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Comment by codethief | Hacker News Reader

The author's explanation is based on how the rationals are defined explicitly but this leaves the question open whether the rationals couldn't be extended somehow, so that the inverse of 0 is suddenly defined in this extension. More generally, couldn't there be other fields where 0 has a multiplicative inverse?

Of course the answer to both questions is negative but it's the consequence of a much more general fact: In any field the additive zero element can never have a multiplicate inverse. (If it did, the field's other algebraic operations would suddenly lead to contradictions.) And in this sense, there is also no difference between the expressions 1/0 = 1 · 0⁻¹ and 0/0 = 0 · 0⁻¹ because 0⁻¹ simply doesn't exist.

I've always found this explanation much more illuminating. No matter what you do, you can never have 0⁻¹!

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I did actually write a separate article on this exact subject, which I think is also pretty good.

https://math.stackexchange.com/a/125208/25554

Thanks for the link! I liked your explanation that

> [the] warped numbers are like a hole that you can fall into but you can't climb out of, and 0/0 is a deeper hole inside the first hole.

I've sometimes used in Alexandroff one-point compactification in scenarios where you really "need the zero to have an inverse" and it works well in certain applications.

Although they aren't a field as such (if I'm remembering well, because they don't form a division algebra), sedenions (16-tuples, and also further applications of the Cayley-Dickson construction) do have zero divisors (84 of them).

This makes giving any kind of sense to 0^(-1) even harder for me (when 0 is a sedenion).

No denominators for zero divisors in the localization of any ring!!! [English: No denominators where you can multiply the denominator by anything nonzero to get zero.]

Technically, many definitions of localization do allow for zero/zero divisors to be included (we can use any multiplicatively closed set) however by the definition of localization all of the elements in our localized ring become equal to each other.

The mathematical meaning of a/b = c/d in a general ring is that there exists some value s in the set of legal denominators where s(ad - c*b) = 0. If 0 is an allowable denominator then a/b = c/d for every a,b,c,d.

You can have a multiplicative inverse of 0 if 0 = 1. This results in a field with only a single element.

Unless your definition of field include the non-triviality axiom that 0 != 1.

Not to be confused with the field with one element :P

https://en.wikipedia.org/wiki/Field_with_one_element

The definition of a field explicitly requires the additive and multiplicative identities to be distinct elements. You can't induce a field structure on a singleton set. You can construct a trivial ring, but the trivial ring does not comprise a trivial field because it fails to be an integral domain.

To be fair this is mathematical pedantry when we're talking about trivial objects. But in principle it's meaningful because every field you can construct will lose critical properties if you allow one element to be both the additive and multiplicative identity. Unfortunately authors don't always make it clear that you need both existence and uniqueness when they state the field axioms.

Yes this seems like the right answer, both results are undefined so they are the same but not equal.

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