Question

Can any one prove that root 2 is a irrational number....

Answer 1

Hi
Hope this helps you....☺

Answer 2

$Heyaa \: folk \:$
Yes √2 can be proved as a irr national number

This proving Term is called
$CONTRADICTION \: \: METHOD \: = >$

=> in contradiction method first we assume that something is oppose than our actual prove number which we have to prove and then last we contradict that they are not as same
Ok now coming with our Question

lets prove √2 as a irrational number

let us assume on contrary that √2 as a rational number then we know that there exist two positive number a and b
such that -

$\sqrt{2} = ( { \frac{a}{b}) }^{2} where \: a \: and \: b \: are \: coprime \: tht \: is \: there \: hcf \: is \: 1$

$= > ( \sqrt{2} ) = ( { \frac{a}{b} )}^{2} \ \\ \\ = > 2 = \frac{ {a}^{2} }{ {b}^{2} } \\ \\ = > 2 {b}^{2} = {a}^{2} \\ \\ = > 2| \: {a}^{2} \\ \\ = > 2| \: a \: \: \: \: \: \: \: \: ( \: using \: divisibility \: \: theorem).......1) \\ \\ = = > a = 2c \: \: (for \: some \: integer \: c) \\ \\ = > {a}^{2} = 4 {c}^{2} \\ \\ = > 2 {b}^{2} = 4 {c}^{2} (since \: 2 {b}^{2} = {a}^{2} )\\ \\ = > {b}^{2} = 2 {c}^{2} \\ \\ = > 2| \: {b}^{2} \\ = > 2| \: b......2)$
from 1) and ..2) we obtained that 2 is a common factor of a and b but this contradicts the fact that a and b have no common factor other than 1 this mean that our supposition is wrong hence √2 is an irrational number

Hope this helps u !!