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# ch1 - The Real And Complex Number Systems Integers 1.1...

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Unformatted text preview: 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 A rchimedes 287-212 B. C. In addition, Euler Leonhard (1707-1783) 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 a n- b n = ( a- b ) n- 1 X k =0 a k b n- 1- k Proof : It suffices to show that x n- 1 = ( x- 1) n- 1 X k =0 x k . 1 Consider the right hand side, we have ( x- 1) n- 1 X k =0 x k = n- 1 X k =0 x k +1- n- 1 X k =0 x k = n X k =1 x k- n- 1 X k =0 x k = x n- 1 . 1.3 If 2 n- 1 is a prime, prove that n is prime. A prime of the form 2 p- 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 2 ab- 1 = (2 a- 1) b- 1 X k =0 (2 a ) 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 prob- lem about it. For example, is there infinitely many Mersenne nem- bers? 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 2 n + 1 is a prime, prove that n is a power of 2 . A prime of the form 2 2 m + 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, 2 a + 1 2 ab + 1 and 2 a + 1 < 2 ab + 1 . It implies that 2 n + 1 is not a prime. So, n must be a power of 2 . Remark : (1) In the proof, we use the identity x 2 n- 1 + 1 = ( x + 1) 2 n- 2 X k =0 (- 1) k x k . 2 Proof : Consider ( x + 1) 2 n- 2 X k =0 (- 1) k x k = 2 n- 2 X k =0 (- 1) k x k +1 + 2 n- 2 X k =0 (- 1) k x k = 2 n- 1 X k =1 (- 1) k +1 x k + 2 n- 2 X k =0 (- 1) k x k = x 2 n +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 recur- sion formula x n +1 = x n + x n- 1 , with x 1 = x 2 = 1 . Prove that ( x n , x n +1 ) = 1 and that x n = ( a n- b n ) / ( a- b ) , where a and b are the roots of the quadratic equation x 2- x- 1 = 0 . Proof : Let d = g.c.d. ( x n , x n +1 ) , then d | x n and d | x n +1 = x n + x n- 1 ....
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ch1 - The Real And Complex Number Systems Integers 1.1...

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