<s>[INST] Prove that $\mathbb{Q}[\sqrt{3}, \sqrt{5}] = \mathbb{Q}[\sqrt{3}+ \sqrt{5}]$ Prove that $\mathbb{Q}[\sqrt{3}, \sqrt{5}] = \mathbb{Q}[\sqrt{3}+  \sqrt{5}]$
Take $x \in \mathbb{Q}[\sqrt{3}+  \sqrt{5}]. x = a_x + b_x( \sqrt{3} + \sqrt{5}) = a_x + b_x\sqrt{3} + b_x\sqrt{5} \in \mathbb{Q}[\sqrt{3}, \sqrt{5}] \Rightarrow \mathbb{Q}[\sqrt{3}+  \sqrt{5}] \subset \mathbb{Q}[\sqrt{3}, \sqrt{5}] $
I am having trouble proving the other inclusion. 
 [/INST] $(\sqrt 3 + \sqrt 5)(\sqrt 3-\sqrt 5) = -2$ so $\sqrt 3-\sqrt 5=\frac{-2}{\sqrt 3 + \sqrt 5}\in \mathbb Q[\sqrt{3} +\sqrt 5]$
$\sqrt 3 =\frac{(\sqrt 3-\sqrt 5)+(\sqrt 3+\sqrt 5)}2\in \mathbb Q[\sqrt{3}+\sqrt 5]$
$\sqrt 5 = (\sqrt 3 + \sqrt 5)-\sqrt 3\in \mathbb Q[\sqrt{3} + \sqrt{5}]$
So $\mathbb Q[\sqrt 3, \sqrt 5]\subset \mathbb Q[\sqrt 3+\sqrt 5]$
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