Let us assume, to the contrary, that √p is
So, we can find coprime integers a and b(b ≠ 0)
such that √p = a/b
=> √p b = a
=> pb2 = a2 ….(i) [Squaring both the sides]
=> a2 is divisible by p
=> a is divisible by p
So, we can write a = pc for some integer c.
Therefore, a2 = p2c2 ….[Squaring both the sides]
=> pb2 = p2c2 ….[From (i)]
=> b2 = pc2
=> b2 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.