Let assume that √p is rational
Therefore it can be expressed in the form of \(\frac{a}{b}\), where a and b are integers and b ≠ 0
Therefore we can write √p = \(\frac{a}{b}\)
(√p)2 = (\(\frac{p}{q}\)) 2
P = \(\frac{a^2}{b^2}\)
a 2 = pb2
Since a2 is divided by b2, therefore a is divisible by b.
Let a = kc
(kc)2 = pb2
K 2c 2 = pb2
Here also b is divided by c, therefore b2 is divisible by c2.
This contradicts that a and b are co - primes.
Hence √p is an irrational number.