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

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

Let, $(\sqrt{2} + \sqrt{3}) = x$

Squaring on both sides, we get :

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

Here, x is rational,

∴ x² is rational ………(1)

⇒ x² - 5 is rational (Difference between two rational numbers is always rational)

So, $\dfrac{x^2 - 5}{2}$ is rational (Dividing two rational numbers results in a rational number)

But $\sqrt{6}$ is irrational,

$\therefore \dfrac{x^2 - 5}{2}$ is irrational

Thus, x² - 5 is irrational and so x² is irrational ……..(2)

(1) and (2) do not match with each other.

∴ We arrive at a contradiction.

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

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

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