Prove that the following is irrational.
$6+\sqrt{2}$

Let us assume that $6+\sqrt{2}$ is rational number

Hence $6+\sqrt{2}$ can be written in the form of $\frac{a}{b}$ where $a,b(b\ne 0)$ are co-prime

$⟹6+\sqrt{2}=\frac{a}{b}$

$⟹\sqrt{2}=\frac{a}{b}-6$

$⟹\frac{a-6b}{b}=\sqrt{2}$

But here $\sqrt{2}$ is irrational and $\frac{a-6b}{b}$ is rational

as $Rational\ne Irrational$

This is a contradiction

so $6+\sqrt{2}$ is a irrational number