On Sunday, November 17, 2013 2:55:58 PM UTC-5, Julio Di Egidio wrote: > "Dan Christensen" <Dan_Christensen@sympatico.ca> wrote in message > > news:firstname.lastname@example.org... > > > On Sunday, November 17, 2013 2:04:57 PM UTC-5, Julio Di Egidio wrote: > > <snip> > > > > >> Yes, but you still do not address any objections to that position. > > >> Namely, > > >> that you have *not* proved that 0^0 *must* be undefined, > > > > > > You can't formally "prove" such an informal statement; you can only > > > provide a rationale as I have done. > > > > That is what *I* would say, while you keep proposing "formal proofs" just at > > the next paragraph! >
I have provided a rationale for leaving 0^0 undefined, one based only formal proofs. No hand-waving, no analogies to various combinatorial or set-theoretic notions. Just ordinary, natural-number arithmetic, as you would expect for a definition of repeated multiplication on N.
> > > >> you have rather at > > >> best provided some formal support to the naive high-school level approach > > >> that simply leaves it undefined. Big difference. > > > > > > I have formally proven that for all x0 in N (including 0), there exists an > > > exponent-like functions ^ such that: > > > > Such that "any number works": exactly what I had explained. >
That's pretty much what we mean by 0^0 being undefined.