A BRIDGE TO HIGHER MATHEMATICS www.TechnicalBooksPDF.com TEXTBOOKS in MATHEMATICS Series Editors: Al Boggess and Ken Rosen PUBLISHED TITLES ABSTRACT ALGEBRA: A GENTLE INTRODUCTION Gary L Mullen and James A Sellers ABSTRACT ALGEBRA: AN INTERACTIVE APPROACH, SECOND EDITION William Paulsen ABSTRACT ALGEBRA: AN INQUIRY-BASED APPROACH Jonathan K Hodge, Steven Schlicker, and Ted Sundstrom ADVANCED LINEAR ALGEBRA Hugo Woerdeman APPLIED ABSTRACT ALGEBRA WITH MAPLE™ AND MATLAB®, THIRD EDITION Richard Klima, Neil Sigmon, and Ernest Stitzinger APPLIED DIFFERENTIAL EQUATIONS: THE PRIMARY COURSE Vladimir Dobrushkin COMPUTATIONAL MATHEMATICS: MODELS, METHODS, AND ANALYSIS WITH MATLAB® AND MPI, SECOND EDITION Robert E White DIFFERENTIAL EQUATIONS: THEORY, TECHNIQUE, AND PRACTICE, SECOND EDITION Steven G Krantz DIFFERENTIAL EQUATIONS: THEORY, TECHNIQUE, AND PRACTICE WITH BOUNDARY VALUE PROBLEMS Steven G Krantz DIFFERENTIAL EQUATIONS WITH APPLICATIONS AND HISTORICAL NOTES, THIRD EDITION George F 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CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2017 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S Government works Printed on acid-free paper Version Date: 20161102 International Standard Book Number-13: 978-1-4987-7525-0 (Paperback) This book contains information obtained from authentic and highly regarded sources Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint Except as permitted under U.S Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400 CCC is a not-for-profit organization that provides licenses and registration for a variety of users For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com www.TechnicalBooksPDF.com Contents Preface xi Elements of logic 1.1 True and false statements 1.2 Logical connectives and truth tables 1.3 Logical equivalence 1.4 Quantifiers 1.5 Exercises 1 12 Proofs: Structures and strategies 2.1 Axioms, theorems and proofs 2.2 Direct proof 2.3 Contrapositive proof 2.4 Proof by contradiction 2.5 Proofs of equivalent statements 2.6 Proof by cases 2.7 Existence proofs 2.8 Proof by counterexample 2.9 Proof by mathematical induction 2.10 Exercises 15 16 17 19 20 21 23 24 25 26 30 Elementary theory of sets 3.1 Axioms for set theory 3.2 Inclusion of sets 3.3 Union and intersection of sets 3.4 Complement, difference and symmetric difference of sets 3.5 Ordered pairs and the Cartesian product 3.6 Exercises 33 33 35 38 41 46 48 Functions 4.1 Definition and examples of functions 4.2 Direct image, inverse image 4.3 Restriction and extension of a function 4.4 One-to-one and onto functions 4.5 Composition and inverse functions 4.6 *Family of sets and the axiom of choice 4.7 Exercises 51 51 56 59 60 62 66 69 vii www.TechnicalBooksPDF.com Contents viii Relations 5.1 General relations and operations with relations 5.2 Equivalence relations and equivalence classes 5.3 Order relations 5.4 *More on ordered sets and Zorn’s lemma 5.5 Exercises 73 73 78 81 82 87 Axiomatic theory of positive integers 6.1 Peano axioms and addition 6.2 The natural order relation and subtraction 6.3 Multiplication and divisibility 6.4 Natural numbers 6.5 Other forms of induction 6.6 Exercises 91 91 94 97 99 100 103 Elementary number theory 7.1 Absolute value and divisibility of integers 7.2 Greatest common divisor and least common 7.3 Integers in base 10 and divisibility tests 7.4 Exercises multiple 105 105 107 112 113 Cardinality: Finite sets, infinite sets 8.1 Equipotent sets 8.2 Finite and infinite sets 8.3 Countable and uncountable sets 8.4 Exercises 117 117 119 120 123 Counting techniques and combinatorics 9.1 Counting principles 9.2 Pigeonhole principle and parity 9.3 Permutations and combinations 9.4 Recursive sequences and recurrence relations 9.5 Exercises 125 125 127 128 132 135 10 The 10.1 10.2 10.3 10.4 10.5 construction of integers and rationals Definition of integers and operations Order relation on integers Definition of rationals, operations and order Decimal representation of rational numbers Exercises 139 139 142 145 149 151 11 The 11.1 11.2 11.3 11.4 construction of real and complex numbers The Dedekind cuts approach The Cauchy sequences approach Decimal representation of real numbers Algebraic and transcendental numbers 153 153 162 170 172 www.TechnicalBooksPDF.com Contents ix 11.5 Complex numbers 11.6 The trigonometric form of a complex number 11.7 Exercises 172 175 178 Bibliography 181 Answers to select exercises 183 Index 201 www.TechnicalBooksPDF.com Answers to select exercises 189 Assume f : X → Y , g : Z → X and let y ∈ Y Since f ◦ g : Z → Y is surjective, there is z ∈ Z with f (g(z)) = y We found x = g(z) ∈ X with f (x) = y, so f is surjective Let f : {1, 2, 3} → {a, b}, f (1) = f (2) = a, f (3) = b and let g : {u, v}, g(u) = 1, g(v) = Then f is surjective, but f ◦ g : {u, v} → {a, b}, (f ◦ g)(u) = f (1) = a, (f ◦ g)(v) = f (2) = a, hence f ◦ g is not surjective Ex 4.89 For example, f |(−∞,3/2) and f |(3/2,∞) are injective y−1 √ g −1 : [0, ∞) → (−∞, 0], g −1 (y) = − y √ h−1 : (−∞, 0] → [2, ∞), h−1 (y) = + −y Ex 4.90 f −1 : R → R, f −1 (y) = k −1 : [−6, 3] → [−3, 1], k −1(y) = t−1 : [−1, 1] → [0, π/2], t−1 (y) = y 3, y 2, 0≤y≤3 −6 ≤ y ≤ π y + arcsin √ Ex 4.91 a f1 : (−∞, 3] → [−7, ∞), f1 (x) = x2 − 6x + has inverse f1−1 : [−7, ∞) → (−∞, 3], f1−1 (y) = − y + b f2 : [3, ∞) → [−7, ∞), f2 (x) = x2 − 6x + has inverse f2−1 : [−7, ∞) → [3, ∞), f2−1 (y) = + y + c f3 : (−∞, 0] ∪ [3, 6) → [−7, ∞), f3 (x) = x2 − 6x + has inverse f3−1 : [−7, ∞) → (−∞, 0] ∪ [3, 6), f3−1 (y) = √ − y + 7, y ≥ √ + y + 7, −7 ≤ y ≤ Ex 4.92 f ◦ g ◦ h : [1, ∞) → [0, ∞), (f ◦ g ◦ h)(x) = 4x − has range [0, 1] x2 −1 ± |2x − 3| Ex 4.96 Indeed, x ∈ lim sup En if and only if for any k ≥ there is n ≥ k with x ∈ En Also, x ∈ lim inf En if and only if there is k ≥ such that x ∈ En for all n ≥ k Ex 4.94 f (x) = Ex 4.97 a) lim sup En = lim inf En = (0, ∞) b) lim sup En = Z, lim inf En = {1} Ex 5.75 R = { 2, , 2, , 2, 12 , 3, , 3, 12 , 4, 12 }, R−1 = { 2, , 6, , 12, , 6, , 12, , 12, }, dom(R) = {2, 3, 4}, ran(R) = {2, 6, 12} A bridge to higher mathematics 190 Ex 5.76 R = { 1, , 2, , 3, , 4, }, R ◦ R = { 1, , 2, , 3, , 4, } Ex 5.78 a dom(R1 ) = (−∞, 3], ran(R1 ) = R y R1 b dom(R2 ) = [0, 1], ran(R2 ) = [−1, 1] y R2 x d dom(R4 ) = {−3}, ran(R4 ) = (−4, 4) R4 y x f dom(R6 ) = [−9, 9], ran(R6 ) = [−9, 9] x Answers to select exercises 191 y x R6 Ex 5.79 a R−1 = { 1, a , b, , 4, , y, x } b S −1 = S c T −1 = { x, y ∈ R × R : 3y − 4x2 = 9} d V −1 = { x, y ∈ R × R : x < 2y − 5} e W −1 = { x, y ∈ R × R : x(y + 3) = y} Ex 5.81 a S ◦ R = {{ x, y ∈ R × R : 2(2x − 1)2 + 3y = 5} b S ◦ R = { x, y ∈ R × R : y = sin x2 } Ex 5.83 a) R is transitive Indeed, if R ◦ R ⊆ R, then for x, z , z, y ∈ R we have x, y ∈ R ◦ R, hence R is transitive Assuming R transitive, given x, y ∈ R ◦ R, there is z with x, z , z, y ∈ R and we get x, y ∈ R b) R′ is symmetric Ex 5.84 a) R is reflexive, symmetric and transitive b) R is not reflexive, is symmetric and is not transitive c) R is reflexive and symmetric but not transitive Ex 5.86 a) R is only symmetric b) R is only symmetric c) R is reflexive, symmetric and transitive d) R is reflexive and transitive e) R is reflexive, symmetric and transitive Ex.5.88 Let R = S = { 1, , 2, , 3, } Ex 5.90 R = idX where idX : X → X, idX (x) = x Ex 5.91 R is reflexive since for all x we have | x + 3x = 4x If xRy, then yRx since y + 3x = 4(x + y) − (x + 3y), hence R is symmetric If xRy and yRz, then xRz since x + 3z = (x + 3y) + (y + 3z) − 4y, hence R is transitive The quotient set Z/R has four elements: [0], [1], [2], [3] A bridge to higher mathematics 192 Ex 5.92 Rg = { x, y ∈ [0, 2π] × [0, 2π] : x + y = π or x + y = 3π} The quotient set [0, 2π]/Rg can be identified with [π/2, 3π/2] Ex 5.93 Indeed, if [x]4 = [y]4 , then | x − y, hence | x − y and [x]2 = [y]2 Ex 5.94 We have [1]3 = [4]3 but [1]5 = [4]5 Ex 5.95 Indeed, if [x] = [y], then | x − y and | (2x + 3) − (2y + 3), hence [2x + 3] = [2y + 3] The function f is one-to-one because [2x + 3] = [2y + 3] implies [x] = [y] It is also onto since the equation [2x + 3] = [y] has solutions for all [y] ∈ Z5 Ex 5.97 The relation is reflexive since f ′ = f ′ for all f ∈ D It is symmetric since f ′ = g ′ implies g ′ = f ′ It is transitive since f ′ = g ′ and g ′ = h′ implies f ′ = h′ We have f ∼ g iff f − g is constant [f ] = {g ∈ D : g = f + c, c ∈ R} Ex 5.100 X/R is obtained from a semicircle with the endpoints identified Ex 5.101 Let xSy and ySz Then xRy, yRz and x = y, y = z Since R is transitive, we get xRz If x = z, then since R is antisymmetric, we get x = y, a contradiction It follows that xSz, so S is transitive Ex 5.102 a) Let X = {2, 3, 6}, let S = | and let R = S −1 Then 2, ∈ R ◦ S because 2, ∈ S, 6, ∈ R Also 3, ∈ R ◦ S but = 3, so R ◦ S is not antisymmetric b) Let X = {1, 2, 3}, let R = { 1, , 1, , 2, , 3, } and let S = { 1, , 2, , 2, , 3, }, which are partial orders on X, but R ∪ S = { 1, , 1, , 2, , 2, , 3, } is not transitive, so R ∪ S is not a partial order on X Ex 5.103 R is not antisymmetric Ex 5.104 There are six total orders Ex 5.106 The minimal elements are 0, n and n, where n ∈ N For example, let A = { 1, , 2, } Ex 5.107 (Y, |) is not a lattice since ∨ = is not in Y Ex 5.108 We have x1 , x2 ∨ y1 , y2 = max(x1 , y1 ), max(x2 , y2 ) and x1 , x2 ∧ y1 , y2 = min(x1 , y1 ), min(x2 , y2 ) The elements 1, , 2, are incomparable, hence is not a total order The dictionary order L is a total order, since given x1 , x2 , y1 , y2 ∈ Z×Z or R×R we have x1 , x2 L y1 , y2 or y1 , y2 L x1 , x2 In particular, 1, L 2, Ex 5.109 Let A ⊆ N × N be nonempty Since N with usual order is wellordered, the set of first components of the elements of A has a smallest element a Similarly, the set of second components has a smallest element b The element a, b has the property that a, b L x, y for all x, y ∈ A Answers to select exercises 193 Ex 5.110 The set { n, n : n ≤ 0} has no smallest element Ex 6.41 a) Assuming x ≤ y, there is u ≥ with y = x + u It follows that y + z = x + z + u, hence x + z ≤ y + z The other implication is proved using cancelation b) It follows from the law of trichotomy Ex 6.42 Let s : E → E, s(n) = n + Ex 6.43 Notice that D(n) = {n, n+1, n+2, }, D(s(n)) = {n+1, n+2, } = s(D(n)) For 5, let k be the smallest element of A Ex 6.44 a) xy+z = x · · · x = x · · · x · x · · · x or use induction on z y+z y z b) For z = we have xy·0 = = (xy )0 Assume xyk = (xy )k Then xy(k+1) = xyk+y = (xy )k · xy = (xy )k+1 Ex 6.45 We have u0 = 20 + = Assume uj = 2j + for ≤ j ≤ k Then uk+1 = 3uk − 2uk−1 = 3(2k + 1) − 2(2k−1 + 1) = 2k+1 + Ex 6.46 This is true for n = = · 20 Assume that any positive integer n ≤ k is a product of an odd integer and a power of Consider k + If this is odd, we are done If k + is even, then k + = · m, where m ≤ k Let m = p · 2q with p odd Then k + = p · 2q+1 Ex 6.47 For n = there is only one subset and · · 1/3! = Assume this is true for n = k and consider a set with k + elements B = {b1 , b2 , , bk , bk+1 } The subsets with three elements of B are of two kinds: subsets with three elements not containing bk+1 and subsets containing bk+1 The total number of subsets is k(k − 1)(k − 2) k(k − 1) k(k − 1)(k − 2) + 3k(k − 1) (k + 1)k(k − 1) + = = 3! 2! 3! 3! Ex 6.48 We first show that this is true for n = 8, 9, 10 Indeed, = + 5, = · 3, 10 = · Assuming k = 3a + 5b with a, b ≥ 0, we have k + = 3(a + 1) + 5b Ex 7.34 b We have |x| = |x − y + y| ≤ |x − y| + |y|, |y| = |y − x + x| ≤ |y − x| + |x|, hence |x| − |y| ≤ |x − y| and |x| − |y| ≥ −|x − y|, so |x| − |y| ≤ |x − y| Ex 7.36 e) Let x = zu, y = yv, z = tw Then x = tuw, y = tvw, so x/t = uw, y/t = vw 194 A bridge to higher mathematics Ex 7.38 We may assume n ≥ If p ≤ x for any prime, then there are only finitely many primes, a contradiction with Example 2.13 Ex 7.40 This is false: | + but ∤ and ∤ Ex 7.42 If d | n and d | n + 6, it follows that d | (n + 6) − n = 6, so d ∈ {1, 2, 3, 6} Ex 7.43 This is true for n = Let u, v ∈ Z with = au + bv Assuming gcd(a, bk ) = 1, let r, s ∈ Z with = ar + bk s Multiplying the two equalities, we get = a(aru + ubk s + rbv) + bk+1 vs, so gcd(a, bk+1 ) = Ex 7.44 Consider S = {ax + by + cz : x, y, z ∈ Z} The set S ∩ P is not empty and it has a smallest element d Using the same method as in the proof of Theorem 7.11 we get d = gcd(a, b, c) and d = as + bt + cu Ex 7.45 By completing the square, x4 = (2y +1)2 −24, so (2y +1)2 −x4 = 24 Factoring, we get (2y + − x2 )(2y + + x2 ) = 24 Considering the divisors of 24, the only systems with integer solutions are 2y + − x2 = 4, 2y + + x2 = and 2y + − x2 = −6, 2y + + x2 = −4 with solutions x = ±1, y = and x = ±1, y = −3 Ex 7.47 Replace 10 ≡ 1(mod 9) with 10 ≡ 1(mod 3) in the proof of Theorem 7.30 Ex 7.48 Let n = 2k + Then n2 − = (2k + 1)2 − = 4k + 4k = 4k(k + 1) is divisible by since for any integer k, the product k(k + 1) is even Ex 7.49 We have n3 −n = n(n−1)(n+1) Now n divided by gives remainders 0, or 2, so n = 3k or n = 3k + or n = 3k + In any case, the product n(n − 1)(n + 1) is a multiple of Since n(n − 1) is always even, it follows that n3 − n is a multiple of · = Ex 7.50 Considering the remainders 0, 1, 2, 3, 4, 5, of x when divided by 7, we see that x = is a solution Considering the remainders 0, 1, 2, 3, of x when divided by 5, there is no solution since x2 is congruent with 0, or modulo Ex 7.52 Suppose that 2n + = x2 and 3n + = y Then x2 + y ≡ 2(mod5) But x2 ≡ 0, 1, 4mod5, so we must have x2 ≡ y ≡ 1(mod5) Therefore | n On the other hand, since x must be odd, n must be even Then y must be odd Let x = 2a+1 and y = 2b+1 Now notice that n = y −x2 = 4(b2 +b−a2 −a) The number in parentheses must be even Hence | n Hence 40 | n Ex 8.30 Let |A| = n and f : A → A an onto function, so f (A) = A If f is not one-to-one, then there are a1 = a2 with f (a1 ) = f (a2 ), so |f (A)| ≤ n − 1, Answers to select exercises 195 a contradiction Conversely, if f : A → A is one-to-one, then |f (A)| = n, so f must be onto Ex 8.31 Using the axiom of choice, we can find a countable subset B = {a1 , a2 , a3 , } of A Consider f : B → B, f (an ) = an+1 and extend f to f˜ : A → A such that f˜(x) = x for x ∈ A \ B Then f˜ is one-to-one, but not onto Ex 8.32 Consider the function F : P(A) → {0, 1}A , F (B) = χB , where χB : A → {0, 1}, χB (x) = if x ∈ B if x ∈ A \ B is the characteristic function of B Then F is one-to-one since χB = χC implies B = C and onto since given f : A → {0, 1}, by taking D = f −1 (1) we have f = χD Ex 8.33 Notice that R is reflexive, antisymmetric and transitive, and so is an order relation Let A ⊆ Q be not empty Then φ(A) ⊆ N is not empty, so it has a smallest element n Then a = φ−1 (n) has the property that φ(a) ≤ φ(x) for all x ∈ A, hence (Q, R) is a well-ordered set Ex 8.34 Consider the countable sets An = {an1 , an2 , } for n ≥ To show that A = ∞ n=1 An is countable, we define a bijection f : P → A using the following pattern f (1) = a11 , f (2) = a12 , f (3) = a21 , f (4) = a31 , f (5) = a22 , f (6) = a13 , f (7) = a14 , f (8) = a23 , Ex 8.35 We know that N2 = N × N is countable By induction it is easy to prove that Nn is countable for any n ≥ Let Sn be the set of subsets of N with n elements We define fn : Sn → Nn by f ({a1 , a2 , , an }) = a1 , a2 , , an where a1 < a2 < · · · < an Then fn is injective, hence Sn is countable Then Pf (N) = {∅} ∪ countable ∞ Sn is a countable union of countable sets, and hence is n=1 Ex 8.36 If R \ Q is countable, then R = (R \ Q) ∪ Q would be countable, a contradiction Ex 8.37 Suppose we can find a bijection f : N → S Construct an element s ∈ S such that if f (n)n = sn = if f (n)n = Then f (n) = s for all n ∈ N, a contradiction 196 A bridge to higher mathematics Ex 8.39 We have |A| ≤ |B| ≤ |C| ≤ |A|, so A ≈ B ≈ C Ex 9.36 100 + 100 + 100 − 50 − 50 − 50 + 25 = 175 Ex 9.39 a) There are ⌊5000/3⌋ = 1666 integers divisible by 3, ⌊5000/4⌋ = 1250 divisible by 4, and 416 divisible by 12 The required number is 5000 − 1666 − 1250 + 416 = 2500 b) ⌊5000/4⌋ + ⌊5000/5⌋ + ⌊5000/6⌋ − ⌊5000/20⌋ − ⌊5000/24⌋ − ⌊5000/30⌋ + ⌊5000/120⌋ = 1250 + 1000 + 833 − 250 − 208 − 166 + 41 = 2500 Ex 9.40 P (26, 3)P (10, 3) = 11, 232, 000 license plates Ex 9.41 There are ten digits, but we exclude the digit at the beginning We get P (10, 3) − P (9, 2) + P (10, 2) − P (9, 1) + P (10, 1) − = 738 integers with distinct digits Ex 9.42 Consider a grid which divides the square into ·7 = 49 little squares By the pigeonhole principle, three points must be√in the same little square Since the diameter of the circumscribed circle is 2/7, a disc of radius 1/7 with the same center as the circumscribed circle will contain the little square in its interior Ex 9.43 By the pigeonhole principle, two of the integers must have difference ±1 Ex 9.45 No Ex 9.51 a = 2n+1 − Ex 9.53 In the identity (1+x)m+n = (1+x)m (1+x)n , identify the coefficients of xk Ex 9.55 a) an = (5 · 2n + 4(−1)n ) 3√ √ b) an = 2n/2−1 ( + + (−1)n (1 − 2)) c) an = − 21−n d) an = 2n+1 − n+2 Ex 9.56 an = 22n+1 Ex 9.57 The recurrence is a3 = 6, an = an−1 + n, n ≥ Ex 9.58 We have b1 = 2, b2 = 3, b3 = The recurrence relation is bn = bn−1 + bn−2 for n ≥ We get a translation of the Fibonacci sequence Ex 10.42 a − 12 = − 14 = −6 b a − b = c − c = ⇒ a = b c c − c − = − b ⇒ b = Ex 10.43 a From a, b ∼ a′ , b′ and c, d ∼ c′ , d′ we get a + b′ = a′ + b and c + d′ = d + c′ By adding these together, we obtain a + d + b′ + c′ = a′ + d′ + b + c Since a + d < b + c, there is u > with b + c = a + d + u We get a + d + b′ + c′ = a′ + d′ + a + d + u and by cancellation, b′ + c′ = a′ + d′ + u, so a′ + d′ < b′ + c′ Answers to select exercises 197 b Not well-defined since [0, 1] < [2, 1] since for example < but [0, 1] = [3, 4] and > Ex 10.45 a No; b No; c Yes; d Yes; e No; f Yes; g Yes Ex 10.47 Not well-defined: [0, 1] = [0, 2] but − = − b−a Ex 10.48 Let a < b be two rational numbers Then a < a + < b for all n n ≥ 7 Ex 10.52 = + + + + + + + · · · = 0.1637 in base 21 9 9 9 Notice that the period starts later since 21 = · has a factor of Ex 11.79 Indeed, ∈ C, < but ∈ / C Ex 11.80 a By definition, C is positive if there is r > with r ∈ C It follows that ∈ C Conversely, if ∈ C, then there is r > with r ∈ C b There is r > with r ∈ C \ D and there is s < with s ∈ / D Since for all x ∈ D we have x < s < r, it follows that D ⊂ C, i.e., D ≺ C c Let r < with r ∈ / C Then r ∈ / D and D is negative d If C ≺ ˆ 0, there is r ∈ ˆ0 such that r ∈ / C It follows that C is negative Conversely, if C is negative, then there is r < with r ∈ / C It follows that x < r for all x ∈ C, hence C ≺ ˆ0 Ex 11.81 For r = or s = the equality is obvious Assume first that r, s > and let x ∈ rˆsˆ By definition there are < c < r and < d < s with x ≤ c · d It follows that x < rs, hence x ∈ rs For the other inclusion, given x ∈ rs we must find < a < r and < b < s with x ≤ a · b Since x < rs, we get x/s < r Let < a < r with x/s ≤ a Since x/a < s, we can find < b < s with x/a < b We get x ≤ a · b The other cases can be reduced to the previous case Ex 11.82 Since x · x < for x > 0, we get D · D ˆ2 For the other inequality, let x ∈ ˆ It suffices to take < x < If x2 < 2, we can write x = x · with x, ∈ D Assume x2 > Choose y, z ∈ D such that x ≤ yz This shows that x ∈ D · D, hence ˆ D · D Ex 11.85 If x = a/b with x3 = 7, we get a3 = 7b3 We may assume that gcd(a, b) = Since | a3 it follows that | a Replacing a by 7c we get b3 = 49c3 , in particular | b, a√contradiction The real number denoted is given by the cut C = {x ∈ Q : x3 < 7} √ √ √ √ √ √ Ex 11.88 Since ( 3+ 2)2 = 5+2 it follows that 2+ 3− + = Ex 11.89 If L1 = L2 are two limits, let ε < |L1 − L2 |/2 By the definition of limit, there is N1 ≥ such that |xn − L1 | < ε for all n ≥ N1 Also, there is N2 ≥ such that |xn − L2 | < ε for all n ≥ N2 For n ≥ N = max(N1 , N2 ), we have xn ∈ (L1 − ε, L1 + ε) ∩ (L2 − ε, L2 + ε) = ∅, a contradiction A bridge to higher mathematics 198 Let xn → L Given ε > 0, there is N ≥ such that |xn − L| < ε/2 for all n ≥ N For m, n ≥ N we get |xn − xm | ≤ |xn − L| + |xm − L| < ε/2 + ε/2 = ε, hence (xn ) is Cauchy Ex 11.90 The sequences are partial sums of convergent series, and therefore are convergent and Cauchy √ Ex 11.92 The number − is a root of x2 − 2x − = The number √ √ − is a root of x4 − 14x2 + = Ex 11.93 Assume α is a root of an xn + an−1 xn−1 + · · · + a1 x + a0 = Then kα is a root of the polynomial an xn + kan−1 xn−1 + · · · + k n−1 a1 + k n a0 = If α = 0, then 1/α is a root of an + an−1 x + · · · + a1 xn−1 + a0 xn = Ex 11.95 We have i4k = 1, i4k+1 = i, i4k+2 = −1 and i4k+3 = −i Ex 11.98 Each polynomial p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 with ∈ Z determines an element in Zn+1 , n ≥ Since Zn+1 is countable, there are countably many equations p(x) = 0, each having deg p roots z1 + z2 z2 + z3 z3 + z1 Ex 11.99 The midpoints of the sides are , , The me2 2 dians intersect at a point dividing one median in the ratio : Its coordinate z1 z2 + z3 is + · = (z1 + z2 + z3 ) 3 Ex 11.100 One square has the other two vertices i(w −z)+z and i(w −z)+w Another square has −i(w − z) + z and −i(w − z) + w For the last one, z and w determine a diagonal, so the other vertices are 12 (i(w − z) + z + w) and (−i(w − z) + z + w) √ √ 1− 1+ Ex 11.101 a) x1 = 2i, x2 = −i b) x1 = (1 + i), x2 = (1 + i) 2 Ex 11.102 The set A = {ni : n ∈ Z} is bounded above by since ni ≺ for all n, but there is no least upper bound t t t Ex 11.104 Write z = cos t + i sin t Then + z = cos (cos + i sin ) and 2 t t t − z = sin (sin + i cos ) It follows that |1 + z|2 + |1 − z|2 = 2 7π 7π Ex 11.105 We have w = 2(cos + i sin ) The roots are zk = 4 √ 2kπ + 7π/4 2kπ + 7π/4 2(cos + i sin ) for k = 0, 1, 2, 4 Ex 11.106 Let ω = cos π3 +i sin π3 Then z3 = (z2 −z1 )ω+z1 and ω +ω+1 = A tedious computation shows that z12 + z22 + z32 = z1 z2 + z1 z3 + z2 z3 Answers to select exercises 199 Ex 11.107 The equation z n − = can be factored as (z − 1)(z n−1 + z n−2 + · · · + z + 1) = 2π k We may assume z0 = and z1 = cos 2π n + i sin n Then zk = z1 for ≤ k ≤ n − and hence z0 + z1 + · · · + zn−1 = One can also use the relations between roots and coefficients π π Ex 11.108 Since z + = 1, we get z − z + = and z = cos ± sin It z 3 nπ n follows that z + n = cos z √ Ex 11.109 Since − i = 2(cos π4 − i sin π4 ), using De Moivre’s formula we get √ −20π −20π (1 − i)−20 = ( 2)−20 (cos − i sin ) = −2−10 4 Similarly, √ π π + i = 2(cos + i sin ), 4 so √ 75π 75π (1 + i)75 = ( 2)75 (cos + i sin ) = 237 (−1 + i) 4 Ex 11.110 Using the binomial formula, (cos t + i sin t)3 = cos3 t + 3i cos2 t sin t − cos t sin2 t − i sin3 t On the other hand, (cos t + i sin t)3 = cos 3t + i sin 3t Identifying the real and imaginary parts, we get cos 3t = cos3 t − cos t sin2 t and sin 3t = cos2 t sin t − sin3 t Ex 11.111 Write z = |z|(cos t + i sin t) Then i sin t = cos t z |z| + = cos t + i sin t + cos t − |z| z Ex 11.113 We have z, w = z, 0, + w, 0, i = z + wj Index absolute value, 17, 105, 149, 160, 167 addition of positive integers, 92 algebraic number, 172 algebraically closed field, 172 antecedent, antisymmetric, 81 Archimedean property, 95 argument of z, 175 at most countable set, 120 axiom, 15, 16 axiom of choice, 68 base, 100 biconditional, bijective, 63 binomial formula, 131 cancelation law, 94 Cantor’s diagonal argument, 123 Cantor–Bernstein theorem, 119 cardinality, 117 Cartesian product, 46 Cartesian product of a family of sets, 67 Catalan numbers, 137 Cauchy sequence, 163, 168 Cayley number, 177 ceiling function, 54 chain, 85 characteristic equation, 133 characteristic function, 55 codomain, 52 coextension, 59 combination, 129 common divisor, 107 common multiple, 107 comparable elements, 81 complement, 41 complete induction, 26 complete ordered field, 161 complex number, 172 composition, 62, 77 conclusion, conditional, congruence modulo n, 79 conjecture, 16 conjunction, consequent, constant function, 53 continuum, 123 continuum hypothesis, 123 contradiction, contrapositive, contrapositive proof, 19 converse, corestriction, 59 corollary, 16 countable set, 120 counterexample, 11, 25 De Moivre formula, 176 De Morgan’s laws, 8, 42 decimal fraction, 149 Dedekind cut, 154 descendent, 103 diagonal, 53 dictionary order, 90 Diophantine equation, 111 direct image, 57 direct proof, 17 directed set, 86 disjoint, 38 disjoint union, 47 disjunction, 201 Index 202 divide, 16 divisibility, 98 division, 99 division algorithm, 106 divisor, 98 domain, 52 duality, 43 empty set, 36 end of proof, 16 equipotent sets, 117 equivalence class, 79 equivalence of statements, equivalence relation, 78 Euclidean algorithm, 109 Euler’s constant, 172 exclusive or, existential quantifier, exponent, 100 exponential principle, 127 extension, 59 factor, 98 factorial, 100 family of sets, 66 Fibonacci sequence, 133 filtered set, 86 finite set, 119 first element, 84 floor function, 54 fractional part, 54 function, 51 Fundamental theorem of arithmetic, 110 generalized induction, 26 generalized sequence, 87 generating function, 134 greatest common divisor, 107 greatest element, 82 greatest lower bound, 83, 84, 160 Hausdorff principle, 85 hypothesis, identity function, 53 imaginary part, 174 implication, inclusion, 35 inclusion-exclusion principle, 125 induction with bigger steps, 102 infimum, 83, 160 infinite set, 119 injective, 60 integer, 141 integer part, 54 intermediate value theorem, 24 intersection, 38 inverse, 63 inverse image, 57 inverse of a relation, 75 invertible function, 63 irrational number, 20 lattice, 84 law of trichotomy, 94 least common multiple, 107 least element, 82 least upper bound, 83, 160 least upper bound property, 161 left inverse, 65 lemma, 16 lexicographic order, 90 logically equivalent, lower bound, 82, 160 lowest terms, 148 mathematical induction, 26 maximal element, 82 maximum, 82 maximum function, 56 minimal element, 82 minimum, 82 minimum function, 56 multiple, 98 multiplication of positive integers, 97 multiplication principle, 126 multiplicative inverse, 147 natural numbers, 91, 99 natural order, 94 negation, Index negative cut, 156 net, 87 octonion, 177 one-to-one, 60 onto, 61 open sentence, 17 operations with functions, 55 opposite, 144 opposite order, 82 order relation, 81 ordered field, 149 ordered pair, 46 p-adic numbers, 170 partition, 79 Pascal’s triangle, 130 Peano axioms, 91 permutation, 128 permutation of size k, 129 positive cut, 156 positive integers, 91 postulate, 15, 16 power, 100 power set, 37 predicate, 17 preorder, 82 prime, 20, 101 principle of mathematical induction, 92 product of cardinals, 123 proof by cases, 23 proof by contradiction, 20 proposition, quaternion, 177 quotient, 99 quotient map, 80 quotient set, 80 range, 52 rational number, 145 real number, 160, 167 real part, 174 recurrence relation, 132 recursive, 132 203 reflexive relation, 78 relation, 73 representative, 80 restriction, 59 retract, 65 right inverse, 65 Russell’s paradox, 37 saturated, 81 section, 65 set difference, 43 sign function, 56 solution set, 34 statement, strict order, 81 strong induction, 26 subset, 35 subtraction, 96 successor, 91 sum of cardinals, 123 superset, 35 supremum, 83, 160 surjective, 61 symmetric difference, 44 symmetric relation, 78 tautology, theorem, 16 total order, 81 transcendental number, 172 transitive relation, 78 triangle inequality, 105 uncountable set, 120 union, 38 universal quantifier, upper bound, 82, 160 well-defined, 80 well-ordered, 84 Zermelo’s theorem, 85 Zorn’s lemma, 85 ... ALGEBRA: LABS AND PROJECTS WITH MATHEMATICA® Crista Arangala GRAPHS & DIGRAPHS, SIXTH EDITION Gary Chartrand, Linda Lesniak, and Ping Zhang INTRODUCTION TO ABSTRACT ALGEBRA, SECOND EDITION Jonathan... L Mullen and James A Sellers ABSTRACT ALGEBRA: AN INTERACTIVE APPROACH, SECOND EDITION William Paulsen ABSTRACT ALGEBRA: AN INQUIRY-BASED APPROACH Jonathan K Hodge, Steven Schlicker, and Ted Sundstrom... in MATHEMATICS A BRIDGE TO HIGHER MATHEMATICS Valentin Deaconu University of Nevada Reno, USA Donald C Pfaff University of Nevada Reno, USA www.TechnicalBooksPDF.com CRC Press Taylor & Francis