Question

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

Answer 1

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

So, it can be written in the form $\frac{a}{b}$
$3+2\sqrt{5}=\frac{a}{b}$

Here $a$ and $b$ are coprime numbers and $b\ne 0$

Solving $3+2\sqrt{5}=\frac{a}{b}$ we get,
$\Rightarrow 2\sqrt{5}=\frac{a}{b}-3$
$\Rightarrow 2\sqrt{5}=\frac{a-3b}{b}$
$\Rightarrow \sqrt{5}=\frac{a-3b}{2b}$

This shows $\frac{a-3b}{2b}$ is a rational number.

But we know that $\sqrt{5}$ is an irrational number.

So, it contradicts our assumption.

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

$3+2\sqrt{5}$ is an irrational number

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