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How many integer factors of 0 are there, and what are they?

I'm just curious, but what counts as a factor of 0? My guess is that there are an infinite number of factors of 0, but is there a proof?

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    $\begingroup$ Everybody is a factor of $0$, since $(k)(0)=0$. $\endgroup$ May 1, 2014 at 4:59
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    $\begingroup$ @AndréNicolas, I didn't want to know I can divide zero by you :-) $\endgroup$
    – vonbrand
    May 1, 2014 at 7:31

3 Answers 3

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The statement

$a$ is a factor of $b$

means

$b=ka$ for some integer $k$.

Take $b=0$: then no matter what $a$ is, the equation $0=ka$ is always true for some value of $k$, namely, $k=0$. So every integer is a factor of $0$.


This is an excellent example of the importance of using definitions precisely. Some people think that a factor of $b$ means

$b$ divided by any integer "which goes exactly",

for example, the factors of $10$ are $10/1$, $10/(-2)$, $10/5$ etc. This would give the factors of $0$ as being $0/1$, $0/2$ etc, in other words, $0$ only, which is not correct.

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    $\begingroup$ Can you give a reference for the definition of "factor" you gave? I'm sure you can define being a factor as you stated, but I'm not sure that there is agreement about this. There are many ways to say more or less the same thing, but this does not mean they have to have the same definition. For instance by your definition there is among the factors of $10$ the number $-5$; everybody considers $10$ to be a multiple of $-5$, but I think many would not want to consider $-5$ a factor of $10$. I'm not even sure you would; what you wrote does not suggest it. $\endgroup$ May 1, 2014 at 5:28
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    $\begingroup$ I would certainly consider $-5$ to be a factor of $10$, hope my revised answer makes this clear. And while I recognise that there can be some disagreement over definitions, I should have thought this one was pretty standard. The only variation I can find in standard number theory textbooks is that some regard the statement $0\mid0$ as being undefined - NB, not false, just undefined. For the first book I pulled off my shelf, see Hardy and Wright. $\endgroup$
    – David
    May 1, 2014 at 5:44
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    $\begingroup$ -5 is not a factor of 10, over $\mathbb{N}$. It is, over $\mathbb{Z}$. These statements are not controversial. $\endgroup$
    – vadim123
    May 1, 2014 at 5:47
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    $\begingroup$ Well. . . the statement "$-5$ is a factor of $10$ over $\Bbb N$" is not so much false as meaningless, since $-5$ is not an element of $\Bbb N$. Nevertheless your point that context matters is correct. A better example would be that if $\phi=\frac{1}{2}(1+\sqrt5)$ and $\alpha=-4+6\phi$ and $\beta=2+8\phi$, then $\alpha\mid\beta$ is true in ${\Bbb Z}[\phi]$ but not in ${\Bbb Z}[\sqrt5]$. $\endgroup$
    – David
    May 1, 2014 at 6:03
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This question is about terminology, and one can only give an answer if you provide the precise definition of "factor" you are using. Curiously the term does not appear to often get a formal definition, contrary to "divisor" which the term more conventionally associated to the divisibility relation. One may then for convenience consider those two terms to be synonyms. However, it should be noted that most of the time "factor" is used in connection with factorisation. For instance it is more common to speak about prime factors than about prime divisors; the two basically mean the same thing, but a given prime factor might be considered to occur multiple times for a same number, while I don't think one would say that for a prime divisor (which is just a divisor of the number that happens to be prime). In this context it is relevant to note that the number $0$ does not have a prime factorisation and is explicitly excluded from the prime factorisation theorem (usually in one sweep with excluding negative numbers and sometimes also the number$~1$; indeed those numbers pose some difficulties too, but these are less fundamental than the problem with$~0$). Therefore talking about factors of $0$ may be considered not very appropriate. Certainly asking how many factors$~17$ the number$~0$ has does not make any sense.

Apart from that, the following remarks can be made

  • The phrases "$a$ divides $b$" (written $a\mid b$), "$a$ is a divisor of $b$", and "$b$ is a multiple of $a$" express grosso modo the same relation, but that does not mean they can always be used interchangeably. Notably divisors are often implicitly assumed to be positive (for instance when talking about the number of divisors or the sum of the divisors of a (positive) number, though this is not required for $a$ in $a\mid b$. Also many authors will not say that $0$ divides $0$, presumably because one cannot divide $0$ by $0$ (I think most of them prefer to leave the divisibility of $0$ by $0$ undefined rather than false). I've cited in this answer an example of authors that make their position in this respect very clear.

  • I think most people would agree at least that all nonzero numbers divide$~0$, which provides an answer of sorts to your question.

  • However in ring theory the term "zero divisor" means something else, and it does not apply to any integer.

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Take any non-zero k, it can be written 0=k.0

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  • $\begingroup$ but does that count as a factor? $\endgroup$
    – Jason Chen
    May 1, 2014 at 5:00
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    $\begingroup$ Why does $k$ have to be non-zero? Isn't $0=0\cdot 0$? $\endgroup$
    – JRN
    May 9, 2014 at 4:36

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