
Re: Leaving 0^0 undefined  A numbertheoretic 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, universallyapplicable 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.

