Re: .99999999999999 = 1 and formally real fields

From: S. Enterprize Company (smart1234_at_aol.com)
Date: 10/27/04 

Date: 27 Oct 2004 10:46:40 GMT

>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/

Relevant Pages

  • Which is bigger? 0.999... or (0.111...) base 2
    ... On 27 Oct 2004, S. Enterprize Company wrote: ... > This is not what I was showing. ... >infinite number of dimensions used in the analysis. ... > I have already shown that Partial Sums is inaccurate like for example with ...
    (sci.math)
  • Re: .99999999999999 = 1 and formally real fields
    ... > This is not what I was showing. ... > infinite number of dimensions used in the analysis. ... > series showing that it could converge but it would be inaccurate or have some ... > I have already shown that Partial Sums is inaccurate like for example with ...
    (sci.math)
  • Re: paths through binary trees in ZF
    ... infinite paths ... depth nodes under 1 root)  has ... only countably many infinite branches through it,  thereby showing  ZF ...
    (sci.logic)
  • Re: paths through binary trees in ZF
    ... only countably many infinite branches through it,  thereby showing  ZF ... inconsistent since it also ... In fact, in any complete infinite binary tree, each node "distinguishes" ...
    (sci.logic)
  • Re: Collatz conjecture
    ... Conjecture must have an infinite number of lines". ... course incorrect, and it's a good example of its kind. ... is a prime' is determined in showing that there exists an m>10^100 ...
    (sci.math)
Loading