Definition of $b|a \implies 0|0$? The definition I'm using for $b|a$ (taken from Elementary Numbery Theory by Jones & Jones):

If $a,b \in \mathbb{Z}$ then $b$ divides $a$ if for some $q \in \mathbb{Z}$ $a = qb$. 

However, I have $0 = q\cdot0$ for any $q$ I choose. So this seems to imply that $0$ divides $0$ which I know is always taken to be undefined. Should the definition be for "unique $q$" rather than for "some $q$"? 
Thank-you. 

In the theory of (unitary) commutative rings the following expressions and statements are more of less equivalent, in many cases by definition, for $a,b\in R$


*

*$\exists c\in R: ac=b$

*$b$ is a multiple of $a$

*$b\in aR$

*$bR\subseteq aR$

*$a$ is a divisor of $b$

*$a\mid b$

*$a$ divides $b$


The first alternative is the defining expression, and it so simple that there are no obvious refinement or coarsening that one could associate with any of the other formulations to make them essentially different. However when talking about "divisibility" as in the last three alternatives, much of the usual type of reasoning is only valid if there is a well defined quotient $b/a$, in other words if the element $c$ in the definition not only exists but is also unique. One condition that will ensure this is when $a$ is a regular element of $R$, i.e., neither zero nor a zero divisor. Indeed is is common to restrict the mention of divisibility to the even more special case where $R$ is an integral domain (so there aren't any zero divisors at all), and both $a$ and $b$ are nonzero. (For $b$ there is no real reason to require this, but forbidding it avoids confusion between "$a$ is a divisor of $0$", which is forbidden but would otherwise be valid for all $a\in R$, and "$a$ is a zero divisor", which means something completely different.) Usually reasoning about divisibility is done only for nonzero elements of an integral domain; I guess it would rather tricky when going beyond these cases.
With this conventional restriction of terminology one shouldn't say that $0$ (or anything else) divides $0$, but one could say that $0$ is a multiple of $0$ (or anything else).
I don't think the initial defining condition but modified to "there exists a unique $c$" is much used in the general context (where the ring may have zero divisors); probably there is no much use that could be made of such a notion.
