Answers

2016-01-31T21:07:39+05:30
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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We can also show that √p + √q is irrational, where p and q are non-distinct primes, i.e. p = q 

We use same method: Assume √p + √q is rational. 
√p + √q = x, where x is rational 
√p + √p = x 
2√p = x 
√p = x/2 

Since both x and 2 are rational, and rational numbers are closed under division, then x/2 is rational. But since p is not a perfect square, then √p is not rational. But this is a contradiction. Original assumption must be wrong. 

So √p + √q is irrational, where p and q are non-distinct primes 

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∴ √p + √q is irrational, where p and q are primes
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