🚀 go-pugleaf

"Virgil" <Virgil@gmale.com> wrote in message
news:Virgil-6FD0F3.13355320042008@comcast.dca.giganews.com...
> In article <QIMOj.3408$ko5.2468@news-server.bigpond.net.au>,
>  "Don H" <donlhumphries@bigpond.com> wrote:
>
> > <Pfsszxt@aol.com> wrote in message
> > news:jlah04drbo3nucjn5n7n438gqbng4auf0r@4ax.com...
> > > On Thu, 17 Apr 2008 19:40:24 GMT, "Don H" <donlhumphries@bigpond.com>
> > > wrote:
> > >
> > > >1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.
> > > >Is this axiomatic, or is there a proof?
> > > >==============================
> > > >
> > >  It isn't a math fact --- it's a definition --- made so as to be
> > > consistant and for convenience when writing infinite series etc.
> >
> > # And so is an arbitrary definition?
> >    What is a number n to the power x?
> >    The x indicates how many n must be multiplied together.
> >    Thus,
> >    n^x = n by n (x times)
> >    2^2 = (2, twice) = 4
> >    2^1 = (2, once) = 2
> >    2^0 = (2, no times) = 1 ?
> >      - but if you multiply 2 by itself, no times, you've still 2 ?
> > Whereas, if you multiple 0 by 0, two times, you've still 0 ?
>
> The standard inductive (or, if  you prefer, recursive) definition of
> powers with positive integer exponents is:
>    For any real/complex number x, x^1 = x  and x^(n+1) = x^n * x.

# Thanks.
   However, consider two rabbits, say, Rob and Ruby, living in a Platonic
relationship (2^0), who then decide to copulate, initially without result
(2^1), but finally generating two kids (Richard, Rachel) (2^2), who all then
continue to propagate, exponentially = 2 x 2 x 2 x 2...
   Likewise, two non-entities, Zero and Zilch, who, no matter how many times
they try, produce nothing - or nothing evident to our eyes.
   Does it depend on what you start out with, as to what the result will be?