Exam s 2-4 Solutions

Exam s 2-4 Solutions - MATH 290 EXAM 2 SOLUTIONS 1(10 pts Prove that for any integers a b and c if a|b and a|c then a|(b c Solution Let a b and c

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MATH 290 – EXAM 2 – SOLUTIONS 1. (10 pts.) Prove that for any integers a , b , and c , if a | b and a | c then a | ( b - c ). Solution: Let a , b , and c be integers. Assume that a | b and a | c . We must show that a | ( b - c ). Since a | b there is an integer n such that b = na , and since a | c there is an integer m such that c = ma . Hence b - c = na - ma = ( n - m ) a . Since n - m is an integer, there is an integer k such that b - c = ka . Therefore a | ( b - c ). 2. (10 pts.) Prove by contrapositive that for all natural numbers n , if n 2 is odd then n is odd. Solution: Let n be a natural number. Assume that n is not odd. We must show that n 2 is not odd. Since n is not odd, n is even. Hence there exists a natural number k such that n = 2 k . Hence n 2 = 4 k 2 = 2(2 k 2 ). Since 2 k 2 is an integer, n 2 is even, and therefore is not odd. 3. (14 pts.) Prove that for all integers a and b , a + b is even if and only if a and b are even or a and b are odd. Solution:
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This note was uploaded on 03/05/2011 for the course MATH 290 taught by Professor Alligood,k during the Fall '08 term at George Mason.

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Exam s 2-4 Solutions - MATH 290 EXAM 2 SOLUTIONS 1(10 pts Prove that for any integers a b and c if a|b and a|c then a|(b c Solution Let a b and c

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