Prove that for any prime positive integer p, √p is an irrational number.
If possible,let √p be a rational number.also a and b is rational.then,√p = a/bon squaring both sides,we get,(√p)²= a²/b²→p = a²/b²→b² = a²/p [p divides a² so,p divides a]Let a= pr for some integer r→b² = (pr)²/p→b² = p²r²/p→b² = pr²→r² = b²/p [p divides b² so, p divides b]Thus p is a common factor of a and b.But this is a contradiction, since a and b have no common factor.This contradiction arises by assuming √p a rational number.Hence,√p is irrational.
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.