Question

Prove that $\sqrt{2}$ is an irrational number.

Answer 1

Let us assume on the contrary that $\sqrt{2}$ is a rational number. Then, there exist positive integers $a$ and $b$ such that
$\sqrt{2}=\frac{a}{b}$ where, $a$ and $b$, are co-prime i.e. their $HCF$ is $1$
$\Rightarrow {\left(\sqrt{2}\right)}^2={\left(\frac{a}{b}\right)}^2$
$\Rightarrow 2=\frac{a^2}{b^2}$
$\Rightarrow 2b^2=a^2$
$\Rightarrow 2|a^2[\because 2|2b^{2}and2b^2=a^2]$
$\Rightarrow 2|a...\left(i\right)$
$\Rightarrow a=2c$ for some integer $c$
$\Rightarrow a^2=4c^2$
$\Rightarrow 2b^2={4c}^2[\because 2b^2=a^2]$
$\Rightarrow b^2={2c}^2$
$\Rightarrow 2|b^2[\because 2|2c^2]$
$\Rightarrow 2|b...\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right),$ we obtain that $2$ is a common factor of $a$ and $b$. But, this contradicts the fact that $a$ and $b$ have no common factor other than $1$. This means that our supposition is wrong.
Hence, $\sqrt{2}$ is an irrational number.