Clarification: I think I should have been more clear. What I'd like to know is whether my conclusions (now bolded, at the bottom) are correct, and whether anyone knows examples of specific structures I mentioned. (Also bolded.)
I'm very comfortable with the idea that you can't divide by zero, and I've seen several different explanations and examples about it that are more than convincing. However, I was wondering what it is exactly that breaks down at a more abstract algebraic level—forget numbers.
Say we have a ring $(A, \star, \circ)$ where $e_\star$ and $e_\circ$ are the neutral elements for $\star$ and $\circ$ respectively. By definition, the two neutral elements behave in the exact same way:
$$a\star e_\star = e_\star \star a = a$$
and
$$a\circ e_\circ = e_\circ \circ a = a$$
for any $a \in A$, so at a glance it is unclear why one of them in particular should not have an inverse element with respect to either operation. However, if we say for example that $\circ$ is the "product" and hence it distributes over $\star$ (seeing as how $A$ is a ring), you can prove that $a \circ e_\star = e_\star$ for any $a \in A$, and then use this fact to show that, if $e_\star$ had an inverse with respect to $\circ$, then $e_\star = e_\circ$ and ultimately you're left with the trivial ring.
So I mused that, if you assume $(A, \star, \circ)$ is a structure such that $e_\star \neq e_\circ$ and both are invertible with respect to both operations, then the distributive law must break down somewhere. I came up with this counterexample
$$e_\star \circ (e_\star \star e_\circ) \neq (e_\star \circ e_\star) \star (e_\star \circ e_\circ)$$
where on the left I wound up with $e_\circ$ and on the right with $e_\star$, which are not the same by assumption. However, in the proof I only used the inverse of $e_\star$; the same reasoning using the inverse of $e_\circ$ doesn't lead to any contradiction.
So, I thought that crux might be that if $\circ$ distributes over $\star$, then $e_\star$ specifically can't be invertible under $\circ$. (Again assuming $e_\star \neq e_\circ$.) If there was a structure $(A, \star, \circ)$ where the distributive law didn't work, both neutral elements might (but won't necessarily) be invertible with respect to both operations. Does this ring correct to you? (Pun not intended, I swear.)
I tried to look up pairs of simple operations that don't distribute over each other to see if that could lead me to an example of a structure where you can "divide by zero" in this sense, but I didn't find any. (I did find out about "wheels" and other exoteric things, but I was looking specifically for simple operations and structures that anyone could understand.) Can anyone think of any such examples?
Finally (I haven't looked into this), I guess it might also be that the distributive law doesn't have to break down if something else does. You might have both neutral elements invertible under both operations and keep the distributive law if you gave up on some other axiom. Correct?
Best Answer
This is one way that can happen:
Let $E$ be the set of closed intervals of $\mathbb R$ containing $0$ (that is, of the form $[a,b]$ with $a,b\in \mathbb R$, and $a\leq 0$ and $b\geq 0$.)
Define $[a,b]\oplus[c,d]=[\min(a,c),\max(b,d)]$ and
$[a,b]\otimes[c,d]=[a+c, b+d]$
It can be checked that this makes $(E, \oplus)$ a commutative monoid with identity $[0,0]$ and $(E,\otimes)$ a monoid with identity also $[0,0]$, and that $\otimes$ distributes over $\oplus$ on both sides.
So $[0,0]$ is indeed invertible with respect to both operations. But, of course, $(E,\oplus)$ is not a group at all: it does not have all additive inverses. This is what permits $[0,0]$ to be nonabsorbing.
This example appears in Section 5.3 of
Those are not hard to find... what if we took $\oplus$ and $\otimes$ to be just regular addition on $\mathbb R$?
Then $\oplus$ does not distribute over $\otimes$, and both make group operations on the set, and they both have the same neutral element, which is invertible under both operations.
Not enforcing distributivity, or something like it to relate the two operations, permits things to break down almost immediately.