Question

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

Answer 1

Let $p:$ If $x$ is an integer and $x^2$ is even and $q:x$ is even.

The given statement is if $p$ then $q$

Method of contrapositive-

By assuming $q$ is not true and prove that $p$ must be true, i.e.,

$\sim q\Rightarrow \sim p$

Let $q$ is not true and prove $p$ is also not true.

$\Rightarrow q$ is not true, i.e., $x$ is not even, i.e., $x$ is odd.

Let $x=2n+1$

Squaring both sides,

$x^2={(2n+1)}^2$

$x^2=4n^2+1+4n$

$x^2=4(n^2+n)+1$

$\Rightarrow x^2$ is odd

$\Rightarrow p$ is also not true.

Hence the given statement is true.