Re: Division by zero. Go ahead and laugh.

From: David W. Cantrell (DWCantrell_at_sigmaxi.org)
Date: 10/09/04 

Date: 09 Oct 2004 01:00:26 GMT

"robert j. kolker" <nowhere@nowhere.net> wrote:
> David W. Cantrell wrote:
>
> > No contradiction follows. But please, by all means, feel free to
> > exhibit here how you think that 1 = 0 could be deduced after defining
> > 1/0 in, say, C*.
>
> Definition a/b = c if and only a = b*c. That is what division is.

Fascinating. So according to _your_ definition:

Since 0 = 0*0, we must have 0/0 = 0. And
since 0 = 0*1, we must also have 0/0 = 1.
Therefore, since 0/0 equals both 0 and 1,
by transitivity, we have 0 = 1.

Seriously now: Of course I know the definition you intended to give. And
there's clearly nothing wrong with using that definition of division in
some systems. But you were responding to a statement concerning C*, for
heaven's sake! The definition you intended to give is therefore
inappropriate, not only to C* but to any system in which division of
nonzero quantities by 0 is defined.

> assume 1/0 = x. This means 1 = 0*x (definition of division).

No, that's obviously _not_ what it means in C*.

> 0*x = 0 (well known theorem for rings)
> hence 1 = 0.
>
> QED
>
> > If you have in mind what I suspect you do, then you will have made
> > an unwarranted assumption in order to get your contradiction. Namely,
> > you will have incorrectly assumed at some point that 0 has a
> > multiplicative inverse,
>
> Division means precisely multiplicative inverse. That is what it is.

No, in C* and other systems in which division of nonzero quantities by 0 is
defined, division does not mean that.

> My proof stands.

It never even got off the floor!

David Cantrell

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