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Re: 0^0=1
Posted:
May 7, 2012 1:27 PM
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Dan Christensen writes:
> On May 7, 10:54 am, Jussi Piitulainen wrote:
> > Dan Christensen writes:
> > > set theory are consistent), but wouldn't a system that defined 0^0=k
> > > for ANY integer k also be consistent?
> >
> > > Presumably, x^(m+n) = x^m * x^n would still hold. Then 0^(m+0) = 0^m
> > > * 0^0 = 0 * k. In this equation, ANY value would work for k. On what
> > > basis can we say that k MUST be 1?
> >
> > The consequences of setting 0^0 to any number other than 0 or 1 are
> > considered highly undesirable. Work it out: use the above rule of
> > exponents to show that 0^0 = 3 implies 3 = 9.
>
> Thanks! We would have 0^0 = 0^(0+0) = 0^0 * 0^0. With 0^0 = 3,
> that's 3 = 3*3.
Exactly.
> So, why not 0^0 = 0? Can this be ruled out using only natural number
> arithmetic?
Not directly, as far as I can see, but almost nothing supports it
while much requires x^0 = 1, either for all x or for all natural x.
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