
Re: The ambiguity of 0^0 on N
Posted:
Sep 19, 2013 12:39 AM


On Wednesday, September 18, 2013 7:20:33 PM UTC4, federat...@netzero.com wrote: > On Wednesday, September 18, 2013 9:53:13 AM UTC5, Dan Christensen wrote: > > > Thus, if the Product of Powers Rule is to hold on N, 0^0 will be ambiguous  being either 0 or 1. > > > > To address the problem of how to define 0^0, one should first set the playing field. For example, one might frame the question this way: what [continuous/differentiable/analytic/C^infinity/or/etc.] function, if any, satisfies f(x, y) = x^y when x and y are not zero, and has (x, y) = (0, 0) in its domain. >
Using only natural number arithmetic and assuming PPR, you can prove that 0^0=/=2. Until you can prove either 0^0=/=0 or 0^0=/=1 in this way, 0^0 remains ambiguous.
> > > If the criterion is "continuity", then the list of candidates for f(0, 0) consists of all the limits of f(x, y) as (x, y) > (0, 0) over (say) all paths. > > > > Then, every number is in the candidate list. That is: every COMPLEX number. > > > > If the problem is framed like this: what choice for 0^0 yields, say, a realization of the CurryHoward correspondence, then the answer is unambiguously 1. The correspondence is best seen by laying out the tautologies of the logic of the "ifthen" operator alongside the rules for exponentials. > > > > Then, one has the following correspondences  valid for both classical propositional logic and intuitionistic logic: > > > > (Let X > Y denote the "if X then Y" conditional statement.) > > A^{B+C} = A^B A^C <==> (B or C) > A iff (B > A) and (C > A) > > A^0 = 1 <==> (false > A) iff true > > A^{CB} = (A^B)^C <==> (C and B > A) iff C > (B > A) > > A^1 = A <==> (true > A) iff A. > > (AB)^C = A^C B^C <==> (C > A and B) iff (C > A) and (C > B) > > 1^C = C <==> (C > true) iff C > > > > The equation 0^0 = 1 then corresponds to the tautology (false > false) iff true.
Sorry, I don't follow your argument.
Dan Download my DC Proof 2.0 software at http://www.dcproof.com

