Date: Oct 18, 2013 9:55 AM
Author: Dan Christensen
Subject: Re: Formal proof of the ambiguity of 0^0
On Friday, October 18, 2013 1:46:36 AM UTC-4, fom wrote:

> On 10/18/2013 12:34 AM, Dan Christensen wrote:

>

> > The Case Against 0^0 = 1

>

> >

>

> > Using Theorem 1 (specifying x0=1) , a binary function ^ on N can be constructed such that:

>

> >

>

> > 1. x^0=1

>

> > 2. x^(y+1) = x^y * x

>

> >

>

> > Using this a definition, the usual Laws of Exponents can be derived:

>

> >

>

> > 1. The Product of Powers Rule: x^y * x^z = x^(y+z)

>

> > 2. The Power of a Power Rule: (x^y)^z = x^(y*z)

>

> > 3. The Power of a Product Rule: (x*y)^z = x^z * y^z

>

> >

>

> > However, using Theorem 1 (specifying x0=0 instead) , another binary function ^ on N can also be constructed such that:

>

> >

>

> > 1. 0^0=0

>

> > 2. x^0=1 for x=/=0

>

> > 3. x^(y+1) = x^y * x

>

> >

>

> > The two functions are almost identical, differing only on the value of 0^0 itself (see Theorem 2).

>

> >

>

> > How to choose? For the obvious reasons, 0^0=1 is much more convenient. But the fact is, there does not seem to be any logically compelling reason to choose either alternative. The only real alternative seems to be to leave 0^0 undefined.

>

> >

>

>

>

> Again, you cannot do that. These are "total functions".

>

Both are indeed total functions. Both, however, share the following characteristics:

1. ^ : NxN --> N

2. x^0=1 for x=/=0

3. x^(y+1) = x^y * x

Both are identical except for the value assigned to 0^0 (see Theorem 2).

In fact, there are infinitely many total functions on N that share each of these characteristics (see Theorem 1). I have been calling them "exponent-like" functions.

From these characteristics that are common to ALL exponent-like functions, we can derive the usual Laws of Exponents:

1. x^y * x^z = x^(y+z) for x=/=0

2. (x^y)^z = x^(y*z) for x=/=0

3. (x*y)^z = x^z * y^z for x, y =/=0

There are no partial functions here. There is no need for any form of modal logic to resolve this problem. And there is no need to dump our old friend, 0.

Dan

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