Proof. Assume that x and y are odd integers. By definition of odd, there are integers a and b such that x = 2a + 1 and y = 2b + 1. By substitution
xy = (2a + 1) (2b + 1)
= 4ab + 2a + 2b + 1
= 2(2ab + a + b) + 1.
Since 2, a, and b are integers and integers have closure with respect to multiplication and addition, then 2ab + a + b is an integer. By definition of odd, it follows that the product of xand y is odd. ☐
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. ☐
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. ☐
