**Solution:**

Let us assume, to the contrary, that √p is

rational.

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

such that √p = a/b

=> √p b = a

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

=> a^{2} is divisible by p

=> a is divisible by p

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

Therefore, a^{2} = p^{2}c^{2} ….[Squaring both the sides]

=> pb^{2} = p^{2}c^{2} ….[From (i)]

=> b^{2} = pc^{2}

=> b^{2} is divisible by p

=> b is divisible by p

=> p divides both a and b.

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

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

This contradiction arises because we have

assumed that √p is rational.

Therefore, √p is irrational.