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Topic: The ambiguity of 0^0 on N
Replies: 106   Last Post: Sep 29, 2013 10:06 AM

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 Michael F. Stemper Posts: 122 Registered: 9/5/13
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 UTC-4, Peter Percival wrote:
>> Dan Christensen wrote:
>>

>>> On Wednesday, September 18, 2013 7:11:53 PM UTC-4, 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 is--would
>>
>> return 0. It didn't, it returned--quite properly--1. 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.

Date Subject Author
9/18/13 Dan Christensen
9/18/13 Peter Percival
9/18/13 Dan Christensen
9/18/13 Peter Percival
9/18/13 Virgil
9/18/13 Dan Christensen
9/18/13 Rotwang
9/18/13 Rock Brentwood
9/18/13 Rotwang
9/19/13 Dan Christensen
9/19/13 Peter Percival
9/19/13 Dan Christensen
9/19/13 Peter Percival
9/19/13 Dan Christensen
9/19/13 fom
9/19/13 Dan Christensen
9/19/13 fom
9/19/13 Dan Christensen
9/20/13 fom
9/19/13 Virgil
9/19/13 Virgil
9/19/13 Rotwang
9/18/13 Virgil
9/18/13 fom
9/18/13 Rotwang
9/28/13 Shmuel (Seymour J.) Metz
9/29/13 Marshall
9/19/13 Dan Christensen
9/19/13 Dan Christensen
9/19/13 Peter Percival
9/19/13 Dan Christensen
9/19/13 Michael F. Stemper
9/19/13 Dan Christensen
9/19/13 Peter Percival
9/19/13 Dan Christensen
9/19/13 fom
9/19/13 Dan Christensen
9/19/13 fom
9/19/13 Dan Christensen
9/19/13 fom
9/19/13 Dan Christensen
9/20/13 fom
9/20/13 Dan Christensen
9/20/13 fom
9/19/13 fom
9/19/13 fom
9/19/13 Dan Christensen
9/19/13 fom
9/19/13 Peter Percival
9/19/13 Dan Christensen
9/19/13 fom
9/19/13 Rotwang
9/19/13 Dan Christensen
9/19/13 Helmut Richter
9/19/13 Dan Christensen
9/19/13 Peter Percival
9/19/13 Dan Christensen
9/19/13 Peter Percival
9/19/13 Dan Christensen
9/19/13 fom
9/19/13 fom
9/19/13 JT
9/19/13 JT
9/19/13 Michael F. Stemper
9/19/13 JT
9/19/13 JT
9/19/13 JT
9/19/13 Helmut Richter
9/28/13 Shmuel (Seymour J.) Metz
9/19/13 fom
9/19/13 Peter Percival
9/19/13 Dan Christensen
9/19/13 Peter Percival
9/19/13 Karl-Olav Nyberg
9/19/13 fom
9/19/13 fom
9/19/13 Rotwang
9/19/13 Dan Christensen
9/19/13 fom
9/25/13 Rotwang
9/26/13 Dan Christensen
9/27/13 Brian Q. Hutchings
9/19/13 fom
9/18/13 Rock Brentwood
9/19/13 Dan Christensen
9/19/13 Dan Christensen
9/19/13 Rotwang
9/19/13 Dan Christensen
9/19/13 fom
9/20/13 Dan Christensen
9/20/13 fom
9/20/13 Dan Christensen
9/20/13 Peter Percival
9/20/13 Peter Percival
9/20/13 Dan Christensen
9/20/13 Virgil
9/20/13 Peter Percival
9/20/13 fom
9/20/13 Michael F. Stemper
9/20/13 LudovicoVan
9/21/13 Michael F. Stemper
9/21/13 LudovicoVan
9/21/13 Richard Tobin
9/20/13 Peter Percival
9/20/13 Peter Percival
9/21/13 Dan Christensen
9/19/13 Karl-Olav Nyberg