Prove that $\Big(\sqrt{3} + \sqrt{7}\Big)$ is irrational.

Let us assume $\sqrt{3} + \sqrt{7}$ is a rational number.

Let, $\sqrt{3} + \sqrt{7} = x$

Squaring both sides, we get :

$\Rightarrow (\sqrt{3} + \sqrt{7})^2 = x^2 \\[1em] \Rightarrow (\sqrt{3})^2 + (\sqrt{7})^2 + 2 \times \sqrt{3} \times \sqrt{7} = x^2 \\[1em] \Rightarrow 3 + 7 + 2\sqrt{21} = x^2 \\[1em] \Rightarrow 10 + 2\sqrt{21} = x^2 \\[1em] \Rightarrow \sqrt{21} = \dfrac{x^2 -10}{2}.$

Here, x is rational,

∴ x² is rational ………(1)

⇒ x² - 10 is rational (As difference between two rational numbers is always rational)

⇒ $\dfrac{x^2 - 10}{2}$ is rational (Dividing two rational numbers results in a rational number)

But, $\sqrt{21}$ is irrational.

$\therefore \dfrac{x^2 - 10}{2}$ is irrational.

Thus, x² - 10 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 $\sqrt{3} + \sqrt{7}$ is a rational number is wrong.

∴ $\sqrt{3} + \sqrt{7}$ is irrational.

Hence, proved that $\sqrt{3} + \sqrt{7}$ is an irrational number.