Does Dividing by Zero Return All Numbers? My understanding assumption is that division inverses multiplication. That may be incorrect or incomplete. I will copy the inversion from WolframMathworld
$$a * b = c$$$$a = c \div b$$
Why is the result of dividing any number by zero not the set of all numbers? Is there an unspoken rule that division must always return a single number? Other functions may have many results, for example $\sqrt4 = \{2, -2\}$.
A common example given for why $0/0$ is nonsense, this one found on Reddit:

If $0/0=7$ then $7*0=0$? Yes
If $0/0=54$ then $54*0=0$? Yes
It certainly seems like every number is going to work...

Says that since any number multiplied by zero is zero, the answer to, "what number did I multiply by zero to get zero?" is nonsense since it could be any number. So, to be more specific with the question, why is the act of dividing by zero labeled as undefined, not solvable, and nonsense? Considering in multiplication all numbers times zero is zero, and division is multiplication's inverse, it seems to me zero divided by zero is all numbers.
 A: A function has a certain domain, and a certain target space. For example addition of real numbers can be thought of as a mapping $\Bbb{R}\times\Bbb{R}\to\Bbb{R}$, $(x,y)\mapsto x+y$. Similarly multiplication of real numbers can be thought of as a mapping $\Bbb{R}\times\Bbb{R}\to\Bbb{R}$, $(x,y)\mapsto x\cdot y$.
Now, here comes the problem with division by zero. The set of real numbers with the usual definitions of addition and multiplication form a field. Let's go systematically in this manner of presentation, since that's what you seem to want. Our (inevitably doomed) goal is to somehow define a division operation $\Bbb{R}\times\Bbb{R}\to\Bbb{R}$, meaning we take an input of two real numbers and output a unique real number $(x,y)\mapsto \frac{x}{y}$. Let us now be more precise about the fraction symbol.
For any $x,y\in\Bbb{R}$, consider the following set of numbers
\begin{align}
D_{x,y}:=\{z\in\Bbb{R}\,|\,\, x=z\cdot y\}
\end{align}
Here are some statements we can make: for any $x\in\Bbb{R}$ and $y\in\Bbb{R}\setminus\{0\}$, the set $D_{x,y}$ consists of a single element. This is because $\Bbb{R}$ is a field, meaning by definition every non-zero element has a (necessarily unique) multiplicative inverse. Next, for any $x\in\Bbb{R}$, we have that $D_{x,0}=\Bbb{R}$.
So, to define "division" we can take two routes. One is the standard way, and the other is not.

*

*We can define the mapping $\Bbb{R}\times (\Bbb{R}\setminus\{0\})\to\Bbb{R}$, where we take $(x,y)$ and map it to the unique element of $D_{x,y}$. This is our usual notion of division.


*We could define a new mapping $\Bbb{R}\times\Bbb{R}\to 2^{\Bbb{R}}$, where $2^{\Bbb{R}}$ refers to the power set of $\Bbb{R}$, i.e the set of all subsets of $\Bbb{R}$. The definition is $(x,y)\mapsto D_{x,y}$. There is no way to modify this function to have a target space of $\Bbb{R}$, i.e what is nonsense is trying to modify this function to have $\Bbb{R}$ as target space.
So, the extra layer of abstractness one has to deal with (dealing with the level of sets of sets) just to take into account $0$ is simply not worth it. The larger target space $2^{\Bbb{R}}$ is just unwieldy, and it is inconvenient for many practical calculations because it is not a field/vector space. Therefore, we rather adopt the first alternative: restrict the domain and deal with a manageable target space as opposed to allowing a larger domain but an ugly target space.
Summarizing: it is our (very reasonable) desire to keep $\Bbb{R}$ as the target space (and the field axioms for $\Bbb{R}$) which forces us to say division by zero is undefined (as a real number).

Coming to your comment about square roots. We can do a similar thing: for any $x\in\Bbb{R}$, we can consider the set $S_x:=\{y\in\Bbb{R}\,|\, y^2=x\}$. If $x<0$, then $S_x$ is empty, if $x=0$ then $S_0=\{0\}$ has a unique element of $0$, and lastly, if $x>0$, then $S_x$ has exactly two elements; one of them is positive, the other negative. So, we could define either of the following functions:

*

*$\sigma:\Bbb{R}\to 2^{\Bbb{R}}$, $\sigma(x):= S_x$. So this is the set-valued real-square root.

*Alternatively, we can define $\sqrt{\cdot}\,\,:[0,\infty)\to [0,\infty)$, by defining $\sqrt{x}$ to be the unique non-negative element of the set $S_x$. So, in this definition $\sqrt{0}=0$, $\sqrt{4}=2$, $\sqrt{(-3)^2}=3$ and so on.

Again, the question is one of compromise. Larger domain here requires a larger target space, which is ugly. Or we could restrict our domain and deal with the much simpler target space $[0,\infty)\subset\Bbb{R}$. We obviously adopt the second because it makes arithmetic and notation easier.
I should of course emphasize that this definition of $\sqrt{\cdot}$ is completely arbitrary. We had to make a choice (but once we made this choice nothing is arbitrary anymore); we decided by hand that $\sqrt{\cdot}$ only outputs non-negative real numbers, because we humans like non-negative numbers.
