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Pretty simple problem but I'm having trouble finding the proof.

Numbers in $\mathbb{Q}[\sqrt3]$ are of the form $r+s\sqrt{3}$; $r,s \in \mathbb{Q}$. I tried proving by contradiction. That is, assume $\sqrt{2} \in \mathbb{Q}[\sqrt3]$ . Then, $\sqrt{2} = r+s\sqrt{3}$ $\rightarrow$ $2=r\sqrt{2}+s\sqrt{6}$. Then, since $\sqrt2$ and $\sqrt6$ are both irrational, while $2,r,s$ are all rational, there will be some sort of contradiction. However, I'm not sure if this is right, and if it is, how to formalize it.

Bill Dubuque
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HotelTrivago
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  • I think you should move $\sqrt{2}$ and $\sqrt{3}$ to the same side and then square the equation, which will leave only $\sqrt{6}$ and other rational parts. A contradiction. Of course, you have to mention that s is not equal to zero. – toronto hrb Sep 08 '20 at 15:13

1 Answers1

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If you want an elementary proof, write $$\sqrt{2} = r +s \sqrt{3}$$ $$\Longrightarrow\sqrt{2} - s \sqrt{3} = r$$ $$\Longrightarrow(\sqrt{2} - s \sqrt{3})^2 = r^2$$ $$\Longrightarrow2+3s^2-2s\sqrt{6} = r^2$$ $$\Longrightarrow \frac{2+3s^2-r^2}{2s}=\sqrt{6} $$

($s \neq 0$ because $\sqrt{2}$ is not an integer) which is absurd because $\sqrt{6}$ is irrationnal.

TheSilverDoe
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