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

(N/A) Assume that $3+2 \sqrt{5}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $3+2 \sqrt{5} = \frac{a}{b}$.
Subtracting $3$ from both sides,we get $2 \sqrt{5} = \frac{a}{b} - 3 = \frac{a-3b}{b}$.
Dividing by $2$,we get $\sqrt{5} = \frac{a-3b}{2b}$.
Since $a$ and $b$ are integers,$\frac{a-3b}{2b}$ is a rational number.
This implies that $\sqrt{5}$ is a rational number.
However,this contradicts the fact that $\sqrt{5}$ is an irrational number.
Therefore,our assumption that $3+2 \sqrt{5}$ is rational is incorrect.
Hence,$3+2 \sqrt{5}$ is an irrational number.