Question

prove that 3+2√5 is irrational number.​

Answer 1

$\huge {\bold {\pink{Proving...}}}$

$<b ><font color ="purple" >$

Let us suppose that 3+2√5 is a rational number

$= > 3 + 5 \sqrt{5} = \frac{p}{q} \\ where \: q is \: not \: equal \: to \: 0 \\ p \: and \: q \: are \: coprime \: nubers \:$

$= > 2 \sqrt{5} = \frac{p}{q} - \frac{3}{1} \\ \\ = > 2 \sqrt{5} = \frac{p - 3q}{q} \\ \\ = > \sqrt{5} = \frac{p - 3q}{2q}$

Since p, q and 3 are integers and subtraction of an integer from an integer is an integer and division of an integer from an integer is a rational number.

But √5 is an irrational number.

Therefore, L. H. S. = R. H. S.

=>Our supposition is wrong.

Hence, 3+2√5 is an irrational number.

$</b ></font >$

$\huge {\bold {\pink{hope \: you \: got \: it}}}$