TEST BANK FOR Mathematical Analysis 2nd Edition By Apostol

Exam (elaborations) TEST BANK FOR Mathematical Analysis 2nd Edition By Apostol The Real And Complex Number Systems Integers 1.1 Prove that there is no largest prime. Proof : Suppose p is the largest prime. Then p!+1 is NOT a prime. So, there exists a prime q such that q |p! + 1 ) q |1 which is impossible. So, there is no largest prime. Remark: There are many and many proofs about it. The proof that we give comes from Archimedes 287-212 B. C. In addition, Euler Leonhard () find another method to show it. The method is important since it develops to study the theory of numbers by analytic method. The reader can see the book, An Introduction To The Theory Of Numbers by Loo-Keng Hua, pp 91-93. (Chinese Version) 1.2 If n is a positive integer, prove the algebraic identity an − bn = (a − b) Xn−1 k=0 akbn−1−k Proof : It suffices to show that xn − 1 = (x − 1) Xn−1 k=0 xk. 1 Consider the right hand side, we have (x − 1) Xn−1 k=0 xk = Xn−1 k=0 xk+1 − Xn−1 k=0 xk = Xn k=1 xk − Xn−1 k=0 xk = xn − 1. 1.3 If 2n − 1 is a prime, prove that n is prime. A prime of the form 2p − 1, where p is prime, is called a Mersenne prime. Proof : If n is not a prime, then say n = ab, where a 1 and b 1. So, we have 2ab − 1 = (2a − 1) Xb−1 k=0 (2a)k which is not a prime by Exercise 1.2. So, n must be a prime. Remark: The study of Mersenne prime is important; it is related with so called Perfect number. In addition, there are some OPEN problem about it. For example, is there infinitely many Mersenne nembers? The reader can see the book, An Introduction To The Theory Of Numbers by Loo-Keng Hua, pp 13-15. (Chinese Version) 1.4 If 2n + 1 is a prime, prove that n is a power of 2. A prime of the form 22m + 1 is called a Fermat prime. Hint. Use exercise 1.2. Proof : If n is a not a power of 2, say n = ab, where b is an odd integer. So, 2a + 1  2ab + 1 and 2a + 1 2ab + 1. It implies that 2n + 1 is not a prime. So, n must be a power of 2. Remark: (1) In the proof, we use the identity x2n−1 + 1 = (x + 1) 2Xn−2 k=0 (−1)k xk. 2 Proof : Consider (x + 1) 2Xn−2 k=0 (−1)k xk = 2Xn−2 k=0 (−1)k xk+1 + 2Xn−2 k=0 (−1)k xk = 2Xn−1 k=1 (−1)k+1 xk + 2Xn−2 k=0 (−1)k xk = x2n+1 + 1. (2) The study of Fermat number is important; for the details the reader can see the book, An Introduction To The Theory Of Numbers by Loo-Keng Hua, pp 15. (Chinese Version) 1.5 The Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, ... are defined by the recursion formula xn+1 = xn + xn−1, with x1 = x2 = 1. Prove that (xn, xn+1) = 1 and that xn = (an − bn) / (a − b) , where a and b are the roots of the quadratic equation x2 − x − 1 = 0. Proof : Let d = g.c.d. (xn, xn+1) , then d |xn and d |xn+1 = xn + xn−1 . So, d |xn−1 . Continue the process, we finally have d |1 . So, d = 1 since d is positive. Observe that xn+1 = xn + xn−1, and thus we consider xn+1 = xn + xn−1, i.e., consider x2 = x + 1 with two roots, a and b. If we let Fn = (an − bn) / (a − b) , 3 then it is clear that F1 = 1, F2 = 1, and Fn+1 = Fn + Fn−1 for n 1. So, Fn = xn for all n. Remark: The study of the Fibonacci numbers is important; the reader can see the book, Fibonacci and Lucas Numbers with Applications by Koshy and Thomas. 1.6 Prove that every nonempty set of positive integers contains a smallest member. This is called the well–ordering Principle. Proof : Given ( 6=) S ( N) , we prove that if S contains an integer k, then S contains the smallest member. We prove it by Mathematical Induction of second form as follows. As k = 1, it trivially holds. Assume that as k = 1, 2, ...,m holds, consider as k = m + 1 as follows. In order to show it, we consider two cases. (1) If there is a member s 2 S such that s m + 1, then by Induction hypothesis, we have proved it. (2) If every s 2 S, s  m + 1, then m + 1 is the smallest member. Hence, by Mathematical Induction, we complete it. Remark: We give a fundamental result to help the reader get more. We will prove the followings are equivalent: (A. Well–ordering Principle) every nonempty set of positive integers contains a smallest member. (B. Mathematical Induction of first form) Suppose that S ( N) , if S satisfies that (1). 1 in S (2). As k 2 S, then k + 1 2 S. Then S = N. (C. Mathematical Induction of second form) Suppose that S ( N) , if S satisfies that (1). 1 in S (2). As 1, ..., k 2 S, then k + 1 2 S. 4 Then S = N. Proof : (A ) B): If S 6= N, then N − S 6= . So, by (A), there exists the smallest integer w such that w 2 N − S. Note that w 1 by (1), so we consider w − 1 as follows. Since w − 1 /2 N − S, we know that w − 1 2 S. By (2), we know that w 2 S which contadicts to w 2 N − S. Hence, S = N. (B ) C): It is obvious. (C ) A): We have proved it by this exercise. Rational and irrational numbers 1.7 Find the rational number whose decimal expansion is 0.4.... Proof: Let x = 0.4..., then x = 3 10 + 3 102 + 4 103 + ... + 4 10n + .., where n  3 = 33 102 + 4 103  1 + 1 10 + ... + 1 10n + ..  = 33 102 + 4 103  1 1 − 1 10  = 33 102 + 4 900 = 301 900 . 1.8 Prove that the decimal expansion of x will end in zeros (or in nines) if, and only if, x is a rational number whose denominator is of the form 2n5m, where m and n are nonnegative integers. Proof: (()Suppose that x = k 2n5m , if n  m, we have k5n−m 2n5n = 5n−mk 10n . So, the decimal expansion of x will end in zeros. Similarly for m  n. ())Suppose that the decimal expansion of x will end in zeros (or in nines). 5 For case x = a0.a1a2 · · · an. Then x = Pn k=0 10n−kak 10n = Pn k=0 10n−kak 2n5n . For case x = a0.a1a2 · · · an · · · . Then x = Pn k=0 10n−kak 2n5n + 9 10n+1 + ... + 9 10n+m + ... = Pn k=0 10n−kak 2n5n + 9 10n+1 X1 j=0 10−j = Pn k=0 10n−kak 2n5n + 1 10n = 1 + Pn k=0 10n−kak 2n5n . So, in both case, we prove that x is a rational number whose denominator is of the form 2n5m, where m and n are nonnegative integers. 1.9 Prove that p 2 + p 3 is irrational. Proof: If p 2 + p 3 is rational, then consider p 3 + p 2  p 3 − p 2  = 1 which implies that p 3 − p 2 is rational. Hence, p 3 would be rational. It is impossible. So, p 2 + p 3 is irrational. Remark: (1) p p is an irrational if p is a prime. Proof : If p p 2 Q, write p p = a b , where g.c.d. (a, b) = 1. Then b2p = a2 ) p  a2 ) p |a (*) Write a = pq. So, b2p = p2q2 ) b2 = pq2 ) p  b2 ) p |b . (*’) By (*) and (*’), we get p |g.c.d. (a, b) = 1 which implies that p = 1, a contradiction. So, p p is an irrational if p is a prime. 6 Note: There are many and many methods to prove it. For example, the reader can see the book, An Introduction To The Theory Of Numbers by Loo-Keng Hua, pp 19-21. (Chinese Version) (2) Suppose a, b 2 N. Prove that p a+ p b is rational if and only if, a = k2 and b = h2 for some h, k 2 N. Proof : (() It is clear. ()) Consider p a + p b  p a − p b  = a2 − b2, then p a 2 Q and p b 2 Q. Then it is clear that a = h2 and b = h2 for some h, k 2 N. 1.10 If a, b, c, d are rational and if x is irrational, prove that (ax + b) / (cx + d) is usually irrational. When do exceptions occur? Proof: We claim that (ax + b) / (cx + d) is rational if and only if ad = bc. ())If (ax + b) / (cx + d) is rational, say (ax + b) / (cx + d) = q/p. We consider two cases as follows. (i) If q = 0, then ax+b = 0. If a 6= 0, then x would be rational. So, a = 0 and b = 0. Hence, we have ad = 0 = bc. (ii) If q 6= 0, then (pa − qc) x+(pb − qd) = 0. If pa−qc 6= 0, then x would be rational. So, pa − qc = 0 and pb − qd = 0. It implies that qcb = qad ) ad = bc. (()Suppose ad = bc. If a = 0, then b = 0 or c = 0. So, ax + b cx + d =  0 if a = 0 and b = 0 b d if a = 0 and c = 0 . If a 6= 0, then d = bc/a. So, ax + b cx + d = ax + b cx + bc/a = a (ax + b) c (ax + b) = a c . Hence, we proved that if ad = bc, then (ax + b) / (cx + d) is rational. 7 1.11 Given any real x 0, prove that there is an irrational number between 0 and x. Proof: If x 2 Qc, we choose y = x/2 2 Qc. Then 0 y x. If x 2 Q, we choose y = x/ p 2 2 Q, then 0 y x. Remark: (1) There are many and many proofs about it. We may prove it by the concept of Perfect set. The reader can see the book, Principles of Mathematical Analysis written by Walter Rudin, Theorem 2.43, pp 41. Also see the textbook, Exercise 3.25. (2) Given a and b 2 R with a b, there exists r 2 Qc, and q 2 Q such that a r b and a q b. Proof : We show it by considering four cases. (i) a 2 Q, b 2 Q. (ii) a 2 Q, b 2 Qc. (iii) a 2 Qc, b 2 Q. (iv) a 2 Qc, b 2 Qc. (i) (a 2 Q, b 2 Q) Choose q = a+b 2 and r = p1 2 a +  1 − p1 2  b. (ii) (a 2 Q, b 2 Qc) Choose r = a+b 2 and let c = 1 2n b−a, then a+c := q. (iii) (a 2 Qc, b 2 Q) Similarly for (iii). (iv) (a 2 Qc, b 2 Qc) It suffices to show that there exists a rational number q 2 (a, b) by (ii). Write b = b0.b1b2 · · · bn · ·· Choose n large enough so that a q = b0.b1b2 · · · bn b. (It works since b − q = 0.000..000bn+1...  1 10n ) 1.12 If a/b c/d with b 0, d 0, prove that (a + c) / (b + d) lies bwtween the two fractions a/b and c/d Proof: It only needs to conisder the substraction. So, we omit it. Remark: The result of this exercise is often used, so we suggest the reader keep it in mind. 1.13 Let a and b be positive integers. Prove that p 2 always lies between the two fractions a/b and (a + 2b) / (a + b) . Which fraction is closer to p 2? Proof : Suppose a/b  p 2, then a  p 2b. So, a + 2b a + b − p 2 =