Well, why not? The usual answers boil down to either an appeal to arithmetic intuition (e.g. treating division as repeated subtraction, you can never subtract zero to get anywhere) or definition (you can't divide by zero because zero is defined not to have a multiplicative inverse, i.e. “because I said so”). But the former is mathematically naïve and the latter is unsatisfyingly circular. So let's dig in to why we want to define zero out of division.

It is entirely possible to coherently define a system in which you can divide by zero. Let's do it! Let's define “the reals but you can divide by zero” (which we'll go ahead and notate “𝔻”, for “division”) as: the real numbers, except for all x, x / 0 = 0.

You can divide by zero in 𝔻!

Let's go ahead and try it:
5 / 0 = 0

But, from this we can gather there's probably a reason we had to amend ℝ to allow this. And indeed, there is. The real numbers form a field, a mathematical structure about which a great many very general properties are known, which makes them automatically well-understood and reasonable to manipulate. (Note that this relationship is ahistorical: understanding of the reals predates the definition of a field, with the latter defined around the properties of arithmetic on the reals. But we're not living through history here, so if it's pedagogically convenient to start with fields, I can do that.) Now, while there are a handful of different ways to define a field, the properties we're interested in boil down to the following:

  • A field is a set of elements, with two operations:
  • the first, conventionally called “addition” (because on the reals it's, well, addition), with an identity element conventionally called “zero” such that x + 0 = x, and an inverse for every element such that x + (-x) = 0; and
  • the second, conventionally called “multiplication” (for exactly the reason you think), with an identity element conventionally called “one” such that x * 1 = x and an inverse for every element other than zero such that x * x-1 = 1.
  • And crucially, multiplication distributes over addition: that is, x * (y + z) = (x * y) + (x * z).
If we imagine an alternative set of axioms that drops the exclusion of zero from the definition of multiplicative inverse, such that 𝔻 could satisfy them, what happens?

Well… nothing good, unfortunately. Distribution actually requires that x * 0 = 0 for all x:

x * 0 = a
0 + 0 = 0 by definition of additive identity
x * (0 + 0) = a
(x * 0) + (x * 0) = a by distribution
(x * 0) = a - (x * 0)
But a = x * 0, so a - (x * 0) = 0 by definition of additive inverse
So x * 0 = 0
And therefore a = 0 for all x.
And if x * 0 = 0 for all x, then we can't have a multiplicative inverse for 0, because
x * 0 = 0
x * 0 * 0-1 = x * 1 = x by definition of multiplicative inverse
But x * 0 = 0, so 0 * 0-1 = x for all x.
This can only possibly be true if there's only one element in your field. For fairly sophisticated reasons, there's no such thing as a field with one element, but ℝ and 𝔻 both have significantly more; so they lack a multiplicative inverse for zero with distribution. (𝔻 turns out to lack both, because 0 / 0 ≠ 1, but there's no possible workaround.)

So there's the answer. Division by zero is undefined because if you define it, you can't distribute multiplication over addition; and distribution turns out to be far more useful than division being total.