Basic Mathematics PDF

BASIC MATHEMATICS SERGE LANG Columbia University BASIC MATHEMATICS A TT ADDISON -WESLEY PUBLISHING COMPANY Reading, Massachusetts Menlo Park, California London Don Mills, Ontario This book is in the ADDISON-WESLEY SERIES IN INTRODUCTORY MATHEMATICS Consulting Editors: Gail S. Young Richa...

BASIC MATHEMATICS SERGE LANG Columbia University BASIC MATHEMATICS A TT ADDISON -WESLEY PUBLISHING COMPANY Reading, Massachusetts Menlo Park, California London Don Mills, Ontario This book is in the ADDISON-WESLEY SERIES IN INTRODUCTORY MATHEMATICS Consulting Editors: Gail S. Young Richard S. Pieters Cover photograph by courtesy of Spencer-Phillips and Green, Kentfield, California. Copyright © 1971 by Addison-Wesley Publishing Company Inc. Philippines copy­ right 1971 by Addison-Wesley Publishing Company, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written per­ mission of the publisher. Printed in the United States of America. Published simultaneously in Canada. Library of Congress Catalog Card No. 75-132055. To Jerry M y publishers, Addison-Wesley, have produced my books for these last eight years. I want it known how much I appreciate their extraordi­ nary performance at all levels. General editorial advice, specific editing of the manuscripts, and essentially flawless typesetting and proof sheets. It is very gratifying to have found such a com­ pany to deal with. New York, 1970 Acknowledgments I am grateful to Peter Lerch, Gene Mur- row, Dick Pieters, and Gail Young for their careful reading of the manuscript and their useful suggestions. I am also indebted to Howard Dolinsky, Bernard Duflos, and Arvin Levine for working out the answers to the exercises. S.L. Foreword The present book is intended as a text in basic mathematics. As such, it can have multiple use: for a one-year course in the high schools during the third or fourth year (if possible the third, so that calculus can be taken during the fourth year); for a complementary reference in earlier high school grades (elementary algebra and geometry are covered); for a one-semester course at the college level, to review or to get a firm foundation in the basic mathematics necessary to go ahead in calculus, linear algebra, or other topics. Years ago, the colleges used to give courses in “ college algebra” and other subjects which should have been covered in high school. More recently, such courses have been thought unnecessary, but some experiences I have had show that they are just as necessary as ever. What is happening is that the colleges are getting a wide variety of students from high schools, ranging from exceedingly well-prepared ones who have had a good first course in calculus, down to very poorly prepared ones. This latter group includes both adults who return to college after several years’ absence in order to improve their technical education, and students from the high schools who were not adequately taught. This is the reason why some material properly belonging to the high-school level must still be offered in the colleges. The topics in this book are covered in such a way as to bring out clearly all the important points which are used afterwards in higher mathematics. I think it is important not to separate arbitrarily in different courses the various topics which involve both algebra and geometry. Analytic geometry and vector geometry should be considered simultaneously with algebra and plane geometry, as natural continuations of these. I think it is much more valuable to go into these topics, especially vector geometry, rather than to go endlessly into more and more refined results concerning triangles or trigonometry, involving more and more complicated technique. A minimum of basic techniques must of course be acquired, but it is better to extend these techniques by applying them to new situations in which they become ix X FOREW ORD motivated, especially when the possible topics are as attractive as vector geometry. In fact, for many years college courses in physics and engineering have faced serious drawbacks in scheduling because they need simultaneously some calculus and also some vector geometry. It is very unfortunate that the most basic operations on vectors are introduced at present only in college. They should appear at least as early as the second year of high school. I cannot write here a text for elementary geometry (although to some extent the parts on intuitive geometry almost constitute such a text), but I hope that the present book will provide considerable impetus to lower considerably the level at which vectors are introduced. Within some foreseeable future, the topics covered in this book should in fact be the standard topics for the second year of high school, so that the third and fourth years can be devoted to calculus and linear algebra. If only preparatory material for calculus is needed, many portions of this book can be omitted, and attention should be directed to the rules of arithmetic, linear equations (Chapter 2), quadratic equations (Chapter 4), coordinates (the first three sections of Chapter 8), trigonometry (Chapter 11), some analytic geometry (Chapter 12), a simple discussion of functions (Chapter 13), and induction (Chapter 16, §1). The other parts of the book can be omitted. Of course, the more preparation a student has, the more easily he will go through more advanced topics. “ More preparation” , however, does not mean an accumulation of technical material in which the basic ideas of a subject are completely drowned. I am always disturbed at seeing endless chains of theorems, most of them of no interest, and without any stress on the main points. As a result, students do not remember the essential features of the subject. I am fully aware that because of the pruning I have done, many will accuse me of not going “ deeply enough” into some subjects. I am quite ready to confront them on that. Besides, as I prune some technical and inessential parts of one topic, I am able to include the essential parts of another topic which would not otherwise be covered. For instance, what better practice is there with negative numbers than to introduce at once coordinates in the plane as a pair of numbers, and then deal with the addition and subtraction of such pairs, componentwise? This introduction could be made as early as the fourth grade, using maps as a motivation. One could do roughly what I have done here in Chapter 8, §1, Chapter 9, §1, and the beginning of Chapter 9, §2 (addition of pairs of numbers, and the geometric interpretation in terms of a parallelogram). At such a level, one can then leave it at that. The same remark applies to the study of this book. The above-mentioned sections can be covered very early, at the same time that you study numbers  FOREW O RD xi and operations with numbers. They give a very nice geometric flavor to a slightly dry algebraic theory. Generally speaking, I hope to induce teachers to leave well enough alone, and to avoid torturing a topic to death. It is easier to advance in one topic by going ahead with the more elementary parts of another topic, where the first one is applied. The brain much prefers to work that way, rather than to concentrate on ugly technical formulas which are obviously unrelated to anything except artificial drilling. Of course, some rote drilling is necessary. The problem is how to strike a balance. Do not regard some lists of exercises as too short. Rather, realize that practice for some notion may come again later in conjunction with another notion. Thus practice with square roots comes not only in the section where they are defined, but also later when the notion of distance between points is discussed, and then in a context where it is more interesting to deal with them. The same principle applies throughout the book. The Interlude on logic and mathematical expression can be read also as an introduction to the book. Because of various examples I put there, and because we are already going through a Foreword, I have chosen to place it physically somewhat later. Take a look at it now, and go back to it whenever you feel the need for such general discussions. Mainly, I would like to make you feel more relaxed in your contact with mathematics than is usually the case. I want to stimulate thought, and do away with the general uptight feelings which people often have about math. If, for instance, you feel that any chapter gets too involved for you, then skip that part until you feel the need for it, and look at another part of the book. In many cases, you don’t necessarily need an earlier part to understand a later one. In most cases, the important thing is to have understood the basic concepts and definitions, to be at ease with the simpler computational aspects of these concepts, and then to go ahead with a more advanced topic. This advice also applies to the book as a whole. If you find that there is not enough material in this book to occupy you for a whole year, then start studying calculus or possibly linear algebra. The book deals with mathematics on both the manipulative (or computa­ tional) level and the theoretical level. You must realize that a mastery of mathematics involves both levels, although your tastes may direct you more strongly to one or the other, or both. Here again, you may wish to vary the emphasis which you place on them, according to your needs or your taste. Be warned that deficiency at either level can ultimately hinder you in your work. Independently of need, however, it should be a source of pleasure to understand why a mathematical result is true, i.e. to understand its proof as well as to understand how to use the result in concrete circumstances. x ii FO REW ORD Try to rely on yourself, and try to develop a trust in your own judgment. There is no “ right” way to do things. Tastes differ, and this book is not meant to suppress yours. It is meant to propose some basic mathematical topics, according to my taste. If I am successful, you will agree with my taste, or you will have developed your own. New York S.L. January 1971 Contents PART I ALGEBRA Chapter 1 Numbers 1 The integers.... 5 2 Rules for addition..... 8 3 Rules for multiplication.... 14 4 Even and odd integers; divisibility. 22 5 Rational numbers..... 26 6 Multiplicative inverses... 42 Chapter 2 Linear Equations 1 Equations in two unknowns. 53 2 Equations in three unknowns 57 Chapter 3 Real Numbers 1 Addition and multiplication. 61 2 Real numbers: positivity. 64 3 Powers and roots.... 70 4 Inequalities... 75 Chapter 4 Quadratic Equations. 83 Interlude On Logic and Mathematical Expressions 1 On reading books. 93 2 Logic.... 94 3 Sets and elements. 99 4 Notation 100 x iii XIV CONTENTS PART II INTUITIVE GEOMETRY Chapter 5 Distance and Angles 1 Distance 107 2 A n g le s............................. 110 3 The Pythagoras theorem. 120 Chapter 6 Isometries 1 Some standard mappings of the plane. 133 2 Isom etries.............................................. 143 3 Composition of isometries.... 150 4 Inverse of isom etries............................. 155 5 Characterization of isometries 163 6 Congruences.... 166 Chapter 7 Area and Applications 1 Area of a disc of radius r.... 173 2 Circumference of a circle of radius r 180 PART III COORDINATE GEOMETRY Chapter 8 Coordinates and Geometry 1 Coordinate systems 191 2 Distance between points. 197 3 Equation of a circle 203 4 Rational points on a circle 206 Chapter 9 Operations on Points 1 Dilations and reflections........................................ 213 2 Addition, subtraction, and the parallelogram law. 218 Chapter 10 Segments, Rays, and Lines 1 Segments 229 2 Rays.... 231 3 L i n e s.................................. 236 4 Ordinary equation for a line. 246 Chapter 11 Trigonometry 1 Radian measure 249 2 Sine and cosine. 252 3 The graphs. 264 4 The tangent 266  CONTENTS XV Addition formulas. 272 Rotations 277 Some Analytic Geometry The straight line again 281 The parabola 291 The ellipse.... 297 The hyperbola. 300 Rotation of hyperbolas 305 MISCELLANEOUS Functions Definition of a function 313 Polynomial functions. 318 Graphs of functions 330 Exponential function. 333 Logarithms 338 Mappings Definition.... 345 Formalism of mappings 351 Permutations 359 Complex Numbers The complex plane 375 Polar form. 380 Induction and Summations Induction 383 Summations 388 Geometric series 396 Determinants M a t r i c e s.................................. 401 Determinants of order 2. 406 Properties of 2 X 2 determinants 409 Determinants of order 3. 414 Properties of 3 X 3 determinants 418 Cramer’s Rule..... 424 Index 429  Part One ALGEBRA  In this part we develop systematically the rules for operations with num­ bers, relations among numbers, and properties of these operations and relations: addition, multiplication, inequalities, positivity, square roots, n-th roots. We find many of them, like commutativity and associativity, which recur frequently in mathematics and apply to other objects. They apply to complex numbers, but also to functions or mappings (in this case, commuta­ tivity does not hold in general and it is always an interesting problem to determine when it does hold). Even when we study geometry afterwards, the rules of algebra are still used, say to compute areas, lengths, etc., which associate numbers with geometric objects. Thus does algebra mix with geometry. The main point of this chapter is to condition you to have efficient reflexes in handling addition, multiplication, and division of numbers. There are many rules for these operations, and the extent to which we choose to assume some, and prove others from the assumed ones, is determined by several factors. We wish to assume those rules which are most basic, and assume enough of them so that the proofs of the others are simple. It also turns out that those which we do assume occur in many contexts in mathe­ matics, so that whenever we meet a situation where they arise, then we already have the training to apply them and use them. Both historical experience and personal experience have gone into the selection of these rules and the order of the list in which they are given. To some extent, you must trust that it is valuable to have fast reflexes when dealing with associativity, commutativity, distributivity, cross-multiplication, and the like, if you do not have the intuition yourself which makes such trust unnecessary. Further­ more, the long list of the rules governing the above operations should be taken in the spirit of a description of how numbers behave. It may be that you are already reasonably familiar with the operations between numbers. In that case, omit the first chapter entirely, and go right 3 4 ALG EB RA ahead to Chapter 2, or start with the geometry or with the study of coordinates in Chapter 7. The whole first part on algebra is much more dry than the rest of the book, and it is good to motivate this algebra through geometry. On the other hand, your brain should also have quick reflexes when faced with a simple problem involving two linear equations or a quadratic equation. Hence it is a good idea to have isolated these topics in special sections in the book for easy reference. In organizing the properties of numbers, I have found it best to look successively at the integers, rational numbers, and real numbers, at the cost of slight repetitions. There are several reasons for this. First, it is a good way of learning certain rules and their consequences in a special context (e.g. associativity and commutativity in the context of integers), and then observ­ ing that they hold in more general contexts. This sort of thing happens very frequently in mathematics. Second, the rational numbers provide a wide class of numbers which are used in computations, and the manipulation of fractions thus deserves special emphasis. Third, to follow the sequence integers- rational numbers-real numbers already plants in your mind a pattern which you will encounter again in mathematics. This pattern is related to the exten­ sion of one system of objects to a larger system, in which more equations can be solved than in the smaller system. For instance, the equation 2x = 3 can be solved in the rational numbers, but not in the integers. The equations x2 = 2 or 10x = 2 can be solved in the real numbers but not in the rational numbers. Similarly, the equations x2 = —1, or x2 = —2, or 10x = —3 can be solved in the complex numbers but not in the real numbers. It will be useful to you to have met the idea of extending mathematical systems at this very basic stage because it exhibits features in common with those in more advanced contexts. 1 Numbers gl. THE INTEGERS The most common numbers are those used for counting, namely the numbers 1, 2, 3, 4, , which are called the positive integers. Even for counting, we need at least one other number, namely, 0 (zero). For instance, we may wish to count the number of right answers you may get on a test for this course, out of a possible 100. If you get 100, then all your answers were correct. If you get 0, then no answer was correct. The positive integers and zero can be represented geometrically on a line, in a manner similar to a ruler or a measuring stick: 0 1 2 3 4 ") Fig. 1-1 For this we first have to select a unit of distance, say the inch, and then on the line we mark off the inches to the right as in the picture. For convenience, it is useful to have a name for the positive integers together with zero, and we shall call these the natural numbers. Thus 0 is a natural number, so is 2, and so is 124,521. The natural numbers can be used to measure distances, as with the ruler. By definition, the point represented by 0 is called the origin. The natural numbers can also be used to measure other things. For example, a thermometer is like a ruler which measures temperature. However, 5 6 NUM BERS [1 , §1] the thermometer shows us that we encounter other types of numbers besides the natural numbers, because there may be temperatures which may go below 0. Thus we encounter naturally what we shall call negative integers which we call m inus 1, m inus 2, m inus 3,... , and which we write as -1 , -2 , -3 , -4 ,.... We represent the negative integers on a line as being on the other side of 0 from the positive integers, like this: -4 -3 - 2 - 1 0 l 2 Fig. 1-2 The positive integers, negative integers, and zero all together are called the integers. Thus —9, 0, 10, —5 are all integers. If we view the line as a thermometer, on which a unit of temperature has been selected, say the degree Fahrenheit, then each integer represents a certain temperature. The negative integers represent temperatures below zero. Our discussion is already typical of many discussions which will occur in this course, concerning mathematical objects and their applicability to physical situations. In the present instance, we have the integers as mathe­ matical objects, which are essentially abstract quantities. We also have different applications for them, for instance measuring distance or tempera­ tures. These are of course not the only applications. Namely, we can use the integers to measure time. We take the origin 0 to represent the year of the birth of Christ. Then the positive integers represent years after the birth of Christ (called a d years), while the negative integers can be used to represent b c years. With this convention, we can say that the year —500 is the year 500 b c. Adding a positive number, say 7, to another number, means that we must move 7 units to the right of the other number. For instance, 5 + 7 = 12. Seven units to the right of 5 yields 12. On the thermometer, we would of course be moving upward instead of right. For instance, if the temperature at a given time is 5° and if it goes up by 7°, then the new temperature is 12°. Observe the very simple rule for addition with 0, namely N l. for any integer a. [1 , §1 ] TH E INTEGERS 7 What about adding negative numbers? Look at the thermometer again. Suppose the temperature at a given time is 10°, and the temperature drops by 15°. The new temperature is then —5°, and we can write 10 - 15 = - 5. Thus —5 is the result of subtracting 15 from 10, or of adding —15 to 10. In terms of points on a line, adding a negative number, say —3, to another number means that we must move 3 units to the left of this other number. For example, 5 + (-3 ) = 2 because starting with 5 and moving 3 units to the left yields 2. Similarly, 7 + (_ 3 ) = 4, and 3 + (-5 ) = -2. Note that we have 3 + (-3 ) = 0 or 5 + ( - 5 ) = 0. We can also write these equations in the form (-3 ) + 3 = 0 or (-5 ) + 5 = 0. For instance, if we start 3 units to the left of 0 and move 3 units to the right, we get 0. Thus, in general, we have the formulas (by assumption): N2. a + ( —a) = 0 and also —a + a = 0. In the representation of integers on the line, this means that a and —a lie on opposite sides of 0 on that line, as shown on the next picture: Fig. 1-3 Thus according to this representation we can now write 3 = -(-3 ) or 5 = -(-5 ). In these special cases, the pictures are: 8 NUM BERS [1 , §2] Remark. We use the name m inus a for —a rather than the words “ negative a” which have found some currency recently. I find the words “ negative a” confusing, because they suggest that —a is a negative number. This is not true unless a itself is positive. For instance, 3 = -(-3 ) is a positive number, but 3 is equal to —a, where a = —3, and a is a negative number. Because of the property a + ( - a ) = 0, one also calls —a the additive inverse of a. The sum and product of integers are also integers, and the next sections are devoted to a description of the rules governing addition and multiplication. §2. RULES FOR ADDITION Integers follow very simple rules for addition. These are: Com m utativity. I f a, b are integers, then a + b — 6 + a. For instance, we have 3 + 5 = 5 + 3 = 8, or in an example with negative numbers, we have _2 + 5 = 3 = 5 + (-2 ). Associativity. I f a, b, c are integers, then (a + b) + c = a + (b + c). [1 , §2] RULES FOR ADDITION 9 In view of this, it is unnecessary to use parentheses in such a simple context, and we write simply cz + 6 + c. For instance, (3 + 5) + 9 = 8 + 9 = 17, 3 + (5 + 9) = 3 + 14 = 17. We write simply 3 + 5 + 9 = 17. Associativity also holds with negative numbers. For example, ( —2 + 5) + 4 = 3 + 4 = 7, - 2 + (5 + 4) = - 2 + 9 = 7. Also, (2 + ( - 5 ) ) + ( - 3 ) = - 3 + ( - 3 ) = - 6 , 2 + ( - 5 + (-3 )) = 2 + (-8 ) = -6. The rules of addition mentioned above will not be proved, but we shall prove other rules from them. To begin with, note that: N3. I f a + b = 0, then b — —a and a = —b. To prove this, add —a to both sides of the equation a + b = 0. We get — cz + cz + 6 = — a + 0 = —c l. Since —a + a + 6 = 0 + 6 = 6, we find b = —a as desired. Similarly, we find a = —6. We could also conclude that —b = —( —a) = a. As a matter of convention, we shall write a —b instead of a + ( — 6 ). 10 NUM BERS [1 , §2] Thus a sum involving three terms may be written in many ways, as follows: (a - b) + c = (a + ( - 6 ) ) + c = a + ( —6 + c) by associativity = a + (c — b) by commutativity = (a + c) — b by associativity, and we can also write this sum as a — b + c = a + c — b, omitting the parentheses. Generally, in taking the sum of integers, we can take the sum in any order by applying associativity and commutativity repeatedly. As a special case of N3, for any integer a we have N4. a = —( —a). This is true because a + ( - a ) = 0, and we can apply N3 with b = —a. Remark that this formula is true whether a is positive, negative, or 0. If a is positive, then —a is negative. If a is negative, then —a is positive. In the geometric representation of numbers on the line, a and —a occur symmetrically on the line on opposite sides of 0. Of course, we can pile up minus signs and get other relationships, like -3 = - ( - ( - 3 ) ) , or 3 = -(-3 ) = -(-(-(-3 ))). Thus when we pile up the minus signs in front of a, we obtain a or —a alternatively. For the general formula with the appropriate notation, cf. Exercises 5 and 6 of §4. From our rules of operation we can now prove: For any integers a, b we have —(a + b) = —a + ( —6) [1 , §2] RULES FOR ADDITION 11 or, in other words, N5. —(a + b) = —a — b. Proof. Remember that if x, y are integers, then x = —y and y = —x mean that x + y = 0. Thus to prove our assertion, we must show that (a -|- b) + ( —a — 6) = 0. But this comes out immediately, namely, (a + b) + ( —a — b) = a + b — a — b by associativity = a —a + b —b by commutativity = 0 + 0 = 0. This proves our formula. Example. We have —(3 + 5) = - 3 - 5 = - 8 , —( —4 + 5) = - ( - 4 ) - 5 = 4 - 5 = - 1 , —(3 — 7) = —3 — ( - 7 ) = - 3 + 7 = 4. You should be very careful when you take the negative of a sum which involves itself in negative numbers, taking into account that —( —a) = a. The following rule concerning positive integers is so natural that you probably would not even think it worth while to take special notice of it. We still state it explicitly. I f a, b are positive integers, then a + b is also a positive integer. For instance, 17 and 45 are positive integers, and their sum, 62, is also a positive integer. We assume this rule concerning positivity. We shall see later that it also applies to positive real numbers. From it we can prove: If a, b are negative integers, then a + b is negative. 12 NUM BERS [1 , §2] Proof. We can write a — —n and b = —m, where m, n are positive. Therefore a + b = —n — m = —(ra + m), which shows that a + b is negative, because n + m is positive. Example. If we have the relationship between three numbers a + b = c, then we can derive other relationships between them. For instance, add —b to both sides of this equation. We get a + b — b = c — b, whence a + 0 = c — b, or in other words, a = c — b. Similarly, we conclude that b = c — a. For instance, if x + 3 = 5, then jc = 5 —3=2. If 4 — a = 3, then adding a to both sides yields 4 = 3 + a, and subtracting 3 from both sides yields 1 = a. [1 , §2] RULES FOR ADDITION 13 If - 2 - y = 5, then —l = y or y = —7. EXERCISES Justify each step, using commutativity and associativity in proving the following identities. 1. (a + b) + (c + d) = (a + d) + (b + c) 2. (a + b) + (c + d) = (a + c) + (b + d) 3. (a — b) + (c — d) = (a + c) + ( —b — d) 4. (a — 6) + (c — d) = (a + c) — (6 + eO 5. (a — 6) + (c — eO = (a — d) + (c — b) 6. ( cl — b) -|- (c — d) = —(6 -|- d) -|- ( cl -|- c) 7. (a — b) + (c — d) = —(b + d) — ( —a — c) 8. ((* + y) + z) + w = (x + z) + (y + w) 9. (x — y) - (z — w) = (z + w) — y — z 10. (x — y) — (z — w) = (x — z) + (w — y) 11. Show that —(a + b + c) = —a + ( —6) + ( —c). 12. Show that —(a — b — c) = —a + b + c. 13. Show that —(a — b) = b — a. Solve for x in the following equations. 14. - 2 + x = 4 15. 2 — x = 5 16. z — 3 = 7 17. —x + 4 = —1 18. 4 - x = 8 19. —5 — x = —2 20. —7 + x = - 1 0 21. - 3 + x = 4 14 NUM BERS [1 , §3] 22. Prove the cancellation law for addition: If a + 6 = a + c, then b = c. 23. Prove: If a + b = a, then 6 = 0. §3. RULES FOR MULTIPLICATION We can multiply integers, and the product of two integers is again an integer. We shall list the rules which apply to multiplication and to its relations with addition. We again have the rules of com m utativity and associativity: ab = ba and (ab)c = a(6c). We emphasize that these apply whether a, 6, c are negative, positive, or zero. Multiplication is also denoted by a dot. For instance 3 7 = 21, and (3 7) 4 = 21 4 = 84, 3 (7 - 4) = 3 28 = 84. For any integer a, the rules of multiplication by 1 and 0 are: N6. la = a and 0a = 0. Example. We have (2a) (36) = 2 (a (36)) = 2 (3a) 6 = (2. 3)a6 = 6a6. [1 , §3] RU LES FOR M ULTIPLICATIO N 15 In this example we have done something which is frequently useful, namely we have moved to one side all the explicit numbers like 2, 3 and put on the other side those numbers denoted by a letter like a or b. Using commutativity and associativity, we can prove similarly (5x)(7y) = 35xy or, with more factors, (2a)(36)(5x) = SOabx. We suggest that you carry out the proof of this equality completely, using associativity and commutativity for multiplication. Finally, we have the rule of distributivity, namely a{b + c) = ab + ac and also on the other side, (6 + c)a = ba + ca. These rules will not be proved, but will be used constantly. We shall, however, make some comments on them, and prove other rules from them. First observe that if we just assume distributivity on one side, and commutativity, then we can prove distributivity on the other side. Namely, assuming distributivity on the left, we have (6 + c)a = a(b + c) = ab + ac = ba + ca, which is the proof of distributivity on the right. Observe also that our rule Oa = 0 can be proved from the other rules concerning multiplication and the properties of addition. We carry out the proof as an example. We have Oa -I- a = Oa -1- la = (0 l)a = la = a. Thus Oa + a = a. Adding —a to both sides, we obtain Oa + a — a = a — a = 0. 16 NUM BERS [1 , §3] The left-hand side is simply 0a a — cl = Oil -|- 0 = Oa, so that we obtain Oa = 0, as desired. We can also prove N7. ( —l)a = —a. Proof. We have ( —l)a + a = ( —l)a + la = ( —1 + l)a = Oa = 0. By definition, ( —l)a + a = 0 means that ( —l)a = —a, as was to be shown. We have N8. —(ab) = ( —a)b. Proof. We must show that ( —a)b is the negative of ab. This amounts to showing that ab + ( —a)b = 0. But we have by distributivity ab + ( —a)b = (a + ( —a))b = 06 = 0, thus proving what we wanted. Similarly, we leave to the reader the proof that N9. —(a6) = a( —6). Example. We have —(3a) = ( —3)a = 3 ( - a ). Also, 4 (a — 56) = 4a — 206. Also, - 3 (5a - 76) = -1 5 a + 216. [1, 53] RULES FOR M ULTIPLICATION 17 In each of the above cases, you should indicate specifically each one of the rules we have used to derive the desired equality. Again, we emphasize that you should be especially careful when working with negative numbers and repeated minus signs. This is one of the most frequent sources of error when we work with multiplication and addition. Example. We have (-2 a ) (36) (4c) = ( - 2 ) - 3 4a6c = —24a6c. Similarly, ( —4*) (530 ( —3c) = ( —4)5(—3)xyc = 60xyc. Note that the product of two minus signs gives a plus sign. Example. We have ( —1) ( —1) = I* To see this, all we have to do is apply our rule —(a6) = ( —a)6 = a( —6). We find ( - l ) ( - l ) = - ( l ( - D ) = - ( - l ) = l. Example. More generally, for any integers a, 6 we have N10. ( —a) ( —6) = a6. We leave the proof as an exercise. From this we see that a product of two negative numbers is positive, because if a, 6 are positive and —a, —6 are therefore negative, then ( —a) ( —6) is the positive number a6. For instance, —3 and —5 are negative, but ( —3 )(—5) = —(3 (—5)) = —( —(3 5)) = 15. Example. A product of a negative number and a positive number is negative. For instance, —4 is negative, 7 is positive, and ( —4) 7 = - ( 4 - 7 ) = -2 8 , so that ( —4) 7 is negative. 18 NUM BERS [1 , §3] When we multiply a number with itself several times, it is convenient to use a notation to abbreviate this operation. Thus we write aa = a2, aaa = a3, aaaa = a4, and in general if n is a positive integer, an = aa a (the product is taken n times). We say that an is the n-th power of a. Thus a2 is the second power of a, and a5 is the fifth power of a. If m, n are positive integers, then N il. am+n = aman. This simply states that if we take the product of a with itself m + n times, then this amounts to taking the product of a with itself m times and multiplying this with the product of a with itself n times. Example a2a3 = (aa) (aaa) = a2+s = aaaaa = a5. Example (Ax)2 = 4x 4x = 4 4xx = 16x2. Example (7x)(2x)(5x) = 1 2 - 5xxx = 70s3. We have another rule for powers, namely N12. (am)n = amn. This means that if we take the product of a with itself m times, and then take the product of amwith itself n times, then we obtain the product of a with itself mn times. Example. We have ( a 3) 4 = a 12. [1 , §3] RU LES FOR M ULTIPLICATION 19 Example. We have (ab)n = anbn because (ab)n = abab -a ft (product of ab with itself n times) = aa abb b Example. We have (2a3)5 = 25(a3)5 = 32a15. Example. The population of a city is 300 thousand in 1930, and doubles every 20 years. What will be the population after 60 years? This is a case of applying powers. After 20 years, the population is 2 300 thousand. After 40 years, the population is 22 300 thousand. After 60 years, the population is 23 300 thousand, which is a correct answer. Of course, we can also say that the population will be 2 million 400 thousand. The following three formulas are used constantly. They are so important that they should be thoroughly memorized by reading them out loud and repeating them like a poemy to get an aural memory of them. (a + ft)2 = a2 + 2ab + ft2, (a — ft)2 = a2 — 2ab + ft2, (a + 6) (a — b) = a2 — ft2. Proofs. The proofs are carried out by applying repeatedly the rules for multiplication. We have: (a -f" b)2 = (a + 6)(a + b) = a(a — |—6) —J—6(a -1- b) = aa -|- ab -I- ba -I- bb = a2 + ab + ab + b2 = a2 + 2ab + ft2, 20 NUM BERS [1 , §3] which proves the first formula. (a — b)2 = (a — b) (a — b) = a (a — b) — b(a — b) = aa — ab — ba + bb = a2 — ab — ab + b2 = a2 - 2ab + ft2, which proves the second formula. (a + b)(a — b) = a(a — 6) + 6(a — b) = aa — ab + ba — bb = a2 — ab + ab — b2 = a2 - ft2, which proves the third formula. Example. We have (2 + 3x)2 = 22 + 2 2 Sx + (3x)2 = 4 + 12x + 9x2. Example. We have (3 - 4x)2 = 32 - 2 3 4x + (4x)2 = 9 — 24* + 16x2. Example. We have ( - 2 a + 5b)2 = 4a2 + 2(-2a)(56) + 2562 = 4a2 - 20a6 + 2562. Example. We have (4a - 6) (4a + 6) = (4a)2 - 36 = 16a2 - 36. We have discussed so far examples of products of two factors. Of course, we can take products of more factors using associativity. Example. Expand the expression (2x + l)(x - 2)(x + 5) as a sum of powers of x multiplied by integers. [1 , §3] RULES FOR M ULTIPLICATIO N 21 We first multiply the first two factors, and obtain (2* + l)(x - 2) = 2x(x - 2) + 1 (x - 2) = 2x2 - 4x + x - 2 = 2x2 - 3x - 2. We now multiply this last expression with x + 5 and obtain (2* + l)(x - 2)(x + 5) = (2x2 - 3x - 2)(x + 5) = (2x2 - 3x - 2)x + (2x2 - 3x - 2)5 = 2x3 — 3x2 — 2x + 10x2 — 15* — 10 = 2x3 + 7x2 - 17* - 10, which is the desired answer. EXERCISES 1. Express each of the following expressions in the form 2m3naTb’ , where m, n, r, s are positive integers. a) 8a263(27o4)(25a6) b) 16i3a2(6ai4)(a6)3 c) 32(2a6)3(16a2b5) (24i2a) d) 24a3(2a62)3(3a6)2 e) (3a6)2(27a36)(16a65) f) 32aib5a3b2(6ai3)4 2. Prove: (a + b)3 = a3 + 3a26 + 3o62 + b3, (a — b)3 = a3 — 3a2b + 3oi2 — b3. 3. Obtain expansions for (a + b)4 and (a — b)4 similar to the expansions for (a + b)3 and (a — b)3 of the preceding exercise. Expand the following expressions as sums of powers of x multiplied by integers. These are in fact called polynomials. You might want to read, or at least look at, the section on polynomials later in the book (Chapter 13, §2). 4. (2 - 4*)2 5. (1 - 2x)2 6. (2* + 5)2 7. (* - l ) 2 22 N UM BERS [ 1 , §4] 8. (x + 1)(* - 1) 9. (2* + 1)(* + 5) 10. (x2 + l)(x 2 — 1) 11. (1 + x3)(l - x3) 12. (x2 + l ) 2 13. (x2 - l ) 2 14. (x2 + 2)2 15. (x2 - 2)2 16. (x3 - 4)2 17. (x3 - 4)(x3 + 4) 18. (2x2 + l)(2x2 - 1) 19. ( - 2 + 3 * ) (- 2 - 3*) 20. (* + 1)(2* + 5)(* - 2) 21. (2x + 1)(1 - *)(3* + 2) 22. (3* - l)(2x + 1)(* + 4) 23. ( - 1 - * ) ( - 2 + *)(1 - 2x) 24. ( - 4 * + 1)(2 - *)(3 + x) 25. (1 - * ) ( 1 + *)(2 - *) 26. (x - l ) 2(3 - x) 27. (1 - *)2(2 - x) 28. (1 - 2jc) 2(3 + 4*) 29. (2x + 1)2(2 - 3*) 30. The population of a city in 1910 was 50,000, and it doubles every 10 years. What will it be (a) in 1970 (b) in 1990 (c) in 2,000? 31. The population of a city in 1905 was 100,000, and it doubles every 25 years. What will it be after (a) 50 years (b) 100 years (c) 150 years? 32. The population of a city was 200 thousand in 1915, and it triples every 50 years. What will be the population a) in the year 2215? b) in the year 2165? 33. The population of a city was 25,000 in 1870, and it triples every 40 years. What will it be a) in 1990? b) in 2030? §4. EVEN AND ODD INTEGERS; DIVISIBILITY We consider the positive integers 1, 2, 3, 4, 5 ,... , and we shall distinguish between two kinds of integers. We call 1, 3, 5, 7, 9, 11, 13,... the odd integers, and we call 2, 4, 6, 8, 10, 12, 14,... [1 , §4] EVEN AND ODD IN TEGERS; D IV ISIB IL ITY 23 the even integers. Thus the odd integers go up by 2 and the even integers go up by 2. The odd integers start with 1, and the even integers start with 2. Another way of describing an even integer is to say that it is a positive integer which can be written in the form 2n for some positive integer n. For instance, we can write 2 =21, 4 = 2-2, 6 = 2-3, 8 = 2-4, and so on. Similarly, an odd integer is an integer which differs from an even integer by 1, and thus can be written in the form 2m — 1 for some positive integer m. For instance, 1 = 2 1 - 1 , 3 = 2 2 - 1, 5 = 2 3 -1 , 7 = 2 * 4 — 1, 9 = 2 5 - 1, and so on. Note that we can also write an odd integer in the form 2n + 1 if we allow n to be a natural number, i.e., allowing n = 0. For instance, we have 1 = 2 0 + 1, 3 = 2 1 + 1, 5 = 2 2 + 1, 7 = 2 3 + 1, 9 = 2 4 + 1, and so on. Theorem 1. Let a, b be positive integers. I f a is even and b is even, then a + b is even. I f a is even and b is oddy then a + b is odd. I f a is odd and b is even, then a + b is odd. I f a is odd and b is odd, then a + b is even. Proof. We shall prove the second statement, and leave the others as exercises. Assume that a is even and that b is odd. Then we can write a = 2n and b = 2k + 1 24 NUM BERS [1 , §4] for some positive integer n and some natural number k. Then a, b — 27i “|“ 2k “I- 1 = 2(n + k) -(-1 = 2m + 1 (letting m = n + k). This proves that a + b is odd. Theorem 2 Let a be a positive integer. I f a is even, then a2 is even. If a is odd, then a2 is odd. Proof. Assume that a is even. This means that a = 2n for some positive integer n. Then a2 = 2n 2n = 2 (2m2) = 2m, where m = 2m2 is a positive integer. Thus a2 is even. Next, assume that a is odd, and write a = 2n + 1 for some natural number n. Then a2 = (2 n + l ) 2 = (2 m ) 2 + 2 ( 2 m )1 + l 2 = 4m2 + 4m + 1 = 2 ( 2 m2 + 2 m) + 1 = 2^ + 1, where k = 2 m2 + 2 m. Hence a2 is odd, thus proving our theorem. Corollary Let a be a positive integer. I f a2 is even, then a is even. If a2 is odd, then a is odd. Proof. This is really only a reformulation of the theorem, taking into account ordinary logic. If a2 is even, then a cannot be odd because the square of an odd number is odd. If a2 is odd, then a cannot be even because the square of an even number is even. We can generalize the property used to define an even integer. Let d be a positive integer and let n be an integer. We shall say that d divides n, or that m is divisible by d if we can write m = dk for some integer k. Thus an even integer is a positive integer which is divisible by 2. According to our definition, the number 9 is divisible by 3 because 9 = 3 -3. [1 , §4] EVEN AND ODD IN TE G E RS; D IV ISIB IL ITY 25 Also, 15 is divisible by 3 because 15 = 3 5. Also, —30 is divisible by 5 because - 3 0 = 5( —6). Note that every integer is divisible by 1, because we can always write n = 1 7i. Furthermore, every positive integer is divisible by itself. EXERCISES 1. Give the proofs for the cases of Theorem 1 which were not proved in the text. 2. Prove: If a is even and b is any positive integer, then ab is even. 3. Prove: If a is even, then a3 is even. 4. Prove: If a is odd, then a3 is odd. 5. Prove: If n is even, then ( —l ) n = 1. 6. Prove: If n is odd, then ( —l ) n = —1. 7. Prove: If m, n are odd, then the product mn is odd. Find the largest power of 2 which divides the following integers. 8. 16 9. 24 10. 32 11. 20 12. 50 13. 64 14. 100 15. 36 Find the largest power of 3 which divides the following integers. 16. 30 17. 27 18. 63 19. 99 20. 60 21. 50 22. 42 23. 45 26 N UM BERS [1 , §5] 24. Let a, b be integers. Define a = b (mod 5), which we read “ a is congruent to b m odulo 5” , to mean that a — b is divisible by 5. Prove: If a = b (mod 5) and x = y (mod 5), then a+ x= b+ y (mod 5) and ax = by (mod 5). 25. Let d be a positive integer. Let a, b be integers. Define a=b (mod d) to mean that a — b is divisible by d. Prove that if a = b (mod d) and x = y (mod d), then a + x= b+ y (mod d) and ax = by (mod d). 26. Assume that every positive integer can be written in one of the forms Sk9 Sk + 1, 3& + 2 for some integer k. Show that if the square of a positive integer is divisible by 3, then so is the integer. §5. RATIONAL NUMBERS By a rational num ber we shall mean simply an ordinary fraction, that is a quotient — also written m/n, n where m, n are integers and n 0. In taking such a quotient m/n, we emphasize that we cannot divide by 0, and thus we must always be sure that n t* 0. For instance, 1 2 3 5 4 3* 4’ 7 are rational numbers. Finite decimals also give us examples of rational numbers. For instance, 1 14 ^ 141 1 4 -10 and 141 - 1 0 0 ' Just as we did with the integers, we can represent the rational numbers on the line. For instance, J lies one-half of the way between 0 and 1, while [1 , §5] R A T IO N A L NUM BERS 27 § lies two-thirds of the way between 0 and 1, as shown on the following picture. The negative rational number —f lies on the opposite side of 0 at a distance | from 0. On the next picture, we have drawn —J and —f. -2 - 3 _ a o i 1 f ? 2 Fig. 1-6 There is no unique representation of a rational number as a quotient of two integers. For instance, we have 1 2 2 —4 We can interpret this geometrically on the line. If we cut up the segment between 0 and 1 into four equal pieces, and we take two-fourths of them, then this is the same as taking one-half of the segment. Picture: 0 i i - 3 i Fig. 1-7 We need a general rule to determine when two expressions of quotients of integers give the same rational numbers. We assume this rule without proof. It is stated as follows. Rule fo r cross-m ultiplying. Let m, n, r, s be integers and assume that n 5* 0 and s ^ 0. Then — = - if and only if ms = m. n s The name “ cross-multiplying” comes from our visualization of the rule in the following diagram: Example. We have 1 = 2 2 " 4 because 1 4 = 2 2. 28 NUM BERS [1 , §5] Also, we have ? = — 7 21 because 3 21 = 9 7 (both sides are equal to 63). We shall make no distinction between an integer m and the rational number m/1. Thus we write m m = m/1 = — With this convention, we see that every integer is also a rational number. For instance, 3 = 3/1 and —4 = —4/1. Observe the special case of cross-multiplying when one side is an integer. For instance: 2n 6 2n _ _ _ 30 _ — - - > — - 6, 2re-30, n - - j - 15 are all equivalent formulations of a relation involving n. Of course, cross-multiplying also works with negative numbers. For instance, -4 8 5 -1 0 because ( _ 4 ) ( —10) = 8 - 5 (both sides are equal to 40). Remark. For the moment, we are dealing with quotients of integers and describing how they behave. In the next section we shall deal with multi­ plicative inverses. There, you can see how the rule for cross-multiplication can in fact be proved from properties of such an inverse. Some people view this proof as the reason why cross-multiplication “ works”. However, in some contexts, one wants to define the multiplicative inverse by using the rule for cross-multiplication. This is the reason for emphasizing it here independently. Cancellation rule fo r fractions. Let a be a non-zero integer. Letm, n be integers, n 0. Then am _ m an n [1 , §5] R A T IO N A L NUM BERS 29 Proof. To test equality, we apply the rule for cross-multiplying. We must verify that ( 0. We shall list the basic properties of positivity from which others will be proved. POS 1. I f a, b are positive, so are the product ab and the sum a + b. POS 2. I f a is a real number, then either a is positive, or a = 0, or —a is positive, and these possibilities are mutually exclusive. If a number is not positive and not 0, then we say that this number is negative. By POS 2, if a is negative, then —a is positive. We know already that the number 1 is positive, but this could be proved from our two properties, and the basic rules for addition and multiplication. It may interest you to see the proof, which runs as follows and is very simple. By POS 2, we know that either 1 or —1 is positive. If 1 is not positive, then — 1 is positive. By POS 1, it must follow that ( —1)( —1) is positive. But this product is equal to 1. Consequently, it must be 1, which is positive and not —1. Using property POS 1, we can now conclude that 1 + 1 = 2 is positive, that 2 + 1 = 3 is positive, and so forth. Thus our calling 1, 2, 3,... the positive integers is compatible with our two rules POS 1 and POS 2. Other basic properties of positivity are easily proved from the two basic ones, and are left as exercises (Exercises 1, 2, 3), namely: I f a is positive and b is negative, then ab is negative. I f a is negative and b is negative, then ab is positive. I f a is positive, then 1 /a is positive. I f a is negative, then 1 /a is negative. One o f the properties of real numbers which we assume without proof is that every positive real number has a square root. This means: I f a > 0, then there exists a number b such that b2 = a. Because o f this, and Theorem 4, §5 of Chapter 1, we now see that a number whose square is 2 is irrational, but exists as a real number. It is a reasonable question to ask right away how many numbers there are whose squares are equal to a given positive number. For instance, what are all the real numbers x such that x2 = 2? [3, §2] RE A L nu m bers: p o s it iv it y 65 This is easily answered. There are precisely two such numbers. One is positive, and the other is negative. Let us prove this. Let b2 = 2, and let x be any real number such that x2 = 2 also. We have ^ x2 — b2 = 0. However, the left-hand side factors, and we find (jc + 6)(jc — b) = 0. Hence we must have jc + 6 = 0 or x — 6 = 0 so that x = —6 or x = 6. On the other hand, the square of —6 is equal to 2, because ( —ft)2 = ( —6) ( —6) = 62 = 2. Thus we have proved our assertion. Of the two numbers whose square is 2, we conclude from POS 2 that precisely one of them is positive. We now adopt a convention, in force throughout all of mathematics. We agree to call the square root o f 2 only the positive number 6 whose square is 2, This positive number will be denoted by V 2. Therefore the two numbers whose square is 2 are y /2 and —\ /2, and we have V2 > 0. Exactly the same arguments show that given any positive number a, there exist precisely two numbers whose square is a. If 6 is one of them, then —6 is the other. Just replace 2 by a in the preceding arguments. Again by convention, we let Va denote the unique positive number whose square is a. The other number whose square is a is therefore —y /a. We shall express this also by saying 66 R E A L NUM BERS [3, §2] that the solutions of the equation x2 = a are x = dby /a. We read this as “ jc equals plus or minus square root of a”. Another way of putting this is: If y ore numbers such that x2 = y 2, then x = y or x = —y. But we cannot conclude that x = y. Furthermore, for any number x , the number y /x 2 is ^ 0. Thus V ( —3)2 = \/9 = 3. There is a special notation for this. We call y/sP the absolute value of x, and denote it by |*| = V * 5. Thus we have |—3| = 3 and also |—5| = 5. Of course, for any positive number a, we have \a\ = a. Thus |3| = 3 and |5| = 5. We won’t work too much with absolute values in this book, and we do not want to overemphasize them here. Occasionally, we need the notion, and we need to know that the absolute value of —3 is 3. In that spirit, we give an example showing how to solve an equation with an absolute value ip it, just to drive the definition home, but not to belabor the point. Example. Find all values of x such that \x + 5| = 2. T o do this, we note that \x + 5| = 2 if and only if x + 5 = 2 or x + 5 = —2. Thus we have two possibilities, namely x = 2 — 5 = —3 and x = —5 — 2 = —7. This solves our problem. [3, §2] REAL num bers: p o s it iv it y 67 Observe that 1 _ V2 V2 ~ 2 ' This is because 2 = V 2V 2, and so our assertion is true because of cross-multiplication. It is a tradition in elementary schools to transform a quotient like 1 V 2 into another one in which the square root sign does not appear in the denominator. As far as we are concerned, doing this is not particularly useful in general. It may be useful in special cases, but neither more nor less than other manipulations with quotients, to be determined ad hoc as the need arises. Actually, in many cases it is useful to have the square root in the denominator. We shall give two examples of how to transform an expression involving square roots in the numerator or denominator. The manipulations of these examples will be based on the old rule (x + y) (x - y) = x2 - y 2. Example. Consider a quotient 2 + V 5 We wish to express it as a quotient where the denominator is a rational number. We multiply both numerator and denominator by 2 - Vs. This yields 3 (2 - V5) = 6 - 3V5 = 6 - 3V5 = _ 6 3v/g (2 + V5) (2 - V5) 22 - (V5)2 -1 Example. This example has the same notation as an actual case which arises in more advanced courses of calculus. Let x and h be numbers such that x and x + h are positive. We wish to write the quotient y j x + h — y /x h 68 R E A L NUM BERS [3, §2] in such a way that the square root signs occur only in the denominator. We multiply numerator and denominator by {y /x + h + y /x ). We obtain: (V s + h — y /x ) {y /x + h + y /x ) _ ( V x + ft)2 — ( V s )2 h {y /x + h + y /x ) h {y / x + h + y /x ) _ x + h —x h { V x + h + y /x ) = ________ h________ h {y / x + h + y /x ) _ 1________ V x -f-~h -f- y /x Thus we find finally: y /x + h — y /x ________ 1_______ ^ y /x + h + y /x In the first example, the procedure we have followed is called rational­ izing the denominator. In the second example, the procedure is called rationalizing the numerator. In a quotient involving square roots, rationalizing the numerator means that we transform this quotient into another one, equal to the first, but such that no square root sign appears in the numerator. Similarly, rationalizing the denominator means that we transform this quotient into another one, equal to the first, but such that no square root appears in the denominator. Both procedures are useful in practice. Square roots will be used when we discuss the Pythagoras Theorem, and the distance between points in Chapter 8, §2. You could very well look up these sections right now to see these applications, especially the section on distanceN EXERCISES 1. Prove: a) If a is a real number, then a2 is positive. b) If a is positive and b is negative, then ab is negative. c) If a is negative and b is negative, then ab is positive. [3, §2] REAL num bers: p o s it iv it y 69 2. Prove: If a is positive, then a~l is positive. 3. Prove: If a is negative, then a~l is negative. 4. Prove: If a, b are positive numbers, then la _ Va I*" vT 5. Prove that -----= - ( 1 + V 2). 1 - V2 6. Prove that the multiplicative inverse of 2 + y/S can be expressed in the form c + d y / 3, where c, d are rational numbers. 7. Prove that the multiplicative inverse of 3 + y /E can be expressed in the form c + dy/E, where c, d are rational numbers. 8. Let a, b be rational numbers. Prove that the multiplicative inverse of a + by/ 2 can be expressed in the form c + d y / 2, where c, d are rational numbers. 9. Same question as in Exercise 8, but replace y /2 by y/S. 10. Let jc, y, z, w be rational numbers. Show that a product (jc + y y /b ) (z + w y/5) can be expressed in the form c + d y /b , where c, d are rational numbers. 11. Generalize Exercise 10, replacing y /5 by y/a for any positive integer a. 12. Rationalize the numerator in the following expressions. N V 2* + 3 + 1 Vl + x - 3 a) --- j--- ---------- b) ----j---. Vx —h —V i V x — h + y fx c ) --------- h--------- -----------------------d ) ----------- *--------- NV x + h + V i V x + 2h — V x e ) ---------- i ---------- f ) -----------i ----------- 13. Find all possible numbers x such that a) |x — 1| = 2, b) |x| = 5, c) |x - 3| = 4, d) |x + 1| = 6, e) |x + 4| = 3, f) |x - 2| = 1. 14. Find all possible numbers x such that a) |2x — 1| = 3 , b) |3x + 1| = 2, c) |2x + 1| = 4, d) |3x - 1| = 1, e) |4x - 5| = 6. 70 R E A L NUM BERS [3, §3] 15. Rationalize the numerator in the following expressions. x Vx + i + V* - i j. V x — 3 + V x c) ------- — — d) , ---------p Vx + 1 — V X — 1 Vx — 3 —Vx 3 + Vx + y 16. Rationalize the denominator in each one of the cases of Exercise 15. 17. Prove that there is no real number x such that y / x — 1 = 3 + y /x. [Hint: Start by squaring both sides.] 18. If y /x — 1 = 3 — y /x , prove that x = 19. Determine in each of the following cases whether there exists a real number x satisfying the indicated relation, and if there is, determine this number. a) y /x — 2 = 3 + 2\fx b) y /x — 2 = 3 — 2 y /x c) y /x + 3 = 1 + y /x d) y /x + 3 = 1 — y /x e) y /x — 4 = 3 + y /x f) y /x — 4 = 3 — y /x 20. If a, b are two numbers, prove that \a — b\ = \b — a\. §3. POWERS AND ROOTS Let n be a positive integer and let a be a real number. As before, we let be the product of a with itself n times. The rule a m +n — a ma n holds as before, if m, n are positive integers. ;3, §3] POW ERS AND ROOTS 71 Let a be a positive number and let n be a positive integer. As part of the properties of real numbers, we assume, but do not prove, that there exists a unique positive real number r such that ^^ rn = a. This number r is called the n -th root of a, and is denoted by l/n ^ n/~ a or va. The n-th root generalizes the existence and uniqueness of the square root discussed in the preceding section. T h eo rem I. Let a, b be positive real numbers. Then (ab)l,n = al/nbl/n. Proof. Let r = a1/n and s = bl/n. This means that rn = a and sn = b. Therefore (rs)n = rnsn = ab. This means that rs is the n-th root of ab, and proves our theorem. The n-th root can be further generalized to fractional powers. Let a be a positive real number. We shall assume without proof the following property of numbers. Fractional p o w e r s. Let a be a positive number. To each rational number x we can associate a positive number denoted by ax, which is the n-th power of a when x is a positive integer n, the n-th root of a when x = l / n , and satisfying the following conditions: POW 1. For all rational numbers x, y we have a x+y = a xa y m POW 2. For all rational numbers x, y we have (ax)v = axy. POW 3. I f a , b are positive, then ( Exam ple. We have (V 2f = V 2 V 2 V 2 ^ = 2 V 2 = 23/2. Exam ple. We have /25 \ 3/2 _ 253/2 _ 125 \9 / ” 93/2 ” 27 We would also like to take powers with irrational exponents, i.e. we would like to define numbers like 2 V2. This is much more difficult, but it can be done in such a way that the two conditions POW 1 and POW 2 are satisfied. We shall not need this, and therefore shall postpone a systematic development for a more advanced course, although we shall make some further comments on the situation in the chapter on functions. However, we are led to make a final comment concerning the real numbers, as distinguished from the rational numbers. Note that the properties of addition, multiplication, and positivity hold for rational numbers. What distinguishes the real numbers from the rationals is the existence of more numbers, like square roots, ra-th roots, general exponents, etc. To make this “ etc.” precise is a more complicated under­ taking. We can ask: Is there a neat way (besides stating that the real numbers consist of all infinite decimals) of expressing a property of the reals which guarantees that any number which we want to exist intuitively can be proved to exist using just this property? The answer is yes, but belongs to a much more advanced course. Thus throughout this course and through­ out elementary calculus, whenever we wish a real number to exist so that we can carry out a certain discussion, our policy is to assume its existence and to postpone the proof to more advanced courses. 74 R E A L NUM BERS [3, §3] EXERCISES 1. Express each one of the following in the form 2k3marb8, where k, m, r, s are integers. a) - a3b ~ 42 5a ~ 2 b) 3_42 5a3b6 \ —A - ; 8 23 a4 9 3o364 16a~36~5 2a566 9b4a72 ~ 3 2. What integer is 811/4 equal to? 3. What integer is (\/2 )6 equal to? 4. Is (V 2 )5 an integer? 5. Is (\ /2 )~ 5 a rational number? Is ( V ^)5 a rational number? 6. In each case, the expression is equal to an integer. Which one? a) 161/4 b) 81/3 c) 93/2 d) 15/4 e) 84/3 f) 642/3 g) 253/2 7. Express each of the following expressions as a simple decimal, a) (.09)1/2 b) (.027)1/3 c) (.125)2/3 d) (1.21) 1/2 8. Express each of the following expressions as a quotient m /n , where m, n are integers > 0. >(ir »(9" i 3f 2 C) e r , * © ' 9. Solve each of the following equations for x. a) (x - 2)3 = 5 b) (x + 3)2 = 4 c) (x - 5) - 2 = 9 d) (x + 3)3 = 27 e) (2x - I ) " 3 = 27 f) (3x + 5)~4 = 64 [Warning: Be careful with possible minus signs when extracting roots.] [3, §4] IN EQ U ALITIES 75 §4. INEQUALITIES We recall that we write a > 0 if a is positive. If a, b are two real numbers, we shall write a > b instead of a — b > 0. We shall write a < 0 instead of —a > 0 ^ and also b < a instead of a > b. Exam ple. We have 3 > 2 because 3 — 2 = 1 > 0. We have - 1 > —2 because - 1 + 2 = 1 > 0. In the geometric representation of numbers on the line, the relation a > b means that a lies to the right of b. We see that —1 lies to the right of “ 2 in our example. -2 - 1 o l 2 Fig. 3-2 We shall write a ^ b to mean a is greater than or equal to b. Thus 3^2 and 3^3 are both true inequalities. Using only our two properties POS 1 and POS 2, we shall prove rules for dealing with inequalities. In what follows, we let a, 6, c be real numbers. IN 1. I f a > b and b > c, then a > c. IN 2. I f a > b and c > 0, then ac > be. IN 3. I f a > b and c < 0, then ac < be. 76 R E A L NUM BERS [3, §4] Rule IN 2 expresses the fact that an inequality which is multiplied by a positive number is preserved. Rule IN 3 tells us that if we multiply both sides of an inequality by a negative number, then the inequality gets reversed. For instance, we have the inequality 1 < 3. But “ 2 is negative, and if we multiply both sides by —2 we get “ 2 > —6. This is represented geometrically by the fact that —2 lies to the right of —6 on the line. Let us now prove the rules for inequalities. To prove IN 1, suppose that a > b and b > c. By definition, this means that a - b > 0 and b - c > 0. Using property POS 1, we conclude that a — b + b — c > 0. Canceling b gives us a — c > 0, which means that a > c, as was to be shown. T o prove IN 2, suppose that a > b and c > 0. By definition, a — 6 > 0. Hence using POS 1 concerning the product of positive numbers, we conclude that (a — b)c > 0. The left-hand side of this inequality is equal to ac — be by distributivity. Therefore ac — be > 0, which means that ac > 6c, thus proving IN 2. [3, §4] IN EQU ALITIES 77 We shall leave the proof of IN 3 as an exercise. Other properties which can easily be proved from the three basic ones will be left as exercises (see Exercises 2 through 5). They will be used constantly without further reference. In particular, we use some of them in the next examples. Exam ple. We wish to show that the inequality 2x - 4 > 5 is equivalent to an inequality of type x > a or x < b. Indeed, it is equivalent to 2x > 5 + 4 = 9, which is equivalent to 9 *> 2 Example. Suppose that x is a number such that (1) < 2- We wish to find equivalent conditions under which this is true, expressed by simpler inequalities like x > a or x < 6. Note that the quotient on the left makes no sense if x = 4. Thus it is natural to consider the two cases separately, x > 4 and x < 4. Suppose that x > 4. Then x — 4 > 0 and hence, in this case, our inequality (1) is equivalent to Sx + 5 < 2{x — 4) = 2x — 8. This in turn is equivalent to Sx — 2x < “ 8 — 5 or, in other words, x < “ 13. However, in our case x > 4, so that x < —13 is impossible. Hence there is no number x > 4 satisfying (1). Now assume that x < 4. Then x — 4 < 0 and x — 4 is negative. We multiply both sides of our inequality (1) by x — 4 and reverse the inequality. Thus inequality (1) is equivalent in the present case to 2) 3x + 5 > 2(x — 4) = 2x — 8. 78 RE A L NUMBERS [3, §4] Furthermore, this inequality is equivalent to (3) 3x - 2x > - 8 - 5 or, in other words, (4) jc > -1 3. However, in our case, x < 4. Thus in this case, we find that the numbers x such that x < 4 and x > —13 are precisely those satisfying inequality (1). This achieves what we wanted to do. Note that the preceding two inequalities holding simultaneously can be written in the form - 1 3 < x < 4. The set of numbers x satisfying such inequalities is called an interval. The numbers —13 and 4 are called the endpoints of the interval. We can represent the interval as in the following figure. -13 0 4 Fig. 3-3 Example. The set of numbers x such that 3 < x < 7 is an interval, shown in the next figure. Example. The set of numbers x such that 3 ^ x ^ 7 is also called an interval. In this case, we include the endpoints, 3 and 7, in the interval. The word “ interval” applies to both cases, whether or not we admit the endpoints. We represent the interval with the endpoints in the next figure. o 3 7 Fig. 3-5 In general, let a, b be numbers with a ^ 6. Any one o f the following sets of numbers is called an interval. The set of numbers x such that a < x < b, called an open interval. The set of numbers x such that a ^ x ^ 6, called a closed interval. The set of numbers x such that a ^ x < b. The set of numbers x such that a < x ^ b. The last two intervals are called half open or half closed. [3, §4] IN EQ U ALITIES 79 Exam ple. Again by convention, it is customary to say that the set of all numbers x such that x > 7 is an infinite interval. Similarly, the set of all numbers x such that x < —3 is an infinite interval. In general, if a is a number, the set of numbers x such that x > a is an infinite interval, and so is the set of numbers x such that x < a. Again by convention, we may wish to include the endpoint. For instance, the set of numbers x such that x ^ 7 is also called an infinite interval. The set of numbers x such that x ^ —3 is called an infinite interval. We illustrate some of these intervals in the next figure.... , r 0 7 Interval of numbers x > 7 1 0 7 Interval of numbers x ^ 7 1 1 1 1 —3 0 Interval of numbers x < —3 1 1 -3 0 Interval of numbers x ^ —3 Fig. 3-6 EXERCISES 1. Prove IN 3. 2. Prove: If 0 ^ a ^ b, if c ^ d, and c ^ 0, then ac < bd. 3. Prove: If a < 6 < 0, if c < d < 0, then ac > bd. 80 R E A L NUM BERS [3, §4] 4. a) If x < y and x > 0, prove that 1 /y < 1 /x. b) Prove a rule of cross-multiplication o f inequalities: If a, 6, c, d are numbers and b > 0, d > 0, and if a c b < d ’ prove that ad < be. Also prove the converse, that if ad < be, then a/b < c/d. 5. Prove: If a < b and c is any real number, then a + c < b + c. Also, a — c < b — c. Thus a number may be subtracted from each side of an inequality without changing the validity of the inequality. 6. Prove: If a < b and a > 0, then a2 < b2. More generally, prove successively that 3. t3 a < o , a4 < b\ a5 < ft5. Proceeding stepwise, we conclude that an < bn for every positive integer n. T o make this stepwise argument formal, one must state explicitly a property of integers which is called induction, and is discussed later in the book. 7. Prove: If 0 < a < 6, then a1/n < b1/n. [Hint: Use Exercise 6.] 8. Let a, 6, c, d be numbers and assume b > 0 and d > 0. Assume that :s, §4] INEQ UALITIES 81 a) Prove that a ^ a + c c b < b + d < d' (There are two inequalities to be proved here, the one on the left and the one on the right.) b) Let r be a number > 0. Prove that a a + rc c b < b + rd ^ d c) If 0 < r < s, prove that a + rc a + sc b + rd < b + sd 9. If 3jc — 1 > 0, prove that x > J. 10. If 4x + 5 < 0 , prove that x < —f. In each of the following cases, find the intervals of numbers x satisfying the stated inequalities. 11. 5x + 2 > - 3 12. - 2 x + 1 < 4 13. 3x + 2 < 1 14. —3x — 2 > 5 15. 3x - 1 < 4x + 5 16. 2x + 7 > - x + 3 h-1 CO to to 1 1 17. —3x - 1 > 5x 18. A X X + 3x - 1 —2x + 5 < i 19. < 1 20. x - 2 x + 3 2 - x 3 - x 21. > 2 22. 2x + 1 x —5 > 3 - 4x 23. > 2 24. 3x + 1 < 3 3x - 1 2x - 6 2-5. x 2 < 1 26. x 2 < 2 27. x2 < 3 28. x 2 < 4 29. x 2 > 1 30. x2 > 2 :i. x2 > 3 32. x 2 > 4 4 Quadratic Equations We know how to solve an equation like 3x - 2 = 0. In such an equation, x appears only in the first power. We shall now consider the next most difficult case, when x appears to the second power. We first deal with examples. Exam ple 1. Consider the equation (1) x2 — Sx + 1 = 0. We wish to solve for x, that is, determine all values for x which satisfy this equation. We shall ultimately derive a general formula for this. Before deriving the formula, we carry out on this special example the method used to derive the general formula. Solving our equation amounts to solving (2 ) x2 - 3x = - 1. We wish to add a number to both sides of this equation so that the left-hand side becomes a square, of the form (x — s)2. We know that (x — s)2 = x2 — 2sx + s2. Thus we need 2s = 3, or s = §. Consequently, adding (§)2 to each side of equation (2), we find 83 84 QUADRATIC EQUATIONS [4; The left-hand side has been adjusted so that it is a square, namely *23*+!=(*I) * and hence solving this equation amounts to solving We can now take the square root, and we find that x is a solution if and only i: Therefore finally we find two possible values for x, namely This is an abbreviation for the two values Exam ple 2. We wish to solve the equation (3) x2 + 2x + 2 = 0. We apply the same method as before. We must solve x 2 + 2x = —2. We add 1 to both sides, so that we are able to express the left-hand side in the form x2 + 2x + 1 = (x + l ) 2. Solving equation (3) is equivalent to solving (x + l )2 = “ 2 + 1 = - 1. But a negative real number cannot possibly be a square of a real number and we conclude that our equation does not have a solution in real numbers Exam ple 3. We wish to solve the equation (4) 2x2 - Sx - 5 = 0. This amounts to solving 2x2 - Sx = 5. This time, we see that x 2 is multiplied by 2. T o reduce our problem to on± similar to those already considered, we divide the whole equation by 2, ani  QUADRATIC EQUATIONS 85 solving (4) is equivalent to solving (5) We can now complete the square on the left as we did before. We need to find a number s such that x 2 — | x = x 2 — 2sx. This means that s = Adding s2 to both sides of (5), we find 2 _ 3 , _ 9 = 5 _ 9 = 49 * 2* 16 — 2 16 — 16 Expressing the left-hand side as a square, this is equivalent to 49 (- - » - 16 We can now solve for x, getting _ 3 = 49 X 4 \16 or equivalently, * - |± which is our answer. Although this answer is correct, it is sometimes con­ venient to watch for possible simplifications. In the present case, we note that ! 16 4’ and hence 3 7 * - 4 ± 4 Therefore 10 x = — ^ and ~7— x = — 4 = —1 4 4 are the two possible solutions of our equation. We are now ready to deal with the general case. T h eo rem. Let a, b, c be real numbers and a 0. The solutions of the quadratic equation 6) ax2 + bx + c = 0 86 QUADRATIC EQUATIONS are given by the formula provided that b2 — 4ac is positive, or 0. I f b2 — 4ac is negative, then the equation has no solution in the real numbers. Proof. Solving our equation amounts to solving ax2 + bx = —c. Dividing by a, we see that this is equivalent to solving 2. b c (7) X + - x ---------- a a T o complete the square on the left, we need x 2 + - x = x 2 + 2 sx, a and hence s = b/2a. We therefore add s2 = b2/4 a 2 to both sides of (7), and find the equivalent equation (*+è) ~~i + r 4a2 _ b — Aojc 4a2 If b2 — 4ac is negative, then the right-hand side b2 — 4ac 4 a2 is negative, and hence cannot be the square of a real number. Thus our equation has no real solution. If b2 — 4ac is positive, or 0, then we can take the square root, and we find y /b 2 — 4ac X + — = zh 2a 2a Solving for x now yields b \ /b 2 — 4ac X 20 ± 2a  QUADRATIC EQUATIONS 87 which can be rewritten as —b ± V b 2 — 4ac x = --------------------------- 2a This proves our theorem. Rem ark. If b2 — 4ac = 0, then we get precisely one solution for the quadratic equation, namely -b X ~ 2a ’ If b2 — 4ac > 0, then we get precisely two solutions, namely —6 + V b 2 — 4ac X ~ 2a and —b — y /b 2 — 4ac x = 2a The quadratic formula is so important that it should be memorized. Read it out loud like a poem, to get an aural memory of it: “ x equals minus b plus or minus square root of b squared minus four ac over two a.” Example 4. Solve the equation 3x2 — 2x + 1 = 0. We use the formula this time, and get - ( - 2 ) ± V (—2)2 - 4 - 3 x = 2 -3 4 ± V^ 8 In this case, we see that the expression b2 — 4ac under the square root sign is negative, and thus our equation has no solution in the real numbers. Example 5. Solve the equation 2x2 + 3x — 4 = 0. 88 QUADRATIC EQUATIONS Again, use the formula, to get - 3 ± \/9 - 4 - 2 - ( - 4 ) * = 2 ^ 2 ------------------------------------- = - 3 ± V 9 + 32 4 = -3 ± V4Ï 4 This is our answer, and we get the usual two values for x, namely -3 + V4Ï. - 3 - V41 x = ---------------- and x = ------------------ 4 4 R em ark. In the proof o f our theorem concerning the solutions of the quadra­ tic equation, we needed to operate with addition, multiplication, and square roots. If we knew that the real numbers could be extended to a larger system of numbers in which these operations were valid, including the possibility of taking square roots of negative real numbers, then our formula would be valid in this bigger system of numbers, and would again give the solutions of the equation in all cases. We shall see in the chapter on complex numbers how to get such a system. EXERCISES Solve the following equations. If there is no solution in the real numbers- say so, and give your reasons why. In each case, however, give the values for x which would solve the equation in a larger system of numbers where negative numbers have a “ square root”. Use the formula. 1. x2 + Sx - 2 = 0 2. jc2 — 3jc — 2 = 0 3. jc2 — 4jc + 5 = 0 4. jc2 — 4jc — 5 = 0 5. Sx2 + 2x — 1 = 0 6. 3x2 — 4x + 1 = 0 7. 3x2 + 3x — 4 = 0 8. - 2 x 2 - 5x = 7 9. - 2 x 2 - 5x = - 7 10. 4x2 + 5x = 6  Q U ADRATIC EQUATIONS 89 11. X2 - V 2 X + 1 = 0 12. X2 + V2 £ — 1 = 0 13. x2 + 3x - V 2 = 0 14. x2 - 3x - V5 = 0 15. x2 —3x + V 5 = 0 16. x2 - 2 x - V 3 = 0 You will solve more quadratic equations when you do the exercises in Chapter 12, finding the intersection of a straight line with a circle, parabola, ellipse, or hyperbola.  Interlude On Logic and Mathematical Expressions §1. ON READING BOOKS This part of the book can really be read at any time. We put it in the middle because that’s as good as any place to start reading a book. Very few books are meant to be read from beginning to end, and there are many ways of reading a book. One of them is to start in the middle, and go simultaneously backwards and forward, looking back for the definitions of any terms you don’t understand, while going ahead to see applications and motivation, which are very hard to put coherently in a systematic development. For instance, although we must do algebra first, it is quite appealing to look simultaneously at the geometry, in which we use algebraic tools to systematize our geometric intuition. In writing the book, the whole subject has to be organized in a totally ordered way, along lines and pages, which is not the way our brain works naturally. But it is unavoidable that some topics have to be placed before others, even though our brain would like to perceive them simultaneously. This simultaneity cannot be achieved in writing, which thus gives a distortion of the subject. It is clear, however, that I cannot substitute for you in perceiving various sections of this book together. You must do that yourself. The book can only help you, and must be organized so that any theorem or definition which you need can be easily found. Another way of reading this book is to start at the beginning, and then skip what you find obvious or skip what you find boring, while going ahead to further sections which appeal to you more. If you meet some term you don’t understand, or if you need some previous theorem to push through the logical development of that section, you can look back to the proper reference, which now becomes more appealing to you because you need it for something which you already find appealing. 93 94 LOGIC AND M ATH EM ATICAL EXPRESSION S [Int., §2] Finally, you may want to skim through the book rapidly from beginning to end, looking just at the statements of theorems, or at the discussions between theorems, to get an overall impression of the whole subject. Then you can go back to cover the material more systematically. Any of these ways is quite valid, and which one you follow depends on your taste. When you take a course, the material will usually be covered in the same order as the book, because that is the safest way to keep going logically. Don’t let that prevent you from experimenting with other ways. §2. LOGIC We always try to keep clearly in mind what we assume and what we prove. By a “ proof” we mean a sequence of statements each of which is either assumed, or follows from the preceding statements by a rule of deduction, which is itself assumed. These rules of deduction are essentially rules of common sense. We use “ I f... , then” sentences when one statement implies another. For instance, we use sentences like: (1) If 2x = 5, then x = - 2 This is a true statement, patterned after the gener