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

Let us assume $3 - \sqrt{2}$ is a rational number.

Let $3 - \sqrt{2} = x$

Squaring on both sides, we get :

$\Rightarrow (3 - \sqrt{2})^2 = x^2 \\[1em] \Rightarrow (3)^2 + (\sqrt{2})^2 - 2 \times 3 \times \sqrt{2} = x^2 \\[1em] \Rightarrow 9 + 2 - 6\sqrt{2} = x^2 \\[1em] \Rightarrow 11 - 6\sqrt{2} = x^2 \\[1em] \Rightarrow 11 - x^2 = 6\sqrt{2} \\[1em] \Rightarrow \sqrt{2} = \dfrac{11 - x^2}{6}.$

Here, x is rational,

∴ x² is rational ………(1)

⇒ 11 - x² is rational

So, $\dfrac{11 - x^2}{6}$ is rational.

$\Rightarrow \dfrac{11 - x^2}{6} = \sqrt{2}$ is rational

But $\sqrt{2}$ is irrational

$\Rightarrow \dfrac{11 - x^2}{6}$ is irrational i.e. 11 - x² is irrational and so x² is irrational ……..(2)

From (1), x² is rational, and

from (2), x² is irrational

∴ We arrive at a contradiction.

So, our assumption that $3 - \sqrt{2}$ is a rational number is wrong.

∴ $3 - \sqrt{2}$ is irrational.

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