Let us assume, to the contrary, that √2 is rational.So integers r and s ≠ 0 such that √2 =r/s.Suppose r and s have a common factor other than 1. Then we divide by the common factor to get √2=a/b, where a and b are coprime.So b√2=a.Squaring on both the sides and rearranging,2b²=a². Therefore,2 divides a².Now, by the theorem 1.3, 2 divides a.So, a=2c for some integer c.Substituting for a, we get 2b²=4c,i.e.,b²=2c².This means that 2 divides b² and so 2 divides b.Therefore,a and b have at least 2 as a common factor.But this contradicts the fact that a and b have no common factors other than 1.This contradiction has arisen because of our incorrect assumption that √2 is rational.So, we conclude that √2 is irrational.
I tell answer of this question tommoro because time is 9 :53