Question

Show that $3\sqrt{2}$ is an irrational number.

Answer 1

Consider that $3\sqrt{2}$ is a rational number. Then,

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

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

Squaring both sides,

$18=\frac{p^2}{q^2}$

$18q^2=p^2$ (1)

This shows that $p^2$ is divisible by 18 and thus $p$ is divisible by 18. Therefore,

$p=18k$

$p^2=324k^2$ (2)

From equation (1) and (2),

$18q^2=324k^2$

$q^2=18k^2$

It can be observed that $p$ and $q$ both are divisible by 18 which is a contradiction to the fact that $p$ and $q$ are co-primes.

Therefore, the assumption is wrong.

Hence, $3\sqrt{2}$ is an irrational number.