solution :- let us assume to the contary,that √2 is a rational number.
so √2=(a/b) {here a and b are intergers and a and b have no common factor other than 1 with b≠0.}
√2b=a
by squaring both sides
(√2b)^2=a^2
2b^2=a^2 ................(i)
here a^2 is divisible by 2,
so a is also divisible by 2.
so we can write a=2k
by squaring both sides
a^2=(2k)^2
a^2=4k^2 ................(ii)
from (i) and (ii)
2b^2=4k^2
b^2=2k^2
here b^2 is divisible by 2,
so b is also divisible by 2.
Therefore a and b have at least 2 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 √2 is irrational
So, we conclude that √2 is irrational.
