Question

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

Answer 1

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.

Answer 2

Hi friend....
For eg: if p=2 which is an even prime number, then
$\sqrt{2} = 1.414........$
therefore 1.414....... is an irrational number.
Hence proved.

Hope it is useful please mark it as brainliest if you find it useful friend :-)