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Topic: 0^0=1
Replies: 11   Last Post: May 7, 2012 10:01 AM

 Messages: [ Previous | Next ]
 William Hale Posts: 49 Registered: 5/2/12
Re: 0^0=1
Posted: May 7, 2012 12:07 AM

In article
Dan Christensen <Dan_Christensen@sympatico.ca> wrote:

> On May 6, 11:38 pm, William Hale <bill...@yahoo.com> wrote:
> > In article
> >  Dan Christensen <Dan_Christen...@sympatico.ca> wrote:
> >
> >
> >
> >
> >
> >
> >
> >
> >

> > > On May 6, 10:44 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> > > > Dan Christensen <Dan_Christen...@sympatico.ca> writes:
> > > > > On May 6, 8:01 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >
> > > > > [snip]
> >
> > > > >> 0^0 is defined to be 1 because it simplifies the definition and use.
> > > > >> It's handy.  Ain't no more reason needed.

> >
> > > > > I see. It's inconvenient to have 0^0 being undefined, so, well... you
> > > > > define it! Simply and easy. One seems to work nicely in some cases, so
> > > > > why not one? It also saves having to consider all those IRKSOME
> > > > > special cases in proofs: if the base and exponent are zero, yada,
> > > > > yada...

> >
> > > > > Well, good for you, Jesse. I hope it continues to work for you and
> > > > > that you don't come up against any IRKSOME inconsistencies along the
> > > > > way (e.g. 0=1). Just don't be surprised if all those IRKSOME writers
> > > > > of algebra textbooks don't see it your way.

> >
> > > > I don't believe it is possible for a definition to yield an
> > > > inconsistency, Dan.

> >
> > > What about the set of all sets that are not elements of themselves? Or
> > > the universal set -- the set of everything. At one time, these would
> > > have been very convenient definitions, Jesse. Now they are a
> > > cautionary tale for all would-be definers of things.

> >
> > The definition of the set of all sets that are not elements of
> > themselves does not yield an inconsistency. The contradiction resulted
> > from the old axiom which yielded that such a set actually existed.

>
> In the same way that defining 0^0=1 may contradict other axioms.

It can't contradict the other axioms (unless the other axioms are

Are you saying that I can't define 0\$0 to be 1, 0\$1 to be 0, etc without

Are you saying that the fibonacci function may contradict other axioms?

Are you saying that any defined function (not just the recursive ones)

Date Subject Author
5/6/12 Jesse F. Hughes
5/6/12 Dan Christensen
5/6/12 William Hale
5/6/12 Dan Christensen
5/7/12 William Hale
5/7/12 Dan Christensen
5/7/12 Frederick Williams
5/7/12 Pubkeybreaker
5/7/12 LudovicoVan
5/6/12 Jesse F. Hughes
5/6/12 Dan Christensen