Question

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

Answer 1

Let us assume on contrary that $3+2\sqrt{5}$ is rational. Then there exists

co-prime positive integers a and b such that

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

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

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

$⟹\sqrt{5}$ is rational.

If $\sqrt{5}$ is rational then, there exist co-prime positive integers a and b such that

$\sqrt{5}=\frac{a}{b}$

$5b^2=a^2$

$5|a^2$ $\left[5|5b^2\right]$

$5|a$ ....(1)

$a=5c$ for some positive integer c.

$a^2=25c^2$

$5b^2=25c^2$

$b^2=5c^2$

$5|b^2$

$5|b$ .....(2)

From equations 1 & 2, we find that a and b have at least 5 as a common factor.

This contradicts the fact that and b are co-prime.

Hence, $\sqrt{5}$ is irrational.

So, our assumption is wrong and $3+2\sqrt{5}$ is irrational.