From Newsgroup: sci.math

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(reR), 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(reR) 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(reR) 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
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