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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: Log i = 0
Date: Tue, 27 May 2025 21:37:33 +0200
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On 27.05.2025 18:05, FromTheRafters wrote:
 > WM wrote :
 >> On 26.05.2025 22:25, efji wrote:
 >>> Le 26/05/2025 à 16:36, WM a écrit :
 >>>> That is wrong. Present mathematics simply assumes that all natural 
numbers can be used for counting. But that is wrong.
 >>>
 >>> What's the point ?
 >>> It is the DEFINITION of "counting". A countable infinite set IS a 
set equipped with a bijection onto \N.
 >>>
 >> This bijection does not exist because most natural numbers cannot be 
distinguished as a simple argument shows.
 >
 > Bijected elements need not be distinguished, it is enough to show a 
bijection.

You mean it is enough to believe in a bijection?

According to Cantor the infinite sequence thus defined must have the 
property to contain the positive rational numbers completely, and each 
of them only once at a determined place.

That requires moe than belief. But your belief can be shattered by the

     X-O-Matrices

All positive fractions

     1/1, 1/2, 1/3, 1/4, ...
     2/1, 2/2, 2/3, 2/4, ...
     3/1, 3/2, 3/3, 3/4, ...
     4/1, 4/2, 4/3, 4/4, ...
     ...

can be indexed by the Cantor function k = (m + n - 1)(m + n - 2)/2 + m 
which attaches the index k to the fraction m/n in Cantor's sequence

1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 
5/1, 1/6, 2/5, 3/4, ... .

Its terms can be represented by matrices. When we attach all indeXes k = 
1, 2, 3, ..., for clarity represented by X, to the integer fractions m/1 
and indicate missing indexes by hOles O, then we get the matrix M(0) as 
starting position:

XOOO...    XXOO...    XXOO...    XXXO...
XOOO...    OOOO...    XOOO...    XOOO...
XOOO...    XOOO...    OOOO...    OOOO...
XOOO...    XOOO...    XOOO...    OOOO...
M(0)       M(2)       M(3)        M(4)      ...

M(1) is the same as M(0) because index 1 remains at 1/1. In M(2) index 2 
from 2/1 has been attached to 1/2. In M(3) index 3 from 3/1 has been 
attached to 2/1. In M(4) index 4 from 4/1 has been attached to 1/3. 
Successively all fractions of the sequence get indexed. In the limit, 
denoted by M(∞), we see no fraction without index remaining. Note that 
the only difference to Cantor's enumeration is that Cantor does not 
render account for the source of the indices.

Every X, representing the index k, when taken from its present fraction 
m/n, is replaced by the O taken from the fraction to be indexed by this 
k. Its last carrier m/n will be indexed later by another index. 
Important is that, when continuing, no O can leave the matrix as long as 
any index X blocks the only possible drain, i.e., the first column. And 
if leaving, where should it settle?

As long as indexes are in the drain, no O has left. The presence of all 
O indicates that almost all fractions are not indexed. And after all 
indexes have been issued and the drain has become free, no indexes are 
available which could index the remaining matrix elements, yet covered by O.

It should go without saying that by rearranging the X of M(0) never a 
complete covering can be realized. Lossless transpositions cannot suffer 
losses. The limit matrix M(∞) only shows what should have happened when 
all fractions were indexed. Logic proves that this cannot have happened 
by exchanges. The only explanation for finally seeing M(∞) is that there 
are invisible matrix positions, existing already at the start. Obviously 
by exchanging O and X no O can leave the matrix, but the O can disappear 
by moving without end, from visible to invisible positions.

Regards, WM