**Solution:**

Let us assume, to the contrary, that √3 is rational.

So, we can find coprime integers a and b(b ≠ 0)

such that √3 = a/b

=> √3b = a

=> 3b^{2} = a^{2} ….(i) [Squaring both the sides]

=> a^{2} is divisible by 3

=> a is divisible by 3

So, we can write a = 3c for some integer c.

=> a^{2} = 9c^{2} ….[Squaring both the sides]

=> 3b^{2} = 9c^{2} ….[From (i)]

=> b^{2} = 3c^{2}

=> b^{2} is divisible by 3

=> b is divisible by 3

=> 3 divides both a and b.

=> a and b have at least 3 as a common factor.

But this contradicts the fact that a and b are coprime.

This contradiction arises because we have assumed that √3 is rational.

Hence, √3 is irrational.