**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}=x^{2}

3y^{2}=x^{2}……..(1)

x^{2} 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

3y^{2}=(3k)^{2}

y^{2}=3k^{2}

y^{2} 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.