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

## Answers ( )

Answer:

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

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