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Let $a, b \in \mathbb{Q}$ be two elements such that $a,b$ and $ab$ are not squares in $\mathbb{Q}$

Prove that $\mathbb{Q}(\sqrt{a}, \sqrt{b})= \mathbb{Q}( \sqrt{a}+\sqrt{b})$

Viktor Vaughn
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Mogul
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1 Answers1

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Usually, to prove equality of sets $A=B$ you show that $A \subseteq B$ and also $B \subseteq A$.

In your case, here are the two parts.

Part I.

Let $x \in \mathbb{Q}\left(\sqrt{a}, \sqrt{b}\right)$. Then, .... and you conclude $x \in \mathbb{Q}\left(\sqrt{a} + \sqrt{b}\right)$. Hence, $\mathbb{Q}\left(\sqrt{a}, \sqrt{b}\right) \subseteq \mathbb{Q}\left(\sqrt{a}+ \sqrt{b}\right)$.

Part II.

Let $x \in \mathbb{Q}\left(\sqrt{a} + \sqrt{b}\right)$. Then, .... and you conclude $x \in \mathbb{Q}\left(\sqrt{a}, \sqrt{b}\right)$. Hence, $\mathbb{Q}\left(\sqrt{a} + \sqrt{b}\right) \subseteq \mathbb{Q}\left(\sqrt{a}, \sqrt{b}\right)$.

Please update your question or post comments about your progress.

gt6989b
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    i think i got it, since $\sqrt{a}, \sqrt{b} \in \mathbb{Q}( \sqrt{a}, \sqrt{b})$ the sum also so $\sqrt{a}+ \sqrt{b} \in \mathbb{Q}(\sqrt{a}, \sqrt{b})$ then Part II its ok. – Mogul Dec 10 '20 at 20:47
  • and for part I i got $\sqrt{a}= \frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})}{2} \in \mathbb{Q}(\sqrt{a}+ \sqrt{b})$ and $\sqrt{b}=(\sqrt{a}+ \sqrt{b})-\sqrt{a} \in \mathbb{Q}(\sqrt{a}+ \sqrt{b}) $ Thank you :)! – Mogul Dec 10 '20 at 20:52
  • @Mogul both make sense but how do you know that $\sqrt{a} - \sqrt{b} \in \mathbb{Q}\left(\sqrt{a} + \sqrt{b}\right)$? – gt6989b Dec 10 '20 at 21:30