Question

prove that under root 5 is an irrational nunber .... full explanation​

Answer 1

Answer:

Answer

$let \: be \: assume \: that \\ \: \sqrt{5} \: \: is \: a \: rational \: number \\ \\ therefore \: it \: can \: be \: written \: in \: \\ the \: form \: of \: \frac{a}{b} \\ where \: b \: not \: equals \: to \: zero \\ and \: a \: and \: b \: are \: co \: prime \\ positive \: integers \\ \\ \sqrt{5} = \frac{a}{b} \\ \\ \sqrt{5} b = a \\ \\ squaring \: both \: side \\ \\ 5 {b}^{2} = {a}^{2} - (1) \\ \\ {a}^{2} \: is \: divisible \: by \: 5 \\ \\ therefore \: a \: is \: also \: divisible \: by \: 5 \\ \\ 5c = a \\ \\ squaring \: both \: side \\ \\ 25 {c}^{2} = {a}^{2} - (2) \\ \\ from \: (1) \: and \: (2) \\ \\ 25 {c}^{2} =5 {b}^{2} \\ \\ 5 {c}^{2} = {b}^{2} \\ \\ {b}^{2} \: is \: divisible \: by \: 5 \\ \\ therefore \: b \: is \: also \: divisible \: by \: 5 \\ \\ therefore \: 5 \: is \: a \: common \: factor \\ of \: a \: and \: b \\ \\ but \: it \: is \: contradict \: that \: a \: and \: b \\ are \: co \: prime \: \\ \\ therefore \: our \: assumption \: is \\ wrong \: and \: \sqrt{5} \: is \: an \: irrational \\ number.$

Answer 2

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