Show that $3 \sqrt{2}$ is irrational.

(N/A) Let us assume,to the contrary,that $3 \sqrt{2}$ is rational.
That is,we can find 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 $3$,$a$,and $b$ are integers,$\frac{a}{3b}$ must be a rational number,which implies that $\sqrt{2}$ is rational.
However,this contradicts the established fact that $\sqrt{2}$ is irrational.
Therefore,our initial assumption is false,and we conclude that $3 \sqrt{2}$ is irrational.