Solution:

Let’s suppose

$p$: $n^{2}$ is an even integer.

$\sim p$: $n$ is not an even integer

$q$: $n$ is also an even integer

$\sim q$ $=$ $n$ is not an even integer.

As, in the contrapositive, a conditional statement is logically equivalent to its contrapositive.

Hence,

$\sim q\rightarrow \sim p =$ If n is not an even integer then n2 is not an even integer.

As a result, $\sim q$ is true $\rightarrow \sim p$ is true.