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 \(n \in \mathbb{Z}\). If \(n^{2}\) is even, then \(n\) is even.

Short answer

The statement 'If \(n^{2}\) is even, then \(n\) is even.' holds true as proven by the contrapositive 'If \(n\) is not even (i.e. \(n\) is odd), then \(n^{2}\) is not even (i.e. \(n^{2}\) is odd).'}

Step-by-step solution

Understand the Contrapositive Concept

In a contrapositive proof, we want to prove the inverse of the statement. In this case, the original statement is 'If \(n^{2}\) is even, then \(n\) is even.' The contrapositive of this would be 'If \(n\) is not even, then \(n^{2}\) is not even.' Therefore, looking at this, in order to prove the original statement, we must prove the contrapositive statement.

Defining Even and Odd Numbers

First, let's remember what defines an even and an odd number. An integer \(n\) is even if it can be written in the form \(2k\) where \(k\) is an integer. Conversely, an integer \(n\) is odd if it can be written in the form \(2k + 1\), where \(k\) is an integer.

Proving the Contrapositive Statement

Now, consider \(n\) is not an even number. This means \(n\) is an odd number and can be written in the form \(2k + 1\), where \(k\) is any integer. Now if we square \(n\), that is \(n^{2}\), we get \((2k + 1)^{2} = 4k^{2} + 4k + 1\). This number can be looked at as \(2*(2k^{2}+2k) + 1\), which is of the form \(2m + 1\), where \(m = 2k^{2} + 2k\). This form represents an odd number. Hence, the square is odd if the number \(n\) is odd.