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in Number System by (56.3k points)

If p, q are prime positive integers, prove that √p + √q is an irrational number.

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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.

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