Unformatted text preview: Solutions homework 2. Page 8 Problem 14 : Note first that the function y = 2 x + x 2 has a minimum at x = 1, so 2 x + x 2 ≥  2 + 1 = 1 for all x . Hence by Problem 13 we have (1 + x ) 2 n = (1 + 2 x + x 2 ) n ≥ 1 + n (2 x + x 2 ) ≥ 1 + 2 nx. Page 8 Problem 17. Assume n = r 2 s 2 with 0 < r,s ∈ N such that gcd( r,s ) = 1. Assume gcd( r 2 ,s 2 ) > 1, then there exist a prime p which divides both r 2 and s 2 . However, if a prime number p divides r 2 , then it divides r . Similarly p divides s , which contradicts gcd( r,s ) = 1. Now ns 2 = r 2 implies that 1 = gcd( r 2 ,s 2 ) = gcd( ns 2 ,s 2 ) ≥ s 2 ≥ 1, which shows that s 2 = 1. Hence n = r 2 is a square of an integer. Page 8 Problem 19. Prove by induction. For n = 1 we have 1 < 2 1 = 2, so the claim is true for n = 1. Now assume the claim is true for k ∈ P , i.e., assume k < 2 k is true. Then it is in particular true that 1 < 2 k as 1 ≤ k . Then k + 1 < 2 k + 1 < 2 k + 2 k = 2 k +1 , so the claim holds for k + 1. Therefore by induction the claim holds for all+ 1....
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This note was uploaded on 02/05/2012 for the course MATH 554 taught by Professor Girardi during the Fall '10 term at South Carolina.
 Fall '10
 Girardi

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