On 5/23/2022 2:01 PM, Jim Burns wrote:
> On 5/23/2022 8:20 AM, WM wrote:
>> Jim Burns schrieb
>> am Sonntag, 22. Mai 2022 um 18:45:38 UTC+2:

>>>> For that assumption you have to claim that the
>>>> sequence +1, -1, +1, -1, +1, -1, ... will acumulate
>>>> infinitely many units. I will not join you. Hardly
>>>> will any mathematician unless he is told that
>>>> otherwise set theory would be wrong.
>>>
>>> If,
>>> for each BEFORE and AFTER =< m
>>
>> There is no relevance of BEFORE and AFTER.
>
> For each entry in sequence 1,2,3,4...
> for each split BEFORE and AFTER =< that entry,
> some j ends BEFORE and j+1 ends AFTER.

some j ends BEFORE and j+1 begins AFTER.

> That describes each of the entries, rows, and columns
> of a matrix in which the first column of X's
> can match the whole matrix of X's and O's
>
> X:X       O:X       O:X       O:X       ...
> X:X       O:X       O:X       O:X       ...
> X:X       O:X       O:X       O:X       ...
> X:X       O:X       O:X       O:X       ...
> X:X       O:X       O:X       O:X       ...
> X:X       O:X       O:X       O:X       ...
> X:X       O:X       O:X       O:X       ...
> X:X       O:X       O:X       O:X       ...
> X:X       O:X       O:X       O:X       ...
> ...       ...       ...       ...
>
> This is how they match:
>
> 1/1:1/1   1/2:2/1   1/3:4/1   1/4:7/1   ...
> 2/1:3/1   2/2:5/1   2/3:8/1   2/4:12/1  ...
> 3/1:6/1   3/2:9/1   3/3:13/1  3/4:18/1  ...
> 4/1:10/1  4/2:14/1  4/3:19/1  4/4:25/1  ...
> 5/1:15/1  5/2:20/1  5/3:26/1  5/4:33/1  ...
> 6/1:21/1  6/2:27/1  6/3:34/1  6/4:42/1  ...
> 7/1:28/1  7/2:35/1  7/3:43/1  7/4:52/1  ...
> 8/1:36/1  8/2:44/1  8/3:53/1  8/4:63/1  ...
> 9/1:45/1  9/2:54/1  9/3:64/1  9/4:75/1  ...
> ...       ...       ...       ...
>
> An entry is m/n:k/1 such that
> k = (m+n-1)*(m+n-2)/2+m
>
> For each m and n in 1,2,3,4,...
> k is in 1,2,3,4,...
>
> This is a consequence of the description I gave of
> being in 1,2,3,4,...
>
> For each entry in sequence 1,2,3,4...
> for each split BEFORE and AFTER =< that entry,
> some j ends BEFORE and j+1 ends AFTER.
>
>> There is relevance only of the fact that
>> every indexed fraction can get an index
>> only from a previously indexd fraction.
>
> Each indexed fraction gets its index from
> the first column.
>
>> And previously most fractions were not indexed.
>
> Each fraction is indexed.
>
> Not all collections have Bob-conservation.
>

måndag 23 maj 2022 kl. 14:13:51 UTC+2 skrev WM:
> Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 17:12:21 UTC+2:
> > On Sunday, May 22, 2022 at 3:32:46 PM UTC+2, WM wrote:
> That means, there is a [distribution] such that the X of the first column
> > >
> > > XOOOO...
> > > XOOOO...
> > > XOOOO...
> > > XOOOO...
> > > XOOOO...
> > > ...
> > >
> > > [...] cover all matrix positions.
> >
> > Indeed!
> >
> > There's a simple distribution of the X to achieve this goal:
> >
> > > 1/1, 1/2, 1/4, 1/7, ...
> > > 1/3, 1/5, 1/8, ...
> > > 1/6, 1/9, ...
> > > 1/10, ...
> > > ...
> >
> Please explain why the sequence -1 + 1 - 1 + 1 - 1 + 1 -+ has sum oo or write down the complete matrix in order to prove your claim.
>
> Regards, WM
it has no sum, it is DIVERGENT

måndag 23 maj 2022 kl. 14:26:04 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Montag, 23. Mai 2022 um 07:06:09 UTC+2:
> > torsdag 19 maj 2022 kl. 15:44:32 UTC+2 skrev WM:
>
> > > > k is subject to induction *because*
> > > > k is finitely-many '+1' from 0.
> > > That does not exclude aleph_0 dark numbers immediately before omega.
> > All of which you fail to define in any meaningful way.
> All natnumbers which you can define have an infinite distance from omega. This distance is existing but undefinable.
>
> Regards, WM

this says nothing as that includes all natural numbers, none of your "dark"

FromTheRafters schrieb am Montag, 23. Mai 2022 um 14:29:49 UTC+2:
> WM was thinking very hard :
> > Jim Burns schrieb am Sonntag, 22. Mai 2022 um 20:49:56 UTC+2:
> >> On 5/22/2022 9:32 AM, WM wrote:
> >
> >>> That means, there is a permutation such that
> >>> the X of the first column
> >>> XOOOO...
> >>> XOOOO...
> >>> XOOOO...
> >>> XOOOO...
> >>> XOOOO...
> >>> ...
> >>>
> >>> by being exchanged with the O's cover all
> >>> matrix positions.
> >> Yes.
> >
> > By exchanging X and O the shares will remain constant.
> > I am not willing to discuss with stubborn stupidity.
> Decades prove otherwise.

But not as obvious as the denial to accept that enumerating all fractions in my model by exchange is tantamount to getting an infinite sum from the series -1 + 1 - 1 + 1 -+ ... .

Regards, WM

Fritz Feldhase schrieb am Montag, 23. Mai 2022 um 16:57:16 UTC+2:
> On Monday, May 23, 2022 at 2:26:04 PM UTC+2, WM wrote:
> > zelos...@gmail.com schrieb am Montag, 23. Mai 2022 um 07:06:09 UTC+2:
> >
> > All natnumbers [...] have an infinite distance from omega. This distance is existing but undefinable.
>
> Oh, it is "undefinable" but you are still talking about "it"? (Since "it" is existing - somehow?)

It is dark. The numbers there cannot be removed individually.
>
> Fascinating - not!

Everybody knows the triviality
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|ℕ \ {1, 2, 3, ...}| = 0
but only few understand its reason.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 24. Mai 2022 um 07:21:24 UTC+2:
> måndag 23 maj 2022 kl. 14:13:51 UTC+2 skrev WM:
> > Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 17:12:21 UTC+2:
> > > On Sunday, May 22, 2022 at 3:32:46 PM UTC+2, WM wrote:
> > That means, there is a [distribution] such that the X of the first column
> > > >
> > > > XOOOO...
> > > > XOOOO...
> > > > XOOOO...
> > > > XOOOO...
> > > > XOOOO...
> > > > ...

> > Please explain why the sequence -1 + 1 - 1 + 1 - 1 + 1 -+ has sum oo or write down the complete matrix in order to prove your claim.
> >
> it has no sum, it is DIVERGENT

Therefore the X cannot cover all places of the matrix by simply exchanging them with O's.

Regards, WM

tisdag 24 maj 2022 kl. 12:51:29 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 24. Mai 2022 um 07:21:24 UTC+2:
> > måndag 23 maj 2022 kl. 14:13:51 UTC+2 skrev WM:
> > > Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 17:12:21 UTC+2:
> > > > On Sunday, May 22, 2022 at 3:32:46 PM UTC+2, WM wrote:
> > > That means, there is a [distribution] such that the X of the first column
> > > > >
> > > > > XOOOO...
> > > > > XOOOO...
> > > > > XOOOO...
> > > > > XOOOO...
> > > > > XOOOO...
> > > > > ...
> > > Please explain why the sequence -1 + 1 - 1 + 1 - 1 + 1 -+ has sum oo or write down the complete matrix in order to prove your claim.
> > >
> > it has no sum, it is DIVERGENT
> Therefore the X cannot cover all places of the matrix by simply exchanging them with O's.
>
> Regards, WM

The sum is irrelevant to your claim there and that claim in turn is irrelevant to the function N-> Q+

tisdag 24 maj 2022 kl. 12:47:51 UTC+2 skrev WM:
> Fritz Feldhase schrieb am Montag, 23. Mai 2022 um 16:57:16 UTC+2:
> > On Monday, May 23, 2022 at 2:26:04 PM UTC+2, WM wrote:
> > > zelos...@gmail.com schrieb am Montag, 23. Mai 2022 um 07:06:09 UTC+2:
> > >
> > > All natnumbers [...] have an infinite distance from omega. This distance is existing but undefinable.
> >
> > Oh, it is "undefinable" but you are still talking about "it"? (Since "it" is existing - somehow?)
> It is dark. The numbers there cannot be removed individually.

That odesn't mean anything! Definite in First Order Logic already! Everything else can in mathematics!

> >
> > Fascinating - not!
>
> Everybody knows the triviality
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> |ℕ \ {1, 2, 3, ...}| = 0
> but only few understand its reason.

No, everyone knows the reasons.
1: your N_def=N
2: Removing finite number of elements from an infinite set changes no cardinality
3: Removing all of a set laeves you the empty set

>
> Regards, WM

zelos...@gmail.com schrieb am Dienstag, 24. Mai 2022 um 13:02:28 UTC+2:
> 2: Removing finite number of elements from an infinite set changes no cardinality
> 3: Removing all of a set laeves you the empty set

But it is imposible to remove all individually.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 24. Mai 2022 um 13:00:29 UTC+2:
> tisdag 24 maj 2022 kl. 12:51:29 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Dienstag, 24. Mai 2022 um 07:21:24 UTC+2:
> > > måndag 23 maj 2022 kl. 14:13:51 UTC+2 skrev WM:
> > > > Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 17:12:21 UTC+2:
> > > > > On Sunday, May 22, 2022 at 3:32:46 PM UTC+2, WM wrote:
> > > > That means, there is a [distribution] such that the X of the first column
> > > > > >
> > > > > > XOOOO...
> > > > > > XOOOO...
> > > > > > XOOOO...
> > > > > > XOOOO...
> > > > > > XOOOO...
> > > > > > ...
> > > > Please explain why the sequence -1 + 1 - 1 + 1 - 1 + 1 -+ has sum oo or write down the complete matrix in order to prove your claim.
> > > >
> > > it has no sum, it is DIVERGENT
> > Therefore the X cannot cover all places of the matrix by simply exchanging them with O's.
> >
> The sum is irrelevant to your claim there and that claim in turn is irrelevant to the function N-> Q+

I am not willing to discuss with stubborn stupidity.
Come back when you will have recovered.
EOD.

Regards, WM

On 5/24/2022 5:51 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 24. Mai 2022 um 07:21:24 UTC+2:
>> måndag 23 maj 2022 kl. 14:13:51 UTC+2 skrev WM:
>>> Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 17:12:21 UTC+2:
>>>> On Sunday, May 22, 2022 at 3:32:46 PM UTC+2, WM wrote:
>>> That means, there is a [distribution] such that the X of the first column
>>>>>
>>>>> XOOOO...
>>>>> XOOOO...
>>>>> XOOOO...
>>>>> XOOOO...
>>>>> XOOOO...
>>>>> ...
>
>>> Please explain why the sequence -1 + 1 - 1 + 1 - 1 + 1 -+ has sum oo or write down the complete matrix in order to prove your claim.
>>>
>> it has no sum, it is DIVERGENT
>
> Therefore the X cannot cover all places of the matrix by simply exchanging them with O's.
>
> Regards, WM

In the ongoing constantly changing story of WM and his Darkies,

now its

"-1 + 1 - 1 + 1 - 1 + 1"

now discarded WM "theories"

1. XOXOXOXOXOXOOOOXXXX

2 framed Achilles and the turtle

3 infinite sets of endsegments

4. "interval"

5. "next to"

6. touched, touching

7 omega as a natural number

8 distance

9 misquoted Cantor

10. Cantor countability

Note: Classic projection by WM : "I am not willing to discuss with stubborn stupidity".

On 5/24/2022 5:45 AM, WM wrote:
> FromTheRafters schrieb am Montag, 23. Mai 2022 um 14:29:49 UTC+2:
>> WM was thinking very hard :
>>> Jim Burns schrieb am Sonntag, 22. Mai 2022 um 20:49:56 UTC+2:
>>>> On 5/22/2022 9:32 AM, WM wrote:
>>>
>>>>> That means, there is a permutation such that
>>>>> the X of the first column
>>>>> XOOOO...
>>>>> XOOOO...
>>>>> XOOOO...
>>>>> XOOOO...
>>>>> XOOOO...
>>>>> ...
>>>>>
>>>>> by being exchanged with the O's cover all
>>>>> matrix positions.
>>>> Yes.
>>>
>>> By exchanging X and O the shares will remain constant.
>>> I am not willing to discuss with stubborn stupidity.
>> Decades prove otherwise.
>
> But not as obvious as the denial to accept that enumerating all fractions in my model by exchange is tantamount to getting an infinite sum from the series -1 + 1 - 1 + 1 -+ ... .
>

"tantamount" = "guessing"

your model is full of mistakes

try writing it down in MATH as a process, can you do that ?

> Regards, WM

On 5/24/2022 5:47 AM, WM wrote:
> Fritz Feldhase schrieb am Montag, 23. Mai 2022 um 16:57:16 UTC+2:
>> On Monday, May 23, 2022 at 2:26:04 PM UTC+2, WM wrote:
>>> zelos...@gmail.com schrieb am Montag, 23. Mai 2022 um 07:06:09 UTC+2:
>>>
>>> All natnumbers [...] have an infinite distance from omega. This distance is existing but undefinable.
>>
>> Oh, it is "undefinable" but you are still talking about "it"? (Since "it" is existing - somehow?)
>
> It is dark. The numbers there cannot be removed individually.
>>
>> Fascinating - not!
>
> Everybody knows the triviality
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> |ℕ \ {1, 2, 3, ...}| = 0
> but only few understand its reason.
>

red herring. your knowledge of math is finite, and grows smaller by the day, loss of elements...

> Regards, WM

On 5/24/2022 6:31 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 24. Mai 2022 um 13:02:28 UTC+2:
>
>> 2: Removing finite number of elements from an infinite set changes no cardinality
>> 3: Removing all of a set laeves you the empty set
>
> But it is imposible to remove all individually.

WM, sets are not variables.

>
> Regards, WM

On Monday, May 23, 2022 at 12:30:27 PM UTC-7, Jim Burns wrote:
> On 5/23/2022 2:01 PM, Jim Burns wrote:
> > On 5/23/2022 8:20 AM, WM wrote:
> >> Jim Burns schrieb
> >> am Sonntag, 22. Mai 2022 um 18:45:38 UTC+2:
>
> >>>> For that assumption you have to claim that the
> >>>> sequence +1, -1, +1, -1, +1, -1, ... will acumulate
> >>>> infinitely many units. I will not join you. Hardly
> >>>> will any mathematician unless he is told that
> >>>> otherwise set theory would be wrong.
> >>>
> >>> If,
> >>> for each BEFORE and AFTER =< m
> >>
> >> There is no relevance of BEFORE and AFTER.
> >
> > For each entry in sequence 1,2,3,4...
> > for each split BEFORE and AFTER =< that entry,
> > some j ends BEFORE and j+1 ends AFTER.
>
> some j ends BEFORE and j+1 begins AFTER.
>
> > That describes each of the entries, rows, and columns
> > of a matrix in which the first column of X's
> > can match the whole matrix of X's and O's
> >
> > X:X O:X O:X O:X ...
> > X:X O:X O:X O:X ...
> > X:X O:X O:X O:X ...
> > X:X O:X O:X O:X ...
> > X:X O:X O:X O:X ...
> > X:X O:X O:X O:X ...
> > X:X O:X O:X O:X ...
> > X:X O:X O:X O:X ...
> > X:X O:X O:X O:X ...
> > ... ... ... ...
> >
> > This is how they match:
> >
> > 1/1:1/1 1/2:2/1 1/3:4/1 1/4:7/1 ...
> > 2/1:3/1 2/2:5/1 2/3:8/1 2/4:12/1 ...
> > 3/1:6/1 3/2:9/1 3/3:13/1 3/4:18/1 ...
> > 4/1:10/1 4/2:14/1 4/3:19/1 4/4:25/1 ...
> > 5/1:15/1 5/2:20/1 5/3:26/1 5/4:33/1 ...
> > 6/1:21/1 6/2:27/1 6/3:34/1 6/4:42/1 ...
> > 7/1:28/1 7/2:35/1 7/3:43/1 7/4:52/1 ...
> > 8/1:36/1 8/2:44/1 8/3:53/1 8/4:63/1 ...
> > 9/1:45/1 9/2:54/1 9/3:64/1 9/4:75/1 ...
> > ... ... ... ...
> >
> > An entry is m/n:k/1 such that
> > k = (m+n-1)*(m+n-2)/2+m
> >
> > For each m and n in 1,2,3,4,...
> > k is in 1,2,3,4,...
> >
> > This is a consequence of the description I gave of
> > being in 1,2,3,4,...
> >
> > For each entry in sequence 1,2,3,4...
> > for each split BEFORE and AFTER =< that entry,
> > some j ends BEFORE and j+1 ends AFTER.
> >
> >> There is relevance only of the fact that
> >> every indexed fraction can get an index
> >> only from a previously indexd fraction.
> >
> > Each indexed fraction gets its index from
> > the first column.
> >
> >> And previously most fractions were not indexed.
> >
> > Each fraction is indexed.
> >
> > Not all collections have Bob-conservation.
> >

Thanks JB: nice of you to explain what it _is_.

On Monday, May 23, 2022 at 2:13:51 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 17:12:21 UTC+2:
> > On Sunday, May 22, 2022 at 3:32:46 PM UTC+2, WM wrote:
> > >
> > > That means, there is a [distribution] such that the X of the first column
> > >
> > > XOOOO...
> > > XOOOO...
> > > XOOOO...
> > > XOOOO...
> > > XOOOO...
> > > ...
> > >
> > > [...] cover all matrix positions.
> >
> > Indeed!
> >
> > There's a simple distribution of the X to achieve this goal:
> >
> > 1/1, 2/1, 4/1, 7/1, ...
> > 3/1, 5/1, 8/1, ...
> > 6/1, 9/1, ...
> > 10/1, ...
> > ...
> >
> Please explain why the [series] -1 + 1 - 1 + 1 - 1 + 1 -+ has sum oo

It DOESN'T have sum "oo". Since it DIVERGES. (Not taking into account Cesàro summation, etc.) [Note that your question is just diversion.]

> or write down the complete matrix in order to prove your claim.

We do not have to WRITE DOWN a "complete matrix" to PROVE its properties, you silly itiot. (Actually, this would already be quite problematic for BIG matrices with, say, more than 10^80 rows and columns.)

We only have to DEFINE it and then PROVE its properties.

The matrix we are talking about (m_ij) is defined by m_ij = <some formula - containing "i" and "j" - mentioned many times already>/1 for all i,j e IN. (This works since for any two natural numbers i,j <some formula - containing "i" and "j" - mentioned many times already> delivers a natural number.)

Now we would have to PROVE that for each and every natural number k there's exactly one pair (i, j) e IN x IN such that k = <some formula - containing "i" and "j" - mentioned many times already>.

You will find a prove here: https://en.wikipedia.org/wiki/Pairing_function#Inverting_the_Cantor_pairing_function

This implies that for each and every fraction of the form k/1 (with k e IN) there is a (unique) pair of indices (i, j) such that m_i,j = k/1. In other words, the matrix contains each and every fraction of the form k/1 (with k e IN) exactly one time.

Hence "there is a distribution of the X of the first column [of the original matrix]" such that the X "cover all matrix positions [of the matrix defined above]". qed

On Monday, May 23, 2022 at 2:13:51 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 17:12:21 UTC+2:
> > On Sunday, May 22, 2022 at 3:32:46 PM UTC+2, WM wrote:
> > >
> > > That means, there is a [distribution] such that the X of the first column
> > >
> > > XOOOO...
> > > XOOOO...
> > > XOOOO...
> > > XOOOO...
> > > XOOOO...
> > > ...
> > >
> > > [...] cover all matrix positions.
> >
> > Indeed!
> >
> > There's a simple distribution of the X to achieve this goal:
> >
> > 1/1, 2/1, 4/1, 7/1, ...
> > 3/1, 5/1, 8/1, ...
> > 6/1, 9/1, ...
> > 10/1, ...
> > ...
> >
> Please explain [...] write down the complete matrix in order to prove your claim.

We do not have to WRITE DOWN a "complete matrix" to PROVE its properties, you silly itiot. (Actually, this task would already be quite problematic for BIG matrices with, say, more than 10^80 rows and columns.)

We only have to DEFINE it and then PROVE its properties.

The matrix we are talking about (m_ij) is defined by m_ij = <some formula - containing "i" and "j" - mentioned many times already>/1 for all i,j e IN. (This works since for any two natural numbers i,j <some formula - containing "i" and "j" - mentioned many times already> delivers a natural number.)

Now we would have to PROVE that for each and every natural number k there's exactly one pair (i, j) e IN x IN such that k = <some formula - containing "i" and "j" - mentioned many times already>.

You will find a proof here: https://en.wikipedia.org/wiki/Pairing_function#Inverting_the_Cantor_pairing_function

This implies that for each and every fraction of the form k/1 (with k e IN) there is a unique pair of indices (i, j) such that m_i,j = k/1. In other words, the matrix contains each and every fraction of the form k/1 (with k e IN) exactly one time.

Hence "there is a distribution of the X of the first column [of the original matrix]" such that the X "cover all matrix positions [of the matrix defined above]". qed

On Monday, May 23, 2022 at 2:13:51 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 17:12:21 UTC+2:
> > On Sunday, May 22, 2022 at 3:32:46 PM UTC+2, WM wrote:
> > >
> > > That means, there is a [distribution] such that the X of the first column
> > >
> > > XOOOO...
> > > XOOOO...
> > > XOOOO...
> > > XOOOO...
> > > XOOOO...
> > > ...
> > >
> > > [...] cover all matrix positions.
> >
> > Indeed!
> >
> > There's a simple distribution of the X to achieve this goal:
> >
> > 1/1, 2/1, 4/1, 7/1, ...
> > 3/1, 5/1, 8/1, ...
> > 6/1, 9/1, ...
> > 10/1, ...
> > ...
> >
> Please [...] write down the complete matrix in order to prove your claim.

We do not have to WRITE DOWN "the complete matrix" to PROVE its properties, you silly itiot. (Actually, this task would already be quite problematic for BIG matrices with, say, more than 10^80 rows and columns.)

We only have to DEFINE it and then PROVE its properties.

The matrix we are talking about (m_ij) is defined by m_ij = <some formula - containing "i" and "j" - mentioned many times already>/1 for all i,j e IN. (This works since for any two natural numbers i,j <some formula - containing "i" and "j" - mentioned many times already> delivers a natural number.)

Now we would have to PROVE that for each and every natural number k there's exactly one pair (i, j) e IN x IN such that k = <some formula - containing "i" and "j" - mentioned many times already>.

You will find a proof here: https://en.wikipedia.org/wiki/Pairing_function#Inverting_the_Cantor_pairing_function

This implies that for each and every fraction of the form k/1 (with k e IN) there is a unique pair of indices (i, j) such that m_ij = k/1. In other words, the matrix contains each and every fraction of the form k/1 (with k e IN) exactly one time.

Hence "there is a distribution of the X of the first column [of the original matrix]" such that the X "cover all matrix positions [of the matrix defined above]". qed

On Tuesday, May 24, 2022 at 6:08:08 PM UTC+2, Fritz Feldhase wrote:
> On Monday, May 23, 2022 at 2:13:51 PM UTC+2, WM wrote:
> > Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 17:12:21 UTC+2:
> > > On Sunday, May 22, 2022 at 3:32:46 PM UTC+2, WM wrote:
> > > >
> > > > That means, there is a [distribution] such that the X of the first column
> > > >
> > > > XOOOO...
> > > > XOOOO...
> > > > XOOOO...
> > > > XOOOO...
> > > > XOOOO...
> > > > ...
> > > >
> > > > [...] cover all matrix positions.
> > >
> > > Indeed!
> > >
> > > There's a simple distribution of the X to achieve this goal:
> > >
> > > 1/1, 2/1, 4/1, 7/1, ...
> > > 3/1, 5/1, 8/1, ...
> > > 6/1, 9/1, ...
> > > 10/1, ...
> > > ...
> > >
> > Please [...] write down the complete matrix in order to prove your claim.
>
> We do not have to WRITE DOWN "the complete matrix" to PROVE its properties, you silly idiot. (Actually, this task would already be quite problematic for BIG matrices with, say, more than 10^80 rows and columns.)
>
> We only have to DEFINE it and then PROVE its properties.
>
> The matrix we are talking about (m_ij) is defined by m_ij = <some formula - containing "i" and "j" - mentioned many times already>/1 for all i,j e IN. (This works since for any two natural numbers i,j <some formula - containing "i" and "j" - mentioned many times already> delivers a natural number.)

From this definition we get (immediately) that the matrix (m_ij) ONLY contains fractions of the form 1/k (n e IN). Now we would have to show/prove that it contains ALL fractions of the form 1/k (n e IN), and each and every such fractions just once.

In other words,

> we would have to PROVE that for each and every natural number k there's exactly one pair (i, j) e IN x IN such that k = <some formula - containing "i" and "j" - mentioned many times already>.
>
> You will find a proof here: https://en.wikipedia.org/wiki/Pairing_function#Inverting_the_Cantor_pairing_function
>
> This implies that for each and every fraction of the form k/1 (with k e IN) there is a unique pair of indices (i, j) such that m_ij = k/1. In other words, the matrix contains each and every fraction of the form k/1 (with k e IN) exactly one time.

This completes the proof of the claim:

> "there is a distribution of the X of the first column [of the original matrix]" such that the X "cover all matrix positions [of the matrix defined above]".

On 5/24/2022 11:08 AM, Fritz Feldhase wrote:
> On Monday, May 23, 2022 at 2:13:51 PM UTC+2, WM wrote:
>> Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 17:12:21 UTC+2:
>>> On Sunday, May 22, 2022 at 3:32:46 PM UTC+2, WM wrote:
>>>>
>>>> That means, there is a [distribution] such that the X of the first column
>>>>
>>>> XOOOO...
>>>> XOOOO...
>>>> XOOOO...
>>>> XOOOO...
>>>> XOOOO...
>>>> ...
>>>>
>>>> [...] cover all matrix positions.
>>>
>>> Indeed!
>>>
>>> There's a simple distribution of the X to achieve this goal:
>>>
>>> 1/1, 2/1, 4/1, 7/1, ...
>>> 3/1, 5/1, 8/1, ...
>>> 6/1, 9/1, ...
>>> 10/1, ...
>>> ...
>>>
>> Please [...] write down the complete matrix in order to prove your claim.
>
> We do not have to WRITE DOWN "the complete matrix" to PROVE its properties, you silly itiot. (Actually, this task would already be quite problematic for BIG matrices with, say, more than 10^80 rows and columns.)
>
> We only have to DEFINE it and then PROVE its properties.
>
> The matrix we are talking about (m_ij) is defined by m_ij = <some formula - containing "i" and "j" - mentioned many times already>/1 for all i,j e IN. (This works since for any two natural numbers i,j <some formula - containing "i" and "j" - mentioned many times already> delivers a natural number.)
>
> Now we would have to PROVE that for each and every natural number k there's exactly one pair (i, j) e IN x IN such that k = <some formula - containing "i" and "j" - mentioned many times already>.
>
> You will find a proof here: https://en.wikipedia.org/wiki/Pairing_function#Inverting_the_Cantor_pairing_function

nice link!

>
> This implies that for each and every fraction of the form k/1 (with k e IN) there is a unique pair of indices (i, j) such that m_ij = k/1. In other words, the matrix contains each and every fraction of the form k/1 (with k e IN) exactly one time.
>
> Hence "there is a distribution of the X of the first column [of the original matrix]" such that the X "cover all matrix positions [of the matrix defined above]". qed

On Tuesday, May 24, 2022 at 6:08:08 PM UTC+2, Fritz Feldhase wrote:
> On Monday, May 23, 2022 at 2:13:51 PM UTC+2, WM wrote:
> > Fritz Feldhase schrieb am Sonntag, 22. Mai 2022 um 17:12:21 UTC+2:
> > > On Sunday, May 22, 2022 at 3:32:46 PM UTC+2, WM wrote:
> > > >
> > > > That means, there is a [distribution] such that the X of the first column
> > > >
> > > > XOOOO...
> > > > XOOOO...
> > > > XOOOO...
> > > > XOOOO...
> > > > XOOOO...
> > > > ...
> > > >
> > > > [...] cover all matrix positions.
> > >
> > > Indeed!
> > >
> > > There's a simple distribution of the X to achieve this goal:
> > >
> > > 1/1, 2/1, 4/1, 7/1, ...
> > > 3/1, 5/1, 8/1, ...
> > > 6/1, 9/1, ...
> > > 10/1, ...
> > > ...
> > >
> > Please [...] write down the complete matrix in order to prove your claim.
>
> We do not have to WRITE DOWN "the complete matrix" to PROVE its properties, you silly idiot. (Actually, this task would already be quite problematic for BIG matrices with, say, more than 10^80 rows and columns.)
>
> We only have to DEFINE it and then PROVE its properties.
>
> The matrix we are talking about (m_ij) is defined by m_ij = <some formula - containing "i" and "j" - mentioned many times already>/1 for all i,j e IN. (This works since for any two natural numbers i,j <some formula - containing "i" and "j" - mentioned many times already> delivers a natural number.)

From this definition we get (immediately) that the matrix (m_ij) ONLY contains fractions of the form k/1 (n e IN). Now we would have to show/prove that it contains ALL fractions of the form k/1 (n e IN), and each and every such fractions just once.

In other words,

> we would have to PROVE that for each and every natural number k there's exactly one pair (i, j) e IN x IN such that k = <some formula - containing "i" and "j" - mentioned many times already>.
>
> You will find a proof here: https://en.wikipedia.org/wiki/Pairing_function#Inverting_the_Cantor_pairing_function
>
> This implies that for each and every fraction of the form k/1 (with k e IN) there is a unique pair of indices (i, j) such that m_ij = k/1. In other words, the matrix contains each and every fraction of the form k/1 (with k e IN) exactly one time.

This completes the proof of the claim: "there is a distribution of the X of the first column [of the original matrix]" such that the X "cover all matrix positions [of the matrix defined above]".

Fritz Feldhase schrieb am Dienstag, 24. Mai 2022 um 18:08:08 UTC+2:
> On Monday, May 23, 2022 at 2:13:51 PM UTC+2, WM wrote:

> > Please [...] write down the complete matrix in order to prove your claim.
>
> We do not have to WRITE DOWN "the complete matrix" to PROVE its properties,

That's why you cannot recognize your failure.
>
> We only have to DEFINE it and then PROVE its properties.

Alas your "proof" is in contradiction with my proof: If the natural numbers first are in bijection with the integer fractions of the first column of the matrix

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, ...
5/1, 5/2, 5/3, 5/4, ...
....

then they must be distributed over the matrix such that no fraction remains without index. That means, there is a permutation such that the X of the first column

XOOOO...
XOOOO...
XOOOO...
XOOOO...
XOOOO...
....

by being exchanged with the O's cover all matrix positions. All O's have to vanish. This is obviously impossible because exchanging cannot reduce them. The number of not indexed fractions, represented by O's, will remain constant forever, in infinity.
>
> The matrix we are talking about

First consider my proof and find an escape.
>
> Hence "there is a distribution of the X of the first column [of the original matrix]" such that the X "cover all matrix positions [of the matrix defined above]". qed

That is impossible since only exchanges take place.

Regards, WM

Fritz Feldhase schrieb am Dienstag, 24. Mai 2022 um 22:22:48 UTC+2:

> > we would have to PROVE that for each and every natural number k there's exactly one pair (i, j) e IN x IN such that k = <some formula - containing "i" and "j" - mentioned many times already>.
> >
> > You will find a proof here: https://en.wikipedia.org/wiki/Pairing_function#Inverting_the_Cantor_pairing_function

This proof is restricted to definable numbers. Alas when you subtract them off |N, then most numbers remain.
> >
> > This implies that for each and every fraction of the form k/1 (with k e IN)

that his an error. All such k subtracted from |N do not exhaust |N. In orer to subtract each and every, you must subtract collectively.

> This completes the proof of the claim: "there is a distribution of the X of the first column [of the original matrix]" such that the X "cover all matrix positions [of the matrix defined above]".

All definable positions which, when removed from the matrix will leave almost all positions.

Regards, WM

On 5/25/2022 8:38 AM, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 24. Mai 2022 um 22:22:48 UTC+2:
>
>>> we would have to PROVE that for each and every natural number k there's exactly one pair (i, j) e IN x IN such that k = <some formula - containing "i" and "j" - mentioned many times already>.
>>>
>>> You will find a proof here: https://en.wikipedia.org/wiki/Pairing_function#Inverting_the_Cantor_pairing_function
>
> This proof is restricted to definable numbers.

Wrong, and totally misleading. Your "definable" is a daffynition of sillyness, with its beeps, raps, taps + flashing lights.

>>>
>>> This implies that for each and every fraction of the form k/1 (with k e IN)
>

<snip crap>

>
> Regards, WM

On 5/25/2022 8:26 AM, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 24. Mai 2022 um 18:08:08 UTC+2:
>> On Monday, May 23, 2022 at 2:13:51 PM UTC+2, WM wrote:
>
>>> Please [...] write down the complete matrix in order to prove your claim.
>>
>> We do not have to WRITE DOWN "the complete matrix" to PROVE its properties,
>
> That's why you cannot recognize your failure.
>>
>> We only have to DEFINE it and then PROVE its properties.
>
> Alas your "proof" is in contradiction with my proof: If the natural numbers first are in bijection with the integer fractions of the first column of the matrix
>
> 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, ...
> 5/1, 5/2, 5/3, 5/4, ...
> ...
>
> then they must be distributed over the matrix such that no fraction remains without index.

Wrong, bogus,