Prove that $\sqrt{2}$ is an irrational number.

Assume that $\sqrt{2}$ = $\frac{p}{q}$ (rational number)

squaring on both the sides,

$2$ =$\frac{p^2}{q^2}$

$p^2=2q^2$ ---$\left(1\right)$

$p^2$ is a perfect square number which can be even or odd.

But $2q^2$ is only even number.

So, $p^2$ is not necessarily equal to $2q^2$ which is contradictory to the equation $\left(1\right)$.

So, our assumption was incorrect.

Hence, root is an irrational number.