Question

Prove that 2√3/5 is an irrational number

Answer 1

Hey friend !!

To prove :-
2√3/5 is an irrational number

Lets assume that 2√3/5 is rational number.

Let ,
2√3/5 = r , where r is rational.

2√3 = 5r
√3 = 5r / 2

Here ,
we can see that RHS is purely rational.
But on the other hand , LHS is irrational.
This leads to a contradiction.
Hence ,
our assumption was wrong.
Therefore , 2√3/5 is an irrational number

Answer 2

SOLUTION :

Let us assume that $\bf \frac{2\sqrt{3} }{5}$ is a rational number.

so,

$\bf \frac{2\sqrt{3} }{5} = \frac{p}{q}$  

∵ In which p and q are integers and q ≠ 0

⇒ $\bf 2\sqrt{3} = \frac{p}{q}\times5$

[Transporting 5 to RHS]

⇒ $\bf 2\sqrt{3} = \frac{5p}{q}$

⇒ $\bf \sqrt{3} = \frac{5p}{q}\times \frac{1}{2}$

[transporting 2 to RHS]

⇒ $\bf \sqrt{3} = \frac{5p}{2q}$

Since p, q, 2, 5 and q ≠ 0 so,

$\bf \sqrt{3} = \frac{5p}{2q}$ is an rational number

but,

as $\bf \sqrt{3} = \frac{5p}{2q}$ = $\bf \sqrt{3}$

and $\bf \sqrt{3}$ is irrational number

Therefore ,

This contradiction arise due to our wrong assumption

Hence,

$\bf \frac{2\sqrt{3} }{5}$ is an irrational number.