Solution :
Let us assume ,that √ 3 is rational
i.e√ 3=x/y (where,x and y are co-primes)
y√ 3=x
Squaring both sides
We get,(y√3)2=x2
3y2=x2……..(1)
x2 is divisible by 3
So, x is also divisible by 3
therefore we can write x=3k (for some values of k)
Substituting ,x=3k in eqn 1
3y2=(3k)2
y2=3k2
y2 is divisible by 3 it means y is divisible by 3
therefore x and y are co-primes.
Since ,our assumption about √ 3 is rational is incorrect .
Hence, √ 3 is irrational number.