Question

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

Answer 1

Consider that $\sqrt{2}$ is a rational number. Then,

$\sqrt{2}=\frac{p}{q}$

Where $p$ and $q$ are co-prime numbers and $q\ne 0$.

Squaring both sides,

$2=\frac{p^2}{q^2}$

$2q^2=p^2$ (1)

This shows that $p^2$ is divisible by 2 and thus $p$ is divisible by 2. Therefore,

$p=2k$

$p^2=4k^2$ (2)

From equation (1) and (2),

$2q^2=4k^2$

$q^2=2k^2$

It can be observed that $p$ and $q$ both are divisible by 2 which is a contradiction to the fact that $p$ and $q$ are co-primes.

Therefore, the assumption is wrong.

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