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Re: 0^0=1
Posted:
Apr 23, 2012 9:37 AM
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On Apr 23, 3:18 am, Jussi Piitulainen <jpiit...@ling.helsinki.fi>
wrote:
> Dan Christensen writes:
> > On Apr 23, 12:45 am, Jussi Piitulainen wrote:
> > > Dan Christensen writes:
> > > > I have always informally justified 0^0 being undefined as
> > > > follows:
>
> > > > x^0 = x^(1-1) = x^1 / x^1 = x/x is undefined for x=0
>
> > > Then x^n is undefined for any n at x = 0. Think x^3 = x^4 / x^1.
>
> > > Then I suppose also 0 * 0 is undefined since it's 0^2. And 0
> > > itself is undefined since it's 0^1. And 0 * x is undefined for all
> > > x since it's (0 * x)^1 = 0^1 * x^1.
>
> > > I suggest you don't want to follow your own argument after all.
>
> > Ooops! Let me rephrase. Only if x =/= 0 can we say that x^0 = x/x = 1.
>
> > Does this mean that x^0 is "undefined" for x=0? It depends how you
> > define exponentiation for the integers. IIRC, the recursive
> > definition is usually given as: x^1=x and x^(n+1)=x^n * x. How do we
> > attach a meaning to x^0? Only if x=/=0 can we write x^n =
> > x^(n+1)/x. Then, for n=0, we have
>
> > x^0 = x^1 / x = 1
>
> > Thus, using the above definition, it seems we cannot attach any
> > meaning to 0^0, i.e. it is undefined.
>
> Yes, if you leave it undefined, then it will be undefined.
>
> If you want it defined, you define it, and you want all other things
> to work well when you do that. You could start with x^0 = 1. It works.
> Nothing else works.
>
Nothing? Why couldn't you just add to the above recursive definition,
say, the "fact" that 0^0=19 (or any other integer)? It would "work" in
the sense that it would be consistent with any statement formally
derived from the original definition.
Dan
Download my DC Proof 2.0 software at http://www.dcproof.com
Also see video demo
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