On 12/2/23 11:52 AM, WM wrote:
> On 02.12.2023 16:37, William wrote:
>> On Saturday, December 2, 2023 at 11:02:18 AM UTC-4, WM wrote:
>>> On 02.12.2023 15:37, William wrote:
>>>> On Saturday, December 2, 2023 at 6:12:58 AM UTC-4, WM wrote:
>>>>
>>>>
>>>>> If B is completed by k = (m + n - 1)(m + n - 2)/2 + m,
>>>>
>>>> Nope. B is defined by this algorithm.
>>> Then it is completed by it.
>>
>>>>
>>>> Nope. "Completed" is nonsense.
>>> Agreed.
>>
>> ??
>
> The claim that all fractions could be indexed by using all indices is
> nonsense.
>
> *If* B is defined by this algorithm, then it is completed by it.
>
> Regards, WM

Then what is wrong with the formula that shows how every fraction can be
assigned a unique index from the natural numbers?

You problem seems to be that you just don't understand the nature of
infinity.

On Saturday, December 2, 2023 at 9:46:10 AM UTC-8, Richard Damon wrote:
> On 12/2/23 11:52 AM, WM wrote:
> > On 02.12.2023 16:37, William wrote:
> >> On Saturday, December 2, 2023 at 11:02:18 AM UTC-4, WM wrote:
> >>> On 02.12.2023 15:37, William wrote:
> >>>> On Saturday, December 2, 2023 at 6:12:58 AM UTC-4, WM wrote:
> >>>>
> >>>>
> >>>>> If B is completed by k = (m + n - 1)(m + n - 2)/2 + m,
> >>>>
> >>>> Nope. B is defined by this algorithm.
> >>> Then it is completed by it.
> >>
> >>>>
> >>>> Nope. "Completed" is nonsense.
> >>> Agreed.
> >>
> >> ??
> >
> > The claim that all fractions could be indexed by using all indices is
> > nonsense.
> >
> > *If* B is defined by this algorithm, then it is completed by it.
> >
> > Regards, WM
> Then what is wrong with the formula that shows how every fraction can be
> assigned a unique index from the natural numbers?
>
> You problem seems to be that you just don't understand the nature of
> infinity.

Infinity what?

Here billiards is mostly pocket billiards, and mostly
pool and not including snooker, which is a great game itself,
and not bumper billiards, mostly straight pool and eight-ball and nine-ball,
and that around here it's ball-in-hand and mostly bank-the-eight.

I suggest that if you've never sunk the eight ball on the break
or the nine ball on the break, thus winning the game on the first shot,
making the break, then you don't much know billiards.

There are otherwise of course an infinity of games, but,
there are only ones of those.

On 12/2/2023 2:12 AM, WM wrote:
> William schrieb am Freitag, 1. Dezember 2023 um 13:27:28 UTC+1:
> > On Friday, December 1, 2023 at 8:16:58 AM UTC-4, WM wrote:
> > >...the fraction 1/2 is not in B but remains in
> > > whatever is defined of A.
> > There is no A. A(0), A(1) , A(2) ,,,, are not "whatever is defined of
> A".
>
> Let (A(n)) be the sequence  A(0), A(1), A(2), ... . All existing
> matrices are of same size. All elements of B are in every term A(n).
> If B is completed by k = (m + n - 1)(m + n - 2)/2 + m, then also A is
> completed.

The mapping maps all naturals to a pair, and back again. Full circle.
Nothing is lost. Why do you have trouble realizing that fact?

On 12/2/2023 8:52 AM, WM wrote:
> On 02.12.2023 16:37, William wrote:
>> On Saturday, December 2, 2023 at 11:02:18 AM UTC-4, WM wrote:
>>> On 02.12.2023 15:37, William wrote:
>>>> On Saturday, December 2, 2023 at 6:12:58 AM UTC-4, WM wrote:
>>>>
>>>>
>>>>> If B is completed by k = (m + n - 1)(m + n - 2)/2 + m,
>>>>
>>>> Nope. B is defined by this algorithm.
>>> Then it is completed by it.
>>
>>>>
>>>> Nope. "Completed" is nonsense.
>>> Agreed.
>>
>> ??
>
> The claim that all fractions could be indexed by using all indices is
> nonsense.

HUH? The pairs generated by Cantor pairing (a, b) can be used in the
form of a fraction a/b, as long as b is not zero. A pair is different
than a fraction...

>
> *If* B is defined by this algorithm, then it is completed by it.
[...]

On 12/2/2023 9:34 AM, William wrote:
> On Saturday, December 2, 2023 at 12:52:38 PM UTC-4, WM wrote:
>
>> *If* B is defined by this algorithm, then it is completed by it.
>
> Nope. The term "completed" is meaningless.

An infinite process is never completed... However, we can say Cantor
Pairing can handle any natural number, therefore it works for all of
them. Fair enough?

On 12/2/2023 9:46 AM, Richard Damon wrote:
> On 12/2/23 11:52 AM, WM wrote:
>> On 02.12.2023 16:37, William wrote:
>>> On Saturday, December 2, 2023 at 11:02:18 AM UTC-4, WM wrote:
>>>> On 02.12.2023 15:37, William wrote:
>>>>> On Saturday, December 2, 2023 at 6:12:58 AM UTC-4, WM wrote:
>>>>>
>>>>>
>>>>>> If B is completed by k = (m + n - 1)(m + n - 2)/2 + m,
>>>>>
>>>>> Nope. B is defined by this algorithm.
>>>> Then it is completed by it.
>>>
>>>>>
>>>>> Nope. "Completed" is nonsense.
>>>> Agreed.
>>>
>>> ??
>>
>> The claim that all fractions could be indexed by using all indices is
>> nonsense.
>>
>> *If* B is defined by this algorithm, then it is completed by it.
>>
>> Regards, WM
>
> Then what is wrong with the formula that shows how every fraction can be
> assigned a unique index from the natural numbers?
>
> You problem seems to be that you just don't understand the nature of
> infinity.

Afaict, he has a hyper finite mind. the mere thought of infinity makes
blood want to shoot out of his eyes...? ;^)

On 02.12.2023 18:46, Richard Damon wrote:

> Then what is wrong with the formula that shows how every fraction can be
> assigned a unique index from the natural numbers?

This formula is correct for all visible numbers, but as we know and even
you admit, almost all natural numbers cannot be used because after every
applied number, which belongs to a finite set, ℵo numbers are not
applied in eternity.

This is proved by the matrix
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, ...
....

If all natural numbers of the first column are used to cover the whole
matrix, only a fool can start to believe that.
To make it even more vivid: Cover the matrix
XOOO...
XOOO...
XOOO...
XOOO...
....
completely with X by shuffling the X as you like.
>
> You problem seems to be that you just don't understand the nature of
> infinity.

Your problem is that you believe in nonsense.

Regards, WM

On 02.12.2023 18:34, William wrote:
> On Saturday, December 2, 2023 at 12:52:38 PM UTC-4, WM wrote:
>
>> *If* B is defined by this algorithm, then it is completed by it.
>
> Nope. The term "completed" is meaningless.

You are amazingly right! It is possible to cover the upper left-hand
corner of the matrix

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

by suitably shuffling the X. But it is impossible to cover the complete
matrix by X.

Regards, WM

Chris M. Thomasson schrieb am Samstag, 2. Dezember 2023 um 20:30:55 UTC+1:
> On 12/2/2023 9:34 AM, William wrote:
> > On Saturday, December 2, 2023 at 12:52:38 PM UTC-4, WM wrote:
> >
> >> *If* B is defined by this algorithm, then it is completed by it.
> >
> > Nope. The term "completed" is meaningless.
> An infinite process is never completed... However, we can say Cantor
> Pairing can handle any natural number, therefore it works for all of
> them. Fair enough?

Only for naive minds. "Any natural number" belongs to a finite set
whereas almost all natural numbers follow upon that set.

Regards, WM

On Sunday, December 3, 2023 at 12:50:20 PM UTC-4, WM wrote:
> On 02.12.2023 18:34, William wrote:
> > On Saturday, December 2, 2023 at 12:52:38 PM UTC-4, WM wrote:
> >
> >> *If* B is defined by this algorithm, then it is completed by it.
> >
> > Nope. The term "completed" is meaningless.
> You are amazingly right!
WM: The term "completed" is meaningless
B is defined, Your putative A is not defined.

On 12/3/2023 8:54 AM, WM wrote:
> Chris M. Thomasson schrieb am Samstag, 2. Dezember 2023 um 20:30:55 UTC+1:
> > On 12/2/2023 9:34 AM, William wrote:
> > > On Saturday, December 2, 2023 at 12:52:38 PM UTC-4, WM wrote:
> > >
> > >> *If* B is defined by this algorithm, then it is completed by it.
> > >
> > > Nope. The term "completed" is meaningless.
> > An infinite process is never completed... However, we can say Cantor
> > Pairing can handle any natural number, therefore it works for all of
> > them. Fair enough?
>
> Only for naive minds. "Any natural number" belongs to a finite set
> whereas almost all natural numbers follow upon that set.

NOT almost all, __ALL__! Damn it, Man!

On 12/2/2023 9:00 AM, Fritz Feldhase wrote:
> On Saturday, December 2, 2023 at 4:02:18 PM UTC+1, WM wrote:
>
>> "The infinite sequence thus defined has
>> the peculiar property to contain the positive rational numbers
>> completely, and each of them only once at a determined place." [G.
>> Cantor, letter to R. Lipschitz (19 Nov 1883)]
>>
>> Note: completely.
>
> Hint: This means (in the context of the quote): ALL.
>
> So Cantor (essentially) states the following her:
>
> | "The infinite sequence thus DEFINED has
> | the peculiar property to contain ALL positive rational numbers
> | and each of them only once"
>
> Got it, you silly crank?

I think the mere thought of this gives WM GI issues... I feel sorry for
that toilet! Wow... ;^o

On 12/2/2023 11:25 AM, Chris M. Thomasson wrote:
> On 12/2/2023 2:12 AM, WM wrote:
>> William schrieb am Freitag, 1. Dezember 2023 um 13:27:28 UTC+1:
>>  > On Friday, December 1, 2023 at 8:16:58 AM UTC-4, WM wrote:
>>  > >...the fraction 1/2 is not in B but remains in
>>  > > whatever is defined of A.
>>  > There is no A. A(0), A(1) , A(2) ,,,, are not "whatever is defined
>> of A".
>>
>> Let (A(n)) be the sequence  A(0), A(1), A(2), ... . All existing
>> matrices are of same size. All elements of B are in every term A(n).
>> If B is completed by k = (m + n - 1)(m + n - 2)/2 + m, then also A is
>> completed.
>
> The mapping maps all naturals to a pair, and back again. Full circle.
> Nothing is lost. Why do you have trouble realizing that fact?
>

Actually, I should say:

The mapping maps all naturals to a _unique_ pair, and back again.

:^)

Chris M. Thomasson submitted this idea :
> On 12/3/2023 8:54 AM, WM wrote:
>> Chris M. Thomasson schrieb am Samstag, 2. Dezember 2023 um 20:30:55 UTC+1:
>> > On 12/2/2023 9:34 AM, William wrote:
>> > > On Saturday, December 2, 2023 at 12:52:38 PM UTC-4, WM wrote:
>> > >
>> > >> *If* B is defined by this algorithm, then it is completed by it.
>> > >
>> > > Nope. The term "completed" is meaningless.
>> > An infinite process is never completed... However, we can say Cantor
>> > Pairing can handle any natural number, therefore it works for all of
>> > them. Fair enough?
>>
>> Only for naive minds. "Any natural number" belongs to a finite set whereas
>> almost all natural numbers follow upon that set.
>
> NOT almost all, __ALL__! Damn it, Man!

Almost all. Any finite number of them excluded, are followed by the
rest, which is almost all.

On 12/3/2023 12:57 PM, FromTheRafters wrote:
> Chris M. Thomasson submitted this idea :
>> On 12/3/2023 8:54 AM, WM wrote:
>>> Chris M. Thomasson schrieb am Samstag, 2. Dezember 2023 um 20:30:55
>>> UTC+1:
>>>  > On 12/2/2023 9:34 AM, William wrote:
>>>  > > On Saturday, December 2, 2023 at 12:52:38 PM UTC-4, WM wrote:
>>>  > >
>>>  > >> *If* B is defined by this algorithm, then it is completed by it.
>>>  > >
>>>  > > Nope. The term "completed" is meaningless.
>>>  > An infinite process is never completed... However, we can say Cantor
>>>  > Pairing can handle any natural number, therefore it works for all of
>>>  > them. Fair enough?
>>>
>>> Only for naive minds. "Any natural number" belongs to a finite set
>>> whereas almost all natural numbers follow upon that set.
>>
>> NOT almost all, __ALL__! Damn it, Man!
>
> Almost all. Any finite number of them excluded, are followed by the
> rest, which is almost all.

Any natural can be mapped to a pair, then taken back again. Well, a
finite snapshot of an infinite process is finite in nature. Of course an
infinity is beyond it. Also, there are infinite snapshots of an infinite
process... ;^)

William schrieb am Sonntag, 3. Dezember 2023 um 18:32:29 UTC+1:
> On Sunday, December 3, 2023 at 12:50:20 PM UTC-4, WM wrote:
> > On 02.12.2023 18:34, William wrote:
> > > On Saturday, December 2, 2023 at 12:52:38 PM UTC-4, WM wrote:
> > >
> > >> *If* B is defined by this algorithm, then it is completed by it.
> > >
> > > Nope. The term "completed" is meaningless.
> > You are amazingly right!
> WM: The term "completed" is meaningless
> B is defined, Your putative A is not defined.

B is not defined because it is not a matrix but only a triangle. B is smaller than All matrices A(n).

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

Covering the matrix by X is impossible.

Regards, WM

On Monday, December 4, 2023 at 5:53:44 AM UTC-4, WM wrote:
> William schrieb am Sonntag, 3. Dezember 2023 um 18:32:29 UTC+1:
> > On Sunday, December 3, 2023 at 12:50:20 PM UTC-4, WM wrote:
> > > On 02.12.2023 18:34, William wrote:
> > > > On Saturday, December 2, 2023 at 12:52:38 PM UTC-4, WM wrote:
> > > >
> > > >> *If* B is defined by this algorithm, then it is completed by it.
> > > >
> > > > Nope. The term "completed" is meaningless.
> > > You are amazingly right!
> > WM: The term "completed" is meaningless
> > B is defined, Your putative A is not defined.
> B is not defined because it is not a matrix but only a triangle.

Nope. The value of B is defined at every (m,n) in |Nx|N.
The value of A(0) is defined at every (m.n) in |Nx|N.
B and A(0) have the same "size".
Your putative A is not defined.

On 04.12.2023 12:16, William wrote:
> On Monday, December 4, 2023 at 5:53:44 AM UTC-4, WM wrote:

>> B is not defined because it is not a matrix but only a triangle.
>
> Nope. The value of B is defined at every (m,n) in |Nx|N.
> The value of A(0) is defined at every (m.n) in |Nx|N.
> B and A(0) have the same "size".

In fact you are right. A(n) and B, both are potentially infinite triangles.
Note that you cannot find any diagonal (from lower left (m, 1) to upper
right (1, n)) that is shorter than its predecessor.

> Your putative A is not defined.

It is as defined as these potentially infinite triangles.

Regards, WM

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. 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 Monday, December 4, 2023 at 8:37:01 AM UTC-4, WM wrote:
> On 04.12.2023 12:16, William wrote:
> > On Monday, December 4, 2023 at 5:53:44 AM UTC-4, WM wrote:
>
> >> B is not defined because it is not a matrix but only a triangle.
> >
> > Nope. The value of B is defined at every (m,n) in |Nx|N.
> > The value of A(0) is defined at every (m.n) in |Nx|N.
> > B and A(0) have the same "size".
> In fact you are right.
WM: B and A(0) are the same size
So, as all the A(n) are the same size. your billiard game deals only with matrices that are the same size as B. The putative matrix A is not defined..

On 04.12.2023 15:21, William wrote:
> On Monday, December 4, 2023 at 8:37:01 AM UTC-4, WM wrote:
>> On 04.12.2023 12:16, William wrote:
>>> On Monday, December 4, 2023 at 5:53:44 AM UTC-4, WM wrote:
>>
>>>> B is not defined because it is not a matrix but only a triangle.
>>>
>>> Nope. The value of B is defined at every (m,n) in |Nx|N.
>>> The value of A(0) is defined at every (m.n) in |Nx|N.
>>> B and A(0) have the same "size".
>> In fact you are right.
> WM: B and A(0) are the same size
> So, as all the A(n) are the same size. your billiard game deals only with matrices that are the same size as B. The putative matrix A is not defined.

In fact there are no visible matrices at all. What we have discussed are
mainly-dark matrices. B is a tiny submatrix of A because in all terms of
the sequence A(n) there are almost all matrix positions not indexed.
This cannot suddenly change "in the limit". And if the limit A does not
exist, then the limit B does not exist either.

But the visible parts are only triangles. Find a diagonal (from lower
left to upper right) that is shorter than its predecessor. Fail. That is
because the temporary maximum m + n is constant.

Regards, WM

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")

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

> 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 11/26/2023 6:38 AM, WM wrote:
> On 26.11.2023 11:20, FromTheRafters wrote:
>> WM brought next idea :
>>> On 25.11.2023 23:41, Chris M. Thomasson wrote:
>>>> On 11/25/2023 5:16 AM, Transfinity wrote:
>>>
>>>>> But B is produced by the algorithm described in the OP, isn't it?
>>>>> And in the course of this agorithm the fraction 1/2 sits at a
>>>>> position which is not belonging to B (but to A(0).
>>>>
>>>> Your mind seems to be stuck on a step-by-step process, instead of
>>>> saying all...
>>>
>>> All is said in the OP. Is there any natural number that is not at a
>>> position determined by Cantor's algorithm
>>> k = (m + n - 1)(m + n - 2)/2 + m ?
>>
>> Does Cantor's algorithm determine positions of natural numbers?
>
> Of course. Insert values for co-ordinates m and n, get number k of that
> position.

We can rather easily map the pairs to (x, y) in order to plot them on a
2-ary plane...

On 12/4/2023 8:12 AM, WM wrote:
> On 04.12.2023 15:21, William wrote:
>> On Monday, December 4, 2023 at 8:37:01 AM UTC-4, WM wrote:
>>> On 04.12.2023 12:16, William wrote:
>>>> On Monday, December 4, 2023 at 5:53:44 AM UTC-4, WM wrote:
>>>
>>>>> B is not defined because it is not a matrix but only a triangle.
>>>>
>>>> Nope. The value of B is defined at every (m,n) in |Nx|N.
>>>> The value of A(0) is defined at every (m.n) in |Nx|N.
>>>> B and A(0) have the same "size".
>>> In fact you are right.
>> WM:   B and A(0) are the same size
>> So, as all the A(n) are the same size. your billiard game deals only
>> with matrices that are the same size as B.   The putative matrix A is
>> not defined.
>
> In fact there are no visible matrices at all. What we have discussed are
> mainly-dark matrices.

LOL! Only dark to the damaged parts of your brain?

B is a tiny submatrix of A because in all terms of
> the sequence A(n) there are almost all matrix positions not indexed.
> This cannot suddenly change "in the limit". And if the limit A does not
> exist, then the limit B does not exist either.
>
> But the visible parts are only triangles. Find a diagonal (from lower
> left to upper right) that is shorter than its predecessor. Fail. That is
> because the temporary maximum m + n is constant.
>
> Regards, WM
>