Question

Given √2 is irrational.prove that √2/3 is irrational

Answer 1

Let us assume that √2/3 is a rational number.

Rational numbers can be expressed in the form a/b.

Such that a and b are co - prime

and

b ≠ 0.

$\implies \dfrac{\sqrt{2} }{3} = \dfrac{a}{b}$

$\implies \sqrt{2} }= \dfrac{a}{3b}$

Now, The RHS is a rational number

=> √2 is a rational number.

But this contradicts to the fact that √2 is an irrational number.

Hence, our assumption is wrong.

$\boxed{\boxed{\bold{Therefore, \ \frac{\sqrt{2} }{3} \ is \ an \ irrational \ number}}}}}$

                                                           

Answer 2

Given:-

• $\sf{ \sqrt{2}}$ is irrational no.

To prove:-

• $\sf{ \dfrac{ \sqrt{2}}{3}}$ is irrational.

Solution:-

Lets $\sf{ \dfrac{ \sqrt{2}}{3}}$ is rational no.

So, it can be written in the form of $\sf{ \dfrac{p}{q}}$

Such that p and q are co - prime numbers where q ≠ 0.

$\implies \sf{ \dfrac{ \sqrt{2}}{3} = \dfrac{p}{q}}$

$\implies \sf{ \sqrt{2} = \dfrac{p}{3q}}$

Here, $\sf{ \dfrac{p}{3q}}$ is a rational no.

So, $\sf{ \sqrt{2}}$ is also a rational no.

Therefore, this become a contradiction.

$\implies \sf{ \sqrt{2}}$ is an irrational no.

Hence, our supposition is wrong.

$\bold{\underline{\underline{\boxed{\sf{\purple{\dag \; Hence, \; \dfrac{ \sqrt{2}}{3} \; is \; an \; irrational \; number.}}}}}}$

$\rule{200}{2}$