Let’s assume on the contrary that √p + √q is a rational number. Then, there exist co prime positive integers a and b such that
√p + √q = \(\frac{a}{b}\)
⇒ √p = (\(\frac{a}{b}\)) – √q
⇒ (√p)2 = ((\(\frac{a}{b}\)) – √q)2 [Squaring on both sides]
⇒ p = \((\frac{a^2}{b^2})\) + q – (2√q \(\frac{a}{b}\))
⇒ \((\frac{a^2}{b^2})\) + (p-q) = (2√q \(\frac{a}{b}\))
⇒ (\(\frac{a}{b}\)) + ((p-q)\(\frac{b}{a}\)) = 2√3
⇒ \(\frac{(a^2 + b^2(p-q))}{2ab}\) = √3
⇒ √3 is rational [∵ a and b are integers ∴ \(\frac{(a^2 + b^2(p-q))}{2ab}\) is rational]
This contradicts the fact that √3 is irrational. So, our assumption is incorrect.
Hence, √p + √q is an irrational number.