solution :- let us assume to the contary,that √3 is a rational number.
so √3=(a/b) {here a and b are intergers and a and b have no common factor other than 1 with b≠0.}
√3b=a
by squaring both sides
(√3b)^2=a^2
3b^2=a^2 ................(i)
here a^2 is divisible by 3 ,
so a is also divisible by 3.
so we can write a=3k
by squaring both sides
a^2=(3k)^2
a^2=9k^2 ................(ii)
from (i) and (ii)
3b^2=9k^2
b^2=3k^2
here b^2 is divisible by 3,
so b is also divisible by 3.
Therefore a and b have at least 3 as a common factor.
But this contradicts the back that a and b are co prime.
This contradiction has Rajendra because of our incorrect a assumption that √3 is irrational
So, we conclude that √3 is irrational.
