
Re: The ambiguity of 0^0 on N
Posted:
Sep 19, 2013 11:40 AM


On Thursday, September 19, 2013 11:34:54 AM UTC4, Michael F. Stemper wrote: > On 09/19/2013 09:51 AM, Dan Christensen wrote: > > > On Thursday, September 19, 2013 7:26:55 AM UTC4, Peter Percival wrote: > > >> Dan Christensen wrote: > > >> > > >>> On Wednesday, September 18, 2013 7:11:53 PM UTC4, Rotwang wrote: > > >> > > >> > > >> > > >>>> > > >> > > >>>> will follow from people defining exponentiation in the usual way? > > >> > > >>> > > >> > > >>> Whatever consequences may arise from a calculation that results in a > > >> > > >>> value of 1 when it should be 0. The result could be catastrophic. > > >> > > >> > > >> > > >> When "should" 0^0 = 0? > > >> > > > > > > The value of 0^0 should not depend on the context. > > > > > >> > > >> > > >> This is what has happened. You were writing a computer program and you > > >> > > >> wrongly believed that exp(0,0)or whatever the notation iswould > > >> > > >> return 0. It didn't, it returnedquite properly1. Your program > > >> > > >> being buggy is not a catastrophe. The thing to do with bugs is to fix > > >> > > >> them[*], _not_ try to "fix" mathematics. > > >> > > > > > > Let's see your proof that 0^0=1. Sorry, simply defining it as such won't do in this context. > > > > > > There is a simple definition of exponentiation on N, based on > > cardinal arithmetic. It doesn't have any special cases, but > > 0^0 = 1 falls out of it quite naturally: > > > > A^B = {f  f:B>A} > > > > In English, the cardinal number of set A raised to the power > > of the cardinal number of set B is the cardinal number of the > > set of all functions from B to A. No exceptions, no special > > cases. >
Thanks, but as with any analogy, it may not be perfect.
Dan Download my DC Proof 2.0 software at http://www.dcproof.com

