Prove that $\sqrt{5} - 2$ is an irrational number.

Let us assume $\sqrt{5} - 2$ is a rational number.

Let $\sqrt{5} - 2 = x$

Squaring on both sides, we get :

$\Rightarrow (\sqrt{5} - 2)^2 = x^2 \\[1em] \Rightarrow (\sqrt{5})^2 + (2)^2 - 2 \times \sqrt{5} \times 2 = x^2 \\[1em] \Rightarrow 5 + 4 - 4\sqrt{5} = x^2 \\[1em] \Rightarrow 9 - 4\sqrt{5} = x^2 \\[1em] \Rightarrow 9 - x^2 = 4\sqrt{5} \\[1em] \Rightarrow \sqrt{5} = \dfrac{9 - x^2}{4}.$

Here, x is rational,

∴ x² is rational ………(1)

⇒ 9 - x² is rational

So, $\dfrac{9 - x^2}{4}$ is rational.

$\Rightarrow \dfrac{9 - x^2}{4} = \sqrt{5}$ is rational

But $\sqrt{5}$ is irrational

$\Rightarrow \dfrac{9 - x^2}{4}$ is irrational i.e., 9 - 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 $\sqrt{5} - 2$ is a rational number is wrong.

∴ $\sqrt{5} - 2$ is irrational.

Hence, proved that $\sqrt{5} - 2$ is an irrational number.