🚀 go-pugleaf

In article <vz4Pj.3714$ko5.1622@news-server.bigpond.net.au>,
 "Don H" <donlhumphries@bigpond.com> wrote:

> "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?

Of course! Otherwise 2^2 would be the same as 3^2.