1.4-18a: Prove that if n is an integer and 3n + 2 is even, then n is even using a proof by contraposition.

Section 1.4 Problem 18a.

Proof. Let n be an integer. Assume that n is odd. By definition of odd, there exists an integer a such that n = 2a + 1. By substitution,

3n + 2 = 3(2a + 1) + 2

= 6a + 3 + 2

= 6a + 5

= 2(3a + 2) + 1.

Since 3, a, and 2 are integers and integers have closure with respect to multiplication, then 3a + 2 is an integer. Therefore by the definition of odd, 3n + 2 is odd. ☐