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.

Hint: $\sqrt{a}-\sqrt{b} = (a-b)(\sqrt{a}+\sqrt{b})^{-1}$
