horand....@gmail.com schrieb am Freitag, 28. Januar 2022 um 21:19:50 UTC+1:
> On Friday, 28 January 2022 at 15:41:34 UTC-4, WM wrote:
> > Now, for every finite term you admit that all O's have a fraction. No fraction is missing, no O is empty. But in the LIMIT, there are all O's covering nothing because all fractions are in the first column.
> What utter rot. The whole point is that the fractions make up the entire square. No point is left uncovered.

At the beginning.
Finally all fractions are sequenced, in this model they are sequences in the first row.
> > That means aleph_0 O's have passed away in a flash. You can imagine the rattle (if you think of coins in a fruit machine).

Regards, WM

On Friday, 28 January 2022 at 18:33:05 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Freitag, 28. Januar 2022 um 21:19:50 UTC+1:
> > On Friday, 28 January 2022 at 15:41:34 UTC-4, WM wrote:
> > > Now, for every finite term you admit that all O's have a fraction. No fraction is missing, no O is empty. But in the LIMIT, there are all O's covering nothing because all fractions are in the first column.
> > What utter rot. The whole point is that the fractions make up the entire square. No point is left uncovered.
> At the beginning.
> Finally all fractions are sequenced, in this model they are sequences in the first row.

So now you are changing your model again? And god forbid you should ever explain carefully what your model is. That is, of course, your typical way of trying to weasel out of admitting you are wrong. Guess what? You are wrong, wrong, WRONG!

On 1/28/2022 2:41 PM, WM wrote:
> Jim Burns schrieb
> am Freitag, 28. Januar 2022 um 13:47:34 UTC+1:

>> We do not have global Bob-conservation,
>> even though we have local Bob-conservation.
>>
>> The explanation is NOT dark fractions.
>> Bob disappears inside ℕ⁺⨯ℕ⁺
>> ℕ⁺ = ⋃{{1,...,k}}
>> Inside ℕ⁺ there are no dark numbers.
>> Inside ℕ⁺⨯ℕ⁺ there are no dark fractions.
>
> There are fractions which cannot be defined.

If undefinable fractions exist, they're not in ℕ⁺⨯ℕ⁺
ℕ⁺ = ⋃{{1,...,k}}

If they're not in ℕ⁺⨯ℕ⁺ they can't affect Bob,
who is never outside of ℕ⁺⨯ℕ⁺

This happens _within the definable fractions_

Every spot which Bob swaps to is a definable fraction.
1/2, 2/1, 3/1, 6/1, 21/1, 231/1, 26797/1, ...
Bob is not anywhere other than a definable fraction.

Every spot that Bob swaps to Bob also swaps away from.
Bob is not anywhere other than a spot he swaps to,
but he's not at any of those spots, either.
Not at 1/2, not at 2/1, not at 3/1, not at 6/1,
not at 21/1, not at 231/1, not at 26797/1, ...

> Look at this model proving dark fractions
> in a simple and lucid way:

Bob never swaps to a dark fraction.
Bob never swaps away from a dark fraction.
Dark fractions do not explain Bob.

....

> But in the LIMIT,

It's impossible to discuss limits with you if
you won't learn what limits are.

Jim Burns schrieb am Samstag, 29. Januar 2022 um 07:02:29 UTC+1:

> This happens _within the definable fractions_
>
> Every spot which Bob swaps to is a definable fraction.

Sorry, you called Bob the O which initially covers the fraction 1/2 and wanders around. Now, in order to make the dark matter more lucid, we have fixed the O's

1OOOOOOOO...
2OOOOOOOO...
3OOOOOOOO...
4OOOOOOOO...
5OOOOOOOO...
6OOOOOOOO...
7OOOOOOOO...
8OOOOOOOO...
9OOOOOOOO...
....

and let the fractions wander.

> 1/2, 2/1, 3/1, 6/1, 21/1, 231/1, 26797/1, ...
> Bob is not anywhere other than a definable fraction.

Bob will stay fixed at his initial place. Its fraction 1/2 will go from the first line

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, ...
....

to the first column

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

and remain there forever.

Then 2/1 receives index 3 by changing positions with 3/1

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

and so on. The fractions 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, ... will stay in this order in the first column. But below the surface

1OOO...
2OOO...
3OOO...
4OOO...
....

no fraction has disappeared after a definable step. This means after all definable steps all O's cover fractions. Agreed?

In the limit however no O covers a fraction because all fractions like 1/2 have settled in the first column.

> > But in the LIMIT,
>
> It's impossible to discuss limits with you if
> you won't learn what limits are.

The limit is when all fractions in this order 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, ... have settled in the first column. That means between all finite steps ad the limit all fractions covered by O's have left their O's.

Regards, WM

horand....@gmail.com schrieb am Samstag, 29. Januar 2022 um 04:13:55 UTC+1:
> On Friday, 28 January 2022 at 18:33:05 UTC-4, WM wrote:
> > horand....@gmail.com schrieb am Freitag, 28. Januar 2022 um 21:19:50 UTC+1:
> > > On Friday, 28 January 2022 at 15:41:34 UTC-4, WM wrote:
> > > > Now, for every finite term you admit that all O's have a fraction. No fraction is missing, no O is empty. But in the LIMIT, there are all O's covering nothing because all fractions are in the first column.
> > > What utter rot. The whole point is that the fractions make up the entire square. No point is left uncovered.
> > At the beginning.
> > Finally all fractions are sequenced, in this model they are sequences in the first row.
> So now you are changing your model again?

Not really, since it remains a bijection. But the improved version shows the existence of dark fractions in a more direct way.

> And god forbid you should ever explain carefully what your model is.

Here you get the detailled explanation. Try to understand it. You will be one of the first humans to see the light of darkness.

Here is the eighth term of Cantor's sequence when it is realized by figures where indices from the first column are attached to the fractions

XXXXOOOOO...
XXXOOOOOO...
XOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
....

Now, in the improved version, instead of moving X and O, let them remain fixed while the fractions

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, ...
....

are changing places.

First term of the sequence: 1/1 remains at its position, indexed by 1. Second term: 1/2 receives Index 2, by changing positions with 2/1:

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

Then 2/1 receives index 3 by changing positions with 3/1

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

and so on. Since we have a bijection, it doesn't matter which sort moves. Every term of the sequence satisfies Cantor's formula k = (m + n - 1)(m + n - 2)/2 + m.

All this happens below the surface where the first column carries the indices and the rest the O's

1OOO...
2OOO...
3OOO...
4OOO...
....

This surface remains rigid and frozen.

Now, for every finite term you admit that all O's cover a fraction. No fraction is missing, no O is empty. But in the LIMIT, there are all O's covering nothing because all fractions are in the first column. That means aleph_0 O's have passed away in a flash. You can imagine the rattle (if you think of coins in a one-armed bandit). This can happen only if they are dark, not identifyable indivdually.

Regards, WM

After serious thinking WM wrote :
> horand....@gmail.com schrieb am Samstag, 29. Januar 2022 um 04:13:55 UTC+1:
>> On Friday, 28 January 2022 at 18:33:05 UTC-4, WM wrote:
>>> horand....@gmail.com schrieb am Freitag, 28. Januar 2022 um 21:19:50 UTC+1:
>>>> On Friday, 28 January 2022 at 15:41:34 UTC-4, WM wrote:
>>>>> Now, for every finite term you admit that all O's have a fraction. No
>>>>> fraction is missing, no O is empty. But in the LIMIT, there are all O's
>>>>> covering nothing because all fractions are in the first column.
>>>> What utter rot. The whole point is that the fractions make up the entire
>>>> square. No point is left uncovered.
>>> At the beginning.
>>> Finally all fractions are sequenced, in this model they are sequences in
>>> the first row.
>> So now you are changing your model again?
>
> Not really, since it remains a bijection. But the improved version shows the
> existence of dark fractions in a more direct way.
>
>> And god forbid you should ever explain carefully what your model is.
>
> Here you get the detailled explanation. Try to understand it. You will be one
> of the first humans to see the light of darkness.
>
> Here is the eighth term of Cantor's sequence when it is realized by figures
> where indices from the first column are attached to the fractions
>
> XXXXOOOOO...
> XXXOOOOOO...
> XOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> ...
>
> Now, in the improved version, instead of moving X and O, let them remain
> fixed while the fractions
>
> 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, ...
> ...
>
> are changing places.
>
> First term of the sequence: 1/1 remains at its position, indexed by 1. Second
> term: 1/2 receives Index 2, by changing positions with 2/1:
>
> 1/1, 2/1, 1/3, 1/4, ...
> 1/2, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> ...
>
> Then 2/1 receives index 3 by changing positions with 3/1
>
> 1/1, 3/1, 1/3, 1/4, ...
> 1/2, 2/2, 2/3, 2/4, ...
> 2/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> ...
>
> and so on. Since we have a bijection, it doesn't matter which sort moves.
> Every term of the sequence satisfies Cantor's formula k = (m + n - 1)(m + n -
> 2)/2 + m.
>
> All this happens below the surface where the first column carries the indices
> and the rest the O's
>
> 1OOO...
> 2OOO...
> 3OOO...
> 4OOO...
> ...
>
> This surface remains rigid and frozen.
>
> Now, for every finite term you admit that all O's cover a fraction. No
> fraction is missing, no O is empty. But in the LIMIT, there are all O's
> covering nothing because all fractions are in the first column. That means
> aleph_0 O's have passed away in a flash. You can imagine the rattle (if you
> think of coins in a one-armed bandit). This can happen only if they are dark,
> not identifyable indivdually.

The fact of bijection is not affected by your changing the order in
which the rational numbers are generated. It only makes it less
obvious. Instead of pretending that you are smarter than Cantor, try
showing another different generated order which *DOES* show the
bijection obviously -- like he did. All you have been doing is showing
how 'not' using Cantor's enumeration hasn't worked for you.

On 1/29/2022 4:20 AM, WM wrote:
> horand....@gmail.com schrieb am Samstag, 29. Januar 2022 um 04:13:55 UTC+1:
>> On Friday, 28 January 2022 at 18:33:05 UTC-4, WM wrote:
>>> horand....@gmail.com schrieb am Freitag, 28. Januar 2022 um 21:19:50 UTC+1:
>>>> On Friday, 28 January 2022 at 15:41:34 UTC-4, WM wrote:
>>>>> Now, for every finite term you admit that all O's have a fraction. No fraction is missing, no O is empty. But in the LIMIT, there are all O's covering nothing because all fractions are in the first column.
>>>> What utter rot. The whole point is that the fractions make up the entire square. No point is left uncovered.
>>> At the beginning.
>>> Finally all fractions are sequenced, in this model they are sequences in the first row.
>> So now you are changing your model again?
>
> Not really, since it remains a bijection. But the improved version shows the existence of dark fractions in a more direct way.
>
>> And god forbid you should ever explain carefully what your model is.
>
> Here you get the detailled explanation. Try to understand it. You will be one of the first humans to see the light of darkness.
>
> Here is the eighth term of Cantor's sequence when it is realized by figures where indices from the first column are attached to the fractions
>
> XXXXOOOOO...
> XXXOOOOOO...
> XOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> ...

Wrong, you have an X in 9th row first column, Fail. this is NOT Cantor at all.

>
> Now, in the improved version, instead of moving X and O, let them remain fixed while the fractions

No, that is all wrong, dont use the tic-tac-toe X and O, they dont work;

it is easier if you take an infinite set of numbered sheep, and another infinite set of numbered Rocks, then trade one rock per sheep from Pen A to Pen
B while piling up the used rocks into Pen C.

When you stop, which you will, whatever rock you are holding is the count of sheeps that were put into pen B, and also the Sheep has the number on his
side, a one to one mapping and bijection

On 1/29/2022 4:10 AM, WM wrote:
> Jim Burns schrieb am Samstag, 29. Januar 2022 um 07:02:29 UTC+1:
>
>> This happens _within the definable fractions_
>>
>> Every spot which Bob swaps to is a definable fraction.
>
<

<Snip obfuscation>

nothing

On 1/29/2022 5:10 AM, WM wrote:

> no fraction has disappeared after a definable step.

Remember that Cantor's point is to show
the way in which some collections
(steps up to a definable step)
are different from other collections
(all the definable steps).

> This means after all definable steps
> all O's cover fractions. Agreed?

No.
Bob covers 1/2, then 2/1, then 3/1, then 6/1,
then 21/1, then 231/1, then 26796/1, ...

Bob does not cover any fractions other than
1/2, 2/1, ..., each of which is definable.

But Bob doesn't cover any of those, either.
1/2 is replaced by 2/1
2/1 is replaced by 3/1
3/1 is replaced by 6/1
6/1 is replaced by 21/1
21/1 is replaced by 231/1
231/1 is replaced by 26796/1
....

> [...] however no O covers a fraction because
> all fractions like 1/2 have settled in the first column.

Yes, no O covers a fraction.

Each O, like Bob, holds some endless sequence of
definable fractions.
Each O holds nothing other than the fractions in
its sequence.
But each of those fractions is replaced by
another of those fractions.
Each O holds nothing.

A two-ended-stepping-order-able collection of swaps
must end and cannot involve more than a
two-ended-stepping-order-able collection of
swapping-candidates. It is with a collection like that
that the O's must keep fractions.

Not all collections are like that.

On Saturday, 29 January 2022 at 06:20:12 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Samstag, 29. Januar 2022 um 04:13:55 UTC+1:
> > On Friday, 28 January 2022 at 18:33:05 UTC-4, WM wrote:
> > > horand....@gmail.com schrieb am Freitag, 28. Januar 2022 um 21:19:50 UTC+1:
> > > > On Friday, 28 January 2022 at 15:41:34 UTC-4, WM wrote:
> > > > > Now, for every finite term you admit that all O's have a fraction.. No fraction is missing, no O is empty. But in the LIMIT, there are all O's covering nothing because all fractions are in the first column.
> > > > What utter rot. The whole point is that the fractions make up the entire square. No point is left uncovered.
> > > At the beginning.
> > > Finally all fractions are sequenced, in this model they are sequences in the first row.
> > So now you are changing your model again?
> Not really, since it remains a bijection. But the improved version shows the existence of dark fractions in a more direct way.

So it *IS* different from before, you lying fucking piece of shit.

> > And god forbid you should ever explain carefully what your model is.
> Here you get the detailled explanation. Try to understand it. You will be one of the first humans to see the light of darkness.
>
> Here is the eighth term of Cantor's sequence when it is realized by figures where indices from the first column are attached to the fractions
> XXXXOOOOO...
> XXXOOOOOO...
> XOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> ...
> Now, in the improved version, instead of moving X and O, let them remain fixed while the fractions
> 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, ...
> ...
>
> are changing places.

This drivel you call "improved" and detailed? It is even worse than the shit you were peddling before. Now, however, you have a starting position that is not homogeneous, so you have *even more* trouble to convince yourself and others that *ANYTHING* you say and do has merit.

> First term of the sequence: 1/1 remains at its position, indexed by 1. Second term: 1/2 receives Index 2, by changing positions with 2/1:
> 1/1, 2/1, 1/3, 1/4, ...
> 1/2, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...

Where, oh exalted one, do you keep track of the indices in this great scheme of yours? I see no 1 or 2 here. It is clear that by now you are a brain-dead idiot who should not be allowed to go to the bathroom unattended, let alone give lectures to students.

Rest of your garbage taken to the curb.

FromTheRafters schrieb am Samstag, 29. Januar 2022 um 13:36:15 UTC+1:
> After serious thinking WM wrote :
>
> The fact of bijection is not affected by your changing the order in
> which the rational numbers are generated.

They are no generated but are all in the matrix from the beginning.

> It only makes it less
> obvious. Instead of pretending that you are smarter than Cantor, try
> showing another different generated order which *DOES* show the
> bijection obviously -- like he did.

I did use his bijection. Try to find a deviation.

> All you have been doing is showing
> how 'not' using Cantor's enumeration hasn't worked for you.

You are completely wrong. I use exclusively Cantor's bijection.

Regards, WM

Jim Burns schrieb am Samstag, 29. Januar 2022 um 16:03:46 UTC+1:
> On 1/29/2022 5:10 AM, WM wrote:
>
> > no fraction has disappeared after a definable step.
> Remember that Cantor's point is to show
> the way in which some collections
> (steps up to a definable step)
> are different from other collections
> (all the definable steps).
> > This means after all definable steps
> > all O's cover fractions. Agreed?
> No.

Show one that does not.
>
> > [...] however no O covers a fraction because
> > all fractions like 1/2 have settled in the first column.
> Yes, no O covers a fraction.

In the limit!
>
> Each O, like Bob, holds some endless sequence of
> definable fractions.
> Each O holds nothing other than the fractions in
> its sequence.
> But each of those fractions is replaced by
> another of those fractions.
> Each O holds nothing.

Each O covers something in every finite term of the sequence.
In the limit each O does not cover anything. Therefore infinitely many fractions have passed between all finite terms and the limit. They are not indexed. They are dark.

Regards, WM

On 1/25/2022 6:52 AM, WM wrote:
On 1/15/2022 6:52 AM, WM wrote:
> In oder to index all positive fractions
>
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, QUACK!!, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> ...
>
> by natural indexes, collect all these indexes 1, 2, 3, ... and put them in the bucket, covering the fractions n/1. All other fractions get 3 indexes.
This is expressed by an XO's.
>
CORRECTED:

1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4...
2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4...
3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4...
4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4...
1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4...
2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4...
3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4...
4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, QUACK!!, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4...
1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4...
2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4...
3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4...
4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4...
1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4...
2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4...
3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4...
4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4...
....

by natural indexes, collect all these indexes 1, 2, 3, ... and put them in the bucket, covering the fractions X/O. All other fractions get no indexes.
This is expressed by an O's.

For brevity the presence of an index is represented by an X since its particular numerical value is relevant. So we get, as the start position:

XOOOOOOOOXOOOOOOOO...XOOOOOOOO...XOOOOOOOO...XOOOOOOOO...XOOOOOOOO...XOOOOOOOO...
XOOOOOOOO...XOOOOOOOOXOOOOOOOO...XOOOOOOOO...XOOOOOOOOXOOOOOOOO...XOOOOXOOOXOOOOOOOO
....XOOOOOOOO...XOOOOOOOO...XOOOOOOOO...XOOOOOOOO...XOOOOOOOO..
XOOOOXOOO...XOOOOOOOO...XOOOOOOOO...XOOOOOOOO...
XOOOOOOOO...XOOOOOOOO...
XOOOOOXOO...
O O
O XOOOOOOOO...XOOOOOOOOXOOOOOOOO
XOOOOOOOO...
XOOOOOOOO...
XOOOOOOOO... XXXXXXXXXXXXXXXXOOOOOOOOOOOOOOOOO
XOOOOOOOO...
XOOOOOOOO...XXXXXXXXX
XOOOOOOOO..
O O
O XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

see? its all mush

On Saturday, 29 January 2022 at 23:59:57 UTC-4, sergio wrote:
> On 1/25/2022 6:52 AM, WM wrote:
> On 1/15/2022 6:52 AM, WM wrote:
> > In oder to index all positive fractions
> >
> > 1/1, 1/2, 1/3, 1/4, ...
> > 2/1, 2/2, 2/3, 2/4, ...
> > 3/1, QUACK!!, 3/3, 3/4, ...
> > 4/1, 4/2, 4/3, 4/4, ...
> > ...
> >
> > by natural indexes, collect all these indexes 1, 2, 3, ... and put them in the bucket, covering the fractions n/1. All other fractions get 3 indexes.
> This is expressed by an XO's.
> >
> CORRECTED:
>
> 1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4...
> 2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4...
> 3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4...
> 4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4...
> 1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4...
> 2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4...
> 3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4...
> 4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, QUACK!!, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4...
> 1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4...
> 2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4...
> 3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4...
> 4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4...
> 1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4,1/1, 1/2, 1/3, 1/4...
> 2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4,2/1, 2/2, 2/3, 2/4...
> 3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4,3/1, 3/2, 3/3, 3/4...
> 4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4,4/1, 4/2, 4/3, 4/4...
> ...
>
> by natural indexes, collect all these indexes 1, 2, 3, ... and put them in the bucket, covering the fractions X/O. All other fractions get no indexes.
> This is expressed by an O's.
>
> For brevity the presence of an index is represented by an X since its particular numerical value is relevant. So we get, as the start position:
>
> XOOOOOOOOXOOOOOOOO...XOOOOOOOO...XOOOOOOOO...XOOOOOOOO...XOOOOOOOO...XOOOOOOOO...
> XOOOOOOOO...XOOOOOOOOXOOOOOOOO...XOOOOOOOO...XOOOOOOOOXOOOOOOOO...XOOOOXOOOXOOOOOOOO
> ...XOOOOOOOO...XOOOOOOOO...XOOOOOOOO...XOOOOOOOO...XOOOOOOOO..
> XOOOOXOOO...XOOOOOOOO...XOOOOOOOO...XOOOOOOOO...
> XOOOOOOOO...XOOOOOOOO...
> XOOOOOXOO...
> O
> O
> O
> XOOOOOOOO...XOOOOOOOOXOOOOOOOO
> XOOOOOOOO...
> XOOOOOOOO...
> XOOOOOOOO... XXXXXXXXXXXXXXXXOOOOOOOOOOOOOOOOO
> XOOOOOOOO...
> XOOOOOOOO...XXXXXXXXX
> XOOOOOOOO..
> O
> O
> O
> XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
>
>
> see? its all mush

Look at the title of the thread. I think this is a reference to the fact that WM's head exploded after thinking too hard about mathematics.

WM explained on 1/29/2022 :
> FromTheRafters schrieb am Samstag, 29. Januar 2022 um 13:36:15 UTC+1:
>> After serious thinking WM wrote :
>>
>> The fact of bijection is not affected by your changing the order in
>> which the rational numbers are generated.
>
> They are no generated but are all in the matrix from the beginning.

Cantor's enumerations enforce a particular order upon them where given
one n/d the "next" n/d is easily obtained from the generating formula
in much the same way as the natural numbers are generated. This is
being 'countable'.

>> It only makes it less
>> obvious. Instead of pretending that you are smarter than Cantor, try
>> showing another different generated order which *DOES* show the
>> bijection obviously -- like he did.
>
> I did use his bijection. Try to find a deviation.

Your swappings destroy the order. There is still a bijection, but it is
not shown by your new ordering. There is no need to show orderings
which don't show the bijection because the order does not affect the
bijection, only the ability to show it.

>> All you have been doing is showing
>> how 'not' using Cantor's enumeration hasn't worked for you.
>
> You are completely wrong. I use exclusively Cantor's bijection.

No, you don't. It is not his bijection, only his enumeration(s) which
show it.

sergio schrieb am Sonntag, 30. Januar 2022 um 04:59:57 UTC+1:
> On 1/25/2022 6:52 AM, WM wrote:

> > In oder to index all positive fractions
> >
> > 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, ...
> > ...
> see? its all mush

Chuckle. If you don't understand, use this simpler model: Take all fractions and throw them around until the Cantor sequence appears in the first column. The matrix is as big as at the start at every step before you take the limit. Then all columns except the first one will be cleared. You cannot however, discern what elements have been residing there. That proves their darkness.

Regards, WM

horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 05:19:41 UTC+1:

> Look at the title of the thread. I think this is a reference to

this fact: Take all fractions
of

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, ...
....
and throw them around until the Cantor sequence appears in the first column. The matrix is as big as at the start at every step before you take the limit. Then all columns except the first one will be cleared.

1/1, O, O, O, ...
1/2, O, O, O, ...
2/1, O, O, O, ...
1/3, O, O, O, ...
....

You cannot however discern what elements have been residing there. That proves their darkness.

Regards, WM

FromTheRafters schrieb am Sonntag, 30. Januar 2022 um 07:54:52 UTC+1:
> WM explained on 1/29/2022 :

> > I did use his bijection. Try to find a deviation.
> Your swappings destroy the order. There is still a bijection, but it is
> not shown by your new ordering.

Not at all. But if you like take your own way. Take all fractions and throw them around as you think it would be correct until the Cantor sequence appears in the first column. The matrix remain as big as at the start at every step before you take the limit. Do you agree? Or do you forbid to consider this fact?

> > I use exclusively Cantor's bijection.
> No, you don't. It is not his bijection, only his enumeration(s) which
> show it.

His enuemration of the rationals is his bijection.

The matrix is as big as at the start at every step before you take the limit. Then all columns except the first one will be cleared. You cannot however, discern what elements have been residing there. That proves their darkness.

Regards, WM

On 1/30/2022 5:35 AM, WM wrote:
> sergio schrieb am Sonntag, 30. Januar 2022 um 04:59:57 UTC+1:
>> On 1/25/2022 6:52 AM, WM wrote:
>
>>> In oder to index all positive fractions
>>>
>>> 1/1, 1/2, 1/3, 1/4, ...
>>> 2/1, 2/2, 2/3, 2/4, ...
>>> 3/1, QuQcK!! 3/3, 3/4, ...
>>> 4/1, 4/2, 4/3, 4/4, ...
>>> ...
>> see? its all mush
>
> Chuckle. If you don't understand, use this simpler model: Take all fractions and throw them around until the Cantor sequence appears in the first column. The matrix is as big as at the start at every step before you take the limit. Then all columns except the first one will be cleared. You cannot however, discern what elements have been residing there. That proves their darkness.
>
> Regards, WM

Chuckle. You try so hard to prove Cantor wrong, but *you end up proving Cantor right*, and also proving your dark numbers cannot exist.
This is your 4th time playing tic-tac-toe with your rubber leaky sets, and your mistakes are clear and intentional.

On 1/30/2022 5:40 AM, WM wrote:
> FromTheRafters schrieb am Sonntag, 30. Januar 2022 um 07:54:52 UTC+1:
>> WM explained on 1/29/2022 :
>
>>> I did use his bijection. Try to find a deviation.
>> Your swappings destroy the order. There is still a bijection, but it is
>> not shown by your new ordering.
>
> Not at all. But if you like take your own way. Take all fractions and throw them around as you think it would be correct until the Cantor sequence appears in the first column. The matrix remain as big as at the start at every step before you take the limit. Do you agree? Or do you forbid to consider this fact?

that is your sloppy math.

>
>>> I use exclusively Cantor's bijection.
>> No, you don't. It is not his bijection, only his enumeration(s) which
>> show it.
>
> His enuemration of the rationals is his bijection.
>
> The matrix is as big as at the start at every step before you take the limit. Then all columns except the first one will be cleared. You cannot however, discern what elements have been residing there. That proves their darkness.

wrong. It is in bijection, or one to one mapping, with the rationals. All elements are indexed. You have failed again.

>
> Regards, WM

On Sunday, 30 January 2022 at 07:35:49 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 05:19:41 UTC+1:
>
> > Look at the title of the thread. I think this is a reference to
> this fact: Take all fractions
> of
> 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, ...
> ...
> and throw them around until the Cantor sequence appears in the first column. The matrix is as big as at the start at every step before you take the limit. Then all columns except the first one will be cleared.
> 1/1, O, O, O, ...
> 1/2, O, O, O, ...
> 2/1, O, O, O, ...
> 1/3, O, O, O, ...
> ...
>
> You cannot however discern what elements have been residing there.

Of course you can, you imbecile. Take any fraction n/m and compute
(n+m)*(n+m-3)/2+n+1.

If you did not cheat (like leave out fractions that are not in lowest terms), that gives you the position of n/m in the first column. You can then, if you are so inclined, take that integer k and write it as k/1.

The only thing dark is your mind. There is no discernible action taking on there any longer.

On 1/30/2022 10:18 AM, Gus Gassmann wrote:
> On Sunday, 30 January 2022 at 07:35:49 UTC-4, WM wrote:
>> horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 05:19:41 UTC+1:
>>
>>> Look at the title of the thread. I think this is a reference to
>> this fact: Take all fractions
>> of
>> 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, ...
>> ...
>> and throw them around until the Cantor sequence appears in the first column. The matrix is as big as at the start at every step before you take the limit. Then all columns except the first one will be cleared.
>> 1/1, O, O, O, ...
>> 1/2, O, O, O, ...
>> 2/1, O, O, O, ...
>> 1/3, O, O, O, ...
>> ...
>>
>> You cannot however discern what elements have been residing there.
>
> Of course you can, you imbecile. Take any fraction n/m and compute
> (n+m)*(n+m-3)/2+n+1.
>
> If you did not cheat (like leave out fractions that are not in lowest terms), that gives you the position of n/m in the first column. You can then, if you are so inclined, take that integer k and write it as k/1.
>
> The only thing dark is your mind. There is no discernible action taking on there any longer.

I dont think he knows how to actually compute (n+m)*(n+m-3)/2+n+1 nor what it means

he only seems to be able to visualize "leaky rubber bathtubs"(sets), things he can write down on paper, like that matrix, this all underscores the fact
that he does not know algebra.

horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 17:18:36 UTC+1:
> On Sunday, 30 January 2022 at 07:35:49 UTC-4, WM wrote:
> > horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 05:19:41 UTC+1:
> >
> > > Look at the title of the thread. I think this is a reference to
> > this fact: Take all fractions
> > of
> > 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, ...
> > ...
> > and throw them around until the Cantor sequence appears in the first column. The matrix is as big as at the start at every step before you take the limit. Then all columns except the first one will be cleared.
> > 1/1, O, O, O, ...
> > 1/2, O, O, O, ...
> > 2/1, O, O, O, ...
> > 1/3, O, O, O, ...
> > ...
> >
> > You cannot however discern what elements have been residing there.
> Of course you can,
>
> If you did not cheat (like leave out fractions that are not in lowest terms)

I use all combinations of two naturals, compare the matrix: 2/2, 3/3, 4/4, ...

>, that gives you the position of n/m in the first column.

For every definable fraction this position can be found. Who did deny this? But nevertheless every frame shows every O occupied by a fraction. Only in the limit, all O's are empty. Do you agree?

Regards, WM

On Sunday, 30 January 2022 at 14:30:51 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 17:18:36 UTC+1:
> > On Sunday, 30 January 2022 at 07:35:49 UTC-4, WM wrote:
> > > horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 05:19:41 UTC+1:
> > >
> > > > Look at the title of the thread. I think this is a reference to
> > > this fact: Take all fractions
> > > of
> > > 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, ...
> > > ...
> > > and throw them around until the Cantor sequence appears in the first column. The matrix is as big as at the start at every step before you take the limit. Then all columns except the first one will be cleared.
> > > 1/1, O, O, O, ...
> > > 1/2, O, O, O, ...
> > > 2/1, O, O, O, ...
> > > 1/3, O, O, O, ...
> > > ...
> > >
> > > You cannot however discern what elements have been residing there.
> > Of course you can,
> >
> > If you did not cheat (like leave out fractions that are not in lowest terms)
> I use all combinations of two naturals, compare the matrix: 2/2, 3/3, 4/4, ...
> >, that gives you the position of n/m in the first column.
> For every definable fraction this position can be found. Who did deny this? But nevertheless every frame shows every O occupied by a fraction. Only in the limit, all O's are empty. Do you agree?

Yeah, so?? For every fraction you can compute where it will end up, you can determine where it came from, and you can determine what element it replaced in the first column. That you cannot comprehend that after a finite number of moves there will still be an infinite number of moves left to go is really not my problem. Neither is the fact that you have no clue about limits. As I said, your comprehension of things infinite is not just asymptotically small, it is zero.

On 1/30/2022 12:30 PM, WM wrote:
> horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 17:18:36 UTC+1:
>> On Sunday, 30 January 2022 at 07:35:49 UTC-4, WM wrote:
>>> horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 05:19:41 UTC+1:
>>>
>>>> Look at the title of the thread. I think this is a reference to
>>> this fact: Take all fractions
>>> of
>>> 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, ...
>>> ...
>>> and throw them around until the Cantor sequence appears in the first column. The matrix is as big as at the start at every step before you take the limit. Then all columns except the first one will be cleared.
>>> 1/1, O, O, O, ...
>>> 1/2, O, O, O, ...
>>> 2/1, O, O, O, ...
>>> 1/3, O, O, O, ...
>>> ...
>>>
>>> You cannot however discern what elements have been residing there.
>> Of course you can,
>>
>> If you did not cheat (like leave out fractions that are not in lowest terms)
>
> I use all combinations of two naturals, compare the matrix: 2/2, 3/3, 4/4, ...
>
>> , that gives you the position of n/m in the first column.
>
> For every definable fraction this position can be found.

think about that for a while.

> Who did deny this?

And you deny it in the next sentence, with Nevertheless

> But nevertheless every frame shows every O occupied by a fraction.

your Os are not required, nor needed.

> Only in the limit, all O's are empty. Do you agree?\

you Os are diversion.

All rationals are enumerated. forget your tic-tac-toe of X and Os they only confuse you.

>
> Regards, WM
>
>