math403 hw1

# math403 hw1 - SOLUTIONS FOR HWK 1 Problem 1.10.8 We will...

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SOLUTIONS FOR HWK 1 Problem 1.10.8 We will prove that 101 50 - 99 50 > 100 50 , i.e. 101 50 > 100 50 + 99 50 . 101 50 - 99 50 = (100 + 1) 50 - (100 - 1) 50 = 50 X k =0 ± 50 k 1 k (100) 50 - k - - 50 X k =0 ± 50 k ( - 1) k (100) 50 - k = 50 X k =0 ± 50 k ( 100 50 - k ) 1 k - ( - 1) k · . Thus the terms with even k 's are canceled and the terms with odd k 's are positive, so we have (starting with k = 1 ): 101 50 - 99 50 = 2 · 50 · 100 49 + 2 · ± 50 2 · 100 47 + ... > 100 50 Problem A Part 1 Prove that if n N and 3 | n 2 then 3 | n . Proof (BC).Let 3 | n 2 . Assume that 3 does not divide n . Then the remainder after dividing n by 3 is either 1 or 2. In the rst case we have n = 3 k + 1 where k ∈ { 0 , 1 , 2 ,... } and n 2 = 9 k 2 + 6 k + 1 = 3(3 k 2 + 2 k ) + 1 . In the second case n = 3 l + 2 where l ∈ { 0 , 1 , 2 ,... } so n 2 = 9 k 2 +12 k +4 = 3(3 k 2 +4 k +1)+1 . In both cases 3 does not divide n 2 , so we obtained a contradiction to the condition of the statement. Part 2

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## This note was uploaded on 10/13/2010 for the course MATH 403 at Penn State.

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math403 hw1 - SOLUTIONS FOR HWK 1 Problem 1.10.8 We will...

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