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 ?
No. The complete sequence is given just as Cantor gives it: 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, ...

Your argument is applied by matheologians only in order to veil the
contradictions resulting from their belief.

Regards, WM

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?

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.

Regards, WM

WM formulated on Sunday :
> 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.

Oh, it's invertible, kind of like a bijection. :)

On Saturday, November 25, 2023 at 4:53:08 PM UTC-4, Transfinity wrote:
> William schrieb am Samstag, 25. November 2023 um 17:40:52 UTC+1:
> > On Saturday, November 25, 2023 at 11:43:50 AM UTC-4, Transfinity wrote:
> > > William schrieb am Samstag, 25. November 2023 um 14:59:53 UTC+1:
> > > > On Saturday, November 25, 2023 at 9:16:08 AM UTC-4, Transfinity wrote:
> > > > > William schrieb am Samstag, 25. November 2023 um 12:50:01 UTC+1:
> > > > > > On Saturday, November 25, 2023 at 6:38:37 AM UTC-4, Transfinity wrote:
> > > > > >
> > > > > >
> > > > > > > That part of B which "exists",
> > > > > > is B
> > > > > > > exists as a submatrix of A.
> > > > > > Your putative matrix A does not exist.
> > > > > But B
> > > > unlike your putative A, exists.
> > > created by the algorithm given in the OP
> > Though you do not do so an algorithm can be defined.
> 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(0) given by the simple algorithm (m,n) -> m/n. Note that B, unlike your putative A. exists and that B is not a submatrix of A(n) for any n
>
> (3) [my billiards game] tells that all A(n) contain all fractions at not
> indexed positions.

Each of these positions is eventually "indexed" at a step
Let p(f) be the position of "fraction" f in A(n) There is an m(p) such that p is "indexed" in A(m(p))

William schrieb am Sonntag, 26. November 2023 um 19:12:27 UTC+1:
> On Saturday, November 25, 2023 at 4:53:08 PM UTC-4, Transfinity wrote:

> > 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(0) given by the simple algorithm (m,n) -> m/n. Note that B, unlike your putative A. exists and that B is not a submatrix of A(n) for any n

Note that all parts of B that exist are submatrices of A(0) and its followers which exist.
> >
> > (3) [my billiards game] tells that all A(n) contain all fractions at not
> > indexed positions.
>
> Each of these positions is eventually "indexed" at a step

Each visible position, i.e., a small minority.

> Let p(f) be the position of "fraction" f in A(n) There is an m(p) such that p is "indexed" in A(m(p))

Yes, every visible position becomes indexed. But the fractions remain in the matrix and at not indexed positions.

Regards, WM

On Sunday, November 26, 2023 at 4:01:57 PM UTC-4, Transfinity wrote:
> William schrieb am Sonntag, 26. November 2023 um 19:12:27 UTC+1:
> > On Saturday, November 25, 2023 at 4:53:08 PM UTC-4, Transfinity wrote:
>
> > > 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(0) given by the simple algorithm (m,n) -> m/n. Note that B, unlike your putative A. exists and that B is not a submatrix of A(n) for any n
> Note that all parts of B that exist are submatrices of A(0)

Nope, B exists and is not a submatrix of A(n) for any n.

On 11/26/2023 7:30 AM, FromTheRafters wrote:
> WM formulated on Sunday :
>> 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.
>
> Oh, it's invertible, kind of like a bijection. :)

:^D

On 26.11.2023 21:12, William wrote:
> On Sunday, November 26, 2023 at 4:01:57 PM UTC-4, Transfinity wrote:
>> William schrieb am Sonntag, 26. November 2023 um 19:12:27 UTC+1:
>>> On Saturday, November 25, 2023 at 4:53:08 PM UTC-4, Transfinity wrote:
>>>> 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(0) given by the simple algorithm (m,n) -> m/n. Note that B, unlike your putative A. exists and that B is not a submatrix of A(n) for any n
>> Note that all parts of B that exist are submatrices of A(0)
> Nope, B exists and is not a submatrix of A(n) for any n.

If B exists in the course of Cantor's algorithm, then A exists too (by
the same algorithm), because B can only be completed if its supematrix A
is completed.

Regards, WM

On Sunday, November 26, 2023 at 6:31:14 PM UTC-4, WM wrote:
> On 26.11.2023 21:12, William wrote:
> > On Sunday, November 26, 2023 at 4:01:57 PM UTC-4, Transfinity wrote:
> >> William schrieb am Sonntag, 26. November 2023 um 19:12:27 UTC+1:
> >>> On Saturday, November 25, 2023 at 4:53:08 PM UTC-4, Transfinity wrote:
> >>>> 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(0) given by the simple algorithm (m,n) -> m/n. Note that B, unlike your putative A. exists and that B is not a submatrix of A(n) for any n
> >> Note that all parts of B that exist are submatrices of A(0)
> > Nope, B exists and is not a submatrix of A(n) for any n.
> If B exists in the course of Cantor's algorithm, then A exists too

Nope B does not "exist in the course" of some algorithm. B exists if an algorithm exists. As you note, there is an algorithm for B, there is no algorithm for the putative A.

On 26.11.2023 16:30, FromTheRafters wrote:
> WM formulated on Sunday :

>>>> 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.
>
> Oh, it's invertible, kind of like a bijection.

It is a one-to-one mapping. It is bijective between the definable
numbers. It is not complete.

Look: If the matrix B
1, 2, 4, ...
3, 5, 8, ...
6, 9,13, ...
....
is produced by attaching natural indices from the first column of matrix A
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, ...
....
to the positions (m, n) of matrix A(0), then B can only be completed
after all indices have found their final positions in the course of the
sequence A(1), A(2), A(3), ... --> A. But then A can no longer change.
Hence it is completed simultaneously. And it contains all fractions at
not indexed positions.

Regards, WM

William schrieb am Montag, 27. November 2023 um 00:55:19 UTC+1:
> On Sunday, November 26, 2023 at 6:31:14 PM UTC-4, WM wrote:
> > On 26.11.2023 21:12, William wrote:

> > If B exists in the course of Cantor's algorithm, then A exists too
> Nope B does not "exist in the course" of some algorithm. B exists if
an algorithm exists.

An algorithm exists in the OP.

> As you note, there is an algorithm for B, there is no algorithm for
the putative A.

Wrong. If Cantor's algorithm k = (m + n - 1)(m + n - 2)/2 + m, applied
in the OP, makes B existing, then it makes all natural numbers from the
first column of matrix A(0)
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, ...
....
settle at positions called B
1, 2, 4, ...
3, 5, 8, ...
6, 9,13, ...
....
If this happens, then B is defined. But then also A is defined because
the sequence A(1), A(2), ... has ceased to go on but has found its final
form A.

It is impossible to complete B, i.e., to describe every position,
without completing B as a submatrix of A which afterwards remains unchanged.

Note that there is no other algorithm to complete B than that used in
the game. Either A and B exist or none of them.

Regards, WM

Fritz Feldhase schrieb am Sonntag, 26. November 2023 um 23:53:05 UTC+1:
> On Sunday, November 26, 2023 at 11:31:14 PM UTC+1, WM wrote:
>
> > B can only be completed if
> There's no need to "complete" B since it already has been *defined*,

If it is defined, then every position is defined by Cantor's formula,
i.e., by the algorithm outlined in the OP. Then also the border between
its positions and the positions occupied by fractions would be defined
too. That is not the case. Therefore B is not defined.

Regards, WM

WM wrote on 11/27/2023 :
> On 26.11.2023 16:30, FromTheRafters wrote:
>> WM formulated on Sunday :
>
>>>>> 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.
>>
>> Oh, it's invertible, kind of like a bijection.
>
> It is a one-to-one mapping. It is bijective between the definable numbers.

Sheeesh, you and your 'definable numbers' again. Both sets having been
already defined, this is just a pairing demonstrating the size of the
sets.

On Monday, November 27, 2023 at 4:43:11 AM UTC-4, WM wrote:
..
> If this happens, then B is defined

After step n(p) the , "index" of p, the value at position p will "settle down".
However there is no step which will work for all p. B is not defined by something that "settles down"

>But then also A is defined

Nope. As you note, there is no algorithm for the putative matrix A.

On 27.11.2023 13:24, FromTheRafters wrote:
> WM wrote on 11/27/2023 :
>> On 26.11.2023 16:30, FromTheRafters wrote:
>>> WM formulated on Sunday :
>>
>>>>>> 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.
>>>
>>> Oh, it's invertible, kind of like a bijection.
>>
>> It is a one-to-one mapping. It is bijective between the definable
>> numbers.
>
> Sheeesh, you and your 'definable numbers' again. Both sets having been
> already defined,

Yes, the sets have been defined, but not all their numbers.

> this is just a pairing demonstrating the size of the sets.

It demonstrates that the pairing covers only potentially infinite
sequences. The actual set ℚ has |ℚ| = 2|ℕ|^2 + 1 elements. That is as
obvious as the fact that the X in
XOOO...
XOOO...
XOOO...
XOOO...
....
will never cover all positions of the matrix.
It is as obvious as the fact that the intersection of infinite
endsegments cannot be empty.

Regards, WM

Fritz Feldhase schrieb am Montag, 27. November 2023 um 12:35:12 UTC+1:
> On Monday, November 27, 2023 at 9:54:32 AM UTC+1, WM wrote:
> > Fritz Feldhase schrieb am Sonntag, 26. November 2023 um 23:53:05
UTC+1:
> > > On Sunday, November 26, 2023 at 11:31:14 PM UTC+1, WM wrote:
> > > >
> > > > B can only be completed if
> > > >
> > > There's no need to "complete" B since it already has been *defined*,
> > >
> > If it is defined
> Nope, the definition is not "conditional". There is no "if".
> Hint: B = (b_n,m) is the INxIN matrix defined with b_n,m = (m + n -
1)(m + n - 2)/2 + m for all n,m e IN.

You are wrong. By applying this very definition in the OP we see that B
is a submatrix of A. The border between B and A however is not defined.
So B is not completely defined.

Regards, WM

William schrieb am Montag, 27. November 2023 um 13:43:42 UTC+1:
> On Monday, November 27, 2023 at 4:43:11 AM UTC-4, WM wrote:
> .
> > If this happens, then B is defined
> After step n(p) the , "index" of p, the value at position p will
"settle down".

And the fraction from there will settle down at a position outside of B
but inside of A.

> However there is no step which will work for all p. B is not defined
by something that "settles down"

For all of B that is defined, the step when k has settled at position
(m, n) is known. Parts of B where this is not possible, are not defined.

> >But then also A is defined
> Nope. As you note, there is no algorithm for the putative matrix A.

There is no algorithm for the dark parts of B either.

Note this condition which is indispensable in mathematics:
********
For every position (m, n) of B that is defined, the step A(n) when k has
settled there is known.
********

If you disagree without proving the contrary, then you are outside of
mathematics, and further discussion is void. In order to prove the
contrary, name a defined positions that is indexed at a definable step A(n).

Regards, WM

On 27.11.2023 19:20, Fritz Feldhase wrote:

> Def.: B = (b_n,m) is the INxIN matrix defined with b_n,m = (m + n - 1)(m + n - 2)/2 + m for all n,m e IN.
>
> Hint: This ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ is the definition for B.

Please name a position (b_n,m) which is not occupied by an index k at a
step n of the sequence A(n).

Note this condition which is indispensable in mathematics:
********
For every position (m, n) of B that is defined, the step A(n) when k has
settled there is known.
********

If you disagree without proving the contrary, then you are outside of
mathematics, and further discussion is void. In order to prove the
contrary, name a defined position that is not indexed at a definable
step A(n).

Regards, WM

Fritz Feldhase schrieb am Montag, 27. November 2023 um 19:41:48 UTC+1:
> On Monday, November 27, 2023 at 7:36:04 PM UTC+1, WM wrote:
> > On 27.11.2023 19:20, Fritz Feldhase wrote:
> > >
> > > Def.: B = (b_n,m)_(n,m e IN) is the matrix defined with b_n,m =
(m + n - 1)(m + n - 2)/2 + m for all n,m e IN.
> > >
> > > Hint: This ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ is the definition for B.

In the same way it defines all elements of A because in the OP the
matrices A and B are connected by simple exchanges.

> We were talking about the matrix (function) B and its definition. See
above.

In the same way A is defined. But note this condition which is
indispensable in mathematics:
********
For every position (m, n) of B that is defined, the step A(n) when k has
settled there is known.
********

If you disagree without proving the contrary, then you are outside of
mathematics, and further discussion is void. In order to prove the
contrary, name a defined position that is not indexed at a definable
step A(n).

Regards, WM

On Monday, November 27, 2023 at 2:30:49 PM UTC-4, WM wrote:
> William schrieb am Montag, 27. November 2023 um 13:43:42 UTC+1:
> > On Monday, November 27, 2023 at 4:43:11 AM UTC-4, WM wrote:
> > .
> > > If this happens, then B is defined
> > After step n(p) the , "index" of p, the value at position p will
> "settle down".
> And the fraction from there will settle down at a position outside of B

Nope. Any time a "fraction" moves it moves to a position inside of B. However, there is no last move of a "fraction" so it does not "settle down".

> but inside of A.

Your putative A does not exist.

William schrieb am Montag, 27. November 2023 um 20:30:25 UTC+1:
> On Monday, November 27, 2023 at 2:30:49 PM UTC-4, WM wrote:
> > William schrieb am Montag, 27. November 2023 um 13:43:42 UTC+1:
> > > On Monday, November 27, 2023 at 4:43:11 AM UTC-4, WM wrote:
> > > .
> > > > If this happens, then B is defined
> > > After step n(p) the , "index" of p, the value at position p will
> > "settle down".
> > And the fraction from there will settle down at a position outside of B
> Nope. Any time a "fraction" moves it moves to a position inside of B.
However, there is no last move of a "fraction" so it does not "settle down".

There is a last move of every number belonging to B. In the same step
A(n) a fraction is removed but remains inside.
>
> > but inside of A.
>
> Your putative A does not exist.

If A does not exist, then B does not exist.

Note this condition which is indispensable in mathematics:
********
For every position (m, n) of B that is defined, the step A(n) when k has
settled there is known.
********

If you disagree without proving the contrary, then you are outside of
mathematics, and further discussion is void. In order to prove the
contrary, name a defined positions that is ***not*** indexed at a
definable step A(n).

Regards, WM

On Tuesday, November 28, 2023 at 6:36:12 AM UTC-4, WM wrote:
> William schrieb am Montag, 27. November 2023 um 20:30:25 UTC+1:
> > On Monday, November 27, 2023 at 2:30:49 PM UTC-4, WM wrote:
> > > William schrieb am Montag, 27. November 2023 um 13:43:42 UTC+1:
> > > > On Monday, November 27, 2023 at 4:43:11 AM UTC-4, WM wrote:
> > > > .
> > > > > If this happens, then B is defined
> > > > After step n(p) the , "index" of p, the value at position p will
> > > "settle down".
> > > And the fraction from there will settle down at a position outside of B
> > Nope. Any time a "fraction" moves it moves to a position inside of B.
> > However, there is no last move of a "fraction" so it does not "settle down".
> There is a last move of every number belonging to B.

Correct. This is one way of defining B. However no "fraction" belongs to B.

> > Your putative A does not exist.
> If A does not exist, then B does not exist.

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

On Tuesday, November 28, 2023 at 11:36:12 AM UTC+1, WM wrote:
> William schrieb am Montag, 27. November 2023 um 20:30:25 UTC+1:
> > On Monday, November 27, 2023 at 2:30:49 PM UTC-4, WM wrote:
> > > William schrieb am Montag, 27. November 2023 um 13:43:42 UTC+1:
> > > > On Monday, November 27, 2023 at 4:43:11 AM UTC-4, WM wrote:
> > > > .
> > > > > If this happens, then B is defined
> > > > After step n(p) the , "index" of p, the value at position p will
> > > "settle down".
> > > And the fraction from there will settle down at a position outside of B
> > Nope. Any time a "fraction" moves it moves to a position inside of B.
> However, there is no last move of a "fraction" so it does not "settle down".
> There is a last move of every number belonging to B. In the same step
> A(n) a fraction is removed but remains inside.
> >
> > > but inside of A.
> >
> > Your putative A does not exist.
> If A does not exist, then B does not exist.
>
> Note this condition which is indispensable in mathematics:
> ********
> For every position (m, n) of B that is defined, the step A(n) when k has
> settled there is known.
> ********
>
> If you disagree without proving the contrary, then you are outside of
> mathematics, and further discussion is void. In order to prove the
> contrary, name a defined positions that is ***not*** indexed at a
> definable step A(n).
>
> Regards, WM

Akka bakka bonka rakka
etla metla sjong dong
fili fong
issa bissa topp
-- from a collection of children's rhymes published by Rikka Deinboll

William schrieb am Dienstag, 28. November 2023 um 14:54:24 UTC+1:
> On Tuesday, November 28, 2023 at 6:36:12 AM UTC-4, WM wrote:

> > If A does not exist, then B does not exist.
> Nope. B is defined. Your putative A is not defined

Both are defined simultaneously in the OP. Find any point of B that is
defined in your dreamworld but not defined according to the game
described in the OP.

Regards, WM