Question

Prove that $\sqrt{3}$ is an irrational number.

Answer 1

Let us assume on the contrary that $\sqrt{3}$ is a rational number.

Then, there exist positive integers $a$ and $b$ such that
$\sqrt{3}=\frac{a}{b}$ where, $a$ and $b$, are co-prime i.e. their $HCF$ is $1$
Now,
$\sqrt{3}=\frac{a}{b}$
$\Rightarrow 3=\frac{a^2}{b^2}$
$\Rightarrow 3b^2=a^2$
$\Rightarrow 3$ divides $a^2[\because 3\text{divides}3b^2]$
$\Rightarrow 3$ divides $a...\left(i\right)$
$\Rightarrow a=3c$ for some integer $c$
$\Rightarrow a^2=9c^2$
$\Rightarrow 3b^2={9c}^2[\because a^2=3b^2]$
$\Rightarrow b^2={3c}^2$

$\Rightarrow 3$ divides $b^2[\because 3\text{divides}3c^2]$

$\Rightarrow 3$ divides $b...\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right),$ we observe that $a$ and $b$ have at least $3$ as a common factor. But, this contradicts the fact that $a$ and $b$ are co-prime. This means that our assumption is not correct.

Hence, $\sqrt{3}$ is an irrational number.