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Re: 0^0=1
Posted:
May 7, 2012 12:07 AM


In article <230e3c4e859c43489ee4bba2163951d0@i8g2000vbm.googlegroups.com>, Dan Christensen <Dan_Christensen@sympatico.ca> wrote:
> On May 6, 11:38 pm, William Hale <bill...@yahoo.com> wrote: > > In article > > <89c349de7d82424b8ccd4dc906835...@e15g2000vba.googlegroups.com>, > > 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 wouldbe 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 contradictory themselves).
Are you saying that I can't define 0$0 to be 1, 0$1 to be 0, etc without maybe contradicting other axioms?
Are you saying that the fibonacci function may contradict other axioms?
Are you saying that any defined function (not just the recursive ones) may contradict other axioms?



