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This question already has an answer here:

Why is $\mathbb Q(\sqrt 2+\sqrt 3)=\mathbb Q(\sqrt2,\sqrt 3)$ ?

I am Having problems understanding why this is true.

Any input would be greatly appreciated!

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marked as duplicate by Zev Chonoles, lulu, Stefan4024, Dietrich Burde, arctic tern Sep 10 '16 at 20:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Note that $(\sqrt3+\sqrt2\,)(\sqrt3-\sqrt2\,)=1$.

Thus we call $\xi=\sqrt3+\sqrt2$ and note that $\sqrt3=\frac12\left(\xi+\frac1\xi\right)$and $\sqrt2=\frac12\left(\xi-\frac1\xi\right)$.

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