Question

Prove that
3 √5
is an irrational number.

Answer 1

Let us suppose3√5 is a rational no.
so, 3√5 = a/b
√5=a/3b
Here in LHS √5 is an irrational no. and hence our supposition is wrong

Therefore 3√5 is an irrational number.

Answer 2

let 3√5 be rational
therefore there exists two coprimes a and b(not = to 0) such that
$3 \sqrt{5 } = \frac{a}{b}$
$\sqrt{5} = \frac{a}{3b}$
here a, b and 3 are all integers and 3 and b both are non zer
therefore
$\frac{a}{3b}$
is a rational so √5 must also be a rational no
But this is a contradiction to the fact that √5 is irrational
These contradiction is due to our wrong assumption that 3√5 rational
Hence 3√5 is irrational