Question

Prove that2 root3 by 5 is irrational

Answer 1

Let us assume $\bold{\dfrac{2\sqrt{3} }{5} }$ is a rational number

Rational numbers are expressed in the form $\bold{\dfrac{p}{q} }$, where p and q and co-prime and q ≠0

$\implies \dfrac{2\sqrt{3} }{5} = \dfrac{p}{q}$

$\implies 2\sqrt{3} = \dfrac{5p}{q}$

$\implies \sqrt{3} = \dfrac{5p}{2q}$

$\dfrac{5p}{2q}\bold{ \ is \ a \ rational \ number}$

$\implies \sqrt{3}\ \bold{ is \ a \ rational \ number}$

But this contradicts to the fact that √3 is an irrational number

Hence, our assumption is wrong.

$\boxed{\boxed{\bold{Therefore, \ \dfrac{2\sqrt{3} }{5} \ is \ an \ irrational \ number.}}}}$

                                                       

Answer 2

$hey$

Let 2root 3 / 5 be a rational no. in the form a/b where, a and b are integers.

Now,

2root3 / 5 = a / b

2root3 = 5a / b

root3 = 5a / 2b

Here. since a/b is a rational no., so 5a / 2b is also a rational no.

Hereby, root3 should also be a rational no.

 

But this contradicts tha fact that root3 is a rational no.

So, our assumption is wrong and hence 2root3 / 5 is an irrational no.