Show that the following statement is true by the method of contrapositive.
$p:$ If $x$ is an integer and $x^{2}$ is even,then $x$ is also even.

(N/A) The statement $p$ is: If $x$ is an integer and $x^{2}$ is even,then $x$ is even.
To prove this by the method of contrapositive,we assume the negation of the conclusion and prove the negation of the hypothesis.
Let $q: x$ is an integer and $x^{2}$ is even.
Let $r: x$ is even.
The contrapositive of the statement $q \implies r$ is $\neg r \implies \neg q$.
Assume $\neg r$: $x$ is not even,i.e.,$x$ is odd.
If $x$ is odd,then $x = 2k + 1$ for some integer $k$.
Then $x^{2} = (2k + 1)^{2} = 4k^{2} + 4k + 1 = 2(2k^{2} + 2k) + 1$.
Since $2(2k^{2} + 2k) + 1$ is of the form $2m + 1$,$x^{2}$ is odd.
Thus,$\neg q$ is true: $x^{2}$ is not even.
Since $\neg r \implies \neg q$ is true,the original statement $p$ is true.