Re: 0^0 = 1?

*To*: mathgroup at smc.vnet.net*Subject*: [mg95644] Re: 0^0 = 1?*From*: Dave Seaman <dseaman at no.such.host>*Date*: Fri, 23 Jan 2009 05:09:45 -0500 (EST)*Organization*: Information Technology at Purdue*References*: <gl7211$c8r$1@smc.vnet.net> <gl9mua$ajr$1@smc.vnet.net>

On Thu, 22 Jan 2009 11:56:58 +0000 (UTC), dh wrote: > Hi, > 0^0 means the limit if both base and exponent go to zero. No, that is not how 0^0 is defined. Does 2+2 mean the limit as both summands go to 2? The value may happen to be the same in that case, but that is not how 2+2 is defined. The value of x^y for cardinal numbers x and y is the cardinality of the set of mappings from y into x. In the case where x and y are the empty set, there is exactly one such mapping. Hence, 0^0 = 1. It's a theorem of ZF (as stated in Suppes, _Axiomatic_Set_Theory_) that m^0 = 1 for every cardinal number m. Another way is to notice that 0^0 represents an empty product, whose value is the identity element in the monoid of the integers (or the reals). In[1]:= Product[0,{k,0}] Out[1]= 1 One might also consider the series expansion for Exp[0], which reduces to 1 = 0^0/0! + (lots of terms that all reduce to zero). Having x^y be discontinuous at (0,0) does not "cause problems" any more than having the Sign function be discontinuous at 0 causes problems. Anyone who works with limits should be aware that you can't just blindly assume continuity when evaluating limits. You have to consider the actual definition of the limit. -- Dave Seaman Third Circuit ignores precedent in Mumia Abu-Jamal ruling. <http://www.indybay.org/newsitems/2008/03/29/18489281.php>

**Follow-Ups**:**Re: Re: 0^0 = 1?***From:*Daniel Lichtblau <danl@wolfram.com>