1.4-26: Prove that if n is a positive integer, then n is even if and only if 7n + 4 is even.

Proof. Let n be a positive integer. Assume that n is even. By definition of even, this means that there exists an integer a such that n = 2a. By substitution

7n + 4 = 7(2a) + 4

= 14a + 4

= 2(7a + 2).

Since 7, a, and 2 are integers and integers have closure under addition and multiplication, then 7a + 2 is an integer. Therefore by the definition of even, 7n + 4 is even.

Now assume that n is odd. By the definition of odd, this means there is an integer b such that n = 2b + 1. By substitution

7n + 4 = 7(2b + 1) + 4

= 14b + 7 + 4

= 14b + 11

= 2(7b + 5) + 1.

Since 7, b, and 5 are integers and integers have closure under addition and multiplication, then 7b + 5 is an integer. Therefore by the definition of odd, it follows that 7n + 4 is odd.

Since n is odd implies that 7n + 4 is also odd, then clearly 7n + 4 is even implies n is even must be equivalently true.

Furthermore, since n is even implies 7n + 4 is even, and conversely 7n + 4 is even implies n is even are true, then by the definition of a biconditional statement it follows that the original proposition n is even if and only if 7n + 4 is even must also be true. ☐