solution :- let us assume to the contary,that √5 is a rational number.
so √5=(a/b) {here a and b are intergers and a and b have no common factor other than 1 with b≠0.}
√5b=a
by squaring both sides
(√5b)^2=a^2
5b^2=a^2 ................(i)
here a^2 is divisible by 5 ,
so a is also divisible by 5.
so we can write a=5k
by squaring both sides
a^2=(5k)^2
a^2=25k^2 ................(ii)
from (i) and (ii)
5b^2=25k^2
b^2=5k^2
here b^2 is divisible by 5 ,
so b is also divisible by 5.
Therefore a and b have at least 5 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 √5 is irrational
So, we conclude that √5 is irrational.
