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.) Suppose \(a, b \in \mathbb{Z}\). If \(a^{2}\left(b^{2}-2 b\right)\) is odd, then \(a\) and \(b\) are odd.

Short answer

The original statement is proven to be true by proving its contrapositive. If \(a\) and \(b\) are even, then \(a^{2}(b^{2}-2b)\) is even.

Step-by-step solution

Formulate the Contrapositive of the Statement

First, write down the contrapositive statement, which is: 'If \(a\) and \(b\) are not odd (so, they are even), then \(a^{2}\left(b^{2}-2 b\right)\) is not odd (so, it is even).'

Define Even Integers

Recall that even integers can be expressed as \(2k\), where \(k \in \mathbb{Z}\). So, we let \(a = 2m\) and \(b = 2n\), where \(m,n \in \mathbb{Z}\).

Substitute \(a\) and \(b\) into the Expression

Replace \(a\) and \(b\) with \(2m\) and \(2n\) respectively in the expression \(a^{2}(b^{2}-2b)\). This would give \(a^{2}(b^{2}-2b) = (2m)^{2}(4n^{2}-4n)\).

Simplify the Expression

Simplify the expression to get \(16m^{2}n^{2}-16m^{2}n\). Now, factor out 2 to show that the expression is even: \(a^{2}(b^{2}-2b) = 2(8m^{2}n^{2}-8m^{2}n)\).

Interpret the Result

As shown, \(2(8m^{2}n^{2}-8m^{2}n)\) is an even number because it is obtained by multiplying 2 (an even number) by another integer. This proves the contrapositive statement, meaning our original statement 'if \(a^{2}\left(b^{2}-2 b\right)\) is odd, then \(a\) and \(b\) are odd,' is true.