Question

Question number 8. Prove that 3 + √5 is an irrational number.​

Answer 1

Answer :

Let's assume the opposite of the statement i.e., 3 + √5 is a rational number.

$\\ \sf \: 3 + \sqrt{5} = \frac{a}{b} \\ \\ \qquad \: \tiny \sf{(where \: \: a \: \: and \: \: b \: \: are \: \: integers \: \: and \: \: b \: \neq \: 0)}$

$\\ \sf \: \sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}$

Since, a, b and 3 are integers. So,

$\\ \sf \: \frac{p - 3b}{b} \\ \\ \qquad \tiny \sf{ \: (is \: \: a \: \: rational \: \: number \:) }$

Here, it contradicts that √5 is an irrational number.

because of the wrong assumption that 3 + √5 is a rational number.

Hence, 3 + √5 is an irrational number.