horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 20:23:35 UTC+1:
> 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.

Determine an instance where one of the O's gets empty. Impossible. All fractions leaving O's move simultaneously.

> 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.

Every move that can be checked happens after a finite number of moves. That means all moves that can be checked leave all O's populated by fractions.

> Neither is the fact that you have no clue about limits.

In the limit all O's are empty. But there is no continuos exit of fractions from the O's to the first column. That means infinitely many fractions must move simultaneously without being counted. They cannot be identified. They are dark.

Regards, WM

sergio schrieb am Sonntag, 30. Januar 2022 um 21:08:12 UTC+1:
> On 1/30/2022 12:30 PM, WM wrote:

> > 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

No. The fractions occupying O's after all definable steps are undefinable.

> > But nevertheless every frame shows every O occupied by a fraction.
> your Os are not required, nor needed.

They are useful.
>
> > Only in the limit, all O's are empty. Do you agree?
>
> you Os are diversion.

We could simply say most matrixelements are not in the furst column during all steps where this can be checked.
>
> All rationals are enumerated. forget your tic-tac-toe of X and Os they only confuse you.

As long as numbers are vailable, almost all rationals are covered by O's and are not enumerated. As long as we can prove it, almost all fractions are not in the first column.

Regards, WM

On 1/30/2022 3:35 PM, WM wrote:
> horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 20:23:35 UTC+1:
>> 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.
>
> Determine an instance where one of the O's gets empty. Impossible. All fractions leaving O's move simultaneously.

disagree with all of that nonsence.

>
>> 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.
>
> Every move that can be checked happens after a finite number of moves. That means all moves that can be checked leave all O's populated by fractions.

there is no need to move numbers, there is no need to check numbers. See how using your haphazard constrution of tic-tac-toe has failed.

>
>> Neither is the fact that you have no clue about limits.
>
> In the limit all O's are empty.

no. In the limit all, each and every, rationals have been indexed by the natural numbers.

your empty Os are littering your home, more than trillions of them.

> But there is no continuos exit of fractions from the O's to the first column. That means infinitely many fractions must move simultaneously without being counted. They cannot be identified. They are dark.

your obvious trickery is QuaCkery of a very poor kind, perhaps all those empty Os are really your brain cells that went "dark".

use an infinite ball of string, where you cut it at k, and then you have a finite piece and an infinite piece, and all those unidentifible dark stringlets.

or how about pigs? there is an infinite number of pigs walking in a line, all prenumbered, which you stop at k. That pig says "Oink". So where are the
dark pigs ?

>
> Regards, WM

On 1/30/2022 3:40 PM, WM wrote:
> sergio schrieb am Sonntag, 30. Januar 2022 um 21:08:12 UTC+1:
>> On 1/30/2022 12:30 PM, WM wrote:
>
>>> 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
>
> No. The fractions occupying O's after all definable steps are undefinable.

no. you cannot identify them, nor where they are at.

>
>>> But nevertheless every frame shows every O occupied by a fraction.
>> your Os are not required, nor needed.
>
> They are useful.

no, diversion.

>>
>>> Only in the limit, all O's are empty. Do you agree?
>>
>> you Os are diversion.
>
> We could simply say most matrixelements are not in the furst column during all steps where this can be checked.

just leave the matrix elements alone and apply Cantors Enummeration, and you will see that is is right.

all your changing of the matrix is folly

>>
>> All rationals are enumerated. forget your tic-tac-toe of X and Os they only confuse you.
>
> As long as numbers are vailable, almost all rationals are covered by O's and are not enumerated. As long as we can prove it, almost all fractions are not in the first column.

that is the problem with doing it your way, by your own admission it does not work, and Cantor's Does.

>
> Regards, WM

On Sunday, 30 January 2022 at 17:35:30 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 20:23:35 UTC+1:
> > 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.
> Determine an instance where one of the O's gets empty. Impossible. All fractions leaving O's move simultaneously.

Again, so?? All you show continually is that you have no clue how infinity works.

> > 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.
> Every move that can be checked happens after a finite number of moves. That means all moves that can be checked leave all O's populated by fractions.

For the third time, so?

> > Neither is the fact that you have no clue about limits.
> In the limit all O's are empty. But there is no continuos exit of fractions from the O's to the first column.

Why ever would you think that there is a *continuous* process at work? I am convinced that continuity is another one of those concepts you are incapable of understanding.

> That means infinitely many fractions must move simultaneously without being counted. They cannot be identified. They are dark.

And again, the only thing dark is your mind. No detectable brain activity. But perhaps that is a prerequisite for teaching at your famous institution of higher learning?

On 1/29/2022 5:50 PM, WM wrote:
> Jim Burns schrieb
> am Samstag, 29. Januar 2022 um 16:03:46 UTC+1:
>> On 1/29/2022 5:10 AM, WM wrote:

>>> [...] 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!

If I wrote "Gift" thinking of the English word
and you read "Gift" thinking of the German word,
then we would not be communicating.

I will continue to ignore your use of "limit"
and avoid using it myself whenever I am trying to
communicate with you -- unless you were to start
using "limit" correctly. However, it looks like
you would need to be struck by lightning out of
a clear sky for that to happen. So, it's possible.

>> 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.

Instead of moving Bob &co. around,
we'll nail each O and each X to its perch,
and move the fractions around according to
the rule k = (n+d-1)*(n+d-2)/2 + n

Initially, each fraction in ℕ⁺⨯{1} is covered by an X
(non-identical X's, one for 1/1, one for 2/1, ... )
and each fraction in ℕ⁺⨯(ℕ⁺\{1}) is covered by
a (non-identical) O.

Move the fractions.
The O or X that _was_ covering n/d
_will be_ covering k/1 _next_
where k = (n+d-1)*(n+d-2)/2 + n
Define
next(n/d) = k/1

> Each O covers something in every finite term of
> the sequence.

Yes.

For each n/d in ℕ⁺⨯(ℕ⁺\{1}) (covered by an O)
next(n/d) is in ℕ⁺⨯ℕ⁺ and
next(next(n/d)) is in ℕ⁺⨯ℕ⁺ and
next(next(next(n/d))) is in ℕ⁺⨯ℕ⁺ and
so on.

"And so on" is a promise to provide a description,
if necessary. Here it is:

| For each n/d in ℕ⁺⨯(ℕ⁺\{1}) and each j" in ℕ⁺
| | a collection { ⟨0,n/d⟩, ..., ⟨j",n"/d"⟩ } of ordered
| pairs exists such that
| | { ⟨1,n/d⟩, ..., ⟨j",n"/d"⟩ } has a two-ended stepping-
| -order (each cut has a step, and there are two ends)
| such that,
| | for each cut, the initial segment ends and
| the final segment begins with two ordered pairs
| ⟨ j', n'/d' ⟩ and ⟨ j'+1, next(n'/d') ⟩
| and
| | the first and second ends are ⟨0,n/d⟩ and ⟨j",n"/d"⟩
| | ( ℕ⁺ = ⋃{{1,...,k}} )

Lemma.
n"/d" as described above is in ℕ⁺⨯ℕ⁺

> Each O covers something in every finite term of
> the sequence.

Yes,
each O covers something in every finite term of
the sequence.

> In the limit each O does not cover anything.

Yes, for each O.
In other news,
in the limit, each X other than the one X
which initially covers 1/1 does not cover anything.

| Define
| next^j"(n/d) = n"/d"
| iff
| the collection { ⟨0,n/d⟩, ..., ⟨j",n"/d"⟩ }
| defined as above exists.
| | next^0(n/d) = n/d
| next^1(n/d) = next(n/d)
| next^(j"+1)(n/d) = next(next^j"(n/d))
| | Define
| an O which initially covers n/d
| covers n"/d" _in the limit_ iff
| there exists some i in ℕ⁺ such that,
| for each j in ℕ⁺
| j >= i -> next^j(n/d) = n"/d"

next(n/d) = ( (n+d-1)*(n+d-2)/2+ n )/1

For each n"/d" /= 1/1 next(n"/d") /= n"/d"

For each n/d in ℕ⁺⨯ℕ⁺
there is no n"/d" in ℕ⁺⨯ℕ⁺ covered in the limit
by the O which initially covers n/d

In the limit each O does not cover anything.

> Therefore infinitely many fractions have passed
> between all finite terms and the limit.

No.
In the limit each O does not cover anything.
Not. Anything.

In order for something to be between a finite term
and a limit, there needs to be a limit.

> Therefore infinitely many fractions have passed
> between all finite terms and the limit.
> They are not indexed. They are dark.

Each fraction n/d in ℕ⁺⨯ℕ⁺ is indexed by k
k = (n+d-1)*(n+d-2)/2 + n

However,
for each fraction n/d in ℕ⁺⨯ℕ⁺
the collection of fractions n"/d" indexed by k" > k
is not two-ended-stepping-order-able
(is infinitely-many).

Also however,
for each fraction n/d in ℕ⁺⨯ℕ⁺ excepting 1/1
the collection of fractions n"/d" = next^k"(n/d)
with k" in ℕ⁺ is not two-ended-stepping-order-able
(is infinitely-many).

----
Here is your whole "argument":
<WM> They are dark. </WM>

Your implicit argument is of the form
| | You all might think that's a running woodchipper
| but you're wrong.
| To prove you're wrong, I (WM) am going to
| jump into the woodchipper.
| Surely, someone as brilliant as I am
| would not jump into a running woodchipper.

Then you (WM) jump into the running woodchipper
and get (metaphorically) shredded.
Then, later, you jump into a running woodchipper again.
Then, later, again.
And again.

While I would prefer that you stop jumping into
running woodchippers, there doesn't seem to be
anything I can do about that.
Have a nice day.

On 1/30/2022 4:11 PM, Jim Burns wrote:
> On 1/29/2022 5:50 PM, WM wrote:
>> Jim Burns schrieb
>> am Samstag, 29. Januar 2022 um 16:03:46 UTC+1:
>>> On 1/29/2022 5:10 AM, WM wrote:
>
>>>> [...] 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!
>
> If I wrote "Gift" thinking of the English word
> and you read "Gift" thinking of the German word,
> then we would not be communicating.
>
> I will continue to ignore your use of "limit"
> and avoid using it myself whenever I am trying to
> communicate with you -- unless you were to start
> using "limit" correctly. However, it looks like
> you would need to be struck by lightning out of
> a clear sky for that to happen. So, it's possible.
>
>>> 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.
>
> Instead of moving Bob &co. around,
> we'll nail each O and each X to its perch,
> and move the fractions around according to
> the rule k = (n+d-1)*(n+d-2)/2 + n
>
> Initially, each fraction in ℕ⁺⨯{1} is covered by an X
> (non-identical X's, one for 1/1, one for 2/1, ... )
> and each fraction in ℕ⁺⨯(ℕ⁺\{1}) is covered by
> a (non-identical) O.
>
> Move the fractions.
> The O or X that _was_ covering n/d
> _will be_ covering k/1 _next_
> where k = (n+d-1)*(n+d-2)/2 + n
> Define
> next(n/d) = k/1
>
>> Each O covers something in every finite term of
>> the sequence.
>
> Yes.
>
> For each n/d in ℕ⁺⨯(ℕ⁺\{1}) (covered by an O)
> next(n/d) is in ℕ⁺⨯ℕ⁺  and
> next(next(n/d)) is in ℕ⁺⨯ℕ⁺  and
> next(next(next(n/d))) is in ℕ⁺⨯ℕ⁺  and
> so on.
>
> "And so on" is a promise to provide a description,
> if necessary. Here it is:
>
> | For each n/d in ℕ⁺⨯(ℕ⁺\{1}) and each j" in ℕ⁺
> |
> | a collection { ⟨0,n/d⟩, ..., ⟨j",n"/d"⟩ } of ordered
> | pairs exists such that
> |
> | { ⟨1,n/d⟩, ..., ⟨j",n"/d"⟩ } has a two-ended stepping-
> | -order (each cut has a step, and there are two ends)
> | such that,
> |
> | for each cut, the initial segment ends and
> | the final segment begins with two ordered pairs
> | ⟨ j', n'/d' ⟩ and ⟨ j'+1, next(n'/d') ⟩
> | and
> |
> | the first and second ends are ⟨0,n/d⟩ and ⟨j",n"/d"⟩
> |
> | ( ℕ⁺ = ⋃{{1,...,k}} )
>
> Lemma.
> n"/d"  as described above  is in ℕ⁺⨯ℕ⁺
>
>> Each O covers something in every finite term of
>> the sequence.
>
> Yes,
> each O covers something in every finite term of
> the sequence.
>
>> In the limit each O does not cover anything.
>
> Yes, for each O.
> In other news,
> in the limit, each X other than the one X
> which initially covers 1/1 does not cover anything.
>
> | Define
> | next^j"(n/d) = n"/d"
> | iff
> | the collection { ⟨0,n/d⟩, ..., ⟨j",n"/d"⟩ }
> | defined as above exists.
> |
> | next^0(n/d) = n/d
> | next^1(n/d) = next(n/d)
> | next^(j"+1)(n/d) = next(next^j"(n/d))
> |
> | Define
> | an O which initially covers n/d
> | covers n"/d" _in the limit_  iff
> | there exists some i in ℕ⁺ such that,
> | for each j in ℕ⁺
> | j >= i  ->  next^j(n/d) = n"/d"
>
> next(n/d) = ( (n+d-1)*(n+d-2)/2+ n )/1
>
> For each n"/d" /= 1/1  next(n"/d") /= n"/d"
>
> For each n/d in ℕ⁺⨯ℕ⁺
> there is no n"/d" in ℕ⁺⨯ℕ⁺ covered in the limit
> by the O which initially covers n/d
>
> In the limit each O does not cover anything.
>
>> Therefore infinitely many fractions have passed
>> between all finite terms and the limit.
>
> No.
> In the limit each O does not cover anything.
> Not. Anything.
>
> In order for something to be between a finite term
> and a limit, there needs to be a limit.
>
>> Therefore infinitely many fractions have passed
>> between all finite terms and the limit.
>> They are not indexed. They are dark.
>
> Each fraction n/d in ℕ⁺⨯ℕ⁺ is indexed by k
> k = (n+d-1)*(n+d-2)/2 + n
>
> However,
> for each fraction n/d in ℕ⁺⨯ℕ⁺
> the collection of fractions n"/d" indexed by k" > k
> is not two-ended-stepping-order-able
> (is infinitely-many).
>
> Also however,
> for each fraction n/d in ℕ⁺⨯ℕ⁺  excepting 1/1
> the collection of fractions n"/d" = next^k"(n/d)
> with k" in ℕ⁺ is not two-ended-stepping-order-able
> (is infinitely-many).
>
> ----
> Here is your whole "argument":
> <WM> They are dark. </WM>
>
> Your implicit argument is of the form
> |
> | You all might think that's a running woodchipper
> | but you're wrong.
> | To prove you're wrong, I (WM) am going to
> | jump into the woodchipper.
> | Surely, someone as brilliant as I am
> | would not jump into a running woodchipper.
>
> Then you (WM) jump into the running woodchipper
> and get (metaphorically) shredded.
> Then, later, you jump into a running woodchipper again.
> Then, later, again.
> And again.
>
> While I would prefer that you stop jumping into
> running woodchippers, there doesn't seem to be
> anything I can do about that.
> Have a nice day.
>

this is the mapping one to one, where k is a natural number, and n is numerator natural number and d is denominator natural number of same fraction

k = (n+d-1)*(n+d-2)/2 + n #1

*nothing else is needed*, no matrix, no Xs, no Os they are all crutches.

EXTRA CREDIT:
How do you adjust the row and column numbers in #1 above ?
Can equation #1 be changed for a different mapping, still one to one ?
If Achilles is an X, and the Turtle is a O, who wins if the goal is 100 miles away ?

On 1/30/2022 8:19 PM, sergio wrote:

>> Have a nice day.
>
> this is the mapping one to one,
> where k is a natural number, and
> n is numerator natural number and
> d is denominator natural number of same fraction
>
> k = (n+d-1)*(n+d-2)/2 + n  #1
>
> *nothing else is needed*, no matrix, no Xs, no Os
> they are all crutches.

Speaking as someone who is merely human,
when I need a crutch, I use a crutch.
YMMV.

> EXTRA CREDIT:
> How do you adjust the row and column numbers in #1 above ?

Adjust what how?

> Can equation #1 be changed for a different mapping,
> still one to one ?

Without having tried to prove it, I think that
there are uncountably-many bijections from ℕ to ℕ
It really looks as though something could be worked out.

> If Achilles is an X, and the Turtle is a O,
> who wins if the goal is 100 miles away ?

It's a tie. They're both killed by Leonidas.

On 1/30/2022 7:42 PM, Jim Burns wrote:
> On 1/30/2022 8:19 PM, sergio wrote:
>
>>> Have a nice day.
>>
>> this is the mapping one to one,
>> where k is a natural number, and
>> n is numerator natural number and
>> d is denominator natural number of same fraction
>>
>> k = (n+d-1)*(n+d-2)/2 + n  #1
>>
>> *nothing else is needed*, no matrix, no Xs, no Os
>> they are all crutches.
>
> Speaking as someone who is merely human,
> when I need a crutch, I use a crutch.
> YMMV.
>
>> EXTRA CREDIT:
>> How do you adjust the row and column numbers in #1 above ?
>
> Adjust what how?

n is a row (or column), and d is a column (or row) just add or subtract a constant.

>
>> Can equation #1 be changed for a different mapping,
>> still one to one ?
>
> Without having tried to prove it, I think that
> there are uncountably-many bijections from ℕ to ℕ
> It really looks as though something could be worked out.

an obvious one,
k = (n+d-1)*(n+d-2)/2 + n #1
k = (n+d-1)*(n+d-2)/2 + d #2

>
>> If Achilles is an X, and the Turtle is a O,
>> who wins if the goal is 100 miles away ?
>
> It's a tie. They're both killed by Leonidas.
>

so in the end they were both shafted.

On 1/30/2022 8:39 PM, sergio wrote:
> On 1/30/2022 7:42 PM, Jim Burns wrote:
>> On 1/30/2022 8:19 PM, sergio wrote:
>>
>>>> Have a nice day.
>>>
>>> this is the mapping one to one,
>>> where k is a natural number, and
>>> n is numerator natural number and
>>> d is denominator natural number of same fraction
>>>
>>> k = (n+d-1)*(n+d-2)/2 + n  #1
>>>
>>> *nothing else is needed*, no matrix, no Xs, no Os
>>> they are all crutches.
>>
>> Speaking as someone who is merely human,
>> when I need a crutch, I use a crutch.
>> YMMV.
>>
>>> EXTRA CREDIT:
>>> How do you adjust the row and column numbers in #1 above ?
>>
>> Adjust what how?
>
>
> n is a row (or column), and d is a column (or row) just add or subtract a constant.
>
>>
>>> Can equation #1 be changed for a different mapping,
>>> still one to one ?
>>
>> Without having tried to prove it, I think that
>> there are uncountably-many bijections from ℕ to ℕ
>> It really looks as though something could be worked out.
>
> an obvious one,
> k = (n+d-1)*(n+d-2)/2 + n  #1

k= (n^2 +n*d -2n + d*n + d^2 -2*d - n - d + 2)/2 + n

k= (n^2 + d^2 + 2*n*d -3*n -3*d + 2)/2 + n

k= ((n+d)^2 - 3*(n+d) +2)/2 + n

cantor style

> k = (n+d-1)*(n+d-2)/2 + d  #2
>
>>
>>> If Achilles is an X, and the Turtle is a O,
>>> who wins if the goal is 100 miles away ?
>>
>> It's a tie. They're both killed by Leonidas.
>>
>
> so in the end they were both shafted.

fredag 28 januari 2022 kl. 11:53:24 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 28. Januar 2022 um 06:09:19 UTC+1:
> > torsdag 27 januari 2022 kl. 14:38:14 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 14:24:15 UTC+1:
> > > > torsdag 27 januari 2022 kl. 12:32:48 UTC+1 skrev WM:
> > >
> > > > > > There is no FINITE steps, there are no steps PERIOD!
> > > > > But there are finite terms of the sequence --- only finite terms. Where, when, and how does the first O exit?
> > > >
> > > > >What is the frist term when an O exits? Where does it go then?
> > > > Irrelevant, you are obfuscating with this and trying to confuse things.
> > > No, I am using mathematics and logic.
> > > > >But there are finite terms of the sequence --- only finite terms. Where, when, and how does the first O exit?
> > > > You are asking "Why are all natural numbers finite but not N?"
> > > No, I am asking where do the O's go?
> > > >
> > >I am using mathematics and logic. I am asking where do the O's go?
> > Irrelevant
>
> Very relevant, because it kills set theory. Note that the O's are there and cannot exit. And if they tried, where should they settle?
> > again all your Os and shit is irrelevant to the fact that there is a bijection
> But the O's are relevant to prove that there is no bijection. If this proof surpasses your intellectual competence, try to use the simpler picture where the O's remain fixed at their places and the fractions try to disappear through the first column.
>
> Regards, WM

your Os prove nothign of the sort as always you go from finite and assume it applies to the infinite, iti s fallacious.

Your Os and shit is not a valid argument and you're using it to obfuscate.

The bijection exists, get over it

sergio schrieb am Sonntag, 30. Januar 2022 um 22:52:43 UTC+1:
> On 1/30/2022 3:35 PM, WM wrote:

> > In the limit all O's are empty.
> no. In the limit all, each and every, rationals have been indexed by the natural numbers.

The indexes remain in th first column:

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

They are occupied at every definable step. They are empty in the limit.

Regards, WM

sergio schrieb am Sonntag, 30. Januar 2022 um 22:59:42 UTC+1:
> On 1/30/2022 3:40 PM, WM wrote:

> > As long as numbers are vailable, almost all rationals are covered by O's and are not enumerated. As long as we can prove it, almost all fractions are not in the first column.
> that is the problem with doing it your way, by your own admission it does not work, and Cantor's Does.

My way shows that Cantor's way does not work. Because I simply do what Cantor did. The only difference is that he forget to ask where the indexes come from.

Regards, WM

horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 23:07:40 UTC+1:
> On Sunday, 30 January 2022 at 17:35:30 UTC-4, WM wrote:

> > Determine an instance where one of the O's gets empty. Impossible. All fractions leaving O's move simultaneously.
> Again, so?? All you show continually is that you have no clue how infinity works.

Infinity does not work.

> > Every move that can be checked happens after a finite number of moves. That means all moves that can be checked leave all O's populated by fractions.
> For the third time, so?

Fact.

> > In the limit all O's are empty. But there is no continuos exit of fractions from the O's to the first column.
> Why ever would you think that there is a *continuous* process at work? I am convinced that continuity is another one of those concepts you are incapable of understanding.

I have proved that there is no continuous process at work. There is a vast amount of fractions which are processed simultaneously, that is: without any chance to distinguish them by indices.

> > That means infinitely many fractions must move simultaneously without being counted. They cannot be identified. They are dark.

> And again, the only thing dark is your mind.

You said not continuous, that means not one by one, that means not countable.

Regards, WM

Jim Burns schrieb am Sonntag, 30. Januar 2022 um 23:11:10 UTC+1:
> On 1/29/2022 5:50 PM, WM wrote:

> >>> [...] 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!

That is, when all indexes have been applied.

Here are the first transpositions:

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

> > Therefore infinitely many fractions have passed
> > between all finite terms and the limit.
> No.

Yes. All finite steps see all O's occupied. That is before all indices have been applied, as long as any index has to be applied. After all indexes have been applied all O's are empty.

> In the limit each O does not cover anything.
> Not. Anything.
>
> In order for something to be between a finite term
> and a limit, there needs to be a limit.

That is when all indices have been applied.

Regards, WM

zelos...@gmail.com schrieb am Montag, 31. Januar 2022 um 06:18:15 UTC+1:
> fredag 28 januari 2022 kl. 11:53:24 UTC+1 skrev WM:

> your Os prove nothign of the sort as always you go from finite and assume it applies to the infinite, iti s fallacious.

As long as any index is available the O's are not empty. After all indices have been applied, all O's are empty. That means infinitely many fractions have changed their places out of any control. They are not indexed.

> The bijection exists, get over it

For all definable numbers yes. But that are not all numbers. Those, which switch after all indexes have been applied, are dark.

Regards, WM

måndag 31 januari 2022 kl. 09:37:47 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 31. Januar 2022 um 06:18:15 UTC+1:
> > fredag 28 januari 2022 kl. 11:53:24 UTC+1 skrev WM:
>
> > your Os prove nothign of the sort as always you go from finite and assume it applies to the infinite, iti s fallacious.
> As long as any index is available the O's are not empty. After all indices have been applied, all O's are empty. That means infinitely many fractions have changed their places out of any control. They are not indexed.
> > The bijection exists, get over it
> For all definable numbers yes. But that are not all numbers. Those, which switch after all indexes have been applied, are dark.
>
> Regards, WM

>As long as any index is available the O's are not empty. After all indices have been applied, all O's are empty. That means infinitely many fractions have changed their places out of any control. They are not indexed.

Blah blah blah "finite stuff"

That is all you're doing, you are arguing about finite stuff when it is an infinite set, stop it.

>For all definable numbers yes. But that are not all numbers. Those, which switch after all indexes have been applied, are dark.

It exists for ALL

zelos...@gmail.com schrieb am Montag, 31. Januar 2022 um 09:56:31 UTC+1:
> måndag 31 januari 2022 kl. 09:37:47 UTC+1 skrev WM:

> >As long as any index is available the O's are not empty. After all indices have been applied, all O's are empty. That means infinitely many fractions have changed their places out of any control. They are not indexed.
> Blah blah blah "finite stuff"

No, after all indexes have been applied, the stuff is infinite.
Before all indexes have been applied, i.e., as long as any index is available to be applied, the stuff is finite. In this case all O's are occupied. Afterwards all fractions exit in an uncontrolled way because there are no further indexes.
>
> That is all you're doing, you are arguing about finite stuff when it is an infinite set, stop it.
> >For all definable numbers yes. But that are not all numbers. Those, which switch after all indexes have been applied, are dark.

> It exists for ALL

Indexes exist only as long as infinitely many fractions remain not indexed.

Regards, WM

måndag 31 januari 2022 kl. 11:36:10 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 31. Januar 2022 um 09:56:31 UTC+1:
> > måndag 31 januari 2022 kl. 09:37:47 UTC+1 skrev WM:
>
> > >As long as any index is available the O's are not empty. After all indices have been applied, all O's are empty. That means infinitely many fractions have changed their places out of any control. They are not indexed.
> > Blah blah blah "finite stuff"
> No, after all indexes have been applied, the stuff is infinite.
> Before all indexes have been applied, i.e., as long as any index is available to be applied, the stuff is finite. In this case all O's are occupied. Afterwards all fractions exit in an uncontrolled way because there are no further indexes.
> >
> > That is all you're doing, you are arguing about finite stuff when it is an infinite set, stop it.
> > >For all definable numbers yes. But that are not all numbers. Those, which switch after all indexes have been applied, are dark.
>
> > It exists for ALL
> Indexes exist only as long as infinitely many fractions remain not indexed.
>
> Regards, WM

You are still arguing "after finite steps" or "This one case" which is entirely irrelevant you crank!

On Monday, 31 January 2022 at 04:23:12 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 23:07:40 UTC+1:
> > On Sunday, 30 January 2022 at 17:35:30 UTC-4, WM wrote:
>
> > > Determine an instance where one of the O's gets empty. Impossible. All fractions leaving O's move simultaneously.
> > Again, so?? All you show continually is that you have no clue how infinity works.
> Infinity does not work.

You've made it abundantly that that is your position. That makes you an ultrafinitist not matter how much you try to deny it. However, all *you* can prove is this: IF there are only finitely many integers, then the cardinality of N is finite. All the rest is bluster, posturing, lies, and false logic..

EOD

On 1/31/2022 2:37 AM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 31. Januar 2022 um 06:18:15 UTC+1:
>> fredag 28 januari 2022 kl. 11:53:24 UTC+1 skrev WM:
>
>> your Os prove nothign of the sort as always you go from finite and assume it applies to the infinite, iti s fallacious.
>
> As long as any index is available the O's are not empty. After all indices have been applied, all O's are empty. That means infinitely many fractions have changed their places out of any control. They are not indexed.

and So you have failed.

Try using Cantors Enumeration, it is a very old proof, but it is solid, simple and holds up.

>
>> The bijection exists, get over it
>
> For all definable numbers yes. But that are not all numbers. Those, which switch after all indexes have been applied, are dark.

Fail. All numbers are numbers.

>
> Regards, WM

On 1/31/2022 2:15 AM, WM wrote:
> sergio schrieb am Sonntag, 30. Januar 2022 um 22:52:43 UTC+1:
>> On 1/30/2022 3:35 PM, WM wrote:
>
>>> In the limit all O's are empty.
>> no. In the limit all, each and every, rationals have been indexed by the natural numbers.
>
> The indexes remain in th first column:
>
> 1OOOOOOOO...
> 2OOOOOOOO...
> 3OOOOOOOO...
> 4OOOOOOOO...
> 5OOOOOOOO...
> 6OOOOOOOO...
> 7OOOOOOOO...
> 8OOOOOOOO...
> 9OOOOOOOO...
> ...
>

that is your problem. You failed to enumerate the rationals.

> They are occupied at every definable step. They are empty in the limit.

that is your problem. Anyone can tell you your construct will fail.

Major mistakes;
1. You manipulated the matrix of the rationals
2. You confused yourself with un needed X and O's
3. You failed to do diagonal threading

But don't feel too bad, beginning students usually make less obvious mistakes.

Try Again. But leave the Matrix of Rationals as is.

>
> Regards, WM

On 1/31/2022 2:15 AM, WM wrote:
> sergio schrieb am Sonntag, 30. Januar 2022 um 22:59:42 UTC+1:
>> On 1/30/2022 3:40 PM, WM wrote:
>
>>> As long as numbers are vailable, almost all rationals are covered by O's and are not enumerated. As long as we can prove it, almost all fractions are not in the first column.
>> that is the problem with doing it your way, by your own admission it does not work, and Cantor's Does.
>
> My way shows that Cantor's way does not work.

wrong, you make 3 major mistakes.

Cantor's works, your does not.

> Because I simply do what Cantor did.

Liar. You manipulate the matrix of the rationals.

> The only difference is that he forget to ask where the indexes come from.

that is red herring.

there is no need to "ask where the indexes come from", you can buy a set of them from eBay.

>
> Regards, WM

you have been caught intentionally misleading people.

zelos...@gmail.com schrieb am Montag, 31. Januar 2022 um 12:41:04 UTC+1:
> måndag 31 januari 2022 kl. 11:36:10 UTC+1 skrev WM:

> > > That is all you're doing, you are arguing about finite stuff when it is an infinite set, stop it.
> > > >For all definable numbers yes. But that are not all numbers. Those, which switch after all indexes have been applied, are dark.
> >
> > > It exists for ALL
> > Indexes exist only as long as infinitely many fractions remain not indexed.

> You are still arguing "after finite steps"

After finite steps = definable transpositions the matrix has provably no empty position

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, ...
....

After all indices have been applied, the matrix, except the first column, is empty.

This shows undefinable steps.

Regards, WM

horand....@gmail.com schrieb am Montag, 31. Januar 2022 um 13:56:21 UTC+1:
> On Monday, 31 January 2022 at 04:23:12 UTC-4, WM wrote:
> > horand....@gmail.com schrieb am Sonntag, 30. Januar 2022 um 23:07:40 UTC+1:
> > > On Sunday, 30 January 2022 at 17:35:30 UTC-4, WM wrote:
> >
> > > > Determine an instance where one of the O's gets empty. Impossible. All fractions leaving O's move simultaneously.
> > > Again, so?? All you show continually is that you have no clue how infinity works.
> > Infinity does not work.
> You've made it abundantly that that is your position. That makes you an ultrafinitist not matter how much you try to deny it.

Not at all. I accept all infinitely many terms of the sequence of transpositions 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, ...
....
namely

1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ... 4/1, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ...
...., and so on.

For *all* *infinitely* many of them I have proved that no matrix position gets empty. How can you accuse me of finitism?!

> IF there are only finitely many integers, then the cardinality of N is finite.

If there are more than all finite terms, then we get

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

I agree. I would only mention that the exit of the fractions happens somewhat panic-stricken. We have no means of individually controlling them.(They have no Passierschein like the inhabitants imprisoned in the former GDR would have needed.)

I think that you share my description of facts.

Regards, WM