Question

Prove that 2+ 3√5 is an irrational number given √5 is an irrational

number​

Answer 1

Let us suppose 2+3√5 = p/q , is a rational number where p and q are primes and q≠0.

$= > \: 2 + 3 \sqrt{5} = \frac{p}{q} \\ \\ = > \: 3 \sqrt{5} = \frac{p}{q} - 2 \\ \\ = > \: \sqrt{5} = \frac{p - 2q}{3q} \\ \\ Here, \: \frac{p}{q} \: is \: a \: raional \\ \\ = > \: \frac{p - 2q}{3q} \: \: is \: also \: a \: rational$

But √5 is an irrational (given)

Therefore, this contradicts the fact that

(p-2q)/3q

is a rational number.

This is due to our wrong supposition.

Hence 2+3√5 is an irrational number.