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All the articles I've read regarding "Division by Zero" the main argument for it being an undefined operation, because all proofs lead to contradictions.

iff (0 / x) = (x / 0) = (0 / 0) = (0)

Irrespective of a proof, if the above rules were observed what field axioms of the real numbers would be violated, and how?

In regards to the multiplicative inverse of zero:

(0 * x) = (0) == (0 * y) = (0)

(0 * x) = (0 * y)  // dividing both sides by 0 using the rules above results in

(0) = (0)

Quick Reasoning:
Graphing Division by Zero shows two limits, as the graph tends towards these limits their combined projected values at these two limits negate each other. ie: -infinity (as x tends towards 0 from below) and +infinity (as x tends towards 0 from above) either summate or negate each other to Zero.

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    $\begingroup$ Math Overflow is not the right place for your question. There are other sites more appropriate for such questions, like Ask Dr. Math (mathforum.org/dr.math). In fact, that site has a whole discussion on this topic: mathforum.org/dr.math/faq/faq.divideby0.html $\endgroup$
    – Anton Geraschenko
    Oct 10, 2009 at 15:54
  • $\begingroup$ @Anton: Cutting and Pasting the Stackoverflow FAQ, making slight amendments for "Programmers" to "Mathematicians" and their associated nomenclature, is a surefire way of stating an intention which differs greatly to your actual intent for this site. I recommend you rectify this! "As long as your question is of interest to at least one other mathematician somewhere, (110 views suggests it is,) it is welcome here. Please make your question detailed and specific, and write clearly and simply. (My question is.)" PS: I've read what you've suggest, and it really DOES NOT answer the Q I have posed. $\endgroup$
    – Anonymous User
    Oct 11, 2009 at 9:38
  • $\begingroup$ @Anton: Oh another quote from your supposed FAQ. "We don't run Math Overflow. The community does." LMAO... On a serious note unless you allow a wide variety of Questions, you might as well only allow membership to an Elite Alumni who can pass on memberships to other of their ilk, because as it stands you, the moderators, are hypocrites and elitists. $\endgroup$
    – Anonymous User
    Oct 12, 2009 at 4:11
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    $\begingroup$ @Anonymous: I'm sorry that the FAQ did not properly communicate the intention of this site, and I'm sorry if you feel you've wasted your time. I really do believe in letting the community run the site, but for the time being, there are no 3000+ rep users to vote to close inappropriate questions, so we moderators have to be somewhat dictatorial. The FAQ is a guideline, but there's no magic formula for what the community members find appropriate. As a Math Overflow community member, I have a right to try to keep this site the kind of site that I want to keep visiting. $\endgroup$
    – Anton Geraschenko
    Oct 14, 2009 at 5:40
  • $\begingroup$ I have what I believe to be a better approach to division by zero. For numbers a, b: a/0 = b ⟺ a = 0. This follows simply from the equivalence of multiplication and division relations, and is not or should not be "research level mathematics". I was surprised to find that some people who have studied mathematics do not agree with me. What do you think, could I ask a question here on this topic without it being down voted or closed? $\endgroup$
    – Sam Watkins
    Dec 17, 2015 at 13:19

1 Answer 1

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There's a thing called a meadow which is a (successful) attempt to make multiplicative inverses globally defined. What it does is instead of defining multiplicative inverses, it defines an operation $M \to M$, $x \mapsto x^{-1}$ with the property not that $xx^{-1} = 1$ but that $xx^{-1}x = x$. For any non-zero element then this agrees with the usual inverse but one can extend the inverse operation by defining $0^{-1} = 0$ and it works. I may be wrong, but I think that the result is that every field embeds in a meadow.

So providing you don't claim that $xx^{-1} = 1$ but rather $xx^{-1}x = x$ then you are absolutely fine with $0^{-1} = 0$.

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    $\begingroup$ I like this. It seems completely analogous to pseudoinverses in linear algebra; in fact, I think it's just the special case in which you (1) consider a one-dimensional vector space over some field, (2) then consider that field as the ring of linear transformations on that vector space. $\endgroup$
    – Darsh Ranjan
    Nov 5, 2009 at 16:23
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    $\begingroup$ Given that this question has a useful answer, perhaps it should be re-opened, and edited if necessary to conform with the site's policies. $\endgroup$
    – Sam Watkins
    Dec 17, 2015 at 13:23
  • $\begingroup$ For those unfamiliar with this terminology, meadows are really just commutative von Neumann regular rings. Btw, the axiom list should also include $x^{-1}xx^{-1}=x^{-1}$, or equivalently, $(x^{-1})^{-1}=x$. $\endgroup$
    – Emil Jeřábek
    Jul 5, 2023 at 14:09

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