<s>[INST] Show that $ a，b, \sqrt{a}+ \sqrt{b} \in\mathbb Q \implies \sqrt{a},\sqrt{b} \in\mathbb Q $ Inspired by this, I was wondering if there is a simple logical argument to 
Show that $ a,b, \sqrt{a}+ \sqrt{b} \in\mathbb Q  \implies  \sqrt{a},\sqrt{b} \in\mathbb Q $
Note that the original link is using a computational method, where as I am looking for a simple logical argument.
I tried (unjutifiably) to argue that if some of two square roots is rational then each one is rational, this is a different than the (incorrect) argument that if sum of two algebraic numbers is rational then each one is rational ( counter example $a=1-\sqrt{2},b= 1+\sqrt{2} $)
 [/INST] Note that $a+b=(\sqrt{a}+\sqrt{b})^2-2\sqrt{ab}$ and since both a+b and $\sqrt{a}+\sqrt{b}$ are rational, we may claim that $\sqrt{ab}$ is also rational. These remind us of the quadratic: $(x-\sqrt{a})(x-\sqrt{b})=x^2-2x(\sqrt{a}+\sqrt{b})+\sqrt{ab}$ Solving for x gives us $x=\frac{\sqrt{a}+\sqrt{b}+-\sqrt{a+b-2\sqrt{ab}}}{2}$ which implies that if $\sqrt{a}-\sqrt{b}$ is rational, then so are $\sqrt{a}$ and $\sqrt{b}$. Since $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$ and both $a-b$ and $\sqrt{a}+\sqrt{b}$ are rational, we may conclude that both $\sqrt{a}$ and $\sqrt{b}$ are rational.
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