Question

Show that $\sqrt{2}$ is not a rational number

Answer 1

Solution:

If possible , let $\sqrt{2}$ be a rational number and its simplest form be $\frac{a}{b}$ then, $a$ and $b$ are integers having no common factor other than $1$ and $b\ne 0.$

Now, $\sqrt{2}=\frac{a}{b}⟹2=\frac{a^2}{b^2}$ (On squaring both sides )

or, $2b^2=a^2$ .......(i)

$⟹2$ divides $a^2$ $(\because 2$ divides $2b^2)$

$⟹2$ divides $a$

Let $a=2c$ for some integer $c$

Putting $a=3c$ in (i), we get

or, $2b^2=4c^2⟹b^2=2c^2$

$⟹2$ divides $b^2$ $(\because 2$ divides $2c^2)$

$⟹2$ divides $a$

Thus $2$ is a common factor of $a$ and $b$

This contradicts the fact that $a$ and $b$ have no common factor other than $1.$

The contradiction arises by assuming $\sqrt{2}$ is a rational.

Hence, $\sqrt{2}$ is irrational.