Prove that $3+2\sqrt{5}$ is irrational.

Suppose that $3+2\sqrt{5}$ is rational number.

i.e. $3+2\sqrt{5}=\frac{a}{b}$ (where, a and b are co-primes)

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

$\sqrt{5}=\frac{1}{2}\left(\frac{a}{b}-3\right)$

Since a and b both are integers, Thus we get $\frac{1}{2}\left(\frac{a}{b}-3\right)$ as a rational number.

Thus $\sqrt{5}$ is rational number.

Since this statement contradicts the fact that  $\sqrt{5}$ is irrational number, $3+2\sqrt{5}$ is an irrational number.