$\sqrt{6}$
let us suppose that $\surd 6$ is rational number.
There exist two co-prime numbers , say p and q
So $\sqrt{6}=\frac{\mathrm{p}}{\mathrm{q}}$
Squaring both sides , we get
$6=\frac{{\mathrm{p}}^2}{{\mathrm{q}}^2}$..(1)
Which shows that ,p² is divisible by 6
this implies , p is divisible by 6
Let p = 6a  for some integer a
Equation (1) implies = 6q² = 36a²
⇒ q² = 6a²
q² is also divisible by 6
⇒ 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