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Re: Matheology § 258
Posted:
May 1, 2013 12:27 PM
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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") .
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