Question

prove that √5 is irrational number​

Answer 1

Step-by-step explanation:

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Answer 2

⭐SOLUTION

$|solution| \\ \\ let \: us \: assume \: \sqrt{5} \: \: is \: an \: \\ rational \: number \: \sqrt{5 } \\ can \: be \: expressed \: in \: the \\ \: form \: of \: \: \frac{a}{b} \: \: where \: a \: \: and \: b \: are \: \\ co - prime \: integers \: \:. \\ \sqrt{5} = \frac{a}{b} \\ square \: on \: both \: side \\ {5} = \frac{ {a}^{2} }{ {b}^{2} } \\ ( 5 \: {b}^{2} = {a}^{2} ) = > |1| \\ 5 \: \: divides \: {a}^{2} \\ = \: 5 \: divides \: a \\ ( \: by \: using \: theorm \: 1.3) \\ let \: a \: = 5m \\ square \: on \: both \: side \\ {a}^{2} = {(5m)}^{2} \\ {a}^{2} = {25m}^{2} = > |2| \\ substitute \: eqn \: |2| \: in \: eqn \: |1| \\ {5b}^{2} = > {a}^{2} \\ {5b}^{2} = {25m}^{2} \\ {b}^{2} = \frac{25}{5} {m}^{2} \\ b = 5 {m}^{2} \\ 5 \: divides {b}^{2} \\ 5 \: divides \: b \\ \\ = > \: a \: and\: b \: have \: 5 \: as \: common \: \\ factor \: and \: this \: is \: contradiction \: \\ to \: the \: fact \: that \: a \: and \: b \: \\ co \: - \: prime. \\ \sqrt{5} \: cannot \: be \: written \: \\ in \: the \: form \: of \: \frac{a}{b} \\ = > so \: \sqrt{5 \: } \: is \: an \: irratioal \\ number..$

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