Re: .99999999999999 = 1 and formally real fields

From: Fenix (vpowpwv_at_hotmail.com)
Date: 10/27/04 

Date: 27 Oct 2004 09:24:14 -0700

smart1234@aol.com (S. Enterprize Company) wrote in message news:<20041027064640.01575.00003271@mb-m28.aol.com>...
> >Carlos Moreno <moreno_at_mochima_dot_com@xx.xxx> wrote in message
> >news:<g3Ted.18427$WY.215258@wagner.videotron.net>...
> >
> >> I've been increasingly thinking that maybe he does; and he
> >> perfectly knows that 0.999.... is equal to 1, and that he's
> >> simply an excellent troll with too much time on his hands and
> >> missing good reasons to have a good laugh (*).
> >
> >Maybe what's in his bean vaguely somewhere is something other than the
> >real numbers. If you take all sequences of real numbers and mod out by
> >a maximal ideal which gives a field not order isomorphic to the real
> >numbers, you get a hyperreal field. In that field, the sequence 9/10,
> >99/100, 999/1000 corresponds to a hyperereal number less than 1 by an
> >infinitesimal quantity. Note this is *not* a sum of a series, but it
>
>
> The sum of the series was shown to be,
>
> .9999999.... =
> .9 + .09 + .009 +......+ 9/10^n
>
> You do point out that an infinitesimal convergence analysis does show that
>
> .999999.... < 1
>
> Then you imply I was showing it as,
>
> .9,.09,.009,.......
>
> This is not what I was showing.
> The sum of the series leads to a certain function which happened to be,
>
> 9/10^n
>
> The question is, will this function converge infinitesimally as it closer and
> gets to it's smallest limit where LIM n-->oo to 1.
>
> And the derivative shows it doesn't. Then multiple derivatives were taken to
> see if at a higher order or higher dimensional state
> if it would reach some maximum or minimum optimum convergence point which
> causing it to converge to 1 and it didn't. Then I went even further to an
> infinite number of dimensions used in the analysis. The closest that .9999...
> gets to one at (DIM oo) was shown to be
>
> 9 * (LOG(10))^oo
> --------------------------
> 10^n
>
> or
>
> 9 * 1^oo
> ------------
> 10^n
>
> This is as close to 1 as it gets. This is the very last number that separates
> .99999.... from 1.
>
> Therefore it doesn't converge to 1.
>
> Partial Sums was shown to be only an approximation ( rounding off) of a
> series showing that it could converge but it would be inaccurate or have some
> degree of error in it.
> In math they just assume that .9999... =1 because it appears to converge in
> a finite way within about 16 significant figures in the series. But in reality
> .99999.... is more than just 16 significant figures.
>
> I have already shown that Partial Sums is inaccurate like for example with
> 1/3. Partial Sums says
>
> .333333...... converges to .3333333333326
>
> If you round off the last two digits, it is 3. So math in general can be used,
> but in a limited way. Then they state how accurate the answer or solution is.
>
> I believe I have shown the most accurate
> solution as to whether or not,
>
> .999999.... converges to one, infinitesimally. After all, it does get
> infinitesimally small as it approaches 1. But it NEVER reaches 1, as can be
> clearly seen.
>
>
>
>
> >is also true that \sum{i=1}{N}9/10^i = 1-10^(-N) is infinitesimally
> >closed to but less than 1 if N is a nonstandard (ie infinite) integer.
> >
> >This argument has been going on for years but would be entirely
> >pointless if the people involved didn't really mean real numbers and
> >limits.
>
>
> Smart's Alt. Physics News Group
> http://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv=1
> S. Enterprize (Science Journal)
> http://smart1234.s-enterprize.com/

Am I the only one that isn't going to read this crap?

Relevant Pages

  • Re: .99999999999999 = 1 and formally real fields
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