and product symbols, and the concept of a convergent sequence of real numbers are presented. The material should thus present an improved foundation for the calculus. Chapter IV consists of an introduction to the purely algebraic theory of polynomials and rational functions. All four of the first chapters are written so as to provide the student not only with new material but with a review ~f the fundamental techniques of elementary algebra. The material usually labeled Mathematical Induction is presented here for what it really is, 8t derivation of summation formulas, by the use of mathematical induction and algebraic identitiel;!. It is then the first part of Chap. V on Identities and Applications. This chapter includes also the identity known as the binomial theorem, and the theories oj progressions, ,which are simply applications of summation formulas. Chapter VI presents the general theory of polynomial equations as a special case of the theory of factorization of polynomials. Chapter VI,I on the Real Roots of Real Equations contains some' techniques usually -left for a special course on the theory of equations. A student understanding of the~ techniques is worth while and the techniques can easily be taught if no attempt is made to derive them. ' Chapter VIII on Vectors in the Plane is intended to be treated as optional material. It is hoped that this chapter will provide a basis for a future unification of college mathematics. The chapter begins with a derivation of the parallelogram law for the Stddition of vectors and the connection of unit vectors with trigonomet~c functions. The parallelogram. law is then used in a derivation of the

viii

PREFACE

formulas for the rotation of axes in plane analytic geometry. These formulas yield the addition laws of·trigonometry as well as De Moivre's theorem for complex numbers, and they are used to complete the student's understanding of the theory o{ radicals which began in Chap. III. The connection of these topics shows how closely knit college algebra, trigonometry, and analytic geometry really are, and it is hoped that the chapter may be used to inspire a more compact and improved treatment of trigonometry and of plane and solid analytic geometry. Chapter IX contains the theory of determinants and linear systems. It presents the inductive definition of a determinant. and the elementary transformation technique for its computation. There is no reason why a college student should be limited to determinants of two and three rows. Systems of linear equations are solved best by elimination rather than by determinants, and the chapter does not place emphasis falsely on the determinant method. The final chapter is a pedagogical experiment. There has long been a demand by mathematical economists and psychologists for an early treatment of matrices and, quadratic forms. It is also very desirable that such a treatment precede a course on solid analytic geometry. The theory is an advanced mathematical topic only because of the abstract nature of its proofs. The author's personal ·experience is that the theorems and· techniques can be taught to immature students if no attempt is made to derive the results, and so such a presentation has been written here. It is hoped that this chapter will fulfill the needs of the social scientists as well as provide a foundation for an improved treatment of solid analytic geometry. ADRIAN ALBERT.

CmCAGO, ILL., May, 1946.

CONTENTS P.t.eB

PREFAClil •

i

i

;

OJLU'TB.

V

•

,

I. NATURAL NUMBERS • • • 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

n.

.

Introduction. • • Section labels. • • The natural numbers . Addition and muUiplica.tion (FuLL COUR8E) • Addition and multiplication J.a.~. • • • • The ordering of natural numbers (FuLL COURSE) • Subtraction (FuLL COUBBE) Division Powers . • • • Summary • • • Finite sequences. Infinite sequences Counting of pairs Factorials . . . Permutations and combinations Cyclic and circular permutations (FuLL COUB8E) Distinguishable permutations (FuLL COUBSE)

...

INTEGERS . . . • . • . . .

1. 2. 3. 4. 5. 6. 7.

The domain o( all integers . Addition and mUltiplication. The law of subtraction Absolute values. .'. . . Divisibility • • . . . . The division algorithm (FuLL COUR8E) The Euclidean greatest-common-divisor process (FuLL COUR8E) • • • • • . • • • .

8. Linear combinations (FuLL COURSE) • . 9. The fundamental theorem of arithmetic (FuLL

1 1

3 4 5 6

8 10

10 11

13 14

17

18

21 25

30 31 35 35 35 36

37 38 39 40 42

. .

44

10. The greatest common divisor and the least common multiple of several integers (FuLL COURSE) 11. The factors of an integer. . . . . . . . . .

45 46

COURSE)

. • . • • • • . .

ix

. • . .

x

CONTENTS

CHAPTER

III.

F4G)!l

RATIONAL, REAL, AND COMPLEX NUMBERS

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

The real line . . Rational numbers . . The division law. . . Fractions in least terms The laws of exponents. Decimals . . . . . The real number system . Positive real powers of positive real numbers Logarithms (FULL COURSE). . Change of base (FULL COURSE) Irrationality of real numbers The sigma and pi symbols . . Infinite series and products: . Cartesian coordinates in the plane Complex numbers . . . Complex units . . . . Fields of complex numbers Rational operations, . .

IV. POLYNOMIALS AND RATIONAL FUNCTIONS

1. 2. 3. 4. 5'. 6. 7. 8. 9. 10. 11. 12. 13.

Polynomials in x . . . . . Sums, differences, and products Computation of polynomials The division algorithm . . Reducibility of polynomials. The factbrization theorems . The Euclidean g.c.d. process (FULL COURSE) Linear combinations (FULL COURSE) The unique factorization theorem (FULL COURSE). Polynomials in several variables . . . . . . . Derivatives (FULL COURSE) . . . . . . . . Determination of multiple factors (FULL COURSE). Rational functions. . . . .

V. IDENTITIES AND ApPLICATIONS.

49 49

.50 53 54 57 59 '60

63 66

70 70

72.

74 77

79

81

83 86

87 87 8!;) 92 94'

95 96 98

102 104 105 108

109.

111 116

1. The binomial theorem. . . . 116 2. Summation formulas . . . . 119 3. The method of undetermined coefficients (FULL COURSE) . . . . . 122, 4. Arithmetic -progressions 124 5. Geometric progressions 128 6. Harmonic progressions 130·

CONTENTS OHAPTER

VI. EQUAT.IONS

1. 2. 3. 4. 5. 6. 7. 8. 9.

Conditional equations. The degree of an equation Linear equations. Quadratic equations The remainder and factor theorems . Synthetic division Factorization into linear factors Expression of coefficients in terms of roots The imaginary roots of real polynomials 10. Multiple roots by derivatives (FULL COURSE) VII. REAL ROOTS OF REAL EQUATIONS

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

Transformations of polynomials Integral roots. Rational roots Upper and lower bounds . Proof of the method of radicals (FULL COURSE) Isolation of the real roots of a real equation . Descartes' rule of signs (FULL COURSE) Sturm's theorem (FULL COURSE) . Horner's method Other methods (FULL COURSE) The function concept . Operations on functions The equations of a curve.

VIII. VECTORS IN THE PLANE (FuLL COURSE).

1. 2. 3. 4. 5. 6. 7. 8. 9.

IX.

Xl PAGE

Vectors. Addition of vectors. Angular measurement. Polar coordinates and trigonometric functions Rotation of axes. Addition formulas and inner products The binomial equation Equations of lines ., . Conic sections

132 132 134 134 136 139 141 143 144 147 149 151 151 156 159 162 165 166 169 171 174 178 180 181 183 185 185 187 188 190 193 195 197 199 202

MATRICES, DETERMINANTS, AND LINEAR SySTEMS .

.

.

.

.

.

.

.

1. Systems of equations . . . 2. Rectangular matrices . . . 3. Elementary transformations

206 206 208 210

m

CONTENTS

0iLUTIIB

PAS.

4. Special matrices. . . . . ..••... 5. Submatrices • . .' . . . . . . . . . • 6. Determinants . • . . . . . . . . . . . 7. Froof of the equality of all expansions (FuLL COUBSJD) • • • • • • • • • • • • • • • 8. Properties of detenninants. . . . . • . . . 9. Proofs of the determinant theorems (FuLL COURSE) 10. The rank of a matrix. . . . . • . . 11. The matrices of a linear system • • • • 12. Solution by elimination and \SUbstitution' . 13. Solution by determinants . • . • . . 14. Partial fraCtions. • . . . . . . . .

X.

MATRICES

AND

QUADRATIC

FORMS

INDEX

219 222

228 224 227 229

235 23·8

(FULL.

COURSE) . . . . . . . . . . 1. 2. 3. '4. 5. 6. 7. 8. 9. 10.

212 218 214

Inner products • . . . . . Matrix multiplication. . . . The inverse of a. square matrix Linear transformations . . . simiia.r matrices. • • • . . Quadratic forms. • • • . . Equivalence of quadratic forms Orthogonal matrices • • • • . • . • OrthogonaJ. reduction of a quadratic form. . • Factorization of a positive symmetric matrix '.

248 248 245 249 251 258 257 259

262 264 268 • 273

College Algebra CHAPTER I NATURAL NUMBERS

1. Introduction. Mathematical formulas may be likened to machine tools which operate on raw materials to yield finished products. The raw materials of mathematics arise from the problems of the physical world and are usually numbers. The finished products are the solutions of problems. The numbers used in our subject are elements of certain 8ets of numbers called number SY8tems. The properties of these sets are the properties of the raw materials. Then it will be natural for us to begin our study with an extensive discussion of the properties of number systems. Formulas are mathematical frameworks in which letter symbols appear~ We put the materials into these mathematical machine tools by replacing the symbols by numbers. Thus the formulas will consist of letters and connecting mathematical symbols rather than of particular numbers. It should be clear that it i8 only by using letter8

to repre8ent unspecified number8 that we can 8et down results For example, the equality (5 + 3)(5 - 3) == (5' 5) - (3' 3) is a true statement, but the formal equality (a + b)(a - b) == (a . 'a) - (b' b) implies a result true for all numbers a and b (read the dot in a . a as " times"). In using letter symbols it is frequently desirable to select them so that they suggest an order. Thus we may use a, b, c to connote that a is our first number, b our second number, and c our third number. Similarly we may use of any general significance.

1

2

NATURAL NUMBERS

[CHAP. 1

a, b, c, d for four numbers, and a, b, c, d, e for five. Other sets of symbols used in this way aref, g, h, k; x, y, Z, t, u, V; i, j, k, l; m; n, p, q, r, s, t, u, v. It is sometimes desirable to use several sets of letters to connote triples of numbers related in some fashion. For example, we might use the letters a, b, c for one triple and A, B, C for a second triple. The number of such suggestive combinations which are available may be increased by the use of Greek letters. We shall give the Greek alphabet at the end of this section. The' notation a = b (read either as " a equals b" or as "a is equal to b") is used to indicate that the symbols a and b represent the same number. When the number represented by a is different from that represented by b we write a ~ ·b (read "a not equal to b," or "a'differentfrom b"). If a = b and b = c then a = c. We indicate. all three of these stateIIl:ents of equality by writing a = b = c (read equals b equals c"). . Many of our statements of mathematical results will contain the phrase if and only if. Such statements really consist of two separate results. They state th~t a conclusion (i.e., consequence) holds if and only if a hypothesis (i.e. assumption) is true. Thi~ means, first, 'that if the assumption is made the conclusion will hold, and we' mean this when we say that the assumption is a sufficient condition. It also means that the conclusion cannot hold without the hypothesis being true also, and·so the hypothesis is a necessary condition. Then the two statements, hypotp,esis __ and conclusion, may be called equivalent conditions. Another phrase which occurs frequently in mathematical discussions is the statement of the existence of a unique element with some prescribed property. This too is a double statement. It says, first, that such an element does exist; second, that only one such element exists. For example, we have the statement that the number zero has the property that if we add it to any other number a we get the same number a. But no other number has this property,

"a

-3

SECTION LABELS

BEC.2]

and we state that there exists a unique number 0 such that a+O=a. The Greek alphabet is given in the accompanjring table. L.o. indicates lower case. i.8•• small letters; c'indioates capitals.

L.c. C

Name

a

A

fJ

B

Alpha Beta

r

Gamma.

A

Delta Epsilon Zeta Eta. Theta

'Y

a

E

r

"e

E Z H

e

L.c.

C

Name

, "A

I K

Iota Kappa Lambda Mu Nu Xi Omicron Pi

I' II

~

A

M N

:e

0

0

'II"

n

L.c.

C

Name

p

P

IT

l}

v

T T

Rho Sigma Tau Upsilon

q;

Phi

X 'iJ!"'

Chi Psi Omega

.,.

•

x

'" IAI

{)

2. Section labels. The time normally· allott~d to a course·on college algebra is not adequate to cover in class all the material to be presented here. Nonetheless it seems very desrrable that our text shall contain this material so as to make it conveniently available to the student who desires to have a clear and sound foundation for his study of more advanced mathematics. We shall therefore add to our full treatment suggestio;ns as to the sections that would not usually be given class time. The headings· of all our sections will begin with titles descriptive of their contents, and those sections whose headings consist only of these titles should be fully discussed in the classroom. However a number of· section headings will have the additional parenthetical label (FULL COURSE) and should be presented, if at all, only" as reading assignments for the class. A number of these full-course sections occur· in the first three chapters where we develop the number systems of algebra. Their subject matter is usually presupposed by most authors of texts oli. our subject, and they may be regarded as being, in a very broad sense, review material. In later chapters some full-course sections contain techniques which may be interesting to students and should be

4

NATURAL NUMBERS

[CHAP. I

given classroom attention if time permits. The selection of such additional topics is left to the instructor. It should be noted that entire Chaps. VIII and X are of this kind. Finally, some full-course sections contain the rigorous and necessarily involved proofs of theorems which are themselves presented, with exercises, 41 other sections. The- author· recommends the omission of these sections. The omission of all full-course sections will not disturb the continuity of our' treatment and should resUlt in a text which is at least as rich in material as the usual text .. 3. The natural numbers. The symbols used in counting are called the natural numbers or natural integers. The set of all these numbers may be conceived of as being an unending line of symbols

(1)

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 •••

(the three dots-being read here and later as "and so on"), and this in-line arrangement is a basis for the intuitive definition of the successor of a natural number as the natural number immediately to the right of it in line (1). We shall distinguish the zero int~ger 0 from all the. others and call these others positive integers. We now proceed to give what is called' (by mathematicians) an axiomatic characterization of the set of all positive integers in terms of the (undefined) concept of successor. We shall regard this characterization here merely as a statement of properties of natural numbers. Every positive integer has a unique positive integer as its successor, and two integers ha,!e the same successor only when they are the same integer. The number 1 differs from all the other positive integers in that it is the only positive integer not the successor of a positive integer. We now complete

our characterization with the very important Principle of Mathematical Induction. PRINCIPLE 1. Let K be a set of positive integers such that 1 is in K and the successors of all the numbers in K are in K. Then K is the set of all positive integers.

SIlO. 4)

5

ADDITION AND MULTIPLICATION

This principle is the formal mathematical statement of the intuitive notion that we will eventually reach any positive integer in the line of natural numbers above if we begin at 1 and repeat the process of taking successors. It states that in a mathematical argument, where we wish to show that all positive integers may be reached, it is sufficient to sh9W that the two processes of beginning with 1 and of taking 8'UCC888DrS are allowable. The principle is used very frequently in mathematics, and its ~e is so natural that one is apt to carry out applications of it without noticing that the principle has been applied. 4. Addition and multiplication (FULL COURSE). The sum a + b of any two natural numbers a and b may be defined inductively. We define a + 0 to be a, a + 1 to be the successor of a. This amounts to the introduction of the notation a + 1 for the successor of a. We complete our definition inductively by defining (2)

a

+ (k + 1)

= (a

+ k) + 1.

,

'

The intuitive application of this definition then states that to find the sum a + b of any two natural numbers a and b, in every case where'b ~ 0 or 1, we use the formula above with k = 1,2 .... (read Ilk equal to 1, 2 and so on") until we arrive at a + (k + 1) with k + i = b. This is, of course, not the elementary arithmetic process for finding sums. That process uses the concept of digit and an addition table of sums of the integers 1, 2, 3, 4, 5, 6, 7, 8, 9. However,.that process is based upon the definition we have given above. The product aO of any natural number a and the number ois defined to be 0, the product al of a and 1 to be a. We then define a unique product ab of any two natural numbers inductively by the following formula: (3)

a(k

+ 1)

= ak

+ a.

This is the basis of the arithmetic process for forming products, which uses digits and a multiplication table.

6

NATURAL NUMBERS

[CHAP, I

5. Addition and multiplication laws. The system of all natural numbers is a set of numbers in which the sum a + b and the product 'ab of any two numbers of the set are uniquely determined numbers of the same set. All 'the number systems of elementary algebra have these two properties, and in all cases the operations of addition and multiplication are such that certain-laws are obeyed. We shall state these laws as properties of our number systems without trying to verify them. ADDITION LAWS

Addition is commutative, that is, every sum a + b is equal to b + a. . II. AsSOCIATIVE LAW. Addition is associative, that is, every sum (a + b) + c is equal to a + (b + c). III. IDENTITY LAW FOR ADDITION. The element 0 has the , property thai, every sum a + 0 = a. IV. CANCELLATION LAW. If a + b = a +.c then b = c. Properties III and IV have analogues which are consequences of I.. From III and I we obtain 0 + a = a for every a, while from IV and I we' see that if b + a = c + a then b = c.

I.

COMMUTATIVE LAW.

MULTIPLICATION LAWS V.' COMMUTATIVE LAW. Multiplication is commf!,tative, . that is, every product ab is equal to ba. VI. AsSOCIATIVE LAW. Multiplication is associative, that is,/every product (ab)c is equal to d(bc). . VII. IDENTITY LAW FOR MULTIPLICATION. The element 1 has the property that every product al = a. VIII. CANCELLATION ,LAW. If ab = ac and a ~ 0 then b=~ . If a is any number and a + b = a, then we use the identity law for addition to obtain a +.b = a + O. By the cancellation law for addition we have b = O. Thus we have the following: TheQrem 1. A ~m a b = a if and only if b = O.

+

SEC.

5]

ADDITION AND MULTIPLICATION LAWS

7

This-result is a uniqueness theorem. It states that there is one and only one number b such that a + b = a. The similar discussion for products is made for a ¢ O. Then, if ab = a,' we have ab = aI, b = 1. Also, if ab = 0 and a ¢ 0, we.have ab = aO and b = O. We state these results as follows: Theorem 2. A product ab = a ¢ 0 if and only if b = 1. A product ab = 0 ~l and only if a = 0 or b = O. Addition and multiplication occ'l:ll" in combination in the last of our laws. IX. DIsTRmUTIVE LAW. Anyproducta(b + c) = ab + ac. The reader who has had even elementary algebraic experi-' ence probably feels that there is no need to state these laws so formally. However; they are the basic properties of all the number systems' of our subject and should be learned formally as such. They require at least the same obedience in the manipulations of algebra as do the rules for moves in a ,game of chess or the rules for play in any other game. ORAL EXERCISES

1. From which laws may we deduce that (b + e)a = ba + ea? 2. Use the laws to show that (a + b)(e + d) = ac + be + ad + bd. 3. Interchange the roles of addition and mqltiplicatipn in the distributive law and obtain an analogous formula not valid for numbers. 4. Use the result of Oral Exercise 2 to compute (a + b)(~ + b) and (a + b)(a + b)(a + b) as sums of products of ~'s and b's. 6. If we express (a + b) (e + d) (e + J) (g + h) as a sum, how many terms does it have? EXERCISES

1. Using the distributive law, express the following sums as products of two factors: (a) (b) (e) (d) (e)

2z

+ 2y

za + ya 3za + 3ya 3ab 6zy

+ 6ac + 8y

(J) (g) (h) (t) {J1

+ 2ay + 7y + 12bz + 3e + 8b + 16e + 32 + y)2a + (z + y)3b + 3y + z)3z + 2(2z + 3y + z)2y

zy 6a:!: 4a (z (2z

2. Use the result of Oral Exercise 2 to express the following sums as products of two factors: .

8

NATURAL NUMBERS

(a) az + by + ay + bz (b) 00 + a + b + 1 (c) 6ab + 4b + 9a + 6 (d) 2z + 2y + zy + 4 (e) 6Bt + Sa + 9t + 12 (J) 4ab+12a+b +3

[CHAP. I

+ be + ad + btl + a + b + c + 2ac + 4ad + 2bd + be + 2a + b +

(g) ac

a+1 (Il)

c+2a+1

(,1 300 + 9ay + bz + ~ + 12a +

4z+2b+6rI+8 . + ay + a8 + bz + by + b8 +

(31 az

cz+cy+cs+a+b+c+z+ y+*+1

6. The ordering of natural numbers (FULL COURSE). If c and a are natural numbers such that c is the sum. a + b of a and a positive integer b, we say. that c is greater than a and write c > a. •

The number c will then appear to the right of a in formula (1) of natural numbers. When a appears to the left of c, we say that a is le88 than c and write

a < c. These notations should be learned by aU 8tudentB. Any two natural numbers a and c are so related ~hat one and only one of the properties a == c, a > c, a <. c can hold. Moreover the relation of being greater than is what is called a tra'Mitive relation, that is, if c > b ~d b > a. then c > a. This remark is verified by observing that c > b means that c == b + g where g is a positive inte~r; b > a means that b == a + h where h is a positive integer. Then c == b + g == (a + h) + g == a + (h +" g). Since h + g is positive, c > a. It is customary to indicate that c > b and b > a by writing

c· > b > a, (read HC greater than b 'greater than a"), and i1r is then important to know that this means also ·that c >. a. A relation haying the 'properties described in the paragraph above is called an order relation and a ~t, for which an order relation holds for all pairs of its elemeJ;l.ts, is called.

SEC.

6]

THE ORDERING OF NATURAL NUMBERS

9

an ordered 8et. All but the final one of the number systems we shall study will be ordered. This last unordered number system will be called the 8et of all complex number8. A notation indicating the property that a is not greater than c is a:l> c (read "a not greater than c"). This' property occurs in the case where a and' c are natural numbers (as well as in the other cases of ordered number systems which we shall study) only when either c = a or c > a. We then write c~a

(read "c greater than or equal to a"). In the case of natural numbers the condition c ~ a is equivalent to the property that c = a +'b where b is also a natural number. The property is identical with a~c

(read "a less than or equal to c"). we have c ~ a and so write

c

~

b

~

Also if c

~

band b

~

a

a.

When we apply the notations c ~ 0 and c > 0 to natural numbers c, they mean only that c is a natural number in the former case and that c is a positive integer in the latter case. The relation c ~ a is called an inequality. When we have such a relation and show that it can be replaced by c > a, we say that we have 8trengthened the inequality. Thus c > a is called a 8trong inequality, c ~ a a weak inequality. The inequalities c ~ b ~ a imply the equality c = a if and only if neither of the inequalities c ~ b, b ~ a is a strong inequality. , The principle of mathematical induction can be stated in two other useful forms which should be learned by the student of the FULL COURSE. They are as follows: PRINCIPLE 2. Let K be a 8et of positive integer8 and 1 be in K. Suppo8e that.whenever an integer n i8 such that all integer~

10

'N ATUR:A:L

N-UM'BERIil

[CHAP. '1

l'688 than n are in K, then n i8 in K. 'i'hen K i8 the 8et of all positive integer8. PRINCIPLE 3. Every 8et of p08itive integer8 contain8 a lea8t integer. Pririciple 3 states that L"l every set K cf positive ip.teger-s there is an integer c such that if k is ail integer of K then k ~ c. We shall not try to show how to obtain each of the three forms of the principle of mathematical induction from anyone of them. 7. Subtraction (FULL COURSE). Whenever c ~ a, there' is a natural number b such that c = a + b. We call this . number the difference of c and a, write (4)

b = c - a,

(read" b equals c minus a"), and say that b is the result of subtracting a from c. If a = c, the difference (5)

c- a

=

O.

Otherwise b = c - a is a positive integer. If c = b + g and b = a + h then c = (g + h) + 'a and c - a =:= g + h. When h > 0, we have g + h > .g, b - a = h, c - a > b - a. In any case c - a ~ b, - a and we have a result which we shall state as follows: Theorem 3. Let a, b, c be natural number.s 8uch tha~ c ~ b ~ a. Then c - a ~ b - a ~ 0 and c - a > b - a ,if

> b.

. It is most important for us to keep the fact in mind that if a > c the difference c - a, as a natural number" dO~8 not exi8t. In Chap. II we shall so extend our number system that in the new system all differences b - a will exist. Remember that c - a'means a number b such that c = a + b. 8. Division. Let a and b be natural numbers. Then we say that b divides a if there exists a natural number c such that

'c

(6)

a :;= bc.

SEc. 9]

POWERS

11

When this oocurs, we call b either a factor or a divisor of a and say that a has the factorization a = bc. Then c is also a factor of a. If b = 0, the only natural number a such that a- = Oc is a = O. Hence 0 does not divide any number but itself. On the other hand if a = 0, then a = bO for every b. Hence every natural number divides zero. It follows that the symbol a

'b' for what we call the quotient c of a = bc by b, has no meaning when b = 0 and a ¢' O. Moreover, when a = b = 0, this quotient represents all natural numbers. In the solution of a problem by division we want our answer both to exist and to be unique, and we do not satisfy one or the other of these requirements when b = O. This is what is meant by the statement that division by zero is impossible. The quotient c is unique when b ¢' O. For the cancellation law for multiplication states that if b ¢' 0 and a = bc = bd then c = d. However c may not exist in the set of natural numbers. - For example, take a ~ 1 and b = 2. This will be our motivation in the extension we shall make in Chap. III of our set of integers to a number system including all quotients and called the set of all rational numbers or all fractions. 9. Powers. The formal distinction between the products ah and ba is in the order of the two factors a and b. The distinction between a(bc) and (ab)c is in the grouping of the factors. The commutative and associative laws may be combined and extended to products with any number of factors and the combined extension -reads as follows: Theorem 4. T,he value of any product is independent both of order and of grouping. We see similarly that the value of any sum is independent both of order and of grouping. If sums and products are both involved, the value depends upon the reading of the expres-

12

[CHAP. 1

sion as a sum. of products (the products might involve sums!) and the use of the distributive law.' . In a product of n factors all equal to a the independence of grouping implies that there is a common value for all such products. We call this value the nth power of a and designate it by an (read "a to the nth"). Here n is any positive integer, and we call n the exponent of the power an. For example, we are stating that the common value of (00) (00) == a[a(oo)] == a[(oo)a] == [a(oo)] a is a4• Every such power of 0 is O. We shall not need to consider powers of zero further, since if even one factor of a product is zero the product is zero. Henceforth let the products of this section be products of positive integers. We extend our definition of powers by defining (not by proving) (7)

aO

== 1

(a

;;&!O

0).

Then we have defined an for every natural number n and all positive integers a. A product be of b· == and e == am is a product of m + n factors all equal to a. By Theorem 4 we have the first of our laws of exponents given simply as

or

(8)

anam == an+tA.

We ~y translate this formula into words by saying that the exponent oj a product oj pOW6r8 oj the 8ame number is the wm oj the e:z;ponenta oj the Jactor8. A product (fA of m factors all equal to an is a product of n + n"+ • • • + n (read "n plus n plus and so on plus n") factors a. There are m terms in this sum and hence its value is nm. Then (9)

(an)m == a-.

Thus, to Jorm the mth power oj the nth power oj a we multipl1/ the exponenta m and n.

BlDC.

10]

-

SUMMARY

13

We notice finally that a"'b'" is a product with n factors a and the same number of factors b. The product (ab)'" has exactly the same factors as a"'b'" but with a different order and grouping. By Theorem 4 we have the third la}V of exponents (10)

a"'b'" == (ab)"'.

In words, the nth power oj a product i8 the product oj the nth power8 oj the Jactor8. This law has been derived here for all natural numbers n. EXBRClSES

1. The following products may be expressed as products of powers of 2, 3, 0, 7 by factoring and the use of the laws of exponents. Give the e",pressions. (a) 2" 5' • 4 • (14)' . (35) (b) (10)4. 8' • 91 (e) (28 • 31 • 4' . 5' . 71)0 (d) (2" 3' . 5 • 71)1 • [2 . 9 • (70)]8 (6) 3· 5 • 7 • 9 • (49) . (25) • 8 • 6 . (24) • (21) • (28) • (105) I. Use formula (9) and find a positive integer z such that (a) 2'3851 = Zl (b) 2'3851& = zI (e) 16(2'3')' = z4 (d) 2'3'(zl)' '= 3(Z8)' (e) 8' 38(y',8)' = z', y' = 71, .' = 25

10. Summary. In the sections just completed we defined the concept of natural numbers and told how the operations of addition, multiplication, subtraction, and division on natural numbers are performed. We also presented the fundamental properties of our number system consisting of all natural numbers as nine laws of addition and multiplication, and derived three consequent laws of exponents. We showed that this number system was ordered and gave the principle of mathematical induction and two consequences. Finally we defined the terms difference and Jactor (or diviaor) and observed that subtraction and division of natural numbers were not always possible. No problem 'Solving techniques were given in these sections except for the bit of drill of the exercise just

completed on the laws of exponents and the use of the' distributive . law. We shall give some drill later on the removal of parentheses, brackets, and braces by the use of our fundamental laws. ORAL REVIEW EXERCISES

1. State the principle of mathematical induction and its two sub.. stitutes. 2. Let K be a set of even integers. From what two hypotheses 'Would it follow, by the principle of mathematical induction, that K is the set of all even integers? S. From what two hypotheses would it follow that a set of odd integers is the set of all odd integers? 4. State the four laws of addition. 6. StQ.te the four laws of multiplication. 6. Give all the possible groupings in a product abed.

11. Finite sequences. The remainder of this chapter will be devoted to topics .involving nothing but counting, and so requiring the use of no numbers but natural numbers. The concepts presented in the present section on counting are of major importance in mathematics. The procedure used in counting the objects of a set of five objects may be described in terms of a J;lotation that has many uses in our subject. We begin by selecting a first object and represent it by the symbol al (read" a sub, one" or simply "a one"). Then we might think of al as a name for the first object. We name a second object a2 and then label a third as, a fourth a" and the remaining oneaG. We shall then have exh~usted the set, that is, used up all its elements (i.e., objects) and so shall have completed the count. We call this counting process that of setting up a correspondence between the five objects and the integers 1, 2, 3, 4,5. The correspondence is one to one, that is, each object has been made to correspond to precisely one positive integer, each of the five positive integers to only o~' object. The notation just used may be employed when we have a set of n o!>jects. Here n is any unspecified positive integer.

Bl!IC.

11J

FINITE SEQUENOES

15

When we do this, the set is labeled so that the names of its objects are (11) (read It aI, aI, to a. "). Here al is the first object, as is the second object, an is the last or nth object. The three dots represent objects which are n~mbered but which cannot be enumerated (i.e., written out) because 11. is umpecified. A line of symbols like formula (11) is usually called a finite sequence (or simply a sequence) and the symbols aI, a.~ and so on, its terms. We could use the letter b instead of a and so em,ploy the notation bl , bl ; • • • , btl or even bl ,' bs, • • • ,b... Of course other letters can also be used. When we wish to refer to an arbitrary term of a sequence (11), we shall speak of the ith term. This is the term as where i is a symbol representing anyone of the integers 1, 2, .•. ,11.. It is also called the general term of the sequence. The letter i need not be used and we might speak of the jth term ai or the kth term all. It is important for us to realize that, althou,gh as is anything, we are using i to represent a counting SUbscript, that is, a positive integer. Symbols that have the connotation of integers are usually limited to i, j, k, m, 11., p, q, r, s, or t. Other letters may be used but x, y, s are rarely employed for this purpose. We shall sometimes refer to the terms of a sequence aI, as, • . • , an as the as. For example; we might wish to' indicate that the terms ale all integers. Then we shall say that the as are aU integers. Later on we shall say that the as 'have a certain property from a certain place on. ThEm we shall mean that there is a positive integer k such that every as' has the prescribed property for all integers i ~ k. , Our sequence notation has the term as in it. However we must agree that if we specify later that 11. = 1 the sequence collapses to its first term al = an. In the theory of polynomials w:hich we treat l8.ter on it will 'be 'desirable to consider sequences that begin with the

16

NATURAL NUMBERS

[CHAP. I

syinbol ao rather than al. Then the notation for such a sequence will be (12)

ao, at, •.. ,

term:

an.

Here ao is the first al is the second term, CIt is the (i + 1)st term, an is. the (n + 1)st term. The sequence has n 1 terms. There is also no reason why we could not write

+

(13)

to represent a sequence in which n ~ 5 and the number of terms is not n but n - 4. For example, we might wish to speak of th~ sequence consisting of all those terms in formula (11) which appear after Th~ notation (14)

CIt

a,. "

(i = 1, . . . , n),

(read "CIt for i equals one to n") may be used instead of ai, a2, • • • , an, and the notation CIt

instead of ao, ai, ..• (15)

,an.

(i = 0, 1, . . . , n),

We could then use the notation CIt (i = 5, ••• ,n)

for the sequence as, as, • • • ,an. Sequence notations are used in many" connections. One example is that of the notation which could be employed for the digits of an n-digit positive integer a. We write a "= (ai, a2, ••• , an). Then al is the extreme left-hand digit, a2 is the next digit, and so on. In this case every CIt is an integer taken from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and al is not zero. In the case abpve a is a symbol for the whole sequence of n objects, and we have indicated this by putting parentheses around the sequence. We sometimes call such an object an n-tuple. If n = 2, it is a pair of objects; it is a triple of objects when n = 3.

smc. 12]

INFINITE SEQUENCES

17

Exercises designed to test the student's underStanding of the sequence notation are given at the close of the next section. 12. Infinite sequences. It may be desirable to consider sequences that do not terminate. Such sequences are called infinite 86quence8. One such sequence is the sequence of all natural numbers given in. formula (1). Infinite sequences represent sets of objects which may not be counted in the sense of a terminating counting process, but which are countable in that they may be put into one-to-one correspondence with the set of all positive integers. Such sets are called denumerable and their objects form infinite sequences (16) al, all .•• (read "al, as, and so on"). We shall usually designate the general term of an infinite sequence by tin (rather than ~). As before we could use hlJ hI ••• , or ao, al . . . , or (17) tin (11, = 1, 2 •.. ) (read "11, equals one, two, and so on"), or finally (18) tin (11, = 0, 1 . . . ). Sequence notations are used so frequently in algebra that it is of great importance that students be perfectly at home in their use. The following exercises have been constructed to give the student· drill work which should result in the desired facility. ORAL EXERCISES

1. Wha.t is n a.il.d wha.t are al, az • • . , tit. for the following finite sequences: (a) 1, 3, 5, 7, 9, 11 (b) 1, 2, 2, 3, 3, 4, 5, 6 (c) '3, 3, 3, 3, 3, 3, 3 (d) 1, 2, 1, 2, 1, 2, 1, 2, 1, 2 I. Give a formula for as for the following infinite sequences: (a) 1, 2, 3, 4, 5 •.. (the sequence of positive integers) (b) 2, 4, 6, 8, 10 .•. (the sequence of positive even integers) (c) 0,2, 4, 6 . . . (the sequence of even natural numbers) (d) 1, 6, 11, 16' •.. (the sequence whose members differ by 5).

18

N ATlJRAL NTiTM:6~'Ri;S

(c.Hap. I

EXERCISES

1. Enumerate the values of a1, a2, aa, a4, ai; in the sequences whose' general terms are given by the following formulas: (a) a,. = 2n (b) a .. = 3n {c) a,. = 3n (d) a.. n(n (e) an (2n

= =

+2 1 + 1) + 1)(3n -

+

(f) a" = £1,,_1 2n - 1 for n > 1, a1 = 1 (g) a" = na.. _1 for' n > 1, a1 = 2 (h) a,.

1)

(i) a"

= =

a2 =

2an _1 a,,_1

2

(3) a,. = a1

for n

> 1, a1 =

2

+ a"_2 for n > 2, a1 =: 1,

+ a2 fo:r n > 2, a1 =

a2

=1

2. Let bo, b1 • • • be a sequence such that bo = a1, b" = a,,+l for each of the sequences of Exercise 1. Write out the corresponding formul9.!! ~or b... 3. Write out the fifth, sixth, seventh, eighth, and ninth terms of a sequence b1, b2 • • • formed from each of the sequences of Exercise 1 by the use of the formulas b1 = a1, b.. = 2a" - a..-1 for every n > 1. 4. The first six terms of a sequence are given below. FInd a formula for a,. in terms of n which will fit these terms. (a) (b) (c) (d)

1,4, 7, 10, 13, 16 5, 9,13, 17, 21, 25 2' 1, 3 . 2, 4 . 3, 5 . 4, 6 . 5 0, :3 . 1, 4 . 2, 5 . 3, 6 . ~, 7'5

(e) 2' 1, 4· 3, 6· 5, 8· 7, 10· 9

12' 11 (f) 6, 6, 6, 6, 6, 6 (g) 1, 3, 5, 7, 9, 11 (h) 3, 5, 7, 9, 11, 13

5. Use the formulas obtained in each part of Exercise 4 and find tlie values of a10 and the formula for aH1. 6. The terms of a finite sequence of six terms are 1, 2, 5, 3, 4, 2. Define a new sequence b1, b2, • • • , ba by b1 = a1, b.. = a" rb.._1 for n > 1. Find the terms of the sequence of b" if

+

(a) r = 1 (b) r = 2

(c) r

(d) r

=4 = 10

(e) r = 0 (f) r = 5

13. Counting of pairs. The sequence notation may be used in counting the number of pair8 of objects if the first ,object in the pair may be anyone of n objects and the second anyone of m objects. Designat,e the objects of the first set by aI, a2, • . . ,an and $.ose of the second set b~ bl, b2 , • • • ,bm • Then the pairs are Ca., bi ) where i h~'S any value from 1 to nand j any value from 1 to m. There are then 11,1, 8et8 of pairs and each consists of the n pairs

SJDC.

13J

.

COUNTING OF PAIRS

19

(al, bi), (a2, bi), •.• , (an, bi)' Hence there are n~ pairs altogether. In a special case we may enumerate the pairs (a, b). Let us do so in the case where a may be anyone of the integers 1, 2, 3, 4, 5 and b anyone of the integel'B 6, 7, 8. Then the pairs occur in three sets, each 'of which consists of five pairs. The first set consists of the pairs (1, 6), (2, 6), (3,6), (4,6), (5,6), the second set of (1,7), (2,7), (3,7), (4, 7), (5, 7), and the last of (1, 8), (2, 8), (3, 8), (4,8), (5, 8). There are 15 == 5 . 3 pairs all told, and we have liste!i them. The result above is used in ordinary experience in counting pairs of events. We state it then in the following form: FUNDAMENTAL COUNTING PRINCIPLE. Let it be p08Bible Jor an event to take place in anyone oj n way8 'and, aJter it take8 place, Jdr a.8econd event to take place in anyone of m waY8. Then the number oj waY8 in which the sequence oj two events can take place is nm. The result may be extended to triples of events, quadruples of events, etc. In every case the number of ways in which the sequence of events can occur is the product of the numbers of ways in which each entry can occur. It cannot be overemphasized that the way we do any .job of counting depends upon what we are counting. If we have two mutually exclusive sets of events and we are merely interested in counting these events, we add the numbers of events in each set. But, if we are counting pair8 of events, we multiply the numbers. In the special case above there are altogether 8 == 5· + 3 integers 1, 2; 3, 4, 5, 6, 7, 8. But there are 15 == 5 . 3 pairs. Some of the examples given below will combine both types of co.unting. mustrative ExamPles I. A man living in Chicago can go to Minneapolis by plane, train, or bus, but can return only by train or bus. He can also go to Milwaukee by train, boa.t, bus, automobile, or plane, and can return by train, automobile, or plane. How many round trips would it be possible for him to take?

20

NATURAL NUMBERS

[CHAP. I

Solution

We are counting the total number of round trips. By the fundamental counting principle there are 6 = 3 . 2 round trips possible to Minneapolis and back and there are 15 = 5 . 3 round trips to Milwaukee and back. • The total number of round trips is 21 ... 6 + 15. II. Ho:w many different trips would it be possible for the man in the example above to take? Here we are agreeing that a trip from Minneapolis to Chicago by train is not the same as one from Chicago to Minneapolis by train. Solution

In this problem we are counting trips, not round trips. = 13 trips.

There are

3+2+5+3

EXERCISES

1. The first' digit of an ,n-digit natural number a may not be zero unless n = 1 and a = O. Use this to find the number of (a) Four-digit numbers (b) Four-digit even numbers (i.e., final digit 0,2,4,6,8) (0) Even numbers with three, two, or one digit 2. A license plate is to have on it two letters followed by a four-digit number, the first digit being not zero. How many plates with different labels can be made? Am. 6,084,000. 8. How many four-letter code words can be made if the first two letters are any consonants and the . last two any vowels? (Consider a, e, i, 0, v, and 1/ as vowels.) " How many five-letter code words can be made if the first, and last letters are any one of the letters b, 0, d, p, q, r, 8, t, the second and fourth letters any vowels, and the middle letter anyone of the letters g, h, j, Te, l, tn. Am. 13,824. 6. A collection ~f pictures contains five nineteenth-century portraits, eight'landscapes, and six modern paintings. It is desired to make an exhibit consisting of one painting of each kind. How many exhibits differing in at least one painting can be selected? 6. A student is taking English, chemistry, and mathematics and can get anyone of the grades A, B, C, D, F in each subject. How many different sets of three grades could he get? Am. 125. 7. A signal device consists of four rows of lights each consisting of a red light, an amber light, and a green light. A si.gna.l consists of. a light in the top row, or a light in each of the top two rows, or a light in each of the top three rows, or one in all four rows. How many different signals can be made ~th this, system?

SEC.

21

FACTORIALS

14J

8. A company comma.nder has under his comma.nd five platoons each of nine men. He details one man out of each platoon for special A.na: 9&. duty. How many different groups can he select? 9. A woman ha.s 15 different handkerchiefs, 8 different hats, and 4 different pairs of gloves. She wishes to select one of each for a given day. How ~y different selections could she make? 10. How many two-digit numbers can be made from the digits 1, 2, 3, 4,7,8 if the two digits must be different? A.na.30. 11. Two men cross a river separately. The first man can use any one of five different boats and the second man anyone of those remaining. How many different choices of boats can the two men make?

14. Factorials. It is useful to have a symbol to indicate the product of the first n positive integers. We therefore define (19) nl = n(n - l)(n - 2) • • . 3' 2 . 1 and read the symbol nl as "nJactoriaZ." Then we have the values . 1!=1, 21=2, 81=6, 41=24, 51=120, and so forth. Some authors use the symbol ~ rather then' nl. It may also be desirable to introduce the convention that zero factorial be defined to have the value 1. Since nl is the product of the first n integers, it should be' clear that if r is a positive integer less than n the product rl of ~he first r integers is a factor of nl. Indeed we have (20)

nl == n(n - 1) .•. (r

+ 1) • rl.

Then the number n(n - 1) ... (r + 1) is the product of the n - r Zarge8t displayed factors of n I. The difference n - r is also less than n and we may replace r by n - r in the formula above to obtain (21) nl = n(n - 1) . . . (n - r

+ 1) . (n -

r)!.

We shall use the symbolP",r (readP, n, r) for the product of the factors to the left of (n - r) 1in formula. (21) and hence define the integer (22)

n!

P",r == (n _ r)l = n(n - 1) ••• (n - r

+ 1).

22

NATURAL NUMBERS

It is the product of the r integers n, n - 1, n - 2, . . . , 'n - (r - 1), and we emphasize that the last 9f these factors

isn-r+1. We have now seen that n! is divisible both by rt and by' (n ....,. r)! for any r < n. Indeed n! is divisible by thejr product, a result we shall derive in the next section. Let us introduce the, notation

c

(23)

_

n,r ~

nl r!(n - r)1

for the integral quotient. If we replace r by n - r in Cn,r we obtain (fn,n-r' becomes (n - r)! and (n - r) I becomes 8! where 8

=

Then r

r

n - (n - r) = r.

Thus, (24)

for all values of n > 1 and r (25)

P n,r

Cn,r

=-, r. =

< n, The formula

n(n - 1) . . . (n - r r.I

+ 1)

is an immediate consequence- of our definitioI!l.S. The numerator is the product of the r integers obtained by beginning with n and diminishing by 1 a total of r - 1 ti~es. It should be compiled by the simple process of , enumerating these r factors, Similarly the denominator is the product of r factors obtained by beginning with rand diminishing by one r - 1 times. For example,

2 3

C11,6 =

11 . 10 . 9 . g . 7 . ~ ~

•g. 2

. is"

=

11 . 7 . 6 = 462 = C11, D·

How then should we compute the 'quotient C80,78 of two products each with 78 factors? We simply use formula (24) to write C 80,78

=

C80,2

80' 79

= 2~

=

40' 79 = 3,160.

SEC.

14]

23,

FACTORIALS

I. Find 11. if C"' 3

nlustrative Examples . = 84.

Remarks: The problem requires that 11. be 11.(11. - 1)(11. - 2) _ 84 '

6.

-,

a natural number such that

11.(11. - 1)(11. - 2) = 504.

Then 11. 3 > 11.(11. - 1)(11. - 2) .> (11. - 2)3 and 11. is that natural number for which 504 lies between (11. - 2)3 and 11.3• Note that a problem such as this need not have a solution. For example, the problem of finding an 11. such that C... 3 = 85 has no solution. Solution

We list the cubes

33

=

43 = 64,

27,

53 = 125, 68 = 216, 88 = 512, 93 = 729 78 = 343,

The only pos81ole solutions of this problem are 11. = 8, 9 and we verify that .

CU• 3 = II. Find 11. if P ...9

9·8'7

~ =

3·4'7

=

84.

= 42P... 7•

Solution The integer P ... 9 has nine factors, the product of the first seven of which is P ... 7• Then P •• 9 = P ... 7(n - 7)(11. - 8). Here the final factor is 11. - 'I' + 1 where 'I' = 9. Then P ... 7(n - 7)(11. - 8) = 42P... 7 , 7)(11. - 8) = 42. The pro<Juct of two consecutive integers equal to 42 is 7' 6 and 11. - 7 = 7,11. = 14. III. Fmd 11. and 'I' if P ... r = 15,120. C... r = 126.

en -

Solution

P ... r = (rl)C... r so that '1'1

15,120 = ---r26 = 120 = 2 . 60 = 2 . 3 • 20 = 2 . 3 • 4 . 5,

= 5. Then P ... 5 = 11.(11. - 1)(11. - 2)(11. - 3)(11. - 4) = 15,120. A table of values of P ... & for 11. ~ 5 yields P s.& = 120; P a.& = 720; P 7•S = 2,520; P a.& = 6,720; P 9• S = 15,120. Hence 11. = 9, 'I' = 5. 'I'

ORAL EXERCISES

Express the ·first of the following numbers in terms of the second (using products or quotients): ·

24

NATURAL NUMBERS

L 81,61 2. 61,81 3. 101, 91 4. 101,81 6. 91, 61

6. 61, 71 7. P •. G, P •. 4

8. P ...., P".•- l 9. nl, P ..."-l

,10. p".s, 0.... & EXERCISES

1. Simplify the following by writing the result as an integral multiple of nl for n as large as possible: (a) (b) (0) (d)

81 - 4' 3' 91 8' 51 12' 81 -

(6) 3' 71 - 2 . 61 (f) 4· 71 - 6 . 51 (g)'3161 71 (h) (41)2 2 . 61

61 12' 81 61 7·61

+ +

Am. Am. Am. Am.

19·61 2,7' 61 13.' 61 84 '41

2. Compute (a) P&'8 (b) o&'& (0) P S.7 P •. 4 (d) 2P10.& P S•8 (6) PU.8 33P7•8 (f) 50s. 8 0S.4

Am.10·

Am. 1.

(g) 70s. 7 0 7•8 (h) 50s.6 3C10•4 (.1.) 91P12 .& P 10.&014.& (~ 80U.aPI.8 :J P 14.•

Am. 1.

Am. 10.

Am.6.

3. Find n if (a) (b) (0) (d) (6)

= 110 = 28 = 60 p".• = 1,716

P ... 2 0"'2 P ....

Am. 8. Am. 13.

0 ... 8 = 364

Am. 12. (f) 0";8 = 220 (g) P ".8 = 42P... 8 Am. 6. (h) P" .• = 120,,_1.& (i) 50".& = 80,,-1.4 W 180..- 1.4 = 15(J".• Am. 9.

4. Find r if (a) P19 .• (b) P17 .•

= 5,814

= 272

Am. 2.

(0) 017 .• = 12,376 (d) 40018 •• = 51016 ••

Am. 2.

6. Find'n and r if (a) P .... = 840, 0 .... = 35 (b) P .... = 11,880, a.... = 495 (0) rl = 6,P.... = 210 (d) nl. = 5,040, 0" .• = 35

Am. n = 7, r = 3. Am. n = 7, r = 3.

BEc.15]

PERMUTATIONS AND COMBINATIONS

25

6. Show that 0 .... + 0 ....- 1 .. 0,,+1 .•• 'I. Compute the values of the expression 21" + 1 for n = 1, 2, 3, 4. (These integers are connected with the number of sides in a regular polygon constructible with ruler and compass.) 8. If the legs of a right triangle have lengths a and b and if c is the length of its hypotenuse, the theorem of Pythagoras states that as + bl = cl • It is known that a.ll cases where a, b, c are positive relatively prime integers are given by the formulas

+

a ... 2mn, b = m' - nl, c ... ml nl, where m and n are positive integers;" one of them is even, and the other odd, m > n. Compute at least four difierent sets of values of a, b, c.

15. Permutations and combinations. The theory of permutations and combinations is concerned with methods of counting the number of ways in which a subset of objects of a given finite set may be selected, or selected and arranged, in a sequence. A combination of n objects r at a time is a 8election of r of the n objects. A permutation of n objects r at a time is an arranged 8election. It is important for the student to be alert to the fact that combination meam 8election (i.e., choice), permutation meam arranged 8election. The number of combinations of n objects n at a time is evidently precisely one. Then we may refer to a permutation of n objects n at a time as simply a permutation of this single choice of n objects. We shall show shortly that the number of permutations of n objects is nl. Let us now consider the simple example of three objects a, b, c. Their permutations are (a, b, c), (a, c, b), (b, a, c), (b, c, a), (c, a, b), (c, b, a). Thus there are 6 = 31 permutations. If we count permutations two at a time, we observe first that we may select a, b or a, c or b, c for our pairs. The corresponding permutations are (a, b), (b, a), (a, c), (c, a), (b, c), (p, b), and there are 3 . 2 = Pa,s such permutations. We proceed to count the number of permutatiom of n object8 r at a time. As first object in the sequence of r objects we may select anyone of the n objects". When it has been selected and used, only n - 1 objects are left to be chosen as a second object. By the fundamental counting principle there are n(n - 1) ways to fill the first two

26

NATURAL NUMBERS

[CHAP. I

places in our sequence. There remain n - 2 objects and so the third term may be selected ip n - 2 = n - (3 - 1) ways. There are then n(n --'- l)(n - 2) ways to select tlie first three terms of our sequence. Mter r such steps we arrive at a final factor of n - (r - 1) = n ~ r + 1 and SQ the number of permutations of n obfects r at a time is the integer P n,r of formula (22). In the special case r = n we have Pn,n =;: n!. Note that nand r are integers and that

o
~

n.

The number of combinations of n obfects r at a time may be computed by grouping the Pn,r permutations into subsets each of r! p'ermutations. We shall first consider the illumi-' nating special case of combinations of four objects three at a time. The selection of three out of four objects is achieved by the omissi0'Ylt of o'fl,e object. Hence there are four such selections. We list them systematically as [a, b, c], [a, b, d], [a, c, d], [b, c, d]. From each selection we obtain 3! = 6 permutations. They are (a, (a, (a, (b,

b, c), (a, c, b), (b, a, c), (b, c, a,) (c, a, b), (c, b, a), b, d), (a, d, b), (b, a, d), (b, d, a), (d, a, b), (d, b, a), c, d), (a, d, c), (c, a, d), (c, d, a), (d, a, c), Cd, c, a), c, d), (b, d, c), (c, b, d), (c, d, b), (d, b, c), (d, c, b) . •

Thus we have the total of 4 . 6 = 24 = P 4,3 permutations. We have now shown that from each selection of three objects from four we may obtain 3! arrangements. In the general case each selection of r from n objects will yield r! permutations. Hence the total number of permutations of n objects r at a time is the product by r! of the number of combinations of n objects r at a time. In formula (25) we defined a quotieRt On,r such that On,r • r! = P n,r' It follows that On,r is the number of combinations of n obfects r at a time. We shall apply these results in the following exerClses. Illustrative Examples I. How many numbers of four different digits can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9?

SEC.

15]

PERMUTATIONS AND COMBINATIONS

27

Remark: T~ problem asks us to count the number of digit arrangements. It is therefore a permutation problem. Am. P9,4 = 9 . 8 . 7 . 6' = 3,024. II. A group of 5 soldiers is to be chosen from a platoon of 15. In how many ways can the group be chosen? Remark: The counting involves selection only and the problem is one

7

..

..

mvolvmg combmatlOns.

Am. 0 16 .& =

3

liS . U . 13 . l2 . 11

is . I: . a . 2

= 1,q01.

III. A regular polygon has 23 vertices and, consequently, 23 sides. How many additional lines need to be drawn so that every pair of vertices will be connected? Remarks: The problem is again one of selection; It is then a problem on combinations. It also illustrates what may be called an exclusion principle. Instead of counting what is called for in the problem, we count all lines joining pairs of points and subtract the number of lines which are sides. 23' 22 Ans. 028.2 - 23 = - 2 - - 23 = 23(11 - 1) = 230. IV. How many odd numbers with four distinct digits can be formed .using the numbers 3, 4, 5, 6, 7, 9 as digits? Remarks: The problem involves arrangement and is a permutation problem. We shall give two solutions one of which uses the exclusion principle.

Solution 1 With the six digits it is possible to form PM four-digit numbers. Of these P&.8 end in 4 and the same amount in 6. They are to be omitted. Hence our answer is P 6.4

-

2P&.8

= (6' 5 . 4 . 3) - (2' 5 . 4 . 3) = (6 - 2)(5' 4 . 3) = 240. Solution 2

The final digit is permitted to be anyone of four numbers. Mter it has been selected, the remaining three digits are to be selected and arranged from the remain,ing five numbers. This can be done in P&.8 ways. By the fundamental principle our answer ~ 4· P&.8 = 240. V. How many word groups of five distinct letters can be made from the letters of the alphabet if at least two of the letters are vowels? Remarks: We shall use the military convention here (and in the exer..: cises) in which the letters a, e, i, 0, u, 11. are regarded as being vowels . . Observe that our problem may be regarded as really a collection of problems. For we may solve it by finding the number of words with two vowels, three vowels, four vowels, and five vowels. We shall give a simpler solution using the exclusion principle.

28

NATURAL NUMBERS

[CHAP. I

Solution The total number of five-letter word groups is P26.a. Of these P20.6 have no vowel. There are six vowels and five places to insert each vowel and P 20 •• choices of consonants in words with exactly one vowel Thus there are 30P 20 •• such words. The answer is 1'16 •• -

P 20•• - 3OP 20•4 = (26' 25 . 24 . 23 • 22) - (20 ·19·18· 17 . 16) - (30' 20 . 19 '18' 17) = (26' 25 . 24 . 23 . 22) - (46 '. 20 • 19 . 18 '17) = 1,443,720. EXERCISES

1. How many arrangements are there of the

~etters

of the words

(d) quadric

(a) equations (b) logarithms (c) number

(e) cleari!l.g (f) mile ."

. S:-How many ways are there of arranging the books on a shelf if the shelf holds (a) 12 books (b) 20 books

(c) 11 books (d) 15 books

3. A license-plate designation is to be a number of six distinct digits with the digit zero not allowed. How many different plates can be formed? . 4. How many different plates can be formed if the designation is to consist of two distinct letters followed' by a number of three distinct nonzero digits? Am. 327,600. 6. How many plates can be formed if the designation consists of three distinct letters followed by (a) a number formed of three arbitrary but distinct digits, (b) a number of two distinct nonzero digits? 6. Solve Exercises 3, 4, 5b in the case where zero is permitted to he a digit. Am. (3) 151,200; (4) 468,000; (5b) 1,404,000. 7. Solve Ex~rcises 3,4,5 in the case where zero may be a digit but not· the first one. S. How many word groups of six distinct lettei-s can be formed in which the first, third, and fifth letters are consonants and the remaining letters vowels? Am. 820,800 9. How many word groups of seven distinct letters can be formed if the second, fourth, and sixth letters are to be vowels and the remaining letters consonants? 10. How many numbers having distinct digits can be formed, of five odd digits followed by either one or two even digits? Am. 3,000.

smc.15]

PERMU,TATIONS AND COMBINATIONS

29

11. How many different pairs of tennis players can be selected from a group of 11 players? From a group of 8 players? lS. In how many ways can a ta.sk force of 5 wa.rships be chosen from a :fleet of 20 warships? Ana. 15,504. 18. In how many ways can a class of 30 students choose a committee of 4 students? Of 6 students? Of 12 students? 14. A. deck of ordinary' pla.ying cards consists of four suits. The suits are spades, hearts, diamonds, and clubs, and each suit consists of ca.rds designated by a.ce, king, queen, ja.ck, 10, 9, 8, 7, 6, 5, 4, 3, 2. Two ca.rds numbered a.like are said to have the same rank. (When your answers to the problems below are la.rge give them in terms of the a.ppropriate symbols P .... and a..... ) (a) In a solitaire game we deal, fa.ce up, seven piles of cards with five cards in ea.ch pile. How many different layouts are possible? (b) A. ca.sino deal consists of four ca.rds to ea.ch of two opponents a.nd four ca.rds fa.ce up on the table. How many different opening deals are possible? Ana. all., . (c) A. gin-rummy deal consists of 11 ca.rds to the dealer's opponent and 10 cards to the dealer. How many different opening deals are possible? (d) A. poker hand consists of five cards. How many different poker hands are there? (6) A. poker hand contains a pair if it contains two cards of the same rank. How many hands are there with at lea.st one pair? , AM. all,&-"a~ •. & .... 1,281,072. (f) A. flush is a poker hand consisting of ca.rds in the same suit. How many different flushes can be formed? (Flushes differing only in suit are to be regarded as different.) , (g) A. full house is a poker hand consisting of a set of three ca.rds of the same rank and a pair of another rank. How many different full houses are there (if a difference in suit only is to be counted as a different full house)? AM. 3,744. (h) A. bridge deal consists of four hands of 13 ca.rds ea.ch. The arrangement of ca.rds in each hand is unimportant but the order of the hands including the question as to which hand is given to the dealer is important. How many different deals are possible? (t') .How many bridge ~ are there in which at least one hand co~ of 13 cards a.1l in the same suit. Ana. 16 • a.8.18 • aH.1 •• , j W How many bridge deals are there in whioh the dealer's hand contains a.ll the a.ces, kings, and queens? ' 16. A group of books consists of five mathematics texts, seven physics texts, and three chemistry texts. A. selection of nine texts conta.ining at least one chemistry text and three mathematics texts is- to be placed on a shelf. How many different arrangements of texts on this shelf are possible? Ana. (3,360)(91).

aa .• . a".,.

,16. How many word groups of six distinct letters can be formed if Itt. least two and at most four of the letters are to be vowels? 1'1. How many numbers of seven distinct digits can be formed if the fir~t digit and at least three of the other digits are to be odd? Am. 180,000.

18. A flag signal is to consist of at least two flags. The color but not the order of the flags is important. How many different signals can' 'be made if seven flags of different colors are available? 1$. Two points determiti.e a line. How many lines can b.e drawn Am. (1",2, through n points no three of which are in the same line? 20.' A plane is determined by three points., How many planes can be passed through 11 points, no 4 of them coplanar? 21. A regular polY!l;on has 17 vertices joined by lines. How many additional lines need to be drawn so that every pair of vertices will be Am. 119. connected? 22. A man has eight coins of different denominations. JIe pays out an amount of money consisting of not less than two nor more than six of the coins. How many diffe,rent amounts is it possible for him to payout with the given coins? 23. A set of exercises consists of three groups each of seven exercises. An assignment is to be made consisting of six exercises with at le8ist one exercise from each of the first two groups and at least two from the final group. How many different assignments are possible? Am. 31,556.

16. Cyclic and circular permutations (FULL COURSE). Up to now we have been counting the number of linear (i.e., in a line or sequential) arrangements of letters. We shall now discuss the number of arrangements in a circle where only the relative position of the elements is important. 'The ordering a2, as, . . . , an, al of the terms of a sequence aI, a2, • . • , an i& said to be obtained from the original (la~ter) sequence by a cyclic permutation. A cyclic permutation of the Iormer sequence will result in as, a4, •.• ; a..., aI, ~2, and after n such steps we shall retUl:n to the origi~' nal order. Thus we can derive n cyclic permutations fromany given permutation of n objects. ' Let us now arrange the n objects of a permutatfon as n equally spaced points on a circle. Then the n cyclic permuta:tJi~ns of this arrangement will not be distinguishable and so will give the same circular permutation. Clearly

SEc,17]

D ISTINGUIS HAB LE PERMUTAT IONS

31

any m distinct circular permutations yield nm distinct linear permutations. It follows immediately that the number of distinct circular permutations of n:objects is P,. = (n - 1)!. n

(26)

It should also be clear that each circular permutation of n objects r at a time will yield r linear permutations each obtained by cyclic permutation of anyone of them. Hence the number of circular permutations of n objects r at a time is (27)

P,..r

-= r

n(n - 1)

(n - r r

+ 1)

n!

= r[(n - r)]!'

EXERCISES

1. How many seating arrangements are possible of nine people about a round table? 2. In how many ways is it possible to arrange five keys on a key ring if the keys are to be selected from a group of seven keys? Ana. 504. , 3. A group of four women is to be seated around a table and a group of six men around a second table. How many arrangements are possible? • 4. A group of six women and six men are to be seated about a round table so that women and men sit alternately. How many arrangements are possible? Ana. 86,400. 5. How many circular permutations ,are there of four letters taken , one, two, three, and four at a time? 6. How many circular permutations are there of seven letters taken five or six at a time? Ana. 1,344. 7. A game wheel is to consist of three relatively fixed rings, each colored in seven differently' colored sections. How many different color arrangements are possible?

17. Distinguishable permutations (FULL COURSE). The study of distinguishable permutations is best begun by a focus of attention on r objects in a sequence of n objects. Let us count the number of permutations derived if we permute these r objects and leave all others fixed. Since the fixed objects might just as well not appear insofar as they affect our count, there are r I such permutations.

32

NATURAL NUMBERS

[CHAP. I

It follows that the n! permutat~ons of n objects n at 'a time may be subdivided into (28)

n' ri =.n(n - 1) . . . (r

+ 1)

groups, of permutations. Each group is such that every permutation in anyone group is obtainable from every other one in the same group by permutating the same set of r prescribed objects,; two permutations in different groups are not obtainable one from the other without permuting some of the remaining n - r objects. Suppose now that the r fixed objects were indistinguishable. Then all the permutations in each of the groups described above cannot be told apart and must be counted as a single permutation. It follows that the number of permutations of n objects (n at a time) of which r are indistinguishable is n(n - 1) . . . (r + 1). Let us next focus our attention on s of the remaining n - r objects. If we permute these s objects and the original r objects; we shall obtain a group of rlsl permutations from eaoh permutation of n objects.· It follows that there Win be n! (29) rlsl such groups. Hence the number of distinguishable permutations of n objects of which r of one kind are alike and s of another kind are alike is given by the quotient in formula (29). In the particular case where s = n - r, the value of formula (29) is C""r and we have shown that the number of distinguishable permutations of n = r + s objects of which r are alike and the remaining s alike is C""r = C",,8' Suppose finally that we write n as a sum (30)

n

of positive integers. of n objects into

= nl +

... + n, Then we may group the permutations

SEC.17]

DISTINGUISHA.BLE PERMUTA.TIONS

33

nl

(31)

sets of permutations and, if the objects in each set of 1I.i objects are alike, formula (31) gives a formula for the number of distinguishable permutations. The principal importance of this formula is that it shows that if we partition n in any way as a sum [formula (30)] the number nl will be divisible by the product nl !n21 . • • n, I. Let us observe, in closing, that there is no formula for the number of distinguishable permutations of n objects r at a time if some of the objects are alike. nlustrative Examples I. Verify that 81 is divisible by 412!21 88 well as 21212121 and 5131. Why is it automatically true that 81 is divisible by 31212111?

Solution 81 8' 7' 6' 5 41~121 = 2.2 = 2' 7 . 6' 5 - 420, 81 8 .7.6.5.4.3.2 212 . 2 . 2 == 8.2 == 7 • 6 • 5 • 4 • 3 == 2,520, 81 5131

8' 7' 6

= ---s:2

== 56.

Sinpe a = 31212111 is a factor of b = 4!2121 and b divides 81, so does a. II. How many distinguishable linear arrangements are there of a group of five red balls, three green balls, and two white balls? Solution 101 513121

=

10' 9 . 8 . 7 . 6 3.2.2

= 10'9·4'7 -

2,520.

BDltCISES

1. How many distinguishable arrangements can be made of the letters of the following words: (a) Illinois (b) Massachusetts (c) Tennessee

(h) distinguishable (i) arrangement

(g) diminish

(d) Mississippi (6) Indiana (J) California

W syzygy (1:) discriminant (l) mathematical

34

• N .A. T'1!T It' A L N'IT M B E'R' S

2. Verify py direct computation that the following quotienii13 84'e integers: 91

(a)

7!21 111

~b) 313141 111 (c) 514121

27! (d) 9141141

271 (e) 81818!31

1301 (I) 101101101

3, .A. group of eleven red books, six green books, five orange books, and one black book are to be arranged in a fine. How maI)Y, di~ tinguishable color combinations are possibie? Am. 23 . 21 ·19·17·16·7·13·11 ·3.

CHAPTER

n

INTEGERS

1. The domain of all integers. The line of natural numbers may be extended by adjoining the symbols -n for all positive integers n. Define -0 = 0 an~ we have a line of symbols . (1) " ' , -7, -6, -5, -4, -3, -2, -1,0, 1,2,3,4, . 5, 6,'7, . . . running out fudefinitely in both directions and containing -n for every natural number n. We shall call the new symbols -1, - 2 . . . the negative int6(l6f'8 and shall speak of all the symboJ.s in formula (1) as int6(l6f'8. Then we shall indicate that an integer a is a positive !nteger by writing a > 0, that a is a natural number by writing a ~ 0, and that a is a negative integer by writing a < O. .' In the next section we shall give the formal definitions of addition and multiplication of integers which results in the processes for these- operations used in high-school algebra. The result will be a number system, the do~n of all integers. 2. Addition and multiplication. Our set of integers consists partly of the set of natural numbers and we shall continue the definitions of sum and product for these integers which .we gave in Chap. I. Suppose then that a' < 0, b < 0 so that a == - c, b = - d where c > 0, d > O. Then we define ab = (-cH -d) to be cd. If a ~ 0 ~nd b < 0, we, write b = - d and define ab = a( - d) to be - ad. Finally, if a < 0 and b ~ 0, we have defined ba. Define ab = 00. ' We define the sum a + b == (-c) + (-d) of any two :p.egative integers a and b to be' -(c + d). If a ~ 0 and b < 0, we write b == - d where d >. 0 and define a + b == a + (-d) to be a - d, in case a ~' d, a + b to be ~

83

36

INTEGERS

fClUP. n

- (d - a) in case d > a. The only remaining case is that where a « 0 and b ~ O. In this case b + a is already defined, and we define a + b to be b + a. As in the case of natural numbers we say that an integer c is greater than a if c == a + b where b is a positive integer. Then c will be to the right of a in the line (1). Principle 3 in the FULL-COURSE Sec. 6 of Chap. I may now be generalized. , Principle 4. EVtII"Jj 8et oj integer8 all oj which are greater . than 80me fixed integer a contai'f1.8 a least integer. ' Principle 6. . Every set of integer8 all oj which, are les8 than 80me,fixed integer a contains a greate8t integer. We shall not try to show how these two principles are de~ved from that of Sec. 6 of Chap.!. S. The law of subtraction. It can be verified that all the nine laws of addition and multiplication are properties' of the domain. of all integers. An additional property is given as the X. LAW OF SUBTRACTION. 1J a and c are any number8, there is a unique number b which we call the cWference oj c and a and write b==c-a 8'Uch that c == a + b. The definition of addition implies that if n is any natural number the sum n + (-n) 0= O. If a < 0 and we write a == -n, we have a + n == O. We may call the number g such that a + g == 0 the mgative of a and have - ( -'7 n) ....; n. Then it may be verified that in all cases (2) -(-a) == a, c - a == c + (-a). We shall also state without verification the properties a(-b) == (-a)b == -Cab), (':"'a)(-b) == ab, -"(a + b) == (-I)(a + b) == -a - b == (-a) + '( -b), -(a - b) == b - a, -(-a - b) == a + b, . -(-a + b) == a - b, (-a)- == a-, (3) , (-a)ln+l= -(a--H),

for all integers a and b and ·positive integers n.'

37

ABSOLUTE VALUES

SEC. 4]

The processes of addition, subtraction, and multiplication will be referred to henceforth as the integral operati0n8. All the symbols used in this text and combined by integral operations will be ~sumed to o~ey the 10 laws we have set down. They will also obey the laws for nonnegative integral exponents of Sec. 9 of Chap. I as well as formulas (2) and (3). . ORAL BDRCISES

1. Compute (-1)&, (-1)8, (-2)8, (-2)'. a. Simplify the following expressions: (a) -2(z - 211) (b) 5z - [-2(71 - 2z)] (c) (-z + 1)(z 1) (d) 1 - (-2)(5 - 3) (e) 1 - 1 1 - 1 + 1 (f) 1 - (1 + 1) - (1

+ + (-z) (z)( -2)

+

3 - 1 1) - 1

+

BDRCISES

.

1. Use the method of the Oral ltxercises to simplify the following , expressIOns : (a) -( -[7(4 - 6 (b) 2[z (c) [3z -

+ 1)] -

[(5

+ 2)(-3 + 1)

+ (3 + 2)(-3 - 1)(-2) + (-3)(-4 + 8)]} (3z - 71)) + 3[ -2(z + 371) + 4(z - 71) + (-2)(2z - By)) (z + 271)2 -. (271 + z) + (-2)( -3y)][z(y + 2z)

+ zy(1 - 2) - zl) (d) (z + 71)1 - (z - 71)1 - (2z)(2y) . (e) -2[Z(ZI - 3z) + (z + 1)(z - 1)] 3z[(z 1)(z - 2) + 1] (f) (z 71)( -1 z)[z Zl - (zy + 71)] - ZI(ZI - 711) , a. Write the following expressions as the difference of the first term. minus a parenthesis enclosing the remaining terms:

+

+

+

+

(a) Zl - 718 - 2z + 371 (b) -a + 2b + 3c - d (c) 2z - 2(71 + ,) + Zl.

+

+

(d) 4z1 - Z - 71 2zy (e) z 2(71' - ,) (f) 2z - (-71 -,) - Zl

+

4. Absolute values. The ab80lute value of any natural number a is defined to be a itself. H a < 0 we define its absolute value to be the positive integer -a. Then the absolute value of a, designated by

lal

38

INTEGERS

[CHAP. IX

is a positive integer unless a = 0, and in this case We shall st,ate without proof the properties

101 = O.

la + bl ~ lal + Ibl· We also define the int~ger c = lal - Ibl and have (5) lei ~ la - bl. (4)

labl = lal Ibl,

OlUL EXERCISES

1. Find 1a:1, if (a) a: = (-1)(-1)(-2)(-3) (b) a: = (-1)(-1)(-3)(-4)

2. Compute

=3

- (4

= (-3)8

+ 6)

la :+ bl~ lal + Ibl in the following cases:

= 7, b = (b) a = 7, b == (c) a = -7, b (d) a = -7, b (a) a

(c) a: (d) a:

(e) a = b = 7 (f) a = b = -7 (g) a = 7 = -b (h) a = -7 = ~b

3

-3 =3 = -3

5. Divisibility. .As in the case of natural numbers we say that an integer b divides an integer a if a = bq where q is Btn integer. Then if b ~ 0 t~e quotient , q =

a

b

is a unique integer.' We !,!hall call b afaetor or divisor of a. If a = bq then a == (-b)( -q). Hence, if. b is a factor of a, so is -'- b. It follows that with every factorization 'of a as It product of integers there are associated other factorizations in which we insert an even number of minus signs. 'Also each factorization a = bq of a results in a factorization -a = b( -q) of -a. It follows that in order to give a complete theory of the factorizations of all integers it is sufficient to give such a theory of the factorizations of posi-' tifJe integers as products of positive factors. Eve:vy positive integer a has the trivial factorization q = a . 1 and we shall call a and 1 the trivial factors of a. An integer p > 1 will be called a prime if it has no nontrivial factors. An integer a > 1 will be called composite if it is not a prime. We now prove the following:

SlIIC.

6]

39

THE DIVISION ALGORITHM

)

Lemma 1. EVfJry nontrivial Jactor b > 0 oj a poBiti1J6 integer a iB less than a. . For if a = bqwhere b ~ a,.b ~ I we have q.~ l~q = 1+ d where d > o. Thena = b(l + rI) = b + bdwhere btl > 0, a> b. AB an immediate corollary of Lemma I we have the following: Lemma~. Let b > a ~ O. Then b divides a onZy if a = o. . Lemma S. The only divisors oJ.1 are 1 and -1. 6. The division algorithm (FULL COURSE). The division process of elementary arithmetic may be extended to hold for all integers. We state the result as the following: \ DIVISION ALGORITHM FOR INTEGERS. Let a and b be any two integers BUch that b ~ o. Th;en there are unit[U6 integers q and r Jor which 0 :ii r < Ibl BUch that (6)

a = qb + r.

To prove this result we consider the set of all of those integral multiples of h = Ibl which are greater then a. By Principle 4 of Sec. 2 there is a least such multiple. Designate it by (t + l)h. . This defines t and, since th < (t + l)h, we have th :ii a < (t + l)h. Then 0 :ii a - th < h. Put r = a - th and JJ.ave a =:= th + r = qb + r. Here q = t if b = Ibl, q = -tif b = -Ibl. To prove q and r unique we suppose that we have a second set of such integers, a = ub + v. Choose the notation so that v ~ r. Then h > v ~ r, h - r > v - r ~ 0, h ~ h - r > v - r ~ o. By Lemma 2 it is possible for h to divide v - r only when v - r = 0, v = r. Since ub + v = a = qb + r we have v - r = qb - ub = (q - u)b, Ibl divides v .:... r, v = r. Also (q - u)b = 0 for b ~ 0 and q = u. I

nlustrative E:J6ample

Compute q and r if a - -249, b = 17.

40

INTEGERS

[CRAP, II

Solution

+ 11. Hence (-14)(17) - 11 + 17 - 17 = (-15)(17) + 6, q = -15, r

By the usual division process 249 = (14)(17) -249

'.

= (-14)(17)

- 11 = .

"" 6.

EXERCISES C~mpute

q and r for the following integers:

(a) a = -180, b = 9

(b) a

= -169, b = 11

(c) a = -982, b = 21 (d) a = 279, b = -11 (e) a = 247, b =- -27

(f) a = 11,692, b = -245 (g) a = -6,942, b = -111 (h) a = -6,721, b = -100 (~') a = -4,127, b = -121 W a = -17;115, b = -489

7. The Euclidean greatest-common-divisor process If a and b are integers and c is an integer dividing a' and b, we say that.c is a common divisor of a and b. If a .;.. b = 0, any integer is a common divisor of a and b. Otherwise there is a largest positive common divisor d of a and b which we shall call their greatest common divisor (abbreviated g.c.d.). It is sometimes referred to as their highest common factor (abbreviated h.c.f.). We first prove the following: Lemma 4. Let b ~ 0 divide a. Then the g.c.d. of a and b is Ibl. , For the largest divisor of b is Ibl and Ibl divides a. We next prove the following: . Lemma 6. Let b ¢ 0 and a = qb + r. Then the com(FULL COURSE).

mon factors of a and b coincide with the common factors of b and r; a and b have the same g.c.d. as band r. For, if c divides a and b, we may write a = if, b = cg and have a - qb = if - qcg = c(J - qg) = r, f - qg is an integer, c divides r. Conversely, if c divides band r, we write b = cg, r = ch, and have a = qcg 4- ch = (qg h)c,

+

c divides a. Lemmas 4 and 5 yield a process for finding the g.c.d. of any two integers a and b not both zero. We suppose that we have labeled our integers so that b is the one not zero. Then Lemma 4 gives their g.c.d. if b divides a. Otherwise a = qb + r where 0 < r < Ibl, and if r divides b it is the g.c.d. of a and b. The process of determining

SIlO.

7]

41

THE EUCLIDEA.N PROCESS

whether or not r divides h is the division process which gives h == qlr + rl for 0 ~ rl < r. H r is not the g.e.d. of a and h, we have 0 < Tl and the g.e.d. of a and h is the g.e.d. of rand Tl. We have thereby described a division process for determining the g.c.d. with a sequence of diminishing positive divisors. Mter at most Ihl steps this· process terminates and the last nonzero remainder is our g.e.d. We shall use the notation {a, hI for the g.c.d. of a and h in the remainder of this chapter. nlustrative Examples I. Fb;I.d th~ g.c.d. of 8,381 and 1,015.

Solution We arrange the computations as follows:

29 -

8 1 3 8 232 261 1015 8381 232 232 ~ 8120 -0- -W 232 261

Am. {8381, 1015} - 29. II. Show that 14233, 884} - 17. Solution By division we find that 4,233 = 249 . 17 and that 884 = 52 • 17. Then 14233,884} - 17 if and only if 1249, 52} = 1. This latter fact follows from the computation

1 2:~' : 1~1~ ~ 24: 2683341208

0 1 2 - 3 81141

This computation may be abbreviated when we reach the remainder 11 whose only factors are 1 and 11. BXERCISBS

1. Find (a) 1191, 78} (b) {l98, 62} (c) {253,29} (d) 184, 276} (6) 192, 876} . (f) 1147,637} (0) 111682, 1072} (It) 14M,2288} (,') 12904, 312}

W 1528, 2343} Am. 2.

Am. 33.

(k) 1648, 2997}

(l) {894,3278}

Am. 298.

Am. 12. (m) 11442, 6489} (~)

Am. 49. Am. 16.

(0) (p) (q) (r)

{l432, 8469} 11134, 8019} 11608,17523} 1768, 9552} 1266664, 877769}

Am. 1. Am. 3. Am. 11,111.

42

[CHAP.

INTEGER,S

'II

2. Use the method of Inustrative Example II to show that (a) {21924,2144) = 4

(d) {5103, 7614) = 81 (e) {4526,9344) = 146 (f) {7992, 17640} = 72

(b) {179,21) = 1 (c) {4078, 814) = 2

8. Linear combinations ar'e integers, the sum

(FULL COUR,SE).

If a, b, m, n,

ma+nb

is called a linear combination of a and b. It is also, of course, a linear combination of m andn. We may prove Lemma 6. Let f and g be linear combinations of a and b. Then any linear combination of f and g is a linear combination of a and b. For we have assumed that f = ma nb, g = sa tb.

+

+

Then

+ kg = h(ma + nb) + k(sa + tb) = hma + hnb + ksa + ktb = (hm + ks)a + (hn + kt)b.

hf

We now prove Theorem 1. The g.c.d. oj a and b is the least positive linear combination

(7) of a and b.

d=ma+nb Every common divisor of a and b divides d.

For we have seen that if b divides a the number Ibl = ± 1 . b + 0 . a = d is such a linear combination. Otherwise a = qb + r, r = 1 '. a + (-q)b is a linear combination of a and b. If r ~ d, we form b = qlr + rl and rl is a linear combination of band r. :Sut rl is a linear combination of band r and hence of a and b by Lemma 6. Thus the process of Euclid gives a' sequence. of remainders each a linear combination of a and b. The last such nonzero remainder is d. Hence d = ma + nb. If w = sa + tb we see that a = aId, b = bi.d, w = (sal + tbl)d is divisible by d . . But then d is the least such positive linear c'ombination.

DC.

8]

LINEAR COMBINATIONS

43

Two integers a and' b are called relatitJely prime if their only positive common divisor is 1. Then we shall say that "a is prime to b" and, of course, that "b is prime to a." The g.c.d. of a ~d b is 1 and, by Theorem 1, there are integers m and n such that ,

(8)

ma+nb==1.

COD,versely, if ma + nb == 1, every common divisor of a and b divides 1, and a and b are relatively prime. , Theorem 2. Two integer8 a and b are relatively prime when and only when there are integer8 m and n 8Uch that ma+nb==1.' . If d is the g.c.d. of two integers. a and b, we may write d == ma + nb. But a == gd, b == hd, and d == mgd

+ nhd ==

(mg

+ nh)d,

mg +.M == 1. We thus have,the following: Theorem S. Let d be the g.c.d. of a and b 80 that a == dg, b == dh~ Then g and h are relatively prime. To find the integers m and n of formula (8) we need only find an integer m such that ma - 1 is divisible by b. rhe quotient will be -no It is only necessary to try the integers 1,2, . . . ,b - 1. When a and ~ are nQt relatively prime and we' wish to find m and n such that ma + nb == d, we use the process above for mg + nh == 1. Theorem 2 may be used to derive the following important lemma: Lemma 7. Let b divide ac and let b be prime to a. Then b divides c. For ma + nb == 1. Multiply by c to obtain mac + nbc == c. Since b divi~es ac, we have ac == bq, c == mbq

+ nbc ==

Hence b is a factor of c,

(mq

+ nc)b.

INTEGERS

[CHAP. II

~ The g.c.d. of any integer a and a prime P divides P and is .either 1 or Ipl. Hence P divides a or is prime to a. By Lemma 7, if P divides ac and does not divide a, it does divide c. If P divides abc and does not divide a it must divide bc and hence either b or c. This result may be extended to any number of factors and we state it as Theorem 4. .A. prime P divide8 a product abo if and only if it divide8 one oj the fact(1f'8 a (1f' b. 9. The fundamental theorem of arithmetic (FULL COURSE) .. The most important theorem on. positive integers may be stated as follows: . Fundamental theorem of arithmetic. Every integer a > 1 can be ~pr688ed aB· a product PiP2 • • • p, of potJitive prime facf(1f'8 Pi. ThiB factorisation i8 unique apart from the qrder oj the facf(1f'8. If a is itself It prime, our notation means that t.= 1 and we stop with the fll:st factor, a = Pl. In any factorization of a we can order our prime factors Pi so that Pi ~ P2 ~ • • • ~ p,.

Suppose then that we have a second factorization such that a = ~l1g2 ••• g. and we have ordered the prime factors so that gl :! g2 • • • ~ g•• Then the fundamental theorem states that t = 8 and that Pi = gi for every value of i from 1 to t. To prove this result we let Pi be the least positive nontrivial factor of a. Every divisor of Pi divides a and, if nontrivial, is less than Pl. Since Pi is the least ~uch divisor, it must be a prime. Then either a = Pi and we have the desired factorization or a = pial where a > al > 1. We factor out the least nontrivial factor P2 of al and have a = PiP2 or ~ = P1Plia2. where al = P2a2, al > a2 >.1. This process must terminate after a finite number of steps and yields the' desired factorization. We now assume two factorizations as above. By Theorem 4 the prime Pi divides gl • • ; g. only if it divides

PC. 10]

THE GREATEST COMMON DIVISOR

45

one of the q's. But then PI is equal to this q. Let the notation be chosen so that PI == ql' Then al == PI . • . p, == q:a • • • q. By the same argument we obtain PI == ql. After a finite number ot such steps we end with the integer 1 on one side of our equality and would end with a product of primes on the other side unless t == 8. This proves our theorem. 10. The greatest common divisor and least common multiple of several integers (JIULL COURSE). The greatest common divisor of n integers ai, aI, • • • ,a,..is defined to be the largest poSitive integer dividing all of them. It is possible to give an inductive proof of the following result: Theorem 5. Let ai, aI, • • • ,tlra be a sequence of n positive integers and designate the g.e.d. of ai, aI, ••• ,alb by dlb. Then dil.j-I = {dlb, ail.j-I} (k == 2 . . . n - 1), .

every common divisor of ai, aI, • • ., tlra divides d,.. The proof is made by observing that the common divisors of dlb and ail.j-I are common divisors of ai, as, • • • ,alb, tIlb.t-l, the common divisors of ai, • • • ,alb, ail.j-I divide dlb and tIlb.t-l. Tl;len d,; and alb+1 have the sam~ g.c.d. We shall not go into further details about the rigorous formulation of the inductive proof. . The least eommon muZtiple (abbreviated l.c.m.) of ai, aI, • • • ,tlra is the least positive integer divisible by every a. We shall prove Theorem 6. The l.e.m. of a and b is the quotient ab (9)

d

of their product by their greatest common divisor. For a == gd, b = hd where g and h are relatively prime. If m is the l.c.m. of a and b, we may write m == sa = tb, sgd == thd, sg == tho Hence h is a factor of sg and is prime to g; h divides s. Since.m == sa, we see that ha = ghd is a factor of m. But ghd is divisible by a == gd and b == hd, m

46' "

must be equal to ghd,. This product is the quotient of' formula (9). As in the case of the g.c.d. we have a result for the l.c.m. which we shall state without proof. Theorem 7. Denote the l.c.m. of aI, 'a~, . . . ak by mk. Then mk+l is equal to the l.c.m. of ak,J-I and mk. fllustrative Examples

I. Find the g.c.d. of 8,381, 1,015, 87. Solution .By a previous illustrative example {8381, 1015} = 29. divides 87, the answer is 29. II. Find the l.c.m. of 528, 792, 132.

Since 29,

Solution By our process we find that \.528, 792) = 264 and that 528 = 264· 2, 792 = 264 . 3. Hence the l.c.m. of 528 and 792 is 264 . 3 . 2 = 1,584. Also 132 is a divisor of 264 and our answer is 1,584. EXERCISES

1. Find the g.c.d. of each of the following sets of integers: (a) 102, 174, 112

(b) 378, 462, 399 • (0-) 1,296, 1,584, 936, 1,558 ~.

(d) 968, 2,057, 1,276, 348 (e) 460, 644, 1,012, 598 (f) 4,329, 8,991, 2,553, 2,109

Find the l.c.m. of each of the following sets of integers:

(a) 2, 6, 8, 10

(b) 3, 81, 15 (c) 87, 58, 174

(d) 108, 90, 135, 315 (e) 81, 54, 135, 225

11. The factors of an integer. If p is a positive prime factor of an integer a 'F 0 and e is the largest positive integer such that pB divides a, we shall say that a is exactly divisible by pB. Then in the factorization of a as plus or minus a product of positive primes, exactly e,of the prime factors will be equal to p. We may group the prime factors of an integer a so as to write ±a as a product of powers of distinct prime~. Then the fundamental theorem of arithmetic states that. these primes and their exponents are unique.

SEC.

11]

THE FACTORS OF AN INTEGER

47

The process of factoring an integer is illustrated in the example below. A systematic method of listing divisors is also given and will be useful in Chap. VI. lll~strative

°

Examples

I. Sh,ow that if n > 1 and b ;oE are integers and p is a prime there is no integer a such that aa = ph-.

SoZution Suppose that there is such an integer a. Then pba ¢

°and a cannot

be zero. Let a be exactly divisible by p" and b by pl. Then a- will be exactly divisible by p"-, b- by pI_, and pb- by pia+!. Since pb- = aa we have en ... In + 1, en - In = (6 - f)n - 1. This is impossible since n > 1 cannot be a factor of 1. . . II. Factor the integer 80,784.

SoZution We list the positive primes less than 50 for use in the exercises below. Thel are 2,3,5,7, 11, 13, 17, 19,23,29,31,37,41,43,47. Testing 2, 4, 8, 16, 3, 9, 27 in turn we find that 80,784 = 4(20,196)

= 2'(5,049} = 2' . 9(561} = 2' . 3 1 • 187

=0

2' • 31 • 11 '17.

III. Let a = plqlr where p, q, and r are distinct primes., List aU positive divisors of a. Solution The divisors of a are 1, p, pi, pI, q, pq, pig, plq, q', pq', plq', p8q', r, pr, plr, par, qr, pqr, plqr, p'qr, qlr, pqlr, p'qlr, plqlr. There are (3 + 1)(2 + 1)(1 + I) = 24 divisors. IV~ List the positive divisors of 1,800.

Solution 1,800 = 2 3 ' 51. The 36 divisors are 1, 2, 4, 8, 3, 6, 12, 24, 9, 18, 36, 72, 5, 10, 20, 40, 15, 30, 60, 120, 45, 90, 180, 360, 25, 50, 100, 200, 75, 150, 300, 600, 225, 450, 900, 1,800. Listing. them in increasing order we have 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 25, 30, 36, 40, 45, 50, 50, 72, 75, 90, 100, 120, 150, 180, 200, 225, 300, 360, 450, 600, 900,1,800. 1

INTEGEBS

[CJUP. n

BDRCIDS i~ :Faotor the following integers:

Ca) 504 (b) 2,310 (0) 1,078 (d) 32,175

(,,") 8,281 65,875 (Tc) 1,120,581 (0 15,750

(e) 5,712 (f) 31,603 (g) 1,271

W

(h) 22,607

t. List all positive integral divisors of the following numbers: (a) 504. (b) 5,712 (0) 21 • 31 • 11

(d) 28 • 31 • 71 (6) 1,084 (f) 512

(g) 1,944 (h) 936

(,1 588

W 420 (Tc) 486 (0 384

CHAPTER

m

RATIONAL, REAL, AND COMPLEX NUMBERS 1. The rea1line. The intuitive idea of ·a .positive real number is that of a magnitude, i.e., a size or a measurement of length. Basically this idea is that of regarding positive real numbers as corresponding, in a one-to-one fashion, to the points on a ray (a half line) such as in Fig. 1. The point marked 0 is called the origin of our system of measurement, i.e., the point from which we measure. The point marked 1

o

1

Q

. FIG. 1.

is our unit point and the distance from 0 to 1 is our unit of

length. The ray is unending on the right and each of its Points determines a length from the origin to the point. The number a of units of length is then the positive real number a corresponding to the point, and we label the point with the symbol a.

o FIG. 2.

1

•

It should be observed that a change in the unit point changes the correspondence between points and positive real numbers. For example, if the unit is an inch and we change it to a foot, the point marked 24 (inches) becomes 2 (feet). However, the real numbers are fixed. OnIy the corresponde~ce between point and number is variable. H we extend our ray to be a line (Fig. 2), we have wh~t is normally the intuitive idea of the set of all real numbers. Then we have both magnitude and direction. Points to the left of the origin now correspond to negative real numbers. 49

50

REAt AND COMPLEX NUMBERS

[CBAP.. m

They are stiU magnitudes but are directed in a direction opposite to. that of positive real numbers. Real nUmbers may now be added graphicaJ.ly. lor suppose that a and b are any two real numbers. Then we add a to b by taking the line segment from 0 to b and laying it off on the line segment from 0 to a so that the point on the

. o

•.

•

11+6

FIG. 3.

former segment which was at 0 is now at a. Then the point which was 'at b will be at a + 'b. For example, when a and b are positive, we have the diagram shown in Fig. 3. In case a is positive and b is negatiye, we have the diagram as in Fig. 4. Similar~y we may form a - b, by laying off the segment from 0 to b on the segment lJ:om 0 to a so that the point which was at b is at a. Then the point foI"ID.6rly at 0

.

b

o

6

G

FIG. 4.

will be at a-b. In case both a and b are positive, the diagram is as in Fig. 5. The result is then the same as that of adding a to -b. Multiplica~on of' real numbers cannot be described by this graphic method, however, and we shaJ.l pass to the , 0'

.

b

t

b G-b FIG.

o.

G

more abstract and rigorous algebraic definition of the real number system. 2. Rational numbers. If b is any positive integer, the intuitive idea. of the rational number (also caJ.led fraction or rational fraction) 1

b

51

RATIONAL NUMBERS

nc.2]

is that of the length of the line segment obtained. by dividing the segment of unit length (i.e., length 1) into b equal parts. Then if a is a natural number, the r~tional number alb is the length of a of these parts. It follows that if a is zero alb is zero. Siniilarly if a > 0 the number -alb is the length of a of these parts but directed to the left. Examples are given in Fig. 6. Then the lengths t and f are the same, - t is the negative of t in the sense that their sum is zero. The intuitive idea above may be regarded. as the inspiration of the general description we shall now present. ,

,

-I

-~

,

~

,

,

-"

o FIG.

,

, 1

6.

A rational number a is a pair of integers a and b where we write the pair as a

a='b

(re8.d. "a equals a over b"). Here a is any integer and is called. the numerator of a; b is any nonzero integer and is called the denominator of a. Our first definition of equality (i.e., of being the same pair) is now extended. and we define (1)

if and only if (2)

ad = be.

If b is a factor of a so that a = bq, the rational number alb

Is identified with the integer q. Thus our set of rational numbers is conceived. of as a number system containing the set of all integers. In particular each integer a may be replaced at any time by the rational number

a

1" Finally our definition of equality is really equivalent to the property that

52

REAL AND COMPLEX' NUMBERS

[CHAP. m

(3)

for every nonzero integer c. . In order to obtain the intuitively desirable property (4)

we define the

BUm

of any two rational numbers by ~

( 5)

b

+ d~ -_ ad bd + bc.

We also define products by (6)

We have now completed the definition of the number system of all rational numbers. It may be verified that all the laws fo:r integral operations hold and that a c ad-bc

(7)

b-a=

bd

Here the rational num1;>er

o

O=b has the property a where

+0 =a

for every a, a .

+ (-a)

=0

-(-i)=i·

(8)

Observe that the definition of addition in formula (5) and that of multiplication in formula. (6) have meaning only because a product bd of denominators b ~ 0 and d ~ 0 is not zero and so may be used as a denominator. Notice also that formula (4) may be obtained from formulas (5) and (3) by writing a +c ab"+ cb (a' + c)b a+c

b

b=

bb

=

bb

= -b-·

SEC.

53

THE DIVISION LAW

3]

Conversely we may obtain formula (5) from formula (3) and (4). For

All rational numbers may be written in one and only one of the forms

a -a 0

(9)

b' T' b ==

0,

a

where and b are positive integers. The numbers of the first kind are called the positive rational numbers (or positive fractions), the numbers of the second kind the negative rational numbers, and the single last number is the rational number zero. The notion of absolute value which we .gave for integers in Chap. II carries over and we define lal == a if a is a positive or zero rational number, lal == I-PI == Pif a is a negative rational number. ORAL EXERCISES

1. Express the following fractions in one of the forms of formula (9): (a)

1 --=3

(b) -

~ := :

(e)!

:= :

(d) 1-~ 3+2

2. What is the absolute value of each of the following fractions?

4-2

(a) 3 - 1

(b) :

:= ~

(e)!

:= :

5-5

(d) 3 - 11

8+1 +2

(e) 1 - 3

8. The division law. The rational number system has one property iIi addition to those which we have already listed in Chaps. I and II as properties of integers. We state this property as a law.

XI. LAW OF DIVISION. If a andp are any two numbers such that P is not zero, there is a unique number 'Y in the number system such that a == fJ'Y. We shall call 'Y the quotient of a by P and write a

-• . 'Y .== P

54

REAL AND COMPLEX NUMBERS

[CRAP. m

Then every rational number is the quotient of two rational numbers which happen also to be integers. Thus the rational number system may be thought of as resulting from the system of integers by adjoining to it all quotients of integers. The verification of our new law is a consequence of the . formula (10) when c ~ O.

In particular b

1

(11)

alb =

a'

for all rational numbers a and fJ ~ 0 and all integers a ~ 0, b ~ O. It is sometimes convenient to use the special case~,

a

(12)

cld

all

= c/~ =

ad

ad

Ie = c'

for 8.l1. integers a, c ~ 0, d ~ O. Formulas (10) to (12) remain correct when we replace these integers by fractions. ORAL EXERCISES

Express the following as quotients alb where a a.nd b are integers: (a)

~

(d)

~

(J)3t

(~)

l

. t

(g) 1 -

1 (0) -1-

t (e) -5t

t

1" . 1 1---

1-+

3+t ,. Fractions .in least terms. Every ·rational number except zero may be expressed as a quotient alb where b Is a positive integer and a js positive or negative according as alb is positive or negative. We say that alb is in lea8t termB if the g.c.d. of a and b ~ 1. If alb is not in least terms

SEC.-4]

55

FRACTIONS IN LEAST TERMS

and d > 1 is the g.c.d. of a and b, then a the g.c.d. of g and h is 1. Then

a

b=

= dg, b =

dh where

dg g dh = h'

and we have expressed the fraction alb as the fraction glh which is in least terms. The expression of a rational number in least terms is unique. mustrative Examl'es I. Express 11,604/15,472 in least terms. Solution We find the g.c.d. by the computation

3

3868

1

I 11604 I 15472 11604 I 11604 3868

and see that 15,472 ... 4(3,868).

Am.... II. Express the following fraction as a quotient alb in least terms: Or!

= 27 _

123

6-

1 5-

1

1 4- 3---t

Solution

We simplify as follows: 1 6-1 5 12 220-218 3 - '2 = -2- = 2' t = 5' 4 - '5 ... - 5 - == 1f' 1 5 5 90 - 5 85 1 18 ¥ = I8' 5 - 18 = ---"IS = 18' it - 85' 18 510 - 18 492 123 85 85 6 - 85 = 85 == "85' W = 123· 492 ... 4' 85 108 - 85 23 27 - 4 4 - TAm. III. Let 110 ... 15, m = 9. List all positive fractions alb where a divides 110 and b divides m.

56

REAL AND COMPLEX lfUMBER8

(OlW', IiJ

Solution

The factors of 15 are 1, 3, 5, 15 and those -of 9 are 1, 3, 9, Hence our answers are 1, 3, 5, 15, t, t, t, .., EXERCISES 1. Write the following fractions in least terms:

(J)

() 812 a 1044 , (b) () c (d) () 6

6

1,038 -2,249 -2,730 -1274 , 1,027 - 832 2,268 812

Ana. - 13'

888 744 62 57 (0) 31 108 (1&) 147 _ 101 21 105 (10') 2,175 _ 872 1,450 1,308

+

91

Ana, 36' 5

Ana'6'

81

Ana, 29'

I. Simplify the following and write in least terms: 9

Ana. - 59' 5

Ana. 28'

1

(0) 1 - 1 _

1

(1&)

Ana. -1.

1

-

1

1 1-1-+

(2-~):(3-~) 2 a 2-

3-

57

THE LAWS OF EXPONENTS

SIlO. 5)

(!_.!X~~l~ (!+-~!+D--- ---+~ 11 1

!.....J! 1

+

~ .1 2

(.' 3 7 ~V \)11 G:V\l~;

.,

_7_ 1

1--

1

.Am. 5.

3. List all positive fractions alb such that a divides n and b divides m in the following cases: (a) (b) (c) (d)

n = 9, m = 15

n = n = n = (6) n =

24, m = 8 16, m = 12 180, m = 30

-48, m

= 20

(f) n = 90, m = -42 (g) n - 504, m = 60 (/l) n = -512, m = 48 (or.") n = -1080, m = -63 W n = 240, m = 143

6. The laws of exponents. The laws of exponents which we have already given for integers hold also for rational numbers. They may be repeated as ..

= a"+tn, (an)m = a"m, an{J" = (afJ)n. Here a and P are any rational numbers, nand m are any (13)

and"

rational numbers, aD is defined to be 1 for any a ~ o. A new phenomenon now arises since every rational number a = alb has an inverse P = bla such that afJ = 1. We define the symbol. 1 a l = -ab = a-, (14) for every a = alb p!i 0 and every natural number n. It follows from our multiplication definition in formula (6) that (15)

or'" = (an)-l,

and that (16)

58

REAL AND COMPLEX NUMBERS

[CHAP.

m

It may now be shown that formulas (13) hold for all integers nand m. We also have the additional laws 1 _an = an-m -- a - --. (17) am It will be convenient to observe that anp p p a.-p (18) -=-, -=-, 'Y a.-'Y an'Y 'Y that is, afactor of the numerator (denominator) may be carried into the denominator (numerator) by changing the sign of its exponent. We shall not carry out the verifications of these properties. EXERCISES

1. Express the following rational -numbers as products of powers (with integer exponents) of the integers 2, 3, 5: (28 • 32 • 5)4 (a) (2 2 • 3 8 • 54)8 Ana. 26 • 3-1 • 5-8• (b)

8' (75) (2 . 4 . 3 . 45)8 (28)4 • 3-8 • 56

-(c) 2- 4 • 86 (3- 10 • 52)4

Ana. 2 . 382 • 514•

2. 8. 4)8

(2 3 5 (d) 2'34.57 (e) 54 (37 • 2 6)8(3-14 • 4-2 • 5-1) (J) (24 • 58 • 3-&)1(2-8 . 5-4 . 36)-3

Ana. 211 . 3 7 • 58.

2. Express the following numbers as quotients 2»7'" /3r 5' for integers n, m, r, t: 24.7-9 Ana. 3-9 . 5-2'

24 • 716 Ana. 38. 5-4

59

DECIMALS

SJIC.6J

6. Decimals. A finite decimal is a finite sequence of digits with a decimal point, such as 1,472.697, or the negative of such a sequence. The decimal point has the property that the integer obtained by deleting it divided by an appropriate power of 10 is the given decimal. For example, -1,472.697 = -1,~~~,697.

Since finite decimals are rational numbers, their sums, products, differences, and q'Llotients are also rational numbers. Their sums, products, and differences are also finite decimals obtainable by the procedures of elementary arithmetic. The quotient of two finite decimals is expressible as the quotient of two integers and, when reduced to least terms, is expressible as a finite decimal if and only if all prime factors of the denominator are 2 or 5. If not, the division process of elementary arithmetic yields an infinite decimal. For example t = 0.333 . . • • An infinite decimal is an infinite sequence of digits with a decimal point and a sign. Infinite decimals may be described in terms of a concept called approximation. We define the first approximation of any decimal a to be the integer that appears to the left of the decimal point in a. It is an r digit integer which may be zero, positive, or negative. Designate this approximation by ao. Define ak to be the finite decimal of r k digits formed by the first r + k digits of a. When a is also a finite decimal such as

+

643.29654,

will

the adjoined digits from some point on be zero and the approximations from some point on will be a. Indeed they are

ao

= 643, a.

al = 643.2,

= 643.296,

a4

a2

= 643.29,

= 643.2965,

G.

= 643.29654,

a=a,=a'···.

60

REAL AND COMPLEX NUMBERS

[cJ:UP. m

Note that our approximation a1 is 643.2 and not 643.3 even though as = 643.29. We are not rounding off digits but taking them as they actually occur. .Any decimal a may now be conceived of as being an infinite sequence

of the finite decimals ak which are its approximations. Moreover this sequence has the property that the first r + k digits of all approximations ak, aJ:+1, aAl+s • • • are exactly the same and all have the same sign. For exain.ple if a = -462.3578965942 . . • the first eight digits of all approximations ai, aa .•. are -462.35789. Let us agree finally that any infinite decimal a, in which all the digits from a certain place on are 9's, will be identified with the finite decimal obtained by deleting all these 9's and adding 1 to the digit to the left of the first in the unending sequence of 9's. For example, . -199.98999 . . . = -199.99;

0.999 . . . = 1.

.7. The real number system. We shall now define a real number to be any positive, zero, or negative infinite decimal. Let us agree that two decimals are equal real numbers if and only if their digits are either the same or one is obtained from the other by the proceSs above of deletion of 9's. The definition of real numbers which we have now given is complete and the set of all real numbers will become what is called the r6(J,l number 8Y8tem when we tell how real numbers are added, subtracted, multiplied, and divided. We shall make these definitions in terms of the following very important concept. Consider an infinite sequence act,

a1 •••

whose members (i.e., terms) are real numbers. Suppose ~hat.for every poBititJe integer k there is' a place in the sequence

SEC.

7]

THE REAL NUMBER SYSTEM:

61

(the place depends upon k) beyond which all the members have the same sign and the same first k digits: Then we shall call this real number sequence a confJergent 8equenc6. Every convergent sequence defines an infinite decimal and hence a real number whose digits are obtainable from the digits of the terms of the convergent sequence sufficiently far out. This decimal is called the limit of the sequence. The simplest example of a convergent sequence is of course the approximation sequence of a real number, and the real number is then the limit. More complicated illustrations are furnished below. Our operations on real numbers are now easily defined. Let ao, a1 ••• be the approximation sequence for a and bo, b1 • • • be the approximation sequence for a real number {3. Then it will be true that the sequence Co, C1 • • • of sums c, = ~. + b, is a convergent sequence and we define a + {3 to be its limit. Similarly a - {3 is the limit of the sequence of differences ~ - b" a(3 is the limit of the sequence of products a,.:fJ,. Finally if {3 is not zero we may delete a finite number of its terms and obtain a convergent sequence with the same limit {3 but in which all terms are not zero. For example the sequence for t is 0, 0.3, 0.33, 0.333, etc. and we would replace it by 0.3, 0.33, 0.333, 0.3333, etc. If the members of the new convergent sequence for {3 are designated by {30, {31 • • • ,the number {3, is a nonzero finite decimal and the sequence of rational numbers ~/b,' is a convergent sequence whose limit is the quotient a/{3. Let us observe that the sequences for a and {3 which we have used are sequences of rational numbers and that we have defined a + {3, a - {3, a(3 and a/{3 in terms of the limit concept and the corresponding operations for rational numbers. It is not at all obvious that the sum, difference, product, and quotient sequences defined above are convergent but we shall not try to give the proofs here. We shall also wish to state without proof that the real number system obeys all the eleven laws we have put down as well as the exponent laws. Moreover the real number

62

REAL AND COMPLEX NUMBERS

[CHAP.

m

system is ordered by the property that a is greater than b (b is less than a) if a - b is a positive real number. mutT«tive EumPles I. Let a "" 3.165 • . • , fJ - 2.153 . . •• Compute as much of a fJ-as the information given p8l'Qlits.

+

Solution We observe that ao at

la,

+ bo = 3 + 2 = 5,

a1

+ b1 = 3.1 + 2.1

- 5.2,

+ bs .. 3.16 + 2.15 "" 5.31, + b. "" 3.165 + 2.153 = 5.318.

Hence from the information given we are almost certain that

= 5.31 •..• Indeed, if a .. 3.1659 • • • and fJ = 2.153999 . . . ,we have a +fJ

a. + b,

= 5.3198

and so 5.318 is false. In the most extreme case where a = 3.165999 • • • , fJ = 2.153999 • . • ,

+

we have a = 3.166, fJ = 2.154 and so a fJ = 5.320 but otherwise 5.31 • • • is correct. Remark: It is true, in general, that the use of the (k l)st approximations of a and fJ will yield the true value of the kth approximation Ci of a fJ or a - fJ. However this is not true for products. Indeed suppose that a "" 300, fJ = t = 0.333 • • • so that afJ = 100. Then ao "" a1'" • • • "" 300, aobo "" 0, a1b1 = 90, asb. = 99, a,ba = 99.9, a.b, = 99.99, and we see that aobo and a1b1 are not close to the correct value of Co = ~ "" 100. n. Define a real number by the formula ao = 0.4,

+

+

"'+1 III ac

+ (0.8)(0.5 -

a,).

Compute the first five approximations of a and estimate its value. Solution a1 = 0.4 (0.8)(0.1) = 0.48, as = 0.48 (0.8)(0.02) = 0.496, a. = 0.496 0.8(0.004) = 0.4992, a. = 0.4992 (0.8)(0.0008) = 0.49984, at = 0.49984 0.8(0.00016) = 0.499968.

+ + + + +

Am. a = 0.5.

Ro.8]

POSITIVE REAL POWERS

63

Remark: If r " 0 is any finite decimal between 0 and 0.5 and we define ao = r, a'+1 = ex. + 2r(0.5 - a,), the sequence is convergent and has 0.5 as limit. It converges slowly when r is near zero, that is, we need to take many terms to get near the limit t. As we have seen above, it converges rapidly when r is close to 0.5. EXERCISES

•

1. Compute the first eight approximations to the number defined by ao'" 0.3, ",+1 = a, + 0.6(0.5 - a,). z. Compute the first eleven approximations to the number defined byao ... 0.1, ",+1 = a, + 0.~(0.5 - a.).

S. Positive real powers of positive real numbers. Every Positive real number a determines a unique positive real number b ==

Va

(read" b equals the nth root of a") for every positive integer n. Then b is defined to be that positive real number whose nth power 1i'is a. We shall give a formula for b in Sec. 13. We shall prove that there is only one such number for every 11. and a in Chap. VII. Note that it is customary to write vb instead of ~b and that ~ == b. The symbol Vb is called a radical. Computations with radicals are not so simple as those with exponents and we prefer to use the fractional exponent form given by (19)

(read the "nth root of a equals a to the one over nth power"). We also define the fractional power (2O)

amlf' == (a1!"}m

(read am!" as "a to the m over nth power") for all rational nl1mbers min, and see that also am!" == (am}1!". (21)

For [(alJf'}m]" == (a1lf')- == [(alJf'}"]m == am and so (a1lf'}m is.the nth root (amp'" of am. We may now define the positive real power a" for every positive real number a and every real number n. Let no,

64

REAL AND COMPLEX NUMBERS

[CHAP. III

nl, n2 • • . be an approximation sequence for n so that every Then every ~ is a rational number and defines a positive real number bi = an, The sequence bo, bl • • • can be shown to be a convergent sequence with a positive limit b. We define an to be this limit . . - The student will have observed by now that, as we have pointed out where they occur, our exposition lacks a number of verifications. The material left out is suitable only for much more advanced courses and it would not be worth while to present it here. However, we do feel that what we have given should do much to give the student a clear picture of the meaning of the basic concept of a positive real power of a real number. We shall now state the following important extension of our laws of exponents without proof: Theorem 1. The positive real powers of positive real numni is a finite decimal.

bers satisfy the laws of exponents

(an)m

= anm ;

anbn

= (ab)n,

for all positive real numbers a and b and all real numbers n andm.

The laws of exponents for the cases where the exponents are fractions lin and where n ¢ 0 is an integer may be expressed as certain formulas for operation with radicals. These are (23)

_" '!'7:.

-V'ab = yaW;

VVa

=

_"I-~r: _ vava -

,,~ r:

va;

"Van+m •.

It is important to observe that

bVa = ',ybna. So, for example, ~ = ~ = 2~ since 8 = 28. Also we may wish to insert a number within a radical and so write' .

SEC.

8)

POSITIVE REAL POWERS

65

2 ~ == -¢'2i ~ == {I2 8a/2 == -¢Ta Of course, in all of these formulas the numbers a and b are positive and we cannot overemphasize that the radical8

are all positive. If a is negative and n is an odd integer, we may write a'== -d where d is positive.

Then if n is odd,

a l '" == (-d)l'" == -d l '" is a unique negative real number. However a l is not a real number since (al )2 is positive whenever al is real and must be negative if a is negative. The meaning of afl for a negative and n real is obviously an exceedingly complicated notion and we shall not discuss it further in this text. We shall discuss the meaning of a l ' " for integers n and negative real numbers a in Chap. VIII. nlustTative Emmples I. Express the number a ... -¢'2 V5 {13 as an integer raised to fractional power. Solution

-¢'2 = 21, V5 = 51, {13 = 31. The least common denominator of the exponents is 12 and we write a = 2ft . 5& . 3& = (2" 51 . 31)0. II. Simplify a =

V8 + 3 V2 -

4 vTs

+ 5 V'4.

Solution a

V8 = Vi V2 = 2 .y2, vTs = 3 V2, -¢'4 = (21)1 = 21 ... V2, = 2V2+3V2-12V2+5V2 = (2+3-12+5)V2= -2V2.

•

EXERCISES

1. Express the following numbers -(a) ..v27 {164 Am. (72)1. (b) V3-¢'2 (c) V2 V3 {/4' 27 Am. (3,888)t. (d) ~ Vi5 -¢'8

-¢'36

(6) V2

Am. (~62)1.

88

powers of a single integer-:

66

REAL AND COMPLEX NUMBERS

l. Write the following in the form

[CB.A.P. m

Va where n is a positive integer:

(a) 2V2 (b) 6 Vi (c) 10~ (d) 6 {1 (e)

~~

S. Simplify the following expressions:

+ v'24

(a) 2 Vi2 - 4 V27 4 (b) -6~+ lO~ + {IIOO (c) (~+ {12)(~ - ~) -?"i2 (d) (VS - VW)(Vi VW) (e) (~-~)~+~

+

+

-

{liS

~

{'6 {II6O -. (J) v'2 + -¢'3 - ~

a

9. Logarithms (FULL COURSE). Let be any fixed real number greater than unity. Then it can be shown that for every positive real nlimber c there exists an exponent n such that

a" = c. We shall give a formula for computing n in Sec. 12. The exponent n is determined uniquely by c (and the fixed a) and is called the logarithm oj c to the baBe a. We shall desip,ate it by log. c and so have the definitive property The most commonly used base for computational purposes is the base a = 10. Thus a table of common logarithms is a table of numbers c and the corresponding exponents n = 10glo c such that 10" = c. For example 101 = 100 so that 10glo 100 = 2. Also the logarithm of 2 to the base 10 is approximately 0.30103, that is, 100.80108 is approximately equal to 2. Since logarithms are exponents, the laws of exponents may be translated into laws for logarithms. Let e arid d

SEC.

9]

LOGARITHMS ,

67

be any positive . real numbers and write c = an, d = am where the n = log" c, m = log" d. Then cd = anam = an+tn. Thus log" cd = n + m and we have the formula log" cd

(24)

=

log" c + log" d.

In words, we have the foUowing property: PROPERTY 1. The logarithm oj a product is the sum oj the logarithms oj its Jactors. We similarly have c/d = an/am = a-. From the meanings of nand m we have (25)

log"

c

d

=

loga c - log" d.

This result may be stated as follows: PROPERTY II. The logarithm oj a quotient is the logarithm oj the numerator minus the logarithm· oj the denominator. If g = r!" and loga c = n then g = (an)m = anm so that log" r!" = mn; i.e., log" r!"

(26)

=

m log" c.

This result may be stated as follows: PROPERTY III. The logarithm oj a power- r!" is the product oj the exponent m by the logarithm oj c. Property III holds for all real exponents and may be· used in particular for fractional exponents. For example, it states that loglo v'2 = loglo 2! = t 10glO 2 = 0.15051 approximately. . . Let: us observe now that every positive real number c may be expressed· as a product c. = 10th, where h is a real number between 1 and 10. Indeed h may be obtained from a decimal expression for c by moving the decimal point to the right of the first (beginning at the left) nonzero digit of c. For example, 145.632

= 102 (1.45632),

0.00435 = lO-a(4.35).

68

REAL AND COMPLEX NUMBERS

Then loglo

1()& = t

m

and loglo c = t

(27)

[CRAP.

+ 10glO h.

It would then be sufficient to construct a table of approximate values of the l~garithms to the base 10 of all k-place decimals between 1 and 10 in order to have a table of approximate values of th,e logarit~ .of all numbers. Such tables are used in trigonometry. The integer t is called the characteristic of loglo c. It may be any positive or negative' integer. The real number loglo h is a positive number between 0 and 1 .called the manti8sa of loglo c. For example, loglo 0.2 = -1 + 0.30103, loglo 2,000,000

loglo 200 = 2.30103,

= 6.30103.

Computation with logarithms is best treated in a course in trigo:Q,ometry and we shall not discuss it further here. In that branch of mathematics called the calculus a base other than 10 is more convenient. The base used is the real number 6

= 2.7182818285

. . . •

We shall give a definition of e in Sec. 12. Logarithms to the base 6 are called N apierian or natural logarithms and are sometimes designated by the convenient notation J.D. c (read simply as "log natural of c"), while common logarithms are usually designated by log a. This latter notation is used in some calculus texts instead of In c. nlustrative Examples

I. What is the characteristic of log 341.621

Solution The characteristic of loglo a is 1 less, algebraically, than the number of digits to the left of the decimal point in a. Hence our answer is 2. II. Give the five-place logarithms of 2, 0.2, 0.02, 0.00002.

Solution The numb'er of digits to the left of the decimal point in 2 is 1. Hence the cl,1aracteristic is O. The digits to the left of the decimal point in

SEC.

69

LOGARITHMS

9]

0.2, 0.02, 0.00002 are, respectively, 0, -1, -4, and their characteristics are thus -1, :""2, -5. Indeed 0.2 = 10-1 (2), 0.02 = 10-'(2), 0.00002 = 10-1(2). Hence log 2 = 0.30103, log 0.2 = 0.30103 - 1, log 0.02 = 0.30103 - 2, log 0.00002 = 0.30103 - 5. ORAL EXERCISES 1. Give the characteristics of the common logarithms of the following numbers: (a) (b) (c) (d)

34,156.2 32,134,569 0.00123 0.142

(e) (I) (g) (h)

1.423 0.0213 0.0000003 56.0000012

(I) (g)

IOg10

2. Give the following logarithms: (a) log, 8

(b) log2 V8 (c) log2~ (d) log, {14 1 ,(e) log2

.y2

1000 IOg10 (0.001) (h) IOg10 (0.01) ViO . (i) loglo (0.001) ~

W log. ev'2

3; Find :!lif (a) log,:lJ = 3 (b) loga:lJ = 2 (c) log,:lJ = -2

(d) logs:lJ = t (e) log,7:lJ = t (f) log,,:lJ = -l

4. Find:lJ if (a) (b) (c) (d)

log., log., log., log.,

100 = 2 e' = 2 e' = 3 100 ,.;, 3

(e) log., 64 = 3 (f) log., 64 = 6 (g) log., 2 = t (h) log., 9 = -i

EXERCISES

1. Given that log 2 ing: (a) (b) (c) (d) (e)

log 4 log 8 log 0.5 = log t log 5 = log ¥ log 9

= 0.301 and log 3 = 0.477, (f) (g) (h) (i)

W

compute the follow-

log 6 log 12 log t log t log-7!'-

2. Using the data computed above, approximate the following to two decimal places:

70

REAL AND COMPLEX !:iUMBERS

= approx.log.yoo (b) log 11 - approx. log (c) log 13 = approx.log -¢'2200 (d) log 17 = approx.log V2oo, if log 29

(a) log 7

[CHAP. m

Vl25

=

1.4624

10. Change of base (FULL COUBSE). If the base of a system of logarithms is replaced by a new base, the logarithms in the new system are related very simply to those in the old system. Indeed, let a be the old base and b be the new base. Then a == b" where r == .10g" a. If e is any positive number and e == a!' == 11" where n == log,. e and m == log" e then e == b"' so that m == m. This gives the formula

10gb e == (log,. e)(log" a),

(28)

that is, the logarithm of e to the base b is the product of the logarithm of e to the base a Iby the logarithm of a to the base b. Since it is true that log" b == 1 no matter what the value of the number b > 1, we have loga b log" a == 1, . that is, the numbers 10gb a and lo~ b are reciproc$. .particular 1 ' loge (29) In 10 == 1 - ' In e == (log e) (In 10) == 1 - '

,

~e

In

-~e

for all positive numbers e. Actual computation gives the approximate formulas . log 6 == 0.4342945,

In 10 == 2.302585.

BXDCISBS

Use the value In 10 = 2.30 and the values given in Sec. 9 to compute the following: (a) In 2 (b) In 3

(c) In I) (d) In 90

(e) In 60 (j) In t

(g) In Tr .(ll) In n

11. Irrati.onality of real numbers. A real number is ea11ed an irrational number if it is not a rationaZ number. ·If

SEC.Il]

IRRATIONALITY OF REAL NUMBERS

*

71

*

n is any integer greater than 1 and p is a positive prime integer, the number is irrational. -For let alb == where a and b are integers. Then a == b( vp), a" == b"p. This was shown to be i.ID.possible in an illustrative example at the end of Chap. II. ' It is not a simple matter to show whether or not a given real number is rational, but we may at least develop a criterion for this property. We call a decimal a repeating decimal oj period n if there is a group of n digits which are repeated from a certain place on to give all of the infinite sequence of digits of the decimal. For example, 361.793842568425684256 . . • is a repeating decimal of period five if we assume that all digits not shown are obtained by repeating the group of digits 84256. We may now give a formal proof of the criterion. . Theorem 2. An infinite decimal i8 a rational number iJ and only iJ it i8 a repeating decimal. IJ n iB the period oj d then there i8 a finite decimal b BUCk that (30)

d

b

== 10" - 1"

If we multiply a repeating decimal of period n by 10, the result is the same sequence of digits but with the decimal point moved one place to the right. Then d is a repeating decimal of period n if and only if n is·the least integer such tbat the digits of lO"d from a certain place on are exactly the same as the correspondingly placed digits of d. Then b == lO"d - d == (10" - l)d is a finite decimal and we have formula (30). . Conversely, let us write a rational number d as a quotient cle of two integers c and e > o. Then d will be a repeating decimal if we can show that there exists an integer n such that (10" - l)d' is a finite decimal. We observe that the only remainders on division of integers by e are the e numbers 0, 1, 2, • . . ,e - 1 and therefore that the e + 1

72

REAL AND COMPLEX NUMBERS

[CHAP. III

numbers 1, 10, lOs, • • . , 108 cannot all have different remainders. Then'there are two integers g and h such that e ~ g > h ~ 0 and such that 1(}11 has the same remainder on division by e as does lOla. Then 1(}11 - lOla will be divisible bye. Write g - h = n for a positive integer n and have 1(}11 - 10" = 10"(10n - 1) = eq where q is an integer. Then (10n

_

l)d - 1QB(lOn - l)e _ qee _.!JE.. - b lO"e - lO"e - 10" - •

Since q and e are integers, the number b is a finite decimal. This proves that d is a repeating decimal. EXERCISES

1. Express the following fractions as repeating decimaJs by actual division and state the length of the smallest period for each: (a) (b)

t t

(0) (d)

t

(e)

-fr

(I> +r

fi

(g) (h)

-it it

2. Express the following repeating decimals as rational numbers in least terms: (a) (b) (0) (d)

1.714285714285 ••• 3.70370 ••• 0.076923076923 ••. 0.1212 •••

(e) 0.120120 ••• Ans.+h. (f) 0.132132 • . . Ans.&. (g) 0.10241024 ••• Ans.tHt. (h) 2.111211121 •.• Ans.-w#-.

12. The sigma and pi symbols. A simple notation for 8" of the :first n terms of a sequence ai, as . . . is

the sum.

8n =at+as+'"

+an

(read It at, plus as, plus and so on, plus an "). For n = 1 this sum reduces to its first term, i.e., 8t = at. Another and more compact notation for the same sum is one which involves the use of the capital Greek letter sigma. It is frequently used in mathematics and is

SEC.

(read "the sum for i equals 1 to n of a/'). n

i

k==

73

T'HE SIGMA AND PI SYMBOLS

121

~ is used for the sum of the first n

The symbol

+ 1 terms of a

0 n

sequence whose first term is ao, and the more general ~ lJi '''III

is the sum of terms beginning with the mt~ term and proceeding through consecutive integer subscripts to the nth 10

term. For example

k ~ = a5 + as + a7 + as + ag + alO. i=5

We shall use this symbol only for m < n. The product Pn of the first n terms of a sequence a1, a2 • • • is representable by (read "al times a2 times and; so on times an"). AI! above Pl = ale We also have a notation using the Greek letter pi, which is

(read "the product for i equals 1 to n of a/'). uct

n

lli

i=O

where m

The prod-

n ai = amam-rl . . . an

= aOa1 • • . an, and

i = m

< n. ORAL EXERCISES 't

1. Read off the meaning of

k a, in the following cases: i .. 1

(a) (b) (c) (d)

a, = i a, = 2i a, = 2i - 1 a, = 2i 1

(e) a, =

+ 3i + 1

(g) a, = C7 •• (h) a, '" box' (i) = ~_.x, W a, = bi-lx7 -,

a,

3

(k)

k j' i" a, = k iyi i

a. =

1

7

(l)

= 1

74

REAL AND COMPLEX NUMBERS

{CHAP.

m

I. Read off· the meaning· of the following sums and products: 6

:t

(a)

(it - 1)

(e)

, .. 1

·5

(b)

10

n (1 +~)' n (!, __ . ,+

(J)

, .. 1

·8

(c)

1 ) 1

i .. 1 10

(d) i

(1

• (1i - i +1) :t. 1

, .. 1

(g)

1)

:

ni~1

i ... 1 10

n 2,-1 2' + 1 •

i .. 1 4. •

n i3 -:

,3-1

~1 i - , + 1

(h)

1

,- 1

18. Infinite series and products. Every infinite sequence aI, az • • • of real numbers determines an accompanying sequence 81, 8z • • • where

(31)

.8.

• = i=l :t a; = a1 +

az + • • ~ +

an

is the sum of the first n terms of the original sequence. Such a sequence of ~ums is called an infinite 8eries or simply a serie8 and it is sometimes designated by the notation a1+ aZ+" • or by the notation

(32)

•

:ta; i-I

(read" the sum for i equals 1 to ihfinity of the ~"). A series (32) is said to oorwerge if the sequence of real numbers 81, 8z • • • defined by formula (31) is a convergent sequence. Then the real number 8 which is the limit of the sequence 81, 8. • • • is called the sum of the series and we write . .

We shall not discuss such nonconvergent series as 1+2+3+" . which diverge nor 1 - 1 + 1 - 1 + . . . which o8cillate~

SEC.

IS]

75

INFINITE SERIES 'AND PRODUCTS

Every infinite sequence a1, all • • • also defines an infinite sequence P 1, PI. • • of products n

(33)

P"

n ~. '-1

=

Then we call the sequence P 1, PI. • • an infinite product and designate it by alaI • •• or, better, by

.

H the sequence Ph PI •.• converges and its limit is the real number P, we write·

.

P= n~.

(34)

i .. 1

Infinite series and products are of importance for the computation of various mathematical constants. For example, the ratio 'Ir of the circumference of any circle to its diameter may be computed by the use of (35)

1 1 1(12 + 31) + 51(12 + 31) - ~(i7 + i7) + .

4 == 2 + 3 - 3

'Ir

so that 'Ir/4 ==

i

k

lID

1

3

5

3

5

~ in which ~ == (-I~~11/m(z-m

+ 3-)

and m = 2i - 1. ' The base e of N apierian logarithms may be computed by the use of (36)

1+ -3!1+ .. . . + -n!1+ . .'. •

e == 1 + 1 + -2!

To compute the natural logarithm of any positive real number a, we proceed as follows. H a > 1 the real number 1 1-(37)

a-I a+l

a

:x;.=--=--

1+! a

76

REAL AND COMPLEX NUMBER·S

[CHAP. In

is positive and less than one, we may compute our logarithm by the formula

(38)

In a = 2 ( x

+ "3X8 + 5"x" :- '7X + . .. ) . 7

If 0 < a < 1 then b = 1/a > 1 and In a = - In b. Evidently the formula loglo a, = In a ·loglo e may be used with formula (38) to compute commop logarithms. The positive nth root of a positive real number may be computed by the use of the so-called binomial series 1

The (r

+ mx + m(m2 ,

1) X2

+

+ l)st term of this series is m~ ~ ~~(m.:.:...---.!.l)_._....,...~(m_---.:...:...n----:+~l)

-

nl

xr

and it will converge and have (1 + x)m as its limit if -1 < x < 1. To compute the nth root of a positive real number a, we write a = bn ' + c = bn(l

+ x),

c x=bn <1.

Then Va = b(l + X)llfl and we compute (1 + x)m by the formula above with m = lin and this valu~ of x. We shall not concern ourselves here with the accuracy of computation by the use of series except for the following simple statement. Let al + a2 + . . . be a convergent infinite series in which the terms alternate in sign. Then if 8 is the sum of this series, it is known that the absolute value of 8 - 8n is less than 1a,.,..~I. nlustrative Example Approximate ~ to two decimal places•.

Sol'lJtwn 37

= 27 + 10,

~ = 3 -V1 + ~~.

77

CARTESIAN COORDINATES

SJDc. 14]

t, Z

Hence m =

=

it, and

i -

..wT = 3(1 + ~ ~ Z2 + =3 +

:1 Z3 -

~ z' +

•.• )

10 1 (10)11 5 (10)' 10 (10)' 27 - 3 27 + 27 27 - 81 27 + . . .

10 1 (10)2 1 (10)4 ... 3+ 27 - 3 27 +2 27

-31'(10)' 27 +

••••

10 100 1 (10)' • Now 27 =- 0.370370 ••• , 2187 = 0.069, 2 27 18 less than but . tely equaI. to 21 (10)' approxuna. 25 mately equal. to

i (~)

I

8 = 001 = 625 . ; 31 (10)1 27

and hence to

6~.

Thus

. approXI.

18

~ ... 3.31 a.pproxi-

mately.

EXERCISES 1. Use formula (35) to compute 'II' correct to two decimal places. S. Use formula (36) to compute the first three digits of 6. 3. Compute the first three digits of e by the use of the definition of e as the limit of the sequence ai,

GI! • • •

in which a,. =

(1 + ~)".

" Compute the coDUtlon logarithms of the following numbers a to two decimal places by the use of formulas (37) and (38), and the value 2 loglO 6 = 0.868. (a) a == 2 (b) a = 3

(c) a (tl) a

=7

= 11

(e) a .. 0.13 (J) a = 0.17

6. Compute,

VIS

(a) (b) ~ (c) V35

(tl)

-¢'iO

(e) ~ (J) ~1005

(g)

-¢'68

(h) ~aiP

(,) {Ii

14•. Cartesian coordinates in the plane. Comple:x: numbers, like negative ~al numbers, are not magnitudes but may be conceived of as being directed magnitutie8. We shall indeed describe them as the points of a plane just as we described all real numbers as the points of a line. In"s. plane· we may construct the diagram shown in Fig. 7. This is the diagram of what is called a rectangular Cartesian coordinate 8YStem. It consists of a pair of mutually perpendicular lines intersecting at a point which we label 0

78'

REAL AND COMPLEX l,iUMBERS

[CHAP. m

and call the origin. The horizontal (west to east) line is called the z'a:tiB and is a real·line with positive direction to the right and having 0 as its zero point. The vertical (south to north) line is called'the 11 a:tiB and is a real line with positive ·direction upward and having 0 as its zero point. The point labeled (1,0) is called the unit point and its distance from 0 is the unit used in alZ measurements in the plane. Then (0, 1) is one unit above 0 on the 11 axis. We now set up a one-to-one correspondence between the points P in the plane and pairs (a, b) of real numbers.

o

\

(0,6) P

n

, (0,1)

I

6

o m

(1,0)

IV

FIG. 7.

or'

If P is given, the first mpnber a is called the ab8ti88a z coordinate of P. It is the properly signed perpendicular distance from the 11 axis to P. In Fig. 7 a is a negative nUmber. The second number b is called the ordinate or 11 coordinate of P. It is the properly signed perpendicular distance from the z axis to P. In the figure, b is positive. The pair of numbers a and b are called the coordinate8 of P. Whenever we give a pair (a, b) we mean that C!- and not b is the z coordinate. Conversely, if a and b are given, the point P with cOOrdinates a and b is uniquely determined as the intersection of two lines. One is the line parallel to the 11 axis which cuts the ·z axis at the real number point a, and the other is the

SEC.

79

COMPLEX NUMBERS

15)

line parallel to the z axis which cuts the 'V axis at the real number point b. The coordinate axes divide the plane into four p~s called the f[U(ldrantB. We label the quadrants I, II, III, IV in a counterclockwise direction. ORAL EXERCISES

1. State, for each quadrant, which coordinates of points in the quadrant are positive and which are negative. 2. How may the points on the :r; axis be described in terms of the coordinates of its points? Of the :r; axis? Of the line through 0 and the point (1, I)? Of the line through 0 and the point (1, -I)? Of the line through (0, 1) and (2, I)? Of the line through (1, 0) and (1, 2)? 3; Describe a construction by measuring distances on the coordinate aXis for the following points, and state the number of the quadrant in which the point lies. (a) (1,2)

(c) (-3,1)

(b) (2,1)

(d) (-1,3)

(e) (-2, -2) -1 -VS (J) 2'-2-

16. Complex numbers. The set consisting of all points in a plane becomes a number system when we define addition, subtraction, multiplication, and division for them. We define (39)

(a, b)

+ (e, d)

+

+

= (a e, b d), (a, b) - (e, d) = (a - e, b - d), (a, b) . (e, d) = (ae - bd, ad

+ be),

and, if (e, d) is not the origin, (40)

(a, b) (e, d) =

(00e· ++ dB' bd be - at!\ e· + dB r

Then it may be shown that all the laws that we have developed for addition, subtraction, multiplication, and division as well as the laws for integral exponents will hold for the number system· thus defined. We call this new number s~em the number 8'g8tem (or field) oj aU complez number8. It is not an ordered· set. .

80

.

REAL AND COMPLEX NUMBERS

[CHAP. m

The point (a, 0) on the 3: axis may be identified with. the co~sponding real numbers a, and the formulas (39) and (40) show that the number system of these points is a mathematical equivalent of the real number system. Then the complex number system may be said to contain the real ~umber system. It is customary then to call the 3: axis the real a:x:i8. Thus we shall write a instead.of (a, 0). The complex number i == (0,1) has the property that i2 == (0, 1) . (0, 1) == (-1, 0) == -1. The products bi == (h,0)(0, 1) == (0, b),

and the points bi are on the y axis. This line is called the imaginary a:z;iB and the numbers bi ~ (0,0) are called pure imaginary number8. Every complex number (a, b) may now be expressed as a sum

a == (a, b) == (a,O)

+ (0, b)

== a

+ bi,

where the real numbers a and b are the uniquely determined coordinates of a. We may now express the definitions given by formula (39) in the form . (a (a

+ bi) + (e + di) + bi) - (e + di) (a

+ In)(e + 00

+ bdi + (ad + bc)i == 2

and formula (40) for e

+ bi (a + In)(e e + di == (e + d~)(e -

a

+ e) + (b + d)i, e) + (b - d)i, .

== (a == (a di) ==

(00 - bd)

+ (ad + be)i,

+ di different from (0, 0), == 0 as d~)

di) ==

(00

+ bd) + (be - ad)i e2 + d'J. ae+bd be-ad. == e2 +d2 + elll+d'J.~·

Thus we may remember these laws as obtainable by adding, subtracting, multiplying, and dividing as is usual with algebraic expressions except that we always employ the property i'J. == -1 when we have a power ~'fIo. For example, il == -i, i' == 1, i& == i, etc.

SEC.

16]

81

COMPLEX UNITS

EXERCISES

1. Express the following complex numbers in the form a 2 + 3i - 3 + 2i (2 + 3i)(2 - 3i) (2 + 3i)(3 - 2i) [(2 i) - (3 - 2~')](3 - i) 1+3i (6) 1 _ 3i

(a) (b) (c) (d)

+

2. Use the definition

+ bi:

18 + i (f) 3 - 4i (1 + i)(1 - i) (g) 1 2i 3+3i (h) (1 + 2i)(1 - 3i)

+

v'=d = Vd i

for all positive real numbers d

to simplify the following expressions and write them in the form a + bi:

vi) + v=I6

+

2i - 3 v'=4 (a) 1 (b) (1 + V'=9)(1 - 3i) (c) (~+ (d)

V -5)(0 - v'=5)

V9 - 3i 1+v=t

(6) V16 + V -9 . (f)

V=4 + 2i vI=9 + 3i

1

16. Complex units.

The nU1l\ber a=a-bi (read a as II a conjugate ") is called the conjugate of the complex number a. If a is also a real number, then b = 0 and a = a. If a is not real, i.e., b ¢ 0, we call a an imaginary number and notice that a ¢ a, and that a-a=2bi¢0 is a pure imaginary number. It can be shown that if we replace i by -i throughout any algebraic expression A obtained from complex numbers a, {J, etc., by a finite number of additions, subtractions, multiplications, and divisions, the result is X. In particular a

+ {J

= a + 'P,

a - {J =

a - p, -

-

afJ = a {J,

(~)

\fl = Iip'

The number aa = a 2 + b2 is a real number which is not negative and is positive except when a is zero. The number

82

Rl!lAL AND COMPLEX NUMBERS

(ClIAP.

m

is a nonnegative real number which is called the ab80lute uaZue' of a. It is the distance from the origin to the point Ca, b). Thus we have the diagram in Fig. 8 and may think

of a as a directed distance of length lal. The angle (J measured in a counterclockwise direction from the ray Ox to the ray Oa determines the position of a and is called the amplitude of a.

" " FIG.S.

We now see that

I-al == lal,

I~I == VC~)(~)

==

v'aatiP == lall.81

~I == ~~)~) ~ ~ == ~.

It may also be shown that

la +.81

~

lal + 1.81,

Iial - 1.81\ ~ la - .81· I

If 'Ial == 1, we call a a complex unit. All complex units lie on a circle with center at· 0 and radius unity. Then every complex number a ~ 0 may be written as

a == a + bi

==

lal . C:I + ~) == Icq . IJ

':'9'here the complex number IJ has the property that

IIJI

all

==

+b

lal ll

ll

== 1.

slle.17]

FIELDS OF COMPLEX NUMBERS

83

Thus every complex number is the product of its absolute value and a cpmplex unit p.. Since allal and bllal are proportional to a and b the amplitude (J of a is also the amplitude of p.. We shall obtain further properties of complex numbers in Sec. 7, Chap. VIII. EXERCISES

1. Find the absolute values of the following complex numbers:

+ 4i 1+i

(a) 3

(e) (5

(b)

(I) (7

(d) (1.+ 2i) (1 -

2'1

+ ~1

(3 - 4i). (-3 - 4i)(7 - ~1 (0) 63 - 16i

~i

(e) 1 -

+ 1~1(1 -

&1

g. Find the absolute value and the amplitude (in degrees) for each of the following complex numbers:

+i

(a) -12 (b) 2

+ 2i + VSi

(e)

-34

(f)

-2 - 2V3i -VS+V3i V2 - V2i

(e) -1

(0)

(d) 2i

(ll)

17. Fields of complex numbers•. A set F of complex numbers is called a field if it has the following properties: 1. The 8'Um a + fJ oj any two number8 a in F and fJ in F i8 inFo 2. The difference a - fJ oj any two number8 a in F and fJ inF ill in F. 3. The product afJ oj any two number8 a in F and fJ in F i8 inFo 4. IJ a i8 in F and fJ i8 in F and i8 not Bero, the quotient alP i8 in F. mustrative Examples (FULL eoUBSE) I. Let' m be any integer without a cube as a factor and m > 1 80 that m has a positive cube root ~. Show that the set F of all numbers of the form a + b ~ + e W, where a, b, e are any rational numbers, is a field.

84

REAL AND COMPLEX NUMBERS

[CRAP. m

• Boiution H a -= a + b -¢m + c ~ 'and fI = d

+ e -¢m + I ~ml (a + d) + (b + e) ~m + (c + Jr ~ml

a

+ fI

a

-.fI = (a.-

is in p,

os

d)

+ (b -

then

_"e) v_'r.= m + (c - f) ~ ml

= r + 8 ~ + t -¢'fiii is in II where -¢'fiii = (~)Ito obtain • r == ad + blm + cern, 8 = ae + bel + cfm,

is in P, afJ

we use the fact that

(~)a = m,

t ... of + cd + be.

A number fI of II is zero unless one of d, e, I is not zero. Then we use our product formula. to compute what is called the norm of fl. It is the product N(fI)

= 'YfI,

'Y

==

(dl

-

me/)

+ (JIm -

de) ~

+ (e

de)/m

+ (el -

df)em .. dB

l -

df) -¢'fiii

and we compute

r 8

t

= (ell -

me/)d

+ (Jim -

+ elm + Jlml

== (dl - me/)e + (Jim - de)d + (el - df)fm = 0, = (dl - me/)I + (el - df)d (Jim - de)e ... 0 so that N(fI) -= d' elm + Jlml - 3mdef.

+

- 3me4f,

+

But then, if N(fI) pE 0, we have

1 'Y ~ ... N(fI)'·

a

~

==

CI'Y N(fI)'

and we need only show that, if fI pE 0, then N(fI) pE O. But if N(fI) = 0 for d, e, I not all zero, we can always multiply by the least common denominator and so have dl + elm + Jlml - 3mdef -= 0 for integers d. e,1 having no factor in common to all of them except 1. H P is a. prime factor of m, it divides d l and hence d, d = gp, p. divides

elm + Jlml - 3mdef. H pi does not divide m, then pi divides elm and hence p divides e, ·pl dividesJlmI, and p divides I, a. contradiction. Similarly, if m is exactly divisible by pi, then pi divides elm, p divides e, 1" divides dl, pi divides d, pi divides Jlml, p divides I, a contradiction. This proves that N(fI) pE 0 if d, e, I are not all zero and also that fI pE 0 if d, e, I are not all zero. It also completes our proof that II is a field.

8lIlC.

17J

FIELDS OF COMPLEX NUMBERS

85

IL Express the quotient 1+3-¢'2-~

8= in the form a

3-{li

+ b -V2 + e -¢'4. Solution

We compute d .. 3, e form

8 = (1

+ 3 -¢'2 -

= 0, J =

+

-1, 'Y" 9

+ 2-¢'2 + 3-¢'4 and

+

-¢'4)(9 2 -¢'2 3 {Ii) (3 - ~)(9 2 -¢'2 3 {Ii)

+

+

= 23 +:: {l2 = 1 + -¢'2. 1. 2. 8. 4.

ORAL BDRCISBS Is the set of all integers a field? Is the set of a.ll positive and negative fractions afield? Is the set of a.ll rationsl numbers afield? Is the set of a.ll real numbers afield?

BDRCISBS Show that the set F of a.ll numbers of the form a fJ v'iii is a field where m '" 0, m is a fixed integer without square factors, and a and b are any rational numbers. HINT: Verify that 1. (FULL co~).

+

1

c - d.ym

e+d.ym= e2 -dlm' and show that el - dim is not zero if e and d are not both Bero. 2. Express the following quotients in the form a b Vm where a and b are rftionsl:

+

( ) -1

+ V2

a 3-2V2 (b) 3

(e)

+ 4 VS

2+'\1'3 -2 - 3V6 4 + '\1'6 1-'\16

+ '\1'6)2 (3 + V5)2

(d) (1 (e)

2+V5

(J) (1 + (1 -

V2)' V2)' ('\1'6 - 3)(V2 + v'3) (g) (VS - 2)(V2 + Vi) (ll) 2 '\1'6 - 8 + 3 V2( VS - v'2) 2 + V6 + VS( V2 - '\1'3) (,") (V8 - 1)8(2 '\1'6 + VS)'

(VS + 1)1. 9 (~ (V2 + V5 + VS)( v'2 - V5 - VS) J

(VS + V5)2

86

REA.L A.ND COMPLEX NUMBERS

8. (lI'ULL

COURSE).

[CRAP. m

Express the following quotients in t1!.e form

a + h ~ + 3 -?Ii where a, h, e are ra.tional, by using mustrative Example I. 1

(a) (h)

1+~-~ 1+~ 1-~

~

(e) ~+i 1

(d) ~ _

9'4

() 6

1+~+2~ ~2 1-~-2-?1i

1-2~

(J)

5-4~-2-?1i

(g) 1

h

+2~ -

3 {Ii

3+2~+-¢'4

() -1+2~+~

18. Rational operations. Any algebraic expression, in which only the integral operations of addition, subtraction, and multiplication appear, is called an integral fUnction or a polynomial in its quantities. Such expressions are always manipulated according to the 10 laws for integral operations. An expression in which the indicated operations are the rational operationB of addition, subtraction, multiplication, and division (with denominators which must never be zero) is called a rational function of its quantities, and we shall always assume that the laws for rational operations hold. We may always simplify rational functions by the use of the 11 laws, the laws of exponents, and the formulas (3) to (11) for rational operations which we listed in the beginnjng of this chapter as properties of rational numbers. Thus every rational function may be expressed as a quotient of two polynomials. We pass on now to a study of rational functions.

CHAPTER IV POL YNOMIALS AND RATIONAL FUNCTIONS 1. Polynomials in x. A polynomial in z is a formal algebraic expression fez) = aoZ"

(1)

+ alz-l + ... . + an

(read "f of z equals a zero z to the nth, plus a one z to the 1, plus and so on, plus a sub 11."). It is a formal sum of 11. + 1 terms, the :first one being aoZ". The symbol 11. represents an integer which is positive or zero. The term ~ is called the general term of the polynomial and is the (i + 1)st term. The last term an is called the constant term of fez) and fits the formula. for the general term if we agree to identify 1 with zo, an with anzo. The symbols ao, aI, . . . , an are called the coejficients off(z). They represent unspecified ordinary complex numbers, and we shall refer to them as constants. In any particular problem they will be specified. When the coefficients of a polynomial are all zero, the polynomial is called the zero polynomial. We shall agree that all such polynomials are equal and are to be identified with the number zero. For example, 11. -

Oz·

+ Oz + 0 =

Oz

+ 0 = o.

H a coefficient tli in a polyUomial fez) is zero, the corresponding term ~,...... = Oz""'" is normally omitted. We shall agree to identify all polynomials that differ only in the presence or absence of terms with zero coefficients. For example, fez)

=

3z + 6

+ 7z -

5z8

2

=

+ 7z - 2 = 3z + Oz' + 5z + Oz· OZT + Oz8 + 3z + 5z8 + Oz· + 7z 6

6

87

8

2.

88

POLYNOMIALS

[CBAP. IV

If J(x) and g(x) are any two polynomials, we may adjoin terms with zero coefficients to one of them, if necessary, and thereby write them both like formula (1) for the 8ame n. For example, if J(x) = 3x6 + 5x3 + 7x - 2 and g(x) = 2X4 - 6x 2 + 2x - 1,

we may write J(x)

= 3x6 + OX4 + 5x3 + OX2 + 7x -

g(x) = Ox6

+ 2X4 + Ox

3 -

2,

6x 2 + 2x - 1.

Then we make th~ definition that two polynomial8 J(x) and g(x) are equal iJ corresponding coefficients are equal number8. If J(x) is not the zero polynomial, it has at least one nonzero coefficient. . We may therefore delete successively those early terms with zero coefficients until we arrive at a new first term with coefficient not zero. This will enable us to write any nonzero polynomial like formula (1) with ao ¢. O. When this is done, we shall call ao the leading coejJieient of J(x) and n the degree of J(xj. For example, Ox& + OX4 + 43;8 - 2x - 1 has degree 3, not 5, and leading coefficient 4. A polynomial like formula (1) with n = 0 reduces to its first (i.e., last) term aQX1Io = ao = an. It is called a constant polynomial. We give it the degree zero even when it is the zero polynomial. We shall sometimes refer to polynomials of degree n > 0 as nonconstant polynomials. A polynomial of degree n may also be called an n-ic polynomial. In the respective cases n = 1, 2, 3, 4, 5 the words linear, quadratic, cubic, quartic, quintic polynomial are used. If we replace the symbol :J by a complex number c wherever x occurs in a polynomial J(x), the result is a number which is usually designated by J(c). We call J(c) the re8'Ult oj sub8tituting c for x in J(x). For example if J(x) = 3x4 - 43;8 - 6x - 4 thenJ(1) = 3(1)4 - 4(1)8 - 6(1) - 4 = -11, J(2) = 3(2)4 - 4(2)8 - 6(2) - 4 = 48 - 32 - 12 - 4

= O.

SEC.

2]

89

SUMS, DIFFERENCES, AND PRODUCTS

The numbers J(c) are called the Value8 of J(z). We refer also to J(c) as the value of J(x) Jor z == c. It should be clear that if J(z) == g(z) then J(c) == g(c) for every complex number c. The valuesJ(O),J(I),J( -1) are computed very frequently, and it is worth while to observe thatJ(O) is the constant term an of J(x), J(I) is the sum of the coefficients of J(x), J( -1) is the sum of the coefficients of all even powers of z minus the sum of the coefficients of all odd powers of x. ORAL EXERCISES 1. Give the degree n and the coefficients ao, ai, . . . , following polynomials: (a) /(z) = 5z4 - az8 + 23;1 - 7z + 1 (b) /(z) = 3zI - Z8 + z (e) /(z) = 6z1 - &;8 + 9z - 7 + z4

a.. for the

(d) /(z) = -5zI + 1 (e) /(z) = Zl + 0:1;4 (f) /(z) = _ZlO + zI

+1

2. Compute/(O), /(1),/( -1) for the polynomials in Exercise 1. Describe the polynomials in (a) - (e) above with the terms "linear," "quadratic," etc. 4. Read off the first five terms of CJoZ" + alZ"-l + . . . + a... 6. Read off the last four terms in Exercise 4. 6. ,What assumptions about the size of n must be made if the results of Exercises 4 and 5 are to have meaning?

a.

2. Sums, difterences, and products. Polynomials are added by adding corresponding coefficients and' are subtracted by subtracting corresponding coefficients. For example, . I

+ 2z + 1) + (-2x' + 5:c8 + 3z - 2) == (3z11 + Oz' + OZ8 - 4:&1 + 2z + 1) + (Ozll - 2z' + 5z + Ozl + 3z - 2) == 3z& - 2z' + 5z 4:&1 + 5z -

(;i:c& - 4:&1

8

8 -

1.

If n is the larger of the degrees of two polynomials J(z) and g(z), we may write J(z) == ac¢'

+ a r + ... + an + . . . + b",. If they have dif1

1

and g{z) == boX'" + b1Zfl-l ferent degrees, one of ao and bo is zero.

~en

we have

90

POLYNOMIALS

written them this way, we have (2) J(:c) + g(:c) == (ao + bo):C" + (al

[CJWl.lV

+ bl):c-l + ... + (a. + b,,).

This is a polynomial whose degree is not larger than n. The result may be extended to sums of several polynomials and stated as follows: Theorem 1. The degree oj a BUm oj polynomial8 i8 not larger than the largeBt degree oj all the polynomial8 that are added.

The product of J(:c) == ao:c" + al:c-l + ... + a. of degree n and g(:c) == bl¢" + bl:tf'l-l + . . . + b", of degree m ~ the sum of all products (a;:c~) (b~l) == (a,:fJj)zff-f-m-+-j. All terms with the same exponen,t on :c are combined by adding their coefficients. If either J(:c) or g(:c) is the zero polynomial, theil· product is zero. Otherwise ao ~ 0, bo ;oS 0 and J(:c)g(:c) has precisely one term of degree n + m, ~ely aob~. This is the term of highest degree and we see that n + m is the degree of J(:c)g(:c). The result just obtained holds for products of several polynomials. We call the polynomials that are multiplied the Jactor8 of the product and state the res~t as follows: Theorem 2. The degree oj a product oJ' norIMrO PQb/nomial8 i8 the BUm oj the degre~8 oj ita Jactor8. The leading coejficient oj any product i8 the product oj the leading fXJ6;ffi,ti,enf,B oj ita Jactor8, and the COnBtant term oj a product is the product oj the comtant termB oj ita Jactor8. This result has a simple consequence. Theorem 3. A product oj fW'TIMrO polynomial8 i8 not sero and i8 a 'nO'TI,S8f'0 COnBtant if and only iJ aU ita Jactor8 are no_o conBtantB. . For the degree of the product is a sum of degrees. These degrees are all natural numbers and their sum can be zero only when the degrees we add are all zero, the factors are all constants. . The product of a polynomial ao:&" + alZ-l + + a. by a constant polynomial b is

SEC.

2)

SUMS, DIFFERENCES, AND PRODUCTS

91

In particular, (-l)f(x)

=

=

-f(x)

(-ao:t")

+ (-alX-l) +

+ (-an)

and any differencef(x) - g(x) = f(x) + (-l)g(x). Moreover, Theorem 1 holds not only for sums but also for ~erences. Indeed it is true for sums of constant multiples of polynomials. We may now state that any algebraic expression obtained by a finite number of applications of the integral operations of addition, subtraction, and multiplication on x and constants is a polynomial in x. Moreover, our definitions 'of polYnomial addition and multiplication are such that our laws for integral operations and for nonnegative integral exponents are satisfied for polynomials. Then it will be true that if we replace x by any number in two formally different expressions for the same polynomial the resulting numbers will be the same. For example, we have the polynomial equality (3)

(3x - 2)2(X

+ 1)

- x(2x

+ 1)(2x -

=

+

1) - 5(x 1)x2 2 - 8x - 7x 4.

+

Then we are stating that, in particular, (3 . 1 - 2)2(1

+ 1)

- 1(2 . 1 + 1)(2 . 1 - 1) - 5(1 + 1)(1)2 = -8(1)2 - 7(1)

+ 4.

ORAL EXERCISES

1. Give the degree n and the coefficients ao, aI, • . • , following polynomials: (a) (&;3 - 4X2 + 2x - 1) + (-2:1: 3 + 2:1;2 + 2x + 3) (b) 3(2x5 + 4:/: 4 - 3x 3 + X2 - 1) ....,. (8x 5 + 11x4 + 2)

+ +

,

+

+

a.. for the

+ 2(x + X + 2) + 2(3x2 - 1) 6

(c) 1 2x 3x2 - 4:/: 4 2 - x - 3x 3 - 3 2X4 (d) 2(2x3 - 3x 1) - 3(x 3 2X2 x) - (x 3 ' - 9:1;)

+

+

+

2. Give the degree, the leading coefficient, and the constant term a.. of the following polynomials/ex) by the use of Theorem 2 and/(O) == a..:

92

POLYNOMIALS

+

+ + +

[CRAP. IV

·(a) (3x& - 43;2 1)(2x - 1) (x 3)3 (b) (Xl - 1)&(2x3 - 3)(a;3)(x 1)2 (C) (X 3 1)(x2 - l)(x' 2)(2x - 1) (X 3 x)(2x2 - 1)(x3 2)xl (d) x&(x - 1)x3(x l)(x 2) - x'(x l l)(x& 1) (x' - 1)3 (6) (3x' - 1) (3x2 - l)(xl 2) - (2x' - 3) - 2x'

+

+

+ +

+

+

+ + + + +

+

3. Computation of polynomials. When it is necessary for a student to compute many polyllomi,us, it becomes clear that certain types of expressions occur so frequently as to warrant the development of formUlas for computing them. Some of these formulas are (4)

(x

+

+ +

(x y)2 = x 2 (x y)8 = x 3 y) (x - y) = X2

+ 2xy + y2, + 3x2y + 3xy2 + y8, y2.

-

For example, we compute f(x)

=

(3x - 2)2(X-+ 1) - x(2x

+ 1)(2x -

1) - 5(x

+ l)x2

by their use to obtain f(x)

=

+

+

(9x 2 - 12x 4) (x 1) - x(4x 2 - 1) - 5(x8 X2) = 9x8 - 12x2 4x 9x2 - 12x 4 - 4xs x - 5x 3 - 5x2 = -8x2 - 7x 4.

+

+

+

+

+

+

There is a procedure slightly different from that just used for computing polynomials. It differs principally in the way we select and order our terms. It is a valuable time- and space-saving procedure and the student will find that it is well worth his while to drill himself in its use. We shall call this process the single eXponent process. We select a fixed exponent i and compute all coefficients of the power x' which occur in the terms of which f(x) is the sum. Then the sum of these coefficients is the coefficient of x' inf(x). The process begins with the determination of the coefficient of the highest power of x which appears in all the terms of f(x). This coefficient will be the leading coefficient of f(x) unless it is zero. The process terminates with the computation of an = f(O). The details of the process are amplified in the examples below. .

SEC.

3]

93

• COMPUTA.TION OF POLYNOMIA.LS IUustrative Examples

I. Compute fez) ... (3z - 2)I(Z

+ 1) :.... z(2z + 1)(2z -

1) - 5(z

+ l)zl.

Remarks: We first write fez)

12z + 4)(z

= (9z1 -

+ 1) -

Z(4z1 - 1) - 5z8 - 5z1l.

The maximum exponent is three, a.nd terms in Zl are leading terms of our products. Terms in Zl are products of z by Z, Zl by constants. Terms in z are products of z by constants. Solution The coefficient of Z8 is 9 - 4 - 5 ... 0, of Zl is 9 - 12 - 5 ... -8, of z is (-12 + 4) + 1 = -7. Also f(O) = 4. Hence

•

fez) ... _Szl - 7z

+ 4.

II. Compute the coefficient of Zl in the polynomial (2z1 - 1)(z + 2) + (2z + 1)(zl - 2z - 2) + 2(3z - l)(z Solution

The coefficient is (2)(2) + 1 + 2( - 2) III. Simplify the expression f( ) - 6(z - 3)(3z + 1) z 10

+ (2)(3)

+

1) -3(z - 1)1.

- 3 = 4.

+ 8(3z + 1)1.

Solution In the numerator the coefficient of Zl is

(6)(3)

+ (8)(9)

that of z is 6(1 - 9)

+ (8)(6)

Hencef(z) = 9zll

1.

+ 72 ... 90,

... 0, and the constant term is

6( - 3) -

... 18

+8 =

-10.

EXBllCISES

1. Use the single exponent method to compute the coefficients of the following polynomials: (a) fez) = (3z' - 4)(z + 1) - (2z + 1)(z - 2) (b) fez) ... 2z(ZI + 2z - 1) - (Zl + 3z + 2)(z - 1) (c) fez) - (22; + 3)(22; - 3)z - (2z1l + 2z + l)(z - 1) (d) fez) ... (2z1 + 3z -,I)(z' - 2z - 2) - z8(2z - 1) (6) fez) ... (z' - z - 1)(z8 + 2z + 1) + (z + 1)(z8 + 2) (f) fez) ... (z + 1)(zl + 1) - 2(2z1 - 1)(z - 1) + 3z1l(Z - 2)

POLYNOMIALS

[ClUP,lV

So Compute the coefficients' of ~a and ~ in the following polynomials: (a) (~. 23; - 1)(~2 - ~) - ~(~ 1)(23;) - ~(~ 1)

+ + + + 1)1 -' (~ + 1)(23; - 2) + (~ + 1)2 (c) ~I(~ + 1)(~ - 1) - (~ + 1)(~ - 1) + 2(~2 + 1)(~) (d) 2(~ + 1)2:1: - ~(~ + 3)(23; - 3) + (~2 + 1)~ & Comp1,lte the coefficient of ~ in the following expressions: (a) '3(~ - m1/)1 + 2(~ - my)(~ + 1/) - 4(~ + 1/)2 (b) (~- m1/)2 + 6(~ - m1/)(~ + 1/) - 9(~ + 1/)1 (c) -2(~ - m1/)1 + (~ - m1/)(~ + 1/) - 3("":" + 1/)1 (b) ~(~

'- The division algorithm. The division algorithm for integers has a polynomial analogue which We state as follows: DIVISION ALGORITHM FOB' POLYNOMIALS. Let J(z) and g(z) be polynomialB BUCk fAat g(z) i8 not a comtant. Then there ezi8t unique polynomial8 q(z) and r(z) BUck Uw,t the def!"ee, fJJ r(z) illle88 than that of g(z) and

(5)

J(z)

== q(z) , g(z)

+ r(z).

When the degree uJ J(z) ill lu8 than that oj g(z) the polynomial q(z) == 0 and I(z) , = r(z). Otherwi8e the degree oj q(z) ill the degree oj J(z) minu8 the degree oj g(z). From the statement of our algorithm it should be clear that we need only prove the existence.of q(z) and r(z) in the case where J(z) = aoz" + alz"-l + . . . + a,., g(z) = boZ~

+ b1:r-1 + ' , . + bm,

ao pi! 0, bo pi! 0 and n ~ m > O. Then the ordinary so-called long-diviBion prOce88 begins with the subtraction J(z) - aoZ-bo-1g(z).

This eliminates the term aoZ", and so yields a polynomial of degree at most n - 1. A finite number of such subtractions of multiples of g(z) reduces the degree' of the final difierence r(i) below that of g(z). Then we have formula (5) where q(z) begins with aobo-lz- and has the degree stated in the algorithm.

SEC.

5)

REDUCIBILITY OF POLYNOMIALS

95

To show that q(x) and rex) are unique we suppose that + t(x) where the degree of t(x) is less than m. Then by Theorem 1 the degree of

f(x) = 8(X) . g(x)

rex) - t(x) = [8(X) - q(x)]g(x)

is less than m.

This is not possible, by Theorem 2, unless For g(x) has degree m. Hence

8(X) - q(x) = O.

8(X)

=

q(x),

= 0, rex) = t(x). We note that when c is any number and g(x) = x - c the remainder polynomial rex) has degree zero and is a constant r,

rex) - t(x)

f(x)

(6)

= q(x)'. (x

- c)

+ r.

When g(x) is a quadratic polynomial x 2 remainder polynomial rex) is linear, (7)

f(x)

=

q(x)(x 2

+ ax + b,

the

+ ax + b) + ex + d

for constants c and d. The student will have had practice in the division process from his elementary algebra experience. If further drill is deemed desirable, it is best provided for by the exercises of Sec. 7 on the g.c.d. process. 6. Reducibility of polynomials. A polynomial f(x) is divisible by a polynomial g(x) if f(x) = g(x)q(x) where q(x) is a polynomial. When g(x) is a nonconstant polYnomial, this occurs only if the remainder rex) of the division algorithm is the zero polynomial. The equality f(x) = g(x)q(x) is called a factorization of f(x) with polynomial factors (or divisors) g(x) and q(x). Every polynomial has the trivial factqrizations (8)

f(x)

= a[(r~(x)]

for any number a ;;t: O. Thus every polynomial has all nonzero constants and all nonzero constant multiples of itself as trivial factor8.

96

PO LYNOMIALS

[CHAP. IV

If a polynomial has no factorizations except its trivial factorizations, we call it an irreducible (or a prime) polynomial. All linear polynomials are irreducible. When a polynomial does have nontrivial factorizations, we call it a reducible polynomial. For example Z2 -

3z

+ 2 ==

(z - 1) (z - 2)

is reducible. The polynomial Z2 - 2 == (z + '\1'2) (z ~ V2) and so is a reducible polynomial if we permit the use of irrational coefficients. However if we insist that only rational coefficients be employed this polynomial is irreducible. For otherwise we Inay use the result about leading coefficients in Theorem 2 to write Z2 - 2 == (z - a)(z - b) for rational numbers a and b. Then a2

-

2 == (a - a)(a - b) == 0,

a2 == 2. This has already been shown to be impossible. It should be clear now that the question of the reducibility or irreducibility of a polynomial depends upon the nature of the coefficients that Inay be used. We shall assume throughout the remainder of this chapter that the coefficients will be restricted to lie in some fixed field F of complex numbers. The most important cases are those where F is the field of all rational numbers or the field of all re8J. numbers or the field of all complex numbers. We shall thus develop results holding for any of these cases. S. The factorization theorems. A reducible polynomial J(z) has degree 11. '> 1 and a nontrivial factorization J(z) == g(z)q(z).

Neither of the nontrivial fagtors g(z) and q(z) is a constant and the degree of J(z) is the sum of the degrees of these factors. We phrase this result as follows: ·Lemma 1. The degree oj a nontrivial Jactor oj J(z) i8 le" than the degree oj J(z).

SEC.

6]

THE FACTORIZATION THEOREMS

97

If f(x) = g(x)q(x) and g(x) = h(x)s(x) are nontrivial factorizations then f(x) = h(x)s(x)q(x) is a nontrivial factorization. in which we have three factors. Lemma 1 implies a lowering of degrees. Such a process must terminate and can terminate only when we arrive at a factorization in which all factors are irreducible. This gives us the first of our factorization theorems. Theorem 4. Every reducible polynomial is a product f(x) = Pl(X)P2(X) ••• Pt(x) of a finite number of irreducible polynomials.

By Theorem 2, the leading coefficient of f(x) is the product of the leading coefficients of its factors. We may thus remove these coefficients and carry their product to the left so as to obtain the following: Theorem 6. Every reducible polynomial is a product (9) of its leading coefficient ao and a finite number t of irreducible polynomials Pi(X) each having leading coefficient unity. We shall present a proof later that there is essentially only one factorization of the kind in formula (9). Of course

the order of the factors is not unique. (x - l)(x - 2)(x

+ l)(x -

l)(x - 3)

= (x - 3)(x - 2)(x

For example

+ l)(x -

l)(x - 1).

However, both the irreducible polynomials Pi(X) with leading coefficient unity and the number of times they occur in formula (9) are unique. If an irreducible polynomial p(x) occurs exactly e times as a factor of f(x) in formula (9), we may group these factors together and write the product as p(X)8. Then p(X)8 will divide f(x) and p(X)8+1 will not. We shall call e the multiplicity of the irreducible factor p(x) of f(x). EXERCISES

1. In the following polynomials the irreducible' factors are linear. Write the polynomials as products of constants by powers of distinct

98

POLYNOMIALS

[CHAP. IV

linear factors with leading coefficient one and state the multiplicity of each factor. (a) (b) (c) (d) (II) (J)

+

+

(z - l}(zl - 33; 2}(Z' - z - 2}(z 2}4 (z - l}'(z l}l(z - l}(z - 2}4 (2z 1}'(33; 1}(2z 1}(4:r;1 - I) . z(z l}8zI(zl - l}'(z' zl) (:r;4 - 33;8 2z1}(Z4 - z8)(ZI - 2z) (z 1}8(z - l}l(zl - 1}4(z' - 9}~(Z' 4:r; 3)' (Zl - 2z - 3}(2:r;1

+ + + + + +

+ +

+ +

+ 4:r; -

6).

I. In the following polynomials the factors, as written, are powers of polynomials irreducible in the field of rational numbers. Give the multiplicity of all the irreducible factors with leading coefficient unity. (a) (b) (c) . (d) (II) (J)

+ 1}8(z - 1}1(2z1 + 2}(z + I) - 2:r; - 2}'(2z' - 4}8z + 2}(z. + 2}(ZI + 2)

(Zl Z8(ZI (Zl (Zl (33; (2z1

2}I(ZI - 2}1(2:r;. - 4)' 6}I(Z - 1}4(z - 2)' 4}(ZI 2}(4:r;1 8)&

+

+

+

7. The EucHdean g.c.d. proc~ss

(FULL COURSE).

A c0m.-

mon diviBor of two or more polynomials is a polynomial

factor of all of them. The g.c.d. of several polynomials, not all zero, is that common divisor of largest degree and leading coefficient unity. It can be shown that there is only one such polynomial. The g.c.d. of two polynomials J(:x;) and g(:x;) ~ 0 may be computed by a polynomial analogue of the process of Sec. 7 of Chap. II. We first prove the following: Lemma I. Let g(:x;) ~ 0 be a divi80r oj J(:x;). Then the g.e.d. oj J(:x;) and g(:x;) is a eomtant multiple oj g(:x;). For every common divisor of J(:x;) and g(:x;) divides g(:x;). Thus the degree of their g.e.d. cannot exceed that of g(:x;). However g(:x;) is a common divisor of J(:x;) and g(:x;) and has

the maximum degree possible. Lemma 3. Let g(:x;) ~ 0, a and b be any nonuro

+

c0n.-

stants, J(:x;) = q(:x;)g(:x;) r(:x;). Then the common divisor8 oj J(:x;) and g(:x;) eoincide with the common· diviBor8 oj ag(:x;) and br(:x;). Hence the g.e.do oj J(:x;) and g(:x;) is that oj ag(:x;) and br(:x;).

800.7]

THE EUCLIDEAN G.C.D. PROCESS

99

For, if c(x) divides ag(x) and br(x), it divides f(x) == q(x)g(x)

1- b-1br(x)

and ag(x). Conversely, if c(x) divides f(x) and g(x), it divides ag(x) and br(x) == b[J(x) - q(x)g(x)]. Lemmas 2 and 3 result in a process for computing the g.c.d. d(x) of f(x) and g(x) ~ O. If g(x) divides f€x), the product of g(x) by the inverse of its leading coefficient is d(x). Otherwise we use the division algorithm to :find the polynomial rex) of Lemma 3. (The constants a and b are used to avoid fractions in the case where we study polynomials with integral coefficients.) If rex) divides g(x), some constant multiple of it is d(x). If not we divide g(x) by rex) and get a new remainder of lower degree. The process terminates with a :final nonzero remainder which is, apart from a constant factor, the required g.c.d. Let hex), f2(X); ••• ,fe(x) be any nonzero polynomials and ~(x) be the g.c.d. of hex): . . . ,h(x). Then it may be shown that the g.c.d. di+l(X) of hex), ••• ,h(x), fi+l(x) is the g.c.d. of d,(x) andfi+l(x). Thus the division pro.cess above may be utilized to :find the g.c.d. of any number t of polynomials. . In the rather rare case where polynomials are given in factored form, their g.c.d. may be computed trivially. Indeed it is the product of all the prime polynomials occurring in the various factorizations but with exponents the least that occur for the individual primes in all the polynomials. For example, let hex) = (x - 1)8(X 1- 1)2(X2 1- 1)(2x - 4)4, hex) == (x - 1)6(X 1- 1)8(2x - 4)4, f8(X) == (2x - 4)3(X - 1)4(X2 1- 1)3(X 1- 1)2.

Then the least exponent for x - 1 is 3, the least for (x 1- 1)2 is 2, the least for X2 1- 1 is 0, the least for x - 2 is 3. Thus the g.c.d. is (x - 1)3(X 1-·1)2(X - 2)3. Thls procedure is of relatively little use and so we shall not dwell on it further.

100

POLYNOMIALS

[CJLU>.

IV

The l.c.m. of two nonzero polynomials f(:r;) and g(:r;) is that polynomial M(:r;) with leading coefficient unity which is divisible by both f(:r;) and g(:r;) and whose degree is the smaJIest of aJl nonzero polynomials divisible by both f(:r;) and g(:r;). It may be shown that, if ao, bo are the respective leading coefficients of f(:r;) and g(:r;) and 'if fl(:r;) is their g.c.d., then f(:r;)g(:r;) aobod(:r;)

is their l.c.m. M(:r;). The l.c.m. of a set of polynomials /l(x), .•. ,f,(:r;) may be computed by using the property that if M,(:r;) is the l.c.m. of /l(:r;), ••• , f.(:r;)· then the l.c.m. of /l(:r;), ••• , f'+I(:r;) is the l.c.m. of M,(:r;) and f,+I(:r;). The computation of the l.c.m. is not particularly important and we shaJl only illustrate it below. DIU8trative Examples

I. Find the g.c.d. of z' - 2:z:1 - 21; - 1 and Zl - 2:z:a - 2:z:1 - 3:z: - 2.

Solution Zl

-2 a + ZZ4 - 2:z:I + lz~.

.+2

- 2:z: - lzl - 2z' - 2:z:1 - 3:z: - 2 za+zz' +Zl zl-2z4 -2zl - z o _2:z:8 zl-2z 2z'-2z' -2z-2 _21;8 -2z 2z'-4:J:a -4:J:-2 -Zl -1 21;8 +2z

Hence the g.c.d. is Zl

+ 1.

Note that since z is not a. factor of

g=z'-2:z:'-2z-1

+

+

the g.c.d of g and Zl z is the same as the g.c.d. of g and Zl 1. So we could save one step in the process. However this follows only when we have the factorization theory of Bee. 9. II. Find the g.c.d. of Zl + 2:z:1 - Z - 2, Z8 - Sa: + 2, .:Z:B + z - 2. Remar1D~ It would seem simplest to find the ,g.c.d. of Zl - 3:z: 2 and Zl + z - 2 and then the g.c.d. of the result of this first step and Zl + 2:z:1 - Z - 2. .

+

SEC.

THE EUCLIDEAN G.C.D. PROCESS

7]

101

Solution _ _--=!Z2+3Z+2 . z2+z+2 1 + z - 2 Z8 z - 1 Z8 + 2Z2 - Z - 2 z - 1 Z8 Z8 - Z2 Z3 - Z2 Z8 --Z2 + Z ' 3z2 - Z 3z2 -3z Z2_ Z 2z-2 --2z-2 2z-2 2z-2

o

' - 3z + 2 + z - 2 - 43: + 4

0 Am. G.c.d. = z - 1.

ITI. Find the l.c.m. of the polynomials in Illustrative Example IT.

Solution

+

= Z8 z - 2 = (Z2 + z + 2)(z Z8 - 3z + 2 = (Z2 + z - 2)(z so that Ml = (Z2 + z + 2)(Z2 + z - 2)(z - 1). fa = Z8 + 2Z2 - Z - 2 = (Z2 + 3z + 2)(z - 1) . = (z Ml = (Zl z + 2)(z + 2)(z -

/I

+

- 1), 1), Now + 1)(z + 2)(z - 1), 1)2

and, by using either this factored form or the g.c.d. process, we find that the g.c.d. of Ml and!8 is (z + 2)(z - 1). Then

M = (Z2 + z + 2)(z + 2)(z - 1)2(z + 1). EXERCISES

1. Find the g.c.d. of each of the following pairs of polynomials: (a) Z8 - 3z2 + 4

z8_2:l: 2 _Z+2

(f) z, + z + 2 z'-z2-1

(b) Z8 - Z2 + 2 Z8 - Z2 z - 1

Am. 1. (g) z, - Z8 - 7z 2 + z + 6 Z8 + Zll - 4:1; - 4

(c) z, - 3z 2 + 2

(k) z8 - 2:/:6

+

Ana. Z2

Z4 - Z2 - 2 (d) Z4 + Z8 - Zll + z - 2

(~)

Z4 + Z8 - 3z 2 - Z + 2

+ 43:2 + 1 z, - 4:1;2 + 8z - 7

(e) z,

W

+ Zl + 2:/:2 -

+ 3z + 2.

Z- 1 Z8 + z& + Z4 - Z - 2 Am. Z4 - 1. Z8 - 2:/: + 4 Z8 - 2:/: 4 + 43:3 + Z2 - 2:1: + 2 Am. Z2 - 2z + 2. z4-2z 3 +z 2 -1 z& - &8 + 2:/:2 - 1

Am. 1.

-102

POLYNOMIALS

[CHAP. IV

2. Find the g.o.d. of each of the following sets of polynomials: (c) Z4 - 23;3 - 2z1 - 2z - 3

(a) Z4 - 1

+

Z8 + 3z2 + 3z + 1 z8-2z 2 -3z

Z8 2z' - z - 2 z4-2:I)8+2z2 -2z+1

.

Ans. z - 1.

. (b) Z4 - 2:1)8 - 5z2 + 6z

Am. 1. (d) z& + 2Z4 - Z8 - 5z2 - 6z - 3

Z4 - 7z2 + 6z z4+43;8+3z2 -43;-4

Z8 + Z2 + 3z + 3 z4+ z 3 - z -1

Am. z - 1.

Am. z

+ 1.

(6) Z6 - 11z2 - 9z - 2

z8-3z 2 +3z-2 Zl1 - 2z& - 121z3 - 198z2 - 100z - 24 (f) Z8 + Z1 - 3z& - 43;4 + z.8 + 3z - 3 :l)6_:I)4+Z2_2:I)+2 z8 + 2:1)1 - 3:1)6 - 7z4 - 2:1)8 + 3z + 6 (g) :1)1 + 6z8 + 20 Z8 - 3z8 - 10 :1)8 - 2:1)4 - Z8 - 43; - 6 . . (h) Zl - :1)8 - 3:1)1 + 3z8 - z& + 43;4 - 5z2 + 43; - 2 :1)4 + :1)8 - :1)1 + 2z - 3 :1)1-2:1)+1

8. Linear combinations (FULL COURSE). A linear combination of two polynomials f(x) and g(x) is an expression a(x)f(x)

1- b(x)g(x),

where a(x) and b(x) are also polynomials in x. Then we have the following: Lemma 4. Let u(x) and v(x) be linear combinatio'M of f(x) and g(x). Then any linear combination of u(x) and v(x) is also a linear combination of f(x) and g(x). For suppose that u == af + bg, v == cj + dg, where all the letters represent polynomials in x. Then 8'U + tv == saf + sbg + tcj + tdg == (sa + tc)f + (sb + td)g is a linear combination of f and g•. We may use the simple result to prove the following.: Theorem 6. The g.c.d. of two nonzero polynomials f(x) and g(x) is their only common divisor of the form (10)

d(x) == a(x)f(x)

+ b(x)g(:e).,

which has leading coejficient one. "

SEC.

8]

LINEAR COMBINATIONS

103

For the first step in our g.c.d. process expresses f(x) in the form f(x) = q(x)g(x) + rex). Then g(x) = 0 . f(x)

+ 1 . g(x)

and rex) = 1 . f(x) + [-q(x)]g(x) are both linear combinations of f(x) and g(x). Each step in our process will then begin with a dividend and divisor polynomial which are linear combinations of f(x) and g(x) and yield a remainder polynomial which is also a linear combination of f(x) and g(x). Thus d(x), which is a constant multiple of the final remainder, has the form (10). Every common divisor c(x) of f(x) and g(x) divides d(x) if formula (10) holds. n c(x) = A(x)f(x) + B(x)g(x), then d(x) will divide c(x). But c(x) and 'd(x) differ by a constant factor and are equal if they both have leading coefficient 1. The result just proved implies that ,the g.c.d. d(x) is unique. For any other common divisor of the same degree aEl d(x) would divide d(x) and would then differ from d(x) by a constant factor just as does c(x) in the proof above. The problem of actually determining the polynomials a(x) and b(x) of formula (10) is easily solved but we shall not emphasize it by a set of exercises. n exercises are wanted, the polynomials of the Exercise 1 of Sec. 7 may be employed. We illustrate the procedure below. . nlu.trative Example , Find the polynomials a(z), b(z) of formula (10) for g(z) = Z4 - 2z8 - 23: - 1, I(z) = z& - 2z' - 2z1 - 3:r; - 2.

Solution By Dlustrative Example I of Sec. 7 we have g = (z a = (z -

a

1 = (z + 2)g + 2r,

+

2)r 80 that 2r = 1 - (z 2)g, 2)r - g = (z - 2)trJ - (z + 2)g] - g = !(z - 2)/

- !(Zl - 4)g - g = i(z - 2)J + i(2 - Zl)g. To check we compute (z - 2)1 + (2 - Zl)g = :r;I - 2z' - 2z' - 8z1 - 2z - 2z& + 4:/:8 + 4:/:1 + 6:r; + 4 + 2z' - 4:/:' - 4:/: - 2 - Zl + 2z& + 2z1 + Zl =2z l +2-2d

104

POLYNOMIALS

and have shown that a(z) of formula (10).

= Hz -

2), b(z)

[CHAP. IV

= t(2 -

Z2)

are solutions

'

9. The unique factoriZation theorem (FULL COURSE). Two nonzero polynomials f(x) and g(x) are called relatively prime if their only common factors are constants. Then their g.c.d. is 1 and, by Theorem 6; there are polynomials a(x) and b(x) such that (11)

a(x)f(x)

+ b(x)g(x)

== 1.

Conversely when formula (11) holds, every common divisor of f(x) and g(x) divides 1 and is a constant. We have proved the following: Theorem 7. Two nonzero polynomia18 f(x) and g(x) are relatively prime if and only if there are polynomia18 a(x) and b(x) 8UCh that formula (11) hold8.

We then have the following: Theorem 8. Let f(x) and g(x) be relatively prime aJ1,d let f(x) divide the product g(x)h(x).

Then f(x) divide8 hex).

For we may use formula _(11) to obtain h(x) == [a(x)f(x)

+ b(x)g(x)]h(x)

==' a(x)h(x)f(x)

+ b(x)g(x)h(x).

By hypothesis, g(x)h(x) == q(x)f(x), hex)

== [a(x)h(x)

+ b(x)q(x)]f(x)

as desired. Theorem 9. Let f(x) be irreduc?'ole.

Then f(x) and g(x) ' are either relatively prime or f(x) divide8 g(x). For the g.c.d., of f(x) and g(x) divides f(x). If it is 1,

the polynomials are relatively prime. Otherwise it is a multiple of f(x) and divides g(x), so doesf(x). Theorem 10. An irreduc?'ole polynomial divide8 a product of polynomia18 if and only if it divide8 one of the factor8. Theorem 8 implies that f(x) divides a product only if f(x) and at least one of the factors of the product are not relatively prime. By Theorem 9, if f(x) is irreducible it

divides this product.

Be. 10] POLYNOMIALS IN SEVERAL SYMB OLS 105

We may now use Theorem 10 to prove that the factor:ization in Theorem 5 is unique. The result is trivial for polynomials of degree 1, and we assume that it is true for po1ynomials of degree less than the degree n of fez) = aOP1(z), . . . ,p,(z) = aoql(z), ••• ,q.(z). Here the polynomials Pl(Z), .•• ,p,(z), q,(z), ••• ,q.(z) are irreducible and have leading coefficient 1. By Theorem 10, Pl(Z) divides one of the q,(z) and, since both are nonconstant irreducible polynomials with leading coefficient 1, they must be equal. This proves that the factorization formula (9) of Theorem 5 is unique apart from the order in which we write the factors. 10. Polynomials in several symbols. A polynomial in z and y is an algebraic expressionf(z, y) (read "f of z and y") obtained by formally applying a finite number of integral operations to z, y and constants. It may be regarded as a polynomial in z and so written uniquely in the form (12)

fez, y) = ao(y):r

+ al(y):r-1 + . . . + a".(y).

.

The coefficients a.(y) will, of course, not be constants but polynomials in y, and m is the degree in z of fez, y); ao(Y) is not the zero polynomial unless fez, y) is the zero polynomial in z and y. The polynomialf(z, y) may also be written in the form (13)

fez, y) = bo(z)y'

+ b1(z)y-l + . . . + b,(z).

The coefficients bo(z) , ••• , b,(z) are then unique polynomials in z, t is the degree in y of fez, y), bo(z) is not zero unless fez, y) is the zero polynomial. Both formulas (12) and (13) ~xpress fez, y) as a sum of constant multiples of power products ziyi. The two expressions differ only in the manner in which we have ordered and grouped the terms. If we igno~ these groupings, we may define the degree of ziyi in both z and y to be i + j and the degree of fez, y) in z and y to be the largest degree of

106

POLYNOMIALS

[CHAP. IV

any of its terms with cOlU!tant coefficient not zero. example, I(x, y) = 3x4 + 5X8y8 + 6x2(y6 + 2y4) + y8

For

I

is'a polynomial of degree four in x, six in' y, seven iIi x a~d y. It can be written also as y6 + 6xly6 + 12x2y4 + 5x8y3 3x4 or as 6x8y6 (5X 8y8 + 12xly4 + y6) + 3x4, the laSt expression being an arrangement according to descending degrees in x and y. If all the terms of \ a polynomial have the same degree in all the symbols x, y, z •.•• , we call the polynomial a homogeneou8 polynomial. For example, 3x2 + 2Xy - 5y2 is homogeneou8 01 degree two in x and y. Every polynomial I(x, y) of degree n can be written as a sum

+

+

(14) I(x, y)

= lo(x, y)

+ /1 (x, y) + ... + In(x, y)

of homogeneous polynomials/i(x, y) of degree n "'7'" i. The leading polynomial lo(x, y) has degree n and is not zero unless I(x, y) = o. An example of such an expression is I(x, y) == (x6y

.

+ 3X8y4 + 6xly6) + (2x4y2 + 3xy6)

in which lo(x, y) /1(x, y) 12(X, y) h(x, y) /&(x, y) 16(X, y)

.

+ (2xly -

+ +

y8)

+ (3x 2 -

xy)

+

== x6y 3X8y4 6x 2y6, == 2x4y2 3xy6, == 18(X, y) = 0,

= 2xly -

y8, 2 = 3x - xy, == 17(X, y) = o.

Note that in our example we have written each homogeneous polynomial as a polynomial in x with coefficients constant multiples· of powers of y. Homogeneoq,s polynomials are sometimes called lorms. . The general polynomial I(x, y) of degree two is Qf considerable importance in other branches of elementary mathematics. It may be written in the fo~ (15) Ax2 + BX'Jj + Cy2 + Dx Ey F

+ +

SEC.

10]

POLYNOMIALS IN SEVERAL SYMBOLS

107

for constants A, B, 0, D, E, F which are usually real numbers. The de~tions above can be extended to polynomials in any number of symbols. When the number of symbols is large, it is customary to use Xl, X2, • • • ,Xn rather than X, y, s, t • .•. We shall be interested later principally in linear functions '

and in quadratic forms allXl 2

+ a22Z2 2 + . . . + a.nXn2 + 2a12XIX2 + + 2a-InXn-IXn. ORAL EXBRCISES

1. Give the degree in z, in y, and in z and y together of the following polynomials: (a)· &;Iy + &;'111 - 2zy8 - 3y' (b) -2:/:1 zlyl - Zl &;'y zill 5 (c) -zly! zy' &;& 2z'y (d) -2:/:iI' + 6z'1/' + &'yB - 2:/:11 (8) 4:/:1 3zy - 2y1 5y - 3z 1 2. Express each of the polynomials above in each of the forms (12), (13), (14). 3. Which of the polynomials above are forms? 4. Give the values of A, B, 0, D, E, F of formula (15) for each of the polynomials: . (a) &;! - 2zy 4y1 &; - 2y1 1 (b) y' 3y 1 2:/:1 - 4:/: 3 (c) (z + 3y)(z - 4y) + 2 (d) (2z + y)(2:/: - y) - (&; + 2y)(&; - 2y) (8) 4(z 3y)1 - 5(z 3y)(3:/l - y) 2(z - By)1 (J) 2(4:/: 3y)! - (4:/: 3y)(3:/l - 4y) (&; - 4y)1 2(3z - 4y) - (z 3y) 1 6. What are the degrees of the following polynomials? (a) Z8 3:/l1y 3s2 - 4ta ",a (b) (z 3y)(z - 21)' ",10 (c) (z + yl)1 - t B • \ (d) (Zl til)' - (z + 11 + S)I

+ +

+

+

+ + +

+

+

+ + + + + + +

+ +

+

+

+ +

+

+

+

+

+ + +

+

+

+

108

[ClLU'.

POLYNOMIALS

IV

6. Verify the following formulas by direct multiplication or use of earlier formulas in the list: (a) Zl - yl = (z y)(z - y) (b) Z8 - 11' = (z - Y)(ZI + zy + yl) . (c) Zl y' = (z Y)(ZI - zy yl)

+ + + + (d) z' - 'Y' = (Zl - yl)(ZI + yl) = (z - y)(z + Y)(ZI + yl) (6) Z8 + 11 = (Zl + yl)(Z' - zlyl + y')
Use the formulas of Oral Exercise 6 above to verify the following factorizations: (a) (2:1: y)1 - (2:1: - y)1 (2y)1 = 411(2:1: y) (b) (3z - 2y)1 + 2~ - (3z + 2y)1 = zy . (c) (z 2y)8 - (z - 2y)8 = 4y(3z1 + 4111) yl) (d) (2:1: + y)8 + (2:1: - y)1 - &UI = 411;(411;1 (6) Zl yl - z(z' - y') = (z y)'Y'. (f) (2:1: y)' - 16:1;(z y)1 - y' = _4a;y(4II;1 6zy 2y1)

+ + + +

+

+

+

+

+

... +

+

11. Derivatives (FULL COURSE). If fez) = aDZ" + alz-1 + . . . + tin is any polynomial in z, the polynomial (16)

fez) == na~l + (n - l)alz- 1I +,' .. + (n - ~)Q,tI:fl-i-l + . . . + 2a..-lIZ

+ tJn-l

is called the derivative of f(z) , In particular (a)' = 0, (azll + bz

(17)

(az + b)' = a, + c)' == 2az + b,

where a, b, c are any numbers. The derivative has the properties (18)

(af

+ bg)' ==af' + bg',

(jg)' == fg'

(/)' = ~1f'

+ f'g,

for all polynomials f and g, all numbers a and b, and all . positive integers t. _ We shall use the properties above without proof. Proofs of these properties are given in the calculus.

109

MULTIPLE FACTORS

12]

SEC.

nlustrative Examples

I. Find the derivative of! = 5z 6 + 6z 4 - 7z2 + 2.

Solution

!' =

II. Find the derivative of! =

f

III.

-

Solution

+ 5z + 4)' = 5(3z2 + 5z + 4)4(6z + 5). Find the derivative of! = (2z - 1)8(z2 + 4)2 by using formula

= 5(3z2

+ 5z +

+ ~8 141;. (3z2 + 5z + 4)6 by using formula (18).

30z6

4)4(3z 2

(18).

f' =

+

(2z - 1)8[2(z2 4)2z] = (2z - 1)2(z2

Sol?Jtion

+ (Zl + 4)23(23: - 1)22 + 4)[(2z - 1)8z + 6(Z2 + 4)] = (23: - 1)2(z2 + 4)(22z2 -

8z

+ 24).

EXERCISES

1. Find the derivatives of the following polynomials by using formula (16) but not (18): 5z2 + 6z - 2 + 7z2 - 8z + 1 1)(2z2 + 1)2 + 1)8(z2 - 1)

(a) 41;6 (b) 2z8 (c) (2Z2 (d) (Z2

2. Find the derivative in (c) and (d) above by using formula (18) as well as, of course, the definition (16). 3. Use formula (18) to find the derivatives of the following polynomials: 1)8(3z + 2)4 (c) (X2 + 1)3 + (Z2 - 1)8 + 1)4(z - 1)8(2z - 1)2 (d) Z3(Z - 1)4(z + 1)6 + 5)2 + (z - 3)2(Z + 2) - (&4 + 7Z2) + z + 1)2 ('i) (Z4 + 5z2 - 3z + 1)8 2z + 1)3(z2 - 1)4 W (Z2 - 1)6 - (Z6 - 1)2 Z)4(Z2 + 1)3

(a) (Z2 (b) (z (e) (3z 2 (f) Z4(Z8 (g) (Z8 (h) (Z8 -

12. Determination of multiple factors (FULL COURSE). The derivative of a polynomial f(x) may be employed to determine whether.or not a prime factor p(x) of f(x) has multiplicity greater than unity. Theorem 11. Let d(x) be the g.c.d. of f(x) and f'(x). Then every irreducible factor p(x) of multiplicity a ~ 1 of

110

POLYNOMIALS

[CBAP.lV

J(z) .1,8 a Jactor oj multiplicity a - I oj J'(z)

all well all oj d(z) .. For let J(z) == p(z)"q(z) where p(z) does not divide q(z) and a il= 1. Then

J'(z) == p(z)~q'(z)

+ q(z)[P(z)"]' == p(:C)"q'(z) + aq(z)p(z)lt-1p'(z)' . := p(Z)lt-1[P(Z)q'(z) + aq(z)p'(z)].

The polynomial p(z) is irreducible and prime to q(z). Hence it could divide aq(z)p'(z) only if it divided p'(:c). But p'(z) has a degree less than p(z) and is not zero. Hence p(z) does not' divide p(z)q'(z) + aq(z)p'(z). It follows that p(z) is a factor of multiplicity a - 1 of f(z).· Then p(:c) is a factor' of multiplicity a - 1 of d(z). It follows from this result that the g.c.d. of a polynomial J(z) and its derivative is the product of the powers p(Z)lt-1 of those prime factors with leading coefficient 1 ofJ(z) which have multiplicity a > 1. Then, if J(z) has no prime factor of multiplicity a > 1, the g.c.d. of J(z) and fez) will be 1. Illustrative Example Compute the multiple factors of and 80 factor J.

J == Zl - 2z' -

SIll' + 1611:1 + 1611: - 32

Solution

+

+

f .. &;, - SIIl8 - 24z1l 32z 16. We multiply J by 25 to avoid fractions and make the following computation: at

rll+1I

l1li-11

... - 411 + 811z4 - . . - .... + u. + 18 .. - IICb' -lIOCbI + toOzI + 400. - 800 • IIz4 - 1l1li -lIQaI + 40. . . - 4Oz4 - 1I11III + 1.... + so. IIaI - 41111IaI- 4111-

o

, oa

811 + 18 811+18

-118(.1 - "" -

- 1... - 8QaI + Il4o.l + 3IIID - 800 -1... + 1... + 4811"- 8411.- 311 - II... + 11'''' + 38411- 768 411 + 8)

It followa that

25J == f(&;

+

- 2) - 9ad = d(&; 2)(5z - 2) - 96d = d[(25a;1 - 4) - 96] .. d(25z1 - 100),

J .... (Zl - 4)d. This could also be computed by direct division. Now f1V8rY factor of d of multiplicity m is a factor of multiplicity m 1 of J

+

nc. 13]

111

RATION AL FUNCTIONS

+

and the only possible factors are 2 z, z - 2, or Zl - 4. Divide d by z 2 to get -d = (z 2)(ZI - 4z 4) = (z + 2)(z - 2)1, 1 - (z 2)I(Z - 2}8.

+ +

+

+

EXERCISES

Use' the process above to obtain the multiple factors of the following polynomials I(z) as well as a factorization in those cases where such a factorization is determined by the existence of multiple factors. The student is warned that computations may be complicated. (a) 1 = Zl - 4z8 2zt z 2 (b) 1 = z& - 5z' + 15:1:1 - 30z + 35 (c) 1 = z, 4z8 - I&; - 16 (d) 1 = z, ,- Z8 - 3z1 z +2 (e) 1 = 3z' - 4z8 - 12:1:' + 24z - 48 (J) 1 ... Z8 4:1;1 21z - 18 (g) 1 = Z8 - &;1 32 (h) 1 ... z, 5z8 &;1 - 4z - 8 (t.') 1 = Z8 - 2:1:& 3z' - 4z8 3zt - 2:1: 1 <:J') 1 = Z8 - ZT - 5z8 3z& - 3z' - 3z8 - 7zt + z 2 (k) 1 = z8 - 4z1 &;, - 8z8 + gzl - 4z + 4 (l) 1 = Z8 - &;1 4z8 - 3z' 5 (m) 1 = ZT 14z8 - 7zt + 2Iz - 56 (n) 1 = ZT - 35z' 115z8 - 8Izt 70s - 15 (0) 1 ... Zl - 6z8 - 1& + 36 (p) 1 = Z10 - 22:1:8 - 2Oz' - 3z1 - 4

+

+

+ +

+

+ +

+

+ + + + + + + +

+

+

+

+

+

13. Rational :fuD.ctions. A rational function of several symbols is any algebraic expression obtainable by formally applying a finite number of the rational operations of addition, subtraction, multiplication, and division to the symbols and numbers. We assume that our laws for rational operations are obeyed and so may then use the processes of Sec. 2 of Chap. III to replace any rational function by a formal quotient of two polynomials. We call this process that of simplifying the rational function. Let us observe that if R(z) is a symbol for a rational function of z and c is any number, the value R(c) obtained by replacing z by c may have no meaning. For example, if Zll -

,R(z)

1

== z - 1

112

.

POLYNOMIALS

[CHAP. IV

and we replace z by 1, then R(l)

1- 1

== 1 - l'

which is meaninglese. However, we have the formal equalityof R(z) and the ratidnal function (which happens to be also a polynomial) S(z)

== z

+ 1,

and S(l) == 2. Thus it may not be true that, if R(z) and S(z) are two formally equal rational functions, their values R(e) and S(e) are always equal. The resulting values R(e) and S(e) are equal if no polynomial denominator J(z) in R(z) or S(z) has the value J(e) ~ o. Fina.lly, we note that for expressions such as R(z)

1

== --1 z-

the value R(l) has the form. t and we cannot rep1a.ce R(z) by a forma.lly equal function S(z) such that S(1) does have an ordinary numerical value. Illustrative Examples I: Simplify 2z-1 D() _ ~

.n.z-

2z

2z+1

:-2Z=1

l'

4a;1-1-2z-1

Solvtion (2$ - 1)1 - (2z + 1)1 . 4a;1 - 1 B(z) == 4a;1 _ 2z - (4a;1 - 1) (4a;1 - 1)(2z - 1) 42;1 - 4a; + 1 - (4a;1 + 4a; = 4a;1 - 1

-Sz

=- -1

+ 1)

.

(4a;1 - 1)(2z - 1) -(2z :.... 1)

. = Sz.

II. Simplify the function above by a multiplication.

SJIIC.

RATION AL FUNCTIONS

13]

, 118

Solution

We multiply the numerator and denominator by

u' -

1 = (2:1: + 1)(21; - 1)

to obtain (21; - 1)' - (21; + 1)' 2:1: - (2:1: + 1) [(2:1: - 1) + (21; + 1)][(2:1: - 1) - (21; + 1)] -1

R(:I:) =

..

U' (-2) -1 .. &1:. BDRCISBS

1. Express tlie following rational functions as polynomials or quotients of two polynomials: 2 :1:-2 1 (k) - - " - : - 1 () z=2 - - 2 :I: - -.....::1:=-----::1 a 1 4 :1:---1 1 :1:+2+:1: -4 :I:-~ _1_+2 :1:' (b) :I: 1 Am. :1:' - :I: - 1 :1:1_1- 4 (i) _ _"",,11_'_-_1_ __ (:I: -11)' _ 1 211+1"1 (c) (:I: + 11)' 1--:1:-11 1· 1-:1:+11- 1 11 Am. 11 - 1. 2 ( :I: + 2a)' _ (:I: + (d) :I: + a :I: + a . 01 :1:+1 + 2:1: :I: :I: :I:+2a_ 2 :1:+:1:-1 :1:-:1:+1 :I:+a &:1 + 2:1:-1 :I:+a+:I:-a Am. :1:' • :I:-a :1:+ a (e) :1:' + a' :1:' - a' 1 1 :1:' - a' + :1:' + at (k) ;;-=-y - z:tI 1 1 :1:'-1-:1:_ 1 :I:+z:tI 1 (J) 1 :1:'-1 :1:'+1 :1:' - 1 -. (:I: - 1) Am.:I:' + 1. :I: :I: - 211

i

2a) _

:I:'-4Y'+-:l:-

411'

. (g) :1:' -

:I:

:I: +:1:

-211

114

POLYNOMIALS

(m)

(.1)2

[CRAP. IV

+ y2)2 _

x2y2

x2y2 X2 -

yl.

Am. x2y2 • ,

2. Simplify the following rational functions: (a)

+ 3) (x- _(x 1)4 3x +2)2(x -

(x - 1)2(23;

(b) (x - 2)2[(x

2 -

+ 1)(x -

1)

Am. X2

1)

. 7 -

23;

+ l'

+ (x + 1)(x + 2) + (x - 1)(x + 2)] (x - 2)4 - 2(x + l)(x + 2) (x - l)(x - 2) (x - 2)'

Am.

_23;2 - 3 X4

•

+ 1)6t - 3t (t + 1)2 2t(t + 1) '- (t + 1)2 (t

(d)

--

,

3(1 (6) (1

+

2

t2

+- t 3)

- 3t(3t2) (1 t)2 t 8)6t.- (3t2)(3t2) (1 t 3)1 •

+ +

Am.

(1 - 2t8)(t2 - t 2t _ t4

+ 1)2

.

8. Write the following rational functions as the sum of a polynomial in x and a rational function whose numerator has degree less than the denominator:

SEC.

RATIONAL FUNCTIONS

13)

z+1

z-1

(i) ~ -;:j:l Z2 - 1 z 1

+

z+1 -Z2=1 1

Z2

W z+1-z-1 2

Am.z - 1 -Z2--1 -· Z8 Z2 + z (k) - - - Z2 - 1 z - 1 (I)

1

1+_1_ 1 z+-z Am. 1 -

z

Z2

+ z + l'

115

CHAPTER V IDENTITIES AND Ap·PLIC:A. TIONS 1. ·The binomial theorem. The formulas for (3; + y)2 and (3; + y)8 which are used in Chfl,p. IV !tre the cases n = 2, 3 of what is called the binomial theorem, which we shall give below. L~t n be any positive integer. Then the expression (3; + y)'" means a product of n factors each· of which is 3; + 'Y. The laws of algebra imply that this product of n factors may be expressed as a sum of 2'" terms each of which is a product of a power of 3; by a power of y. The degree of the polynomial in 3; and y is n, and the polynomial is homogeneous. Hence the degree of each term of the sum of 2'" terms is n. If we collect and arrange the terms according to descendirig powers of 3;, we obtain a sum which we may write as

+ y)'" = aca'" + al~-ly + . . . + atC"'-iy' + ... + a",y"'. It should be clear that we have expressed (3; + y)'" as a sum of n + 1 terms, that (1)

(3;

ao

= a", = 1,

and that all the coefficients ~ are positive integers. Note that the,term which has the coefficient a, and which involves y', is not the ith term of our sum but is its (i l)st term. The binomial theorem states that

+

(2)

~

= 0", i = n(n ,

1) . . :, (n - i

+ 1).

~

The coefficient ao fits this formula if we agree that shall mean 1. 116

0",,0

SEC.

11

THE BINOMIAL THEOREM

117

The proof of the binomial theorem may be made most easily by considering a product (Xl + Y1) (XS + YS) • • • (Xn + Yn) of n different binomials. Thi~ too is a sum of 2n terms, each term of which is a product of n - i of the letters Xl, • • • ,Xn and i of the letters Y1, • • • ,Yn' H we keep i fixed, we see that the number of terms with i faetors from Y1, • • • , Yn is the number of combinations of. n letters i at a time. H we put Xl = Xs = . • . = x~ =' X and Y1 = Y2 = . • . = Yn = y, - each of these terms becomes z-iy', there are On,i terms, and the product becomes (x + y)n, ai = On,i' The coefficients of xn-iyi and yn-ixi in the binon;tj.al expansion (1) must be equal since the expression (x + 'y)n is symmetrical in X and y. This follows from formula (2) and the known equality On,i = On,n-i- H n is even, ·we write n = 2m and see that formula (1) is a sum of an odd number 2m + 1 of terms. Then there is a middle term and it is the (m + l)st term. The term is (3) On,maf"ytn, (n = 2m). H n is odd, we write n = 2m + 1. Then the expansion has 2(m + 1) terms and there are two terms which could be called middle terms. These are evidently the final term of the first m + 1 terms and the initial term of the last m + 1 terms. Hence they are the (m + l)st and (m + 2)4 terms, and they are On,maf"ym+1

(4)

Let us note that, if we put expansion becomes (1

+ l)n

= 2n

=

1

X =

(n

= 2m + 1).

Y = 1, the binomial -

+ On,l + an,s + . . . + On,n'

Hence the tota} pos81,ole number of combinations of n letters 1. Let us note in closing that the binomial series (which was used in Chapter III) is obtained by fowlly usin,; the

is 2n

118

IDENTITIES AND APPLICATIONS

[CBAP.T

binomial expansion as if it held for the case where n is not a positive integer. IIlu,trativ, ExampZII I. Give the binomial expansion of (3a - b/2)1.

Solution The exp8.llSion is

5 . 4 (3a)8 (b)' 5.4' 3 (b)8 (- 2b) +""""2 - 2 + 3""='2 (3a)1 - 2 •3 •2 (b)' ( b)' 405 + 5 4'• 4 3' 2 (3a) - 2 + - 2' = 243a -""2 a'b

(3a)I, + 5(3a)'

l

+ -135 alb 2 Do· Find the eighth term of (2a + b/2)1&.

l -

45

-

4

15 bl alba + - abe - -. 16

32

SoZution We see that n ... 15 and i ... 7. The term is

o

II.'

(2a)'

(!)' ... 15 • 147 ••613. 5• 12• 4 ••113 •.210 • 9 2a8b" = 2

ID. Find the middle terms of (2:1: - 31/)' and (2:1:

12 870a'b' ,.

+ 31/)10,

BoZutiona In the firSt case, n ... 7 .. 2 . 3 + 1 so that m - 3. The terms are 0, ••(2:1:)'( -31/)8 ... (35)(16)(-27):1:'11' and (35) (2:I:)1( -31/)' - (35)(8)(81):1:'11'.

In the second case, m .. 5 and the middle term. is 0 10••(2:1:)1(31/)& .. (252)(2&)(3&):l:i/'. JmmCISBS

1. Give the binomi8J. expansions of

1)'

(a) (:I: - 1)' (b) (:I: - 1)1 (e) (:.: 1/)8

:I: (J) ( 2-i

+

(d)

(2a -1)'

(8)

~I

1

r

- 2:l:a

(g)

(;a

-lr

(h) (al - ~)' (i) (~+ 2bl)1 (;1 (2:I:l - 3y1)'

BlIIC.

2]

119

SUMMATION FORMULAS

Z. Give the mth term of the following binomial expansions for the given value of m: (a) (1 - b)1', m = 9 (e) (a' - 2bl)18, m = 7 (b)

(e)

{at + ~r', m - 7

(&: -1)11, m _ 11

(at - ia- t) 18, m = 4

(d)

(2z -:- ~r', m = 8 Am.

1~1 :.;liyl0.

Am. -102a'. l. Am• (1&) (:.; - 1/)", m .. 20. Am. -OI&..zSyU. a. Find the middle terms of the binomial expansions in Exercise 2. " Express the following numbers as powers of binomials 1()1o ± f or 1()1o ± 2 and 80 compute them by the use of the binomial theorem. (a), (99)' (d) (9999)' Am. 999,700,029,999. (b) (998)' (e) (998)' Am. 992,023,968,016. (e) (99)' (J) (98)1 Am. 9,039,207,968. (J)

(g) (at

+ ia-t )17, m = 6

.u.ua

2. Summation formulas. The binomial expansion (1) is what is called an algebraic identity. It states that two algebraic expressions involving :1:, y, and n are identical in the sense that, if the indicated operations are carried

out, the two expressions become precisely the same.' The expressions are polynomials in :I: and y but are not polynomials in the exponent n. Let us refer in this section to algebraic ezpre88iom involving several letters as JUndiom of these letters, and to relations expressing the fact that two such functions may be made identical, by ca.rrying out indicated operations, as algebraic identities. We have already studied a number of polynomial identities and shall next pass on to a new class of nonalgebraic identities which are called BUmmation Jormulas. These are formulas which express the sum al + all + . . . + a. of the first n terms of special sequences al, all • • • as algebraic expression in the symbol n. The formulas may be proved by use of the following consequence of the principle of mathematical induction: . Summation theorem. Let al, all • • • be a sequence whose general term a. is a Junction oj n, and let J(n) be a Junction oj n BUM that ; . (5) . J(I) = al, J(n) - J(n - 1) - a.

120

IDENTITIES AND APPLICATIONS

i8 an algebraic identity.

i8

v

Then the formula.

fen) == al

(6)

CCHAP..

+ . . . + a.

true for all val'U68> of n.

For let K be the set of all positive integers for which formula (6) holds. Since f(l) == al the integer 1 is in K. If Ie is in K thenf(le) == al + ... + a. and f(le)

+ a.....l

Replace n by Ie . f(1e and so Ie

== al

+ . . . + a. + a.....l.

+ 1 in formula (5) to obtain

+ 1) == f(le) + a.....l == al + ... + tlJl.f-l +1

is in K. The principle of mathematicol induction states that K is the set of all positive integers, that is, formula (6) is true for all values of n. .;

. lUustrative EumjJlu I. Prove that 1 2 n - n(n 1)/2. . ..R6fnmo1c: We are trying to prove that the sum of the first n integers is eq~a1 to n(n 1)/2,'

+ + ... +

+

+

SoZution In this sequence a. - n and J(n) = n(n J(I) ... 1 . 2/2 -.1 ... 1.11. .Also

+

1) n(n 1) (n - l)n J( ). J( n -. n == 2 2.

==

+ 1)/2.

n(n

Then

1.11

== 1,

+ 1 2- n -+ 1) 2ft

-2==n==a..,

and our result follows from the summation theorem. II. Prove that if r " 1 then 1 r == (,... - 1)/(r - 1).

+ + ... + ,...-1

In this sequence a.. ..

r·- 1,

SoZution J(n)

1.11 -

= (,... -

r 1- 1

1)/(r - 1). Then

1,

Also (,... - 1) - (,...-1 - 1) r.- 1(r - 1) J(n) - J(n - 1) = r _ 1 .. r _ 1 == a..

andJ(I) ... (r - 1)/(r - 1) - 1 ... ..

...

1.11.

as desired.

m. Prove that 1 • 3 + 2 • 4: + . . . + n(n + 2) .-

.. in(n

+ 1)(2n + 7).

121

SUMMATION FORMULAS

BEc.2]

Solution HereJ(1) -= 1(2·9) =- 3 ... al and J(n) - J(n - 1) =- i[n(n + 1)(2n + 7) - (n --1)n(2n + 5)] = 1n(2n1 + 9ft + 7 - 2n1 + 5) =- ..",(6n + 12) ... n(n + 2)

an

= a-

as desired. EXERCISBS 1. Prove the following summation formulas by the use of the summation theorem: -

+ + ...+ + + ... +

(a) 1 3 (2n - 1) = n l (b) 1 5 (4n - 3) = n(2n - 1) (c) l' + 21 + . . . + nl = in(n + 1)(2n + 1) (d) 11 + 31 + ... + (2n - 1)1 = In(4n1 - 1) (6) 21 + 41 + ... + (2n)l = fn(n + 1)(2n + 1)

(J) l' + 2' + ... + n' = [n(n:- 1)T

= nl(2n1 - 1) 1 1 1 n (h) 1. 2 + N + ... + nCn + 1) = n + 1 1 1 1 n (,") 1· 3 + 3 . 5 + . . . + (2n - 1)(2n + 1) =- 2n + 1 (J1 1· 2 . 3 + 2 . 3 . 4 + • . . + n(n + 1)(n + 2) (g) l' + 3' + . . . + (2n - 1)'

n(n

+ 1)(n + 2)(n + 3) 4

(k) 2·4 + 3 . 5 + . . . + (1

n

-t n)(3 + n) = 6 (2n1 +

(Z) 2· 10 + 3 . 11 + . . . + (1 + n)(9 + n) =

15ft + 31)

n

6 (2nl +

33n + 85)

(m) (6 + 1)(J + 1) + (6 + 2)(1 + 2) + . . . + (6 + n'(J + n) = "6n (n - 1)(2n - 1) + n(n 2- 1) (6 + 1 + 2) + n(6 + 1)([ + 1) (n) 11 .2+21 .3+ ... +nl (n+1) =-nn(n+1)(n+2)(3n+1) (0) 1· 4 + 4 . 7 + ... + (3n - 2)(3n + 1) = n(3nl + 3n - 2) 1 2 n (P) 3. 3 . 5 + 4 . 5 . 6 + ... + (n + 2)(n + 3)(n + 4) n(n 1) = 6(n 3)(n 4)

+

(q) (r)

(8) (t)

1 1 1 n 1· 4 + 4·7 + ..• + (3n - 2)(3n + 1) = 3n + 1 1· 2 + 2 . 21 + . . . + n l = (n - 1)2..+1 + 2 ! + t + .... + =13 + 31 + • • • + 3" = 1(3..+1 - 3)

m..

m..

+

+

122 '

IDENTITIES AND APPLICATIONS

[ClL\P.v

I. Prove (a) in Exercise 1 by subtraction of two formulas derived . from IDustrative Example L 8. Prove (d) in Exercise 1 by subtraction of two formulas derived from (0). '" Derive the formula. 1/;"

+ 1/;"-'1/ + . . . + :1;"-'11' + . . . + 1/.

0=

1/;,,+1 - 1/,,+1 ~_L..1/;-1/

by replaCing r by 1/;/1/ in mustrative Example n. .. I. Derive a. pairof corresponding formulas in the ca.ses where n 9r 2m 1 by replacing r by -1/;/1/.

+

= 2m

S. The method of unc:letermined coe:fJlcjents (FUI40 COt:llJSE). The student may- become Qurious as to how the l!Q.IIUDf'tion formulas he has proved were -obtained. We sh8.u .,atisfy this curiosity in the case where a. is a polynomial in n and shall thereby introduce the so-called method oj undet8rmined coejficientB. Let a. = g(n) be a polynomial of degree m. Then the exercises of Sec. 2 suggest that

+ cln'" + · .. + c......l is a polynomial of degree m + 1 whose coefficients are to be J(n) = conm+l

determined. By the summation theorem the expression (7)

co[nm+l - (n - l)m+l]

+' Cl[~'" -

(n - 1)"']

+ .. · + c".[n -

(n - 1)] - g(n)

must be identically zero and we must also have (8)

J(l)

= Co + . . . + c......l = gel).

The difference has degree at most m. By using the binomial theorem we obtain the term eo(m + l)n'" and may select Co so that the degree of the expression in formula (7) is at m9st m - 1. We then use this value of Co and select Cl in clmn--1 so that the degre'e of the expression in formula (7) is at most m - 2. The sequence of determinations terminates with the determination of c". so that the constant term of the expression in formula (7) is zero. We then select c......l = g(l) - (co + . . . c".) and have the desired summation formula.

+

BlBc.3]

123

UNDETERMINED COEFFICIENTS

It is really only necessary to use the process above to obtain formUlas for the sums of powers '

+ ~ + ... + nil. Indeed, if a,. == bon'" + b n-- + . . . + b"" then al + all + ... + am == bo(I'" + 2'" + ... + 11,"') + b1(1--1 + 2--1 + . . . + 11,--1) + . . . + b1(1 + 2 + . . . + 11,) 111

1

1

nb", and we may substitute the polynomial expressions for 111 + ~ + ... + nil, taken where ~ == 1, ••• , m, if they have already been obtained. nlustrative Example, I. Find a formula for 1· 5 + 2 . 9 + ••. + n(4:1& + 1) by the method of undetermined coefficients.

Solution Here a. == 4:1&1 + n and we wish a polynomial such that a + b + c + d = 5 and a[Z8 - (z - 1)']

+ b[Z' -

(z -'1)2]

+c=

tJII:I

+

ba;1

+

+ z. 1)1 ... raz -

c:J)

+ d

4z1

Then Zl - (z - 1)' == 3z1 - 3z + 1, Zl - (z 3a == 4:, ~ == t, -3(t) + 2b .. 1, 2b -= 5, b == t, t that c ... ¥ - t .. t. Finally, .

1 and 0 so

t +c =

a+b+c+d=t+¥+t+d=d+5=~

d == O. Hence!(n) ... In(Snt + 100 + 7) == in(n + I)(Sn + 7). II. Find !(n) in mustrative Example I above by the use of the formulaa for 1 + ... + n and 11 + ... + nt.

a. ==

4:1&'

Solution

+n and so

!(n) == 4(1 + ••• + nl) + (1 + ••• + n) ...

I + n(n:- 1) == n(n

t 1)

[(Sn

III. Find a formula for l' + 2' +

+ 4:) +

64: n(n +

1)(2n + 1)

3] == n(n + I~Sn + 7).

. . ~ + n'.

Solution We wish a[zl - (z - I)&]

+ b[z' -

(z - 1)4] + C[ZI - (z - 1)1] + d[Z' - (z - I),] + e[z - (z - 1)] - z'.

124

IDENTITIES AND ,APPLICATIONS

[ClU.P.v

,

Thus a(5:r;' - lOs-

Hence a ==

+ 10s1 -

1,

+ 1) + b(~a - &;1 + ~ - 1) + O(SsI - Ss + 1) + d(2:.; - 1) + e - :.;'. -2 + 4b == 0 and b ... ~; 2 - 3 + 30 = 0 and c ... ~,

-1 e

=

~

111

+ ~ - 1 + 2d == 0 and d == 0, 5 - 2 + 3 + e ... 0,

15 - 10 - 6 1 • 30 == - 30· Finally

a +b+0

+ d + e +1== 6 + 15

t,

10 - 1

+1,1" - :~.

We have

shown that l'

+ . . . + n' == anI + 100' +30IOn' -

n - 27

.

BDRCISBS 1. Use the method of undetermined coefficients to derive for al + ... + a.. in the following cases: (a) (b)

(0) (d) (e)

a.. - (n - l)(n' + 2)

(J)

a.. = (2n + 1)(2n - 3) . a.. ... n(3n - 1) a.. = 2n1 - nl a.. = (nl + l)(n - 1)

(g) (n) (i)

&

formula

a.. == n(n + l~(n + 2) a.. ... nl + 3n - 1 a.. - nl + 2n - 1 a.. ... n' - n

W a..

== nl

2. Derive the formulas in (a)-($) of Exercise 1 by the use of previously determined formulas for sums of powers.

4. Arithmetic progressions. Our sum.pl&tion formulas may be applied to obtain the sum of n terms of two kinds of special sequence called progre8BiO'fUl. The first of these is the arithmetic progresBion in which the difference d = ~l

(9)

-

a.

oj any two consecutive t6Nn8 is a jig;ed constant. We shall call d the common difference in the' progression. Let us now obtain a formula for the general term of an arithmetic progression. H al is the first term, the second term is all == al + d, the third term is as = al + 2d, and so forth. Thus it follows that the general term is (10)

CJ,a

== al

+ (n -

l)d.

no. 4]

ARITHMETIC PROGRESSIONS

125

The sum of n terms of an arithmetic progression will be denoted by B.. Since each term has a summand ai, we have B. == nal + [0 + 1 + 2 + ... + (n - l)]d. By using the illustrative Example I of Sec. 2 with n replaced by n - 1, we have . n(n - 1) n (11) B. == 001 + 2 d == 2 [2a l + (n - 1)d]. Since an == al yield

+ (n -

l)d, this formula may be modified to

(12) The finite sequence al; ..• , an may be thought of as having (al + an)/2 as an average term and formula (12) states that the sum is the number of terms multiplied by the average term. An arithmetic progression may be used to define other arithmetic progressions with the same common difference but with a different starting place. For example, if ai, at . • • is an arithmetic progression, so is bl , bs . . • where bl == ali, bt == as, . . . , b. == a.....4 • • • • We may also extend an arithmetic progression by adjoining the terms al + (n - l)d in which n takes on negative integral values. These are the terms al - d, as - 2d • • • and form an arithmetic progression by themselves with common difference - d. In the study of finite arithmetic progressions of n terms the number n is the number of term8 and it is usual to call al the fir8t term and a. the laat term. In cases where we wish to compute B., but a. and d are given, it is simplest to reverse the progression. Then it will become a progr.ession bl , • • • , b. in which bl == a. and the common difference is --d. Since b, == ~l we have

B. ==

al

+ . . . + a. == bl + . . . + b., _ n(n - 1) d 8 .-~ 2 .

126

IDE NTITIES AND APPLICATIONS

[CRAP. V

If two numbers a and b and a positive integer m are given, we can construct an arithmetic progression of n = m + 2 terms in which a = al is the first term and b = llm+. is th~ last term. The m intermediate terms are called a set of m arithmetic means between a and b. The common difference is

b-a

b-a

n-1

m+l

d=--=--,

(13)

and the means are (14)

b-a b-a a + m + l' a + 2 m + l'

When m

=

b-a

,a+m m +l"

1, the single mean is called the arithmetic mean

of a and b, its value is a

t b., It is the average of a and b'.

ntustrative Examples I. Find d and SlO if al alO

= al +

= -8 and alO = 19.

Solution 9d = -8 + 9d = 19, 9d = 27, d

SlO = ¥(al

= 3.

Also

+ a..) = 5 . 11 = 55.

II. Find'd and a8 if al == 15, SlO = 60. Solution SlO = ¥(2 . 15 + 9d) = 5(30 + 9d) = 60. 'Hence 9d = -18, d = -2. Then a. = 15+ 7d = 1. III. Find al, alO, SlO if al = -3 and au = 17.

9d + 30 = 12.

Solution Write a& = bl and au = bu. Then bu = bl lOd = -3 lOd = 17, d = 2. This is the same common difference as in (it, a2, • • • , and 80 al = al 4d = -3 = al + 8, al = -11 9d ='7, SlO = ¥( ..... 11 + 7) = -20. IV. Insert six arithmetic means between -5 and 23.

+ +

+

+

Solution al = -5, a. = 23 80 that 23 = -5 + 7d and 'd = 4. The means are -1,3,7, 11, 15, 19. V. In an arithmetic pro~on SlO = 115 and d == 3. Find alO.

ARITHMETIC PROGRESSIONS

127

Bolution

a

The reversed :progression has alO .. bit = -3, and the same sum + 9d) = 5(2bl - 27). "Then 23 = 2bl - 27, alD = bl ... 25.

BlO - 115 - 5(2b1

BDRCISBS 1. In an arithmetic progression find the following: (a)

B., as if al = 3, a = -4

Ana. B. B11, as if al = 5, a ... 2 (e) Bu, al' if al = -3", a - 3 Ana. B" (d) Bu , au if al .. -6, as = 6

0=

-88, as ... -17.

(b)

= 231,

au ... 36.

(e) Be, a, if /J7 ... 13, au ... 28

Ana. B, = 63, a. - 1.

M B., au if at = 12, as = -12 (g) a, as if Ba ... 108, al == 24

Ana. a ... -3, as ... 3. a, a7 if Bl = 25, al ... 20 ('1 a, al if B• ... 90, at = 14 Ana. a ... 1, al - 6. W a, au if B• ... -108, a, ... -15 (1:) al, at if B• ... 117, a = 6 Ana. al = -11, at ... 37. (I) al, au if B1I =- 138, a = 5 (h)

O

(m) n, al if B.. = 153, tl - 2, al integral

I. Insert m arithmetic means between a and b in the following cases: (a) m .. 6, a = 3, b ... -11 Ana. 1, -1, -3, -5, -7, -9. " (b) m = 5, a ... -2, b = 4 (e) m = 7, a .. 6, b = 10 A"". 6t, 7, 7t, 8, st, 9, 9t. (d) m = 11, a = 18, b = 14 (e) m ... 8, a = 10, b - -17 , Ana. 7, 4,1, -2, -5, -8; -11, -14. m ... 9, a ... "-9, b = -7 (U) m = 3, a ... -6, b ... 6 Ana. -3, 0, 3. (1&) m = 14, a = 0, b ... -5 ('1 m = 5, a = 16, b = -8 Ana. 12, 8, 4, 0, -4. W m=8,a-200,b=20

(n

.

128

InlllNTITUIS AND APPLICATIONS

.[CIIAP. V

6. Geometric progressions. A geometric progre8sion is a sequence aI, aI, • • • in which the rati(,)

r = ~ tli of any two consecutive terms is a constant. It is called the common ratio for the progression. Then tli.tl = rtli, al == ral, aa = ral· =. rIal, and we see that . (15)

(16)

By Illustrative Example II of Sec. 2 we see that the sum of n terms of the progression is given by

fA - 1 = antI - 1ale B. = al r - 1 r-

(17)

Just as finite arithmetic progressions may be reversed so may finite geometric progressions be reversed. This is done by writing bl = a. and Writing 11r for the common ratio instead of r. Thus, for the reversed progression, (llr)· - 1 1 - r" B. = a. (llr) _ 1 = a. '-1(1 - ·r)· A set of m nlimbers aI, • • • , a.n.t-l is called a set of m geometric meam between. a and b if a = aI, aI, • • • , tJn...rl, tJn...r1 = b form a geometric progression. Then ,.-+1

=£ a

and, if we restrict our attention to the case where a and b are positive real numbers, r is the positive (m + l)st root ef b/ a. The means are then ar, ar·, . • • , arm. nlustrative Examples I. Find al-and.S. in a·geometric pro~n if al == 2 and r == 1. Solution 1

a. == 2(1)' == 8'

1 at =- 16;

a,-al

-h-2

S. == 1 - 1 == 1 - 1

=2 (2 -

1) 31 31 16 == 2 16 ... S.

II. Find al and S. in a geometric progression if Ii., == 192 and r == 2.

GEOMETRIC PROGRESSIONS

SEc. 5]

0Ir

= a128 = 3 • 2~, al

129

Solution = 3 . 2& = 96 so that

= 3 and a.

S& = 96 - 3 = 93.

III. Find r in a geometric progression if al

=i

and S6"1= ¥-.

Solution S&

=i

e& ~ i) = 3i.

Evidently we cannot hope to solve the

equation r8 - 1 = 31(r - 1) except for integral values by substitution and we obtain r = 2 as a solution. IV. Find r and n in a geometric progression if al = 2, a.. = - 54,

.

&=-~

Solution -

a.. = 2r,,-1 ... -54, r,,-l == -27, r" =

-27r and

-40(r - 1) = al(r" - 1) = (-27r - 1)2.

Then -27r - 1 = -2Or + 20 and -7r = 21, r = 3, n = 4.

= -3, (-3),,-1 =

-27,

n- 1

EXERCISES

1. In a geometric progreBBion find (a) S., a, if al = 8, r = i Am. S, (b) S8, a& if al = 16, r = -i (c) Sa, a4 if al = t, r = 2 Am. S. (el) al, S& if aa = t, r = -! (6) a7, S, if a4 .. iT, r = -i Am. a7 (/) at if as = 42, a7 = 5,250 (g) as if au = 1,080, a14 = 233,280

(h) r, n if al = 2, a" = 8, Sn = 3 (1.) n if al = -1, r = 2, S" ... -63

= ¥, a. = t. = ¥, a4 = 1. = - vh, S, = - tn.

Am. as

= 5.

Am. n

=

6.

(3) r, n if al = 3, a" = 48, S,. = 93

2. Insert ~ geometric means between a and b in the following~: (a) m = 4, a = 1, b = 32 (b) m = 5, a = 2, b = iT (c) m = 3, a = 18, b = i

(el) m

= 7, a = ..;."

b

=3

(6) m = 5, a ... 2, b =

t

(J) ~ ... 2,(J==3,b ... 41

Am. 2, 4, 8, 16. Am. 6, 2, Am.

t.

V22, 1, 2 V21 1 ' 2' 2 V2'

lao:

IDENTITIES AND APPLICATIONS

[ClIAP.v

8. Harmonic progressions. A S'equen'ce (h, all • • • is said to form a harmonic progr68Bion if aU the numbers a. are not zero, and the sequence' of reciprucals 1 1

(18)

is an arithmetic progression. Then 'd = ~ _

(19)

a..rl

!

a.

==

1. _.! == (h all

al

- flll alai

for every i and

1.. == 1.. + (n _ 1) (a l tin

al

-

all) ==,all + (n -alaIl)(fll -

al ),

,

alaI

so that (20)

The terms of a hal'Illonic progression may be computed by the use of this formula, but it is prefera.ble to pass to the cottesponding arithmetic progression and make the computations directly by the Use of the theory of arithmetic progressions. See the illustrative examples below for this procedure. There is no formula for the sum-or n te1'mS of a. harmonic progression. However, harmonic m6an8 are defined in a fashion ~na1ogous to the definition for arithmetic means. Thus, if a and b are any two nonzero numbers, we may form a harmonic progression of n ==. m + 2 terms in which al == a, a....t-li == b,

a-b

d == (m + 1)00' The m arithmetic means between 11a and lib are then computed and their reciprocals are the m harmonic means between a and b. . lUustrtltiv. EZtmlpl.B I. In a harmonic progression find Ga, Gl0 if GI == -

t, Gil ... ir.

BEc.6]

HARMONIC PROGRESSIONS

131

Solution

The elements c, = -:-3, Cle = 17 form. the fifth and sixteenth terms of an arithmetic progression in which d = 2 by the IDustrative Example III of Sec. 4. Then in that example Cl = -11 and

=

Cs

-11

+ 2d = -7,

= -11 + 9d =

CIO

7.

Hence a, = -t, alO = t. II. Insert six harmonic means between -i and

n.

Solution We form. an arithmetic progression of eight terms with al == -5, as = 23. Then 23 == -5 + 7d and d ... 4, the arithmetic means are -1,3,7,11,15, 19. The required harmonic means are -1, i, t,

n.

~,n.

BDRCISBS

1. In a harmonic progression find

=-

(a) ai, ai, a, if al i, a, = t (b) a" a, if a., = -h, all = iT

(e) as, all if as = -h, as = (d) at, au if a. = 1, a, = 3 (e) al, a., if as = t, a, = h (f) au = if al = -h, a7 0= i (g) lJI8 if al = -h, as =

"*

Tr:

Am. -

i, 1,t.

Am. ae

= - i, all == -

n.

Am. --Ir.

S. Insert m harmonic means between a and b in the following cases: (a) . (b) (c) (d) (e)

m

= 2, a = 1, b ... 3

Am.

t, t .

m = 5, a = i, b = 1 m == 7, a = 1, b ... 2 Am. tt, tt, tt, tt, tt, tt, JoI. m == 4, a = i, b = -h m = 11, a = iT, b = t S. Describe each of the following progressions by the appropriate term "arithmetic," "geometric," "harmonic" or none of these: (a) 1, (b) 1, (e) 1, (d) i,

(e) (J) (g)

(1&) (t.')

i, t, t, T • . • i, t, T, n ... i, 0, - i, - 1 •••

t, n, n

.. .

i, t, -h, n .. . i, t, - t, i, 1, t, JoI, H .. . i, 1, t, i, t .. . 3, I, 1, i, i, i .. ·

*.. .

.

CHAPTER VI EQUATIONS

1. Conditional equations. Up to the present point in our study of polynomials we have written f(x) = 0 only in the case where f(x) is the zero polynomial. We have also written f(x) =:g(x) only in cases where f(x) and g(x) are equal polynomials in the sense of Sec. 1 of Chap. IV. We'shall henceforth use the notationsf(x) = 0, f(x) ='g(x) (read "f of x is identically equal to g of x") for the two properties above and shall give a new meaning to our earlier notation. If we replace x, by ,a complex number c in a polynomial f(x) whose coefficients are complex numbers, the resulting number: has been called the value of f(x) at x = c and has been designated by fCc). If fCc) is the complex number zero,- we call c a root or a zero of f(x). We shall devote this chapter to a study of the properties of the roots of polynomials, that is, to the roots of what we shall call conditional (polynomial) equations. A conditional equation (1) f(x) = 0 is not a statement but is a question. It asks: What are the complex numbers r such that fer) = O? It thus asks: What are the roots of f(x)? We shall call the answers to this question the roota (or 80lutions) of the (conditional) equation. An answer to formula (1) is, not a complete answer unless all roots are given. We shall say that we have 80lved the equation of formula (1) when we have found all its roots. If f(x) and g(x) are two polynomials in x, the conditional equation . (2)

f(x)

= g(x) 132

..

BEc.11

COND ITIO N AL EQUATIONS

133

is also a question. It asks: What are the complex numbers r such that fer) and g(r) are the same complex number? It should be clear that the answers to this question are precisely the same as the answers to the question. (3) fez) - g(z) == 0 in which the left member is the difference polynomial defined as in Sec. 2 of Chap. IV. Formula. (3) mayor may not be more convenient to solve than formula (2).' We call them equivalent equations and make th.e following general statement: Definition. Two equatiom are called equivalent if they have precisely the same roots.

Let us now observe that the equation fez) == g(z) is equivalent to the equations (4) aj(z) == ag(z), fez) + h(z) == g(z) + h(z) (a F 0) for any complex number a and any polynomial h(z). Thus, if we multiply (or divide) both sides of an equation by a nonzero number, we obtain an equivalent equation. H we add (or subtract) the same polynomial to (from) both sides of an equation, we obt~in an equivalent equation. The polynomial fez) is called the left member and the polynomial g(z) the right member of the equation fez) == g(z).

Both are sums of terms. .Any term of either member may be carried over as a term of the other, provided its sign is changed. The result will be an equivalent equation. This procedure is called the trampoBition process. It is the procedure by means of which formula (3) may be seen to arise from (2) and it is an immediate consequence of the second relation of formula (4). H a term (i.e., a summand) of f(z) coincides with a term of g(z), both terms may be cancelled, i.e., deleted. This is a consequence of formula, (4) or of the transposition process. For example, in the equation 9zS + 5z 3 - 7zS + 3z - 8 == 5z 3 + z - 2 + 2zs,

134

EQU ATIONS

[CHAP. VI

we cancel5x8 , add 9:& - 7X2 = 2:&, cancel 2x2, transpose to obtain 3x - x = - 2 + 8, 2x = 6, x = 3. 2. The degree of an equation. Every polynomial equation g(x) = hex) is equivalent to an equation f(x) = 0 where f(x) is a polynomial. Then we shall write (5) f(x) = aoXn + a l r1 + ... + an = 0, in which ao ~ O. The degree n of f(x) will be called the degree of the equation of formula (5) and the coefficients of f(x) will be called the coefficients of the equation. Then ao is the leading coefficient of the equation and an is the constant term. If we multiply formula (5) by a nonzero number a, we .obtain an equivalent equation af(x) = 0 with leading coefficient aao and other coefficients aQ,j. In case the coefficients are real this process is· usually employed to make ao positive. In case the coefficients are integerS the equation is usually divided by their g.c.d. We shall call the equation of formula (5) a linear equation if n = 1, a quadratic equation if n = 2, a cubic equation if n = 3, a quartic equation if n = 4, a quintic equatIon if n = 5. An equation of degree n may be properly referred to as an n-ic equation. "Note that an equation such as OX4

+ x + 2X2 8

3x

+ 5 = x + 2:& 8

4x

+6

is not a quartic, a cubic, or a quadratic equation but is a linear equation. For it is equivalent to x - 1 = O. ORAL EXERCISES

Give the degrees, the leading coefficients ao F 0, and the constant terms of the following equations: (a)· ~2 - 3~8 + 43; - 2 = 0 (b) ~8 - 3~2 + ~ - 1 = ~8 - 43;1 + 5~ - 1 (c) (~- 1)8(2~ + 1)2 = ~6 - 3~2 + ~ - 1

3. Linear equations. A linear equation is an equation = g(x) in which f(x) and g(x) are polynomials whose difference is a linear polynomial ax + b. Thusf(x) = g(x) is equivalent in this case to

f(x)

00

ax+b=~

~~~

135

LINEAR EQUATIONS

SJDC.3]

as well as to az = -b, and to -b a

z = --.

(7)

It follows that a linear equation has a unique solution and that it is given by formula (7). The standard technique for solving the most complicated of linear equations is to transpose all term8 involving z as a factor to the left member of the equation, and all constants to the right member. The equation may then be put in the form cz = d,

(8)

where d is a constant, and will be a linear equation only when c is a nonzero constant. The unique solution will then be z = die. nlustrative ExamPle Solve the equation

aySz

+ yl -

2ay

+ 2tJ:1l =

a'IP - 4ay

+ 2a + tJ:Il + 2ayl + y -

1

for z.

Remark: We give this example as an illustration of the transposition procedure. Exercises of this type may be found in texts on elementary algebra. They have little value and will not be given here. Solution

We transpose all z terms to the left and all terms not involving z to the right so as to obtain

(ay'

+ 2a -

ay - a)z

= _yl + 2ay -

4ay

+ 2a + 2ayl + 11 -

1,

and the solution is Z ..

+ (2a -

yl(2a - 1) - y(2a - 1) a(yl - y 1)

+

1)

2a - 1 .. -a-·

I

ORAL BDRCISBS Drill until you can solve tJ:te following equations very rapidly: (a) 2z (b) 3z

+3 = 0 +4 = 0

(0) 2z - 4 = 0 (d) 2z - 3 - z - 4

(e) 3z - 4 = 5z - 2 (J) Szl 4z - 3 - Szl 2z - 11 (g) (z 1)(2z 3) = 2z' 4z I) (h) 2az 3a = 3z 2a1

+ + +

+

+

+

+ +

136

EQU ATIONS'

[CBAP. VI

4. Quadratic equations. An equation is a quadratic equation if it is equivalent to an equation (9)

f(x)

a:cl

EE

+ bx + c =

(a ~ 0),

0

where we use the notation a, b, c for the coefficients in this simple case rather than ao, al, al. We solve quadratic equations by the use of the polynomial identity (x

+ 1/)1 =

Xl

+ 2zy + 1/1

which becomes (10)

when we replace 1/ by t/2. Note the following principle: COMPLETING THE SQUARE. The e:tpre8Bion Xl + t:c i8 macle into a 8quare if we add to it the 8quare of half the coeiftcient of x. The re8ult of thi8 completion i8 the 8quare of x plU8 hall thi8 coejficient. Any quadratic equation may now be solved by writing f(x)

EE a (XI + ~ X+ ~) == a [ (x +

:ar + ~ - !:2}

We compute

and use the identity Xl

-

_bl(2a)2 - 4ac] [(x + !.)I 2a

1/1

= (x + 1/)(x -

:::; (x + _b + Vb=-=I------:-4a-c') (x + _b _ 2a

1/) to see that

Vb l

-

2a.

2a

2a

4ac

-b - Vb 2

4ac).

Then (11)

:

where (12)

.. ' -!J +. :Vb 2 rl = ... :

2q

-

,

, .2a

-

4ac

BEC.4]

137

QUADRATIC EQUATIONS

Let us suppose that r is a root of the equation f(x) == O. Thenf(r) == a(r - rl)(r - r.) == O. But a product of com. plex numbers is not zero unless one factor is zero. Hence r == rl or r == rs. Conversely f(rl) == a(rl - rl) (rl - r.) == 0, f(r.) == a(r. - rl)(r. - r.l == O.

Hence rl and r. are the roots of our quadratic equation and are its only roots. • Formula (12) for the two roots of the equation of formula (9) is called the quadratic formula. It reduces the problem of finding the roots of a quadratic equation to that of finding the square root of a == b· - 400 and to rational processes. We shall calla the discriminant of the equation of formula (9). Note that A (13) (rl - r.)S == "2' a and so the equation of formula (9) has equal roots if and only if a == O. If a, b, c are all real, so is a. When a is positive, its square root VA' is a perfectly well-defined positive real number and rl and r. are distinct and real. When a is negative, we write

a == where

-A,

VA == VAi,

VA is positive and the roots -b -b VA. rl == 2a

+

2a ~,

VA.

r.==---~

2a

2a

are a pair of conjugate imaginary numbers. If a, b, c are complex numbers, the number a might be an imaginary number, and we have not yet discussed the meaning of YK. We shall do so in Seo. 7 of Chap. VIII. Let us note that -b c (14) rlrS == -, rl + r. == -a , a

138

EQUATIONS

[CJl4P. VI

a;nd that these relations may be computed by explicit computation from formula (12) or from the identity

a(a: - rl)(a: - rl) == a[a:1

(rl

-

+ 1-1)a: + rlr.]

+ ba: + c.

= azl

We shall close our study of quadratic equations by a proof of an irreducibility theorem. Theorem 1. A polynomial f(a:) IE azl ba: c with real coejficient8 i8 irreducibk in the field of all real numberB if and only if 11 = bl - 4ac < O. For we have shown how to factor f(a:) if 11 ~ 0 as a product a(a: - rl) (a: - r.) where rl and rl are real. H 11 < 0 and f(a:) is not irreducible, it has a factorization f(a:) = (ga: h)(Ba: t) where g, h, 8, t are real and Bg = a ~ O. Then rl = -h/g, r. = -t/B are real roots of f(a:) contrary to our proof that f(a:) ~ a pair of conjugate imaginary numbers as its only roots. Formulas for solving cubic and quartic equations exist but are not of great value. We shall not study"them here.

+ +

+

+

nlustrative ExamPles I. Solve the equation

23;1 -

4z

+ 8 = Szt + 23; -

5.

Solution

This equation is equivalent to :n;t + &: - 13 = 0 and a == 3, h == 6, 12 '13 = 12 '16 == 3· 4 '16 == 81 • 3. c = -13, ht - 4ac = 36 The roots are

+

-6 ±68 v'3 == -1 ±

~v'3.

II. Determine Ie so that the equation 9bt equal roots. "

-

60z

+ 61e + 1 == 0 has

Solution

The disoriminant is

hi - 4ac = 3600 :- 4 . 9Ji(61e The equation 6k1

+ Ie -

+ 1) == 36(100 -

6let

-

100 = 0 has as roots

Ie== -1±V1+2400= -1±49==4 -25.

12

12'

6

Ie).

SEC.

5]

REMAINDER AND FACTOR THEOREMS

139

+ +

The value k = 4 gives the equation 363;1 - 60z 25 = (6z - 5)2 = 0, and the value k = - ¥ gives the equation - l/zl - 60z - 24 == 0 which is equivalent to 25z1 402: 16 = (52: 4)2 == O.

+

+

ORAL EXERCISES 1•. What are the sums and the products of the roots of the following equations?

+ + + + v=a +

+

(e) (1 0)xl z 0 (f) iz 2 - Z = 0 (g) (i I)ZI - Z = 0 (n) V3 Zl - 9 = 0

(a) Z2 - 3z 27 == 0 (b) Z2 3z - 27 ... 0 (/:) 2z2 4z - 27 = 0

(d)

+ + 3z + 12z + 8 = 0 1

=0

2. Give the discriminant of each of the equations (a) to (d) of O~al

Exercise 1. EXERCISES 1. Give the radical form of the solution of all the equations of Oral Exercise 1. 2. Solve the following equations by the use of the quadratic formula.: (a) Zl

+ 2z -

=0 52 = 0

63

(b) Zl - 9z (/:) Z2

+ 6z -

40

=0

(d) Z2 - 63; - 40 = 0 (e) Zl z 4 =0 (f) Zl - Z - 24 == 0

+ +

(g) 3z1

+ 2z -

1

5

== Z2 - 4z + 1

5

(n)

Z2 - ;; - 84 == 0 \

(i)

!? + 2: -

21

+

=0

+

W (z 1)1(z - 1) = Z8 - 3z 8 (k) 2(z - 1)8 = (2z l)(z ,1)2 3z - 5 (l) 12:1:2 4z - 1 = 0 (m) Zl = 10z 75

+

+

+

+

+

3. Determine k so that each of the following equations has two equoJ. roots:

+9 = 0 +k +3 = 0 + 9k + 2 = 0

(a) 2kz2 - & (b) kz2 - 14:1: (/:) kz2 - 12kz

4. Find k if 2 is a root of k 2z 2 6. Find k if 1 is a root of 2kz2

+ + +

(d) kzl kz 1 ,.. 0 (e) (2k + I)ZI - 2(k (J) (41c - 8)Z2 4kz

+ 2)z + k = 0 + (k + 3) = 0

+ z - 9k = O. + k z - 15 = O. 2

5. The remainder and factor theorems. If c is a number, the remainder on division of a polynomial/(x) by the linear polynomial x - c is a number. Its value is given by the following: Remainder theorem. The remainder on diui8ion oll(x) by x - c is/(c).

140

EQUATIONS

[CHAP. VI

For the division algorithm states that

+ r.

f(z) = q(z) • (z - c)

Then f(a) == q(a) • (a - c) + r for any number a. In particular f(e) == q(e) • (e - c) + r = r. There is an almost automatic consequence of our definitions and the remainder theorem which we state as the following: Factor theorem. A number e i8 a root of f(z) if and only if z - e i8 a factor of f(z) • • For by the remainder theorem (15)

f(z) = (z - c) • q(z)

+ f(e)

for any e. The statement that e is a root of f(z) means thatf(e) = 0 and thus thatf(z) = (z - c) • q(z), z - e is a factor of f(z). The statement that z - e is a factor of f(z) means that f(z) == (z - c) • q(z) and we conclude that f(e) = 0 either from formula (15) and the uniqueness of the remainder in the division algorithm, or by computing f(e) == (e - c) • q(e) == O. nlustratille E%amples I. Compute the remainder on division of :1:71

-

7:&,8

+ 1 bY:I: + 1.

Solution

The remainder is/( -1) .. 1 - 7 + 1 == -5. II. Show that:t + 2 is a factor of:tl + 7:t' + &:.

+ 2:t -

Solution 1(-2) ... -32

+ 112 -

64 - 4 -12 "" O.

ORAL EXERCISES

1. Tell why:t - r is a factor of 1(:1:) in the following cases: (a) r = -1,/(:t) "" :t' +:t8 - 2:&1 + 2:t + 4 (b) r .. 1,/(:t) == :tl - 7:t' + 2:t8 - 2:&1 - 1O:t + 16 (c) r = -2,/(:t) .. :tl + 3:t1 + 5:t + 6 (el) r = 3,/(:t) = :t8 - :tl - 4:& - 6 (6) r .. 2,/(:t) .... :tl - :t - 6 (f) r = -4,/(:t) = :t8 + 4:&1 + :t + 4 (g) r .. 3,/(:t) = :t8 - 3:t1 + 2:t - 6

12.

SEC.

SNYTHETIC DIVISION

6)

141

2. Give the remainder on division of the following: (a) x 8 - 3x 2 - 3x - 3 by x + 1 (b) x 3 + x2 + 5z + 7 by x + 2 (c) x 6 + 6x 4 - 7x2 + 23: - 1 by x-I (d) x 3 - x 2 - 3x + 1 by x - 3 (6) x 3 + 4:/: 2 + 23; + 4 by x + 4 (f) x 3 -x-3byx-2

6. Synthetic division. f(x) = aoX'"

If c is any number and

+ aIx"'-1 + ... + an

is any polynomial, we may write

f(x)

= q(x)(x

- c)

+ b""

where q(x) = boX"'-1 + bIx"'-2 + ... + bn-l, b", = f(c). Then (x - c)q(x) = f(x) - b", = xq(x) - cq(x), so that

+ b", = f(x) + cq(x), that is, boX'" + bIx"'-1 + . . . + b", = (aoX'" + aIx"'-1 + ... + an) + CboX"'-1 + cb Ix"'-2 + . . . + Cbn-l.

xq(x)

Comparing coefficients we have

+ cbo, + Cbi-I,

bl = al b. = a.

Thus the coefficients bo, . . . ,bn-l of the quotient q(x) and the remainder b", may be computed by the tabular form

+ + a2 + + an-I + an ~ + cao + cbl + . . . + Cbn-2 + Cbn-l bo + b + b2 + . . . + bn-I + b",

ao

al

i

This procedure is a modification of the ordinary division process and is called 8ynthetic division. It is not applicable without the use of special devices for the division of a polynomial by other divisors than those of the form x-c. Synthetic division may be used to compute f(c) when direct substitution of c for x might prove tedious. For

142

EQUATIONS

[CHAP. VI

example, if f(z) == z, - 7z 8 + &;2 - 100z - 31, then the computation of f(8) == 8' - 7(8)8 + 5(8)2 - 800 - 31 is complicated. But the synthetic division 1 - 7 + 5 - 100 - 31~ + 8 + .8 + 104 + 32 1 t 1 + 13 + 4 + 1 gives f(8) == 1. II c is a root of f(z) , the computation of fCc) by synthetic division gives the value fCc) == 0 as well as the quotient q(z) in f(z) == (z - c)q(z) •. Then the remaining roots of f(z) are the roots of the equation q(z) == O. This equation of degree n - 1 is called the depre88ed equation. nlustrative Examples I. Compute the quotient and remainder on division of &:' + 142;8 - 22:1;1 - 50 by z + 6. Solution 3+14-22+ 0-501-6 - 18 + 24 - 12 + 72 3 - 4 + 2 - 12 + 22 The quotient is &:' - 4z1 + 2:e - 12 and the remainder is +22. II. Verify that 8 is a root of the equation z' - 6Oz1 - 29:e - 24. Solution 1+0-60-29-24~

+8+64+32+24 1+8+ 4+ 3+ 0 EXERCISES

1. Use synthetic division to compute the quotient and remainder on division of the following: (a) ~I - 3z1 + 2:e - 1 by z - 2 (b) z' - 142;i + 2:e' + 49:& - 36 by z + 2 . Am. q(z) = Zl - 16:&1 + 34z - 19, r = 2. (c) z, + Zl + z - 2 by z + 3 (d) Zl - Z8 - 32:e by z - 3 Am. q = z, + 3z8 + &1 + 24z + 4O,'r = 120. (e) &:8 + 2zI + 5:1: + 10 by z + 2 (J) ZT - 6:&1 + &a - 3z1 - 7z + 1 by z + 1 Am. q - ~I - ~I - 5:1:' + 5:1:8 + &:1 - 6:& - 1, r ... 2.

SEC.

143

LINEAR FACTORS

7]

(g) :1:4 + 10:1:3 + 22:1:2 - 7:1:

+

+ 5 by :r: + 4

(h) :1: 4 + :1:8 - 22:1:2 15:1: - 32 by :r: - 4 Am. q = :1:8 + 5:1:2 - 2z + 7, r = -4 (i) :r: 4 + 2z8 - 13:r:2 + 13:1: - 21 bY:l: + 3 W 2z4 + 9:1:8 + 14:r: + 8 by :I: l Am. q = 2:1:8 8:1:2 - 4:r: + 16, r = O. (k) 6:1:4 + 5:1:8 10:1: - 4 by :I: - i

+

+

+ + 5:1:8 +:1: + 17:1: - 6 bY:l: - i Am. q = 3:1: 8 + 6:1:2 + 3:r: + 18, r = O. (m) 3:1: 4 - 14:r:8 - 57:1: 2 + 65:1: - 56 by :I: - 7 (n) 3:1:4 + 40:1:8 + 85:1: 2 + 97:1: + 99 by :I: + 11 Am. q = 3:1:8 + 7:1:2 + 8:1: + 9, r = O. (0) :1: 8Iz8 - 2:r: 8 + 18:1:2 + 8:1: - 72 bY:l: - 9 (p) 18:1:6 + 20:1: 4 + 29:1:8 + 57:r:2 + 6:r: by :r: + t Am. q = 9(2z4 + 2:1:8 + 3:1:2 + 6:1:), r = O. (q) :1:8 - 3:1:2 + 16:r: - 44 bY:l: - -nr (r) :1:8 + 0.4:r: 2 - 0.18:1: + 0.33 by :I: - 0.2 Am. q(:I:) = :1:2 + 0.6:1: - 0.06, r =: 0.318. (8) :1:8 - 0.2z2 + O.~Iz - 0.424 by :r: - 0.9 2. Use synthetio division to show that (a) (:I: - 2)2 is a faotor of :1: 4 - 4:r: 8 + 5:1:2 - 4:r: + 4 (b) (:I: + 3)2 is a faotor of :1: 4 - 17:1:2 + 6:1: + 90 (c) (:I: + 1)8 is a faotor of 2z6 + 6:1:4 + 5:1:8 - :1:2 -=:.j:r: (d) (:I: - 5)2 is a faotor of :1: 8 - 75:1: + 250 (e) (:I: + 4)8 is a faotor of :1: 4 + 8:1: 8 - 128:r:. - 256 (l) 3:1:4

2

8 -

1

7. Factorization into linear factors. There is a theorem called the fundamental theorem 0/ algebra which states that every polynomial/(x) with complex number coefficients has a complex root.. The pI:oof of this theorem is a complicated part of graduate mathematics· and will not even be suggested here. The result of this theorem and the factor theorem is that every polynomial/ex) with complex number coefficients has a linear factor x - c where c is a complex number. We state this result: Theorem 2. The only polynomials that are irreducible in the field 0/ all complex numbers are the linear polynomials. We may now apply our theory of the factorization of polynomials as products of irreducible factors. Theorem 2 states that the factors are linear and, since any linear polynomial ax + b = a(x - r), where r = -bfa, we may

144

EQUATIONS

[OBAl'. VI

take the factors to be polynomials z - rio H r is any root of f(z), then f(z) = (z - r)q(z) and we thus have a factorization of f(z) in which z - r is one of the linear factors. However the (linear) factors z - ri of f(z) were proved unique in Sec. 9 of Chap. IV. Hence r is one of the rio By the factor theorem, if z - ri is a factor of f(z), then r, is a root of f(z). We combine these results: Theorem 8. Let f(z) = ao:en + alzft-l + . . . + an where ao, ai, • • • , an are comple:x; number8, n > 0, ao ~ O. Then f(z) = ao(z - r0 (z - rll) ••• (z - rn),

(16)

where rl, • . • ,rn are comple:x; number8 which are the rootB of f(z). Thefactor8 z - ri are unique apartfrom their poBition in the factorization above. H r is. a root of f(z) and z - r occurs exactly m times in the factorization (16) of f(z) then f(z) is divisible by (z - r)m and not by (z - r)m+l. Then we call r an m-tupk root or a root of multiplicity m of the polynomial f(z) and of th~quationf(z) = '0. We" ~all r a Bimpk root if m = 1, a doubk root if m = 2, and a triple root if m = 3. ORAL EDRCISES

Give ities:

the roots of each of the following equations with their multiplic-

(a) (~- 8)1(~

+ 2)~ ... 0

+ 1) = 0

(~+ 8)1(~ - 2)'(~ - 1)(~ (c) (~+ 2)'(~~ - 4)1(~ - 2)8 ...

(b)

0

+ ~ + 2)'(~ + 2) ... 0 (~I - 4)1(~8 + ~)a(~1 + ~ + 1)1 = 0

(d) (~I

(e) (f)

(~I

-

~

+ 2)(~1I' - 4)(~1 - &: + 4)(~1 -~) = 0 + 3~ - 1)'(~5 - ~)(~I + ~I) ... 0

(g) (~I - ~.

8. Expression of coeftlcients in terms of roots. The factored form of a (17)

f(z)

pol~omi.al

==:x;"

+ C1Z-l + ... + en

with leading coefficient unity is (18)

f(z)

E

(z - rl)(z - r2)

(z ;- rn),

SEC.

8)

COEFFICIENTS IN TERMS OF ROOTS

145

where T1, T2 • • • T" are the roots of f(x). The factored form may be multiplied out and the coefficients in formula (17) compared with those in (18). We shall not give the details of the argument and recommend that the student carry out the details for n = 3, 4. The results may be expressed by the form~as

+ T2 + ... + T" = -C1, T1T2 + T1T a + . . . + Tn-1T" = C2, • Ti + T1T2 • • • T....1Ti+1 + .. . + Tn-{ ....t>Tn-{....2) • • • T" T1

(19) T1T2'

=

(_1)iCi,

Thus the sum of the Toots of f(x) is the negative of the coefficient C1 of its' second highest poweT X"-l. The sum of all possible formally distinct products of two roots is (-1)2c2 .. The sum of all possible formally distinct products of i roots is (-1)'c.. There is only one product of n roots and it is equal to (-1)"c". We have already noted these relations in the case n = 2. For n = 3 they become T1T2

T1 + T2 + Ta = -C1, + T1Ta + T2Ta = C2, T1T2T a

=

-Ca,

and for n = 4 they become T1 + 1"2 + Ta + T4 = -C1, + T1TS" + T.1T4 + T2Ta + T2T4 + TaT4 = C2, T1T2Ta + T1T2T4 + T1TaT 4 + T2TaT 4 = -Ca, T1T2TaT4 = C4. T1T2

The relations just given depend upon the hypothesis that the leading coefficient of f(x) is 1. If it is not, we may write f(x)

=

ao3!'

+ a1x"-1 + ... + an = ao(x" + C1x"-1 " + ... + c"),,

146

EQUATIONS

where the coefficients

Ci

[CHAP. VI

are the quotients.

.

The roots of I(x) are the same as those of

xn' + CIXn-

1

+ . . . + Cn

and so we may obtain the coefficients Ci in terms of the roots of I(x). Then I(x) is determined by its roots if we prescribe ao. mustrative Examples I. Find the equation with leading coefficients unity whose roots are 2, -1

+ 0,

-1 -

O.

Solution Let rl = -1 + 0, rs = -1 - 0, r8 = 2. Then rl + r2 = -2, rlr2 = (-1)2 - (O)2 = 1 - 3 = -2, (rl + r2) + r8 = -2 + 2 = 0, rlrS

+ rlr8 + rsr8 = rlrS + (rl + r2)r8 = (rlrS)r8

-2

= (-2)2 = -4.

+ (-2)2 =

-6,

The answer is:r:8 - (O):r:s + (-6):r: - (-4) = :r: 8 - 6:r: + 4 = O. II. Find the equation with leading coefficient 1 whose roots are -3 + y'2, -3 - 0, 2 + i, 2 - i. Solution

If rl = -3 + y'2, r2 = -3 - y'2, r8 = 2 + i, r, = 2 -i, we have r1 + r2 = -6, rlrS = 7, r8 + r, = 4, rar, = 5. Then -C1 = (r1

+ rs) + (r8 + r,)

= -2,

= ri.rs + (r1 + rS)(r8 + r,) + r8r, = 7 -C8 = rlr2(r8 + r,) + (r1 + rs)rar, = 7·4 c, = (rlr2)(rar,) = 35. Cs

Am. :r:'

+ 2:1:8 -

24 + 5 = -12, 6' 5 = -2,

12:r:2

+ 2:r: + 35 = O.

ORAL EXERCISES

1. Give the sum and product of the roots of· the following equations: (a) 3~s - 2:r: + 1 = 0 (b) 2:1:8 - 4:1)2 + 5:r: - 6 (c) 3:1)8 4:r; - 18 = 0

.

(d) 2:1:' + 2:r:8 + 7:1) = 0 (e) (:1)8 - 2:r:s + 1)(:r: + 3)

=0 (J) (:1)8 - 4:r;2 + 3)(:r:S - 2) = 0 (g) (2:1:8 - 3:1) + 1)(:r:2 + 4:r; - 7), ,= 0 (h) (:r: 2 - 3:1) + 2)(:r:8 - 2:1: 2 + 5)(:r:3 - 3:r;S + 6) = 0

+

=0

BlIIC.

9)

147

ROOTS OF REAL POLYNOMIALS

I. The sum of two of the roots of the polynomial2:l:1 is 3. What is the third root? S.· The product of two of the roots of the equation 2a;8

-

7~1

- 5a; + 4

+ 11:' - 1111: - 10 .. 0

is 2. What is the third root? BDRCISBS

Use the relations between roots and coefficients to find the polynomialJ(II:) = II:" + Cl/P-l + ... + c.. whose roots are (a) -3, 1 '\1'=3, 1 - '\1'=3 (c) 5, -I V2, - 1 - v'2 (b) 4, -2 + V -5, -2 - v'::5 (d) -2, i + " -3/2, i - ,,-3/2 (e) 1 V3, 1 - V3, -2 v'=2, -2 - v'=2 (J) 1+2i,I-2i,2-3i,2+3i (g) 2i - 1, -2i - 1, v'2 - 1, - v'2 - 1 (1&) 0, 3i + 2, -3i + 2, -2 + V5, -2 - V5

+

+

+

+

9. The imaginary roots of real polynomials. If r is a complex root of a polynomial J(:c) with real coefficients, the expression :c - r is a factor of J(:c). When r is real and the degree of J(:c) is greater than 1, the polynomial J(:c) cannot be irreducible in the field of all real numbers. Suppose next that r is imaginary and so write r =

where 8 and't are real, t

~

8

+ ti,

0, i" == -1. Then

'=8-ti is the conjugate imaginary, and (20)

g(:c) = (z - r)(:c - ,) = :c" - 28:&

has real coefficients. write

+ 8" + t"

We apply the division algorithm to

J(:c) = q(:c)g(:c)

+ a:c + b.

Since J(:c) and g(:c) have real coefficients, so does the remainder a:c + b. But J(r) = q(r)g(r)

+ ar + b' == ar + b =

0

148

EQUATIONS

[CHAP. VI

since fer) = g(r) = O. Hence either a = b =: 0 and g(x) divides f(x), or a ~ 0, r = -b/a is real. This is contrary to hypothesis and we have proved the following: Lemma. Let r be an imaginary root of f(:&). Then, i8 an imaginary root of f(x) , (x - r)(x - ,) i8 a factor of f(x). As a consequence of the lemma we see that every real polynomial (i.e., polynomial with real coefficients) has a real linear or a real quadratic factor. Using Theorem 1 we may state the following: Theorem 4. A real polynomial is irredUcible if and only if it is linear or is a quadratic axil + bx + c with bll - 4ac < O. We now apply the factorization theorem of Sec. 9 of Chap. IV with Theorem 4 and obtain the following: Theorem 6. Every real polynomial is ewpresBible uniquely, apart from arrange~ and constant factor8, as a product of real linear and real irreducible quadratic factors. The roots of the real linear factor8 are the real roots of f(x), and the roots of the quadratic factor8 are pairs of conjugate imaginary roots of f(x). . . H r is an imaginary root of multiplicity m of f(x), the corresponding polynomial (x - r)(x - ,) must occur exactly m times in the factorization of f(x). Then,. has multiplicity m. Theorem 6. Let a and b ~ 0 be real and a + bi be a root of multiplicity m of a real polynomial f(x). Then a - bi is a root of multiplicity m of f(x). ORAL BDllCISBS

1. The number 3 + 2i is a. root of the equation Z8 -

9z1

+ 31z -

39 = O•.

What are its other roots? So If 2 - 3i is a. root of Zl + Zl + ba: + c = 0, what are its other roots and what is the value of c? . 3. Let 2 + i be a double root of Zl + az4 + bz8 + CZI + d:& + 25 -0. What are its other roots and what is the value of a? " Give the factored form of the equation of degree six whioh has

SlDc.l0] ..

149

MULTIPLE ROOTS BY DEIUVATIVES .

real coefficients, has 2 simple root.

v=s as

&

double root, and has 1

+i

as

&

10. Multiple roots by derivatives (FULL COURSE). We may compute the derivative polynomial J'(z) of J(z) and the g.c.d. d(z) of J(z) and J'(z). The roots of the factor d(z) of J(z) are roots of J(z) and Sec. 13 of Chap. IV implies that, if r is a root of multiplicity m of J(z), then it is a root of multiplicity m - 1 of d(z). When d(z) == 1, the polynomial J(z) has no multiple roots. Otherwise we may find the multiple roots of J(z) by solving the equation d(z) == O. A nonconstant polynomial J(z), which is irreducible in a field F of numbers containing its coefficients, factors with linear factors in the field of all complex numbers but has no multiple root. For f(z) is not the zero polynomial and has coefficients in F, so does d(z); d(z) divides f(z) and has degree at most m - 1. Then d(z) is not a constant multiple of J(z) and can divide J(z) only if it is a constant. Let us apply the process above to any quintic polynomial J(z) with a root of multiplicity m > 1. H J(z) has only one mUltiple root r and this is a double root, the polynomial d(z) == z - r, (z - rr' is a factor of J(z) == (z - r)lIq(z). Then the problem of solving the equation J(z) == 0 is reduced to the solution of the cubic q(z) == O. H J(z) has two double roots or a triple root, the polynomial d(z) is a quadratic and may be solved by the quadratic formula. It determines three (or four) roots ofJ(z) and the determination of .the rema,;njng two roots'is a trivial matter. When i(z) has a double and a triple root, or a root of multiplicity four or five, the polynomial d(z) has degree three or four and a multiple root. The equation d(z) == 0 niay be solved .by inspection or the derivative process, and' the solution of J(z) == 0 completed by dividing J(z) by d(z) or by comparing coefficients and roots. lUustrative Examples I. Find the roots of J(z) - z' - 242;1 + 64z - 48.

150

EQUA.TIONS

[CRAP. VI

Solution /'(S) .. 42;1 - 4&

+ 16). 24:1:1 + 64s 12:1:1 + 1& 12:1:1 + 4& -

+ 64 = 4(SI -

+ 41~+ 4 - 12:1: + 16~ j;. - 42;1 + 42; s' 42;1 - lOa: + 16 42;1 - lOa: + 16 .. .. Sl - 42; + 4

12:1:

'Sl - 42;

des)

.. (s - 2)1. I(s) ... (s - 2)I(S - r),

-12(SI - 42;

( -8)( -r) - -48,

48 48

+ 4)

r - -6.

Am. 2 is a triple root, -6 is a simple root. Ii. Find the roots of Sl'- 7S1 - 2:1:1 + 12:1: +~. Solution

/'(s) - 5s' - 21s1

-

"'-11

1,_+10

42; + 12.

18'-.... -"+11 + .. -_-.1,.. + "'-.-10 MIz'+ MIz'+I01\11- 1NOII-7OCb , .. -

, .. -1M

lO.III-lo.-1O lO.III-lo.-1O

- 10l\Il -10&81-

-... - .... -lillif.+'o

1188 18'-1"'-'-'+1111 -1'-'- . . +.+~ lW + IiIHII + 1188 - -2('1"+"'-.-10)

OIl' - 10l11J111- 1 .

... +8.+800 , - 1"-'+ 1"- + lI88 ... 1~-_-J)

des) _ Sl - S - 2 t - (s - 2)(s + 1). I(s) - (s - 2)I(S + l)l(s - r), (-2)11'( -r) ... 8, r - -2. Am. Double roots 2, -1. Simple root -2. BIlmCISBS

Each of the following equations has at least one multiple root. Use this property to solve them. (a) s' - 0a:1 - 8s - 3 - 0 (b) s' 2:J:I Sl - 12:1: 8 - 0 Am. I, I, -2 2i, -2, -2i.· (e) s' - 24:1:1 64s - 48 - 0 (d) s' 4:J:I - 16:1: - 16 - 0 Am. -2, -2, -2, 2. (e) s' ~ 3:1:1 ~ U 4- 0 Am. I, I, -2, -2. (J) s' - u l - 2:1:1 12:1: 9 == 0 (g) Sl lOs' 15s 6 - 0 , Am. -I, -I, -1, t ± l V -15. (A) Sl 8Os1 240s 192 = 0 (i) sa - 6:1:' 5s' 12:1:1 42; .. 0 W Sl - 10s1 - 20:1:1 - 15s - 4 - 0

+

+ +

+ +

+ + -

+

+ + + + + + + + + +

+

CHAPTER VU REAL ROOTS OF REAL EQUATIONS 1. Transformations of polynomials. The principal tools used in the study of the real roots of equations J(x) = 0 are certain transformations which replace a polynomial (1) J(x)!5 aoX" + alxtl-l + . . . -t an =' ao(x - rl)(x - rs) ••• (x - r,,) by a polynomial g(x) whose roots are related to those of J(x) in a prescribed manner. We shall first prove the' following: Theorem 1. Let g(x) = J(x + a) and rl, rs, • • • , r" be the roots oj J(x). Then rl - a, rs - a, •••. , r" - a are the root8 oj g(x).

For J(x

+ a) lEI'ao(x + a -

But x

+a-

r.

rl)(x

!5

E!

ao(x

+a -

r,,).

ao(x - Bl) • • • (x - B,,)

where Ii ='r. - a. The coefficients bo, bl ,

.

rs) ••• (x

= x - (r. - a), and

g(x)

g(x)

+a -

• •

• ,

btl of

+ a)" + al(x + a)tl-l + . . . + a......l(X + a) + an boX" + blztl-l + . . . + btl E!

may be computed by a process of repeated synthetic division. We observe that since J(x + a) = g(x) then g(x - a)

= J(x)

&

bo(x - a)"

+ bl(x -

a)"-l + ... + btl-l(x - a) + btl.

Then btl is the remainder on division ofJ(x) by x - a and bo(x - a)tl-l + ... + btl-s(x - a) bn-lls the quotient. If we divide this quotient by x - a, the remainder is bn-l

+

151

152

REAL ROOTS OF. DEAL EQUATIONS

[CJLU>. VII

+ ... +

and the new quotient is bo(z - a)ft-I btl-2. We use this second quotient to compute btl-I as a third remainder, and so succ~BSive synthetic divisions will yield the desired coefficients bn, btl-I, . . • , bl, bo = ao as a sequence of remainders. The reader will have realized from the start that bo = ao. Our second transformation is that which replacesf(z) by

g(z)

~ tt1(I) ,

E

(I -

tnao ~ - rl) r2) ••. rn) - ao(z - WI)(Z - tr2) • • . (z - wn ).

(r -

The general term of fez) is a¢'-1 and that of g(z) is tnlSi

G)"""""'

E

lSirZ"""""'.

We have proved the following: Theorem 2. Let fez) be given by formula (1), t be any

1l.O'Tl.I67'0 number, and g(z)

5!!!

aoZ"

+ ta1z"-1 + t alr 2 + ... + t-llln-IZ + tnan. l

Then, if rl, . . • , rn are the roots of fez), the roots of g(z) are trl, WI • • • trn. If an - 0, then z is a factor of some multiplicity m of f(z) , zero is a root of multiplicity m of fez), and we may remove the factor zm from fez). Let us then restrict our attention to polynomials for which an ;oS 0, so that no root r, of fez) is zero and every rr l exists. We define

g(z) = z"f (~) - anz"

+ lln-lrl + . . . + ao,

and see that

g(z)

i5

aoZ" (~ - rl) (~ - r2) ••• E

(~ -

rn)

ao(l - rlz)(l - rlZ) ••• (1 - rnZ).

Then, if 8, = l/r" we ~ve g(z) = aorlrl .•• rn(81 - Z)(81 - z) ••• (8n - z).

SEC.

1]

TRANSFORMATIONS O.F POLYNOMIALS

153

Since TIT! • • • T" = (-I)"an/ao, we have g(x) = an(x - SI) • • • (x - B,,)

and have proved the following: 1).eorem 3. LetJ(x) Eii ao:c" +

... + + ... +

an with aoan ~ O. Then the TOOts oj g(x) = an:J!' ao aTe the Teciprocals oj the Toots oj J(x). This last transformation is used less frequently than the

others. The sum Tl + T! + . . . + T" of the roots of the equationJ(x) = ao:c" + a1xtl-l + ... + an = Ois -al/aO' Diminish the roots by - adnao and obtain an equation g(x) E!! bo:J!' + b1Xtl-l + ... + btl = 0 whose roots are T, + al/naO' Then the sum of the roots of g(x) = 0 is zero and b1 = O. Such an equation is called a Teduced equation. The solution of any equation J(x) = 0 may be achieved if we can solve the related reduced equation obtained above. A reduced cubic equation may be divided by its leading coefficient and so put in the form. h(x) ;;s x' + px + q = O. The number.6. = _4p' - 27q! is called the discriminant of h(x) = 0, and it can be shown that h(x) = 0 has three real roots except when.6. < O. We shall not try to prove this result. . nlustrative Examples I. Increase the roots of :1:' + 83:8 + 2:1:1 - 143: + 2 by 2. Solution 1 + 8 + 2 - 14 + 21-2 - 2 - 12 + 20 - 12 6 - 10 + 6 - 10 - 2 - 8+36 4 - 18 + 42 -2 - 4 2 - 22

-2 The desired

equa.ti~n

1+0 is :1:' - 22:1:1

+ 42:1: -

10

= O.

154

REAL ROOTS OF REAL EQUATIONS

[CHAP.VII

II. Find a polynomial whose roots are related to the roots r of 27z8 + 542;2 + 30z + 5 by the formula 8 = '3r + 2.

Solution A polynomial whose roots are 3r is 27z. + 3 . 543;1 + 9 • 30z + 27 • 5 .. 27(Z8 + &:2 + 10z + 5). Diminish the roots by -2, using the synthetic division process.

1 + 6 + 10 + 51-2 - 2 - 8-4 4+ 2+1 -2- 4 2- 2 -2 1+0

Ana. Z8 - 2z

+ 1.

III. Find an equation whose roots are related to the roots r of 4z· - 38z2 + 120z - 127 = 0

by the formula

8

= 2(r - 3).

Solution 4 - 38 + 12 -26+ 12 -

-14+

120 - 12713 78 12642- 1 42

+

0

12 4- 2

An equation with roots r - 3 is 43;1 - 2z1 - 1 .. 0, and we multiply the roots by 2 to obtain.4z8 - 43;1 - 8 .. O. Ana. Zl - Zl - 2 =0. IV. Determine whether or not the roots of Z8 - 2z1 - 2 = 0 are all real.

Solution Here ao = 1, al = 2, and -al/3ao = f. To avoid fractions we multiply the roots by 3 and so obtain Z8 - 6zll - 54. Diminish the roots by 2. 1- 6+ 0 - 54~ +2 - 8 -16 - 4 - 8-70 2- 4 -2-12. +2

1+0

SEC.

1]

TRANSFORMATIONS OF POLYNOMIALS

The reduced cubic is Zl

-

155

12z - 70 = 0 and

A = -4( -12)' - 27(70)1

= 27(256 -

4,9(0)

is negative. The equation has only one real root.. ORAL BXERCISES

1. Use Theorem 2 to give an equation whose roots are the negatives of the roots of

+

+

(a) 3z4 4z1 - 2z1 7z - 1 (b) 2Z1 - Szl + 7z1 + 2z - 8 (c) Zl - Szl - 9 (g) Zl 4z' - Sz - 1 (h) 7:1:' Sz' - 9z1 - 10:1:1 - 2z

+ +

(d) Zl - 7z8 - Szl - 1 (e) Z8 - 9z' 10:1: (f) z, 6z1 - 7z'

+

+

- 1

2. Give polynomials whose roots are t times the roots of the following polynomials for the given values of t: (a) 32;1 - 2z1 4z - 5;.1 = 2, 3 (b) 2z' &1 - iz 1; t .. -2, 2 (c) 3z' - &1 + ISz - 54; t = i, i (d) z, - Szl - 24z - 64; t = i, i 3. Give!J. for each of tbe following equations and state whether or not the roots are all real:

+

(a) (b)

Zl Zl -

+

+

Z - 1 = 0 (c) 2z - 1 = 0 (d)

2 =0 + 2z + 1 = 0

Zl -

(e)

Zl -

Z8

(f)

Zl -

4:/: - 2 - 0 3z - 1 = 0

" Give polynomials whose roots are the reciprocals of the nonzero roots of the polynomials of Oral Exercise 1. EXERCISES

1. Diminish the roots of the following polynomials by the given n~bera:

+ 4z - 7, a = 1 Am. 3z1 + '32;1 + z - 6. 2z' + 3z1 - 5z + 8, a = 3 Zl -·1& + 17, a = 4 Am. Zl + 12z1 + 32z + 17. Zl - 32;1 + 2z - 4, a = -5 z, - 4:/:1 - 2z + 1, a = 2 Am. z' + Szl + 20:1:1 + I4:/:- 3.

(a) 32;1 - &1

"(b) (c) (d) (e)

(f)

Zl

+ 32;1 -

Z -

1, a = 3

S. Increase the roots by a in each case of Exercise 1. " 3. Find a polynomial whose roots B are related to the roots r of each of the following polynomials by the corresponding formula:

156 (a)

(b) (c) (d) (6)

(j) (0) (1&) (i)

REAL ROOTS OF REAL EQUATIONS

[CHAP.VII

+ 56:1; - 40, 8 = it' - 2 Am. Zl + 2z + 7. Z8 - 6:1;1 + 831 + 16, 8 = l(r - 2) 4z1 - Z + 1, 8 == 2r + 1 Am. Zl - 3z1 + 2z + 2. 1 1 729:1: + 243z + 9:1; - 1, 8 = 3r + i 9:1;1 - 9:1;1 + 4z, 8 = 3(r - i) Am. Zl + z + 2. 8zI - 36:1;1 + 54z - 35, 8 == 2r - 3 Zl - 16z1 + 64z + 48, 8 = it' - 6 Am. z· + l(l:/;I + 28z + 30. 8zI - 12z1 - 1831 - 23, 8 -= 4r + 2 Z8 -

12z1

Zl -

9:1;1

(J) z·

+ 45z -

54,

8 ==

+ 21z1 + 14k + 54,

ir - 1

Am. Zl

8

= i(r

+ 6)

+ 2z + 1

" Check your answer g(z) = 0 in each case of Exercise 3 by solving for r in terms of 8 and applying the corresponding transformations to g(z) to obtain the original/(z). 5. Apply transformations that will repJace each of the following • equations by a reduced equation: 6:1;1 + 2 ... 0 3z1 + 4 == 0 (c) 2z' - 12z1 + 5z - 9°= 0 (d) Z4 + 8zI - 3z - 1 = 0 (6) 3z4 + 12z' + 7z1 + 2z 4

(a) (b)

Z8 -

Zl -

Am. Zl - 3z + 2 ... O. Am.

+ = O.

Z4 -

243;1

+ 61z -

43

= O.

'

6. Determine whether the roots of the following equations are or are not all real: (a),3z' -

Zl

+ 1 ... 0 == 0

(b)

Z8 -

Zl -

2

(c)

Zl -

2z1

+1

(d)

Zl -

6:1;1

+1 =

0=

0 0

(6) z· - 4z1

+ 3z -

1 == 0 Am. Not all real.

+ 6z 2z + 2 - 0 Am. All real. (0) 2z1 + 6z1 - 3z - 4 ... 0 Am. All real. (1&) 2z' + Zl - Z - 1 - 0 (J)

Zl

1 -

2. Integral roots. The study of the roots of equations with rational coefficients is equivalent to the study of the roots of equations with integral coefficients. For if we multiply an equation with ra.tional coefficients by the l.c.m. of the denominators of these coefficients, we obtain an equiva.lent equation with integral coefficients. Let us first study 8. process for finding the integral roots of such equations.

SEC.

157

INTEGRAL ROOTS

2]

If the coefficients of f(x) 5 aoXn + a1x"'-1 + + an are ordinary integers and r is an integral root of f(x) = 0, then fer) = br + an = 0 where b = aorn- 1 + ... + a-l is an integer. Then an = (-b)r is divisible by r. Since the integer an has only a finite number of divisors, it is possible to determine all integral roots of f(x) = 0 by listing these divisors d and by computing fed). We may, however, improve this procedure as follows. Let a be any integer and g(x) 5 f(x + a). By the process of Sec. 1 the coefficients

bo = ao,

bn

= g(O) = f(a)

are obtainable by addition and "multiplication of integers. Then they are integers. The roots Si of g(x) = 0 are related to the roots Ti of f(x) = 0 by the relation Si = ri - a and a is an integer. Then Si is an integer if and only if ri is an integer. It follows that r can be an integral root of f(x) = 0 only if S = r - a divides f(a). We have proved the following:

Theorem 4. The integral roots of a polynomial equation f(x) = 0 with integral coefficients are integers r such that r - a divide~ f(a) for every integer a. It is usually undesirable to utilize more than a few

values of a in applying Theorem 4 to sift out the integral roots of f(x). A systematic procedure might begin with an, the computation of f(O) = an, f(l) = ao + al + . . . and f( -1) equal to the sum of all of the coefficients of the even powers of x minus the sum of the coefficients of all the odd powers. We are assuming that f(O) ;;z!! o. If f(l) = 0, we ~vide f(x) by x - I to obtain a quotient equation cf>(x) = 0 of degree n - 1 which has already been called the depressed equation. Then we begin over again and study the integral roots of this new equation. We proceed similarly if f( -1) = O. Let us then assume that f(O), f(I), !( -1) are all not zero. Factor these integers. The three factorizations

-+

158

REAL ROOTS OF REAL EQUATIONS

[CHAP. VII

will indicate which integer has fewest divisors and we list the divisors of this integer. Suppose first that we have listed the divisors d of !(O) omitting the values d == 1, d == .-1 which have already been tested. Then a particular d will not be a root of !(z) == 0 unless d - 1 divides !(1) and d + 1 divides !( -1). These divisibility conditions will sift out most divisors d that are not roots and we complete our solution by computing !(d) for the rema.ining values. . Our second possibility is that we have listed the divisors e of!(I). Then any particular e will arise from an integral root d of !(z) = 0 only if e == d - 1, d == e + 1. We omit d == 0, -1 and so e == -1, -2 from our divisor list in this case, and we see that e + 1 is an integral root of !(z) == 0 only if e + 1 divides !(O) and e + 2 == d + 1 divides !( -1). . Our final possibility is that of a list of divisors c of !( -1). Since c == d + 1 and d " 0, 1 we omit the values c == 1, 2 from this list. Then d = c - 1 will be a root of !(z) == 0 only if c - 1 divides !(O) and c - 2 == d - 1 divides !(1). We shall illustrate the procedure just described in the examples given below. However we shall delay the exercises until after the next section in which the results are given additional significance. mutTGtive E%Gmples I. Determine the integral roots of /(~) 51 ~,

-

49~1

+ &: + 56 .. O.

Solution Weseethat/(O) division

= 56,/(1) -

16,/(-1) .. O. Carryoutthesynthetic

1 + 0 - 49 + 8 + 561-1 - 1 + 1 + 48 - 561-1-48+56+ 0 Hence /(~) E (~ + 1)4>(~) where 4>(~) E ~. - ~. - 4&: + 56. Then 4>(0) ... 56 .. 8·7, 4>(1) = 8, 4>(-1) = 102 ... 2·3 ·17. The divisors 6 of 4>(1) which we list are 1, 2, 4, -4, 8, :-8. The eorrespondirig

SEC.

3]

159

RATION AL ROOTS

values of a = 6 + 1 are 2, 3, 5, -3, 9, -7 and we retain only 2,-7. Then a + 1 = 3, -6 are both retained. By the synthetic divisions 1-1 - 48 + 56~ 2+ 2-92 1+1-46-36

1 - 1 - 48 + 561-7 -7+56-561-8+ 8+ 0

we see that the integral roots of J(z) are -1, -7. Note that we may now compute the remaining roots of J(z). They are 4 + 2 0,

4-2'\1'2. II. Determine the integral roots ofJ(z)

E

Z4 -

3z1

-

63z - 72

=

o.

Solution WecomputeJ(O) = -72,J(I) = -137,J(-I) = -11. Thedivisors of J( -1) which we list are c = -1, 11, -11 and a = c - 1 = -2, 10, -12. We delete 10 and see that a-I = -3, -13 do not divide J(I). Thus J(z) has no integral roots.

3. Rational roots. A rational root r ~ 0 of an equation (1) with integral coefficients may be expressed as a fraction

r=l!. q.

where p and q are nonzero integers whose g.c.d. is unity; q may be taken to be positive. By Theorem 2 the integer p is a root of the equation aoXn + qalXn-1 + . . . + qnan = 0 and, by Sec. 2, p divides gnan. Since p and q are relatively prime, so are p and qn. By Lemma 7 in Sec. 8 of Chap. II we conclude that p divides an. The number q/ p is a root of anxn + . ... + ao = 0 and q is a root of anxn + PCln.-1Xn- 1 + . . . + pnao. Then the positive integer q divides pnao and by the lemma referred to above we see that q divides ao. We have proved the following: Theorem 5. Let p/ q be a nonzero rational root, in least terms, oj the equation (1) with integer coefficients. Then the integer p divides an and the positive integer q divides ao. In the case where ao = 1 the positive integer q divides 1 and q = 1. This yields the ·following:

160

REAL ROOTS OF REAL EQUATIONS

[CBAP.vn

Theorem 8. Every rational root oj an equation with inte-

-ural coejficientIJ and leading C06jficient 1 is an integer. Theorem 5 may be used to find the rational roots of We list the divisors p of CIn, the divisors q of ao, the corresponding quotients plq and then use synthetic division to compute J(plq). However it is simpler to select a number t such that the equation

J(z) == O.

(2)

has integral coefficients. The rational roots of this new equation are integers u and may be determined by the method of Sec. 2. The rational roots of J(z) are then the numbers ult. Observe that a value of t which will suffice is t == ao. However it is preferable to use the least positive integer t for which formula (2) has integral coefficients. lUustrative Examples I. Find the rational roots of 1(1e) 55 &;8

+ 4z1 + 5z -

6 = O.

Solution The positive divisors of 6 are 1, 2, 3, 6 and those of 3 are 1, 3 so that the possible rational roots are 1, -1,2, -2, 3, -3, 6, - 6, i, - i, i, - i. Apply Theorem 2 with t = 3 to C?btain an equation g(1/) a i(3y8 + 121/1 + 45y - 6 . 27) 1/8 + 4y1l + 151/ - 54 = O. We multiply our previous possibilities by 3 to obtain 3, -3,6, -6,9, -9, 18, -18, 1, -1, 2, -2 as possible rational roots of g(1/). Note that all numbers divide 54 but some divisors of a.. == 54 do not occur in this list. We compute g(l) ,;.. -34, g( -1) = -66, omit d = 1, -1, and see that d - 1 divides 34 only for d = 3, 18, 2. Then d + 1 divides 66 only if d == 2. By synthetic division g(1/) = (yi + 6y + 27)(1/ - 2) and the only rational root of 1(1e) is f. Its other two roots are -1 + ~ ..-2,. ...., 1 - v'=2, since the polynomial whose roots are i those of 1/1 + 6y + 27 is leI + 24; + 3 -= (Ie + 1)1 + 2. n. Find the rational rOots of 1(1e) = 48z8 + 1&1 - 31e - 1 .. 0.

=

Solution Since a.. = 1, the rational roots of this equation are the reciprocals of the integral roots of g(1/) = 1/8 + 31/1 -.l6y - 48 = O. Now g(l) ... -60, g( -1) = -30. The divisors h of -30 .. -3' 5 . 2 are 1, -1,2, -2,3, -3,6, -6,5, - 5, 10, -10,15, -15,30, -30. Then d - h - 1 divides g(O) only if d = -2, -3, 2, -4, 4, -6, -16, and

161

RATION AL ROOTS

SEC. 3)

d - 1 divide.s -60 only if d = -2, -3,2, -4,4. We now compute = 8 + 12 - 32 - 48 ~ 0, g( -2) = -8 + 12 + 32 - 48 ~ 0,

g(2)

1 + 3 - 16 - 481-3 -3+ 0+48 0- 16 0

+

=0

and see that g(y) = (y2 - 16)(y -.\- 3). The roots of J(:c)

- i,l, -1. III. Determine the rational roots of :c 3

-

48:c

are

+ 64 = o.

Solution

+

We employ Theorem 2 with t = 1 to obtain g(y) = yS - 3y 1 = o. Evidently g(l) ~ 0, g( -1) ~ 0 and the equation has no rational roots. EXERCISES

1. Determine the integral roots of the following polynomials: 15:&2 + 54:& - 54. 8:&2 + 9:c - 36 19:c2 + 82:& + 60 17:c2 + 78:& - 56 4:&2 - 95:c + 198 :c 2 - 64:& - 72 2:&8 + 39:&2 - 188:& - 176 :c8 - 14:&2 + 9:& + 324 :c8 + 15:&2 + 84:c + 196 3:&3 + 2O:c2 - 131:c - 132 :c4 + :c 8 - 3O:c2 - 1O:c + 200 2:c4 - 47:c 2 - 42:c - 135 :c4 - 2:c 8 - 6x 2 - 2:& - 24 :c 8 2:c 2 - 15:c - 52 :c4 + 15:&8 - 2:c 2 - 34:& - 60 :c 8 - 9:&2 - 24:& + 216 :c4 + 4:&8 - 75:&2 - 324:& - 486 :c5 - 24:&4 149:c3 - 352:&2 + 294:& - 608

(a) :c 8

(b) (c) (d) (e) (J)

(g) (h) (i)

(;) (Te) (l) (m)

(n) (0) (p) (q) (r)

-

2:c8 :c 8 + :c 8 :c 8 :c 8 +

+

+

Am. 3. Am. -5. Am. 2, 11, -9. Am. None. Am. -7. Am. 4, -5. Am. 4. Am. 2, -15. Am. 9, -9.

s. What can be said about the rational roots of the polynomials above in all cases except (g), (i), and (l)? 3. Find all rational roots of the following:

+

X 2= 0 6z8 - 9:c2 + 2:& - 3 = 0 2O:c3 + 11:c 2 - 8:& + 1 = o· :c 3 + 3:&2 - 2:c + 15 = 0 3:c 8 - 37:c 2 93:c - 27 = 0 203:c 8 + 351:c2 + 39:& - 1 = 0

(a) 2z8 - 4:&2 -

(b) (c) (d) (e) (J)

+

Am. 2. Am. -1,1, i. Am.

i, 3, 9.

162

REAL ROOTS OF REAL EQUAT,IONS

(g) 2:1;' - z

+

+5=

Am. ,None.

0

+

(/I) 15:1;1 7z1 - 7z 1 = 0 (1.1 2:!;t + 16z' + 19z1 + 001: + 27

(J1

(I:)

(I) (m) (ft,)

[CHAP.VII

27z1 + 2z - 3 .. 0 9z8 + 9z1 - Z - 6 .. 0 6zI + 21z· + 26z - 13 - 0 4r - u 8 - 3z1 - Z - 6 - 0 2:1;8 - 15:1;1 + 42z 44 .. 0

=0

Am. None.

54z8 -

Am. f. Am. -1.

+

4. Upper and lower bounds. Let a and b be real numbers and a be less than b. Then we, designate the set of all real numbers z between a and b by a ~ z ~ b. If we omit a from this set, we designate the result by a <; z ~ b and, if we omit b, by a ~ z < b. If we omit both a and b we designate the result by a < z < b. All these sets of real numbers are called real intenJal8. An upper bound for the real roots of an equationf(z) = 0 with reaJ coefficients is a real number U such that f(b) F 0 for every b ~ U. A lower bound is a real number L such that f(a) F 0 for every a ~ L. Then' the real roots of f(z) = 0 are in the interval L < z < U. Let us restrict our attenti0!l to polynomials f.
+ . . . + (Ita

in which ao is positive. Then, if no coefficient of f(z) is negative, there is no positive root of f(z). Indeed when b > 0 the number f(b) is a sum of positive numbers and is positive. Thus every positive real number is an upper bound. If a. F 0 an upper bound is U == O. We shall assume now that f(z) does have a negative coefficient. Let Ie be the least integer for which ail is negative. Then k is the difference between the degree n of f(z) and the exponent of the highest power of z with negative coefficient. For example, in &:& - 23:4 + 7z1 - 1 we have Ie == 5 - 4 == 1, and in &7

+ 23:8 -

Z, -

1623:8 + 23:1

-

4z - 2

we have Ie == 7 - 4 == 3. We let G be the largest absolute value of the negative coefficients of f(z). In the second

BBlC.4]

UPPER AND LOWER BOUNDS

163

example just considered G - 162. Then an upper bound is

U-l+

(3)

G ~ -. ao

In the second example above G/ao - 27 and U-1+~-4.

Note that in the first example above k - 1 and

U-l+~-I+*. We shall actually show not only that U is an upper bound but that/(b) > 0 for every b ~ U. It is desirable to have a second procedure for obtaining an upper bound. We divide the absolute value of every negative coefficient of /(:c) by the sum of an positive coefficients that precede it. In the second example above, the fractions obtained are 1

6

162

4

2

+ 2' 6 + 2' 6 + 2 + 2' 6 + 2 + 2'

Add 1 to the largest fraction so computed. The number U obtained is an upper bound. In the special case the value is 1 + 20t = 21-1. . 1 162 Note that the fractIOns 6 + 2 and 6 + 2 have the same sum as denominator, and so the one with smaller numerator need not be listed. Similarly we would normally not list 2 4 6 + 2 + 2 but only 6 + 2 + 2' To obtain lower bounds we study the polynomial g(:c) = (-1)"/( -:c). This polynomial has the same leading coefficient ao > 0 as does /(:c) and its roots are the negatives of those of /(:c). Then, if every root of g(:c) is less than a real number U o, every root of /(:c) will be greater than L - - U o. Hence the negative of any upper

164

Rl!IAL ROOTS OF Rl!IAL l!IQ U ATIONS

[CIlA.P.

VII

bound of the real roots of (-1)"!( -x) is a lower bound of the real roots of J(x). The coefficients of J(x) are said to alternate in aign when the coefficients of (-1)"!( -x) all have the same sign. This occurs, of course, if all even powers of x have coefficients with the same sign and if this sign is opposite to the sign of the coefficients of all odd powers of x. For example, according to this definition, the signs of 63;7

+ 33;6 + 2x

8 -

+ 43; 63;8 + 2:1: 2:1:1

1

alternate-but the signs of 7x" 1 do not alternate. When the coefficients of J(x) alternatel no negative number is a root of J(x) == O. 1£ a. ~ 0 the number 0 is a lower bound. . We have now given two methods for obtaining upper and lower bounds. Let us call the first method the method oj radicals and the second the method oj jractiom. We shall call the lesser of the two upper bounds obtained a best upper bound and the greater of the two lower bounds a best lower bound. It may happen that in a Single exercise one of the methods will give a best upper bound and the other a best lower bound. 1llu.tratiue Example. I. Find beet bounds for I(a;) ;;;;;; :r - 14a;1

+ 51a;1I -

14a; - SO

= 0:

Solution

Here G = 80, k

= 1, and the method of radicals gives U1

We compute the fractions

-

1 + SO = 81.

¥, H,

and the method of fractions gives

. Us = 1 + 14 == 15. Hence the best upper bound is U = 15. Also

+ 143;8 + 51a;1I + 14:1: - 80 and 80 G - SO, k = 4 - 0 = 4, - Ll = 1 + {ISO is approximately 4. The method of fractions gives - Ll = 1 + = 2. Hence the best I( -a;) = :r

*

lower bound is L = -2. We indicate the result by stating that the real roots r of I(a;) - 0 lie in the interval -2 < r < 15. II. Find best bounds for:r + 2a;1 l1a;1I a; - 81 = O.

+

+

SEC.

5] PROOF OF THE METHOD OF RAD ICALS

165

Solution Here G = 81, k = 4, and U1

U2

= 1+~

4. Also

= 1 + it> 6,

+

Also /(-z) ==:c4 - 2z3 11Z2 - Z - 81 and k = 1, = 82. We examine the fractions 2, and -L2 = 1 ~f. = 1.;-. Hence - 1lf < r < 4.

and so U

= 4.

G = 81, so that -L1 = 1 + 81

-tt = ¥

=

+

ORAL EXERCISES

1. What can be said immediately about the real roots of (a) (b) (c) (d)

+ + +

+ 2:1; + 1 = 0 + 2Z2 = 0 + 2Z2 - 4:1; + 3 =

z' 3z2 Zl 3z3 z, - 3z3 Z6 2z' -

3z8 - 631 = 0

0

(e) (J) (g) (h)

-Z3 - 6z 2 - 53l - 4 = 0 -z' 2:1;3 - Z2 = 0 -Z2 2z - 1 = 0 Z3 - 3z2 3z - 1 = 0

+ +

+

2. What are the numbers k and G of the two methods for the following equations? (a) 2z3 - 4b2 320z - 800 = 0 (b) 4:1;6 - Sz' 20:1:8 100z2 - 73:1: 5 = 0 (c) z' - 36:1;2 49z 50 = 0 (d) 2:1;6 - 6:1;2 - 7z 14 = 0 (e) 4:1;8 6z7 - 18z' - 6z3 3z - 20 = 0 (I) Z7 4:1;6 4:1;' - 9z - 64 = 0 (g) z' 4:1;8 6z2 - 22z - 27 = 0 (h) 3z' - 12z2 15:1: - 8 = 0 (~) Z8 - 16z 6 24:1:' - 25:1: 2 3z 2 = 0 S. What are the fractions we compute in (e) of Oral Exercise 2? 4. What are the upper bounds in (e)?

+ + + + + + + + + + + + +

+

+

+ +

EXERCISES

1. Compute best upper and lower bounds for the polynomials of Oral Exercise 2. 2. Compute best bounds for the polynomials of the exercises of Sec.3.

6. Proof of the method of radicals (FULL roots of J(x) coincide with the roots of F(x) ==

COURSE).

..!. f(x) == x" + C1Xn - 1 + . . . + en. ao -

The

160

~EAL

ROOTS OF REAL EQUATIO'NS

We define k and G as in Sec. 4 and oDserve that first negative coefficient of F(x). Also define

!CHAP.VtI

Ck

is the

G d =-.

ao

Then -d is a negative coefficient of.F(x) such that d ~ 1c.;1 for every negative coefficient c.; of F(x). We let U = 1 + {Id and suppose that b ~ U. Then b - 1 ~ {Id and (b - 1)k ~ d~' But U > 1 and so b > 1, bk-l ~ (b - 1)k-l. Then bk-l(b - 1) ~ (b - 1)k ~ d and the number

The terms in F(b) with coefficients Cl, C2, • • • negative and so F(b) ~ bn + ckbn-k + .. every coefficient c.; ~ - d and so F(b)

~ 8,

8 =

bn - d(bn- k + bn-k-l

,CTc-l

are not But

+ en.

+ ... + 1).

However 8(b - 1) = bn(b - 1) - d(bn-k+l - 1) = bn+l - b'" - dbn-k+l + d = bn-k+l(bk - b'-l - d) + d = ebn-k+l + d

> 0.

Thus 8(b - 1) > 0, 8 > 0, F(b) > O. Since ao > 0, the number !(b) > 0, for every b ~ U, U is an upper bound for the real roots of lex) = 0. We have given the proof above mainly as a sample of a type of argument in the study of polynomials with real coefficients. We shall not give a proof here of the method of fractions. ~t may be found in L. E. Dickson's "First Course in the Theory of Equations." 6. Isolation of the real roots of a real equation. As we observed in Chap. VI, any polynomial lex) of degree n with real coefficients is expressible as a product

SEC.

6]

ISOLATION OF THE REAL ROOTS

167

where rl, ••• , rt are the real roots of f(x) = 0 and q,(x) is a product of factors of the type X2 - 2dx

+ d2 + e2,

for real numbers d and e such that e ~ O. The roots of q,(x) are the imaginary roots of f(x) = 0, the degree n - t of q,(x) is an even integer, and (_I)n-t = I, (-I)" = (-I)t. If c is any real number, then c2 - 200

+ d2 + e2 = (c -

d) 2

+ e2 > O.

Hence if c is real, the value q,(c) is a positive real number. If c > r, for i = I, • • • , t the factors c - r, are all positive and f(c) has the same sign as ao. If c < ri for i = I, ••• , t the numbers c - r, are all negative and f(c) .has the same sign as (-I)tao = (-I)"ao. We state this result. Theorem 7. Let U be any upper bound and L be any lower bound to the real roots of f(x) == aoX" + ... + a". Then f(c) has the 8ame sign as ao for every c ~ U and f(c) has the 8ame sign as (-I)"ao for every c ~ L. . We also prove the following: Theorem 8. Let a < b where a and b are real number8 and f(a) ~ 0, feb) ~ O. Then there is an even or an odd number of real roots between a and b according as the signs of J(a) and f(b) are the 8ame or opposite. For the sign of f(b)/f(a) is the same as the sign of b -- rl b - r2 ... b --rt Q= . a - rl a - r2 a - rt

If r, > b > a, then b - ri and a - r, are both negative so that their quotient is positive. If b > a > rj, then b - ri and a - ri are both positive and their quotient is positive. But if b > rk > a, the quotient (b - rk)/(a - rk) is negative. Thus the sign of Q is (_1)8 where 8 is the number of real roots rk between a and b. It follows that 8 is even or odd according as the sign of f(b) and f( a) are the same or opposite.

168

REAL· ROOTS OF REAL EQUATIONS

[CHAP. VB

The length 01 the interval between a and b is Ib - al. Evel'Y real root r of I(z) lies in some interval a < r < b such that b - a :i! 1, and such tha~ no root of I(z) distinct from r is in this same interval. When we find such an interval, we shall say that we have i80lated r. H m is any integer such that I(m) and I(m 1) have opposite signs, there is at least one real root of I(z) in the interval m < z < m + 1. Thus a procedure for isolating all real roots of I(z) is that of a computation of 1(0), 1(1), I( -1), 1(2), I( -2), etc. It is effective if no interval m
+

+

IUustrative Examples

I. Isolate the real roots of J(a;)

I!I

a;4 - 23a;B

+ 3()a; + 34.

. Solution We oomputeJ(O) - 34,J(I) = 42,J(2) .. 18,J(3) = -22,J(4) = 42, J(-I) = -18, . . . , J(-5) ... :-66, J( -6) = 322. Hence J(a;) has foliT real roots which are isolated by the inequalities

-6 < rl < -5, -1 < rl < 0, 2 < r8 < 3, 3 < r4 < 4. II. Isolate the real roots of J(a;) ... Sa;8 .... Sa; - 1 .. O. Remark: SinceJ(a;) a Sa;(a;B - 1) - 1, we see tha.tJ(O) = J( -1) - -1 and J(n) < 0 for ev~ negative integer n. Yet J(a;) .. 0 has two negative rootsl .

So1:ution Multiply the roots by 3 and 80 obtain Sa;8 - 27a; - 27 = 3g(a;) whereg(a;) .. a;8 - 9:1; - 9. Theng(O) = -9, g(l) - -17, g(2) = -19, g(3) .. -9, g(4) - 19, g( -1) ... -1, g( -2) = 1, g( -3) = -9. Hence 3 <'1 < 4, -2 < •• < -1, -3 <" < -2 where " - Sr,. Thus the roots of J(a;) are isolated by the relations 1

< rl < t,

- f < rl < - i,.

-1

< ra < - f:

SEC.

71

169

DESCARTES' RULE OF SIGNS

EXERCISES

1. Isolate the single real root of the following equations: (a) :.;8

+ 33; -

+

1= 0 0 0

+ +2 = + :.; - 1 =

(b) :.;8 3:.; (0) 2:.;8 (tl) :.;8 - 3:.;1

+ 4:.; -

(e) :.;8 - 33; 2= 0 2:.;1 4= 0 (g) :.;8 :.;1 4 =0 (1&) :.;8 - :.;1 - :.; - 2 .. 0

(I) :.;8

=0

1

+ + + +

51. Isolate the three real roots of the following equations: (a) :.;8 - 4:.; - 1 == 0 (b) :.;8 - In 2=0 (0) 2:.;8 - 33; - 1 - 0 (tl) 2:.;8 - 4:.; 1 = 0

(I) :.;8 - 7:.;

+

+

(e) :.;8 - 3:.;1 - 2:.;

+' 5 = 0

+7

(g) :.;8 - 33;1 (1&) :.;8 - 33;1 (i) :.;1 33;1 (J) :.;8 - In -

+

:=

0

+

2:.; 2=0 133; 7=0 33; - 3 == 0 1 0

+

=

3. Isola.te the two real roots of the following: (a) :.;' - 4:.;

+2-

+

0

(b) :.;' - 33; 1 = 0, (0) :.;' - 7:.;1 - &; - 2 = 0 (tl) :.;' - &1 - 203; - 21 = 0 (e) :.;' - 7:.;1 - 26:.; - 40 = 0 (f) :.;' - 14:.;1 41n - 50 = 0

+

=

(g) :.;' - :.;1 - 4:.;1 - 3:.; - 1 0 :.;' - 4:.;1 Inl - 2:.; - 2 ... 0 :.;' 33;1 - & - 6 == 0 :.;' :.;8 - &;1 - 10:1: - 12 - 0 :.;' 33;8 2:.;1 2:.; - 4 == 0 :.;' - 32:.; 40 = 0 .

(1&) (.l) (J) (k) (Z)

+

+ :.;' + + +

+

+

4. Isola.te the four real roots of the quartics: 2= 0

(a) :.;' - Ilnl - 12:.; (b) :.;' - 22:J:!I & (0) :.;' - 40:.;1 - 64:J: (d) :.;' - 11:.;1 - 6:J: + (e) 4:.;' -

+ +8 = 0 + 80 = 0 10 == 0 24:.;1 + & + 3 = 0

(f) 4:.;' - 28:1;1 + 12:.; + 3 = 0 (g) :.;' + 4:.;8 - :.;1 - & - 2 = 0 (1&):.;' - 4:.;1 - 4:.;1 + 12:.; + 3 = 0 (i):.;' - 4:.;' - 3:.;1 + & + 2 = 0 (J) 4:.;' 7" 32:.;1 + 24:.; - 3 = 0 .

7. Descartes' rule of signs (FULL COURSE). An estimate of the number P of positive roots of an equation f(x) == 0 may be made by counting the number V of variations of sign in the sequence of nonzero coefficients of f(x). Consecutive terms of like sign may always be grouped together and the result may be written as (5)

f(x)

= fo(x)

+ /lex) + . . . + /vex).

For example, if J(x)

a;

(3x 12

+ 4:1:10)

-

(2x9

+ 4:1:8 + x + (3x&) - (2x' + 6X2) + (11x), 6)

170

REA L ROO T S 0 F REA L E QUA T ION S -

-

[CHAP,

VII

the polynomials arefo == 3X 12 + 4x 10,h == -2x 9 - 4x 8 - x 6, 12 == 3x5, fa == _2X4 - 6x2, f4 == 11x, and V = 4. The factors x of I(x) and corresponding zero roots may be removed without altering the value of V, and we may assume that f(x) == aox" + ... + a" in which ao and an are both not zero. Then V is the number of times we change sign if we start with ao and pass through the coefficient sequence to an. Hence V is even if ao and an have the same sign and is odd otherwise. The number P is th~ number of roots between 0 and an upper bound U and 1(0) = an, feU) has the same sign as ao. By Theorem 8, P is even precisely when V is even. It follows that V - P is an even integer. Let us now introduce a notation bix'" for the term of highest degree in the polynomial Ii == fi(X) of formula (5), Ci:Jf'" for the term of least degree. The)).

bo = ao,

no = n,

Cy =

an,

my

=

O.

Note that in the special example above f 2 == 3x 5 and so b2 = C2, m2 = n2. We let r be any positive number and form the product (x - r)f(x) == (x - r)/o + (x - r)f2 + ... + (x - r)fy. The term of highest degree in (x - r)fi is biX"~1 and the term of least degree is -rCix'fIU. These terms have opposite signs. The leading coefficient of (x - r)f(x) is the leading coefficient ao of f(x) , and so we start the count of variations of sign of (x - r)f(x) with the same real number ao. The term of least degree in every (x - r)/" for i = 0, 1, .•• , V-I, has the same sign as the term of highest degree in (x - r)/i+i and a sign opposite to that of the term of highest degree in (x - r)/i. It follows that the number of variations in the sequence of coefficients beginning with arr and passing on to the term of highest degree in (x - r)fy is at least V. The constant term -a"r has sign opposite to the leading term of (x - r)ly, and so the number of variations in sign for the coefficient sequence of (x - r)f(x) is at least V 1. .

+

SEC.

8]

171

STURM'S THEOREM

Let us now suppose that f(x) is a polynomial with P positive ro~ts rl, r2, . . . ,rp. Then f(x)

-=

(x - rp) ••• (x - r2)(x - rl)g(x).

The result just proved implies that, if Vo is the number of variations for g(x), then (x - rl)g(x) has at least Vo + 1 variations, (x - r2)[(x - rl)g(x)] at least Vo + 2 variations, . . . ,f(x) at least Yo + Pvariations. Then V ~ P. We have derived the following: " DESCARTES' RULE OF SIGNS. Let V be the number of variations of sign in the sequence of nonzero coefficients of f(x) and P be the number of positive roots of f(x). Then

V-P is a nonnegative even integer.

This rule states that there can never be more positive roots than variations of sign. It gives a precise result only when V = 1 and so P = 1, or when V = 0 and so P = O. It may be applied to f( -x) to provide an estimate of the number of negative roots of f(x). ORAL EXERCISES

Use Descartes' rule to estimate the number of positive and negative real roots of the following polynomials: (a) 3z1i - 2X4 - 43;2 - 1 (c) 2x7 (b) 2:&7 4x 5 - 3x 8 - 2x (d) 6x7 (e) 5x 6 6x 5 - 43;8 - 3x 2 2x - 1 (f) x7 3x 5 - 43;2 - 2x 1 (g) x 10 9x8 6x 4 - 3x 8 - 2x - 1 (h) x10 - 9x9 8x7 - 7x 8 8x 4 1 (i) X4 x8 X2 - 2x 1 (3) x 8 x 5 - X4 - x 8 X2 x-I

+ + + + + + + + +

+ + + + + + +

+ 3x4 + 5x + 6 2

-

5x 4 - 1

8. Sturm's theorem (FULL COURSE). Let us designate any given polynomial f(x) by fo and its derivative f'(x) by Ix. Apply the division algorithm to write fo

-= qJl -

f2'

Note that we have designated the remainder polynomial by -f2' "The minus sign is important.

172

REAL ROOTS OF REAL EQUATIONS

[C.ILU'. VII

Divide h bY!2 to obtain!l E5 q2/; - !a. We have again defined a polynomial fa E5 !3(X) to be the n~gative of a remainder polynomial. We continue these divisions, the .general one being

! i-2 == qi-x!i-l

-

Ii,

and the process terminates when we arrive at !i-2 5! qt-x!t-l - It, !t-l E5 qJt. Actually this is only a minor modification of the g.c.d. process of Chap. IV, and!t E5 !t(x) is a constant multiple of the g.c.d. of!(x) and!'(x). The polynomials in sequence, to == !(x),

It

E5

!'(x),

12 == /2(x),

,

!t == !t(x), are called Sturm's!unctions. The usual device of multiplying a function !i(X) at any step by a positive number (to avoid fractions) will yield a sequence of functions differing from those just defined by positive numerical factors. These modified Sturm's functions may also be used. If a is any real number, we may compute the sequence !o(a), h(a), •.• , !t(a) of real numbers and count the number of variations of sign in the nonzero members of this sequence. Let us designate this number by V(a). We then have the following: Sturm's theorem. Let a and b be real numbers such that !(a) ~ 0, !(b) ~ 0, a'< b. Then V(a) - V(b) is the number o! distinct real roots between a and b. This theorem states that if a is less than b the integer V(a) is greater than or equal to V(b). When!(x) has multiple roots, our count ignores the multiplicity. The information given by the fact that !t(x) is not a constant is of more importance for polynomials of low degree than is Sturm's theorem, and we can always divide !(x) bY!t(x) and obtain a polynomial of lower degree which may then be studied by Sturm's theorem. Hence we shall restrict ·exercises to the case where !(x) has no multiple roots and so !t(x) = ± 1.

SEC.

8]

173

STURM'S THEOREM

We shall not attempt to give any proof of Sturm's theorem in this text. nlustrative Example Isolate the real roots of Z4 - 24z· + 16:1; + 12 Sturm's theorem.

=0

by

~e

use of

Solution 1(z) = 4z8 - 48z + 16 = 4(Z8 - 12:1: -'- 4). /1 arrange the computation as follows:

= Zl -

12:1: + 4. We

2:1: -1 z z+l 2z -1 Zl- z - l Z8 - 12:1: + 4 Z4 - 24z 1 + 16:1: + 12 Z, - 12z1 + 4z 4z1 -4z-4 ZI_Z2- z 4z1- 2:1: Z2 - 11z + 4 - 12:1:1 + 12:1: + 12 Z2- z - l -2:1:- 4 -10z 5 - 2:1:+ 1

+

-5 The Sturm's functions are/o

= z, -

24:1:1 + 16z + 12, /1

= Z8 -

12:1: + 4,

Is = Zl - Z - 1, /8 = 2z - 1, /4 = 1. We use Theorem 7 to compute the signs at an upper and a lower bound and have the signs + - + - + at Land +++++ at U. Hence the polynomial has four real roots. The signs at z = 0 are + - - + and V(O) = 2. So there are two positive and two negative roots. We complete the work and list the results in a table.

+

z L /0+

II -

0 + +

I~+

la V

!c +

4

+ 2

1 U + + + + + + + + 0 '. 2

2

4

-1

5 +

-5

-6 +

+ + + 3

Note that /(z) = 0 has an odd number of real roots between 1 and 2 and so we do not compute /1, Is, /., /, at z = 2 but know there is only one real root in the interval 1 < z < 2. Also /(z)

= ZI(ZI -

24) + 16z + 12

for z - 3, 4, and trivially positive for z = 5,/(z) -3, -4 and/( -5) = 25 - 60 + 12 < 0, /( -6) = 36 '12 - 16' 6 + 12 Hence 1

<'1 < 2,4 <'2 < 5,

-1

<'. < 0,

<0 < 0 for z =

> O.

-6

<" <

-5.

-1, -2.

174

~REAL ROOTS OF REAL EQUATIONS

[OHAP.VII

EXERCISES

Isolate the real roots of the following equations by the use of Sturm's. theorem: (i) x + 3x 3x - 3 = 0 + 2 =~ 0 8x 20x - 21 = 0 W 2:1: 3 - 4x + 1 = 0 32x + 40 = 0 (k) x + 3x + 2 = 0 10x 8x + 5 = 0 (Z) x + ax - 1 = 0 4x 6x 2 + 8x + 2 = 0 (m) x + 3x + 4 = 0 (n) x + 3x 3x - 3 = 0 8x ·-16x + 12 = 0 4x2 - 8x - 4 = 0 \ (0) x 3 - 18x + 12 = 0 24:1: + 84x - 13 = 0 (p) x + 32: 1=0 .

(a) x' - 4x

(b) x' (c) x' (d) x' (6) x' (f) x' (g) x' (h) x' -

3

2 -

2 -

3

2 -

3

3 -

3

2

2

3

2 -

2

3

2 -

9. Homer's method. The irrational real roots of an equation f(x) = 0 may be approximated, to as many decimal places as is desired, by a number of methods. In all the methods the roots are first isolated. It is also customary to study any particular root r by replacing f(x) by (-I)"f( -x), if necessary, and so assume that r is positive. Let us suppose that we have found a real interval a < x < b in which one and only one real root r of f(x) lies. -Let us also suppose that r is a simple root and so f(a) and feb) have opposite signs. Finally assume that b - a ;i! 1, and so r lies in the interval a < r < a + 1. In the method of Horner we first diminish our roots by the number a and so obtain a polynomial, g(x) = box"

+ b1x"-1 + . . . + b",

which will have a root 8 = r - a between 0 and 1. Note that b" = f(a) and that a value x = e + a near r is less than the root r only if gee) has the same sign as b". "In the usual case, a is an integer which is the integral part of the root r. To compute the first digit after the decimal point we write g(x) == b_2X2

+ bn-1X + go(x),

where go(x) = box" + b1x"-1 + . . . + bn-aX 3 + b". If s is a small number, the value of goes) will lie between the two

SEC.

HORNER'S METHOD

9)

175

values go(O) = bn and go(1) = bo + bl

Then

8

+ ... + bn-3 + bn = bn'.

will usually lie between a root 81 of the quadratic bn-2x2 + bn-IX + bn = 0 and a root 82 of

eq~ation

bn-2Z2 + bn-IX

+ bn' = O.

We solve these equations and so determine an integer d = 0, 1, 2, . . . , 9 such that d/1O is the :first digit of 8. We next diminish the roots of g(x) by d/1O to obtain an equation h(x)

==

coXn

+ . . . + Cn-IX + Cn'

The sign of Cn will be the same as that of bn [but not necessarily that of the unrelated constant term an of our original f(x)] if a + d/1O ~ r. Thus an overestimate of d will be detected at this point. An underestimate will result in an approximate root t > of h(x) = O. If d has been correctly determined, the equation h(x) = 0 will have a root t between 0 and itr. The powers t2, t3, ... , tn are small real numbers and t will normally lie between two numbers

-nr

h=

Cn -, Cn-l

where

Moreover the difference It2 - til will be small and we shall be able to determine a digit 6 such that

r = a

+ d/1O + 6/100

to two decimal places. The process of diminishing roots is continued and it will be true in all cases that, after several steps, our process of squeezing the root between the roots of two linear equa-

176

REAL ROOTS OF REAL EQU .A.TIQNS

[CHAP. VII

tions will become valid and yield several digits instead of one. We illustrate this fact below. nlustrative Examples .1. Compute the positive real root of :1:3 transformations.

-

63; - 1 = 0, making three

Solution The equation has one positive root and no rational root. Also /(2) = "':'5,/(3) = 8 so that 2 < r < 3. 1 + 0 - 6 - 11~ 2+4-4 2-2-5 2+8 4+6 2

~ + 63;2 + 6:1: - 5 682 +68-5=0 -6 + "';f":36""'"+""'--;;-'12=0 -6 + VI56 8 = 12 = 12 12 < VI56 < 13, O.d = 0.5

g(:I:) =

1+6 1 + 6.0 + 6.00 - 5.000~ 0.5 + 3.25 + 4.625 6.5 + 9.25 -0.375 11,(:1:) = :1:8 + 7.5:1: 2 + 12.75:1: - 0.375 0.5 + 3.50 0.375 002 . I t = 12.75 = • 9 approXlmate y 7.0 + 12.75 0.5 . 1 + 7.5 1 + 7.50 + 12.7500 - '0.37500010.02 0.02 + 0.1504 + 0.258008 7.52 + 12.9004 - 0.116992 0.02 + 0.1508 7.54 + 13.0512 0.02 1 + 7.56 k(:I:) =;0 :1:3 + 7.563;2 + 13.0512:1: - 0.116992 0.116992 'Ul = 13.0512 = 0.0089+ _ 0.116992 - (0.01)8 - 7.56(0.01)2 OOOB 'U2 13.0512 = . 9+ The correct value 'U definitely lies between 'Ul and 'U2 in this case and so . II. Find the negative root of :1:8 - 3:1:2 - 4:Il + 11 =- O.

r =,2.5289, approximately.

Solution -/( -:I:) == :1:8 + 3:1: 2 - 4:Il - 11 == F(:I:) and F(l) Hence 1 < -r < 2. - . -.-

= -11, F(2) = 1.

SEC.

1 + 3 - 4 - 111!. 1 +4+ 0 4+0-11 1+5 5+5 1

g(z)

= Z8 + 6z1 +

1

" =

+ 6.0 + 5.00 0.9 + 6.21 + 6.9 + 11.21 0.9 + 7.02 7.8 + 18.23

5z - 11

-5 + 12

~5 + V25 + 240

= 11+, d = 9

+ V25 + 264 = 12

11.000/0.9 10.089 0.911

v28l) = 1

12

= -5

8

1+6 1

177

HORNER'S :METHOD

9]

12

h(z) = Zl + 8.7z' + 18.23z - 0.911 t _ 0.911 - 0 049+ - 18.23 - • e= 4

0.9 1 +8.7 1 + 8.70 + 18.2300 - 0.91100010.04 0.04 + 0.3496 - 0.743184 8.74 + 18.5796 - 0.167816 k(z) = Zl + 8.82z1 + 18.9308z - 0.167816 0.04 + 0.3512 8.78 + 18.9308 0.167816 00088 1£1 = 18.9308 = . 0.04 1 8.82 0.166933 _ 0 0088 1£, == 18.9308 . -r = 1.9488, r = - (1.9488) a.pproximately

+

EXERCISES

1. Use Horner's method to find all the real roots of the following equations. Make at most three transformations and obtain/our decimal places in each case.

(a) Zl .. 2 (b) Zl = 3 (c) Zl = 7 (d) Z8 - 5z - 1 ... 0 (e) Z8 + Zl - 2z - 1 (J) Zl + 3z' - 1 == 0 (g) Z8 + 3z - 1 = 0

(h) Z8 - 2z - 6 = 0

=

(i) 2z8 - 6z1 - 1 = 0 (J) Z8 + 9z - 6 ... 0 (1:) Z8 + 3z + 6 0 0 (l) Z8 + 9z + 2 = 0 (m) Z8 - Zl - 3 = 0 (tt) Z8 - Zl - 5 = 0

=

Am. 2.1799 Am. 3.0536 Am. 0.6378 Am. -1.2879. Am. -0.2208. Am. 1.8637 Am. 2.1154

Z. Use Horner's method to find all the real roots of the following equations, making two transformations: (a) Z4 - 3z + 1 (b) z, - 4z 2

=0

+ =0

(c) z' + Z8 - 2 = 0 (d) z, + 2z1 - Z + 1

=0

(e) Z4 (J) z, (g) Z4 (h) z, -

+

Z8 - 3z - 1 6z - 1 == 0 Zl - 1 = 0 7z' - 2 .. 0

=0

10. Other methods (FULL COURSE). The derivative of I'(x) is a polyp.omial f"(x). Let us suppose that we have isolated a simple real root r of f(x) in an interval a ~ x ~ b and that the sign of f"(x) is the same throughout the interval. Then one of- f(a)- and- feb) will have the same sign as f"(x) in this interval. Let us call this number g. It can be proved that f(g) c = g - f'(g) is closer to r than is g. Mter we compute c, we make a second such computation and determine a number f(c) Cl = C - f'(c)' which will be closer to r than is c. The method may be continued as far as desired. This procedure for approximating r is known as Newton' 8 method.

FIG. 9.

Another method for computing r is called the method of interpolation. The theory of similar triangles ~nd Fig. 9 may be used to show that d - af(b) - bf(a). - feb) - f(a)

Moreover in the case described for Newton's method dis closer to r than is either a or b and, indeed, if we compute c as above it can be shown that r is between c and d. The method of interpolation involves the least theory of our three methods for computing roots, but the computations themselvEls are not simple. We recommend that no' exer-

BEC.10]

179

OTHER METHODS

cises on this and Newton's method be assigned. If exercises are desired, those of Sec. 8 may be used. mustrative Examples I. Use Newton's method to approximate the positive root of :1:8 - 63; - 1 = O. Solutiun The root lies between 2 and 3. AlsoJ'(:I:) = 3:1:1 - 6,

>0 = 8; so g = 3. Then

/"(:1:) = 6:1: for 2 ;:ii

:I:

;:ii 3. C

=

But/(2) = -5,/(3) 1(3) 8 . 3 - 1'(3) = 3 - 21 = 2.7, approxuna.tely.

Also Cl

= 2.7 -

1(2.7) 2 7 2.483 2 55 • tel 1'(2.7) == • - 15.87 = . , appro:nm.a y.

Solution 2 In this solution we shall diminish the roots in each case but use the Newton method of approximation.

+0 -

6 - 11~ g(:I:) = :1:8 + 9:1:1 + 213: + 8 ' 3 + 9+9 g'(:I:) = 3:1:1 18:1: 21 3+ 3+8 g(O) -8 3 + 18 C = 0 - 0'(0) = 2f = -0.3 6+21 3 1+9 9.0 + 21.00 + 8.0001-0.3 k(:I:) = :1:3 + 8.1z1 + 15.873: + 2.483! 0.3 - 2.61 - 5.517 k(O) -2.483 8.7 + 18.39 + 2.483 - k'(O) = """I5.87 == -0.15 0.3 - 2.52 8.4 + 15.87 0.3 8.1 Am. r = 3 - 0.3 - 0.15 = 2.55.· 1

+

1+ 1+

:1: 3

+

II. Use the. method of interpolation to find the positive root of - 63; - 1 = O.

Solution /(2)

= -5,/(3) =

8, and so

d _ 3/(2) - ~(3) _ -15 - 16 = 31 = 237 - 1(2) - 1(3) -13 13 .

180

REAL ROOTS OF REAL EQUATIONS

[CHAP.VII

In the next step we must compute 3/(2.37) - 2.37/(3) 1(2.37) - 1(3) . Such a oomputation may be conveniently made only with a computing machine. If instead we diminish the roots by 2 to obtain g(z) =

then g(I) = 8,

a=

Z8

+ 6z + 6z 2

5,

0, b = 1, and

d

= g(O)

g(O) _ g(I)

=

-5

-13

= 0.37

+

is neceBBarily the same value that we obtained above. The value 0.3 and even the value 0.4 are not so good as the 0.5 obtained by Horner's m~thod and so we shall not continue our solution directly. If we diminish the roots by 0.5 instead, to obtain h(z) =

Z8

+ 7.5z + 12.75:1: 2

C).375,

the next approximation according to the method of interpolation is

O.lh(O) 0.0375 _ 0 027 h(O) - h(I) -= 1.351 - . , which is, in this example, also Dot so good as the value obtained at the corresponding stage in Homer's method. '

11. The function concept. We shall close the present chapter with a discussion of the mathematical notation called the function concept. The germ of this idea has already been used here in our process of isolating the roots of an equation f(x) = ,0 by the q.se of a table consisting of values x == a and corresponding values f(a). We define a variable to be a symbol together with a set of objects which we call its range. If the range consists of a single object, we call the symbol a constant. The function concept involves two variables and a correspondence. We make the following: Definition. Let x and y, be two variable8 which are 80 related that to every element of the x range there corre8ponds one or more elements of the y range. Then y is called a function of x.

SliIC.

12]

OPERATIONS ON FUNCTIONS

lSl

The part played by the variable y in this relationship is different from that played by x. It is customary then to refer to x as the independent variable and to y as the depend-

ent variable. If it is true that to each element of the x range there corresponds exactly one element of the y range, we say that y is a Bingle-valued Junction of x. In all other cases we call y a multiple-valued or many-valued function of x. Most of the functions which the student will study :in plane a.naJ.ytic geometry are many valued, and some of the functions of plane trigonometry are infinitely many valued. We shall indicate the fact that y is a function of x by writing y = J(x). Of course we use other letters than J or, for that matter, than x and y. If we replace x in a single-valued function 'U = J(x) by an element a from the x range, the corresponding element of the y range is designated by J(a) and is called the value oj J(x) Jor x ....·a. This ~ the source of the language we have already used for the values J(a) of polynomia~ J(x) at x = a. Our definition of a function may be extended to functions s = J(x, y) of two independent variables x and 'U or, more generally, to functions of n variables. We let Xl, X2, ••.., x" be n independent variables and y be a variable: Suppose that every sequence aI, aI, • • • , an of elements al of the Xl range, al of the XI range, . • • , an of the x. range corresponds to one or more elements J(al, ••• , an) of the y range. Then we call the dependent variable y a Junction oj Xl, X2, • • • , x. and call y a Bingle-valued Junction oj Xl, X2, •.• , X" if every J(al, a2, ••. , an) is unique. The concept whose description we have given is of vital importance in all mathematics. We recommend that the student and the instructor construct illustrations of this concept from life as well as from the mathematics of this text. 12. Operations. on functions. The symbols n, a, b, c, i, j, . . . of Chap. I were varia.bles whose common range

182

REAL ROOTS O~ REAL EQUATIONS

[ClU.P. VD

was the set of all natural numbers. All the variables of Cha.p. II had the set of all integers as their common range and the ranges in Chap. III were other number systems. The range of a variable need not be a number system at ail. However most of the variables emphasized in elementary mathematics have as their ranges subsets of the complex number system. We shall restrict our attention to such variables. H'Y - f(:I:.) and s = g(z) are two functions of the same variable z, their sum 'Y + S' = h(z) = f(z) + g(z) is that function whose values h(a) are defined by h,(a) = f(a)

,

+ g(a)

'

for every a of the z range.. The difierence f(z) - g(z) and the product f(z)g(z) are functions defined similarly. The' quotient J(z) g(z)

~ a funcUon'defined only for those numbers a of the ~ range for which g(a) is not zero. We observe now that every polynomial J(z) defines a func,tion 'Y = f(z) where the z range and the 'Y range are subsets of the complex number system. H g(z) is a polynomial and s = g(z) is the corresponding function, the fQrmal sum polynomial J(z) + g(z) defined in Chap. IV defines the sum function 'Y + s. Similarly 'Y - s and 'YS are defined, respectively,. by the difierence and product polynomials. The function defined by' their formal quotient is the quotient function mentioned above, and the term roJ,ional function, which we have already used for the formal quotient, has its origin in the function concept. , The zero function is the function all of whose values are .zero. H f(z) and g(z) are two functions having the same z and 'Y ranges, we say that J(z) and g(z) are identical and ~te ...... 1' ~.

,"

J(z)

!E!

g(z)

8EC.

13J

THE EQUATIONS OF A CURVE

183

(read "f of x is identically equal to g of x") if it is true that f(a) and g(a) are the same elements of the y range for every x of the x range. Then f(x) ;;e g(x) if and only if f(x) - g(x) is the zero function, that is, f(x) - g(x) O. This is the inspiration of our definition of polynomial equality, the reason why we have regarded as equal all polynomials with coefficients all zero, ~nd the motivation of our use of the notationf(x) g(x) for equal pol~omials. The functions defined by polynomials and rational functions are functions defined by an algebraic formula. We shall have other examples in our next section. In Chap. VIII we shall define the functions of trigonometry by a geometric procedure involving angular measurement and the use of a coordinate system. 13. The equations of a curve. If the ranges of an independent variable x and a dependent variable y = f(x) are subsets of the real number system, we call y a real function of a real variable x. Then there is a corresponding graph of the function which involves the use of a coordinate system in the plane and is the set of all points P with coordinates x = a and y == b such that b == f(a). The graph might be a line, a circle, or some other geometric object which belongs to the family of what the mathematician calls curves. Let a curve C be defined geometrically. For example, a circle is the curve of all points at a :fixed distance r from a • :fixed point called its center. Let us also suppose that we have a polynomial F(x, y) with real coefficients. Then we shall say that the equation

=

=

F(x, y)

== 0

is an equation of the curve C if 1. Every point on C has coordinates x == a, y == b, such that F(a, b) == O. . 2. Every real number pair (a, b) BUch that F(a, b) == 0 defines a point with coordinates x = a, y == b on C.

184

RlilAL ROOTS OF REAL "EQUATIO-XS

[CHAP.VII

Plane analytic geometry is concerned with the study of curves whose equations are polynomial equations F(z, '1/) = 0,

and we shall return briefly to this study in the final two sections of Chap. VIII. " An equation F(z, '1/) = 0 of a curv:e 0 defines a function '1/- =" f(z) whose graph is O. This function is defined by the correspondence in which each real number z = a corresponds to the real roots '1/ ~ b of the equation -

F(a, '1/) = 0

and is called an algebraic Junction of z. When the degree in '1/ of F(z, '1/) is greater than 1, an equation F(a, '1/) may have several real roots and then '1/ will be a multiple-valued function of z. For example, the function defined by :x;I + '1/" = 1 is a real function of a real variable z whose range is the interval -1 ~ z ~ 1. Each value z = a such that -1 < a < 1 corresponds to the two values v1 - a", - v1 - a l of '1/, and hence to two points on the circle.

CHAPTER

vm

VECTORS IN THE PLANE

(FULL COURSE)

1. Vectors. In the present chapter we shall tie up the algebraic theory of vectors and linear equations with trigonometry and analytic geometry. Our presentation should' provide a strong foundation for these related subjects as well as a condensation of material that is common to all of them. We shall regard this entire chapter as optional material. A vector in the plane is a real number pair (z, 'II). We may interpret it geometrically either as the point P in the

,

~----------=---~~~----~~----~---.

FIG. 10.

plane whose coordinates relative to a fixed rectangular Cartesian coordinate system are z and y, or as the directed line segment OP from the origin 0 to P. We may also interpret it algebraically as the complex number z + yi. We shall usually write P = (z, y) here and shall use the words vector and point interchangeably. H t is any real number, the scalar product t(z, y) of the number t and the vector (z, y) is the vector (tz, ty). We construct Fig. 10 in which OHP, OH 1P 1, OH,}'" are evidently similar right triangles. Then, if P is not the origin, and P = (z, y), PI = (Z1, Y1), P s - (zs, Ys), we use similar 185

186

VECTORS IN THE PLA.NE

(CIUP. VIII

triangle properties to obtain Zl Z

= YIY = tl ~ 0 ,

.Z. z

= y.Y = #~

~ O.

"l_

It follows that if (Z, y) ~ (0, 0) every point on the same line through 0 as (Z, y) is a vector t(z, y). Also t ~ 0 or ;:iii 0 according as t(z, y) is or is not on the same ray from 0 as (z, y). Conversely, every vector t(z, y) defines a point on the same line through 0 as (z, y). For if PI = (tz, ty) the legs of the riglit trl&ngJ.e OPIHI and OPH are proportional, the right tri.angles are similar, the points 0, P, and PI are collinear. The norm of a vector (z, y) is the nonnegative real number z· 'V.. The length of (z, y) is

+

r =

V z'l. + y.

ie;

O.

We shall call a vector of length 1 a unit vector. All unit vectors are points on a circle with center at the origin and radius 1. If P is any point except 0, there are precisely two unit vectors on the line through 0 and P. They are ."

Uo

-z, -y) = (r r ,.

and P == rU == -rUo. Thus if we select any unit vector on a line through 0 and U, all vectors P on the line have the form tU where t is a real number. Then tU is on the same ray from 0 through U if and only if t !l: 0, and the nonnegative real number is the length of P. ORAL EXERCISES

1. Give the norms and the lengths of each of the vectors (0, 0), (2, 0), (0, -2), (3,4), (1, 1), (5, -12), (6,8), (3, -3), (-6, -8), (-10, -24). I. Give a unit vector on the same ray as each nonzero vector of Oral Exercise 1. 8. Give a unit vector on the same line but not on the. same ~y as each nonzero vector of Oral Exercise 1.

187

ADDITION OF VECTORS

BEo.21

4. Express each nonzero vector of Oral Exercise 1 in the form r(u, fJ) where r > 0 and (u, fJ) is a. unit vector. 6. Which vectors in Oral Exercise 1 are collinear and which are on the same ra.y?

2. Addition of vectors. The

BUm

== (Xl, Y1) + (X", y,,) of two vectors P 1(X1, Y1) and P" == (x", y,,) is defined to be PI

+ P"

the vector (Xl + X'" Y1 + y,,). In Fig. 11 PI and P" are two given vectors. We complete a parallelogram by drawing a line through PI parallel to OP" and a line through P" 11

--------~~--~~~--~-----L-------M

FIG. 11.

parallel to OP 1• These lines intersect in a point Q, and we show below that Q == (Xl + X., Y1 + y,,) == PI + P". This graphic method of constructing the suni of two vectors is called the parallelogram law for the addition of vectors. In the case of the di.agram. we see that the right triangles OP1H 1 and P .QR are equal and so P~ ==

H.Ha == X'" OHa == OH" + 1lJfa == Xl + X'" HaQ == HaR + RQ == H,p" + RQ == Y.

+ Y1.

A complete derivation merely requires the construction of additional diagrams and will be omitted. The vector (0,0) has the property that (0,0)

+ (x, y)

== (x, y)

+ (0, 0)

== (x, y).

188

[CHAP. vm

VECTORS IN THE PLANE

We call (0,0) the zero vector. The vector (-z, -Y) - -1(z, y) has the property that (z, y) + (-z, -y) = (0,0)

.

and so is called the negative, of (z, y) and is designated by - (z, y). It is a vector of the same length and on the same line through 0 as (z, y). It is not on the same ray. The difference P'J - P l of two vecto~ is given by (z'J, Y'l.) - (Zl, Yl) -, (z'J - Zl, YI -1/1). In Fig. 11 08 is parallel to P lP I and the length of 08 is the length P lPI. Then PI = P l 8 so that 8 == P'I. - Pl. It follows that the undirected distance d between P l and PI is given by the diBtance formula

+

(1)

d =

V (ZI -

Zl)'J

+ (YI -

Yl)'J.

ORAL BDRCISBS 1. Compute the coordinates of P 1 PI a.nd PI - P1 for each of the following pairs of points PlJ p.:

+

(d) (1, 2), (-4, -10) (e) (0, -1), (5,11) (f) (3, -2), (5, -1)

(a) (2, -1), (5, 3) (b) (-1,0), (1,1) (c) (-1, -2), (2, 2)

I. Compute the length of .

P,p.

for each line segment

P,p. in Oral

Exercise 1.

3. Angular measurement. An angle (J may be defined to be a rotation in the plane of a ray about its end point. The ray then starts its rotation as what we shall call the initial Bide of (J and ends its rotation as what we shall call the terminal Bide of (J. Angles may be compared by selecting the same ray as the initial side for all angles. Let us call this ray the polar azUl and its end point the pole. In Fig. 12 we have made a counterclockwi8e rotation of revolutions (complete rotations). We shall always measure counterclockwise rotations by positive real numbers and clockwise rotations by negative real numbers. The zero angle is that in which no rotation takes place. It should be clear now that,

In

smc • .3]

ANGULAR MEASUREMENT

189

if we use the revolution as a unit of measure, we shall have set up a one-to-one correspondence between the real num, ber system and the set of all angles. The use of the revolution as a unit of measurement is inconvenient and it is customary to divide it into 360 equal parts called degree8. Thus 3600 == 1 revolution, and the angle ill Fig. 12 is 3900 • Our definition of angular measurement, using the revolution as a unit of measure, has used the property of plane geometry which states that an angle at t'M center of a circle iB meaBUrea by the arc it, BUbtendB on t'M circumference.

~m FIG. 12.

We may use this property a bit more explicitly in introducing a new unit of measurement. We define an angle of one radian to be that angle which BUbtendB an arc wh08e length iB one unit on the circumference of a circle who8e radiU8 i8 one unit. Since the circumference of this circle is 2r units of length we have the fundamental relations 1 r6fJolution == 27r radiam == 3600 between our units. We shall use the radian as our standard unit in the next sections. The angles in' the accompanying table occur fmquently in the exercises of trigonometry. RevolutioDS Degrees Radians 0'

~

t I i i t

1

0

0

30·

.../6 .../4 .../3

45 60 90

180

'IT/2 'IT

270

3w-/2

360

2r

190

VECTORS IN THE J»LANE

[CIIAP.vm

BDRCISB

Make a table like the accompanying one for the angles obtained by adding 11'/2, 11', 3r/2, -11'/2 to each of the first four angles of that table.

,. Polar coordinates and trigonometric functions. The representation of a vector as a directed line segment of a given length suggests the introduction of a new type of coordinate system in the plane. We select a unit of measurement as. well as a pole 0 and a p~lar axis as defined in Sec. 3. Then every real number pair [r,8] such that r ~ 0 will define a ray which is the terminal side of an angle 11

W FIG. IS.

of 8 radians whose initial side is the polar axis. There is precisely one point P on this ray at a distance of r units from.J) and we call [r, 8] a set of polar coordinate8 oj P. Conversely, let the point P be given. Then the distance r = OP is unique and we may select 8 to be any angle whose terminal side is the ray from 0 through P. If r = 0, the angle 8 is arbitrary. Otherwise any two determinationB of 8 diifer by an integral multiple 0/ 21r radianB. Polar coordinates may be related to rectangular coordinates by rrw.king the polar a:r:i8 coincide with the poBif:ive ~ a:ti8 8UCh that the pole i8 the origin. The connection between coordinates in the two systems then involves the introduction of what are called trigonometric Juncti0n8.

SEC.

4]

191

POLAR COORDINATES

In Fig. 13 we suppose :first that only 6 is given. Then the terminal side of 6 determines a unique unit vector U on it. The x coordinate u of U is called the coBine of 6, its y coordinate v is called the Bine of 6 and we write u == COB 6, v == Bin 6. Now cos 6 and sin 6 are uniquely determined real numbers, for every real number 6, and we have defined two trigonometric functions. They are real functions of a real variable 6 defined for all real values of 6. We observe that since U is a unit vector we have the relation sinl 6 + cosl 6 == 1, and thus that -1 ~ sin 6 ~ 1, -1 ~ cos 6 ~ 1. Convel'Bely, if g is any real number such that -1 ~ g ~ 1, the vectors (g, h) and (h, g) defined for h == gl are unit vectors and g is the sine of an angle, and the cosine of another angle. We now pass. on to our connection between rectangular and polar coordinates. If P is a point whose polar coordinates are [r,6], its rectangular coordinates (x, y) have the property tl].at r == vxl + yl and (x, y) == r(u, v) so that x == fU, Y == ro. This is the same as x == r cos 6, y == r sin 6. Conversely, if x and y are given and (x, y) ~ (0, 0), the numbers r and 6 are determined by the relations

vI -

x r

• 6 ==-, Y sm

cos 6 == -,

(2)

r

The ratios (3)

tan 6 .

sec 6

sin 6 y = - - =-, cos 6

x

1

r

=cos-6 =-, x

cos 6 cot 6 ==sin6 1

x

=1/

esc 6 = sin 6 ==

r

1i

192

VECTORS IN THE PLANE

[CRAP. VIII

are called the tangent, the secant, the cotangent," and the cosecant of 8, respectively. Note that the first p~ of these functions are defined for all real numbers 8 eXcept odd integral multiples of 1r/2. For at the latter values cos 8 = 0, sin 8 = ± 1. The second pair are defined for all real numbers 8 except integral multiples of 1r since at the latter values sin 8 = 0, cos 8 = ± 1. Two angles are said to be coterminal if they have the same initial and terminal sides. Then they differ by integral multiple~ of 21r radians. It should be evident that such angles have the same trigonometric functions, i.e., f(8 + 2n1r) f(8) for all integers n, all real numbers 8, andf(8) interpreted as sin 8, cos 8, tan 8, cot 8, sec 8, or csc 8. We shall close our discussion of trigonometric functions with a further reference to Fig. 13. We let U = (u, v) be any unit vector so that U is on the terminal side of an angle 8, and let W = (uo, vo) be on the terminal side of cJ> = 8 + 1r/2. Drop a perpendicular from U to the z axis and let its foot be H. Similarly let the foot of the perpendicular from'W to the z axis be K. Then the right triangles" UOH and KOW have the same" l-ength 1 as hypotenuse and the sum of their angles at 0 is 1r/2. They must then be equal triangles and Uo = ± v, Vo = ± u. Since the terminal side of cJ> is in the quadrant following the terminal side of 8, we see that Uo and v have opposite signs, Vo and u have the same sign. Let the student draw three diagrams and verify this for all possible positions of the terminal side of 8. It follows that Uo = -v, Vo = u, and that

=

(4)

cos

(8 + ~) = - sin 8

sin

(8 +~) = cos 8

for all real numbers 8. EXERCISES

1. Use the properties of right triangles to construct the following table:

SEC.

51

193

ROTATION OF AXES

"

sin"

0

0

! 6

1

tan "

cot"

sec "

0

...

1

"2

a-

-va

"2

"2

1

1

V2

"2

1

-va a-

2

1

0

-va .

V2

T

i

-va

T

3 T

2

cos" 1

-va

-va

V2

"

0

csc "

-va

f

"

2

V2 f

.

-va 1

I. Extend the table above to angles 8 + 71'/2, 6 + 71', 8 + 37r/2, 8 - 71'/2 where 8 = 0,71'/2,71'/4,71'/3. 3. Determine the values of those trigonometric functions whose values are not explicitly given, if (a) (b) (c) (d) (6)

sin 8 = t, 0 < 8 < 71'/2 (J) sec 8 = ¥, 37r/2 < 8 < 2r sin 8 = t,7I'/2 < 8 < 71' (g) esc 8 = ti,7I'/2 < 8 < 71' tan 8 =- 1,71' < 8 < 3r/2 (h) sin 8 = i,O < 8 < 71'/2 cot 8 = - -h, 3r/2 < 8 < 21r (i) cos 8 = i,O < 8 < 71'/2 cot 8 = 1, sin 8 < 0 (3) sin 8 = i, cos 8 < 0

6. Rotation of axes. If we rotate the coordinate axes through an angle 8 as in Fig. 14, the. unit vector (1, 0) is rotated into a unit vector (u, v) where u = cos 8, v = sin 8. By formula (4) the unit vector (0, 1) is rotated into (-v, u). Let P = (x, y) be any point in the plane and regard the rotated axes as defining a new coordinate system in which OH = x' is the new ab8ci88a of P and OK = y' is the new ordinate of P. Then the (x, y) coordinates 'of H are x'u, x'v, and those of K are -y'v, y'u. But OHPK is a rectangle and the parallelogram law implies that P

= H + K = (x'u, x'v) + (-y'v,

y'u).

This yields the formulas (5)

{

X

y

= x'u - y/v, = x'v + y/U.

Formulas (5) are called the formulas for a rotation ofaxe8. They express the (x, y) coordinates of any point in the

194

[0lLU'. VIII

VECTORS IN THE PLANE

plane in terms of its. (:e', y') coordinates and the (:e, y) coordinates of the unit point on the :e' axis. Every point P = (:e, y) defines a complex number:e yi and formulas (5) are· equivalent to the complex number equation

+

(6)

x

+ yi =

(x'

+ y'i)(u + vi).

'I'

------~~~~--~~------~---.

FIa.14.

+

+

+

For formula (6) is :e 'Vi = :e'u - y'o :e'm y'ui. Since (u tn)(u - tn) = u· vi ,;,.. 1, formula (6) is also equiv. alent to

+

(7)

x'

+

+ y'i =

(x

+ yi)(u -

and hence to (8)

{

X' =

l'

xu

vi),

+ yv, + yu.

= -xv

These final formulas are called the 8oZtJed!qrm of the rotation formulas (5). ORAL BXEltCISBS

1. Give formulas (5) "and (8) with u replaced by cos (J and 11 by sin (J. I.' Give formulas (5) and (8) for (J = '11'/6, '11'/4, '11'/3, '11'/2. 8. The coordinates of a. vector on the positive :e' axis are (3,4). What are the equations (5)1

SEC.

6]

195

ADDITION FORMULAS

4. The coordinates of a vector on the positive y' axis are (3,4). What are the equations (5)1 6. Let the equations of a rotation be a;

3y' = &Il' -5 4y'' Y = 42;' + 5 •

Give the (a;', y') coordinates of P if its (a;, y) coordin8.tes are (a) (1, -1) (b) (0, 1)

(e) (-1,0) (d) (2,3)

(e) (3,4) (f) (-4, -3)

8. Let the number pairs given in Oral Exercise 5 be the (a;',1/) coordinates of P. Give the (a;, y) coordinates.

. 6. Addition formulas and inner products. Every angle 8 determines a rotation of axes and a unit vector (cos 8, sin 8) on the positive :c' axis. Let the initial side of a second angle q, be the positive :c' axis and ('Us, vs) be the unit vector on the terminal side of q,. Then the angle 8 + q, has the positive :c axis as initial side and the terminal side q,), of q, as terminal side. It follows that 'Us = cos (8 Vs = sin (8 q,), and that 'Us' = cos q" v,,' = sin q,. We use formulas (5) to obtain

+

+

(9)

{

COS (8 + q,) = cos 8 cos q, - sin 8 sin q, sin (8 + q,) = sin 8 cos q, + cos 8 sin q,.

These formulas are called the addition JfYI'm'Ula8 of trigonometry. The study of them and of their consequences makes up too large a part of trigonometry to be;' considered here. Any two nonzero vectors PI '7 (:C1' 'U1) and P s = (:Cs, 'Us) define angles 8 and q, as above, where the initial side of 8 is the positive :c axis, the terminal side of 8 and the initial side of q, coincide with the ray from 0 through P, and the terminal side of q, is the ray from 0 through P s. Then we shall call q, the angle between the vectors PI and P s. Now (:Cl, 'U1) = r1(cos 8, sin 8), (:Cs, 'Us) = rs('Us, vs) and cos q, = 'Us', We use formulas (8) with

VECTORS IN THE PLANE

[CHAP. VIlI

and. obtain (10)

XIX2 + YIY2 , 2 . VXi Yl 2 V X22 + Y2 2 • ,/;. YIX2 - XIYl SID¥,= • : VXl2 + Yl 2 VX22.+ Y2 2 COS

A.

~

=

+

T4e inner product of any two vectors (XII YI) and (X2, Y2) is : defined to be the real number XIX2 YIY2' Then

+

formula (10) states that the cosine of the angle between two nonzero vector8 i8 their inner product divided by the product of their lengths. If cos q, = 0 then q, = 7r' /2, or 37r'/2, and the ray through PI is perpendicular to that through P 2 • In this case we say that our vectors are orthogonal. Since rl > 0, r2 > 0, we s~ that two nonzero vectors (Xl, YI) and (X2' Y2) are orthog.onal if and only if their inner product XIX.

+ YIY2

o.

=

We observe finally that if (x, y) is any nonzero vector the vector (-y, x) is orthogonal to (x, y). Then the vectors

are orthogonal unit vectors which may be used, respectively, as the x' and y' unit points after a rotation of axes . such that (x, y) is on the positive x' axis. ORAL EXERCISES

1. Compute the inner products of the following pairs of vectors: (a) (1, 1), (2, -1) (b) (3, -1~, (1,3) (c) (-2, 1), (2, 1)

(d) (3, -4), (1, 2) (e) (-3,4), (7,12) (f) (1, ~ 1), (7, 12)

(g)

(va, -1), (1, -1)

(h)

(V2, - V2),

(i) (-1,

(1,0)

va), ,0, 1)

2. Give' the cosine of the angle between each pair of, vectors in Oral . Exercise 1. 3. Give a. unit vector orthogonal to P and a. unit vector on the same ray asP for each of the vectors P = (1, -1),P = (3,4),P = (-7,12), P = (-1,.2), P = (-1, - Va), P = (1).

va,

SEC.

THE BINOMIAL EQUATION

7]

197

7. The binomial equation. Every nonzero complex number :x: + yi corresponds to a vector (:x:, y) == r(u, v), where u == :x:/r == cos 8 and v == y/r == sin 8. Then :x: + yi == r(cos 8 + i sin 8). This is called the polar Jorm of a complex number. We shall write

e"

(12)

== cos 8

+i

sin 8

for the complex unit cos.8 + i sin 8 and call 8 the amplitude of :x: + yi. If :x: + yi == re" and :X:o + Yoi == 8e" then (:x: + yi)(:X:o + Yoi') == r8e"e~. But formula (6) implies that (cos 8 + i sin 8)(cos q,

+ i sin q,)

+ 4» + i sin (8 + q,).

== cos (8

This also follows by direct.multiplication from formula (9). It may be written as (13) and we see that (:x: + yi)(:X:o + y~) == r8e'(I#). We have multiplied the absolute values r of :x: + yi and 8 of :X:o + Yoi and have added their amplitudes 8 and q,. Formula (13) may be generalized to a product of any number of factol"! and states that the amplitude oj a product oj comple:x: number8 i8 the BUm oj the amplitude8 oj itB Jactor8. Th~n if n is any positive integer, we have

(re")" == r"e'''·.

(14)

This result is known as De M oivre' 8 theorem. If c == a + bi is any nonzero complex number and n is a positive integer, the equation :x:" == c is called a binomial equation. If we write c == rete where r == via' + b', the nnumbers

(15)

:X:i ==

V; e"',

8i == 8

+ (j n-

1)2'R' (j ==

.

all have the property that :X:i" == re'"'' == re£I+'''(i-1») == rete == c.

i . . . n)

198

. VECTORS IN THE PLANE

[CRAP. VIII

They are then all roots of the binomial equation :&" == c. For the angles 8 + 21r(j - 1) are all coterminal. By formula (13) . (16)

:&/

==

.yr eUl"ti-l,

t == e'"ftl" (j == 1 . . . 11.).

The complex unit t has amplitude 2r/11. and we see that each is obtained by drawing a circle whose radius is vr, locating the point· on this circle whose polar coordinates are [{IT,8/n], and then dividing the circumference of the circle into 11. equal parts by division points the first of which is [Vr, 8/11.]. It follows that the 11. numbers of formula (16) are all distinct and are the 11. roots of the equation :&~ == c. We may then pass on to the question as to what we ihall mean by the 81/rnbol where 11. > 1 iB an imeger and c i8 any compl6:& number. Let us define this symbol to pe that one of our numbers in formula {16) for which 8/ is the least nonnegative angle. Of course other definitions are possible. But we insist that they be such that, if c is a positive real number, and so 8 == 0, then shall be positive. Our definition satisfies this condition S;ll).ce ,8/11. == '0 if 8 == O. Now it is not true that == VCd for every c and d. Indeed, if 11. == 2 and c:= d == -1, the only possible /I , and in either definitions for v'='i are r l l or e'8r/2 == case (v'='i)(v'"=i) == -1 whereas

:&/

.yc

vc

vc.ytl

-e-

v( -1)( -1) == vI == 1. EXBRCISES

1. Use the values of the trigonometric functions to find r and 8 for the following complex numbers c: (a) c - 8 (b) c = -5

(Il) c

(c) c = 3i

( :'I

(d) c =

(t.") 3,

-7i

+

(8) c = 1 i (f) c = -2 2i (g) c = 3 --3i

+

=<

-4 - 4i

= 1 + Vi i -1 + v'3i c= 4

C

(Al) c (Z) c (m) c

= - Vi + i = -2 Vi - 2i = 4 va - 4i

199

EQUATIONS OF LINES

SEC. 8)

2. Find the following products in polar form. giving rand product in each case: (a) (1

+ va i)(l + i)7 + + va +

(b) (-2 2i}(3 - 3i)8 (c) (1 ",j1S( 1)10 (d)

(-1 ~ va i)

va va -

(e) (-2 2i)I(1 (f) (-2 - 2i)120 (g) (2 2)17 (h) (-

3. Find the solutions in polar form. of :/:"

= 3, c = 4 V2 (b) n = 3, c = eli...

(a) n

(c) n

(d) n (e) n

= 6, c = = 8, c = = 4, c =

-1 64

+ 4 V2 i

va + i

(J

for the

iF

Va + 1)18( -1 + i)l& = c in the following cases:

(I) n = 3, c = (g) n = 3, c = (h) n = 5, c = (i) n = 6, c = W n = 8, c =

-8i -

v'2 + V2i

(-1

+ ",j'

va - i -1 - vai -

8. Equations of lines. An equation (17) with real coefficients is called a linear equation if (a, b) is not the zero vector. Such an equation is an equation of a line. Indeed let (Xl, Yl) be any real number pair such that axl + byl + C = 0 and let L be the line through (Xli Yl) perpendicular to the ray from 0 through (a, b) as in Fig. 15. Then P = (x, y) satisfies formula (17) if and only if ax + by + c - (axl + bYl + c) = 0, that is, if and only. if (18) But formula (18) is satisfied if and only if P - P l is orthogonal to (a, b). This means that the ray from 0 through P - P 1 is perpendicular to the ray from 0 through (a, b) and is parallel to L; the line through P and P 1 is parallel to L. Then P is on L. This proves that formula (17) is an equation of L. " Formula (18) involves a, b and a point P = (Xl, Yl) on L. The numbers a, b are called direction number8 of L, and formula (18) is called the point and direction nu?n1Jer form of an equation of L.

200

V1llCTORS IN THE PLANE

(CBAP.vm

H (Zl, 1/1) .and (z.,1/.) are two distinct points on L, the difference vector (ZI - Zl, 1/1 - 1/1) is orthogoIiaJ. to (1/1 - 1/1, -ZI

+ Zl) ~ (0,0).

Thus an equation of the line L through, P 1 and PI is

(19)

(1/1 -. 1/1)(Z - Zl) - (ZI - Zl)(1/ - 1/1) =

o.

This is called the two-point form of an equation of L.

FIG. 16.

Any point P on the line through two distinct points P 1 andP. defines avectorP - P 1 = t(P I - P 1): Heretis~he ratio of lengths

- 1\P

t=--·

PJSs

The numb~r t > 0 if and only if P is on the ray from 0 through PI - Pl. Let us make Pl the zero point of a real number system on the.line'L, the direction from P l to p. the positive direction, and the distance from P 1 to PI the unit ·distance. Then it should be evident that t is the coordinate of P in this coordinate system on L. Also

P - P 1 = •(z - Zl,1/ - 1/1) = t(z. - Zl,1/1 - 1/1),

that is, (20)

{:1: = Zl 1/ = 1/1

+ t(ZI + t(1/1 -

:1:1), 1/1).

S:::C.

8J

EQU ATIONS OF LINES

201

Formulas (20) express the coordinates (z, y) of any point P on L in terms of Zl, Yl, Zll, Yll and the ratio t of the length P lP to PIP lI' They are called a pair oj parametric equatiOnB of the line L, and t is called the parameter. Note that if o ::!i t ::!i 1 the point P lies between PI and P 2• The midpoint of PIPII is the point where t = t so that its coordinates are Zl + (21)

Yl

21 (Z2

) Zl+Z2 - Zl = 2 '

+ 21 (Y2 - . Yl)

=

Yl + Yll

2

.

Formula (17) and its equivalent forms (18), (19) are referred to below as nonparametric Jorms of line equations, and we close by observing that formula (19) is obtainable from (20) by writing (Z -

zJ (Y2

- Yl) = t(Z2 - Zl) (Yll - Yl) = (y - Yl)(ZlI - Zl). BXERCISBS

1. Give nonparametric equations (a) (5,6) and (4, -2) (b) (5,3) and (5,2) (c) (5, -3) and (7, -3) (d) (-6, 2) and (-6, -3) (e) (4,1) and (6, 1) (J) (5, -3) and (6, 2)

of the lines through

VI) (:-9, -5) and (-6, -1) (k) (8, -2) and (-1, -8)

(i) (9, 7) and (6, 3) (J) (2, 0) and (0, 6) (k) (-1,0) and (0, -5) (l) (-5, 0) and- (0, 3)

I. Give parametric equations of the lines of Exercise 1. 8. Give formulas (17) of the lines L through PI = (Zl, 'U1) and orthogonal to Q = .(a, b) in each of the cases of Exercise 1 where we take PI as the first point and Q as the second point. 4. Replace each equation of Exercise 3 by an equivalent equation in which (a, b) is replaced by a unit vector (v, fI) = (air, blr). Ii. Find the coordinates of the mid-point of each line segment " ' where PI is the first given point in each part of Exercise 1 and PI is the second point. 6. Find the coordinates of a point P for PI, PI as in Exercise 5 such that 'FJiIPJ'1 = t where (a) t -= ti (b) t = ti (c) t = ti (d) t = - t. 'I. Put'U = 0 in formula (19) and solve for z to obtain the formula we used in Chap. VII to solve an equation by the method of interpolaiion.:

202

VECTORS IN THE PLANE

9. Conic sections. A quadratic equation (22) F(:c, y) = a:cl + 2b:cy + cyl + d:J; + ey

[CHAP.vm

+ 9 =0

defined for real numbers a, b, c, d, e, 9 such that a, b, c are not all zero is said to define a conic 86ction. If we replace (:c, y) by :c'u - y'v and :c'v + y'u, respectively, as in formula (5), we obtain a new equation (23)

~(:c,

y)

= a:c'~ + 2{J:c'y' + 'Yy'l + 8:c' + fI/I' + r = 0,

in which a, {J, 'Y, 8, E, r are real n.umbers. This new equation is the equation of the same curve with respect to the (:c', y,)-coordinate system. We shall show that u = cos (J and v = sin (J may be ·selected so that fJ = o. The second-degree terms of F(,;, y) make up a quadratic form Q(:c, y)" = a:cl

.(24)

+ 2b:cy + cyl,

and we call a + c the #:race of Q(:c, y) and ac - bl its determinant. These numbers determine an equation ~~

~-~+~+oo-~=~-~~-~=~

called the characteri8tic equation of Q(:c, y). The roots a and 'Y of this· equation are called the characteri8tic roo't8 of Q(x, y) and we already have seen that (26)

.

a"(

= 00 -

bl,

a

+ 'Y = a + c.

The discrimjnant of the quadratic equation. (25) is (a

+ C)I -

4(00 - bl) = (a -

C)I

+ 4bl

and is positive. Hence a and'Y are real and distinct. We nowprovethefollo~:

Theorem. Let Q(:c, y) = a:cl + 2b:cy + cyl where b ;o!! 0, and let a, 'Y be· the characteri8tic roo't8 oj Q(x, y). Then the rotation oj a:ce8 defined by a-a b v = -vr.b===.=;:+=G7"'a==a~)I (27) vbl + (a -

,,=

replaC68 Q(:c, y) byax'l

a/

+ 'Yy'l.

SlUe. 9]

.

203

CONIC SECTIONS

For the first of the equations (28)

+ bv == au -'YV + cv == av

au - av bu

+ bu ==

- bv

+ C'U

== 'YU

is satisfied if formula (27) holds and (27) is indeed derivable from the first equation if we use the equivalent form a-a

(29)

v == -b-u,

u2

+ vi == 1.

The second equation is equivalent, in view of formula' (29), to (a - 'Y)(a - a)u = b2u, which is true since, by formula (26), we have b2

+ (a -

a) (a - 'Y) = bll

+

all -

{a

+ c)a + (ae -

b2)

== O.

The third equation results when we use formula (29) to write its equivalent form (a - c)(a - a)u == b2u and thus [a 2 - (a - c)a + (ae - b2)]u == O. For a is a root of the equation of formula (25). Finally the fourth equation is a consequence of the third equation when we replace u by (a - a/b)v and so have [(c - 'Y)(a - 'Y) - b2]v

== ['Y 2 - (a

+ ch + ae -

b2]v == O.

The first and second equations of formula (28) may be multiplied respectively by u and -v and added to yield au2 + 'YV2 = a. Similarly we may use the third and fourth equations to obtain av2 + 'Yu2 == c. We multiply the first equation by v, the second by u and add to obtain (a - 'Y )uv == b.

Then ax'2

+ 'Y1I'2 E!!

+

55 a(zu + YV)2 + 'Y( -zv you)!! z2(au 2 + 'YV2) 2zy(a - 'Y)uv + y2(av 2 + 'Yu 2) Ei az2 + 2bzy cy2.

+

This proves our theorem.

+

204

(OIW'. vm

VECTORS IN THE PLANE

nlustrative Examples I. Remove the :cy .term from the equation 2z2 a rotation of axes.

Solution

-

4:cy

+ 5112 -

3 by

+

Here a = 2, b = -2, 0 = 5 so that a 0 = 7, ac - b2 ... 6 and the cha.ra.cteristic equation is al - 7a + 6 == (8 - 6)(8 - 1). We take a = 1, 'Y = 6 and the (z'y') equation is zl2 6y'2 = 3.

+

Also au

+ bv = au becomes 2u r(u,

II)

= r(2, 1),

u

211 =

so that u = 211 and

U

= 2/v'5,

" = l/VS.

The equations of rotation are

z, + 2y~

2z' - y' Z=

v'5'

v'5'

11 =:

n. Remove the:cy term from 9z2 + 2Uy + 16yl - 5z + lOtI + 3 by a rotation of axes.

=

0

Solution Here a = 9, b ... 12, 0 = 16 so that we solve

a2 Thus a

25a + (9 • 16 - 144) = al

-

258 = O.

= 0, 25. Take a ... 0 and so solve 9u + 1211 = 0, 3u + 411

= t, II = 511 = -3z'

U

-

t.

The equations for rotation are 5z = 43;' and

+ 411', -5z + lOtI = -43;' -

3y' - &'

= 0,

+ 3y',

+ Sri' = -10z' + 511'.

The rotation of axes replaces our equation by

2511'2 - 10z' + 5y' + 3 = O. EXERCISES

Remove the zy term from each of the following equations by a rota.tion of axes : (a) 4zy - 16 = 0 Am. Z'2 - V2 = 8; u = II = 1/0. (b) 8z2 4zy - 24y2 = 5 (0) Z2 2:cy y2 - 3 = 0 Am. 2:&'2 = 3, u = II .;,,'1/V2. (d) 1&2 - 24zy 9y2 = 17 (e) Z2 - 1O:cy + yl = 5

+ + +

+

Am. 12:&'2 - 10t1'2

=

5, u

=

-5/v'I46, II = 11/-vm.

BEC.9]

CONIC SECTIONS

205

(f) ~:l:1 + 120:1:1/ + 144y2 = 1 (g) 43;2 + 43;y + 4y2 = 12 Am. 3:1:'2 - y'2 = 6, U = II = 1/0. (h) 53;1 - 20:1:1/ + 2Oy2 = 4 (i) :1:1 + 12:1:1/ + y2 = 8 Am. 7:1:'2 - 5y'2 = 8, U = II = 1/0. W 25:1:2 +30:l:1/+9y2+20:1:+6y+1=0 (k) :1:1 - 4:1:11 + 4y2 + 6:1: + 3y - 9 = 0 Am. 5:1:'2 - 3 V5y' = 9, U = -l/V5, II = 2/V5. (l) Z7:1:2 - 36:1:1/ + 12y2 - 20:1: + 9y = 16 (m) :1:1 + :1:1/ + yS + 3y - 4 = 0 Am. 3:1:'2 - y'2 - 3 V2(a:' - y') = 8, U = II = 1/0. (n) 23;2 12:1:1/ 18y2 = 0 (0) 53;2 - 6:1:1/ + 5y1 - 10:1: - lOy - 15 = 0 Am. 2:1:'1 + By'2 + 5~' = 15, U = II = 1/0. (p) 43;2 + 12:1:Y + 9y2 - 2:1: - 3y + 1 = 0 (q) 4:1:11 - 3y2 + 6:1: + 4y = 5 (r) 11:1:2 - 243;y + 4y2 - 6:1: - 8y = 20 (8) 43;1 + 12:1:Y + 2Oy2 + 10:1: - 8y = 6 (t) 53;2 + 6:1:1/ - 3yl + 6:1: + 2y = 1 (U) 53;2 -,12:1:1/ + lOy! +:1: - Y = 4 (II) 53;2 + 8:1:1/ + 5y2 + 2:1: + 3y = 5 (w) 7:1:1 + 8:J:y - 8y2 + 9:1: - 6y = 11 (:I:) 43;1 + 24:1:11 - 3y2 + 8:1: + 8y = 10 (y) 7:1: 2 + 6:1:Y - yl - 76:1: - 60y - 100 =- 0 (z) 853;2 + 96:1:1/ + 45y1 + 1,0643; + 7441/ - 1,308 = 0

+

+

CHAPTER IX MATRICES, DETERMINANTS, AND LINEAR SYSTEMS 1. Systems of equations. Let f(3h, ••• ; Zn) be a complex-number valued. function of 11, complex variables Zl, . • • ,Zn. Then we have said that to each complexnumber sequence CI, • • • , Cn there corresponds a value'of our function which we designate by f(CI, ••• , c,~) and call the value of f(ZI, • • • , :en) at Zl = CI, Zi = Ci, • • • , Zn == Cn. We shall call (CI' • • • , en) a 'solution of the conditional equation f(ZI, • • • , Zn) ='0 if the value f( CI, • • • ,en) is the number zero. The set of conditional equations (1)

fl(ZI, . • •

,:en) == 0,

fi(ZI,···,

,.

:en) == 0,

fm(ZI,"', Zn) == 0,

defined by m functions of n variables, is called a system of equations. A system of equations (1) asks: What are all number sequences (CI' • • • , en) which are simultaneously solutions of all m of the equations? If the functions /1, fi, ••• , fm of formula (1) are the functions defined by linear polynomials in Zl, • • • , Zn, we call formula (1) a linear system. We shall concentrate our study of systems of equations in this text on linear systems. The theory of determinants is a tool for the solution of linear systems and will be presented. in this chapter. The theory of matrices, of which the theory of determinants is a part, also arises from the study of linear systems, and we shall present some of the elements of that theory here. ' We begin our presentation with a generalization of the vector concept. An n-dimensional vector is a sequence 206

SlIIC.

1]

207

SYSTEMS OF EQUATIONS

a == (ai, . . . , a.) of n numbers which are called the element8 oj a. A vector may be interpreted as the sequence of coordinates of a point in an n-dimensional geometric space, as the coefficient sequence of a linear equation alXl + azZlI + . . . + a,.z. == k, or as a value sequence Xl == ai, . . . , z. == a. of n variables. Our algebraic study will not use any of these interpretations but will treat the vector purely as a sequence. "The BUm a + ~ of two vectors a == (ai, . . . ", a.) and ~ == (b l , • • • ,b,,) is the vector

a. + b,,), obtained by adding corresponding elements. The sero vector is the vector (0, 0, .•• , 0) and will be designated simply by the symbol o. It has the property (2)

a

+~

== (al

+ bl, all + b

l,

••• ,

a+O==O+a==a for every vector a. The negative -:-a of a == (ai, . • • ,a.) is that vector with the property that a + (-a) == o. Thus -a == (-ai, . . . ,-a.). The difference a - ~ is defined to be a + (-~) and is (al - bl, • • • , a. - b,,). Addition of vectors may be shown to be commutative and associative. We leave the amplification of the meaning of this statement and its verification to the student. The 8caZar product da of a vector a == (ai, . . . , a.) by a number d is the vector da == (dal , • • • ,da.). We call a sum dial + dsal + . . . + dmam, of products ~ of numbers tis by vectors ai, a linear combination of ai, • • • ,

am· OlU.L EDR.CISES

1. Leta = (3, (a) a

-5,4,2),~

+~

(6) a - ~

2. Let a = (2, -3, 1), the linear combina.tions (a) a +~ +'Y (6) 2a - ~ ~ 'Y

= (-6,2, -3, -1).

+ 2fJ

(c) Sa (d) 2a ~

(6) 4a (f) a -

s.B

= (1,2, -2),

(c) 3a -"2fJ (d) 2a - s.B

Compute

+ 'Y + 'Y

+~

4fJ

'Y - (3, -1',4). Compute (6) 2(a +~) - 3-y (f) 3(2a 4-.y

+ m-

208

IeHAl'. IX

LiNEAR SYSTEMS

2. Rectangular matrices. A rectangular array of number8 is called a matrix.

The array

-5 -1 6 -2

1 2

o

"""

I'

1 1

3

-3

of the coefficients of Xl, Xs, X3, X4, Xli in the following system of four linear equations in five variables

'1-ax1 +,+ -

X2 2X2

5X3 -

X4

+ 6x

1i

=1

+ 6X8 - 2X4 =3 2x1 + 'X8 + X4 + 6x = -2 4x1 + 3X2 + X3 - 3X4 + 2x1i = 7 Xl

1i

is called a 4 by 5 matrix. The horizontal lines of numbers in a inatrix are vectors called its row~. The vertical lines also form finite sequences called its columna. In the ,example above (2

o

1

1

6)

is the third row of A and

is its fourth column. It is customary to call the rows of A row vector8 and the columns of A column vector8. We shall later add rows (columns) of A to other rows (columns) and shall multiply rows (columns) by numbers. 'We note that there is no need to separate the sequence elements in a column vector by commas. It is customary to omit commas also in the rows of a nmtrix and we shall henc~forth do so in all our vectors. A matrix which has m rows and n columns is called an m by n matrix. The vectors which are rows are 1 by n

SEC.

2]

RECTANGULAR MATRICES

209

matrices, and those which are its columns are m by 1 matrices. The numbers which appear in a matrix are called its elements. If all the elements of a matrix A are zero, we shall call it a zero matri:i; and shall write.A. == 0 regardle88 oj the number oj rOW8 and coZumm in A. We shall also say that two matrices .A. and B are equal, and write A == B; if they have the same size m by 11. and are such that correspondingly placed elements are equal. The position of an element of a matrix is at least as important as its value. It may be described by a statement as to the ~w and column in which the element appears. For example, in the matrix above the number 3 is in the fourth row and second column. It is convenient then to introduce a notation for the general element of any matrix. This is the symbol D.ii·

The subscript i is the row 8Ub8cMpt.· Its value is the label of the row' of our matrix A in which D.ii appears. The subscript j is the coZumn 8Ub8cMpt. Its value is the label of tb.e column in which a.j appears. When A is an m by 11. matrix the values of i range from 1 to m and those of j from 1 to n. In the matrix above au == 3, au == a" == -3, alii = aaa = au == au == 1, au == -1, au == -5, au == 6, etc. The third row of the matrix consists of all elements with first subscript 3 and they are all, aal, aaa, au, alii in our case. Let the reader give their values as well.as the notations and values of the elements in the fourth column. . We may now use the notation (3)

210

LINEAR

SYSTEMS

[CHAP. IX

for the arbitrary m by 11, matrix A. The ~otation (4) A· = (o,;i) '(i = 1, . . . , m; j- = 1, . . . ,

11,)

will also be used here. ORAL EXERCISES

1. The matrix at the beginning of this section is a 4 by 5 matrix. Make the corresponding statements about the following matrices. (aj

(_~

(.j

(J) (b) (1 (c)

3-5

(3 2) -1 3

(d)

(7 4 3 0 4 -2

6

2)

(_! ~ =v 7.

1

1

0

-1

/1

2

0

3 -4

1 0

2. Read off ~he elements au, au, au, alB, aas in those of the matriceS above 'Where theae elementa e:m8t. 3. Read off the first row a.nd the first column of each matrix in Oral Exercise 1. 4. Read. off the third row and the second column of those matrices where they exist. 6. What kind of matrix is a row of an m by n matrix? A column? An element?

S. Elementary transformations. We shall be concerned with certain processes for altering a matrix A which are called elementary tramJormation8 on A. They arise in the study of linear systems of equatiQD.S when the equations are interchanged, added, or multiplied by a number. There are three types of elementary transformations on the rows of a matrix and three co~esponding typ~s on its columns. The.first type is that of the interchange oj two rOW8 or two columns. By a finite sequence of such transformations we may obtain any desired permutation of the rows and columns of the matrix. The second type of transformation is that of the addition to a row (column) oj any numeriCal multiple oj another row (column). Our final type is that of multiplication oj a

SEC.

31

211

ELEMENTARY TRANSFORMATIONS

row (column) by any nonzero number. Observe that all our transformations are reversible. The result of applying a finite number of elementary transformations to an m by n matrix A is 'another m by n matrix B. We shall say that A and B are equivalent matrices and shall write A ,...., B. Observe that A :: B does not mean'that A and Bare equal. They are merely related in the manner we have described and the student should not try to read anything else into our definition. Illustrative Examples I. Find the matrix B which is the result of adding the second row plus twice the third row of

A

=

( 5 -4 4 2) -2 -1

0 2

-8 5 2-3

to;ts first row. Solution ,;' The new first row is (5

-4

4

2)

+ (-2 + (-2

-8

0

4

5) -6) = (1

4

0

0

1)

Thus

B =

(-~ -1

0 0 2

0

-8 2

-D

II. Subtract multiples of the columns of the matrix A from other columns and replace A by a matrix B ~ A which has all but one element in the first row zero. Solution Let the columns of A be represented by Cl, C2, Ca, C4. Form Cl - 2C4, C2 2C4, Ca - 2c 4 and then C4 - 2(Cl - 2C4) to obtain

+

A::::::: (-1251

1~ -1~8 -3~) ~ (-1~5 -41~ -1~8 -13 ,2~)'

-4

EXERCISES

1. Let Tl,

Ta, Ta

be the rows of A

=

(-~

2

-: -1

212

LINEAR SYSTEMS

[OBAl'. IX

Find the matrix B obtained by (a) Adding 2rs + r. to rl (b) Adding rl + rt to r. (c) Adding r. - 2rl to r. (d) Subtracting 3rl r. from r. (6) Subtracting r. - 2rl from rl I. Let Cl, CI, Ca, c,.be the columns of the matrix A above. Find the -matrix B obtained by (a) Adding CI Ca 2c, to Cl (b) Adding Cl c. c, to Cs (c) Adding 2cl Cs c, to Ca (d) Subtracting 2cl from c. (6) Subtracting 2cl Ca from c, (fJ Subtracting c, - 3Cl from c. a. Simplify the first row of each of the following matrices by the procedure used in Illustrative Example II.

+

+ + + + + + +

(a)

(6)

G G

-1

3 -3 4 -1 2 -1 4 -3 4 -5 1 2

(.) G

2

0

-1 1 -1

-V

D V

(d)

C5 -1

(.) (

:)

3 2

-2

1

-1

2

1

-1 1

D 2

-1 4 5

o1 3

-V 2

5

-3

'- Simplify the second row of each matrix of Example 4 by a like procedure. 6. Simplify the first column of each matrix of Example 4 by using row transformations. _ 8. Simplify the second column by using row transformations.

4. Special matrices. The elements a;i of a matrix A - (a;i) are called the diagoruil elements of ,A and the elements 'a;i with i ~ j the nondiagoruil elements. - The line all, au, . . . of diagonal elements of A is called the diagoruil of A. Then the elements tI;J with j > i are said to lie above the diagoruil of A and those with i < i to lie below the diagoruil. We shall say that ,A. has triangular form if either all elements below the diagonal of A, or all ele-

81110.

0]

213

SUB KATR I OlD S

ments above the diagona.I of A, are zero. the matrices

(j

0 4 2

0 0 3

~ G

G ~) (~ 0 3 1

1 4 0 0 1 0

For example,

4 5 3

~ 8.

V

have triangular form.. A matrix A is caJIed a Bquare matrix if it has exactly as many ro~ as columns, that is, A is an n by n matrix. We shaJI speak of such a matrix as an n-T0'W6d square matri:e or as a square matri:c oj order n. A square matriX in tria.ngula.r form. is caJIed a trianguZar matrix. If the nondiagonal elements of a square matrix A = (tlii) are aJI zero, we caJI A a diagonal matrix and write A = diag {d 1, • • • ,dn}. Here ~ = tli, is the ith diagonal element of A. A scalar matrix is a diagonal matrix with equal diagonal elements. For example, the first of the matrices

o 3

o

o 3

o

is a diagonal matrix which is not a scalar matrix while the second 'matrix is a scalar matrix. A scalar matrix whose diagonal elements are aJI1 is caJIed an identity matrix. It is customary to designate every identity matrix by the symbol I regardless of its order. 5. Submatrices. If we omit certain of the rows and columns of a rectangular matr:i.x A, the elements in the reIDa.;n;ng rows and columns also form. a matrix caJIed a submatri:e of A. For example, the omission of the first row and the second and fifth columns of the 4 by 5 matrix of Sec. 2 is the 3 by 3 submatrix

214

LINEAR SYSTEMS

[cauo.

6-2} 1

1

1

-3

IX

We shall be particularly interested in the result of deleting one row and one column of an n-rowed square matrix A. Each element D.ii of A determines the row -and column in which the element appears and we shall designate the submatrix obtamed by deleting this row and column by A'S. For example, in the 3 by 3 matrix above we have

Au

= (~

-~}

AI2

=

(-12 -2)1 .

ORAL BDRCISBS

1. Read off A 4I, A.a, As. in the following matrix: 1

-5 -1

2 4

2 0 3

-1

1

6 -2 1 1 1 -3 2 3

C -1

v·

I. In what row and column of Au does the element'as. of A appear? 8. In what row and column of Au do the elements au and a14 of A appear? ' 4r. In what row and column of A18 does all appear? '

8. Determinants. We shall associate with every square matrix A = (aai) a number which we shall call the determinant of .A and shall write IAI for this number. We shall also write IAI - laail as well as (5)

IAI-

au

au

au

a22

ani

'an2

altai

..... .

]'

and shall call this symbol an n-rowed determinant, or a determinant of order n. Matrices that are not, square do not have determinants• ..,

uc. 6)

215

. DETERMINANTS

The determinant of a 1 by 1 matrix A = (a) is defined to be the number a. We define

I: and au an an

al2 a21

au

:1

=

a18 au = au' la u au

aaa

ad - be,

aul - au' Ian aaa an

+ a18' Ian an

aul aas

au /. au

Then we have defined three-rowed determinants in terms of two-rowed determinants. We now proceed inductively. Let us suppose that (11, - I)-rowed determinants have been defined. Then every n-rowed square matrix A = (Q,ji) has elements Q,ji with corresponding (11, - I)-rowed square submatrices A#. We call the determinant IA~il the minor of Q,ji and the number

B# = (-I)C+i IA#1 the co/acInr of Q,ji' If we select any row of A, multiply the elements of this row by their cofactors, and add the resulting products we obtain the number

Q,jlBSl

+ Q,jIB,. + . . . + Q,j"Bi ".

Similarly, if we select any column, we obtain a sum aljBlj

+ a2iBli + . . . + aniB"i'

It is proved in the next section that all these numbers are equal. We 8hall call their common value the determinant 0/ A •. We have now given one expression of 114.1 for every row and one for every column of A. Each expression is called the e:tpamion of IA I according to the corresponding row or column. Each expansion is a sum of 11, terms. Every term is the product of· an element in a :fixed row by its

216

LINEAR SYSTEMS

(CHAP. IX

cofactor. The sign (-1)*"1, which we prefix to each minor /Ail/ to obtain the cofactor of 4ii, is plus if the sum of th~ row and column subscripts is even and is minus if this sum is odd. Thus the signs alternate and may be presented in a checker-board form. For n - 2, 3, 4 the checkerboards are

I~

+

-+1, /

+ +

+ + +

+

+ + +

,

+ + +

The following results are simple consequences of our definition. We shall not give their formal proofs. Theorem 1. Let all the element8 in a rO'lD or in a column oj A be sero. Then /A / - o. Theorem 2. The determinant oj a triangular rnatriz A i8 the product oj the diagonal elementB oj A. The determinant oj the n-r0'lD6d identity ~ is 1. Dlustratille ExamPles

I. What are the minors and cofactors of the elements of the second column of

IAI-

-1

2

_~

2 1 1

-5

: 2

-~4? 3

6

SoZtd.ion We use the technique of first deleting the second column of A simplify the eye work and so write the submatrix

to

DC.

217

DETERMINANTS

6]

The minors are then obtained by omitting the rows in turn and are 4 -1 -5 -1 4 = -5

al ...

a.

1 3, 2 -4 1, 2 "6 0 2 6 3 0

a2 =

-1 -1 -5

a. =

1-14 -I"

3 2 6 3 0 2

-4 3, 2 -41 l' 3

Since 1 + 2 is odd, the cofactors are -aI, at, -aI, a•. n. Give the terms in the expansion of the determinant above according to its third row.

Solution The row is (-1 matrix

1

3). If we delete it, we obtain the

2

Since 1 + 3 is even, the terms are 3

-4 1-1 3-4 1 - (1) 4 0" 1 6" 2 -5 6 2 1-1 2 -4 -1 " - + (2) 4"" 2 - 1 - (3) 4 ". ~5 1 2 -5"

o

2 2 1

3

O· 6

In. Complete the e~on of the determinant above ~cording to its third row. "" Solution We expand the first determinant according to its second row. The resuit is (":"2)(6 + 24) - (12 - 3) = -69. The second determirui.nt is expanded similarly to obtain -4(6

+ 24)

- (-6

+ 15) = -120 -

9

= -129.

Expand the third determinant according ~ its first row ~o obtain (-1)(4 -1) - 2(8 + 5) - 4(4 + 10) = -3 - 26.,... 56 == -:85. The last determinant is (-4)(12 - 3) + 2(-6 + 15) = -36 + 18 = -18. He~ce JAI = 69 + 129 - 170 + 54 = 82. "

218

[CHAP.

LINEAR SYSTEMS

]X

EXERCISES

1. Give the terms in the expansion of the following determinants according to the iP'st row: (a) 2

-1 1 3-1 6-2 -1 4

o

2 3

2

(b)

4

-1 3

o

(d) 1

4 8 2

3 4

0

(e)

2

-1 -1

3

-1

o

1 1

-3

o

-1 2

tn

,2

o

2

4 2 3 2 1

2 4 1 -1 4 000

o

1-1

1

2

-2 1 6-1

480 (c) 0

-2 3 1 -1 -1 2 4 -4

o

2-3

~I~ =~ -~ ~~I21 1

-1

2. Give the terms in the exPansion of each determinant of Exercise 1 according to the thi!d column. S. Compute the determinants of (c), (e), tn of Exercise 1, using a selection of a row 'or a column to make the computation 'as simple as possible. 4. Compute the following determinants:

(b) 1

o

(g)

A.m. -45.

2

-1

0

2 4 3

~I

A.m. -3.

1

1

-1

1

-1 4 -2 1 -4

5 5

0 1

1-2

2 1

0-1 1 ·-1 1 o 1-2

1

o

4

1 -1 2'

1 3 4 -1

1

-1

-7 2

o

2

I:

(d) 1

-3 2

tn

(c)

-1 2 -2 4 2-3

3

(e)

'-!I

-~

-1 2 1 -2 3 306 143

A.m. O.

A.m. 21.'

(h) -3

2 1 -5 6

-1 2 1

1 4 1 2 -4 -10 0-1

(i) -4 -2

o

1 0 2

0 2 -3

1 1 0

-7

2

0

2

W -3

2 3 -1 4

-2

o

-5

(k)

f:

-1

0

-1

2

1

1 2

0 1

-1 -1

2 0

1

Ana.S.

2 0 -1

-1

1

1 2

-:-1

2 3 -1 0

0-1 -1 0 2 4 4-1

(n)

3 -2 2 -4

-2 -1 0 6

2-4 0 6 5-3 -3 -2

(0)

1 -1 3 -4

-1 2 0 5

3-4 0 4 4-2 -2 0

(m)

o

Ana. 1.

1-2 3-1 -3 0 2 9

1 1 3

(l)

219

EQUALITY OF ALL EXPANSIONS

SEc.7J

-3 -1 1.

4 9 0

0

2

Ana. -2.

Ana. 1M.

7. Proof of the equality of -all expansions (FULL COURSE). We wish to consider n-rowed determinants, for n > 1, and to show that the expansion of any such determinant according to any row gives the same value as its expansion according to any other row. This is true for the least value n = 2 by trivial direct computation. We therefore have the first step in an inductive proof. .Ai?, our next step, w~

220

LINEAR SYSTEMS

[CHAP, IX

assume that all row expansions of. any (n - I)-rowed determinant yield the same number. Select any two rows of an n-rowed matrix A and choose their labels i and k so that i is less than k. Expand IA I according to the ith row to obtain the sum (-I)'+ltlilIAill (_I)i+2~2IAi21 + ...

+

.

+ (-I)i+n~nIAinl·

The kth row of A is the (k - l)st row of each Aii since each such submatrix has been obtained by omitting a row of A above the kth row. Expand each IAijl according to its (k - l)st row. The hypothesis we have made states that the results will be the correct determinant values of IA#I. The elements in the (k - l)st row of Aij are elements akh of A ·where h F j. Substitute these expansions in our original sum to obtain a sum of n(n - 1) signed products tli;alrhlGI where the matrix G which goes with each term is the result of deleting the ith and kth rows and the jth and hth columns of A. . Every one of our terms tliialchlGI may be associated with exactly one other term whose position is such as to give the same matriX G. This is the term ~hakilGI and we may always select our notation so that j < h. If we ~o this, we shall see that the element akh is in the (k - l)st row and (h - l)st column of IAiil, and the sign. adjoined to. ~iakhlGI is then (-1)8 = (-1)' where

+ j + (k - 1) + (h - 1) and t = i + j + k + h. The element aki is in the (k - l)st row:.:a.nd .jth column of IAihl and the sign adjoined to ashakilGI is ( -1)" where u = i + h + (k '- 1) + j = t - 1. 8

= i

It follows that our pair of terms has the value (6)

( _1),+iH-rrc ltlii akj

tlihl' IGI. akh

We have proved that IAI can be expanded as a sum of n(n - 1)/2 terms. Each of these is a product one factor af· which is a two-rowed determinant obtained from two

PC. 7]

EQUALITY OF ALL EXPANSIONS

221

fixed rows by all possible choices of a pair of columns. There are then evidently 0,.,1 = n(n - 1)/2 possible terms. The second factor is the (n - 2)-rowed determinant 101 obtained by deleting our two rows and columns, and we prefix the sign. (-1)' where t is the sum of the two row and the two column subscripts which define the term. We now use the expansion of IA I according to its kth row. Expand each IAIIAI according to its ith row. This is the ith row of IAI and we obtain precisely the same 0,.,1 pairs of terms allAaulOI, alGia",IOI. The element Q,;J is in the ith row and jth column of AM and so we prefix the sign. ( -1)' where t = k + h + i + j as before. The element am is in the ith row and (h - l)st column of AlGi and we prefix the sign (-1)" where u = k + j + i + h - 1 = t - 1. Hence the pair of terms is the same, we have expressed our second expansion as the sum of the terms in formula "(6), and it has the same value as before. This completes our induction and proves that all row expamionB are equal. To prove that all the column expanBiOnB are equal to any row expamion we notice, first, that all the arguments above can be phrased symmetrically so as to prove that any two column expansions are equal. Let us omit the details. Then our proof will be complete if we show that the expansion of IA I according to its :first row is equal to its expansion according to its :first column. These two expansions have the term. aulAul in common. The other terms in the row expansion: are ( - '1)i-l'JaliIA~1 for j = 2, ••• , n, and those in the column expansion are ( _1)'+l~lIAill for i = 2, . . ., n. Expand IAlil according to its first column and obtain a sum of terms (-1)'~1101 since ~l is in the (i - l)st row of Ali. Thus the corresponding term of our expansion is , (_I)·+l+i~liIOI. Expand IAill according to its first row }. and obtain terms (-1)iali I01. Here 0 is obtained by omitting the first and ith rows and the first and jth columns of A, and is exactly the same matrix in both expansions. Moreover the corresponding terms ( -1)fti+ laUa liI01 I

222·

LINBAR SYSTEMS

[OHAP. IX

are equal. Thus we have shown how to expand certain of our (n - I)-rowed determinants in the expansion of IAI according to its first row, as well as in the expansion according to its first column. We have also shown that the results may then be paired so that cOITesponding terms will be identical. The expansions clearly give the same value. This completes our proof of the fact that all row and all column expansions of a determinant are equal numbers. 8. Properties of determinants. The n by m matrix whose ith column is the ith row of the m by n matrix A is called the traruJpoBe of A and is designated by A I (read If A prime" or If A transpose"). For example, .

3 ( A == -1 -2

-5 2 1

6

o 4

A' _ -

( 3-1 -2) -5 6 l'

2 ·0 4

1. 4 5

In the next section we shall prove the following: Theorem 8. The determinant oj the transp08e oj a Bf[UQ,re inatriz A is equal to the determinant oj A. . We shall also' prove-the following properties: • Theorem 4. Let B be the mo:I:rix obtained Jrom a Bf[UQ,re rnatriz A by multiplying any row or column oj A by anumb61' c. Then IBI = ciAI. _ , Theorem 6. Let B be the mo:I:rix obtained Jrom a Bf[UQ,re mo:I:rix A by interchanging two oj the rOWB or two oj t~ columruJ oj A. Then IBI == - IAI. . Corollary. Let two roo}B or two c~lumruJ orA have proportional elet1'l.6nt8. Then IAI == O. Theorem 6. Let B be obtained Jrom a Bquare matrix A by adding a numerical multiple oj a rOO} (column) 'oJ A to another rOO} (column) oj A. Then IBI == IAI. . . The corolJ.ary of Theorem 4' and 5 given above will be used in the proof of Theorem 6. It follows from Theorems 4 and 5 since by Theorem 4 we have IAI == dlDI where two rows or columns of D are equal. Interchange these two

SlIIC.

223

DETERMINANT TH"EOREMS

9]

rows to obtain a matrix E such that lEI == -IDI. But E and D are equal matrices IDI == - IDI, 21DI == 0, IDI == 0,

IAI

==

o.

.

We shall use the determinant theorems in the following example to indicate a procedure by means of which we may simplify the computation of determinants. Rlustrative ExamPle Compute

IAI if 1 A _ ( -1 - -3 -1

-2 -3 4 :a 1 12 2-3

D o.

-8

Solution

IAI

2 1 -2 1 -2 -3 2 2 -1 4 0 0 =2 -3 1 12 -4 = 20 -5 -1 1 0 0 2 -3 -1 2 -1 22 = 6-5 2 3. 3 =2 -5 3 0 -2 0 -6 = 6(14 + 15) = 174.

I

-3 -1 3 -6

2 2 2 3 2 2 2 = 6 -5 1 0

3

7 0

'~I

EXERCISES

Use the method a.bove to compute the determinants of Exercise 4 of Sec.6.

9. Proofs of the determinant theorems (FULL COURSE). == i., the transpose of A is A and IAI == IA'I. Let us assume Theorem 3 for all (n - 1)-rowed square matrices and suppose that A is any n-rowed square matrix. Each element ali of the first row of A determines a submatrix Alj of A . . Then ali is in thejth row and first column of A' and the corresponding submatrix is All. 'By our hypothesis each IAIll == IAlil, the cofactor of the element alj of A' is p~cisely the same as its cofactor in A. It follows that the expansion of IA'I according to its first column is identical with the expansion of IA I according to its first row. This completes our inductive proof of Theorem 3.

H n

224

LINEAR SYSTEMS

[CHAP.

]X

If we prove any determinant theorem involving the rows of square matrices, we may apply the theorem to the rows of a matrix A'. These will be the columns of A and Theorem 3 states that IA'I = IAI. We shall then have a corresponding theorem involving columns. It remains then to prove Theorems 4, 5, 6, for rows. To prove Theorem 4 we let the ith row of B be Ca;l, ca;2, • • • ,ca;n' Then we expand IA I according to its ith row to obtain IA I = a;18i1 + . . . + a;n8in. The cofactor of Ca;j in B is 8ij and B = (ca;1)8il + ... + (ca;n)8 in = ciAI. This proves Theorem 4.Theorem 5 has meaning only for matrices A of order n > 1. Let B be the result of interchanging the ith and kth rows of A. We use the expansion of IAI as given in formula (6) and see that the corresponding expansion of IBI is the sum of the On,2 terms (7)

(-1)'/:: :/. IGI

= (-1)'(akja;h - a;jakh)IGI.

These are the negatives of the terms (6) and IBI = - IAI as desired. We finally let B be obtained from A by replacing the ith row of A by its i row plus c times its kth row. Expand both IA I = a;18i1 + . . . + a;n8in and IBI

= (a;1 + cakl)8i1

+ . . : + (a;n + cakn)8in

according to their ith rows and use the fact that the cofactors 8ij are the same. Then

IBI

=

IAI + c(ak18il + ... + ah

8in)

= IAI

+ clNI

where N is the matrix obtained from A by replacing the ith row of A by its kth row. T.hen N has two equal rows, INI- = 0, IBI = IAI· 10. The rank of a matrix. Let t be any integer and A be any m by n matrix. Then the elements of A which occur in t rows and in t columns of A form at-rowed submatrix N of A whose ,determiil.ant is called a t-roweq minor

SEC.

10]

225

THE RANK OF A MATRIX

of A. The elements of A :are its one-rowed minors and the possible values of t are from 1 to the lesser of m and n. The rank r of a matrix A is the largest integer for which an r-rowed nonzero minor of A exists. We may compute the rank of A by computing the values of all of its minors, but it may be computed best by the u~ of the following theorem: Theorem 7. Elementary transformatilJ!Ul do not alter the rank of a matrix.

We shall not try to prove this theorem. 'it may be used to compute the rank of A as follows. We use elementary row or column transformations to carry A into a matrix B in triangular form and with bi; = 0 for j = 1, . . . , n wherever bi; = O. Then the number r of nonzero' diagonal elements bll , • • • ; brr' of Beis the rank of A. For all but'r rows of B are zero and every (r + k)-rowed minor of B is zero. Moreover, B has an r-rowed triangular (square) submatrix T such that ITI = bl l • • '. brr F-,' O. nlustrative Examples

I. Compute the rank of

A

2

=(-~

4

-3 -1 . 0 1 7 -2 -2 0\ 6 Solution

AoJ~ -\g

2

4

1 1

7 7

2

14

~ =D~Gl 1 2

-4 -8 ,

0

2 1

o o

4 7

o o

o1 o o

-D

-4,

0

,0

Am. r = 2. II. Compute the rank of

A=

-2

3

-4

6 -1

o 2

-8 -2

3

5-4

1 2 3 -2 4 11 2 10

2-4 4-3 -3 0 8-11 17-11

o o 2 o 2 2

226

LIN,EAR SYSTEMS

[CHAP. IX

Solution

-2 0 0 0 0 0 -2 0 0 0 0 0

AO:!

~

3 5 -4 3 '0 -5 -8 4 -1 2 4 -3 6 1 2 -4 -1 -8 -12 5 -1 7 12 -7 3 3 5 -4 -1 2 4 -3 4 0 -5 -8 0 13 26 -22 o -lO -16 8 0 5 8 -4

0 0 2 0 2 2 0 2 0 12 0 0

We interchange the third a.nd sixth columns and then the third and 'fourth rows of A to obtain -2 3 0 -1 0 0 0 0 0 0 0 0 -2 3 0 -1 0 0 0 0 0 0 0 0

A~

~

0 2 12 0

5 -4 4 -3

26 -22 -8 4 o -16 8 0 8 -4 0 5 -4 2 4 -3 12 26 -22 4 0 -8 0 0 0 0 0 0

3 2 13 -5 -10 5 3 2 13 -·5 0 0

Ana. r EXERCISES

Compute the ra.nks of the following matrices.

G =D

3 5 1 -2 (b) ( ~ 3 -1 3

(a)

~)

G J) (~ (-~ D (c)

1 -3 -1

1 2 3

= 4.

SEc.,ll]

3 13

1~)

1 -1 4

1~)

5

tn(: ~) (~)

(Q

227

MATRICES OF A LINEAR SYSTEM

G

r

1 2 2

1 2 3

(1) ( '

2 2 4 2

ro ( :

-D -2

-1 2 146 7 -11- -6

G

W(j

D

2 -1 2 3 2 -13 4 0 -1 1 -2 -2

-V 0 0 0 1

(m)

-2

1

~

-4 2 -3 -1 -7 1 -3 -1 0 3 1 2 -2 -1 -1 -2 -1 0 4 -1 0 2 0 -1 3 1 0 1 -1 2 -1 3 5 1 0 1 3 -6 0 7 3 7 -15 0 1 -2 1 -3 -2 -2 -1 -5 -3 -4

n

(a)

(~

(0)

(~

-~ 5

1 0 0

D

J)

~

D

11. The matrices of a linear system. If we transpoSe and collect the terms of a. single linear equation in variables Xl, • • • ,z., we may write the equation as

+ . . . + ~Xft == k. The left m~mber alXI + . . . + ~ft of this equation is alxl

then a linear f,orm in Xl, • • • ,Xft , the right member k is a number, and the coefficients ai, • • • , aft are numbers. We sha.ll say that a. particular variable x, appears ezpZicitly in this equation if its coefficient a. is not zero. A linear system is a set of several linear equations in several variables. . A general notation for such a system requires a notation such as Xl, • • • , Xft for the variables

228

LINEAR SYSTEMS

[CHAP. IX

as well as a notation for the coefficients. The latter notation must indicate both the equation in which the coefficient appears and the variable of which it is'the coefficient. We shall use the symbol ati to indicate the coefficient of the variable Zi in the ith equation. A system of m linear equations in the n variables Z1, ... , z" is then given by

+ auzs + . . . + a1nZta = k1 + assZs + . . . + as-z" == ks lZm1Z1 + a.sZs + ... + lZm-Z" == k". allZ1

(8) .

aUZ1

in which the sYmbols ati and k, represent ordinary complex numbers. The numbers ati determine a matrix A == (ati) in which ati is the element of the ith row and jth column. We call A the coejficient matriz of the system (8). If we adjoin to A the coZumn oj constant8 k 1, • • • , k", we obtain the m by n + 1 matrix

AO =

(fi, .. :.

which is called the augmented coefficient matrix of formula (8).

A. linear system is called homogeneou8 or nonhomogeneou8 according as the numbers in its column of constants are or ~ not all zero. A homogeneous system always has the solution Zl == Zs == • • • == z" == O. We shall call this solution the trivial 80lution of a homogeneous system. Let us note the convention that if the number n of variables is not greater than four we shall use z for Z1, 1/ for Zs, • !or Za, and 'ID for z,. A convention of this kind is necessary in order that the question asking for the matrix of a system of equations, say in z, 1/ or z, 1/, ., shall have a unique answer.

SEC.

12]

SOLUTION BY ELUUN ATION

229

ORAL EXBRCISES

1. What is the coefficient matrix of each of the following linear systems?

+ 41 - W = 3 (c) 3z - 4y + • - w == 0 ·z+2y-.+w = 1 4z-2f!-.=5to-2y -2z + 71 + • - 2w = 4 z=5y+.+6w y=.-w-z 7z - 271 + 3. - 8w = -6 (d) 9z - 371 = 2s + 1 (b) z + 71 - • - w = 2 + z 2z-y+3.+w=l-y 5 - 6z = 771 - 2 -z + 271 + • - w = 0 z - 1 = 2(71 - 1) + 3(. - 2) (e) 3z - 2 + 2(71 + 1) = 0 z - 2y + 3(. + 1) = 3 2(z + 1) - 5(71 - 2) + 6(. - 1) = 6 (f) 2+z=y+.+2z 3-z=2y-.-3z 5y - 3z = -71 + 2s + w (g) 2(z + 1) - 3(71 - 1) = 5(. + 1) 3z - 271 = 2(z + .)

(a) 3z - 271

~-2.=y

+ 3(. 3(z - 1)

+ 1) = 6(71 + i) 4(z + 1) + 71 - 2 = 2 .

(Il) 2z

+ 2(71 + 1) = • -

1

2. What is the augmented matrix of each system of Oral Exercise I? 8. Which systems of Oral Exercise 1 are homogeneous?

12. Solution by elimination and substitution. A linear system may have a unique solution, many solutions, or no solution. In all cases we may solve the sYstem by a process known as solution by elimination and substitution. In this process we select one of our equations and a variable Xi appearing explicitly in it. We then subtract multiples of this equation from nonzero multiples of the rem.ajning equations so as to remove the term in Xi from all of these m - 1 equations. We shall then have obtained a linear system consisting of m - 1 equations in which Xi does not appear, and our original equation in which Xi doeii appear. This new system of m equations in n varial?les has exactly the same solutions as our original system and is said to be equivalent to it. . ;.

230

LINEAR SYS'TEMS

[CHAP. IX

We now proceed to modify the m - 1 equations in which

x. does not appear so as to obtain m - 2 equations in which a second variable Xi does not appear by using one equation in which Xi appears explicitly. This process of eliminating variables will ultimately terminate. Let us examine the end result and speak of its equations as the end equations. It may happen that at some stage in the elimination all variables in certain equations will not appear and we shall be led to an impossible numencal statement such as o = 5 - 4. In this case the system of equations has no solution and we shall say that it is inconsistent. When our system has a unique solution, we shall always have m ~ n, and m - n ·of our end equations will have zero constant terms and left members identically zero. The remaining n end equations will be such that- only one variable appears explicitly in a first equation, only one additional variable explicitly in a second equation, an ith additional variable explicitly in an ith equation, the- final nth variable explicitly in the last equation. We may then solve our first equation for the first variable and substitute the result in the remaining equations so as to get rid of this variable. The process of solving and substituting then continues and will yield the unique solution as the final result. In the only remaining case the process of elimination will yield a system of r ~ m significant end equations such that the remaining end equations have zero constant terms and left members identically zero. In one of these equations a variable will appear explicitly which has been eliminated from all the other equations. In a .second equation a second variable will appear explicitly which MS been eliminated from the remaining m - 2 equations. The ith equation will contain an ith variable explicitly which has been eliminated from all remaining m - i equations and the rth equation an rth variable whose eJiroina.tion from the remaining m - r equations (if there are any left) makes these equations identically zero. W~

SEC.

12]

231

SOLUTION BY ELIMINATION

may then solve our system of r significant end equations for r of the variables as linear functions of the remaining n - r variables, and shall say that our equations are dependent if n > r. We have now a new kind of solution for a system of equations. It is not a numerical solution which is a number sequence (Cl, ••• , en) satisfying the equations, but is a set of formulas for r of the variables expressed as linear functions of the remaining n - r variables. These results are solutions in the sep.se that, if we replac:e the n.- r last variables by numbers and compute corresponding values of the first r variables, the resulting number sequence will invariably be a solution of our system. Let us note that the decision as to which r variables we solve for is not, in general, dictated by the problem but may be entirely up to the equation solver. It can be shown that the number r above is the rank of the coefficient matrix of the linear system, and that the system is consistent if and only if r is also the rank of the augmented matrix of the system. We shall not prove this result but shall use it in Illustrative Example IV below. It is to be applied by the student in Exercise 2. nlustrative Examples 1. Solve the linear system:

x- y+3s= 3x - 2y

x

+ 721 =

+ 3y -

321

=

6 14 -4.

Solution

Designate the equations by E l , E 2, Ea. Form 3El - E2 to obtain -y + 221 = 4, and El - Ea to obtain -4y + 621 = 10. Then 4( -y

+ 221)

- (-4y

+ 621) == 221 = 16 -

Also y = 221 - 4 = 2 and x = y - 321

10 = 6,

+6= 2-

II. Solve the linear system:

x- y+3s= 6 x + 3y - 321 = -4 5x + 3y + 321 = 10.

9

+6 =

21 = 3. -1.

Ans,. (-1, 2, 3).

232

LINEAR SYSTE:U:S

[CRAP. IX

Solttl,ion We form HI - HI to obtain 4u - a. = -10, and H. - 5111 to obtain 8y - 12. - -20. The coefficients of the two eq!l&tions are proportional, the original equations are dependent, and. we may express our .38 - 5 -38 + 7 . . solutions 88 11 .. -2-' :z: ... 2 .

Ill. Solve the linear system: :z:- 11+38 ... 6 :z: 311 - 38 = -4 5:z: + 311 + 38 = 11.

+

Solttl,ion We proceed 88 in illustrative Example II to obtain 4y - a. .. -10 and use H. - 5H1 to obtain 8y - 121 = -19. Subtract the double of 4u - a. = -10 from 811 - 121 ...; -19 to obtain 0 = 1. The equa.tions are inconsistent.. IV. Show that the equations of the linear system of illustrative Example III are inconsistent by computing the ranks of the two matrices. Remarka: We may compute the ranks of A and A * simultaneously by using elementary row transformations only. We note that we are actually doing exactly what we did above but are working with the rows of a matrix rather than with equations.

Solttl,ion

-!) e!(~ 11

0

V -1 3' 6} -1 3 4 -6 -10 8 -12 -1

4· -6 -10.. o .. 0 1 Then A has rank: r = 2 and A * has rank: r* ... 3, so that the equations are inconsistent.

==G

BDllCIUS

1. Each of the following linear systems has a unique solution. Find it. (a) :z: - 11 21 = -2 3:z: - 2y 4s = -5 211-38= 2 (b) 2:z: 11 3. ... 6 4:1:- 11 = 9 -7:z: 2y = -15 Am. (2, -I, 1). (0) 3:z: - 11 - 21 = 7 4:1: 1 2:z: 5rJ 38 - 3·

+ + + + + +• + .... + +

SEC.

12] ,

(d) z

SOLUTION BY ELIMINATION

+

y - 5z =

26

z + 2y + z = -4 z + 3y + 6z = -29

(6)

z

+

2z+

y

z z 2z

(1.)

z 2z

-z

W -4z+ y -2z

Am. (2, -4, 7).

- z+ t = + 2z - t = + z+2t=

-2 5 3 +y - z = -6 + 2y - t = -2 + 3y - z = -5 -y + 2z + 4t = 2 + 4z - t = 6

-y

Am. (1, 0, -5).

= 2

y-z=5 z+2y-z=1 z+3y =0 (f) 2z + 3y - z = -15 3z+ 5y+2z = 0 z+ 3y+3z= 11 7z + 11y = -30 (g) -3z + 2y + z - 2t = 6 -2z + 3y + 3z - t = 14 -y - 3z = -11 -5z + 4y + 2z + 9t = 0 (h)

233

Am. (-1, -2,2,1).

+t=-10 = -4

+ 2z + t

2y-3z = 1 -7z+2y+2z = -15 (k)' 4z - 2y + 5z + t = 6 -4z + y - t = -2 2z+y-z+t= 1 z +z-2t= 4 3z +5z- t= 9(l) z + 2y - 3z + 4t = 9 z -z+t= 1 3z - y + z = -1 -z + y + 2t = 9 3z+ y +3t= 9

Am. (3,2,1,0).

Am. (-1,0,2,4).

2. Use the method of Illustrative Example IV to determine which of the following systems of equations are consistent:

+ 3y - z = 4 z+2y+2z= 5 4z + 7y + 3z = 14

(a) 2z

234

LlNEAR SYSTEMS

(b) 3z -

(c)

(d)

(6)

(J)

(g)

(h)

y + 2z = 3 2z+2y+ z=2 z - 3y + z = 4 z - 4y + 6z = 1 2z+ 9y+ 5z=9 -z + 21y - 13z = 7 4a; - 6y + 7z = 8 z- 2y+6z=4 8z - lOy - 3z = 8 2z - Y + t= 2 -3z + z - 2t = -4 z+y- z+ t= 2 2z-y+5z = 6 :1:+ y+2z- t= 3 2z- y+ z+ t= 1 z - 5y - 4z + 5t = -7 4a; - 5y - z + 5t = 3 z+ y =2 2z- y+ z=l 5z+2y+ z=7 3z-3y+2z=4 z + y + z+ t = 0 2z-2y+z- t=O 3z - 4y + z + 2t = 1

[CHAP. IX

Am. r = 2, r* = 3, inconsistent.

An8. r = r* = 2, consistent.

Am. r = 2, r* = 3, inconsistent.

Am. r = r* = 3, consistent.

S. Solve the following systems: (a) z

+y + z =

10 -6 - Z= 2 (b) 2z - 7y - 6z = 0 &;+5y-2z=0 4a; - 2y - 7z = 0 (c) z - y 5z = 0

zz

y - 3z

=

+

2z+3y =0 4a;-5y+22z=0 (d) z + y + z = 1 3z+4y+5z=2 2z+3y+4z=0 (6) 2z + y + 3z = 0 4a;+2y- z=O 2z + y + 10z = 0 (J) 3z-2y =0 &;+ y-2z=0 z+ y+ z=O "':"2z + z =0

Am. (0,0,0).

Am. Inconsistent.

Am. (0, 0, 0).

SEC.

13]

SOLUTION BY DETERMINANTS

235

+

(g) 2:1; 3y - • = 0 33:+2y-3.... 0 33:+ y+38=O &+6y(1) 211: + 3y - • - t ... 0 11:- y-2.-4t=0 33:+ y+S.-2t=0 6z + 3y - 7t = 0

.-0

(i)

+•+

Ana. (tt, -t, -ft, t).

t...

1 az-2y +2t= 3 11:+ y+_ =-1 -II: + y +_ t= 0 II: -

Y

11:+ y-38+ t=l 211:-4y +2t=2 33:-4y-2z ... 0 II: +2_+3t=3 (k) II: + y + .... 0 az + .=0 1211: + 9y + 10. ... 0 2:1;+ y+ .=1 (1) 11:+ y-2z-0 2:1;+ y-3.=0 42:-2y-2.... 0 6z- y-58=O 711: - 3y - 48 ... 1 (J)

Ana. (0, 0, 0, 1).

Ana. Inconsistent.

13. Solution by determinants. The matrix of a system (9)

allXl a21Xl

+ . . . + altax. == kl + . . . + asnXn == ks

anlXl

+ . . . + annXn

== k n

of n linear equations in n variables is a square matrix A, and it has a determinant d == IAI. Let Ai be the matrix obtained from A by replacing its jth column by the column of constants and let di == IAil. Then if rl, . . . , rn is a solution of the system of formula (9) we may prove that (10)

For let Bi be the matrix obtained from A by multiplying the jth column of A by ri' Then IBil == dri' Alter Bi by adding to its jth column rl timee its ~t column, ra tim.~1f

236

LINEAR SYSTEMS

[CRAP, IX

its second column, . . . , Ti-l times its (j - l)st column, Ti+l times its (j + l)st column, . . . , Tn times its nth column. Call the resulting matrix Ai' By Theorem 6 IBil = IAil. But the jth element mthe ith row of Ai is a;lTl

+ lli2T2 + ... + llin~n = k;,

and Ai is the matrix defined. above.

This proves formula (10). If d ~ 0 the solutions of the system of formula (9) are uniquely determined by formula (10), that is, we see that the system is satisfied only by Xi = di al. When d = 0 and some di ~ 0, the equations are inconsistent. However they may even be inconsistent if d = dl = d~ = . . . = dn = O. We shall show how formula (10) may be used even when d = 0 in the illustrative examples below. 'Note that the method we shall use is .applicable to m equations in n variables. , In closing, we observe that a homogeneous ~ystem always has the trivial solution Xl = X2 = . . . = Xn = 0 and that this is the only solution when d ~ O. For every di has a column of. zeros and every di = 0, dXi = 0, d ~ 0, Xi = O. mustrative Examples I. Solve the linear system

z- 11+3z= 3z - 2y

+ 7z =

z + 3y -

6 14

3z = ~4

by using determinants.

Solution 1

=

d

'.

d. ,

=

I~1 =~3 -3~I = ~1 4~ --~I = 2, 6

II! =~ ~ I~ -~ -4

3

-3

=

14

0

~

6

=

-2,

SEC.

80

13]

that ~

=

-1.

Similarly,

6 . 3

1

1 6 7 = 0 -4 -3 0 -10

14 1 -4 1 -1 d. = 3 -2 1 3

til - 3

80

237

SOLUTION BY DETERMINANTS

1:

=

-4

3 -2 = 4, -6

I~ -~ _~I = 0

4

-1~

6,

that 1/ = 2 and • = 3. Solution 2

We compute ~ ... -1 as above and substitu~ to obtain the equations -1/

+ 31 =

7

-211 + 7. = 17. Solving by determina.:dts, we obtain

11

=

11; 1-1 -2

+

Then 3. = 7 1/ = 9 and .... 3. II. Solve the linear system ~-,I1/+3.=

6

+ 31/ - 3. = -4 52:, + 31/ + 3... 10. ~

Solution

Compute d

=

1 -1 3 5 3

1

3 -3 3

=

1 0 0

-1 3 4 -6 = 4 (-6) 8 -12 2

11

~1 = o.

We delete the last equation and write the first two equations as ~ 1/ = ~+31/=

-3. + 6 31-4.

Solving by determinants, we' obtain

~

..

-11

+6 1-31 31 - 4 3 I~ ~I

- -

-9. + 18 4

+ 31 -

4

-31

=

+7

2

'

238

LINEAR SYSTEMS

as well as y

= 4'1 + (3s -

4) - (3s

[CHAP.

+ 6) = -3s2+ 5.

Dc

The solutions

satisfy the third equation identically. III. Solve the system

x- y+3s= 6 x + 3,y - 3s = -4 5x + 3y + 3s = 11. Solution

=0

As before d y = 3s

t 9.

. and we proceed as above to obtam x

=

-3s + 7 2 '

We substitute in the last equation to obtain

a contradiction. Hence the equations are inconsistent. IV. Solve the system

2x- y- s=O y-11s=O x - 3y + 7s = O.

4:1:+

Solution

We compute 2

d

-1

-1

1 -3

7

=4

1-11

2 6 -5

1 -2\ =

-1 -1 0 -12 = -30 1 0 10 1-2

O.

Write the first and last equations as

2x- y= s x - 3y = -7s and solve by determinants to obtain

I

s -3 -1\

-7s

x

= 12 -1\ = 1

[1 -1[ -3

s -7

-5

= 23,

-3 . EXERCISES

Solve the exercises of Sec. 11 by using determinants.

14. Partial fractions. Any rational function p(x) may be expressed as a quotient of two polynomials in x. By

SEc.14J

PARTIAL FRACTIONS

239

using the division algorithm we may write p(x) in the form p(x) = cJ>(x)

+ f(x) g(x)'

where cJ>(x), f(x), g(x) are all polynomials in x and the degree of the numerator f(x) is less than the degree of the denominator g(x). Let us suppose now that it is possible to factor g(x) and so to write it as a product of powers of distinct irreducible factors. Then we may express p(x) as the sum of cJ>(x) and a number of what are called partial fraction8. The procedure used employs nothing but the solution of a system of linear equations and so is an application of the results of this chapter. If p(x) is an irreducible factor of g(x) and m is its multiplicity, there will be a corresponding sum of m partial fractions h1(x) p(x)

+ h2(X) + ... + h".(x) p(X)2

p(X)m

Each of the polynomials hi(x) is to be determined so that its degree is less than the degree of p(x), and the linear equations we solve are linear equation!!! in the coefficients of the polynomials hi(x). We shall not try to prove that solutions of these exist nor shall we give a general discus.sion of the process of finding the equations. The methods involved are adequately presented in the examples below. Note that in the case where our coefficient8 are permitted to be any real number8 the polynomials p(x) are either quadratic or linear, the polynomials hi(x) are correspondingly linear or constant. Illustrative Examples I. Express the rational function _2:1:4 + 15:1:2 - 20:1: + 13 (:I: - 1)3(:1: + 2)

as the sum of JVLrtial fractions.

240

LINEAR SYSTEMS

[CHAP. IX

Solution

By multiplication we find that the denominator is X4 - x 3 - 3x 2 + 5x - 2.

By division p(x1 == -2

+

-2x 3 9x 2 - lOx (x _ 1)3(x 2)

+

+

+9

,

and we use the theory above to write -2x 3 +9x 2 -10x+17 abc (x - 1)3(x 2) == (x - 1)3 (x - 1)2 x - I

+

+

for numbers a, b, c, d to be determined. -2x 3

+ 9x2 -

lOx

+

d

+x +2

Then

+ 9 = a(x + 2) + b(x -

+ + + d(x -

l)(x 2) 2) + c(x - 1)2(x

1)3.

This is to be a polynomial identity and so we equate coefficients of the powers x 3, X2, x, 1 of x on the two sides, using (x -1)3 = x 3- 3x 2+ 3x -1, (x - 1)2(x + 2) = x 3 - 3x + 2, (x - l)(x + 2) = X2 + X - 2, to obtain the linear system

+

c d = -2 b -3d= 9 a+ b-3c+3d= -10 2a - 2b 2c - d = 9.

+

The solutions are then a = 2, b = 0, c = 1, d = -3. Remark: We ~ay simplify our work by using values of x. vttlue x = 1 to obtain •

3a =

6,

We use the

a = 2,

the value x = -2 to obtain (-2)( -8)

+ 9(4) + 20 + 9 = 81 =

-27d,

and so have d = -3. The solution is completed by using the leading coefficient to obtain -2 ,;" c + d, c = 1, and the value x = 0 to obtain \J = 2a - 2b + 2c - d, 9 = 4 - 2b + 2 + 3 = 9 - 2b, b = O. 2 1 3 Ans. p(x) = -2 + (x - 1)3 + --1 x- - x- ,2'

II. Express p(x) =

x6

3 + 3x 2 - 4x +, 5x(x2++3x1)2(x3 _ 2)

as a sum of partial fractions.

4

5

SEC.

241

PARTIAL FRACTIONS

14J'

Solution

We write _ p(x)

=

ax + b (X2 j- 1)2

ex + d ex 2 + fx + g + X2 + 1 + x 2 ' 3 -

and obtain x6

+ 5x + 3x + 3X2 - 4x - 5 "'" (ax + b)(X3 - 2) + (ex + d)(x 2 + l)(x 2) + (ex2 + fx + O)(X2 + 1)2. 4

3

3 -

However, the factorization we are using permits rational numbers only as coefficients and so a, b, e, d, e, f, g are to be rational numbers. We put x 2 = -1, x = i, and obtain -1

+ 5-

3 - 5 - 3i - 4i = -4

+ 7i =

(ai

+ b)( -i -

2)

= a - 2b - (b Then a - 2b = -4, 2a then become

+ 3x + 3x 2 + 1 Then ci + d = 0, e = x6

4

+b= 7

+ 2a)i.

and a = 2, b = 3. Our equations

+ d)(x 2 + 1) + (ex2 + fx + g)(x 2 + 1j2. d = 0, ex 2 + fx + g == X2 + 1. 2x + 3 X2 + 1 \, "'" (ex

Ans. p(x) "'" (x2

+ 1)2 + x3 _

EXERCISE

Express the following as sums of partial fractions: 3(x 2

+ x)

(a) (x - 2)(x + 1)2 x 2 -.:x (b) (x + 1)(x2 + 1) 2X2 + 8x - 8 (e) (x + 2)(X2 + 4)

(d) x3

Ans. x

1

4x

Ans. X2

4x - 2 X2 - 2x 12 + 6x 2 (g) x 3 + 4X2 + 3x (f) x 3

x

~ 4x

+1 x3 + 8

2'

2

+4 -

17x

X4

(e)

2

+1+x -

+ 2' + 28 17 + 4) + 12(x + 2)'

Ans. x - 12(x2 _ 2x

-

(h) 5x 2

x3

(i) 43:3

-

-

4 'x

Ans - -

9 x+1

11 --+ x+3 -_.

3 x

+ ~X2 + 3x

Ans.

131 2x + 1 + 2x + 3'

x-

2'

LINEAR SYSTJUUiI

&;1 + 21;1 + 1 4m1 -:r: 21;4_:r:1 _&;1+21;+4 (Ie) (:r:1 - 4)(21; - 1) (3)

A

1

1.

1

Am. :r: + :r: _ 2 + :r: + 2 - 21; - 1· 24m1 - IO:r: + 5 (Z) (21; + 1)(21; - 1)1 4m :r: :r:1 - 3:r: (m) (:r:1 + 1)1 Am. - (:r:1 + 1)2 + ~I + 1· :r:1 - l (n) :r;I + 9:r: :r: 1 + 1 1& + 1 8:r: (0) (:r:1 + 4)1 Am. :r: + (:r:1 + 4)1 - :';1 + 4· 21;'-3:r:+2 (p) (:r: + 2)(2:r;' + 2:r; + 2)' (:r: - 3)1 -IO:r: 4 1 (g) (:r:1 + 4m + 5)1 Am. (:r:1 + 4m + 5)1 + :r:1 + 4m + 5· :r:1 + &; (r) (:r: + 2)(:r:' + 4) :r: 4 -3 :r:+1 1 (,), (:r: .....: I)(:r:1 + 1) Am. :r: + 1 + :r:1 + 1 - :r: - f • '3:r;8 - 3:r: 1 :r:4 + 3:r:1 (t) . :r: -10 21;-3 2 (u) :r:8 - 2:r;1 + &; Am. :r:1 - 2:r: + 5 -

+

+

z·

CHAPTER X MATRICES AND QUADRATIC FORMS (FULL COURSE)

1. Inner produCts. The theory of matrices is usually presented in more advanced courses. It! techniques are as simple as many techniques already presented here and are of sufficient importance to make an exposition of them very much worth while. We shaJI give such an exposition in this chapter without proofs of the principal theorems and shall regard the entire chapter as optional. We call a vector a = (ai, . . . , an) an rwlime'n8ional vector and its elements ai, . . . , an the coordinate8 of a. Then we may interpret a as representing a point in an n-dimensional geometric space whose origin is the uro vector (0, 0, . • • ,0). The inner product of a and another n-dimensional vector fJ - (b l , • • • , bn) is the number

(1)

otfJ'

= albl

+ aib! + . . . + anbn•

Our definition is such that otfJ' = fJa'. the inner pro!iuct (2)

aa' =

all

The norm of a is

+ ... + an!,

and we call a a unit vector if ota' - 1. Note that the vector E" whose ith- coordinate is 1 and whose other coordinates are zero, is a unit vector. Two vectors a and fJ are said to be orthogonal ~ otfJ' = o. We call a set ai, .. '. , ae of t distinct vectors a set of pairwise orthogonal vectors if 04.01/ = 0 for every i p6 j. We shall call a vector a with real coordinates a real vector. A real nonzero vector a has a positive nonn ota', and we c~ll the positive square root r = v'"ii.i2 the length of 243

244

MATRICES AND QUADRATIC FORMS

[CBAP.x

, /

Then the scalar product of the vector a by the number

a.

l/r is the unit vector (3)

Hence a ;:::: rp., that is, every real nonz~ro vector is the scalar product of its length r and a unit vector p.. If p. == (th, . . . , 'Un) and 71 == (VI, • • • , Vn ) are unit vectors, their inner product satisfies the relation (4)

-1

~ P.7I' ~

1.

,We shall not try to prove this but observe that it implies that P.7I' is the cosine of an angle 8 between 0 and '1r radians. We shall call 8 the angle between the unit vectors p. and 71. Then if a == rp. and {3 == 871 are any two nonzero ,vectors, we define the angle 8 between a and {3 to be that between the unit vector p. and 71. This yields the formula (5)

cos 8 ==

a{3'

.yaa; -v'ifPt' '

When a{3' == 0 the cosine of 8 is zero and 8 == '1r/2 radians. Then the vectors a and {3 may be thought of as being perpendicular and this is our reason for calling a 'and. {3 orthogonal if a{3' == O. nlustrative Examples I. Compute afJl where a = (1

-2

3

4)

and

(3 = (2

1

-3

2).

Solution

a(31 = (1 . 2)

+ (-2' 1) + (3' -3) + (4' 2) =

-1.

Remark: In cases where the coordinates of a and fJ are decimals it is desirable to superimpose the vectors to facilitate computation. For our simple case this would be a

,

= (1 -2 3 4) (3 = (2 1 -3 2) afJl = ""=2,----2=----"""'9-+:--:S"'"= -1.

II. Write a = (2

-6

IS

6) = rp. where p. is a unit vector.

SEC.

21

MATRIX MULTIPLICATION

SoZttI,ion

aa'=4+36+324+36=400 and -3 9 (1 a - 20 10 10 10

,. - 20,

:0)·

ORAL BDRCISBS

1. (a) Give the normS of the vectors (2), (3 0 2), (2 -1 3), -1 4 2), (4 1 3 -1), (t 1 - it). (b) Give the lengths of the vectors above. (c) Express each vector above as the product of its length by a unit vector. 2. Show that the following vectors are pairwJse orthogonal: (a) (1 -1 2), (1 3 1), (-7 1 4) (b) (1 1 1), (1 -1 0), (1 1 -2) (c) (1 2 3), (1 -5 3), (3 0 -1) (d) (2 1 0), (-1 2 1), (1 -2 5) (3

2. Matrix multiplication. The product AB of two matrices A and B i8 not defl,ned unless the nuniber of columri8 in the left-hand factor A is equal to the number of rOW8 in the right-hand factor B. . We define the product AB of an m by n matrix A == (~i)

(i == 1, .•• ,mj j ==

1, ...

,n), \

by an n by t matrix B

=

(bi ,,)

(j == 1, •.. ,

nj

Ie· == 1, ••• ,t),

; to be the m by t matrix (6)

a == AB

== (c.-.)

(i == 1, • • • ,mj Ie == 1, ••• , t),

in which (7)

c.-.

== ~lbl"

+ ~sbS1l + . . · + llMab"".

Thus every element of.the product matrix AB is the sum of the n products of the n elements in a row of A by the correspondingly placed elements in a column, of B. Our definition is called the row by column rule for matrix multiplication.

246

MATRICES AND QUADRATIC FORMS

[CBAP.x

We note thatifa == (ala2 ••• an) and,8 == (b l b2 ••• bn ) are 1 by n matrices the matrix product 0

b1

al

bn

an

is the inner product we defined in Sec. 1. This is the reason for the notation we used. We observed also that the element in the ith ..row and kth column oj our matriz product AB i8 the inner product oj tlJ,e ith row oj A by the kth column oj B.

Nothing we have said about the definition of the matrix product AB implies that when it is defined it is equal to BA. Indeed BA need not even be defined. However, even when ABo and BA are both matrices of the same size, they rp.ay be different. For example,

1) (0 o 1 ~)

o

The

matric~s

0\ _'/1

0) - \0

~}

(00 .. 01) -_10\0 0)1 .

whose elements

ar~

all zero are all called o. They have the property that OA == AO == 0 for any m by n matrix A and zero matrices of proper size. It can also ha.ppen -that A. ·and .B are nonzero inatrices and that AB == o. For example,

zero malrice8 aIid are represented by the symbol

The identity matrice8 are square matrices whose diagonal ele:m.ents are all 1 and whose nondiagonal elements are all zero. ' We represent all such matrices by I. Then

lA- = AI == A

0

0

SEC.

247

MATRIX MULTIPLICATION

2]

m

for all by n matrices A and a corresponding m-rowed I on the left, and n-rowed I on the right. We shall finally state the following results without proof: Theorem 1. Matrix multiplication i8 as80ciative, that i8, if AB.and BO are defined we have (AB)O = A(BO). Theorem 2. The transp08e of a product AlAI' .. At of matrice8 i8 the product At'At-I' ••• AI'AI' of the transpo8e8 of the factor8 in rever8e order. Theorem 3. The determinant of a product of 8f[Uare matrice8 i8 the product of the determinants of the factor8. Theorem 4. The rank of a product of matrices does not exceed the rank of any factor •. nlustrative Examples I. Compute

o=

(-:

_! _~) (_:

100 Cll Cll CIl CII

C8l

Cal Cn Cn

+

+

-~).

12

Sol'Ution

= (0, 2) (1 • 3) (2' -I), = (0' -1) (1 . 0) (2 . 2), = (-2' 2) (1 . 3) (1 . -I),

+ +

+

+ +

+ +( + +

= (-2' -1) (1'0) (1'2), = (2'2).+ (-4'3) + (-1' -I), = (2' -1) (-4 . 0) -1 . 2), = (1' 2) (0' 3) (0' -I), = (1 . -1) (0 . 0) (0 . 2).

+

II. Compute Y

x =

+ + =

(~)

-2

Ana. 0 = ( -71 2

AX where

y = (;)

A=

(-~ _~. =~)

Solution u=2z+2y+z tI = -z + 2y - 2z w=2z-y-2z. III. Compute AB and BA where

A =

(_=)

B = (4

2

3).

D 4

-4' -1

248'

MATRIOES AND QUADRATIO FORMS

Soltdion 2

3)

=(

: -12

BA = (4

2

[OHAP. x

2 3) 4.

6,

-6 -9

3) (_:) = (-1).

IV. Compute AB and BA where

Sol'l.lJ:lon

(0 1) -!) (-~ -2) 1 = 1 -2' -2) (-11 -4 "3) = (5 1) -2 -7"' 1 V. Compute AB and BA where B is the transpose of A and A =

( 2 2 1) -1

2

2

-1

-1'

-2

Solution

AA' =

(-~2 -1~ -2 -~)( -1~ -2 -~ -2 -~) = to \0 = (

~ ~~ _~) (_~

-1

-2

-2

2

i

~ _~).

o 9 o

~)

-1"-2

Hence AB = BA in this case. EXERCISES

1. Compute A', B', A' = (AB)' as well as B'A' for the products in Illustrative Examples I, II, IV. 2. Compute (AB)O and A(BO) where

A =

(~-A 0=

~),

(-~

"B "

1 1

o

= (~ -1

~ -~).

3

2

~

1

SJIC.

INVE'RBE OF A BQU ARE MATRIX

3)

249

3. Compute the following matrix products:

(a) (a b) (a -b) -b a b a (b)

-!be -(alb ~ ) (: c

(:00

l)

(c) (:

;)

(_~

~ _~ -~) (-1~

(d) (

-2 -4 (6)

-;)

(~

: 2

-2

5

-8

!) ( ~. =: 0

-1

~) (_~

(J) ( : -1

~

1

(g)

G-~){-~

(A)

(=~

(i)

fa

~

~

-!6-5_~)

~

2

:3)

D D

!)G-D ~) fo _~

~) (~

~ -~),

0 1 \~ 0 1 -1 0 2 4. The product of two diagonal matrices is a diagonal matrix. What are its diagonal elements?

\i

3. The inverse of a square matrix. A square matrix A is said to be nonsingular if its determinant is not zero. Then there exists a unique matrix A-I (read" A inverse ") called the inv6'f86.of A such that the product AA-l == A-IA == I

is the identity matrix of the same number of rows as A. By Theorem 3 we have III == lAllA-II. But III == 1 and so we have the following:' Theorem 8. The determinant oj A-I i8 the inv6'f86 oj the determinant oj A.

250

:MATRICES AND QUADR~TIC FOR:M&

[ClIAP.x

We also h~ve (AA-I), == (A-I)' A' == I' == I. This result is stated as a theorem. Theorem 6. The inver8e of the transp08e of a matriz A is the transp086 of A-I. Since (AB)(B"""IA-I) == A(BB"""I)A-I == AA-I == I, we have a special case of a result which we state without additional proof. Theorem 7. The inver8e of a product AlA • ... A, of nonsingtilar 8quare matrice8 is the product A,-l • . . A.-IAI-I

of their inver868 in r61J6r8e order. Two m by n matrices A and B are called equivalent if it i8 p08sible to carry A into' B by a finite 8equence of elementary transformations. Then we state the following: Theorem S. Two matrices A and B are equivalent if and only if there ~8t nonsingtilar matrice8 P and Q 8'UCh that B == PAQ. Indeed, let a sequence of row and column transformations CaITY an m by n matrix A into B. Apply all the row transformations tp the m-rowed identity matrix to obtainP and all the column transformations to the n-rowed identity matrix to obtain Q such that B == P AQ. The inverse of a two-rowed square matrix is given by the formulas: (8)

A == (:

t == ad - bc ~ 0,

A-' -

(~! I~.

The inverse of an n-rowed square matrix A with n > 2 may also be given by a formula involving the determinant ~ A and the cofactors of the elements of A, but it is preferable to compute A-I by solving a system of linear equations. We shall give this solution in our next section.

SlDC.4J

LINEAR TRANSFORMATIONS

251

ORAL EXERCISES

1. Give the inverses of the matrices

(-25 -2) l'

(-~

-D, (_!:5 ~_4:).

i. What is the inyerse of a diagonal matrix whose diagonal elements are all not zero?

4. Linear transformations. H A == (asi) is an n-rowed square matrix, we may form a corresponding linear system

+ a12XZ + . . . + alnXn == YI a21XI + azsZz + . . . + asnZn == Y2 allXI

(9)

This linear system is equivalent to the matrix equation AX == Y,

where

X==

(10)

y==

H IAI '" 0, so that A-I exists, we multiply AX == Yon the left by A-I to obtain X == A-IY. Thus if we solve' formula (9) for Xl, • • • ,:en we will obtain equations Xl

(11)

X2

== PllYI + PUY2 + == P21YI + P22YZ +

. . . + plnYn . . . + P2nYn

equiValent to the matrix equation X == PY, where

252 (12)

MATRICES AND QUADRATIC FORKS

P =

G:~"1 .. ~~ . :... ~) p""

•••

[CHAP.x

= A-1.

p""

This shows how to compute A-1 by solving a linear system. The equations of fomiulas (9) or (11) are called the equations of a noMingular linear (homogeneous) tramff1l'1Tl,Q,f,ion. If we interpret the n~bers :t1, • • • ,:en as the coordinates of a point then formula (9) expresses the coordinates 'V1, • • • , 'V". of thiS same point relative to a new coordinate system in terms of the old coordinates. The equations of formula (11) are called the 80lved form of the equations of formula (9). They express the old coordinates :t~, • • • ,:en in terms of the coordinates 'VI' • • • , 'V". Dlustrativ. E%am,z.

Compute A-I if

A -

~ -~ =~).

\i

0-1

8oh1tion The system of equations (9) becomes

211:-2y-,=u z+1/-28=" Z -,

=

tD.

E1imina1i!'l , to obtain· Z - 211 = u - w, -z + 1/ ... " - 2tD. Then - (z - 211) - 2( - Z + 1/) ... z ... -~u - w)- 2(" - 2tD) ... -u - 2v + 5w. Also1/-z+I1-2tD= -u-v+8tD,'-z-w= -u-2v+4tDj SO that

-1 -2

A-I... ( -1 -1 -1 -2

V·

We can check the answer by computing the product AA-l or A-IA. BDRCISES

Compute the inverses of the following matrices A and check your answers by computing AA-l 1. . .

=

SEC.

(a) (

(b)

5

2 0

4 1 0 0

~

0 0 2 1 1 0 2

-1

G

3 3

(_!

(g) (

-n

1

4 -4

V

3 -3 -4 7 3 -3

a

-3 7 -4

D

-D

G

4 -6 -1 -13) 0 1 -2 0 1 2 -5 -2 1 -5 1 0 1 1 1 ( 1 -2 1 -1 -4 -2 -3 1 1 1 -1 1 1 2 1 -2 1 0 -1 ( 2 -1 1 1 2 1 1 6 1 -2 ":'3 2 ( 2 -1 -4 3 -1 1 2 -1 0 -1 0 1 0 0 0 1 3 1 2 -1 1 5

W(~ (k)

D

-1

(Il)

(i)

a D

(d) (

tn

~)

1 2 -3 -2 1 4

(.) G (e)

253

SIMILAR MATRICES

5]

(ij

(..)

D -D G ~ -D V

(o)

(o)

n

!)

5. Similar matrices.

-)

-D -

If A == (o,;i) is any n-rowed square

matrix, we shall designate by A - xl the result of formally subtracting x from all the diagonal elements. Then (13)

A -xl

au - x . au

-

. a81

anI

au

alB

alta

au - x au

au

a2ta

ala - x

a8ta'

an2

an.

ann - x

. . . . . .....

254

MATRICES AND ~UADRATIC FORMS

[CHAP.x

The matrix of numbers ,obtained by replacing the v~riable x in formula (13) by a number d will be denoted by A - dl. The characteri8tic determinant of any square matrix A is .the polynomial in x with leading coefficient unity given by (14)

F(x) .= (-l)"IA - xII.

The equation F(x) = 0 is called the characteri8tic equaftio'(t of A, and the roots d 1, • • • , dn of this equation are called the characteri8tic root8 of A. They are a set consisting of all numbers di such that (15)

IA - dill =

o.

A s,uare matrix B is said to be similar to A if there exists a nonsingular matrix P such that B = P-IAP. We shall assume the following results without proof: Theorem 9., Similar matrice8 have the same characteri8tic determinant8 and therej01re the 8ame characteristic roota. Theorem 10. The characten8tic roota of a diagonal maf:!iix D are the diagonal elements of D. Theorem 11. Every real symmetric matrix i8 8imilar to a diagonal matrix. Theorem 12. Every matrix with di8tinct characteri8tic root8 i8 8imilar to a diagonal matrix. Theorem 10 implies that if a matrix A is similar to It diagonal matrix D the diagonal elements of D are the characteristic roots of A. The arrangement of these roots in the diagonal of D is actually arbitrary. Let us suppose' then that A is given and that we wish to determine whethe:r or not A is similar to a diagonal matrix p. We first determine the characteristic roots of A and hence a diagonal matrix D. We then seek a nonsingular matrixP such that (~6)

AP = PD.

If Pi is.the jth column of P, the jth column of PD is the s'calar mqltiple diPi . The jth column of AP is APi ~nd, formula (16) is equivalent to n linear homogeneous syste~s,

BBC.

5]

255

SIMILAR MATRICES

each of 11, eq1.1&tions in 11, variables. These systems are expressible in matrix form as (17) (A - diI)Pi = 0 (j = 1, . . . ,11,). It is alway, possible to solve these systems and determine P. When A is not similar to D no set of solutions of the equations of formula (17) will yield a nonsingular matrix p. mustrGtive EZGm;les I. Find the characteristic roots of

A

7 -6)

=(-!

4 2

8oltll.ion

-:e F(s) .. (-1)1-1

o

7

-6

O· -2

I

0

4- s 0 =-1 2 -s - 2 0

sI-4s+7

I,

4-s

o6

2

s+2

so that J(s) .. (s

+ 2)(SIl -

4s

= (s -

+

+ 7)

- 12 = sa - 2s1 - S 2 2)(SI - 1). Ana. ell = 1, ell = 2, ela = -1.

II. Show that the matrix of Illustrative Example I is similar to a diagonal matrix. We first solve the homogeneous 'system

-s +71/- & = 0 -s +31/ . = 0 2y - 3. = 0, whose matrix is A - I. Then s = 31/, 3s = 2y and the first column is a sca.la.r multiple ,of 9, 3, 2. The system with matrix A - 21 gives -s + 2y = 0, 21/ - 4s = 0 so that s = 2y - 4s and the second column may be taken to be 4, 2, 1. Finally, the system with matrix A - 41 gives -s + 411 = 0, 211 - • = 0 and so s = 411, • = 211. Hence we may take

IPI

-3

-4

= 1 3

2

-4

-3

III. Show that the matrix

A=G is not similar to a diagonal matrix.

1 .~).

o 1 "" O. o

256

MATRICES" AND QUADR'ATIC FORMS' [CHAP.x

Solution The characteristic determinant of A is (:I: - 1)2(:1; - 2) and d1 = ds = 1. Thus the first and second columns of P are obtained by solving the linear system with matrix A - I. The system is 'Y = 0"= s and the first two columns of P are necessarily proportional. Thus AP = .PD implies that IPI = 0, A and D are not simils.r. ORAL EXERCISES

Give the "chs.ra.cteristic determinants and roots of the following matrices: 0 (g) (~ (-~ 0 0 0 (6) ( 3 (b) (h) -4 6 0 4 1 3 (c) (~ t n (-4 3 o _ 4 0 -9 3 " 0 o 0 0 5 0 0 0

(a)

G

~)

G D'

~)

G ~)

i)

(~

D

°r~G

~)

D

EXERCISES

1. Compute the_characteristic roots of the following matrices:

(a):(_~ (b)

G

(.) G C~

1 0 5 1 0 0 0 0 1 0

!) o o

-4

2

-1

(d)

-V

-4 -6 1 -

-6 -6 (g) ( 9 -6 -6 8 2 -4 ,1 (h) ( ~ 0 tn

D

1

(c) (

3 -2

f 2 3

o

-2 0) 2-2 -2 1

2 -1 -4

D

Ana. -I, -I, -I, -1.

~) -!)

'D

Ana. I, 4, 16.

SEC.

6]

(i)

(!

010 -1

{il (

{ij

0 -2 4 9

~

257

QUADRATIC FORMS

0 0 1

-D. -1

1 0, 0 4 1 3 6 10 1 1 0 0

AM. -1, 2, 3.

0 1 0 '-6 0 0 0 1

D

~

,3

5

6 0 1

-V

AM. 1, -1,2, -3.

2. Find a nonzero matrix P such that AP = PD for each matrix of Exercise 1 where D is a diagonal matrix whose diagonal ele~ents are the chara.cteristic roots of A. State in each case whether or not A is similar to D. 1 3 3 1

AM. (6) P = (: : ·11

,

AM. (g) P

AM.

=

(1 2-2) 2 2

=: =~,i) (=~! -i : =r, V'

(~1 P = (-:1

AM. (0 P

=

2, A similar to D. -1

1 -2

1

A simila.r to D.

1

0

A similar to D.

0

6. Quadratic forms. An '1I/-O,ry iJua,dratic form. is a polynomial ,in n variables which is sum of terms all of degree two. The most general unary quadratic form. is ax l where a is a number and x a variable. The general binary quadratic form. may be represented as

I(x, y) = axl + 2bxy

+ C!l1

258

MATRIeE SAND QUAD RATIe FORMS

[CBAP.x

'where a, b, c are numbers. Any polynomial may be regarded as being a 1 by 1 matrix. Thenf(x, y) is expressible as the matrix product f(x, y) == (x

The general ternary quadratic form is f(x, y, z)

== ax! + by! + cz! + 2rxy + 28XZ + 2tyz.,

It may be expressed as the matrix product f(x~ y, z) == (x

r

y

b

t A quadratic form in Xl, X2, • • • , x A is a sum of terms for i, j = 1, .•. , n. The coefficient of Xi! is the number au. The total fJoefficient of-xiXi = XiXi is ~i aji. It is customary then to write this total coefficient as twice a' numbe~ which we now designate as ~i and so have ~i = aii· Then our quadratic form will be expres~ible as a matrix product f(X1, ••• , xA ) = X'AX. Here A = (~i) is an n-rowed symmetric matrix, that is aii = aii for every i and j. Also ~X'-Xi

+

so that X' is the transpose of X. We call the syn;unetric matrix A the matrix of the quadratic form f(X1, . • • , x A ) = X' AX and its rank the rank of f(X1, ••• , xA ). • nlustrative Examples I. Give the matrix of 3:1)2 + 23;y + 3yz + 51$2.

SEC.

7]

E Q U IV ALE NeE

0 F QUA D RAT I C FOR M S

259

Solution

We are using x

f(lfJ: Xl,

Y for X2, Z for

X3.

A=(! ~ II. Give the matrix of 2X2

Then

!}

+ 5xy + 6y2 + 3xz + yt + 5z2 + XZ + 2tz. Solution

f(x, y, z, t) = 2X2 + 2(l)yt 2zt and

+

+ 6y2 + 5z2 + Ot2 + 2(i)xy + 2(t)xz + Oxt + Oyz i

6

o i

t 0 5

V o :a

•

1

1

ORAL EXERCISES

Give the matrices of the following quadratic forms:

+

(a) 2X2 y2 - 4xy - 4yz (b) _2X2 4y2 6z 2 + 2xy 6xz 6yz (e) 4x 2 y2 - 8z 2 4xy - 4xz 8yz (d) 3x 2 - 3y2 - 5z 2 - 2xy - 4xz - 6yz (e) 3x 2 - y2 Z2 - xy xz + 3yz (f) 2X2 y2 - fuy - 3yz 6xz (g) 3xy - X2 - yz - 5xz (h) 4zx y2 zy Z2 3yx (i) x 2 - y2 + Z2 - 2xy 3xt 4yt (j) xy+xz+xt+3yz+5zt (k) 2xy 2xt - y2 yz - zx - t2 (l) X2 y2 Z2 t2 - 2xy + 2xz"- 2zt (m) (x - t)2 - (y - Z)2 + (2z - t)2 (n) 4(x y)2 - (2x ~ t)2 - x(y - z)

+

+

+

+

+

+ + + + + + + + + + + + + + + +

+

+

+ xt"

7. Equivalence of quadratic forms. ~et f(:lh, ••• -, xn) be a quadratic form so that f(Xl, . • . ,xn ) = X'AX where X is a column vector and A is a symmetric matrix. Write

I

(18)

y=

x = py, Yn

260

·MATRIOES AND QUADRATIO FORMS

[CRAP.x·

where P = (Pii) is a nonsingular matrix. Then J(:lh, •• • , z.)

= CPY)'.A.(PY)

= Y'(P'AP)Y ~ Y'BY = g(Yl, . • . ,Yta)

is a quadratic form in Yl, • . . , Yta with matrix (19) B = P'AP. Hence a linear tram/ormation with matrix P replaC68 a quadratic Jorm with matrix .A. by a quadratic Jorm 'With mat:ri:I: P'AP. We shall call two quadratic forms J(Xl, • • • , z.) and g(Xl, ••• , z.) equivalent if there exists a nonsingular linear traD.sformation (18) which replacesJ(xl, . . . ,xta ) by g(Yl, .•• ,Yta). We say that two' symmetric matrices .A. and B are CO'fI,(J'iouent if they are related as in formula (19). Then two quadratic JO'I'1Y/,8 are equivalent if and only iJ their matrice8 are congruent. It can be shown that .A. and B are congruent if and only if we can find a sequence of elementary transformations on the rows of .A. such ,that it, followed by the sequence of curresponding column transiQrmations, will carry .A. into B. We shall assume this as well as the following: Theorem 13. Every quadratic Jorm J ill equivalent to a diagonal quadraticJorm b1xl1 + ... + b,.xtal where.' bib. ••. br ~ 0, br+l -_.. . . = btJ ;", 0 and r i8 the rank oj the matrix oj J. In the study of quadratic forms with real coefficients we permit only linear transfo~tions with matrices of real elements. Let uS .then carry a real quadratic form J into an equivalent diagonal form b1xl 1 + ... + brXrl. Then the number of positive coefficients b1, • • • , br i8 the 8ame 1UJ matter what diagonal Jorm we obtain, and. we 'call this natural number the index of J. . It follows that two real quadratic JO'I'1Y/,8 are equivalent iJ and only iJ ·they have. the 8ame rank and index. ' . . A value of a real quadratic form is a real number J(Cl, ... , c,.).

SIIO.7]

EQUIVALENCE OF QUADRATIC FORMS

261

We call !(:Ih, '. . . , XII) a po8itive form if every value I(Cl, ••• , c,.) ~ 0, and call!(xl, . . . ~ x,.) a negative' form if -!(Xl, ••. , x,.) is positive. It can be shown that !(Xl, '. . . , XII) is po8itive i! and only i! its index i8 its rank. A quadratic form !(Xl, . . .., x,.) is said to be positive definite if !(Cl, . . . , c,.) > 0 unless Cl

= Ct = . . . = c,. = O.

This occurs when and only when n is the rank and index of !(Xl, . • • , x,.). When all the values of !(Xl, . . . , x,.) except !(O, 0, • • • ,0) are negative, we call I(Xl, • • • ,x,.) negative definite and this occurs when -!(Xl, . . . , x,.) is positive definite. Illu8trative Example

Compute the index of the quadratic form I(~,

1/,')

I!!!

43;1

+ 1/1 + 211 -

2zy

+ 10= -

4yz.

8oZw,Um The ma.trix of the quadratic form is

A

=

(-! =i

-0' .

We interchange the first and second rows and then the first and second columns to obtain

-~ -! -:). ( -2 5 2 Add the first row to the second and twice the first row to the third to obtain

(~

o

-! -!\. 3

-;J

Subtract the second row from the third to obtain (

- 001

-! -:). o

-5 '

262

MATRICES AND QUADRATIC FORMS

[ClIAP.x

The general theory then states that the corresp,onding ,column' trans.. formatione will carry this matrix into the matrix

~ ~),

o

-5

which is congruent to A. Then the rank of j(x, y, s) is three, its index is two, and j(x, y, s) is equivalent to the diagonal quadratic form x 2 - Sy2' - 5s2• EXERCISES

Use elementary transformations involving rational numbers oruy to find diagonal quadratic forms to which each of the following quadratic forms are equivalent. Give the rank and index in each case. (a) 6x 2 + 14y2 + Z2 + 2xy - 2xs - 6ys . (b) 2X2 + 9y2 + 6s2 + 8xy + 2xs + 8ys (c) 2x 2 + Sy2 + 16s2 + 4xy - 8xs (d) 2xy - 4yz (j) 4xy - 4ys - 6xz W~-_+~-~-~ W2~+yz+~ (h) 5:£2 lSy2 S2 16xy - 2ys

+ + + + 6y2 - 2s t2 W 5x 2 - 2xs + 4xt + y2 (i) lSx 2

2 -

2yt + 8xy - 2st 4ys + 2yt -' 2s 2 - 2st - t2

8. Orthogonal matrices. A matrix P is called an orthogonal matrix if PF:' =, I. Then the transpose of P is the inverse of P. It should be clear that the rows of an orthogonal matrix are unit veators which are pairwise orthogonal. Since PP' = I implies that P'P = I the columns of an orthogonal matrix are also pairwise orthogonal unit vectors. ' If P is an orthogonal matrix and D is a diagonal matrix whose diagonal elements are dl , • • • , dft, the jth column of P D is dj times the jth' column of P. Then the length of the jth column of P D is dj and the colum,ns of P D are still pairwise orthogonal vectors. Hbwever the rows of P D' wi:ll ,usually no longer be pairwise orthogonal. Theorem 14. Let1al, ••• , at be t pairwise orthogonal n-dimensional real vectors. Then t ;;! n and there exist n - t vectors at+l, . • • , aft such that vectors aI, • • • , aft fJ,1'11 pairwise orthogonal.

BEc.8]

ORTHOGONAL MATRICES

263

Corollary. Let al, . . . , an be real vector8 selected as in Theorem 14. Then the matrix P, whose ith row is 1 (20) is an orthogonal matrix. Theorem 15. The product of two orthogonal matrice8 is an orthogonal matrix. Rlustrative Examples I. Find an orthogonal matrix P whose first row is a scalar multiple of(1 -1 2}. Solution The equation te - Y 2B = 0 has te = Y = 1 and z = 0 as a solu1 O) is a vector orthogonal to (1 -1 2). A tion. Thus (1 y z) orthogonal to both (1 -1 2) and (1 1 O) vector (te satisfies the -equations

+

te-y+2z=0 te+y=O. We see that y = -te, 2:e + 2z = 0, z = y = -te, and the rows of the matrix

~ -~

\i

~)

-1

-1

are pairwise orthogonal. If we replace the rows by unit vectors which are scalar multiples of them, we obtain an orthogonal matrix

p=

1

-1

2

v'6

V6

V6

1

1

v'2

v'2

1

0

-1

-1

va va va which is a solution of our problem. II. Find an orthogonal matrix P whose first two roW8 are scalar multiples Of (2 1 -2), (1 2 2). Solution

We solve the equations 2:e+y-2z=O te+2y+2z=O.

264

MATIUCES AND QUADRATIC FORMS

+

Then &; 31/ ... 0, Z = -1/, 2z II "'" 1 is a. solution. The ro~ of

Z

-=

Z

= 21 so tha.t Z

..

[CRAP,

2, 11

=

x

-2,

G-21. -2) 122 'I

are pairwise orthogonal, and a solution-of our problem is the orthogonal matrix '

(I I -I)I':

P=l 1 1 -I

i

:EXERCISES ' 1. Find an orthogonal matrix whose first row is a. scaJa.r multiple of (a) (1 (b) (2'

-1) 1)

(0) (-4

(d) (1 (e) (1

0)

(f) (1

-1 1 2

CJJ) (-3

1) 0) 1)

(h) (2

(,1

1 0

(-1

1) 1)

1

4)

i. Find an orthogonal matrix whose first two rows are scalar multiples, respectively, of the rows of the following: (a) (2

\0

(b) (1

\2

-1 -2) 2-1 1 -1) -1

1

(1 02 1) -1 1 (d) (1 2 3) -1...:1 1 (0)

(e)

(1

o

(f)(224) -2 0 1 CJJ) (1 1 -1 01) 1 -1 0 (h) (' 1 2 -1 01) . -1 1 1

(,1 (-1 -2 1

1

1 1

,

2\ II

2-2\ 1 1J

9. Orthogonal reduction of a quadratic form. An orthogonal liMar trarurjormation· is a linear transformation X == PY, where P is an orthogonal matrix. Then the solved form. of this transformation is Y = P'X, and so is obtained from X == PY simply by the interchange of rows with colmims in the coefficient matrix P. If Xl and X" are any two (column) vectors, their inner product is XI'X". The transformed vectors are Y1 == P'XI , Y" = P'X" and YlY" = X1'PP'X" == XI'X". Thus an orthogonal li'n6Gf 'rom/ormation pr886M188 tM inMr proa'1.1d8

SlilC.

9]

ORTHOGONAL REDUCTION'

265

of aU pair8 of f)ftctor8. It therefore preserves the length of a vector as well as the angle between two vectors. H P is an orthogonal matrix, we have PP' - I, IPP'I - 1 7 IPls and IPI = ± 1. We shall call an orthogonal transformation a rotation of tue8 if IPI = 1, and shall say that a quadratic formf(3h, ••• ,x,,) i8 equivalent under a rotation of (U68 to g(Xt, . . • , 3:,.) if· f(xt, ... ,3:,.)

5&

X'AX == g(Yt, ••• ,y,,) = Y'BY

where B - 'p'AP, PP' - I, IPI - 1. Then we state the following: Theorem 16. . Every real quadratic fqrm

f(xt, . . . ,x,,) - X' AX is equivalent under a rotation of (U68 to a diagonal quadratic fqrm dtxt S + ... + AAs, where d t , ••• ,d,. are the characteristic root8,of the real symmetric matrix A. The matrix P is determined as in Sec. 5 by the property that the jth column Pi of P is a unit vector such that (A - dJI)Pi = O. The property IPI = 1 may be obtained by changing the sign of all elements in a column of P if necessary. In the case n - 3 we shall write

f(x, y, z) = axS + bys + CZS + 2rxy + 28XZ + 2tyz and the equations 'of the rotation of axes will be

+ AsY' + AaZ' JJ.tX' + PosY' + JJ.aZ' 'IItX' + 'IIsy' + 'IIaZ'.

x - AtX' (21)

Y Z-

The solved form of these equations is then x' = AtX , y' = AsX z' - AaX

(22)

Here Xls

+ JJ.tY + 'IItZ . + JJ.sY + 'IIaZ + JJ.aY + 'IIaZ.

+ Pls + 'illS = Aa s + JJ.a s + 'IIz s =

Aa s

+ JJ.as + 'II. s = 1

MATRIC'ES AND QU,ADRATIC FO'~,;~ns

266

and we sdlve the systems (23) (A - d,I)

00 ~

0,

d,I)G~ ~ 0,

(A -

IA -

where d1, d2 , d3 are the roots of: F(x) = -

[OHAP.)f

xII =

o.

fllustrative Example Find the diagonal quadratic form to which we may carry lex, y, z)

== 3x2 + 4y2 -

+ 4yz

Z2 - 12xy - 8xz

by a rotation of axes, and give the equations of the rotation of axes. Solution The matrix of lex, y, z) is

A

=

(-!

-~ -~).

-4

2

-1

Its characteristic function is

, F(x)

== -

3 - x '-6 -4 -6 4- x 2 -4 2 -1 - x

-x - 9

-6

4

4 - x -2

-2x+2

'0

1+x

2

and

+ 9) - 4( -2x + 2)] + (1 + x)[(x + 9)(x - 4) + 6( -2x + 2)i == -2(10x + 10) + (x + 1){x2 - 7x - 24) == (:c + 1)(x2 - 7:c - 44) = (x + 1)(x + 4)(x - 11). Then!(x, y, z) will go into -X'2 - 4y'2 + l1z'2 under the required,rotation of axes. We solve the system of equations 4A1 - 61'1 - 4"1= 0; -6A1 + 51'1 + 2"1 = 0, -4A1 + 21'1 = 0 with matrix A + I to obtain 1'1 = 2A1, -6A1 + 101'1 + 2"1 = 0, "1 = -2A1' Thus our first column

F(x)

==

-2[2(x

is a unit vector which is a scalar multiple of 1, 2, -2. equations with matrix A + 41 is 7A2 - 61'2 - 4"2 = -6A2

+ 81'2 + 2"2 =

-4A2

and -oA2 + 101'2 = 0, A2 = 21'2, "2 = 3A2 - 41'2 second column is a multiple of 2, 1, 2. Finally --BAa - 6pa - 4"a

= -6},,8 -

7113

+ 2"8 9'

-4A3

The sys,tem,of

+ 21'2 + 3"2 = =

0

21'2, $0 that our

+ 21'3 -

12"8 = 0,

267

ORTHOGON AL REDUCTION

Sl!Io.9]

(-41\. - 31'. - 2".) + (-61\. - 71'. + 2".) = -101\8 - 101'. = 0 and 1\. = - 1'., 61'. - 71'. + 2"8 = 0, 1'. = 2"8. The determinant

2

1 2

-2

100 2 -3 6 = -27, -2 6-3

12=

2

-2

1

and so we must change the sign of the elements in one column. We change the signs of the elements in the last column. The length of each column vector is V1 + 4 + 4 = 3, and the matrix of the rotation of axes is

i i t -i' I -t

i

I -I

P=

The equations of the rotation of axes in matrix form are X = PY, where X and Y are column vectors, and so the equations become Z

=

z' + 2y' 3

+ 28'

'

2z'

y=

+ y' 3

2s' '

s =

-2z'

+32y' -

s'

.

We have used the rows of P as coefficients. The column denominators in this exercise are all the same but the denominators in different columns in other problems will uSually be different. The columns of P are the coefficients in the solved form

z'

=

z-2y-2s 3 '

,_2z-y-2s 3 '

Y -

s'

=

2z - 2y - s' 3 '

and it may be preferable to give the answers in this form. .Note, finally, that if p-1AP = D the sum oj the diagonal elements ot A is the sum oj the diagonal elements oj D. For the negative of this sum is the coefficient of 3:"-1 in their common characteristic equation. This result is useful as a check on answers. The matrix P is not unique if the characteristic roots of A are not all distinct. When these roots are all distinct, P is uniq~e apart from a change of sign of any two of its columns. EXD.CISBS Find a rotation of axes which will carry J(:e, y, s) into a diagonal 'quadratic form g(:e', y', s') in each of the following cases. Check the results in the cases where answers are not given by showing that 'AP = PD, where P is the matrix of the rotation of axes, A is the matrix 'of J(:e, y, s), and D is the matrix of g(:e', y', s'). (a)

J=

2:e 2 + y2

.~:e'

(b)

J=

-

- 4:ey - 4ys. Am. g = Z'2 - 2y'2 + 4z'2; + y - 28, 3y' = :e + 2y + 28, 3s' = 2:e - 2y + s. 2z2 + 4y2 + 6s + 2:ey + 6zz + 6ya.

=

2z

2

268 (co)

.M AT RIC E 8 A l'i D QUA D RAT I C FOR.M 8

[CHAP. X

J = 43:2 + y2 - Sz' + 4zy - 43:z +Syz.

Ana. g = 53:'2 + 2y'2 - 1Q~'2; -3: + 2y + z, v'3O z' = 3: ..:... .2y .5z. (d) '1 = 33: 2 - 3y2 - 5z 2 - 23:Y - 6zz - 6yz. (e) J = 33:2 -t y2 + 22 - 23:Y + 23:z - 211z. Ana. g = 43:'2 + Z'2; V63:' == 2x - y + z, v'2 y' = y + z, Z' = -3: - Y + z. (J) J = 23: 2 - y2 + 2Z2 - 63:Y - 6yz. • (g) = .53:2 + 5y2 + 3z2 - 23:Y + 23:Z + 2yz .. , Ana. g = 53:'2 + 6y'2 + 2Z'2; 3:' = 3: + y + z, v'2 y' = 3: - y, v'6 Z' = 3: + y - 2z. (h) J == 23: 2 - Z2 + 2:J:Y - 4yz. (t.') J = X2 + 23:Y - 2xz - 4yz. ' Ana. g = 3y'2 - 2Z'2; V6 x' = 23: - Y + z, y' = x + y - Z, v'2 Z' = Y + z. ' :(3') J = 5:r;2.+ 2y2 + 2Z2 + 23:y - 23:z - 4yz. ~k) I::; 2X2'+ 2y2 - Z2 + 8:J:Y - 43:z - 4yz. Ana. g. = -2X'2 - 2y'2 + 7Z'2;. 33:' = 2:1) - Y + 2z, 3y' = -3: + 2y + 2z, 3z' = 2x + 2y'';;'" z. (1) J = 2;2 + y2 2Z2 + 23:Y - 2yz. (m) J = 123: 2 - 6y2 - 4z2 - 123:y + 12yz. Ana. g = 143:'2 - 12y'2; '\I'9i x' = 9x - 3y - z, VI4 y' - x + 2y + 3z, V26z' = x + 4y - 3z. (n) J = X2 + y2 +. 9z 2 + 163:Y + Syz :- Su. ' (0) J = 43: 2 + 6y2 + 4z2 - 43:z. Ana. g = 2X'2 + 6y'2 + 6Z'2; v'2 x' = 3: + 2, y' = x + y - z, v'6 z' = - 3: + 2y + z. (p) 23: 2 + 3y2 + 2z2 + 2yz + 23:Y.

+

vA = 23: + y, v'6 y' = [1;'

va

r

va

va

+

va

10. Factoriz~tion of a positive symmetric_matrix. Every positive symmetric Ill{l.trix A has a factorization

A

=

FF',

where the number of columns in the rectangular matrix F is the rank: r of A. A method of determining F called the, diagonal method of factorization is presente<;i in the' illustrative Example at the end of this section. ' We shall state {but not prove) the following: Theorem 11. Let A = FF', where F is an r-columned ,matrix and r is the rank 'of A. Then A ::;: GG' for an r-"Columned matrix G if and only if G = FU where U is an ' r-rowed ortho(Jonal m a t r i x . , , !fA = FF', the matrix B = F'F is an r-rowed symmetric matrix a:Q.d, by Theorem 2, there is an r-ro~ed orthogonal(

SIIc.l0]

269

FACTORIZATION

matrix V such that

(24)

V'BV = Do

is a diagonal matrix.

But if H = FV the matrix

=

Do

(25)

V'F'FV

= H'H.

It follows that if 'Y1, • • • ,'Y, are the columns of H, then (26)

'Y/'Yi

= 0,

'Yl'Yi

=~

(i;:06 j; i, j

= 1,

. . . , r).

Also VV' is the r-rowed identity matrix and

HH' = FVV'F' ~ FF' = A.

(27) I

We have obtained a factorization A = HH' s:uch that the columns of H are pairwise orthogonal. We may write (28)

.

A

= HH' = PDP', '

where P is an n-row~ orthogonal matrix whose first r columns are Oll = v'a,.-1 'Y and D is the n-rowed diagonal matrix whose :first r diagonal elements ~re d 1, • • • , d, 'and whose remaining diagonal elements are zero. For a1, • • • ,a, are pairwise orthogonal unit vectors, P exists, PDP' only involves the first r columns of P. Thus the numbers 'Yl'Y;, which are the sums of the squares of the elements in the columns of H, are the nonzero characteristic roots of A = FF' as well as the characteristic roots of F'F. The theory above states that to obtain the nonzero characteristic roots of an n-rowed symmetric mat~ A of rank r < tt, we may factor A = FF' and form the r-rowed syDim.etric nonsingular matrix F'F. Then the characteristic roots of F'F will be the nonzero characteristic roots of A. It also states that, if we find an orthogonal matrix V such as V'BV is a diagonal matrix, the columns orH = FV will be pairwise orthogonal and A = HH'. The procedure outlined above yields what is called an orthogonal factorization HH' of A. It is of practical value

270

MATRICES AND QUADRATIC FORMS

[CHAP. x

only in case r is considerably less than n,' and we shall provide no exercises to illustrate the process. nlustrative Example Factor the matrix

5 7 3 A _ ( 7 17 -3 - 3 -3 9 P 6 -6

-v

Remark8: ,If A is an n-rowed real positive symmetric matrix of rank r, there exists an n-rowed real square matrix B of rank r such that A = BB'. The matrix B can be taken to be· a matrix obtained from a triangular matrix by any desired permutation of its rows and columns. Then exactly r columns of B will be nonzerO' and these columns, in any desired order, form a matrix F such that A = FF'. In our present example and in all the exercises below there is a solution F whose elements are integers. We focus our attention on the diagonal elements of ,A and locate a row of 'a possible B with only one nonzero element by finding a corresponding diagonal element :1)1• . This will determine a 'column of B and we then pass to a row 'Of B with at most two nonzero elements and such that one of these is known. The corresponding diagonal element of A will determine the other. Sol'Ution Since a88 = 9" we take the third row of B to be (3 0 0 0). The third row of A = BB' is (3 -3 9 -6) and so the first row of B' is (1 -1 3 -2). We have thus determined ihe first column of B and may take

BB'

-(-l

-2

~~

7 17 -3 6

:I)

0

'1/ 0 '1.£

s 0 3 -3 9

-6

"

D~ V

-1 11

s

0

3 0 0 0

-~

-6

+

This yields 1 +:1)2 = 5 ~d :I) = 2. Then -1 2'1/ = 7 and 11 = 4, -2 + 2'1.£ = Oandu = 1. We next see that 17 = 1 + 16 +Slsothat ,s = .0, 4 + 1 + "I + WI = 5 whence" = w = 0 and A = FF' where

F

2) 1 ( = -1

4

3. 0 • -2 1

271

FACTORIZATION·

BEc.10]

The.similar results in the exercises below should be verified by com-

putingFF'. EXERCISES

Factor the following matrices:

2 -4) 2 13-11 -4 -11 17

(J) (29

4 -2 -2 17 1 6

Vi)

(a) ( 1

(b) (

(c)

c:

(
(.)

0 2 -2

J)

D

4o -3 -4 2 -3 -4 13 -14 2 -14 17 4

(

~

-2 -3

0 -2 1 -2 -2 8 -3 12

(4 6-2 -V

(l!) (

-!)

~

17-22

17 10 -13 -22 -13 17 3-3 6

=~

(I) (

6 11 -2 -2 -2 2 8 8-6

1 -1 -3

-1 -3 -2

4 -2 2

5 1 4

1 14 13

-2 2 2-1 -1 1 o• -2 0 6 -6 3

-.v ~D 21

-~o -~ 4 6

3 6 1

INDEX A

Abscissa, 78 Absolute value, 37, 82 Addition, of complex numbers, 79 of functions, 182 ' of integers, 35 of polynomials, 80 of rational numbers, 52 of real numbers, 61 of vectors, 187 Addition formulas, 195 Addition laws, 6 Algebra, fundamental theorem of, 143 Algebraic function, 184 Algebraic identities, 119 Amplitude, 82, 197 Angle, 188 between vectors, 196,244 Approximation sequence, 59 Approximations to roots, 174 Arithmetic, fundamental theorem of, 44 Arithmetic means, 125 Arithmetic progressions, 124 Associative law, 6 Augmented matrix, 228 Average term, 125 Axes, coordinate, 78 of imaginaries, 80 Axiomatic characterization of natural numbers, 4

B Base of a logarithm, 66 change of, 70 Binary quadratic form, 259 Binomial coefficient 0 ...., 22 Binomial equation, 197

Binomial series, 76, 117 Binomial theorem, 116 Bounds on roots, 162

C

0 ...., definition of, 22

Cancellation, 6, 133 Cartesian coordinates, 77 Center of a circle, 82 Change of base, 70 Characteristic of a logarithm, 68 Characteristic equation, 202, 254 Circular permutations, 30 Coefficient matrix, 228 Coefficients, of a polynomial, 87 in terms of roots, 145 Cofactor, 215 Columns of a matrix, 208 Combinations, 25 number of, 26 Common difference, 124 Common ratio, 128 Commutative law, 6 Completing the square, 136 Complex numbers 9, 79 Complex units, 82 Composite integer, 38 Conclusion, 2 Conditional equation, 132 Congruent matrices, 260 Conic section, 202. . Conjugate complex number, 81 Consistent equations, 230 Constant polynomial, 87 Constant term, 89 Convergent sequence, 61 Convergent series, 74 Coordinates, 77 Correspondence, 14 Cosine, 191 273

274

COLLEGE ALGEBRA

Counting, 14 Counting psira, 18 Cubes, list of, 28 Cubic polynomial, 88 Cyolio permutation, 30

Division law, 58 Divisor, 11,88 Divisors, listing of, 47 Double root, 144 E

D

Decimals, 59 repeating, 71 Degree, of an equation, 134

of a polynomial, 88 of a; product, 90 of a sum, 90 Degrees in an angle, 189 De Moivre's theorem, 197 Denominator, 51 Denumerable set, 17 Dependent equations, 281 Dependent :vllriable, 181 Depressed. equation, 142, 157 Derivatives, 108 multiple roots by, 149 Descartes' rule of signs, 171 Determinant, 214 , of a quadratio form, 202 of the transpose, 216 Determination of multiple f~

Elementary transformations, 210 Elements, of a matrix, 209 of a vector, 2fY1 Elimination and 'substitution, 229 End equations, 230 Equal polynomials, 88 Equality, of expansions, 221 of fractions, 51 . symbol of, 2 Equations, 132 of a ourve, 182 with given roots, 147 of a line, 199 Equivalent conditions, 2 Equivalent equations, 133 Equivalent quadratic forms, 260 Euclidean greatest-oommon-divisor . prooess, 40, 45, 98 Exact divisibility, 46 Expansions of a determinant, 215 Exponents, laws of, 12, 13, 57, 64

109 Diagonal element, 212 Diagonal matrix, 213 Diagonal method of factorization, 268

Dlagonal quadratic form, 206 Di1ferenoe, 10, 36 of vectors, 188, 207 Digit, 16 Diminishing roots of an equation, 151 Direction. n.umbers, 199 Dhmiminant, of a cubic, 158 of a quadratic, 137 Distinguishable permutations, 31 'Distributive law, 7 Divisibility, 38 Division, 10, 38 by lI8l'O, 11 Division algorithm, 39, 94

F Factor theorem, 140 Factorials, 21 Factorization, 11, 38, 97,

105,

148 Factors, 11, 12, 46, 95 of a matrix, 268 Field of all oomplex numbers, 79 Fields of oomplex numbers, 88 Finite sequenoe, 15 Forms, 106 Fractions, 50 Function concept, 180 Funda.men.tal counting principle, 19 Fundamental theorem, of algebra,

148 of arithmetic, 44

275

INDEX G General term, 16 Geoll).etrio addition of oomplex numbers,187 Geometric progressions, 128 Graphical addition, 50, 187 "Greater than" symbol, 8 Greatest common divisor, of integers,45 . of polynomials, 98 Greek alphabet, 3 Grouping of faotora, 11

H Harmonic progressions, 130 Highest common factor, 45 Homogeneous polynomial, 106 Homogeneous system, 228 Homer's method, 174 . Hypothesis, 2

I Identioally equal functions, 182 Identity matrix, 246 Imaginary axes, 80 Imaginary numbers, 81 Imaginary roots, 147 Independent variable, 181 Index of a quadratic form, 260 Induction, mathematical, 4:, 10, 36, 119 Inequalities, 9 Inequality sign, 8 Infinite products and series, 74 Infinite sequenoes, 17 Initial side of an 1mgIe, 188 Inner product, 196, 243 Integers, 4, 35 Integral function, 86 Integral operations, 37 Integral roots, 157 Interval, real, 162 Inverse of a matrix, 249 Irrational number, 70

Irreducibility of a quadratic, 138 Irreducible polynomial, 96 Isolation of rea:I roots, 168

L

Law, of division, 53 of subtraction, 36

Laws, of ad~n and multiplication, 6 of exponents, 12, 13, 57, 64 Leading coefficient, 88,. 134 Least pommon multiple, 45, 100 Least terms, fractions in, 54 Length of a vector, 1~ Less than, 8 Limit, 61 Linear combinations 42, 102, 207 Linear equations, 1~ Linear functions, 107 Linear polynomial, 88 Linear system, ~ Linear transformation, 252 Listing of divisors, 47 Lower bound, 162 M Mantissa,.68 MathematiOaI induction, 4, 10, 36,119 Matrix,208 of a linear system, 227 of & quadratio form, 258 in triangular form, 212 Matrix muitiplication, 245 Means, 125, 128, 130 Members of an equation, 133 Method, of frpoctions, 163 of interpolation, 118 of radioals, 163 of undetermined coeffioients, 122 Minor, 215 Multiple roots, 144 by derivatives, 149 , Multiple-valued funotion, 181, Multiplication, of functions, 182

276

COLL]!lGE ALGEBRA

Multiplication, of-integers, 35. of ma:trices, 245 of natural numbers, 5 of polynomials, 90 of real numbers, 61 Multiplication laws, 6 Multiplicity, of a factor, 109 of a root, 144

Order relation, 8 Ordered set, 9 Ordinate, 78 0rigin,78 Ort¥ogonal factorization, 269 Orthogonal matrices, 262 Orthogonal,reduction, 265 Orthogonal vectors, 196,243

N

P

Nat1:U'allogarithm, 68 Natural number system, 6 Natural numbers, 4 n-dimensional ,vector, 206 Negative of a vector, 188, 207 Negative exponent, 57 Nega.tive form, 261 Negative integer, 35 Negative rational number, 53 Negative real number, 49 Negative root!! by Homer's method, 174 ' Newton's method, 178 Nonhomogeneous linear system, 228 Nonsingula.r matrix, 249 Nontrivial solution, 228 NolllD., 84 of a vector, 186 Notation for a'sequence, 15 n-tuple,16 Number of pairs, 18 Number syStem, 1 of com,plex numbers, 79 of fractions, 52 of integers, 35 of n"tural numbers, 6 of real numbers, 60

P ...., definition of, 21 Parametric equations of a line, 201 Parallelogram law, 187 Partial fractions, 239 Permutations, 25 circular, 30. distinguishable, 31 Polar coordinates, 190 Polar form of a complex number; 197 Polynomials, 87 homogeneous, 106 in several variables, 105 Positive form, 261 ,Positive integer, 4 Positive rational number, 53 Positive real numbers, 50 Powers, 11 real,63 Prime integers, 38 Prime polynomial, 96 Primes, list of, 47 Principle of, mathematical induction, 4, 10, 36, 119 Products, 5, 35, 61, 74, 90, 182, 196, 243, 245 Progressions, 124, 128, 130 Properties of a determinant, 222

o Q Operations, integral, a7 on functions, 182 rational, 111 Order, of a determinant, 215 of a square ;matrix, 213 of ,symbols, 1

Quadrants, 79 Quadratic equations, 136 Quadratic forms, 107, 25!1 Quadratic formula, 36 Quotient, 11, 38, 53

277

INDEX Quotient, and remainder (see Division algorithm)

R Radian, 189 Radical, 63 Range of a variable, 180 Rank of a matrix, 225 Rational functions, 111 Rational numbers, 50 Rational operations, 86 Rational roots, 159 Real axis, 80 Real functions, 182 Real line, 49 Real number system, 60 Real numbers, 49 Real powers, 63 .Reciprocals of roots, 153 Rectangular coordinates, 77 Rectangular matrix, 208 Reduced equation, 153 Reducible polynomial, 95 Relation, transitive, 8 Relations between roots and coefficients, 145 Relatively prime, 43, 101 Remainder polynomial (see Division algorithm) Remainder ,theorem, 139 Right triangle with integral sides, 25 , Root, of a complex number, 198 of an equation, 132 Rotation of axes, 193, 265 Row of a matrix, 208

S Scalar matrix, 213 Scalar product, 185, 207 Sequence, 15, 17 Series, 74 Similar matrix, 254 Simple roots, 144 I Simplification of fractions, 56

Sine, 191 Single exponent process, 92 Solution, by determinants, 235 of an equation, 132 Square, completing the, 136 Square matrix, 213 Strong inequality, 9 Sturm's theorem, 172 Sub matrix, 213 Subtraction, 10 Sum, 5, 52, 61, 72, 74, 90, 187, 207 of terms in a progression, 125, 128 Summation theorem, 119 T Terminal side, 188 Ternary quadratic form, 258 Total number of combinations, 117 Transformations of polynomials, 151 Transitive relation, 8 Transpose, inverse of, 250 of a matrix, 222 Transposition process, 133 Triangular matrix, 213 Trigonometric functions, 191 Trivial factors, 38, 95 Trivial solution, 225 Two-point formula, 200 U

Undetermined coefficients, 122 Unique factorization, 44, 105 Unit, complex, 82 Unit point, 78 Unit vector, 186, 243 Upper bound, 162

v Value, absolute, 37, 82 of a function, 181

·278

COLLEGE: AL:GEBRA

Value, of a. polynomial, 89 Variable, 180 Vector, in .the plane, 185

tWlimensional, 206

w Weak inequality, 9

Z

Zero,3,6 Zero angle, 188 Zero function, 182 Zero point, 78 Z~ polynomial, 87 Zero vector, 243 Zeros of a. polyno~, 132