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For $a, b\in\mathbb{Q}$ I have to show that $\mathbb{Q}(\sqrt{a},\sqrt{b})= \mathbb{Q}(\sqrt{a}+\sqrt{b})$.

Unfortunately I dont even know where to start. We never used these brackets $(-)$ before, so I dont know how these sets are actually defined.

I would be grateful for the definitions (:D) and any kind of help or advice! Thank you!

TwoStones
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2 Answers2

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Hint: ${1\over{\sqrt a+\sqrt b}}$ $={{\sqrt a-\sqrt b}\over {a-b}}$ implies that $\sqrt a-\sqrt b\in\mathbb{Q}(\sqrt a+\sqrt b)$.

Tsemo Aristide
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  • Warning: Not valid if $a = b$. – Arthur Jun 13 '19 at 11:40
  • Ok, for $a\neq b$ I got that $\sqrt{a} = \frac{1}{2}(\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b})\in\mathbb{Q}(\sqrt{a}+\sqrt{b})$ and then $\sqrt{b} = (\sqrt{a}+\sqrt{b})-\sqrt{a}\in\mathbb{Q}(\sqrt{a}+\sqrt{b})$. That means $\mathbb{Q}(\sqrt{a},\sqrt{b})\subset\mathbb{Q}(\sqrt{a}+\sqrt{b})$ – TwoStones Jun 13 '19 at 11:46
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If $a=b$ it's obvious.

Now,let $a\neq b$.

$$\sqrt{ab}=\frac{(\sqrt{a}+\sqrt{b})^2-a-b}{2}\in\mathbb Q(\sqrt{a}+\sqrt{b}).$$ Thus, $$(\sqrt{a}+\sqrt{b})\sqrt{ab}=a\sqrt{b}+b\sqrt{a}=a(\sqrt{a}+\sqrt{b})+(b-a)\sqrt{a}\in\mathbb Q(\sqrt{a}+\sqrt{b}).$$ Can you end it now?

Michael Rozenberg
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