Question

If p is a prime number, then prove that √p is irrational.

Answer 1

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
=> 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.