4 ms·
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 someh
by codethief 7y ago
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!
- sunstone 7y agoYes this seems like the right answer, both results are undefined so they are the same but not equal.
- daveFNbuck 7y agoYou can have a multiplicative inverse of 0 if 0 = 1. This results in a field with only a single element.
- gizmo686 7y agoUnless your definition of field include the non-triviality axiom that 0 != 1.
- Certhas 7y agoNot to be confused with the field with one element :P https://en.wikipedia.org/wiki/Field_with_one_element https://en.wikipedia.org/wiki/Field_with_one_element
- throwawaymath 7y agoThe 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.
- mjd 7y agoI did actually write a separate article on this exact subject, which I think is also pretty good. https://math.stackexchange.com/a/125208/25554 https://math.stackexchange.com/a/125208/25554
- codethief 7y agoThanks 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.
- pstch 7y agoAlthough 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).
- chess93 7y agoNo 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.
- kccqzy 7y agoI'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.