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

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

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

Squaring on both sides, we get :

$\Rightarrow (\sqrt{3} + \sqrt{2})^2 = x^2 \\[1em] \Rightarrow (\sqrt{3})^2 + (\sqrt{2})^2 + 2 \times \sqrt{3} \times \sqrt{2} = x^2 \\[1em] \Rightarrow 3 + 2 + 2\sqrt{6} = x^2 \\[1em] \Rightarrow 5 + 2\sqrt{6} = x^2 \\[1em] \Rightarrow \sqrt{6} = \dfrac{x^2 - 5}{2}.$

Here, x is rational,

∴ x² is rational ………(1)

⇒ x² - 5 is rational

So, $\dfrac{x^2 - 5}{2}$ is rational.

$\Rightarrow \dfrac{x^2 - 5}{2} = \sqrt{6}$ is rational

But $\sqrt{6}$ is irrational

$\Rightarrow \dfrac{x^2 - 5}{2}$ is irrational i.e. x² - 5 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 $\sqrt{3} + \sqrt{2}$ is a rational number is wrong.

∴ $\sqrt{3} + \sqrt{2}$ is irrational.

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