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From: William Elliot on 23 Mar 2010 06:38
On Tue, 23 Mar 2010, Virgil wrote:
> William Elliot <marsh(a)rdrop.remove.com> wrote:
>
>> On Mon, 22 Mar 2010, elmerturnipseed wrote:
>>
>>> Please forgive my ignorance. I failed math in school, but I just read
>>> an article about how 0^0=1, and I'm wondering if maybe there's
>>> something wrong the way that exponential notation is traditionally
>>> expressed.
>>>
>> It doesn't and the article is wrong.
>> 0^0 isn't defined, doesn't exist.
>
> As a matter of convenience, to avoid having to deal with exceptions, it
> quite often is defined, either as equalling 1, or, a bit less commonly,
> as equalling 0.
>
> The function f(x) = x^0 becomes continuous for all real x provided one
> has defined 0^0 to have the value 1, and that is quite useful when
> dealing with functions of the form g(x,y) = x^y.
>
If x is allowed to be zero, then y can't be negative.
If g is real valued, then x can't be negative.
So g would have to be defined from [0,oo)^2 into R.

The more prudent way is to define g from (0,oo)xR into R.
Dealing with the endpoint 0 is a real inconvenience.

Isn't (0,0) a non-removable singularity?



From: Marc Olschok on 23 Mar 2010 15:37
William Elliot <marsh(a)rdrop.remove.com> wrote:
> On Mon, 22 Mar 2010, elmerturnipseed wrote:
>
> > Please forgive my ignorance. I failed math in school, but I just read
> > an article about how 0^0=1, and I'm wondering if maybe there's
> > something wrong the way that exponential notation is traditionally
> > expressed.
> >
> It doesn't and the article is wrong.
> 0^0 isn't defined, doesn't exist.

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.
You do have exponentiation for sets, with B^E the set of all
maps from E to B, and in particular {}^{} has exactly one element.
Therefore 0^0 = 1 is quite reasonable, even more so because it
does not need some arbitrary choice.

--
Marc
From: Transfer Principle on 23 Mar 2010 19:57
On Mar 22, 10:04 pm, elmerturnipseed <sterlingram...(a)gmail.com> wrote:
> We think of 5*5 as 5^2, two fives multiplied by each other, and 5^3 as
> three fives multiplied.
> What if the act of multiplying a number by itself was considered as
> one occurrence of this function?
> So 5^1 would be 5 multiplied by itself once, 25. And 5^2 would be
> 5*5*5. 5^0 would be 5, with the function performed 0 times.

As an enthusiast in the operation of tetration, I find threads
which discuss the hierarchy of operations (such as here, how
to define exponentiation from multiplication) interesting.

We see that Elmer Turnipseed appears to be recursively defining
exponentiation as follows (in order to distinguish standard
exponentiation from Turnipseed's exponentiation, we shall use
the symbol "v" an upside-down caret to denote the latter):

bv0 = b
bv(S(n)) = (bvn)*b

whereas standard exponentiation is defined as:

b^0 = 1
b^(S(n)) = (b^n)*b

Thus, it appears that bvn = b^(S(n)) -- so Turnipseed's
exponentiation is just standard exponentiation multiplied by
an extra factor of b. (Notice that here, S denotes the
successor function S(n) = n+1 as used in Peano arithmetic.)

> Would this new system of notation change the result of 0^0?

Notice that here 0v0 = 0^1 = 0, just as Turnipseed asserts.

Although Turnipseed's exponentiation may seem awkward, we do
notice the similarity between his exponentiation and standard
addition in terms of their recursive definitions:

b+0 = b
b+S(n) = S(b+n)

Indeed, using the little known operation of _zeration_, which
we write using the symbol "o", we may define addition as:

b+0 = b
b+S(n) = (b+n)ob

and we see that the definitions are parallel, with + and o
being replaced by v and * respectively. (There is some dispute
as how best to define zeration, but here I choose to define it
as aob = S(max(a,b)).)

> (I seem to
> recall that negative exponents are dividers. Please correct me if I'm
> mistaken.)

Using negative exponents in Turnipseed's notation gives:

5v(-1) = 5/5 = 1
5v(-2) = (5/5)/5 = 1/5
5v(-3) = ((5/5)/5)/5 = 1/25
5v(-4) = (((5/5)/5)/5)/5 = 1/125
5v(-5) = ((((5/5)/5)/5)/5)/5 = 1/625

and so on. Since in standard exponentation, we have that
b^(-x) = 1/(b^x), we still have bv(-x) = b^(-x+1) as before.
From: William Elliot on 24 Mar 2010 06:00
On Tue, 23 Mar 2010, Marc Olschok wrote:
> William Elliot <marsh(a)rdrop.remove.com> wrote:
>> On Mon, 22 Mar 2010, elmerturnipseed wrote:
>>
>>> Please forgive my ignorance. I failed math in school, but I just read
>>> an article about how 0^0=1, and I'm wondering if maybe there's
>>> something wrong the way that exponential notation is traditionally
>>> expressed.
>>>
>> It doesn't and the article is wrong.
>> 0^0 isn't defined, doesn't exist.
>
> 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?
Anyway if 0 lived in Z, then 0^0 = 0^(1 - 1) = 0^1 * 0^-1 = 0 * oo.

> You do have exponentiation for sets, with B^E the set of all
> maps from E to B, and in particular {}^{} has exactly one element.
> Therefore 0^0 = 1 is quite reasonable, even more so because it
> does not need some arbitrary choice.
>
Fortunately, even thought some one invented a space with -1
dimension, no one has yet invented a set negative number of elements.

How soon will it be before someone, like the invention of fractional
dimensions, invent sets with fractional number of elements?

If I've a set with one electron in it, how much of an electron is in the
set and how much isn't? Can the electron be both and in the set at the
same time?

From: Jussi Piitulainen on 24 Mar 2010 06:47
William Elliot writes:
> On Tue, 23 Mar 2010, Marc Olschok wrote:
> > William Elliot <marsh(a)rdrop.remove.com> wrote:
> >> It doesn't and the article is wrong.
> >> 0^0 isn't defined, doesn't exist.
> >
> > 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?
> Anyway if 0 lived in Z, then 0^0 = 0^(1 - 1) = 0^1 * 0^-1 = 0 * oo.

Would you say that 0^3 = 0^(7 - 4) = 0^7 * 0^-4 = 0 * oo? Or would you
rather admit that this particular argument for leaving 0^0 undefined
is bogus?
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