Prove that

Let us assume that $2-3 \sqrt{5}$ is rational .Then, there exist positive co primes $a$ and $b$ such that

$2-3 \sqrt{5}=\frac{a}{b}$

$3 \sqrt{5}=\frac{a}{b}-2$

$3 \sqrt{5}=\frac{\frac{a}{b}-2}{3}$

$\sqrt{5}=\frac{a-2 b}{3 b}$

This contradicts the fact that $\sqrt{5}$ is an irrational number

Hence $2-3 \sqrt{5}$ is irrational