If \(x\) is even, then there exists an integer \(k\) such that \(x=2k\). This means:
$$\begin{align}x^2 &= (2k)^2 \\ &= 4k^2 = 2 *(2k^2) \\ &= 2h \end{align}$$
Therefore \(x^2\) is also even. It's contrapositive states that when \(x^2\) is odd, then \(x\) is odd.
If \(x\) is odd, then \(x=2k+1\). This means:
$$\begin{align}x^2 &= (2k+1)^2 \\ &= 4k^2 + 4k + 1 = 2 *(2k^2 + 2k) + 1 \\ &= 2h + 1 \end{align}$$
Therefore \(x^2\) is also odd. It's contrapositive states that when \(x^2\) is even, then \(x\) is even.