<s>[INST] Prove that any field $F$ containing $\sqrt{a}+\sqrt{b}$ also contains $\sqrt{a}$ and $\sqrt{b}$. 
Prove that any field $F$ containing $\sqrt{a}+\sqrt{b}$ also contains $\sqrt{a}$ and $\sqrt{b}$.

I started by taking $(\sqrt{a}+\sqrt{b})^2=2\sqrt{ab}+a+b$ and want to conclude that $\sqrt{ab}\in F$ but am unsure of this. Then I took $\sqrt{ab}(\sqrt{a}+\sqrt{b})=a\sqrt{b}+\sqrt{a}b\rightarrow b\sqrt{a}+a\sqrt{b}\in F$ and I know this one is because it is of the form $x\sqrt{a}+y\sqrt{b}\in F$. Thus $1\cdot \sqrt{a}+0\cdot \sqrt{b}=\sqrt{a}\in F$ and$ 0\cdot \sqrt{a}+1\cdot \sqrt{b}=\sqrt{b}\in F$. I think I have shown what is needed.
 [/INST] You have a good start by showing that $a \sqrt{b} + b\sqrt{a} \in F$, but this does not show that $x\sqrt{a} + y\sqrt{b} \in F$ for an arbitrary choice of $x$ and $y$ ($a$ and $b$ are fixed constants).  
What you want to do is add something to $a \sqrt{b} + b\sqrt{a}$ to cancel out one of the roots. Like maybe a multiple $x(\sqrt{a}+\sqrt{b})$ (here you can choose $x$ to be what you need).
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