Question

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

Answer 1

Let us assume, to the contrary, that $3\sqrt{2}$ is

rational. Then, there exist co-prime positive integers $a$ and $b$ such that

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

$\Rightarrow$ $\sqrt{2}=\frac{a}{3b}$

$\Rightarrow$ $\sqrt{2}$ is rational ...$[$$\because 3,a$ and $b$ are integers$\therefore \frac{a}{3b}$is a rational number$]$

This contradicts the fact that $\sqrt{2}$ is irrational.

So, our assumption is not correct.

Hence, $3\sqrt{2}$ is an irrational number.