
Re: The ambiguity of 0^0 on N
Posted:
Sep 19, 2013 11:34 AM


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.
When A = {} = B, A = 0 = B. What functions are there from B to A in this case?
Let's look at what a function is. A function from X to Y is a subset of X*Y, where: 1. For each x in X, the subset contains an element (x,y) 2. If (x,y) and (x,z) are in the subset, then y=z.
The empty set is the only subset of {}*{}. It satisfies both of these rules. Therefore, there is exactly one function from {} to {}. 0^0 = {}^{} = {{}} = 1.
It would be possible to change the definitions to exclude the 0^0 case, but doing so would make many theorems more complicated without providing any benefit.
References: 1. _Axiomatic Set Theory_, Patrick Suppes, Sections 3.4 and 4.1 2. <http://mathworld.wolfram.com/CardinalExponentiation.html>
>>> The most well known theorem that makes use of 0^0 is probably the >>> Binomial Theorem. And it can easily be restated to avoid its use, >>> e.g. by stating at the outset that (x+0)^n = x^n, etc.
What benefit is to be gained from replacing a single, uniform statement of the theorem with "it's <foo>, except when b=0, in which case it's still <foo>"?
>> Note, btw, >> that the definition of + begins by saying what x+0 is, it doesn't start >> with x+1 or x+2. > > Technically, + on N is not simply defined. It is a construction based Peano's Axioms (or their modern equivalent) and set theory. You could probably construct an exponent function on N with 0^0=1, but you could also construct one with 0^0=0. And they would agree on every value but that assigned to 0^0.
Their modern equivalent is cardinal arithmetic, which maps directly and consistently to Peano's axioms without the need of any special cases.
 Michael F. Stemper This sentence no verb.

