Question

Prove that 2-3√5 is an irrational number.​

Answer 1

Step-by-step explanation:

$\bold\pink{given}⇒ 2 - \sqrt[3]{5} \\ \bold\red{to \: prove} ⇒it \: is \: an \: \\ irrational \: number.. \\ let\: number \: = x \\ ⇒x = 2 - \sqrt[3]{5} \\ ⇒\sqrt[3]{5} = 2 - x \\ ⇒ \sqrt{5} = \frac{2 - x}{3} \\ \bold{since \: x \: and \: 2 - x \: is \: rational \: } \\ \bold{and \: hence \: \frac{2 - x}{3} } \: is \: also \\ \: rational \: number.. \\ \bold{ \sqrt{5} } \: is \: rational \: number.. \\ \bold{hence \: 2 - \sqrt[3]{5} \: must \: be \: an \: irrational \: number.. }$