Question

Q6. If p,q are prime positive integers, prove that √p+√q is an irrational number​

Answer 1

Answer:

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 = abab ⇒ √p = (abab) – √q ⇒

(√p)2 = ((abab) – √q)2 [Squaring on both sides]

⇒

p = (a2b2)(a2b2) + q – (2√q abab) ⇒

(a2b2)(a2b2) + (p-q) = (2√q abab) ⇒

(abab) + ((p-q)baba) = 2√3 ⇒ (a2+b2(p−q))2ab(a2+b2(p−q))2ab = √3 ⇒

√3 is rational

[∵ a and b are integers ∴ (a2+b2(p−q))2ab(a2+b2(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.