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by throwawaymath 2869 days ago
Because a divisor cannot exist unless it is a multiplicative inverse. Therefore 0 is not a divisor.

This is getting to be Kafkaesque...it breaks the field axioms themselves. How many different explanations and external resources do I need to provide in this thread for you to be convinced that this is not a controversial point in modern mathematics? I just explained it in the comment you responded to.

You have exactly two options here.

If you define x/0, that definition must interact with the rest of the definitions and elements of the field. To maintain multiplicative closure (a field axiom!) there must be a unique y element equal to x/0. So tell me how you will define x/0 such that x is not equal to the product of 0 and y. Regardless of what you think the author has shown, the burden of proof is not on me at this point to show that you can't do it, because it follows directly from the field axioms. Trying to impose a one-off bizarro divisor function defined only on {0} is not only mathematically inelegant, it immediately eliminates the uniqueness of all field elements. Therefore your "field" just becomes {0}, and since it lacks a multiplicative identity it ceases to be a field. There is your contradiction. Why don't you tell me how you're going to prove any equation defined over a field that relies on the uniqueness or cancellation properties of fields?

On the other hand, let's say you tell me you want define x/0 so that the definition doesn't interact with any of the field definitions or elements. Then you haven't actually introduced any new operation or definition, you've just designed a notation that looks a lot like division but has no mathematical bearing on the field itself (i.e. absolutely nothing changes, including for 0). That's not a divisor function, it's just a confusing shorthand. You can't just add another axiom to a field and call it a field.

If you believe I'm stubborn, that's fine. I might be a poor teacher! There are ample resources online which will patiently explain this concept in mind numbing detail. It boggles my mind that there are people in this thread still fighting an idea in earnest which has been settled for over a century.

1 comments

> Because a divisor cannot exist unless it is a multiplicative inverse. Therefore 0 is not a divisor.

So there are two separate issues here. One is whether we can extend the definition of the "/" operator, and the other is whether we call it "division".

I'm not interested in what we call it. I'm interested in the claim that extending "/" will break the field.

The dichotomy you're talking about is wrong. The two options are not "multiplicative inverse" and "does not interact with anything". "1/0 = 0" interacts with plenty! If I make a system where it's an axiom, I can calculate things like "1/0 + 5" or "sqrt(1/0)" or "7/0 + x = 7". I can't use it to cancel out a 0, but I can do a lot with it.

> It boggles my mind that there are people in this thread still fighting an idea in earnest which has been settled for over a century.

Remember, the question is not "should this be an axiom in 'normal' math?", the question is "does this actually conflict with the axioms of a field?"

> You can't just add another axiom to a field and call it a field.

Yes you can. There is an entire hierarchy of algebraic structures. Adding non-conflicting axioms to an X does not make it stop being an X.