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'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…
 A: 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. 
A: 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 $b$.
A: 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.
A: 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.
