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 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 n n a b a b n 1 X ak bn 1 k k 0 Proof It suffices to show that n x 1 x 1 n 1 X k 0 1 xk Consider the right hand side we have x 1 n 1 X k x k 0 n 1 X x k 1 k 0 n X n 1 X xk k 0 xk k 1 n n 1 X xk k 0 x 1 1 3 If 2 n 1 is a prime prove that n is prime A prime of the form 2 1 where p is prime is called a Mersenne prime p Proof If n is not a prime then say n ab where a 1 and b 1 So we have b 1 X 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 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 2 n 2m form 2 1 is a prime prove that n is a power of 2 A prime of the 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 x 2n 1 1 x 1 2n 2 X k 0 2 1 k xk Proof Consider x 1 2n 2 X 1 k xk k 0 2n 2 X k 0 2n 1 X 1 k xk 1 1 k 1 xk k 1 2n 1 x 2n 2 X k 0 2n 2 X 1 k xk 1 k xk k 0 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 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 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 N S Note that w 1 by 1 so we consider w 1 as follows Since w 1 N S we know that w 1 S By 2 we know that w S which contadicts to w 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 3344444444 Proof Let x 0 3344444444 then x 3 4 4 3 2 3 n where n 3 10 10 10 10 4 1 33 1 n 1 102 103 10 10 33 4 1 1 102 103 1 10 4 33 2 10 900 301 900 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 2n 5m where m and n are nonnegative integers Proof Suppose that x k 2n 5m if n m we have k5n m 5n m k 2n 5n 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 a1 a2 an Then Pn Pn n k 10 a 10n k ak k x k 0 n k 0 n n 10 2 5 For case x a0 a1 a2 an 999999 Then Pn n k ak …