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Re: 0^0=1
Posted:
May 6, 2012 4:21 PM
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In article <jo6kjs$ceo$1@speranza.aioe.org>,
"LudovicoVan" <julio@diegidio.name> wrote:
> "Frederick Williams" <freddywilliams@btinternet.com> wrote in message
> news:4FA6D39D.9C8F63C2@btinternet.com...
> > LudovicoVan wrote:
> >> "Frederick Williams" <freddywilliams@btinternet.com> wrote in message
> >> news:4FA6CCAA.2FEA538C@btinternet.com...
> >
> >> > Not at all, 0^0 = 1. To be sure, if you assume 0/0 = x (any real
> >> > number
> >> > of your choice) then problems arise. None arise with the assumption
> >> > that 0^0 = 1.
> >>
> >> Among other things, I have just shown that from 0^0=1 follows that 1=0.
> >
> > Show it again. Don't bother to type it, just cut and paste.
>
> It's here:
> <news:jo6egf$qp6$1@speranza.aioe.org>
> or here:
> <http://groups.google.com/d/msg/sci.math/kPhUbtrUoSU/YyMA5hoZ1ZkJ>
>
> To flesh out the very first step: given that the meaning of x^0 is x/x
> (what else?), we have that 0^0=1 implies 0/0=1.
You want to x^0 to be x / x. You say "what else?".
Some comments:
A) If you define x^0 to mean x / x, then 0^0 would be undefined since
the right hand side is undefined (as we would all agree if you used this
definition).
B) If you are working in the natural numbers N = {0, 1, 2, 3, ...}, then
that definition won't have any meaning since "/" is not usually defined
for the natural numbers. The definition won't work in the natural
numbers because "/" is not usually defined for N. If you do define "/"
for the natural numbers, then we are back to comment A.
C) You ask "What else?" could the definition be. One could give a
recursive definition as follows: x^0 = 1; for n in N and n > 0, x^n = x
* x ^ (n - 1).
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