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math336_hw1_solns

math336_hw1_solns - 39 112 n 13-267 n is a solution for all...

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Homework 1 Partial Solutions Question 2 (2A: # 2) We will prove 1 3 + 2 3 + ... + n 3 = n ( n +1) 2 2 for all n 1 by induction. Base case (when n = 1): (1)(1+1) 2 2 = 1 = 1 3 , so the statement holds for n = 1. Assume that for k 1, 1 3 + 2 3 + ...k 3 = k ( k +1) 2 2 . (This is the inductive hypothesis.) Then 1 3 + 2 3 + ...k 3 + ( k + 1) 3 = k ( k + 1) 2 2 + ( k + 1) 3 = k 2 ( k + 1) 2 + 4( k + 1) 3 4 = ( k + 1) 2 ( k 2 + 4( k + 1)) 4 = ( k + 1) 2 ( k + 2) 2 4 = ( k + 1)( k + 2) 2 2 Therefore, by induction 1 3 + 2 3 + ...n 3 = n ( n +1) 2 2 1 3 + 2 3 + ...n 3 = n ( n +1) 2 2 for all n 1. Question 7: 3C # 7i From question 4 i in 3C we know that gcd(267 , 112) = 1 and that 267( - 13) + 112(31) = 1. So if we multiply both sides by 3 we get 267( - 13 × 3) + 112(31 × 3) = 1 × 3 267( - 39) + 112(63) = 3 So ( x, y ) = ( - 39 , 63) is a solution for the equation 267 x + 112 y = 3. To find the other solutions, we use prop. 5. It says that ( x 0 , y 0 ) is a solution exactly when x 0 = - 39 + n 112 1 = - 39 + 112 n and y 0 = 13 - n 267 1 = 13 - 267 n for some n Z . Therefore, ( x, y ) = ( -
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Unformatted text preview: 39 + 112 n, 13-267 n ) is a solution for all n ∈ Z . Question 8 This could be proven several di²erent ways. Below is one method. 1 Let a and b be natural numbers, and let gcd( a, b ) = d . By defnition, we always have d ≥ 1. Let S = { ar + bs ≥ 1 | r, s ∈ Z } . Then by Bezout’s thereom, there exists an r , s ∈ Z such that ar + bs = d . ThereFore, d ∈ S . Let e = ar + bs be any element in S . So r and s are integers, and hence by prop. 4, d | e . This means e = dn For some n ∈ Z . But e, d ≥ 1, so we must also have n ≥ 1. ThereFore the smallest possible value For e is d · 1 = d . Since d ∈ S , gcd ( a, b ) is indeed the least element in the set. 2...
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