Question

prove that root 2 is irrational number

Answer 1

$\red{questions} \\ \\ prove \: that \sqrt{2} \: \\ is \: irrational \: number.$

$Proof$

Let us assume that √2 is a rational number.

$The \: properties \: of \: rational \: \\ number \: is \: it \: will \: be \: of \: \\ the \: form \frac{a}{b} where \: a \: b \: are \: \\ integers \: .and \: b ≠ 0$


Now

Let a , and b have no common factor other than 1 .


$Now , \sqrt{2} = \frac{a}{b}$
On squaring both side , we get

$2 = \frac{ {a}^{2} }{ {b}^{2} } = > {2b}^{2} = {a}^{2} \\ \\ = > 2 \: divides \: {a}^{2} = > 2 \: divides \: a .$
Then a can be written as 2m , where m is an integer.

in putting a = 2m in eq 1 we get.


${2b}^{2} = ( {2m})^{2} = > {2b}^{2} = {4m}^{2} \\ \\ = > {b}^{2} = {2m}^{2}$

$such \: like \: 2 \: divides \: { b}^{2} = > 2 \: divides \: b \: .$

since

2 is a common factor other than a. and b.



But here our supposition is wrong the facts that a , b have Jo common factor other than 1 . that meant the supposition is arises by assuming that √2 is a rational number .


therefore prove that √ 2 is a irrational.



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