Exam (elaborations) TEST BANK FOR Mathematical analysis 2nd Edition By Apostol (solution manual)

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