Question

Prove that $\sqrt{5}$ is an irrational number. Hence show that $3+2\sqrt{5}$ is also an irrational number.

Answer 1

Let $\sqrt{5}$ be a rational number.
So, $\sqrt{5}=\frac{p}{q}$
On squaring both sides
$5=\frac{p^2}{q^2}$
$q^2=\frac{p^2}{5}$

$\Rightarrow 5$ is a factor of $p^2$
$\Rightarrow 5$ is a factor of p.
Now, again let p = 5c.
So, $\sqrt{5}=\frac{5c}{q}$
On squaring both sides
$5=\frac{25c^2}{q^2}$
$q^2=5c^2$
$c^2=\frac{q^2}{5}$
$\Rightarrow 5$ is factor of $q^2$
$\Rightarrow 5$ is a factor of q.
Here 5 is a common factor of p, q which contradicts the fact that p, q are co-prime.
Hence our assumption is wrong, $\sqrt{5}$ is an irrational number.
Now we have to show that $3+2\sqrt{5}$ is an irrational number. So let us assume
$3+2\sqrt{5}$ is a rational number.
$\Rightarrow 3+2\sqrt{5}=\frac{p}{q}\Rightarrow 2\sqrt{5}=\frac{p}{q}-3\Rightarrow 2\sqrt{5}=\frac{p-3q}{q}\Rightarrow \sqrt{5}=\frac{p-3q}{2q}$
$\frac{p-3q}{2q}$ is in the rational form of $\frac{p}{q}$ so $\sqrt{5}$ is a rational number but we have already proved that $\sqrt{5}$ is an irrational number so contradiction arises because we supposed wrong that $3+2\sqrt{5}$ is a rational number. So we can say that $3+2\sqrt{5}$ is an irrational number.