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From: William Elliot on 26 Mar 2010 01:54
On Thu, 25 Mar 2010, Jay Belanger wrote:
> The conclusion should be that 0^n shouldn't be defined for all integers
> n, but it doesn't say anything about 0^0.
>
In set theory it's easy to prove 0^0 = 1.
For real numbers, x^y is defined for x in (0,oo) and y in R.

Opinions may vary how to handled x^n for x in R, n in Z.
The easiest is to limit x to R\0. For x^q, q in Q, what's
the range for x? My opinion it's not worth defining x^q when
x < 0 and the reduced demonator of q is odd.

All exponentation is consider to be real valued (excludes extended reals).
From: William Elliot on 26 Mar 2010 02:14
On Thu, 25 Mar 2010, Marc Olschok wrote:
> William Elliot <marsh(a)rdrop.remove.com> wrote:
>> On Tue, 23 Mar 2010, Marc Olschok wrote:

>>> This depends on where your 0 lives.
>>> If you only deal with natural numbers, it makes sense to
>>> view them as cardinalities of finite sets.
>>
>> Are negative numbers somehow unnatural?
>
> You really did not now the expression "natural number" as synonym
> for "nonnegative integer" ?
>
No. Usage varies. Z is the set of integers, Z^+ (Z+) the set of
nonnegative integers and N the set of natural positive integers.

>> Anyway if 0 lived in Z, then 0^0 = 0^(1 - 1) = 0^1 * 0^-1 = 0 * oo.
>
> One has to be more precise. In fact exponentiation is an operation
> of two variables, so for general discussions like this one should
> make clear where both the base b and the exponent e in an expression
> like "b^e" are supposed to live, and also specify whether "^" should
> be total or not.
>
The choices are to totally define ^ and note all the exceptions to
exponentation laws or to restrict the domain of ^ and not have to bother
with exceptions.

> For the quite common case of ^ : N x N --> N
> (i.e. base and exponent nonnegative integers)
> I have already outlined why 0^0 = 1 makes sense.
>
You're N is Z+? Yes, ^ with that domain requires no notification
of expection to the exponentation laws.

> For the related case of ^ : Z x N --> Z
> (base and exponent integer, exponent nonnegative)
> or ^ : A x N --> A where A is some ring, it also makes sense
> to have 0^0 = 1 in order for the the exponent laws to work properly.
>
Ok.

> In all the above situations 0^0 = 1 is well established
> so your "0^0 isn't defined, doesn't exist." misses large
> parts of Mathematics.
>
Do you want negative exponents or not?

> Now let's turn to your above reasoning with 0^0 = 0^(1 - 1) = 0^1 * 0^-1.
> If exponentiation is of type ^ : Z x Z --> Z then one could as well
> use 2^0 = 2^(1 - 1) = 2^1 * 2^-1 as an argument against 2^0 = 1,
> which is hopefully not your intention.

Negative exponents and 0 base don't mix.

> But let's extend the type to ^ : Z x Z --> Q (with Q the rationals).
> Then your example still applies and illustrates that the laws of exponents
> do not work properly if one insists on ^ being everywhere defined.

> But your example also illustrates that the _negative_ exponents
> pose the problem, not the exponent 0. So you need to allow ^ to
> be a partial operation from Z x Z to Q which is not defined
> at the subset {0} x { z in Z | z < 0 }.

(0,oo)xR allows the laws of real valued exponentation without exceptions.

> The law of exponents then reads:
> | the equation x^(i+j) = x^i * x^j holds whenever both sides are defined.
> and your examples is no more. In particular 0^0 = 1 still makes sense.
>
You're claiming 0^-1 isn't defined?

> Of course other types of exponentiation may require different
> restrictions.
>
The choice of domain for smooth exponentation, ie without exceptions to
the laws of exponentation, depends upon the usage. (0,oo)xR for analysis.

Let's exponentate functions, f^2 = f o f.
How far would you want to push it?
What's f^-1? What's 0^1? What's f^sqr 2? ...

From: Marc Olschok on 26 Mar 2010 15:04
William Elliot <marsh(a)rdrop.remove.com> wrote:
> On Thu, 25 Mar 2010, Marc Olschok wrote:
> > William Elliot <marsh(a)rdrop.remove.com> wrote:
> >> On Tue, 23 Mar 2010, Marc Olschok wrote:
>[...]
> > One has to be more precise. In fact exponentiation is an operation
> > of two variables, so for general discussions like this one should
> > make clear where both the base b and the exponent e in an expression
> > like "b^e" are supposed to live, and also specify whether "^" should
> > be total or not.
>[...]
> > Now let's turn to your above reasoning with 0^0 = 0^(1 - 1) = 0^1 * 0^-1.
> > If exponentiation is of type ^ : Z x Z --> Z then one could as well
> > use 2^0 = 2^(1 - 1) = 2^1 * 2^-1 as an argument against 2^0 = 1,
> > which is hopefully not your intention.
>
> Negative exponents and 0 base don't mix.

In the above, negative exponents do not go well with _any_ noninvertible,
not just 0.

>
> > But let's extend the type to ^ : Z x Z --> Q (with Q the rationals).
> > Then your example still applies and illustrates that the laws of exponents
> > do not work properly if one insists on ^ being everywhere defined.
>
> > But your example also illustrates that the _negative_ exponents
> > pose the problem, not the exponent 0. So you need to allow ^ to
> > be a partial operation from Z x Z to Q which is not defined
> > at the subset {0} x { z in Z | z < 0 }.
>
> (0,oo)xR allows the laws of real valued exponentation without exceptions.

So you do not want to have negative bases any more?

>
> > The law of exponents then reads:
> > | the equation x^(i+j) = x^i * x^j holds whenever both sides are defined.
> > and your examples is no more. In particular 0^0 = 1 still makes sense.
> >
> You're claiming 0^-1 isn't defined?

Yes. More general for the case of an arbitrary (commutative) ring A (with 1)
and exponentiation of type ^: A x Z --> A , once you settled on the idea
that exponentiation with natural exponents should be given by iterated
multiplication, it is pretty straightforward what do leave undefined.

Set a^0 = 1, a^(n+1) = a*a^n for n >= 0
and set a^(-n) = (1/a)^n whenever a is invertible.

So the domain of definition is (A x N) union (U x Z) where U is the
set of all invertible elements of A.

>
> > Of course other types of exponentiation may require different
> > restrictions.
> >
> The choice of domain for smooth exponentation, ie without exceptions to
> the laws of exponentation, depends upon the usage. (0,oo)xR for analysis.
>
> Let's exponentate functions, f^2 = f o f.
> How far would you want to push it?
> What's f^-1? What's 0^1? What's f^sqr 2? ...

As for integer exponents, for invertible f I would let f^-1 be
the inverse map of f, and leave it undefined if f is not invertible.
Surely you must have seen such notation in the special case of
quadratic matrices (as linear maps).

As for noninteger exponents, I do not know any sensible defnition
for such a situation.

--
Marc
From: William Elliot on 27 Mar 2010 02:44
On Fri, 26 Mar 2010, Marc Olschok wrote:
> William Elliot <marsh(a)rdrop.remove.com> wrote:

>> Negative exponents and 0 base don't mix.
>
> In the above, negative exponents do not go well with _any_ noninvertible,
> not just 0.
>
>> (0,oo)xR allows the laws of real valued exponentation without exceptions.
>
> So you do not want to have negative bases any more?
>
You're welcome to them. R\0 x Z; RxN; Rx(N\/{0}).

>>> The law of exponents then reads:
>>> | the equation x^(i+j) = x^i * x^j holds whenever both sides are defined.
>>> and your examples is no more. In particular 0^0 = 1 still makes sense.
>>>
>> You're claiming 0^-1 isn't defined?
>
> Yes. More general for the case of an arbitrary (commutative) ring A (with 1)
> and exponentiation of type ^: A x Z --> A , once you settled on the idea
> that exponentiation with natural exponents should be given by iterated
> multiplication, it is pretty straightforward what do leave undefined.
>
> Set a^0 = 1, a^(n+1) = a*a^n for n >= 0
> and set a^(-n) = (1/a)^n whenever a is invertible.
>
> So the domain of definition is (A x N) union (U x Z) where U is the
> set of all invertible elements of A.
>
>> The choice of domain for smooth exponentation, ie without exceptions to
>> the laws of exponentation, depends upon the usage. (0,oo)xR for analysis.
>>
>> Let's exponentate functions, f^2 = f o f.
>> How far would you want to push it?
>> What's f^-1? What's 0^1? What's f^sqr 2? ...
>
> As for integer exponents, for invertible f I would let f^-1 be
> the inverse map of f, and leave it undefined if f is not invertible.
> Surely you must have seen such notation in the special case of
> quadratic matrices (as linear maps).
>
> As for noninteger exponents, I do not know any sensible defnition
> for such a situation.
>
The same as for real numbers. f^(1/2) o f^(1/2 = f
Just complete an ordered field of functions.
From: Tim Little on 31 Mar 2010 21:34
On 2010-03-26, William Elliot <marsh(a)rdrop.remove.com> wrote:
> No. Usage varies. Z is the set of integers, Z^+ (Z+) the set of
> nonnegative integers and N the set of natural positive integers.

Usage does indeed vary. In standard coursework while I was at
university, our lecturers' conventions were that Z^+ denotes the set
of strictly positive integers, and N includes 0 (being defined in
terms of the set of finite ordinals including the empty set). Exactly
the reverse of your convention!


- Tim
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