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Re: 0^0=1
Posted:
Apr 25, 2012 10:19 PM
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On Apr 25, 4:34 pm, Jussi Piitulainen <jpiit...@ling.helsinki.fi>
wrote:
> Dan Christensen writes:
> > On Apr 25, 12:45 pm, Rotwang wrote:
> > > The first book I checked, namely Goldrei's /Classic Set Theory/,
> > > defines natural number exponentiation by the following recursion:
>
> > > x^0 = 1
>
> > Odd. In Wiki, they have, as I would expect:
>
> > b^1 = b
> > b^(n+1) = b^n * b
>
> >http://en.wikipedia.org/wiki/Exponentiation
>
> > > x^{n + 1} = x^n*x
>
> That Wikipedia article has a lot more specifically on 0^0. Could you
> live with this: "If the exponent is zero, some authors define 0^0=1,
> whereas others leave it undefined"?
>
> Or even this: "In most settings not involving continuity in the
> exponent, interpreting 0^0 as 1 simplifies formulas and eliminates the
> need for special cases in theorems"?
>
How about a disclaimer along the lines of: "Purely for convenience and
without any real theoretical justification, we assume here that 0^0=1.
It just seems to work in this case."
That's how it seems to me anyway.
Dan
Download my DC Proof 2.0 software at http://www.dcproof.com
Also see video demo
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