How to prove root 2 irrational

1.	How to prove root 2 irrational
Answer» Suppose\xa0{tex}\\sqrt2{/tex}\xa0is a rational number. That is ,\xa0{tex}\\sqrt2{/tex}\xa0=\xa0{tex}\\frac{p}{q}{/tex}\xa0for some p{tex}\\in{/tex}Z and q {tex}\\in{/tex}Z.\xa0We can assume the fraction is in lowest fraction, That is p and q shares no common factors.Then {tex}\\sqrt2q=p{/tex}\xa0Squaring both side we get,\xa0{tex}2q^2=p^2{/tex}So\xa0{tex}p^2{/tex}\xa0is a multiple of 2,let\'s assume\xa0{tex}p=2m{/tex}\xa0Then,\xa0{tex}2q^2=\\left(2m\\right)^2{/tex}\xa0{tex}2q^2=4m^2{/tex}Or {tex}q^2=2m^2{/tex}So {tex}q^2{/tex}\xa0is a multiple of 2,{tex}\\therefore{/tex} q is multiple of 2Thus p and q shares a common factor.This is contradiction.{tex}\\Rightarrow {/tex}{tex}\\sqrt { 2 }{/tex}\xa0is an irrational number.

Discussion