Prove that √p+√q is irrational ,if p and q are prime numbers . Explain it with easy method

Prove that √p+√q is irrational ,if p and q are prime numbers . Exp

1.	Prove that √p+√q is irrational ,if p and q are prime numbers . Explain it with easy method
Answer» Search ResultsFeatured snippet from the webNow x, x², p, q and 2 are all rational, and rational numbers are closed under subtraction and division. So (x² - p - q) / 2 is rational. But since p and q are both primes, then pq is not a perfect square and therefore √(pq) is not rational. But this is a contradiction. lets assume that √p is rational,⇒ √p = a/b ( where \'a\' and \'b\' are co primes, meaning they don\'t have any common factors except for 1)From squaring both sides,p = a²/b²⇒pb² = a²⇒ b² = a²/pSince \'p\' divides a², it also divides \'a\' meaning \'a\' has a factor of pLet \'a\' = pm (where m is a positive integer) ⇒ a² = p²m²Now, pb² = a²pb² = p²m²pb²/p²= m²b²/p =m²∴ \'p\' divides \'b\' ⇒ \'b\' also has a factor \'p\'∴ \'a\' and \'b\' are not co primes and our assumption was wrong⇒ √p is irrationalSimilarly √q is irrational∴⇒ √p + √q is irrational Let us suppose that\xa0√p + √q\xa0is rational.\xa0Let\xa0√p + √q\xa0= a, where a is rational.\xa0=\xa0√q\xa0= a –\xa0√p\xa0Squaring on both sides, we get\xa0q = a2\xa0+ p -\xa02a√p=>\xa0√p\xa0= (a2\xa0+ p - q)/2a, which is a contradiction as the right hand side is rational number, while√p\xa0is irrational.\xa0Hence,\xa0√p + √q\xa0is irrational.