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 rationalTherefore 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}= pb^{2}Since a

^{2}is divided by b^{2}, therefore a is divisible by b.Let a = kc

(kc)

^{2}= pb^{2}K

^{2}c^{ 2}= pb^{2}Here also b is divided by c, therefore b

^{2}is divisible by c^{2}.This contradicts that a and b are co – primes.

Hence √p is an irrational number.