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I'm trying to show that

If $a,b$ are positive rational numbers and $\sqrt a+\sqrt b$ is rational, then both of $\sqrt a,\sqrt b$ are rational numbers.

I squared the number $\sqrt a+\sqrt b$ and found that $\sqrt {ab}$ is rational…

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marked as duplicate by Watson, Adam Hughes, Namaste, Henrik - stop hurting Monica, Davide Giraudo Dec 25 '16 at 20:39

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Suppose that both $\sqrt a$ and $\sqrt b$ are irrational.

Since you have $\sqrt a + \sqrt b$ rational, then $\sqrt a - \sqrt b$ is irrational.

$(\sqrt a + \sqrt b)(\sqrt a - \sqrt b)=a-b$

LHS is irrational, RHS is rational. Contradiction!

You can't have one rational and one irrational, hence they are both rational.

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We have $(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=a-b$, hence $\sqrt a-\sqrt b$ is rational, so $2\sqrt a=(\sqrt a+\sqrt b)+(\sqrt a-\sqrt b)$ is rational.
By the same method we can deduce that $\sqrt b$ is rational.

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Another possible approach is the following. Since $\sqrt{a}+\sqrt{b}$ is rational, $(\sqrt{a}+\sqrt{b})^2 = a+b+2\sqrt{ab}$ is rational, hence $\sqrt{ab}$ is rational, so $ab$ is the square of a rational number and $\sqrt{b}$ is a rational number times $\sqrt{a}$, say $q\sqrt{a}$. That gives $\sqrt{a}+\sqrt{b}=(1+q)\sqrt{a}$, hence $\sqrt{a}$ is rational and $a$ is the square of a rational number, so the same holds for $b$.

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Now all you have to do is multiply $\sqrt{ab}$ and $\sqrt a +\sqrt b$. Then we get $b\sqrt a+a\sqrt b$ is rational. If $a = b$ there is nothing to prove . If $a \neq b$ then we can subtract $b\sqrt a+b\sqrt b$ to obtain the result.

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