On Tuesday, December 12, 2023 at 3:19:10 PM UTC-4, WM wrote:
> 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.

Indeed, the "fractions" never leave the A(n) which are the only matrices produced by your billiards game. Since your billiards game never produces the putative matrix A, it does not show the existence of "dark numbers".

On 12.12.2023 21:20, William wrote:
> On Tuesday, December 12, 2023 at 3:19:10 PM UTC-4, WM wrote:
>> 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.
>
> Indeed, the "fractions" never leave the A(n) which are the only matrices produced by your billiards game.

All B(n) are also produced.

> Since your billiards game never produces the putative matrix A, it does not show the existence of "dark numbers".

All produced matrices A(n) contain all fractions p/q with q > 1 in all
produced differences A(n) \ B(n). If A is not produced by using
Cantor's formula, then also B is not produced by Cantor's formula.

In no case any fraction leaves the difference A(n) \ B(n). A mystery
might exorcize them in the limit. I do not believe in mysteries in
mathematics.

Regards, WM

On Tuesday, December 12, 2023 at 5:46:27 PM UTC-4, WM wrote:

> If A is not produced by using
> Cantor's formula, then also B is not produced by Cantor's formula.

Nope. B is defined. Your putative A is not defined.

William schrieb am Dienstag, 12. Dezember 2023 um 22:54:13 UTC+1:
> On Tuesday, December 12, 2023 at 5:46:27 PM UTC-4, WM wrote:
>
> > If A is not produced by using
> > Cantor's formula, then also B is not produced by Cantor's formula.
> Nope. B is defined.
By the formula used in the OP.

> Your putative A is not defined.

All matrices A(n) are defined and contain all fractions. They cannot
leave the matrix. If they are invisible, they are dark.

By the way, same with descending from omega. Every step ends always at a
visible natnumber which has ℵo dark successors which cannot be chosen as
destination.

Regards, WM

On Wednesday, December 13, 2023 at 7:40:43 AM UTC-4, WM wrote:
> William schrieb am Dienstag, 12. Dezember 2023 um 22:54:13 UTC+1:

> > Your putative A is not defined.
> All matrices A(n) are defined

Correct. So what?. Your putative A is not defined.

On 13.12.2023 13:12, William wrote:
> On Wednesday, December 13, 2023 at 7:40:43 AM UTC-4, WM wrote:
>> William schrieb am Dienstag, 12. Dezember 2023 um 22:54:13 UTC+1:
>
>>> Your putative A is not defined.
>> All matrices A(n) are defined
>
> Correct. So what?. Your putative A is not defined.

So what? It is irrelevant. There is no matrix which has lost any
fraction. Hence all are there and occupy not indexed positions. In no
instance any fractions is abolished.

Regards, WM

On Wednesday, December 13, 2023 at 12:40:05 PM UTC-4, WM wrote:
> On 13.12.2023 13:12, William wrote:
> > On Wednesday, December 13, 2023 at 7:40:43 AM UTC-4, WM wrote:
> >> William schrieb am Dienstag, 12. Dezember 2023 um 22:54:13 UTC+1:
> >
> >>> Your putative A is not defined.
> >> All matrices A(n) are defined
> >
> > Correct. So what?. Your putative A is not defined.
> So what? It is irrelevant. There is no matrix which has lost any
> fraction. Hence all are there and occupy not indexed positions.

Nope, In each A(n) only a finite number of elements have been moved, and each of them has only been moved a finite number of positions. Only in the putative A, can we conclude that the fractions must "occupy not indexed positions". However, your putative A is not defined.

On 13.12.2023 19:36, William wrote:
> On Wednesday, December 13, 2023 at 12:40:05 PM UTC-4, WM wrote:
>> On 13.12.2023 13:12, William wrote:
>>> On Wednesday, December 13, 2023 at 7:40:43 AM UTC-4, WM wrote:
>>>> William schrieb am Dienstag, 12. Dezember 2023 um 22:54:13 UTC+1:
>>>
>>>>> Your putative A is not defined.
>>>> All matrices A(n) are defined
>>>
>>> Correct. So what?. Your putative A is not defined.
>> So what? It is irrelevant. There is no matrix which has lost any
>> fraction. Hence all are there and occupy not indexed positions.
>
> Nope, In each A(n) only a finite number of elements have been moved, and each of them has only been moved a finite number of positions.

All that can be moved has been moved, because "in the limit" nothing can
happen. And because you cannot name any index that has failed to cover
its final destination.
>
> However, your putative A is not defined.

All steps where something is moved are defined. Never the whole matrix
is covered by indices.

Regards, WM

On Wednesday, December 13, 2023 at 2:58:14 PM UTC-4, WM wrote:
> On 13.12.2023 19:36, William wrote:
> > On Wednesday, December 13, 2023 at 12:40:05 PM UTC-4, WM wrote:
> >> On 13.12.2023 13:12, William wrote:
> >>> On Wednesday, December 13, 2023 at 7:40:43 AM UTC-4, WM wrote:
> >>>> William schrieb am Dienstag, 12. Dezember 2023 um 22:54:13 UTC+1:
> >>>
> >>>>> Your putative A is not defined.
> >>>> All matrices A(n) are defined
> >>>
> >>> Correct. So what?. Your putative A is not defined.
> >> So what? It is irrelevant. There is no matrix which has lost any
> >> fraction. Hence all are there and occupy not indexed positions.
> >
> > Nope, In each A(n) only a finite number of elements have been moved, and each of them has only been moved a finite number of positions.
> All that can be moved has been moved,

This does not change the fact that in each A(n) only a finite number of elements have been moved and each of them has only been moved a finite number of times. Hence producing any A(n) does not show that "dark" numbers are needed. Only producing the putative A shows that "dark" numbers are needed.

On 12/12/2023 2:08 PM, WM wrote:
> 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.

Then that A that B and that A\B aren't
what I've been referring to.

Consider

B′ which covers only
all locations ⟨i,j⟩ for which
i is in ⟨1,…,i⟩
j is in ⟨1,…,j⟩

A′ which covers
what B′ covers and
possibly covers more

Initially,
for each B′.location ⟨i,j⟩
B′.fraction i/j is at ⟨i,j⟩
i is in ⟨1,…,i⟩
j is in ⟨1,…,j⟩

Initially,
any possibly.existing locations covered by A′\B′
are not B′.locations
any possibly.existing fractions at
any possibly.existing locations covered by A′\B′
are not B′.fractions

For each B′.location ⟨i,j⟩
there exists swap ⟨i,j⟩↔⟨kᵢⱼ,1⟩
There are no other swaps than these.

The swaps ⟨i,j⟩↔⟨k,1⟩ are ordered by k
For swaps ⟨i′,j′⟩↔⟨k′,1⟩ ⟨i″,j″⟩↔⟨k″,1⟩
⟨i′,j′⟩↔⟨k′,1⟩ is before ⟨i″,j″⟩↔⟨k″,1⟩
iff k′ < k″

for ⟨i,j⟩↔⟨k,1⟩
if j>1
then ⟨i+1,j-1⟩↔⟨k+1,1⟩ is next.after
else ⟨1,i+1⟩↔⟨k+1,1⟩ is next.after

for ⟨i,j⟩↔⟨kᵢⱼ,1⟩
kᵢⱼ = i+(i+j-1)(i+j-2)/2

for ⟨iₖ,jₖ⟩↔⟨k,1⟩
sₖ = max{h| (h-1)(h-2)/2 < k }
iₖ+jₖ = sₖ
iₖ = k-(sₖ-1)(sₖ-2)/2
jₖ = sₖ(sₖ-1)/2-k+1

Consider B′ A′ A′\B′
B′.locations, B′.fractions and B′.swaps

>>>> 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.

Initially,
B′.fractions are at B′.locations
and no B′.fraction is elsewhere

Each swap is from‖to a B′.location.

After all swaps each from‖to a B′.location
only index B′.fractions k/1 are at B′.locations.

If, after all swaps,
non-index B′.fractions i/jᙾ¹ are elsewhere,
then it wasn't by those swaps

....because
each swap is from‖to a B′.location.

>> 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).

After all B′.swaps ⟨i,j⟩↔⟨kᵢⱼ,1⟩
for all B′.locations ⟨i,j⟩
all non-index B′.fractions i/jᙾ¹ are
not anywhere in B′

You (WM) and Cantor have
different explanations for that.

>>> 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.

You (WM) say
there is no visibleᵂᴹ n with darkᵂᴹ n+1

Under B′
each location is ⟨i,j⟩ such that
i is in ⟨1,…,i⟩
j is in ⟨1,…,j⟩

Each of those i and j are visibleᵂᴹ

I assume that visibleᵂᴹ i j means that
⟨i,j⟩ is visibleᵂᴹ

| Assume otherwise.
| Assume i in ⟨1,…,i⟩ is darkᵂᴹ
| | A non.∅ split F₁ ᣔ<ᣔ H₁ exists between
| the super.visiblesᵂᴹ F₁ and the others H₁
| | By super.visibleᵂᴹ I mean that
| each number ≤ super.visibleᵂᴹ n is visibleᵂᴹ
| | ⟨1,…,i⟩ is what it is, thus
| i₁‖i₁+1 exists last‖first in part F₁‖H₁
| | Each number < i₁+1 is in F₁ and visibleᵂᴹ
| If i₁+1 is visibleᵂᴹ
| then i₁+1 is super.visibleᵂᴹ and in F₁
| But i₁+1 is in H₁
| Thus i₁+1 is darkᵂᴹ
| | i₁ is in F₁ and super.visibleᵂᴹ
| | i₁ is visibleᵂᴹ and i₁+1 is darkᵂᴹ
| | However,
| there is no visibleᵂᴹ i₁ with darkᵂᴹ i₁+1
| Contradiction.

Therefore,
i in ⟨1,…,i⟩ is visibleᵂᴹ

>>> 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.

No B′.location is darkᵂᴹ

Each B′.fraction is initially at
a visibleᵂᴹ location.

No B′.swap moves anything to
a darkᵂᴹ location.

Nothing else moves anything anywhere.

After all B′.swaps ⟨i,j⟩↔⟨kᵢⱼ,1⟩
for all B′.locations ⟨i,j⟩
all non-index B′.fractions i/jᙾ¹
which are initially in B′
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.
>
> In fact most of all matrices, including B
> is dark.

We know, by seeing claims which are
descriptive or not.first.false,
that all of B′ is visibleᵂᴹ

....because
there is no visibleᵂᴹ i₁ with darkᵂᴹ i₁+1
and
for non.∅ split F₁ ᣔ<ᣔ H₁ of ⟨1,…,i⟩
i₁‖i₁+1 exists last‖first in F₁‖H₁

> All that can be defined of any infinite matrix,
> including B, is
> a triangle in the upper left corner.

We can say _once_ and know in
_infinitely.many_ ways that
⟨i,j⟩ such that
i is in ⟨1,…,i⟩ and j is in ⟨1,…,j⟩
is a location in B′

>> 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.

|B| = ℵ₀ means
B is 1.by.1;1.end.able

Even if ℵ₀ didn't mean that,
for each split of any 1.by.1;1.end.ed,
one side is 1.by.1;1.end.able and
the other side isn't.

1 original end + 2 ends of a split =
1 end of one part and 2 ends of the other.

On 13.12.2023 20:41, William wrote:
> On Wednesday, December 13, 2023 at 2:58:14 PM UTC-4, WM wrote:
>> On 13.12.2023 19:36, William wrote:
>>> On Wednesday, December 13, 2023 at 12:40:05 PM UTC-4, WM wrote:
>>>> On 13.12.2023 13:12, William wrote:
>>>>> On Wednesday, December 13, 2023 at 7:40:43 AM UTC-4, WM wrote:
>>>>>> William schrieb am Dienstag, 12. Dezember 2023 um 22:54:13 UTC+1:
>>>>>
>>>>>>> Your putative A is not defined.
>>>>>> All matrices A(n) are defined
>>>>>
>>>>> Correct. So what?. Your putative A is not defined.
>>>> So what? It is irrelevant. There is no matrix which has lost any
>>>> fraction. Hence all are there and occupy not indexed positions.
>>>
>>> Nope, In each A(n) only a finite number of elements have been moved, and each of them has only been moved a finite number of positions.
>> All that can be moved has been moved,
>
> This does not change the fact that in each A(n) only a finite number of elements have been moved and each of them has only been moved a finite number of times.

*Everything* that you can do and check lies within a finite number of steps.
Note: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

> Hence producing any A(n) does not show that "dark" numbers are needed.

Everything and all that can be done with individual numbers does not
need dark numbers.

> Only producing the putative A shows that "dark" numbers are needed.

No. Producing B shows it because in all definable steps A(n) \ B(n) is
not empty. And in the limit nothing happens anymore.

Fact is that exchanging X and O in

XOOO...
XOOO...
XOOO...
XOOO...
....

will never cover the matrix with X. Simple geometry.

Regards, WM

On 13.12.2023 23:14, Jim Burns wrote:

nothing that could disprove the fact that by exchanging X and O in

XOOO...
XOOO...
XOOO...
XOOO...
....

the matrix will never be covered with X. Simple geometry.

Regards, WM

On Thursday, December 14, 2023 at 6:08:40 AM UTC-4, WM wrote:
> On 13.12.2023 20:41, William wrote:
> need dark numbers.
> > Only producing the putative A shows that "dark" numbers are needed.
> No. Producing B

does not involve a limit

Fritz Feldhase schrieb am Donnerstag, 14. Dezember 2023 um 11:48:26 UTC+1:
> die Matrix B = (b_n,m) wird nicht "produziert", sondern ist (wie
folgt) _definiert_.
>
> Note: The matrix B = (b_n,m) is defined by b_n,m = (m + n - 1)(m + n
- 2)/2 + m for all n,m e IN.

Matheolgians may believe that.

b_n,m = (m + n - 1)(m + n - 2)/2 + m is a word without any meaning. It
can be applied to describe matrix places. But it cannot be applied to
describe all matrix places of B because
∀n,m ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n or m}| = ℵo.
Only "in the limit" B can be established.

Regards, WM

Profilbild von WM
WM
ungelesen,
16:22 (vor 9 Minuten)
an
On 14.12.2023 15:23, William wrote:
> On Thursday, December 14, 2023 at 6:08:40 AM UTC-4, WM wrote:
>> On 13.12.2023 20:41, William wrote:
>> need dark numbers.
>>> Only producing the putative A shows that "dark" numbers are needed.
>> No. Producing B
>
> does not involve a limit

It involves the realization of all natural numbers
k = (m + n - 1)(m + n - 2)/2 + m
at matrix places
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, 4/3, 5/2, 6/1, ...

This requires a limit or a last step. Your belief alone is not sufficient.

Regards, WM

On Thursday, December 14, 2023 at 11:32:42 AM UTC-4, WM wrote:
> Profilbild von WM
> WM
> ungelesen,
> 16:22 (vor 9 Minuten)
> an
> On 14.12.2023 15:23, William wrote:
> > On Thursday, December 14, 2023 at 6:08:40 AM UTC-4, WM wrote:
> >> On 13.12.2023 20:41, William wrote:
> >> need dark numbers.
> >>> Only producing the putative A shows that "dark" numbers are needed.
> >> No. Producing B
> >
> > does not involve a limit
> It involves the realization of all natural numbers

Nope it does not involve "realization"

> k = (m + n - 1)(m + n - 2)/2 + m
An algorithm which defines B. A matrix is defined if there is an algorithm which gives the value, given an element of |Nx|N. There is no algorithm for your putative A.

On 14.12.2023 16:45, William wrote:
> On Thursday, December 14, 2023 at 11:32:42 AM UTC-4, WM wrote:
>> Profilbild von WM
>> WM
>> ungelesen,
>> 16:22 (vor 9 Minuten)
>> an
>> On 14.12.2023 15:23, William wrote:
>>> On Thursday, December 14, 2023 at 6:08:40 AM UTC-4, WM wrote:
>>>> On 13.12.2023 20:41, William wrote:
>>>> need dark numbers.
>>>>> Only producing the putative A shows that "dark" numbers are needed.
>>>> No. Producing B
>>>
>>> does not involve a limit
>> It involves the realization of all natural numbers
>
> Nope it does not involve "realization"
>
>> k = (m + n - 1)(m + n - 2)/2 + m
> An algorithm which defines B. A matrix is defined if there is an algorithm which gives the value, given an element of |Nx|N. There is no algorithm for your putative A.

You are wrong. The algorithm is given in the OP.

Regards, WM
>

On Thursday, December 14, 2023 at 12:23:01 PM UTC-4, WM wrote:
> On 14.12.2023 16:45, William wrote:
> > A matrix is defined if there is an algorithm which gives the value, given an element of |Nx|N. There is no algorithm for your putative A.
> You are wrong. The algorithm is given in the OP.
Nope, your billiards game defines A(n) for each n. Your billiards game does not define A,

On 12/14/2023 5:16 AM, WM wrote:
> On 13.12.2023 23:14, Jim Burns wrote:

> nothing that
> could disprove the fact that
> by exchanging X and O in
>
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
>
> the matrix will never be covered with X.
> Simple geometry.

"Geometry", he says.
Geometry is thronging with
sets you deny exist.

Two lines CA and CB intersect at point C
with angle ∡ACB = 45° between them.

For each Aₙ in CA
there is a point Bₙ in CB such that
segment AₙBₙ ⟂ CB

The distance d(CAₙ) = aₙ
The distance d(CBₙ) = aₙ/2¹ᐟ²

For each Bₙ in CB
there is a point Aₙ₊₁ in CA such that
segment BₙAₙ₊₁ ⟂ CA

The distance d(CAₙ) = aₙ
The distance d(CBₙ) = aₙ/2¹ᐟ²
The distance d(CAₙ₊₁) = aₙ/2

There is a point.sequence in CA
⟨ A₀ A₁ A₂ A₃ ... ⟩
with a corresponding distance.sequence to C
⟨ a₀ a₀/2 a₀/2² a₀/2³ ... ⟩

Each point, each distance is not
the second end of its sequence.
A second end not.exists.

The sequences are 1.by.1;1.end.ed
They are 1.by.1;1.end.able and ℵ₀.many.

The distances are 1.by.1;1.end.ed
They are 1.by.1;1.end.able and ℵ₀.many.

Consider the swap.sequence
⟨ A₀↔A₁ A₁↔A₂ A₂↔A₃ .. ⟩

Initially,
Bob occupies A₀ in CA
After the first n swaps, before the next,
Bob occupies Aₙ in CA

However,
no swap is the last swap.
After Aₙ↔Aₙ₊₁ Bob never occupies Aₙ

After all swaps Aₙ↔Aₙ₊₁
each Aₙ is not.occupied by Bob

Bob never leaves ⟨ A₀ A1 A₂ A3 ... ⟩
But, after all swaps,
Bob isn't in ⟨ A₀ A1 A₂ A3 ... ⟩

Darknessᵂᴹ can't be the explanation.
Watchᵂᴹ Bob travel within the visibleᵂᴹ

Sets which are the same "size" as
proper subsets is the explanation.
It is Cantor's explanation.

Some sets don't have same.sized proper subsets.
⟨1,…,n⟩ are sets like that.

Other sets are the sets from which
Bob can disappear,
sets which have same.sized proper subsets.
The least upper bound of all the ⟨1,…,n⟩
is a set like that.

On 14.12.2023 17:34, William wrote:
> On Thursday, December 14, 2023 at 12:23:01 PM UTC-4, WM wrote:
>> On 14.12.2023 16:45, William wrote:
>>> A matrix is defined if there is an algorithm which gives the value, given an element of |Nx|N. There is no algorithm for your putative A.
>> You are wrong. The algorithm is given in the OP.
> Nope, your billiards game defines A(n) for each n. Your billiards game does not define A,

What defines then the complete enumeration of the algebraic numbers? You
know Dedekind's approach? If not you can learn it here:
https://www.hs-augsburg.de/~mueckenh/HI/HI11.PPT, p.19.

Regards, WM

On 14.12.2023 20:00, Jim Burns wrote:
> On 12/14/2023 5:16 AM, WM wrote:
>> On 13.12.2023 23:14, Jim Burns wrote:
>
>> nothing that
>> could disprove the fact that
>> by exchanging X and O in
>>
>> XOOO...
>> XOOO...
>> XOOO...
>> XOOO...
>> ...
>>
>> the matrix will never be covered with X.
>> Simple geometry.
>
> "Geometry", he says.
> Geometry is thronging with
> sets you deny exist.

Sets exist and their elements remain existing when redistributing them.

> Sets which are the same "size" as
> proper subsets is the explanation.
> It is Cantor's explanation.

It is wrong. Consider the unit fractions. For every x > 0 that you
choose you know that infinitely many unit fractions are smaller. What
can you do to distinguish them in order to enumerate them? Nothing! No
chosen x will separate them.

Regards, WM

On 12/14/23 10:32 AM, WM wrote:
> Profilbild von WM
> WM
> ungelesen,
> 16:22 (vor 9 Minuten)
> an
> On 14.12.2023 15:23, William wrote:
> > On Thursday, December 14, 2023 at 6:08:40 AM UTC-4, WM wrote:
> >> On 13.12.2023 20:41, William wrote:
> >> need dark numbers.
> >>> Only producing the putative A shows that "dark" numbers are needed.
> >> No. Producing B
> >
> > does not involve a limit
>
> It involves the realization of all natural numbers
> k = (m + n - 1)(m + n - 2)/2 + m
> at matrix places
> 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, 4/3, 5/2, 6/1, ...
>
> This requires a limit or a last step. Your belief alone is not sufficient.
>
> Regards, WM

So, what IS the "last Step" that is required?

You don't seem to understand that Natural Numbers are Unbounded, and
thus there isn't a "last" one.

On 12/14/2023 5:00 PM, Richard Damon wrote:
> On 12/14/23 10:32 AM, WM wrote:
>> Profilbild von WM
>> WM
>> ungelesen,
>> 16:22 (vor 9 Minuten)
>> an
>> On 14.12.2023 15:23, William wrote:
>>  > On Thursday, December 14, 2023 at 6:08:40 AM UTC-4, WM wrote:
>>  >> On 13.12.2023 20:41, William wrote:
>>  >> need dark numbers.
>>  >>> Only producing the putative A shows that "dark" numbers are needed.
>>  >> No. Producing B
>>  >
>>  > does not involve a limit
>>
>> It involves the realization of all natural numbers
>> k = (m + n - 1)(m + n - 2)/2 + m
>> at matrix places
>> 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, 4/3, 5/2, 6/1, ...
>>
>> This requires a limit or a last step. Your belief alone is not
>> sufficient.
>>
>> Regards, WM
>
> So, what IS the "last Step" that is required?
>
> You don't seem to understand that Natural Numbers are Unbounded, and
> thus there isn't a "last" one.

WM cannot grasp the fact that infinity means that there is no last step
wrt the process.

On 12/7/2023 4:20 AM, FromTheRafters wrote:
> 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?

And vise versa. It's the full circle mapping... From natural number, to
unique pair and back again; nothing is lost! So they basically, both
define themselves in and of themselves? Is that fair enough in a sense?

On 12/14/2023 11:31 AM, WM wrote:
> On 14.12.2023 20:00, Jim Burns wrote:
>> On 12/14/2023 5:16 AM, WM wrote:
>>> On 13.12.2023 23:14, Jim Burns wrote:
>>
>>> nothing that
>>> could disprove the fact that
>>> by exchanging X and O in
>>>
>>> XOOO...
>>> XOOO...
>>> XOOO...
>>> XOOO...
>>> ...
>>>
>>> the matrix will never be covered with X.
>>> Simple geometry.
>>
>> "Geometry", he says.
>> Geometry is thronging with
>> sets you deny exist.
>
> Sets exist and their elements remain existing when redistributing them.
>
>> Sets which are the same "size" as
>> proper subsets is the explanation.
>> It is Cantor's explanation.
>
> It is wrong. Consider the unit fractions. For every x > 0 that you
> choose you know that infinitely many unit fractions are smaller.

1/16

a smaller one would be:

1/32

So?

What
> can you do to distinguish them in order to enumerate them? Nothing! No
> chosen x will separate them.
>
> Regards, WM
>