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Theorem 1.4 that root 2is irrational proof of real number chapter 1 of class 10 |
| Answer» Let us assume that {tex}\\sqrt 2{/tex} is rationalso let {tex}\\sqrt 2{/tex} ={tex}\\frac pq{/tex}where p and q are integers and also let p and q are not having any common factor\xa0{tex}\\begin{array}{l}2=\\frac{\\mathrm p^2}{\\mathrm q^2}\\\\\\mathrm p^2=2\\mathrm q^2---(1)\\\\\\mathrm{Hence}\\;\\mathrm p^{\\;2}\\;\\mathrm{is}\\;\\mathrm{divisble}\\;\\mathrm{by}\\;2\\\\\\mathrm{so}\\;\\mathrm p^\\;\\;\\mathrm{is}\\;\\mathrm{divisble}\\;\\mathrm{by}\\;2---(2)\\\\\\mathrm{let}\\;\\mathrm p=2\\mathrm t\\;(\\;\\mathrm t\\;\\;\\mathrm{is}\\;\\mathrm a\\;\\mathrm{positive}\\;\\mathrm{integer})\\\\\\mathrm{So}\\;\\mathrm{from}(1)\\\\2\\mathrm q^2=4\\mathrm t^2\\\\\\mathrm q^2=2\\mathrm t^2\\\\\\mathrm{Hence}\\;\\mathrm q^{\\;2}\\;\\mathrm{is}\\;\\mathrm{divisble}\\;\\mathrm{by}\\;2\\\\\\mathrm{so}\\;\\mathrm q^\\;\\;\\mathrm{is}\\;\\mathrm{divisble}\\;\\mathrm{by}\\;2---(3)\\end{array}{/tex}From (2) and (3) it follows thatp and q both having 2 as the common factorBut this makes wrong our assumption that p and q are not having any common factorThis has come as we assumed that {tex}\\sqrt 2{/tex} is rational.Hence {tex}\\sqrt 2{/tex} is a irrational number | |