Question

State: True or False

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

• A. True
• B. False

Answer 1

The correct option is A True

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

That is, we can find coprime integers $a$ and $b$ (b does not equal to 0) such that $3+2\sqrt{5}=\frac{a}{b}$.

Therefore, $\frac{a}{b}-3=2\sqrt{5}$

$\frac{a-3b}{b}=2\sqrt{5}$

$\frac{a-3b}{2b}=\sqrt{5}$

$\frac{a}{2b}-\frac{3}{2}=\sqrt{5}$

Since $a$ and $b$ are integers and rational, we get $\frac{a}{2b}-\frac{3}{2}$ is rational, and so $\frac{a-3b}{2b}=\sqrt{5}$ is rational.

But this contradicts the fact that $\sqrt{5}$ is irrational.

This contradiction has arisen because of our incorrect assumption that $3+2\sqrt{5}$ is rational.

So, we conclude that $3+2\sqrt{5}$ is irrational.