solutionshw2-554-2010

solutionshw2-554-2010 - Solutions homework 2. Page 8...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

Ask a homework question - tutors are online