Calculo Apostol Vol 2

The Real And Complex Number Systems

Consider the right hand side, we have
n−1 n−1 n−1

(x − 1)

xk =
k=0 k=0 n

xk+1 − xk −

xk x

=

k=0 n−1 k k=0

Integers

= x − 1. 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
b−1k=1 n

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 (1707-1783) find another method to showit. 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)

1.3

2ab − 1 = (2a − 1)

(2a )k
k=0

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 calledPerfect 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) If 2n + 1 is a prime, prove that n is a power of 2. A prime of the m form 22 + 1 is called a Fermat prime. Hint. Use exercise 1.2. Proof : If n is a not a power of2, 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
2n−2

1.2 If n is a positive integer, prove the algebraic identity
n−1

1.4

an − bn = (a − b) Proof : It suffices to show that

ak bn−1−k
k=0

n−1

x − 1 = (x − 1)

n

x .
k=0

kx2n−1 + 1 = (x + 1)
k=0

(−1)k xk .

1

2

Proof : Consider
2n−2 2n−2 2n−2

then it is clear that (−1)k xk = (−1)k xk+1 +
k=0 2n−1 k=0 2n−2

(x + 1)
k=0

(−1)k xk (−1) x
k=0 k k

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 withApplications by Koshy and Thomas. Prove that every nonempty set of positive integers contains a smallest member. This is called the well–ordering Principle. Proof : Given (φ =) 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, consideras k = m + 1 as follows. In order to show it, we consider two cases. (1) If there is a member s ∈ S such that s < m + 1, then by Induction hypothesis, we have proved it. (2) If every s ∈ 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 ∈ S, then k + 1 ∈ 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 ∈ S, then k + 1 ∈ S. 4

= =x

(−1)
k=1 2n+1k+1

x +

k

+ 1.

1.6

(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) 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...