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Topic: Matheology � 258
Replies: 104   Last Post: May 5, 2013 2:26 PM

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 dan.ms.chaos@gmail.com Posts: 409 Registered: 3/1/08
Re: Matheology § 258
Posted: May 1, 2013 12:27 PM

On May 1, 5:37 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 1 Mai, 14:26, Dan <dan.ms.ch...@gmail.com> wrote:
>
>
>
>
>
>
>
>
>

> > On May 1, 2:55 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > On 1 Mai, 13:15, Dan <dan.ms.ch...@gmail.com> wrote:
>
> > > > Do you think the natural numbers *can be enumerated*?
>
> > > No. Look here for a detailled exposition:
>
> > >http://arxiv.org/ftp/math/papers/0305/0305310.pdf
>
> > If I knew it were possible for 4 to follow 2 in the natural numbers ,
> > I would be worried to use infinity . The difference between a 'unknown
> > infinity' and 'the infinity of natural numbers' , is that the latter
> > is determined, by thought , and within thought  . There is no bluing,
> > no information missing, no indeterminacy .
> > 2 is always followed by 3 .
> > 999 is always followed by 1000 .

>
> Correct.
>
>
>

> > If I thought of a formula for an infinite sequence of digits, I can't
> > pretend "not to know the digits" , or the formula , when communicating
> > with myself .

>
> There are many such formuas: 1/9, pi.
>
>
>

> > I've honestly tried to follow the argument in section 5 , but it's
> > incomprehensible .

>
> It is easy. Look at the sequence
> 0.1
> 0.11
> 0.111
> ...
> Each term has only natural indices.
> The limit 0.111... has not only natural indices, because all natural
> indices are in the terms of the sequence. So no index or set of
> indices remains to distinguish 0.111... from all terms of the
> sequence.
>
> Of course you can distinguish every term of the sequence from a larger
> one. And this is somethimes used as an argument, that 0.111... can be
> distinguished from every term of the sequence. Erroneously! Since, if
> the limit could be written by digits at finite places only, it was a
> term of the list.
>
> The same argument is used in paragraph 5.
> Every term like 0.111 can be reflected at the decimal point, yielding
> 111.0, since every power 10^-n has a reciprocal 10^n, *as lomg as n is
> a natural number*. 0.111... cannot be reflected at the decimal point.
> This imples that there are not only natural numbers as indices or
> exponents.
>
> I use only this theorem:
>
> As long as n is a natural number, 10^-n is as well defined as 10^n.
>
> Regards, WM

a_k = (10^k - 1)/9 = 1111 .... 1 (k'th ones)
b_k = (1 - 10^(-k))/ 9 = 0.111111 .... 1 (k'th ones)

lim 10^k when k goes to infinity is infinity , not a well behaved
limit .
On the other hand , lim 10^(-k) when k goes to infinity is 1/ (10^k)
= 0

The first limit diverges (infinity is not part of the topology of
natural numbers , or real numbers)
While the second one converges simply to 1/9 .

They have similar formulas,but behave in different ways .
You would be correct in affirming that b_inf = 1/9 is not part of the
list .
If it were part of the list , then a_inf would be a well-behaved
natural number , but it isn't .
Limits only work properly with real numbers, not natural numbers .
Natural numbers tend to be gappy, unlike real numbers ("the
continuum") .

Date Subject Author
4/29/13 Virgil
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4/30/13 dan.ms.chaos@gmail.com
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