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Topic: Leaving 0^0 undefined -- A number-theoretic rationale
Replies: 48   Last Post: Sep 15, 2013 1:06 PM

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 Michael F. Stemper Posts: 125 Registered: 9/5/13
Re: Leaving 0^0 undefined -- A number-theoretic rationale
Posted: Sep 15, 2013 12:22 PM

On 09/14/2013 12:51 PM, rfermat wrote:
> I will echo what some others have posted here.
>
> 0^0 is undefined.

It would be more accurate to say "0^0 is undefined in some subjects."

Take a look, for instance, at the definitions of cardinal and ordinal
exponentiation given by MathWorld:
<http://mathworld.wolfram.com/CardinalExponentiation.html>

"[...] cardinal exponentiation is defined by
|A|^|B| = |set of all functions from B to into A|"

No exception is made for the case when A = B = {}.

Since cardinal arithmetic maps directly into Peano arithmetic, it
would seem strange to exclude this from the mapping.

And leaving 0^0 defined as 1 makes it simpler to express things
such as the binomial theorem. Why have the binomial theorem work
for (x+y)^n -- unless x=0 or y=0? Why introduce a singularity?

<http://mathworld.wolfram.com/OrdinalExponentiation.html>

"Let alpha and beta be any ordinal numbers, then ordinal exponentiation
is defined so that if beta=0 then alpha^beta=1."

Again, no exception for the case when alpha=0.

Now, when you get to analysis, Lim(x,y->0) x^y depends very strongly
on *how* x and y go to zero. There's no single, universally-applicable
answer there. But, for number theory, combinatorics, and pretty much
anything dealing with the naturals or the integers, it's not that
complicated.

> Yes, in some subjects, conventions are adopted in those subjects. The practitioners know the convention. Good for them. That doesn't in any way change the fact that 0^0 is undefined.

Would you be willing to discuss the distinction between "conventions"
and "definitions"?

--
Michael F. Stemper
This sentence no verb.