Question

Prove that 3+2√5 is irrational number​

Answer 1

Answer:

Let us assume that 3 + 2√5 is an irrational number.

Therefore,

$:\implies\sf 3 + 2 \sqrt{5} = \dfrac{p}{q} \: \: \: \bigg\lgroup \bf where \: p \: and \: q \: are \: co \: prime \bigg \rgroup$

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

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

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

LHS is √5 which is irrational number whereas RHS is p - 3q/2q which is rational beacuse p and q are the integers.

• This contradicts the fact that 3 + 2√5 is rational. So, our assumption is wrong.

Therefore, 3 + 2√5 is irrational number .

Hence Proved !