Let’s assume on the contrary that √5 + √3 is a rational number. Then, there exist co prime positive integers a and b such that
√5 + √3 = \(\frac{a}{b}\)
⇒ √5 = (\(\frac{a}{b}\)) – √3
⇒ (√5)2 = ((\(\frac{a}{b}\)) – √3)2 [Squaring on both sides]
⇒ 5 = (\(\frac{a^2}{b^2}\)) + 3 – (2√3\(\frac{a}{b}\))
⇒ (\(\frac{a^2}{b^2}\)) – 2 = (2√3\(\frac{a}{b}\))
⇒ (\(\frac{a}{b}\)) – (2\(\frac{b}{a}\)) = 2√3
⇒ \(\frac{(a^2 – 2b^2)}{2ab}\) = √3
⇒ √3 is rational [∵ a and b are integers ∴ \(\frac{(a^2 – 2b^2)}{2ab}\) is rational]
This contradicts the fact that √3 is irrational. So, our assumption is incorrect.
Hence, √5 + √3 is an irrational number.
