Prove that each of the following numbers in irrational  :$\sqrt{6}$

$\sqrt{6}$
let us suppose that $\surd 6$ is rational number.

There exist two co-prime numbers , say p and q
So $\sqrt{6}=\frac{p}{q}$

Squaring both sides , we get
$6=\frac{p^2}{q^2}$..(1)

Which shows that ,$p^2$ is divisible by 6
this implies , $p$ is divisible by $6$
Let $p=6a$ for some integer $a$

Equation (1) implies = $6q^2=36a^2$
$\Rightarrow q^2=6a^2$
$q^2$ is also divisible by 6
$\Rightarrow q$ is divisible by 6

$6$ is common factors of $p$ and $q$ 
but this contradicts the fact that $p$ and $q$ have no common factor.
our assumption is wrong thus $\sqrt{6}$ is irrational