Prove that the number $3\sqrt{2}$ is irrational.

(N/A) Let us assume,to the contrary,that $3\sqrt{2}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $3\sqrt{2} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{2} = \frac{a}{3b}$.
Since $a$ and $b$ are integers,$\frac{a}{3b}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However,this contradicts the fact that $\sqrt{2}$ is an irrational number.
Our assumption that $3\sqrt{2}$ is rational is incorrect.
Therefore,$3\sqrt{2}$ is an irrational number.