Does an irrational number $C$ exist such that $C \cdot \sqrt 2 \in \Bbb{Q}$, where $\sqrt2 \not\mid C$?
I just thought of this, I'm trying to find answers that aren't of the form $C=a\sqrt2, a\in\Bbb{Q}$.
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What does $\sqrt 2\nmid C$ mean? – Hagen von Eitzen May 14 '15 at 20:21
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$\sqrt2$ does not divide $C$. – YoTengoUnLCD May 14 '15 at 20:22
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What does ‘$ \sqrt{2} $ does not divide $ C $’ mean? – Berrick Caleb Fillmore May 14 '15 at 20:28
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@BerrickCalebFillmore That $\frac{C} {\sqrt 2} \not \in \Bbb {Z}$ – YoTengoUnLCD May 14 '15 at 20:29
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1Well, according to your definition, $ \sqrt{2} \nmid \dfrac{1}{2} \sqrt{2} $, but $ \dfrac{1}{2} \sqrt{2} \cdot \sqrt{2} = 1 $, and as the others have explained, it isn’t possible to find a $ C \notin \Bbb{Q} \cdot \sqrt{2} $. – Berrick Caleb Fillmore May 14 '15 at 20:37
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If $C\cdot\sqrt 2\in\mathbb Q$, then $C\cdot\sqrt 2=q$ for some $q\in\mathbb Q$, hence $C=\frac q{\sqrt 2}=\frac q2\cdot\sqrt 2$.
Hagen von Eitzen
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Obviously not: if $C \cdot \sqrt 2 \in \mathbb Q$, then there exist integers $p,q$ such that $C \cdot \sqrt 2 = \frac p q$, i.e. $C=\frac p {2q} \sqrt 2$, which is exactly what you said you do not want.
Alex M.
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