Re: Log i = 0

On 29.05.2025 17:37, Ross Finlayson wrote:

On 05/28/2025 04:22 PM, Chris M. Thomasson wrote:

On 5/27/2025 1:13 PM, WM wrote:

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
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.

It sounds like you are trying to say, even the following map does not work:
>
map_to(x) return x + 1
map_from(x) return x - 1
>
map_to(0) = 1
map_to(1) = 2
...
>
map_from(1) = 0
map_from(2) = 1
...
>
?

Didn't you already have this conversation,

Yes.

and why are you having it here?

I repeated this argument because many matheologians are lying to have rejected it.

 It seems you're describing a simple book-keeping of an integer continuum
in areal terms.
 Also called a geometrization sometimes.

Right. Do you understand the result?
Regards, WM