This might seem like a stupid question but I've been stuck with it all day. It seems to me that
$c|\frac{a}{b}$ iff $c|a$ & ${c\not{|}{b}}$
but I can't prove it.
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3How do you define $c\mid\frac1b$? – José Carlos Santos Feb 23 '21 at 16:53
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1Where are you encountering the notion of an integer dividing (or not dividing) a fraction? The "$\mid$" notation is usually limited to settings with integers on both sides. – Barry Cipra Feb 23 '21 at 16:55
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1I think we can define $\frac ab=nc$ with $n$ integer. Even for reals I'm not overly shocked by $\frac {\pi}4$ being a divisor of $\frac {\pi}2$. – zwim Feb 23 '21 at 16:58
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@zwim In that case, $b=1$ when the fraction is reduced, so... – Rushabh Mehta Feb 23 '21 at 17:00
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hmm, yeah I was thinking $c=\frac a{nb}$ rational, but for $c$ integer that makes no sense. – zwim Feb 23 '21 at 17:04
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I have seen the notion of $p\mid q$ when $p,q\in\mathbb{Q}$ if $q=np$ for some $n\in\mathbb{Z}$. However, if $p\in\mathbb{Z}$, then that forces $q\in\mathbb{Z}$. – robjohn Feb 23 '21 at 17:07
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@BarryCipra group theory – Othman El Hammouchi Feb 23 '21 at 17:46
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@JoséCarlosSantos why does that matter? – Othman El Hammouchi Feb 23 '21 at 17:47
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It matters because I do not know the meaning of that expression and, without knowing it, I am unable to think about your question. – José Carlos Santos Feb 23 '21 at 17:56
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@JoséCarlosSantos the context is the following: for the proof of Syllow theorem, at one point we need to establish that $p|{{p}\choose{i}}$. The book observes that this is the case because p divides the denominator but not the numerator. – Othman El Hammouchi Feb 23 '21 at 18:35
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I did not ask for context. I asked for a definition of $c\mid\frac ab$. – José Carlos Santos Feb 23 '21 at 18:36
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The same as always, because we assume $a/b \in \mathbb{Z}$ – Othman El Hammouchi Feb 23 '21 at 18:37