Waldek,
Thanks for your patient explanation. So we define
1/0 = 0
but we do not call this a multiplicative inverse. Should we say then that
0^(-1)
is still undefined?
Regards,
Bill Page.
On Mon, Mar 28, 2011 at 10:07 AM, you wrote:
>
> Let us recall field axioms:
>
> 1) field is abelian group with respect to addition. Defining 1/0 = 0
> does not affect this at all.
> 2) _nonzero_ elements of the field form an abelian group with
> respect to multiplication. Again defining 1/0 = 0, because
> here we consider only nonzero elements (using 1/0 = x with
> nonzero x would be more tricky)
> 3) distributive law:
>
> \forall_{a, b, c} a*(b+c) = a*b + a*c
>
> Since distributive law does not involve division it is satisfied
> if 1/0 = 0.
>
> The axioms imply \forall_a 0*a = 0. This is fundamental property
> of fields and means that it is impossible to define 1/0 in such a
> way that _all_ elements of field form multiplicative group.
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