Re: .99999... still=/= 1

From: Tapio (hurmecom_at_dlc.fi)
Date: 12/21/04 

Date: Tue, 21 Dec 2004 15:45:34 GMT

"Dave Seaman" <dseaman@no.such.host> wrote in message
news:cq7ohu$vlf$1@mailhub227.itcs.purdue.edu...
> On Mon, 20 Dec 2004 20:46:26 GMT, Tapio wrote:
>
>> "Dave Seaman" <dseaman@no.such.host> wrote in message
>> news:cq4d8t$4vm$1@mailhub227.itcs.purdue.edu...
>
(snip)
>
>>> but there is no such thing as a
>>> smallest infinite hypernatural. If n is an infinitely large integer,
>>> then so is n-1.
>
>> Omega was defined in this discussion earlier as follows:
>> The number that is greater than any infinite integer. That is the
>> smallest
>> transinfinite number. Omega is the successor of ...999, i.e. n+1. The
>> precessor (n-1) of ...999 is ...998.
>
> It's possible to define the hyperordinals and to talk about *omega in
> NSA, but this *omega has no relation to anything you said in that
> paragraph. For one thing, *omega is not a member of *N. (In fact,
> *omega is identical to the set *N itself).

Correct, because that omega above (maybe some other name is better - let's
use your symbol *w ?) is in the next infinite area. The reasons are:
Every number has a successor, i.e. you can always add 1. All the
placeholders in the infinite long area *N were already occupied.

> For another, there is no
> connection between the hyperordinals and strings (even hyperstrings!) of
> decimal digits. I don't accept your definition.

I could not follow you above, sorry. :-(

> Are you under the impression that ...999 is a hyperinteger? It isn't,
> unless you explain which equivalence class of sequences of integers you
> are talking about. I can think of ways to make such a correspondence,
> but you haven't said what you mean.

OK, let's try again. First of all, I think we should start a new thread from
the beginning so that everything is constructed and we use the same concepts
and definitions.
1) ...999 is not N as it is not in a classic way finite integer. Let's call
it infinite integer (N_inf) over one infinity.
2) I define it as I have done earlier sum (k 0 --> oo) 9*10^k. It does not
have classic limit as the k refers now the standard point of reference, but
it has a limit as you hopefully noticed as we have another point of
reference.
3) It is defined and it exists, now - how do You like to name it?
Hyperinteger or something else?
(snip)

>>> I'm trying to guess what you might mean by "the real part", since you
>>> have not defined your meaning.
>
>> the real part was the decimal part - as you certainly knew.
>
> No, I didn't know what you meant, and I still am not sure. What is the
> "real part" of sqrt(2), for example. It sounds like you trying to say
> the "real part" is sqrt(2) - 1, or approximately 0.41421. That was not
> one of my first two guesses. I would call that the "fractional part."

Ok, that suits for me.

(snip)

> In a similar fashion, you keep assuming that I must know what you mean by
> ...999. However, I assure you that I don't.

Uh! I have tried to explain so simple as possible. I had once a fealing that
You understood very well.
There must be some miscommunication.

>>>> 2)Yes, the limit is exactly 1only if you count the limit including the
>>>> standard part as N --->oo. Then your reference set is N.
>
>>> Are you talking about standard analysis here, or nonstandard?
>
>> Standard analysis as You correctly observed.
>
> Then you are not discussing the value of 0.999... in NSA at all, as I
> previously thought. Looks like yet another example of miscommunication.
> Let's summarize:
>
> (1) We have 0.999... = 1 in standard analysis, because the
> sum is over N.

Yes, the limit is over N. I'm voluntary to point You that the sum and the
limit are not the same thing.

> (2) We have 0.999... = 1 in NSA, because the sum is over
> *N.

Yes, the limit is over *N.
>
> (3) It is not possible to sum over *N in standard analysis,
> because *N is not a set in standard analysis.

Yes. I consider *N is the extension of N.

> (4) It is not possible to sum over N in NSA, because N is
> not a set in NSA.
>
> Do you agree?

I agree 1,2 and 3 but in the case 4 I disagree. Maybe we have to discuss
about that topic more accurate. I consider finite integers are a subset of
infinite integers over the first one infinity.

>>> The set N
>>> (consisting of the finite naturals) is not a set in the internal set
>>> theory of NSA.
>
>> Coorect, but it should be as *N is the extension of the set N.
>
> No, it should not be. There is an important principle involved, known as
> the transfer principle. This says that every theorem of standard
> analysis is also a theorem of NSA, provided we make the appropriate
> substitutions (such as substituting *N for each occurence of N).
>
> The transfer principle is extremely important. Without it, NSA loses
> most of its value as a tool of analysis.
>
> It turns out that if all sets in standard analysis are allowed to be
> internal sets in NSA, then we lose the transfer principle. That's why
> things are defined the way they are.

I cannot see where the transfer principle fails, except in the case of sum
and limit. But in this case there are natural reasons as I have tried to
explain. And the reason is not NSA.

(snip)

>> Your opinion, because You could not consider the successor of ...999.
>> What
>> is the successor of ...999 as all the placeholders are infinitely
>> occupied
>> with the maximal digit 9?
>
> You haven't said what ...999 is. How am I supposed to answer questions
> about it if you don't define it?

I assume You can now answer after the definition - above. I propose anyway
to start a new thread.

Tapio
>
> --
> Dave Seaman
> Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
> <http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>

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