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In the topic of divisibility. Suppose I starts with $a \mid n$, then I manipulate it into $a \mid m$, then $a \mid s$, and finally $a \mid t$. Should I write my train of thought as

$$ \begin{align} a \mid n &\implies a \mid m \\ &\implies a \mid s \\ &\implies a \mid t. \end{align} $$

Or can I just shorthanded it (in the same manner as equality) into

$$ \begin{align} a &\mid n \\ &\mid m \\ &\mid s \\ &\mid t. \end{align} $$

Can I do that? Should I do that? Anyone done that before?

neizod
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    This is just a matter of notation. If you think that your intended audience would be able to understand it (with help from the surrounding context), then you’re good to go. – Soham Saha Mar 13 '24 at 04:44
  • $a \mid n \iff a \mid m$ just means that $n = m$. – J.-E. Pin Mar 13 '24 at 04:53
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    I've never seen anyone do it before (and I hope I never see anyone do it again). – Gerry Myerson Mar 13 '24 at 06:00
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    Equivalence ($\iff$) is usually wrong here. More often than not implication ($\implies$) is what you want. When you write $(a\mid n)\iff (a\mid m)$, you claim that $a$ divides $m$ whenever it divides $n$, and conversely $a$ divides $n$ whenever it divides $m$. But, like Greg, I'm happy with $a\mid m\mid n\mid t$, if all those divisibility relations hold. There are other occasions, when you can make deductions. For example, if $a$ is a prime number known to divide $n=mr$, where we have extra information telling that $a\nmid r$, we can deduce that $a\mid m$. – Jyrki Lahtonen Mar 13 '24 at 06:04
  • Following up with J.-E. Pin's comment, if you are really manipulating $a\mid n$ into an equivalent $a\mid m$, then you often should just be showing $m=n$ or $m=-n$ directly, without the $a\mid$ in the way (however, maybe you have a special context like Jyrki mentioned; in that case, do you really need both directions of the implication?). As pointed out in the comments, showing a single directional implication rather than the bidirectional ones you've written would be a different context. – Mark S. Mar 13 '24 at 10:17
  • I think you are right. It should be $\implies$, not $\iff$. – neizod Mar 13 '24 at 11:24
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    @Mark, manipulating $a\mid n$ into an equivalent $a\mid m$ need not imply $m=\pm n$. I already gave the example $7\mid x+3$ is equivalent to $7\mid x+10$ where this does not imply $x+3=\pm(x+10)$. – Gerry Myerson Mar 13 '24 at 12:13

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If it is the case that $a$ divides $n$ and $n$ divides $m$ and $m$ divides $s$ and $s$ divides $t$, then you can write $a\mid n\mid m\mid s\mid t$ and that's perfectly understandable. (Chaining like this really works for transitive binary relations.) But I wouldn't write this vertically as you've done in the second equation.

On the other hand, if you're proving that all of $n$, $m$, $s$, and $t$ are multiples of $a$ without saying that they divide one another, then the first way (with implication signs) seems reasonable.

Greg Martin
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    I don't think OP is claiming each term divides the next. E.g., $7\mid x+3$ if and only if $7\mid x+10$, but $x+3$ doesn't (necessarily) divide $x+10$. – Gerry Myerson Mar 13 '24 at 06:05
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    You have a point @Gerry, and the question is unclear. My gut feeling, too, was that the expression chain divisibility in the title means using transitivity of "divides". Of course, there is little reason to think that the OP's choice of phrase would align with my intestines. – Jyrki Lahtonen Mar 13 '24 at 06:16