Question

Prove that $(3-\sqrt{5})$ is an irrational number

Answer 1

Let us assume that $3-\sqrt{5}$ is a rational number
Then. there exist coprime integers $p$, $q$,$q\ne 0$ such that
$3-\sqrt{5}=\frac{p}{q}$
$=>\sqrt{5}=3-\frac{p}{q}$
Here, $3-\frac{p}{q}$ is a rational number, but $\sqrt{5}$ is a irrational number.
But, a irrational cannot be equal to a rational number.This is a contradiction.
Thus, our assumption is wrong.
Therefore $3-\sqrt{5}$ is a irrational number.