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by imh 2877 days ago
>Since y = x/0, it follows that the product of y and 0 is equal to x, because division is the inverse of multiplication.

Can you explain how this follows? I thought division was only the inverse of multiplication for all nonzero denominators, which would mean we can't use that definition for deduction in x/0.

It might hinge on your next sentence:

>By the field axioms, division does not exist if there is no multiplicative inverse with which to multiply.

but I don't understand why that's necessarily true. I don't understand how the field axioms require division by x to require the existence of a multiplicative inverse of x when x is zero. Sorry to take a bunch of your Friday, but I'm very curious now. Explanation much appreciated.

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edit:

Come to think of it, couldn't I define x/y as cotton candy for all x,y in field F and still satisfy the field axioms? They just don't refer to division.

Any connection between x/y and y's multiplicative inverse is just a nice convention. That convention states that x/y = x * mult_inv(y) when y != 0, but nothing else. That definition has nothing to do with the field axioms and changing it doesn't require that I change anything about multiplicative inverses. That means I don't touch the field axioms and my field is still a field.

2 comments

Your edit is starting to get it. If by the argument of the article x/y = cotton candy for all x,y, then probably the argument of the article isn't good. And the reason is precisely that division in a field is taken to be nothing but a notational shorthand for multiplication by the multiplicative inverse.
> Come to think of it, couldn't I define x/y as cotton candy

Yes. That's the thesis of the article. Make an arbitrary choice for 1 / 0 = ?, and if it helps you, use it. It's mathematically, rigorously fine.