For any integer a, what integers are divisors of 0?

Question

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.

Answer 1

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 $0$.

Answer 2

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.