Question

Prove that root 2 is an irrational number

Answer 1

$\underline{\red{\bold{QUESTION}}}$ ➡

$\bold{Prove \: \: that \: \: \sqrt{2} \: \: is \: \: an \: \: irrational \: \: number.}$

$\underline{\red{\bold{SOLUTION}}}$ ➡

$\bold{Suppose,} \\ \\ \bold{ \sqrt{2} \: \: is \: \: a \: \: rational \: \:number. } \\ \\ \bold{So,} \\ \\ \bold{We \: \: can \: \: expressed \: \: \sqrt{2} \: \: as \: \: a \: \: rational} \\ \bold{fraction \: \: of \: \: the \: \: form \: \: \frac{a}{b} ,\: \: where} \\ \bold{a \: \: and \: \: b \: \: are \: \: two \: \: relatively \: \: prime} \\ \bold{integers \: \: and \: \: have \: \: no \: \: common \: \: factor.} \\ \bold{and \: \: b \: \: not \: \: equal \: \: to \: \: 0.}$

$\bold{Now,} \\ \\ \bold{ \sqrt{2} = \frac{a}{b} } \\ \\ \bold{ = > 2 = \frac{a {}^{2} }{b {}^{2} } \: \: \: \: [Squaring\: \: both \: \:sides] } \\ \\ \bold{ = > a {}^{2} = 2b {}^{2} \: \: - - - - - > (1)} \\ \\ \bold{ So ,\: \: \: a {}^{2} \: \: is \: \: an \: \: even \: \: integers.} \\ \\ \bold{Therefore ,\: \: a \: \: is \: \: an \: \: even \: \: integers.}$

$\bold{Again,} \\ \\ \: \: \: \: \: \: \: \bold{Let,} \\ \\ \bold{a = 2m \: \: \: [As \: \: square \: \: of \: \: any \: \: odd}\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{integers \: \: is \: \: always \: \: odd.]} \\ \\ = > \bold{ a {}^{2} = (2m) {}^{2} } \\ \\ \bold{ = > a {}^{2} = 4 m {}^{2} \: \: - - - - - - > (2)}$

$\bold{From \: \: (1) \: \: and \: \: (2) ,\: \: we \: \: get} \\ \\ \bold{2b {}^{2} = 4m {}^{2} } \\ \\ = > \bold{ b {}^{2} = 2m {}^{2} } \\ \\ \bold{So, \: \: b {}^{2} \: \: is \: \: an \: even \: \: integer} \\ \\ \bold{Therefore, \: \: b \: \: is \: \: an \: \: even \: \: integers.}$

$\bold{From \: \: the \: \: above ,\: \: we \: \: get} \\ \\ \bold{Both \: \: a \: \: and \: \: b \: \: are \: \: even \: \: integers \: \: and} \\ \bold{have \: \: a \: \: common \: \: factor \: \: i.e. \: \: 2.} \\ \\ \bold{But, \: \: \: this \: \: contradicts \: \: the \: \: fact \: \: that} \\ \bold{a \: \: and \: \: b \: \: have \: \: no \: \: common \: \: factor.} \\ \\ \\ \bold{Hence, \: \: \sqrt{2} \: \: is \: \: not \: \: a \: \: rational \: \: number. } \\ \\ \bold{So, \: \: \sqrt{2} \: \: is \: \: an \: \: irrational \: \: number.}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{ \underline{Proved.}}$

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✝$\red{\mathfrak{THANKS...}}$ ✝

Answer 2

thank you mate hope its clear