Question

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

Answer 1

Consider that $\sqrt{3}$ is a rational number. Then,

$\sqrt{3}=\frac{p}{q}$

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

Squaring both sides,

$3=\frac{p^2}{q^2}$

$3q^2=p^2$ (1)

This shows that $p^2$ is divisible by 3 and thus $p$ is divisible by 3. Therefore,

$p=3k$

$p^2=9k^2$ (2)

From equation (1) and (2),

$3q^2=9k^2$

$q^2=3k^2$

It can be observed that $p$ and $q$ both are divisible by 3 which is a contradiction to the fact that $p$ and $q$ are co-primes.

Therefore, the assumption is wrong.

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