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

$p$ : If $x$ is an integer and $x^2$ is even then $x$ is also even.
Let $q$ : $x$ is an integer and $x^2$ is even 
$r$ : $x$ is even
To prove that $p$ is true by contrapositive method we assume that $r$ is false and prove that $q$ is also false.
Let $x$ is not even
To prove that $q$ is false, it has to be proved that $x$ is not an integer or $x^2$ is not even.
$x$ is not even implies that $x^2$ is also not even.
Therefore statement $q$ is false.
Thus the given statement $p$ is true.