Prove that each of the following numbers is irrational:

(i) $\sqrt{3} + \sqrt{2}$

(ii) 3 - $\sqrt{2}$

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

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

Squaring both sides, we get;

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

Here, x is rational,

∴ x² is rational ………………(1)

⇒ x² - 5 is rational

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

But $\sqrt{6}$ is irrational, as it is square root of non-perfect square.

⇒ $\dfrac{x^2 - 5}{2}$ 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.

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

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

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

Squaring both sides, we get;

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

Here, x is rational,

∴ x² is rational ………………(1)

⇒ 11 - x² is rational

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

But $\sqrt{2}$ is irrational, as it is a square root of non-perfect square.

⇒ $\dfrac{11 - x^2}{6}$ 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.

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