Question

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.

Answer 1

Given statement ,

$p:$ If $x$ is an integer and $x^2$ is even, then $x$ is also even.

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

$r:x$ is even.

The given statement is,

if $q$ then $r$

Using method of contrapositive,

By assuming $\sim r$ is true & prove that $\sim q$ is also true.

i.e., $\sim r\Rightarrow \sim q$

let $\sim r$ is true,

i.e., $x$ is not even

i.e., $x$ is odd

i.e., $x=2n+1,n\in Z$

Squaring on both the sides,

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

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

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

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

$\Rightarrow x^2=2(2n^2+2n)+1$

$\Rightarrow x^2=2(2n^2+2n)+1$

$\Rightarrow x^2$ is odd

$\Rightarrow x^2$ is not even

$\Rightarrow \sim q$ is true

Hence, the given statement is true.