Prove that 3 + $2\sqrt{5}$ is irrational.

Let us assume, to the contrary, that 3 + $2\sqrt{5}$ is a rational number.

So, it can be written in the form $\dfrac{a}{b}$.

3 + $2\sqrt{5} = \dfrac{a}{b}$ ……….(1)

Here a and b are coprime numbers and b ≠ 0.

Solving equation (1), we get :

$\Rightarrow 3 + 2\sqrt{5} = \dfrac{a}{b} \\[1em] \Rightarrow 2\sqrt{5} = \dfrac{a}{b} - 3 \\[1em] \Rightarrow 2\sqrt{5} = \dfrac{a - 3b}{b} \\[1em] \Rightarrow \sqrt{5} = \dfrac{a - 3b}{2b}.$

This shows that $\dfrac{a - 3b}{2b}$ is a rational number, but we know that $\sqrt{5}$ is an irrational number.

∴ Our assumption of $3 + 2\sqrt{5}$ is a rational number is incorrect.

Hence, proved that $3 + 2\sqrt{5}$ is irrational.