Question

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

Answer 1

Sol:

Let $\frac{\sqrt{2}}{3}$ is a rational number.

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

where p and q are some integers and HCF(p, q) = 1 … ….(1)

⇒⇒ $\sqrt{2}q$ =$3p$

⇒⇒
${\sqrt{2}}^2\times q^2$=$3^2{p}^2$

⇒⇒ $2q^2$=$9p^2$

⇒⇒ $p^2$ is divisible by 2

⇒⇒ p is divisible by 2 …….(2)

Let p = 2m, where in is some integer.

Therefore,$\sqrt{2}q$ =$3p$

⇒⇒ $\sqrt{2}q$ =$3\left(2m\right)$

⇒⇒ $2q^2$=$9(2m{)}^2$

⇒⇒ $q^2$=$9\times 2m^2$

⇒⇒ $q^2$ is divisible by 2

⇒⇒ q is divisible by 2 …….(3)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).

Hence, our assumption is wrong.

Thus, $\frac{\sqrt{2}}{3}$ is irrational.