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Okay so here is my thinking please correct me if I'm wrong:

By definition, if $0|a$ then $a=0k$, for any integer $k$ (assuming $a$ is a divisor of $0$).

Since any integer, $k$ (including $0$), multiplied by $0$ results as $0$, every integer including $0$ is a divisor of $0$.

However, if $a=0$, then $0|0$ is undefined, so $0$ can't be a divisor of itself.

In conclusion, all integers except $0$ is a divisor of $0$.

Please correct me in the way I have written my proof, as well as any incorrect statements I have written. Thanks.

Bill Dubuque
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quadingle the 2nd
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    If divisibility means $a \mid b \iff a=kb$ for some $k \in \mathbb{Z}$, and you notice that $0 = 0 \cdot 1$ ($a=b=0, k =1$), then it is only natural to conclude that $0$ is a divisor of $0$ (if the divisors of $b$ are considered all the $a$ such that $a \mid b$ is true). Perhaps the division $0/0$ is undefined, but the definition stated here is not about whether the division is a whole number or even defined. – PrincessEev Mar 14 '23 at 04:13
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    I agree with your argument that every integer is a divisor of $0$. You have written "$0\mid 0$ is undefined", but I think you mean "$0/0$ is undefined", which is true but not relevant to the first statement. – Greg Martin Mar 14 '23 at 04:13
  • @GregMartin I thought that 0|0 and 0/0 mean the same thing? What is the difference? – quadingle the 2nd Mar 14 '23 at 04:24
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    "$a\mid b$" is pronounced "$a$ divides $b$" (or "$a$ is a divisor of $b$") and is a statement that is either true or false. "$a/b$" is pronounced "$a$ divided by $b$" (or "$a$ over $b$") and is a number. – Greg Martin Mar 14 '23 at 04:27
  • "By definition, if $0|a$ then $a=0k$, for any integer $k$ (assuming $a$ is a divisor of $0$)." This makes no sense. Be more careful about your definitions. – David K Mar 14 '23 at 05:25
  • @DavidK Hey do you mind telling what's wrong with the definition I wrote? I'm trying to say that 0 divides a such that a=0k, for any integer k. Am I missing something? – quadingle the 2nd Mar 14 '23 at 06:10
  • The main reason it made no sense was that every other part of the question is about numbers that divide $0$, but this line is about $0$ dividing some other number. So it seemed that the whole statement was backwards. But if you really meant to say $0$ divides $a,$ then you still should say "for some integer $k$," not "for any integer $k$." – David K Mar 14 '23 at 06:17
  • For a `solution-verification` question to be on topic you must specify *precisely* which step in the proof you question, and *why so*. This site is not meant to be a proof checking machine. – Bill Dubuque Mar 14 '23 at 09:41
  • I disagree with Bill's tag edits (removing Discrete Mathematics, Logic; adding Elementary Number Theory), because the available context indicates that this is a proof-writing question (from an introductory Discrete Mathematics class); that is, the OP is concerned about mathematical/proof writing as much as (probably more so than) about number-theoretic facts. – ryang Mar 14 '23 at 13:28
  • Hey Bill, for future reference, I'm searching Meta trying to understand your comment (second time I've seen you write it), but all I'm coming across are answers like [this](https://math.meta.stackexchange.com/a/25340/21813) ("it is not appropriate to flag as a duplicate of a question asking for a proof, because the asker is making explicit that she/he is interested in feedback on the work they did. Good answers then, address the asker's proof, point out possible improvements; offer feedback"); could you provide a Meta link backing your assertion? Thanks. – ryang Mar 14 '23 at 13:49
  • @ryang Some discussions can be found [here](https://math.meta.stackexchange.com/questions/29119/the-problem-with-proof-verification). In particular, user21820 states that: – Arctic Char Mar 14 '23 at 14:07
  • "I also think that a good proof-verification question will avoid the problem of insisting "But I want my proof evaluated..." as follows. After a bit of research one should be able to (or learn to be able to) identify in what key ways one's proof attempt is different compared to existing proofs, and thus ask for verification about those specific new parts. If one cannot identify any part of the proof that is different, then there is no new question to be answered." – Arctic Char Mar 14 '23 at 14:07
  • @ArcticChar Bill (who is certainly much more involved with and knowledgeable about site meta matters than I)'s template assertion suggests (asserts?) a community consensus, which I was unable to verify both times I tried (granted, each time, I merely skimmed the Meta search results). So, to explicitly ask: is there actually some consensus on this, or are users rather divided? – ryang Mar 17 '23 at 06:14

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There are several definitions of "divides", but a commonly accepted one is as follows:

For $a, b \in \mathbb{Z}$, $a$ divides $b$ (denoted $a | b$) if there exists a unique $c \in \mathbb{Z}$ such that $b = c a$.

Let's assume $0 | a$. That means there exists a unique integer $c$ such that $a = 0 c$. But $0$ multiplied by any integer is $0$, so it must be that $a = 0$. But $0 \nmid 0$, since there is no unique integer $c$ with $0 = 0 c$; indeed, any integer satisfies this. What we've shown is that $0$ is not a divisor of any integer (not even itself).

Let's assume now that $a | 0$, where $a \ne 0$. By defintion, there exists a unique integer $c$ such that $0 = a c$. Since $a \ne 0$, it is clear that $c = 0$, and this solution is unique. What we've shown here is that every nonzero integer is a divisor of $0$.

K. Jiang
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  • Your definition of divides is *nonstandard* since usually it is not required that $c$ is unique (so it *is* true that $0\mid 0)$ – Bill Dubuque Mar 14 '23 at 17:13
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Please correct me in the way I have written my proof, as well as any incorrect statements I have written. Thanks.

By definition

  • if $0|a$ then $a=0k$, for any integer $k$ (assuming $a$ is a divisor of $0$).

Rewriting your definition more clearly, retaining its full meaning:

  • (assuming $a$ is a divisor of $0$) for any integer $k,$ if $0|a$ then $a=0k.$
  • if $a$ is a divisor of $0,$ then if $0$ divides $a,$ then for each integer $k,$ $a=0k.$
  • if $a$ is a divisor of $0$ and $0$ divides $a,$ then for each integer $k,$ $a=0k.$

This is problematic, because

  1. are you wanting to define $a$ dividing $0,$ or $0$ dividing $a,$ or both together?
  2. in $a|0,$ $a$ is conventionally taken to be a positive integer (or at least a nonzero integer), while $0|a$ is conventionally undefined
  3. you actually mean 'some' (not 'any', which in this context means 'each') integer $k$
  4. the definition "object if meaning" implicitly means "object if and only if meaning"; your "if...then" phrasing seems to emphasise the single-directionality of the definition.

Since any integer, $k$ (including $0$), multiplied by $0$ results as $0$, every integer including $0$ is a divisor of $0$.

If we accept that $0|a$ is undefined, then this argument is invalid. In any case, it does not follow from your above (mis)definition.

However, if $a=0$, then $0|0$ is undefined, so $0$ can't be a divisor of itself.

More clearly (omit the "if" part): "However, $0|0$ is undefined, so $0$ can't be a divisor of itself." This statement is correct.

ryang
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