Prove that the following number is irrational: $3 \sqrt{5}$.

(N/A) Assume,to the contrary,that $3 \sqrt{5}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $3 \sqrt{5} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{5} = \frac{a}{3b}$.
Since $a$ and $b$ are integers,$\frac{a}{3b}$ 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 is incorrect,and $3 \sqrt{5}$ must be an irrational number.