# hw6solsmc - Solutions to Homework 6 Math 2000 The solutions...

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Solutions to Homework 6 - Math 2000 The solutions to 3.7,3.15,3.21 may be found in the book. Exercises 3.4,3.15 were done during the tutorial. (# 3.2) Let n N . Prove that if | n - 1 | + | n + 1 | ≤ 1, then | n 2 - 1 | ≤ 4. Solution . Let S = Z and P ( n ) , Q ( n ) be the open statements: P ( n ) : | n - 1 | + | n + 1 | ≤ 1 Q ( n ) : | n 2 - 1 | ≤ 4 . We now give a vacuous proof. Let n N . Thus n 1. It follows that n + 1 2 and so that | n + 1 | ≥ 2 . Since | n - 1 | ≥ 0 we deduce that | n - 1 | + | n + 1 | ≥ 2 > 1 Therefore P ( n ) is false for all n N and hence it follows that if | n - 1 | + | n + 1 | ≤ 1, then | n 2 - 1 | ≤ 4. (# 3.8) Prove that if a and c are odd integers and b is an integer, then ab + bc is even. Solution . Since a and c are odd integers, there exist integers k, l Z such that a = 2 k + 1 and c = 2 l + 1 . It follows that ab + bc = b ( a + c ) = b (2 k + 1 + 2 l + 1) = b (2 k + 2 l + 2) = 2 b ( k + l + 1) is even since b ( k + l + 1)

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hw6solsmc - Solutions to Homework 6 Math 2000 The solutions...

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