Question

prove that root 5 is irrational number ​

Answer 1

Answer:

➡️Answer

$let \: be \: assume \: that \: \sqrt{5} \: is \\ rational \\ \\ therefore \: it \: can \: be \: written \\ in \: the \: form \: of \: \frac{a}{b} \\ \\ \sqrt{5} = \frac{a}{b} \\ \\ where \: a \: and \: b \: areco \: prime \\ \\ positive \: integers\\ \\ \sqrt{5} b = a \\ \\ squaring \: both \: side \\ \\ 5 {b}^{2} = {a}^{2} - (1) \\ \\ {a}^{2} \: is \: divisible \: by \: 5 \\ \\ therefore \: a \: is \: also \: divisible \: by \\ 5 \\ \\ 5c = a \\ \\ squaring \: both \: side \\ \\25 {c}^{2} = {a}^{2} - (2) \\ \\ from \: (1) \: and \: (2) \\ \\ 25 {c}^{2} = 5 {b}^{2} \\ \\ 5 {c}^{2} = {b}^{2} \\ \\ {b}^{2} \: is \: divisible \: by \: 5 \\ \\ therefore \: b \: is \: also \: divisible \: by \: \\ 5 \\ \\ therefore \: 5 \: is \: a \: common \\ factor \: of \: a \: and \: b \\ \\ but \: it \: is \: contradict \: that \: a \\ and \: b \: are \: co \: prime \: numbers \\ \\ therefore \: our \: assumption \: is \: \\ wrong \: and \: \sqrt{5} \: is \: an \: \\ irrational \: number.$

Answer 2

Let us assume that √5 is a rational number.

we know that the rational numbers are in the form of p/q form where p,q are intezers.

so, √5 = p/q

    p = √5q

we know that 'p' is a rational number. so √5 q must be rational since it equals to p

but it doesnt occurs with √5 since its not an intezer

therefore, p =/= √5q

this contradicts the fact that √5 is an irrational number

hence our assumption is wrong and √5 is an irrational number.

$\rule{200}{2}$