Theorem 1.4 that root 2is irrational …

Let us assume that {tex}\sqrt 2{/tex} is rational

so 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

 {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 that

p and q both having 2 as the common factor

But this makes wrong our assumption that p and q are not having any common factor

This has come as we assumed that {tex}\sqrt 2{/tex} is rational.

Hence {tex}\sqrt 2{/tex} is a irrational number