Question

Prove that 2√7/5 is an irrational number

Answer 1

Hey there !

To prove :-
$\frac{ 2\sqrt{7} } {5}$ is irrational

Lets assume that $\frac{ 2\sqrt{7} } {5}$ is rational 

let ,
$\frac{ 2\sqrt{7} } {5}$ = r , where r is rational 

2√7 = 5r

√7 = $\frac{5r}{2}$

here ,

RHS is purely rational ,whereas on the other hand , 

LHS is irrational

this is a contradiction

hence , our assumption was wrong .

therefore , $\frac{ 2\sqrt{7} } {5}$ is irrational

Answer 2

Let 2√7/5 is a rational number.

A rational number can be written in the form of p/q where p,q are integers.

2√7/5 = p/q

2√7 = 5p/q

√7 = 5p/2q

p,q are integers then 5p/2q is a rational number.

Then √7 is also a rational number.

But this contradicts the fact that √7 is an irrational numb3r.

So,our supposition is false.

Therefore, 2√7/2 is an irrational number.

Hence proved.

Hope it helps