prove that
is irrational
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First, we'll assume that √p + √q is rational, where p and q are distinct primes
√p + √q = x, where x is rational
Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides.
(√p + √q)² = x²
p + 2√(pq) + q = x²
2√(pq) = x² - p - q
√(pq) = (x² - p - q) / 2
Now 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. Original assumption must be wrong.
So √p + √q is irrational, where p and q are distinct primes .
First, we'll assume that √p + √q is rational, where p and q are distinct primes
√p + √q = x, where x is rational
Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides.
(√p + √q)² = x²
p + 2√(pq) + q = x²
2√(pq) = x² - p - q
√(pq) = (x² - p - q) / 2
Now 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. Original assumption must be wrong.
So √p + √q is irrational, where p and q are distinct primes .
puja77:
thanks bhai it helps me a lot
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