Prove that √5 + √3 is irrational.

1.	Prove that √5 + √3 is irrational.
Answer» 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.

Answer»

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.

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