LudovicoVan
Posts:
4,033
From:
London
Registered:
2/8/08


Re: The ambiguity of 0^0 on N
Posted:
Sep 21, 2013 12:15 PM


"Michael F. Stemper" <michael.stemper@gmail.com> wrote in message news:l1kfqh$gde$1@dontemail.me... > On 09/20/2013 04:48 PM, Julio Di Egidio wrote: >> "Michael F. Stemper" <michael.stemper@gmail.com> wrote in message >> news:l1ie0g$rpg$1@dontemail.me... >>> On 09/20/2013 02:42 PM, Virgil wrote: > >>>> I can see that >>>> f(x,y) = x^y is continuous on domain {(x,y): x > 0 and y > 0} >>>> and can even be extended continuously to both >>>> {(0,y): y > 0} and >>>> {(x,0): x > 0} >>>> but there is no way to extend that function CONTINUOUSLY to >>>> {(x,y) x > 0 and y > 0} to {(x,y) x >= 0 and y >= 0}. >>>> >>>> This suggest that for real functions like >>>> f(x,y) = x^y on {(x,y): x > 0 and y > 0} >>>> there is no universally acceptable limit value at f(0,0). >>> >>> That's true for the reals, yes. I believe that the topic under >>> discussion is "N", the naturals. >> >> Aren't the naturals a subset of the reals? > > Well, they're isomorphic to a subset of the reals, anyway.
Right.
> However, what metric on N would you use that would give a > singular, welldefined limit of any function at any point?
I am on the side that thinks 0^0 must be undefined in (a pointlike number) arithmetic.
Here is my *proof*, using the usual rules:
0^0 = 0^(11) = (0^1)/(0^1) = 0/0 = undefined
Julio

