Question

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

Answer 1

Let us assume, to the contrary, that $3\sqrt{2}$ is rational.
That is, we can find coprime a and b (b $\ne$ 0) such that $3\sqrt{2}$ = $\frac{a}{b}$⋅
Rearranging, we get $\sqrt{2}$ = $\frac{a}{3b}$ ⋅
Since 3, a and b are integers, $\frac{a}{3b}$ is rational, and so $\sqrt{2}$ is rational.
But this contradicts the fact that $\sqrt{2}$ is irrational.
So, we conclude that $3\sqrt{2}$ is irrational.