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| 1. |
Prove that √7 is an irratonal number |
| Answer» Suppose {tex}\\sqrt7{/tex} is a rational number. That is , {tex}\\sqrt7{/tex} = {tex}\\frac{p}{q}{/tex} for some p{tex}\\in{/tex}Z and q {tex}\\in{/tex}Z. We can assume the fraction is in lowest fraction, That is p and q shares no common factors.Then {tex}\\sqrt7q=p{/tex}Squaring both side we get,{tex}7q^2=p^2{/tex}So {tex}p^2{/tex} is a multiple of 7,let\'s assume {tex}p=7m{/tex}Then, {tex}7q^2=\\left(7m\\right)^2{/tex}{tex}7q^2=49m^2{/tex}Or {tex}q^2=7m^2{/tex}So {tex}q^2{/tex} is a multiple of 7,{tex}\\therefore{/tex} q is multiple of 7Thus p and q shares a common factor.This is contradiction.{tex}\\Rightarrow {/tex}{tex}\\sqrt { 7}{/tex} is an irrational number. | |