On 12/4/2023 3:17 PM, Chris M. Thomasson wrote:
> On 12/4/2023 5:10 AM, WM wrote:
>> Fritz Feldhase schrieb am Montag, 4. Dezember 2023 um 12:46:02 UTC+1:
>>  > On Monday, December 4, 2023 at 10:53:44 AM UTC+1, WM wrote:
>>  > > B is not defined because it is not a matrix
>>  >
>>  > Some quotes:
>>  >
>>  > "B, the matrix
>>  > 1, 2, 4, ...
>>  > 3, 5, 8, ...
>>  > 6, 9,13, ...
>>  > ...
>>  > containing only natural numbers does exist." (WM)
>>
>> I only wanted to butter William up a bit. There are many dark numbers
>> which cannot be defined individually.
>
> Name one? Just one out of the many... LOL!
>
> ;^D

Lets say, I have a number... It's dark to you, right? Using your strange
dark logic? ;^)

>
>
>> That is well known and has to be considered in any case.
>>
>> The "matrix" is a potentially infinite triangle. Can you find a
>> diagonal from (m, 1) to (1, n) which is shorter than any of its
>> successors?
>>
>> Regards, WM
>

On 12/4/2023 4:16 PM, Chris M. Thomasson wrote:
> On 12/4/2023 3:17 PM, Chris M. Thomasson wrote:
>> On 12/4/2023 5:10 AM, WM wrote:
>>> Fritz Feldhase schrieb am Montag, 4. Dezember 2023 um 12:46:02 UTC+1:
>>>  > On Monday, December 4, 2023 at 10:53:44 AM UTC+1, WM wrote:
>>>  > > B is not defined because it is not a matrix
>>>  >
>>>  > Some quotes:
>>>  >
>>>  > "B, the matrix
>>>  > 1, 2, 4, ...
>>>  > 3, 5, 8, ...
>>>  > 6, 9,13, ...
>>>  > ...
>>>  > containing only natural numbers does exist." (WM)
>>>
>>> I only wanted to butter William up a bit. There are many dark numbers
>>> which cannot be defined individually.
>>
>> Name one? Just one out of the many... LOL!
>>
>> ;^D
>
> Lets say, I have a number... It's dark to you, right? Using your strange
> dark logic? ;^)

A question just for you, I have a pair:

(42, 422)

What does this pair index back into wrt the natural number it was
originally derived from?

>
>
>>
>>
>>> That is well known and has to be considered in any case.
>>>
>>> The "matrix" is a potentially infinite triangle. Can you find a
>>> diagonal from (m, 1) to (1, n) which is shorter than any of its
>>> successors?
>>>
>>> Regards, WM
>>
>

On 10/29/2023 10:36 AM, Jim Burns wrote:
> On 10/28/2023 3:47 PM, Chris M. Thomasson wrote:
>> On 10/26/2023 2:40 PM, Jim Burns wrote:
>>> On 10/26/2023 4:37 PM, Chris M. Thomasson wrote:
>>>> On 10/26/2023 11:05 AM, FromTheRafters wrote:
>>>>> WM presented the following explanation :
>>>>>> FromTheRafters schrieb am Mittwoch,
>>>>>> 25. Oktober 2023 um 17:08:24 UTC+2:
>>>>>>> WM presented the following explanation :
>>>>>>>> Fritz Feldhase schrieb am Dienstag,
>>>>>>>> 24. Oktober 2023 um 21:26:05 UTC+2:
>>>>>>>>>> Fritz Feldhase schrieb am Montag,
>>>>>>>>>> 23. Oktober 2023 um 18:02:11 UTC+2:
>
>>>>>>>>>>> 2. A bijection is not a process.
>>>>>>>>>
>>>>>>>>> A BIJECTION IS NOT A PROCESS.
>>>>>>>>
>>>>>>>> A bijection with |N can be executed as a process.
>>>>>>
>>>>>> According to Cantor. Of course not in reality.
>>>>>>
>>>>>>> But that doesn't mean it *is* a process.
>>>>>>
>>>>>> A process is what can be executed as a process.
>>>>>
>>>>> Explain.
>>>>
>>>> WM:
>>>> I am a ultra hardcore finite type of person...
>>>> ;^o
>>>
>>> WM:
>>> I reject arithmetic
>>> in order to keep two-ended-ness.
>>
>> WM, I cannot count to seven because
>> I have not thought of it yet...
>> Its dark.
>
> Wittgenstein:
> Whereof one cannot speak,
> thereof one must be silent.
>
> Mückenheim:
> Hold my beer, while
> I speak about that which
> thereof one cannot speak:
> dark actual-infinity.
>
>

For some reason, WM must be able to know that if he keeps drinking
beers, his mind might go dark, and he would not be able to accurately
keep track of the number of beers he consumed to render himself into a
blackout like state... Or, shall I say, dark state of mind?

I am thinking of a number... ;^)

William schrieb am Montag, 4. Dezember 2023 um 17:42:43 UTC+1:
> On Monday, December 4, 2023 at 12:12:38 PM UTC-4, WM wrote:
> > B is a tiny submatrix of A
> Your putative A is not defined by your game of billiards. B is
defined (not as a "limit")

Since every step k of the game is defined by Cantor's formula
k = (m + n - 1)(m + n - 2)/2 + m, the sequence producing the elements of
B is completely reproduced by the exchanges. There is no
step missing that is given by the formula for B alone that you call the
definition of B. Therefore your alleged distinction between B alone and
and B as a submtrix of A does not exist and is not justified. Both lead
to the same result. But my procedure shows that B is smaller than A.

Regards, WM

On Tuesday, December 5, 2023 at 5:55:35 AM UTC-4, WM wrote:
>B as a submtrix of A

Your game of billiards does not define the putative matrix A

William schrieb am Dienstag, 5. Dezember 2023 um 11:04:50 UTC+1:
> On Tuesday, December 5, 2023 at 5:55:35 AM UTC-4, WM wrote:
> >B as a submtrix of A
> Your game of billiards does not define the putative matrix A

Why should that be relevant???

My game of billiards defines and places every natural number that is
defined by
k = (m + n - 1)(m + n - 2)/2 + m.
Can you find a difference (besides your matheologial belief that the
formula would be "greater")? In addition it shows that if Cantor issues
more natnumbers, then there is no bijection n <__> 1/n.

Regards, WM

William schrieb am Mittwoch, 6. Dezember 2023 um 18:43:56 UTC+1:
> On Wednesday, December 6, 2023 at 12:08:01 PM UTC-4, WM wrote:
>
> > > Your game of billiards does not define the putative matrix A
> > Why should that be relevant?
> You repeatedly claim that B is a proper submatrix of the putative A.

Irrelevant claim. Only this is relevant: My game of billiards defines
and places every natural number that is defined by
k = (m + n - 1)(m + n - 2)/2 + m.

Regards, WM

On Thursday, December 7, 2023 at 6:01:25 AM UTC-4, WM wrote:
> William schrieb am Mittwoch, 6. Dezember 2023 um 18:43:56 UTC+1:
> > On Wednesday, December 6, 2023 at 12:08:01 PM UTC-4, WM wrote:
> >
> > > > Your game of billiards does not define the putative matrix A
> > > Why should that be relevant?
> > You repeatedly claim that B is a proper submatrix of the putative A.
> Irrelevant claim. Only this is relevant: My game of billiards defines
> and places every natural number that is defined by
> k = (m + n - 1)(m + n - 2)/2 + m.
Your claim that your billiards game means there must be "dark numbers" depends on the existence of a proper supermatrix of B, the putative A.

on 12/7/2023, WM supposed :
> William schrieb am Mittwoch, 6. Dezember 2023 um 18:43:56 UTC+1:
> > On Wednesday, December 6, 2023 at 12:08:01 PM UTC-4, WM wrote:
> >
> > > > Your game of billiards does not define the putative matrix A
> > > Why should that be relevant?
> > You repeatedly claim that B is a proper submatrix of the putative A.
>
> Irrelevant claim. Only this is relevant: My game of billiards defines and
> places every natural number that is defined by
> k = (m + n - 1)(m + n - 2)/2 + m.

Are you saying that the pairing formula 'defines' the natural numbers?

On 07.12.2023 11:59, William wrote:

> Your claim that your billiards game means there must be "dark numbers" depends on the existence of a proper supermatrix of B, the putative A.

No, it depends on the fact that never any A(n) has less positions than
A(0) which is only in the first column covered by all elements of B.
If B gets its final form, then this implies that A gets its final form.
But that is irrelevant. Relevant is only that the difference between
A(n) and B(n) never shrinks.

Regards, WM

William schrieb am Freitag, 8. Dezember 2023 um 22:02:44 UTC+1:
> On Friday, December 8, 2023 at 10:59:24 AM UTC-4, WM wrote:
>
> > If B gets its final form, then this implies that A gets its final form.
> Nope. B is defined and exists. Your putative A does not exist,
>
> > ...Relevant is only that the difference between A(n) and B(n) never
shrinks.
>
> Nope. The set "positions of A(n)\positions of B(n)" shrinks.

Please show where the first event happens.

Regards, WM

William schrieb am Samstag, 9. Dezember 2023 um 13:32:13 UTC+1:
> On Saturday, December 9, 2023 at 7:43:53 AM UTC-4, WM wrote:
> > William schrieb am Freitag, 8. Dezember 2023 um 22:02:44 UTC+1:

> > >
> > > Nope. The set "positions of A(n)\positions of B(n)" shrinks.
> > Please show where the first event happens.
> The set of elements of (|N x |N) { Positions of A(1)\Positions of
B(1) } shrinks by one element to the set of elements of (|N x |N)
> { Positions of A(2)\Positions of B(2) }

Which element is that?

Regards, WM

On Saturday, December 9, 2023 at 12:32:31 PM UTC-4, WM wrote:
> William schrieb am Samstag, 9. Dezember 2023 um 13:32:13 UTC+1:
> > On Saturday, December 9, 2023 at 7:43:53 AM UTC-4, WM wrote:
> > > William schrieb am Freitag, 8. Dezember 2023 um 22:02:44 UTC+1:
>
> > > >
> > > > Nope. The set "positions of A(n)\positions of B(n)" shrinks.
> > > Please show where the first event happens.
> > The set of elements of (|N x |N) { Positions of A(1)\Positions of
> B(1) } shrinks by one element to the set of elements of (|N x |N)
> > { Positions of A(2)\Positions of B(2) }
> Which element is that?
>

The element of the set |Nx|N, (1,0) .

On 09.12.2023 17:43, William wrote:

>> Which element is that?
>>
> The element of the set |Nx|N, (1,0) .

No, all elements of all A(n) remain. All elements of all B(n) remain.
All elements of the difference remain.

Regards, WM

On Sunday, December 10, 2023 at 4:57:50 AM UTC-4, WM wrote:
> On 09.12.2023 17:43, William wrote:
>
> >> Which element is that?
> >>
> > The element of the set |Nx|N, (1,0) .
> No, all elements of all A(n) remain. All elements of all B(n) remain.
> All elements of the difference remain.

let S be the positions of A(1) (also the positions of A(2))
(1,0) is an element of S

P(1) = the positions of B(1) = {(0,0),(0,1)}
P(2) = the positions of B(2) = {(0,0),(0,1),(1,0)|
(1,0) is an element of S\P(1)
(1,0) is not an element of S\P(2)

On 12/10/2023 3:48 AM, WM wrote:
> On 09.12.2023 17:43, William wrote:

>>> Which element is that?
>>
>> The element of the set |Nx|N, (1,0) .
>
> No, all elements of all A(n) remain.
> All elements of all B(n) remain.
> All elements of the difference remain.

Initially,
A\B is unpopulated by B.elements.

Each swap i/j↔kᵢⱼ/1 does not populate A\B
with B.elements

For each B.location i/j
swap i/j↔kᵢⱼ/1 leaves kᵢⱼ/1 at i/j
and i/j↔kᵢⱼ/1 does not populate A\B
with any B.element.

After all and only swaps i/j↔kᵢⱼ/1
each B.location i/j holds kᵢⱼ/1
and A\B is still not populated
by B.elements.

You (WM) deny that and claim that
after all and only non-populating swaps,
A\B has become populated.

(I guess) your explanationᵂᴹ is that
it's darkᵂᴹ
we can't seeᵂᴹ
so (I guess) we should trust you on this?

However, we know A\B is unpopulated
despite not.seeingᵂᴹ A\B
from the description of
⟨1,…,i⟩ ⟨1,…,j⟩ ⟨1,…,kᵢⱼ⟩ i/j↔kᵢⱼ/1

On 10.12.2023 14:07, William wrote:
> On Sunday, December 10, 2023 at 4:57:50 AM UTC-4, WM wrote:
>> On 09.12.2023 17:43, William wrote:
>>
>>>> Which element is that?
>>>>
>>> The element of the set |Nx|N, (1,0) .
>> No, all elements of all A(n) remain. All elements of all B(n) remain.
>> All elements of the difference remain.
>
> let S be the positions of A(1) (also the positions of A(2))
> (1,0) is an element of S
>
> P(1) = the positions of B(1) = {(0,0),(0,1)}
> P(2) = the positions of B(2) = {(0,0),(0,1),(1,0)|

Why do you start with 0? The first position of the matrix is usually
called (1, 1). But independent of notation: All elements of B are
already in the matrix A, namely in the first column. They will not
become fewer or more but will only be reordered.

> (1,0) is an element of S\P(1)
> (1,0) is not an element of S\P(2)

The elements of A(0) replaced by B(n) remain in the matrix. Nothing shrinks.

Regards, WM

On 10.12.2023 21:05, Jim Burns wrote:
> On 12/10/2023 3:48 AM, WM wrote:
>> On 09.12.2023 17:43, William wrote:
>
>>>> Which element is that?
>>>
>>> The element of the set |Nx|N, (1,0) .
>>
>> No, all elements of all A(n) remain.
>> All elements of all B(n) remain. All elements of the difference remain.
>
> Initially,
> A\B is unpopulated by B.elements.

All elements of B are initially in the first column. When the upper left
corner of A(n) is populated, the elements sitting initially there remain
in the matrix - forever!
>
> You (WM) deny that and claim that
> after all and only non-populating swaps,
> A\B has become populated.

The elements of A\B are never changed. First B is only a column B(0).
Later it becomes a triangle (never a square).
> s
> (I guess) your explanationᵂᴹ is that
> it's darkᵂᴹ

Fact is that B never disappears. My explanation is he is in the dark
part below the Cantor-triangle.

Regards, WM

William schrieb am Montag, 11. Dezember 2023 um 12:38:27 UTC+1:
> On Monday, December 11, 2023 at 6:21:09 AM UTC-4, WM wrote:
> > On 10.12.2023 14:07, William wrote:
> > > (1,0) is an element of S\P(1)
> > > (1,0) is not an element of S\P(2)
> > The elements of A(0) replaced by B(n) remain in the matrix.
> So what? The elements of A(0) do not remain in A(n)\B(n) ("\" does
not mean replace)

B(0) = B(1) is the first column of A(0) = A(1):
1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
....

A(2) =
1, 2, 1/3, 1/4, ...
1/2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
....

B(2) =
1, 2

3
4 5
....

A(2) \ B(2) has precisely the same elements as A(0) \ B(0).

Regards, WM

On Tuesday, December 12, 2023 at 6:02:41 AM UTC-4, WM wrote:
> William schrieb am Montag, 11. Dezember 2023 um 12:38:27 UTC+1:
> > On Monday, December 11, 2023 at 6:21:09 AM UTC-4, WM wrote:
> > > On 10.12.2023 14:07, William wrote:
> > > > (1,0) is an element of S\P(1)
> > > > (1,0) is not an element of S\P(2)
> > > The elements of A(0) replaced by B(n) remain in the matrix.
> > So what? The elements of A(0) do not remain in A(n)\B(n) ("\" does
> not mean replace)
> B(0) = B(1) is the first column of A(0) = A(1):

Nope B(0)=B(1) has one element
B(0)=B(1) =
1

A(0)=A(1)> 1, 1/2, 1/3, 1/4, ...
> 2, 2/2, 2/3, 2/4, ...
> 3, 3/2, 3/3, 3/4, ...
> 4, 4/2, 4/3, 4/4, ...
> 5, 5/2, 5/3, 5/4, ...
> ...
>
> A(2) =
> 1, 2, 1/3, 1/4, ...
> 1/2, 2/2, 2/3, 2/4, ...
> 3, 3/2, 3/3, 3/4, ...
> 4, 4/2, 4/3, 4/4, ...
> 5, 5/2, 5/3, 5/4, ...
> ...
>
> B(2) =
> 1, 2
>
> 3
> 4
> 5
> ...
Nope. B(2) has two elements
B(2) =
1, 2
>
> A(2) \ B(2) has precisely the same elements as A(0) \ B(0).
Nope. A(2) and A(0) have the same elements
B(0) and B(2) do not have the same elements.

On 12.12.2023 15:29, William wrote:
> On Tuesday, December 12, 2023 at 6:02:41 AM UTC-4, WM wrote:
>> William schrieb am Montag, 11. Dezember 2023 um 12:38:27 UTC+1:
>>> On Monday, December 11, 2023 at 6:21:09 AM UTC-4, WM wrote:
>>>> On 10.12.2023 14:07, William wrote:
>>>>> (1,0) is an element of S\P(1)
>>>>> (1,0) is not an element of S\P(2)
>>>> The elements of A(0) replaced by B(n) remain in the matrix.
>>> So what? The elements of A(0) do not remain in A(n)\B(n) ("\" does
>> not mean replace)
>> B(0) = B(1) is the first column of A(0) = A(1):
>
> Nope B(0)=B(1) has one element
> B(0)=B(1) =
> 1
>
> A(0)=A(1)=
>> 1, 1/2, 1/3, 1/4, ...
>> 2, 2/2, 2/3, 2/4, ...
>> 3, 3/2, 3/3, 3/4, ...
>> 4, 4/2, 4/3, 4/4, ...
>> 5, 5/2, 5/3, 5/4, ...
>> ...
>>
>> A(2) =
>> 1, 2, 1/3, 1/4, ...
>> 1/2, 2/2, 2/3, 2/4, ...
>> 3, 3/2, 3/3, 3/4, ...
>> 4, 4/2, 4/3, 4/4, ...
>> 5, 5/2, 5/3, 5/4, ...
>> ...
>>
>> B(2) =
>> 1, 2
>>
>> 3
>> 4
>> 5
>> ...
> Nope. B(2) has two elements
> B(2) =
> 1, 2

Please adhere to my notation.
>>
>> A(2) \ B(2) has precisely the same elements as A(0) \ B(0).
> Nope. A(2) and A(0) have the same elements
> B(0) and B(2) do not have the same elements.

Please adhere to my notation.
Further please denote the matrix elements by
(1, 1), (1, 2), (1, 3), ...
(2, 1), (2, 2), (2, 3), ...
(3, 1), (3, 2), (3, 3), ...
....
not starting with (0, 1) which would be incosistent. (0, 0) would be
consistent but confusing.

Regards, WM

On 12/11/2023 5:28 AM, WM wrote:
> On 10.12.2023 21:05, Jim Burns wrote:

>> You (WM) deny that and claim that
>> after all and only non-populating swaps,
>> A\B has become populated.
>
> The elements of A\B are never changed.

Yes.

For each B.location ⟨i,j⟩
there is a swap ⟨i,j⟩↔⟨kᵢⱼ,1⟩
[1]

⟨i,j⟩ and ⟨kᵢⱼ,1⟩ are B.locations.
B.swap ⟨i,j⟩↔⟨kᵢⱼ,1⟩ does not change A\B

> First B is only a column B(0).
> Later it becomes a triangle (never a square).

For each B.location ⟨i,j⟩
there is a B.swap ⟨i,j⟩↔⟨kᵢⱼ,1⟩ inside B

After ⟨i,j⟩↔⟨kᵢⱼ,1⟩, fraction kᵢⱼ/1 is at ⟨i,j⟩
No exceptions.
[2]

After all B.swaps ⟨i,j⟩↔⟨kᵢⱼ,1⟩ inside B
for all B.locations ⟨i,j⟩ in B
all fractions i/j (j≠1) are not anywhere in B

>> (I guess) your explanationᵂᴹ is that
>> it's darkᵂᴹ
>
> Fact is that B never disappears.

After all B.swaps ⟨i,j⟩↔⟨kᵢⱼ,1⟩ inside B
for all B.locations ⟨i,j⟩ in B
all fractions i/j (j≠1) are not anywhere in B

I'm willing to call that disappearing

But YMMV, you may not be willing to. Fine.
Whatever it's _called_
after all B.swaps ⟨i,j⟩↔⟨kᵢⱼ,1⟩ inside B
for all B.locations ⟨i,j⟩ in B
all fractions i/j (j≠1) are not anywhere in B

----
The problem with explaining
that so-called "disappearance" as
not-a-disappearance
but as darkᵂᴹ events happening unseenᵂᴹ
is that
we don't seeᵂᴹ or not-seeᵂᴹ any of the
fractions or B.locations.
We see
descriptive and not-first-false claims
about fractions and B.locations.

Declaring numbers darkᵂᴹ will not unsee
the claims we've seen.

The Cantorian explanation is that
only _some_ sets cannot match proper subsets,
and that
B and the first column of B are
not among those sets.

We can see (by seeing claims)
that B matches its first column.
Yes.
Its first column can be spread over
all of B,
for example, by kᵢⱼ = i+(i+j-1)(i+j-2)/2
That's how Bob "disappears".

[1]
For each B.location ⟨i,j⟩
i is in ⟨1,…,i⟩
j is in ⟨1,…,j⟩
and,
for swap ⟨i,j⟩↔⟨kᵢⱼ,1⟩
kᵢⱼ = i+(i+j-1)(i+j-2)/2
kᵢⱼ is in ⟨1,…,kᵢⱼ⟩
⟨kᵢⱼ,1⟩ is in B

For each B.location ⟨k,1⟩
k is in ⟨1,…,k⟩
and,
for swap ⟨iₖ,jₖ⟩↔⟨k,1⟩
sₖ = max{h| (h-1)(h-2)/2 < k }
sₖ = iₖ+jₖ
iₖ = k-(sₖ-1)(sₖ-2)/2
jₖ = sₖ(sₖ-1)/2-k+1
iₖ is in ⟨1,…,iₖ⟩
jₖ is in ⟨1,…,jₖ⟩
⟨iₖ,jₖ⟩ is in B

[2]
For each B.swap ⟨i,j⟩↔⟨kᵢⱼ,1⟩
before‖after by kᵢⱼ

Swaps ⟨i′,j′⟩↔⟨k′,1⟩ before ⟨i,j⟩↔⟨kᵢⱼ,1⟩
(that is, k′ < kᵢⱼ)
do not move kᵢⱼ/1 from ⟨kᵢⱼ,1⟩

Swap ⟨i,j⟩↔⟨kᵢⱼ,1⟩ _does_ move kᵢⱼ/1
from ⟨kᵢⱼ,1⟩ to ⟨i,j⟩

Swaps ⟨i″,j″⟩↔⟨k″,1⟩ after ⟨i,j⟩↔⟨kᵢⱼ,1⟩
(that is, kᵢⱼ < k″)
do not move kᵢⱼ/1 from ⟨i,j⟩

After ⟨i,j⟩↔⟨kᵢⱼ,1⟩, kᵢⱼ/1 is at ⟨i,j⟩
No exceptions.

On Tuesday, December 12, 2023 at 1:18:03 PM UTC-4, WM wrote:
> On 12.12.2023 15:29, William wrote:

> Please adhere to my notation

Whatever notation you use, you only need dark numbers to get a location for a "fraction" in your putative matrix A.

On 12.12.2023 18:55, Jim Burns wrote:
> On 12/11/2023 5:28 AM, WM wrote:
>> On 10.12.2023 21:05, Jim Burns wrote:
>
>>> You (WM) deny that and claim that
>>> after all and only non-populating swaps,
>>> A\B has become populated.
>>
>> The elements of A\B are never changed.
>
> Yes.

They are all positive fractions with exception of unit fractions.

> After all B.swaps ⟨i,j⟩↔⟨kᵢⱼ,1⟩ inside B
> for all B.locations ⟨i,j⟩ in B
> all fractions i/j (j≠1) are not anywhere in B

No, they are not in B but also not outside of A (which exists after all
swaps as well as B).
>
>> Fact is that B never disappears.
>
> After all B.swaps ⟨i,j⟩↔⟨kᵢⱼ,1⟩ inside B
> for all B.locations ⟨i,j⟩ in B
> all fractions i/j (j≠1) are not anywhere in B
>
> I'm willing to call that disappearing

They are in the darkness.
>
> The problem with explaining
> that so-called "disappearance" as
> not-a-disappearance
> but as darkᵂᴹ events happening unseenᵂᴹ
> is that
> we don't seeᵂᴹ or not-seeᵂᴹ any of the
> fractions or B.locations.
> We see
> descriptive and not-first-false claims
> about fractions and B.locations.

In fact most of all matrices, including B is dark. All that can be
defined of any infinite matrix, including B, is a triangle in the upper
left corner.
> We can see (by seeing claims)
> that B matches its first column.

The first column can be seen until some defined n. But almost all is dark:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

Regards, WM

On 12.12.2023 18:56, William wrote:
> On Tuesday, December 12, 2023 at 1:18:03 PM UTC-4, WM wrote:
>> On 12.12.2023 15:29, William wrote:
>
>> Please adhere to my notation
>
> Whatever notation you use, you only need dark numbers to get a location for a "fraction" in your putative matrix A.

I need locations for all fractions p/q with q > 1 because I know that
they never can leave the matrix. You must know: my billiard table has
high walls and no holes.

Regards, WM