Question

Prove the following statements with contrapositive proof. (In each case, think about how a direct proof would work. In most cases contrapositive is easier.) If \(a, b \in \mathbb{Z}\) and \(a\) and \(b\) have the same parity, then \(3 a+7\) and \(7 b-4\) do not.

Short answer

The original statement is true as proven by validating its contrapositive.

Step-by-step solution

Understanding Parity

First, be aware, that the parity of a number refers to whether it is even (when it can be divided by 2 with a remainder of 0) or odd (when it doesn't meet this criteria). The expression \(3a+7\) or \(7b-4\) has the same parity means it is either both even or both odd.

Defining the Contrapositive

A contrapositive is an implication when both the hypothesis and conclusion are negated and reversed. Given the implication, if 'a' and 'b' have same parity, then \(3a+7\) and \(7b-4\) do not - the contrapositive is - if \(3a+7\) and \(7b-4\) have the same parity, then 'a' and 'b' do not.

Proving the Contrapositive is True

Assume \(3a+7\) and \(7b-4\) have the same parity. This implies that their difference \(3a+7 - (7b-4) = 3a - 7b +11\) is an even number, since the difference of two numbers with the same parity (both even or both odd) is always even. But \(3a - 7b +11\) cannot be even as the sum/difference of any three odd numbers (since 'a' and 'b' are odd) is odd, which is a contradiction. Therefore, the contrapositive is correct, and so is the original statement.