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

## Solution :

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

Now, let $$3 + 2\sqrt{5}$$ = $$a\over b$$, where a and b are coprime and b $$ne$$ 0.

So, $$2\sqrt{5}$$ = $$a\over b$$ – 3  or  $$\sqrt{5}$$ = $$a\over 2b$$ – $$3\over 2$$

Since a and b are integers, therefore

$$a\over 2b$$ – $$3\over 2$$  is a rational number.

$$\therefore$$  $$\sqrt{5}$$  is an irrational number.

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

This shows that our assumption is incorrect.

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