On 1/18/2022 1:14 PM, WM wrote:
> JVR schrieb am Dienstag, 18. Januar 2022 um 19:50:24 UTC+1:
>> On Tuesday, January 18, 2022 at 3:25:55 PM UTC+1, WM wrote:
>>> zelos...@gmail.com schrieb am Dienstag, 18. Januar 2022 um 13:41:26 UTC+1:
>>>
>>>>> To explain it better I have to go a long way back: In the OP I have shown that the Cantor function k = (m + n - 1)(m + n - 2)/2 + m used for enumerating all positive fractions
>>>>> As long as indexes are availabale, the presence of all O's indicates that almost all fractions are not indexed. And after the drain has become free, there are no indexes available which could index the remaining matrix elements covered by O's.
>>>>>
>>>>> This proves there remain many fractions without indexes. But we cannot find them. They are dark.
>>>>>
>>>> proves nothing of the sort
>>> Where does my model fail to follow Cantor's prescription k = (m + n - 1)(m + n - 2)/2 + m ?
>>>
>> I don't know. I can't bring myself to read any more of your repetitive nonsense.
>> The function k = (m + n - 1)(m + n - 2)/2 + m is biunique as every child can verify.
>> That's all there is to the countability of the rationals.
>
> No, there is a contradiction. No O can leave before all indexes have been issued. No O can leave after all indexes have been issued. (Note that I don't claim that this will ever happen.)

No, your model is inherently flawed. Fix your mistake.

>
> Regards, WM

On Tuesday, January 18, 2022 at 6:08:45 PM UTC-4, WM wrote:
> William schrieb am Dienstag, 18. Januar 2022 um 20:51:23 UTC+1:
> > On Tuesday, January 18, 2022 at 3:21:53 PM UTC-4, WM wrote:
> > > William schrieb am Dienstag, 18. Januar 2022 um 19:38:46 UTC+1:
> >
> > > > If n is not dark it is the largest element of a FISON (which of course may have the property that it cannot be written down.)
> > > I agree.
> > > > So n+1 is the largest element of a FISON and is thus not dark. So if n is not dark n+1 is not dark. This is true whether or not you can write n down. By induction we have that no element of N_P is dark.
> > > But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's
> > >
> > > o
> > > oo
> > > ooo
> > > ...
> > >
> > > than are o's avaible.
> > Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available.
> FISONs have less than aleph_0 o's.

Nothing you can say will change the fact that here are aleph_0 lines. Thus there are aleph_0 o's available.

--
William Hughes

On 1/18/2022 2:26 PM, WM wrote:
> Jim Burns schrieb
> am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:

>> Without ending somewhere, they're not the first kind.
>
> But without ending all O's will remain.

If the O on n/d remains, the replacement has ended
before step i = (n+d-1)*(n+d-2)/2 + n

tisdag 18 januari 2022 kl. 15:25:55 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 18. Januar 2022 um 13:41:26 UTC+1:
>
> > > To explain it better I have to go a long way back: In the OP I have shown that the Cantor function k = (m + n - 1)(m + n - 2)/2 + m used for enumerating all positive fractions
> > > As long as indexes are availabale, the presence of all O's indicates that almost all fractions are not indexed. And after the drain has become free, there are no indexes available which could index the remaining matrix elements covered by O's.
> > >
> > > This proves there remain many fractions without indexes. But we cannot find them. They are dark.
> > >
> > proves nothing of the sort
> Where does my model fail to follow Cantor's prescription k = (m + n - 1)(m + n - 2)/2 + m ?
>
> Regards, WM

It fails because you moving shit around is of no relevants.

William schrieb am Mittwoch, 19. Januar 2022 um 00:39:06 UTC+1:
> On Tuesday, January 18, 2022 at 6:08:45 PM UTC-4, WM wrote:

> > > > But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's
> > > >
> > > > o
> > > > oo
> > > > ooo
> > > > ...
> > > >
> > > > than are o's avaible.
> > > Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available..
> > FISONs have less than aleph_0 o's.
> Nothing you can say will change the fact that here are aleph_0 lines. Thus there are aleph_0 o's available.

OK, if you believe in infinite FISONs, then I cannot win the argument.

Regards, WM

Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 02:03:36 UTC+1:
> On 1/18/2022 2:26 PM, WM wrote:
> > Jim Burns schrieb
> > am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:
> >> Without ending somewhere, they're not the first kind.
> >
> > But without ending all O's will remain.
> If the O on n/d remains,

Every O moves only once and then remains fixed --- in the first column --- above the never empty rest of X's.

> the replacement has ended
> before step i = (n+d-1)*(n+d-2)/2 + n

No. We can agree that the replacement will never end and therefore never all fractions will be indexed, but the fact remains: before the replacement has ended, no O will leave the matrixboard.

Look, I did nothing else but ask where Cantor's indexes come from. Matheologians never did so and thus remained in ignorance. It is clear that the elements of one column will never flood the whole matrix because for every taken X (necessarily in the finite) there opens a gap receiving an O (necessarily in the finite). It is simply unbelievable that people could believe that this could change!!!

Regards, WM

onsdag 19 januari 2022 kl. 10:38:59 UTC+1 skrev WM:
> Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 02:03:36 UTC+1:
> > On 1/18/2022 2:26 PM, WM wrote:
> > > Jim Burns schrieb
> > > am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:
> > >> Without ending somewhere, they're not the first kind.
> > >
> > > But without ending all O's will remain.
> > If the O on n/d remains,
> Every O moves only once and then remains fixed --- in the first column --- above the never empty rest of X's.
> > the replacement has ended
> > before step i = (n+d-1)*(n+d-2)/2 + n
> No. We can agree that the replacement will never end and therefore never all fractions will be indexed, but the fact remains: before the replacement has ended, no O will leave the matrixboard.
>
> Look, I did nothing else but ask where Cantor's indexes come from. Matheologians never did so and thus remained in ignorance. It is clear that the elements of one column will never flood the whole matrix because for every taken X (necessarily in the finite) there opens a gap receiving an O (necessarily in the finite). It is simply unbelievable that people could believe that this could change!!!
>
> Regards, WM

Again, your re-arranging is entirely fuckign IRRELEVANT

WM expressed precisely :
> Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 02:03:36 UTC+1:
>> On 1/18/2022 2:26 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:
>>>> Without ending somewhere, they're not the first kind.
>>>
>>> But without ending all O's will remain.
>> If the O on n/d remains,
>
> Every O moves only once and then remains fixed --- in the first column ---
> above the never empty rest of X's.
>
>> the replacement has ended
>> before step i = (n+d-1)*(n+d-2)/2 + n
>
> No. We can agree that the replacement will never end and therefore never all
> fractions will be indexed, but the fact remains: before the replacement has
> ended, no O will leave the matrixboard.
>
> Look, I did nothing else but ask where Cantor's indexes come from.
> Matheologians never did so and thus remained in ignorance. It is clear that
> the elements of one column will never flood the whole matrix because for
> every taken X (necessarily in the finite) there opens a gap receiving an O
> (necessarily in the finite). It is simply unbelievable that people could
> believe that this could change!!!

If it doesn't work the wrong way, try doing it the right way.

On 1/19/2022 3:38 AM, WM wrote:
> Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 02:03:36 UTC+1:
>> On 1/18/2022 2:26 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:
>>>> Without ending somewhere, they're not the first kind.
>>>
>>> But without ending all O's will remain.
>> If the O on n/d remains,
>
> Every O moves only once and then remains fixed --- in the first column --- above the never empty rest of X's.

Using a boogered Tic-Tac-Toe as a substitute for Cantor's Enumeration is silly.

>
>> the replacement has ended
>> before step i = (n+d-1)*(n+d-2)/2 + n
>
> No. We can agree that the replacement will never end and therefore never all fractions will be indexed, but the fact remains: before the replacement has ended, no O will leave the matrixboard.

because you are doing it wrong.

>
> Look, I did nothing else but ask where Cantor's indexes come from.

no, you put your index in a column of the matrix, wrong move.

>Matheologians never did so and thus remained in ignorance. It is clear that the elements of one column will never flood the whole matrix because for every taken X (necessarily in the finite) there opens a gap receiving an O (necessarily in the finite). It is simply unbelievable that people could believe that this could change!!!
>

both intentionally wrong and silly.

Just wiki Cantor's Enumeration and use that, and see where your mistakes are.

> Regards, WM

On 1/19/2022 3:25 AM, WM wrote:
> William schrieb am Mittwoch, 19. Januar 2022 um 00:39:06 UTC+1:
>> On Tuesday, January 18, 2022 at 6:08:45 PM UTC-4, WM wrote:
>
>>>>> But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's
>>>>>
>>>>> o
>>>>> oo
>>>>> ooo
>>>>> ...
>>>>>
>>>>> than are o's avaible.
>>>> Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available.
>>> FISONs have less than aleph_0 o's.
>> Nothing you can say will change the fact that here are aleph_0 lines. Thus there are aleph_0 o's available.
>
> OK, if you believe in infinite FISONs, then I cannot win the argument.
>
> Regards, WM
>
>

the # of FISONs is infinite, so you cannot win the argument.

On Wednesday, January 19, 2022 at 5:25:44 AM UTC-4, WM wrote:
> William schrieb am Mittwoch, 19. Januar 2022 um 00:39:06 UTC+1:
> > On Tuesday, January 18, 2022 at 6:08:45 PM UTC-4, WM wrote:
>
> > > > > But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's
> > > > >
> > > > > o
> > > > > oo
> > > > > ooo
> > > > > ...
> > > > >
> > > > > than are o's avaible.
> > > > Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available.
> > > FISONs have less than aleph_0 o's.
> > Nothing you can say will change the fact that here are aleph_0 lines. Thus there are aleph_0 o's available.
> OK, if you believe in infinite FISONs,
Nope. I do not believe in an infinite FISON. Each line is finite. Each FISON is finite.
Does not change the fact that there are aleph_0 lines. Thus there are aleph_0 o's availiable.

--
William Hughes

On Wednesday, 19 January 2022 at 09:40:38 UTC-4, sergio wrote:
> On 1/19/2022 3:25 AM, WM wrote:
> > OK, if you believe in infinite FISONs, then I cannot win the argument.
> >
> the # of FISONs is infinite, so you cannot win the argument.

The premise of WM's statement is, of course, entirely redundant. In order to win any argument, one has to be able to reason logically, and WM has proven himself utterly incapable of this for at least the last fourteen years. His premise is also ambiguous, likely intentionally so. Nobody, except perhaps WM himself, thinks that there could be FISONs that have infinite cardinality (the 'F' is there for a reason!). But, of course, there are infinitely many of them. And of course WM also believes that there cannot be infinitely many natural numbers unless omega is counted as a natural number.

On 1/19/2022 4:38 AM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 19. Januar 2022 um 02:03:36 UTC+1:
>> On 1/18/2022 2:26 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:

>>>> Without ending somewhere, they're not the first kind.
>>>
>>> But without ending all O's will remain.
>> If the O on n/d remains,
>
> Every O moves only once and then remains fixed ---
> in the first column --- above the never empty rest of X's.

No.

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

The O at 1/2 moves to 2/1
because (1+2-1)(1+2-2)/2 + 1 = 2

The O at 2/1 moves to 3/1
because (2+1-1)*(2+1-2)/2 + 2 = 3

The O at 3/1 moves to 6/1
because (3+1-1)*(3+1-2)/2 + 3 = 6

The O at 6/1 moves to 21/1
because (6+1-1)*(6+1-2)/2 + 6 = 21

The O at 21/1 moves to 231/1
because (21+1-1)*(21+1-2)/2 + 21 = 231

....

No O remains fixed.
Other than 1/1 n/d moves to a location i/1 with an
index later than n/d. That later-indexed location
moves to a still later-indexed location, ad infinitum.

No definable O moves to a dark location.

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

definable n, definable d
definable n+d-1, definable n+d-2
either 2 divides n+d-1 or 2 divides n+d-2
2 divides (n+d-1)*(n+d-2)
definable (n+d-1)*(n+d-2)/2 + n
i = (n+d-1)*(n+2-2)/2 + n
definable i

Define {1,...,k} to be a collection with a counting-order
which begins at 1 and ends somewhere. Name where it ends k.

Define ℕ⁺ = ⋃{{1,...,k}} union all the {1,...,k}

Define 𝔽⁺ = { n/d | n,d ∈ ℕ⁺ }

n/d ---> i/1
moves the X's and O's within the defined 𝔽⁺

You want the reason to be existence of dark numbers.
But i = (n+d-1)*(n+d-2)/2 + n
Whether or not dark numbers exist, no O moves to
a dark number.

Dark numbers don't explain how all of
𝔽⁺ = { n/d | n,d ∈ ⋃{{1,...,k}} }
fits in only its first column.

The reason that all of 𝔽⁺ can move to only its first column
is that 𝔽⁺ is NOT the first kind of collection,
the sheep-kind, the pebble-kind.

You:
Look! 𝔽⁺ does not behave like the first kind of collection.

Me:
Right. Because 𝔽⁺ is not the first kind of collection.

zelos...@gmail.com schrieb am Mittwoch, 19. Januar 2022 um 12:31:53 UTC+1:
> onsdag 19 januari 2022 kl. 10:38:59 UTC+1 skrev WM:
> > Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 02:03:36 UTC+1:
> > > On 1/18/2022 2:26 PM, WM wrote:
> > > > Jim Burns schrieb
> > > > am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:
> > > >> Without ending somewhere, they're not the first kind.
> > > >
> > > > But without ending all O's will remain.
> > > If the O on n/d remains,
> > Every O moves only once and then remains fixed --- in the first column --- above the never empty rest of X's.
> > > the replacement has ended
> > > before step i = (n+d-1)*(n+d-2)/2 + n
> > No. We can agree that the replacement will never end and therefore never all fractions will be indexed, but the fact remains: before the replacement has ended, no O will leave the matrixboard.
> >
> > Look, I did nothing else but ask where Cantor's indexes come from. Matheologians never did so and thus remained in ignorance. It is clear that the elements of one column will never flood the whole matrix because for every taken X (necessarily in the finite) there opens a gap receiving an O (necessarily in the finite). It is simply unbelievable that people could believe that this could change!!!
> >
> Again, your re-arranging is entirely fuckign IRRELEVANT

Why do you think so?

Of course nobody need be interested in my model. But for those beein interested I can state that my model applies Cantor's first diagonal arguments meticulously. Every fraction is indexed as he did. The only difference is that I ask where the indexes come from.

Since only transpositions are applied, always exchanging two items, an index and a void card at *finite places*, but never creating indices, it is clear that not all fractions can be indexed. And there is absolutely no mistake in my model because it is so simple. Everybody not being prejudiced by inertia and othodox belief, should easily be able to understand this most important correction of mathematics since Hippasos.

Regards, WM

FromTheRafters schrieb am Mittwoch, 19. Januar 2022 um 13:23:08 UTC+1:
> WM expressed precisely :

> > Look, I did nothing else but ask where Cantor's indexes come from.
> > Matheologians never did so and thus remained in ignorance. It is clear that
> > the elements of one column will never flood the whole matrix because for
> > every taken X (necessarily in the finite) there opens a gap receiving an O
> > (necessarily in the finite). It is simply unbelievable that people could
> > believe that this could change!!!
> If it doesn't work the wrong way, try doing it the right way.

I did. My simple precondition is: Every Index is taken from a finite place. Hence every O lands at a finite place. No magic, no limits, simply application of
k = (m + n - 1)(m + n - 2)/2 + m.
The only violation of Cantor's way is to ask where the indexes come from.

If you don't like my result, simply don't ask, and you may stay happy with set theory.

Regards, WM

sergio schrieb am Mittwoch, 19. Januar 2022 um 14:39:49 UTC+1:
> On 1/19/2022 3:38 AM, WM wrote:

> > Every O moves only once and then remains fixed --- in the first column --- above the never empty rest of X's.
> Using a boogered Tic-Tac-Toe as a substitute for Cantor's Enumeration is silly.

It is allowed to use diagrams in order to get the overview, in particular to know where the indxes come from.
> >
> >> the replacement has ended
> >> before step i = (n+d-1)*(n+d-2)/2 + n
> >
> > No. We can agree that the replacement will never end and therefore never all fractions will be indexed, but the fact remains: before the replacement has ended, no O will leave the matrixboard.
> because you are doing it wrong.

What is done wrong?

> Just wiki Cantor's Enumeration and use that,

I do: k = (m + n - 1)(m + n - 2)/2 + m

> and see where your mistakes are.

My "mistake" is to show where the indexes come from. That's the only deviation from Cantor-

Regards, WM

William schrieb am Mittwoch, 19. Januar 2022 um 16:06:05 UTC+1:
> On Wednesday, January 19, 2022 at 5:25:44 AM UTC-4, WM wrote:
> > William schrieb am Mittwoch, 19. Januar 2022 um 00:39:06 UTC+1:
> > > On Tuesday, January 18, 2022 at 6:08:45 PM UTC-4, WM wrote:
> >
> > > > > > But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's
> > > > > >
> > > > > > o
> > > > > > oo
> > > > > > ooo
> > > > > > ...
> > > > > >
> > > > > > than are o's avaible.
> > > > > Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available.
> > > > FISONs have less than aleph_0 o's.
> > > Nothing you can say will change the fact that here are aleph_0 lines. Thus there are aleph_0 o's available.
> > OK, if you believe in infinite FISONs,
> Nope. I do not believe in an infinite FISON. Each line is finite.

Each line is the union of all preceding lines + o. If there is an actually infinite number of o's, ℵo o's, then there is a preceding FISON with one o less than ℵo.

> Each FISON is finite.

The addition of finite FISONs is a fiite FISON.
> Does not change the fact that there are aleph_0 lines. Thus there are aleph_0 o's availiable.

o
oo
ooo
oooo
ooooo
....

Never. Width = Height.

Regards, WM

Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 19:08:11 UTC+1:
> On 1/19/2022 4:38 AM, WM wrote:
> > Jim Burns schrieb
> > am Mittwoch, 19. Januar 2022 um 02:03:36 UTC+1:
> >> On 1/18/2022 2:26 PM, WM wrote:

> > Every O moves only once and then remains fixed ---
> > in the first column --- above the never empty rest of X's.
> No.

It is fact by construction that all X are taken from finite places and O's are landing at finite places. The matrix has no other places.
>
> i = (n+d-1)*(n+2-2)/2 + n
Slightly corrected: i = (n+d-1)*(n+d-2)/2 + n
>
> The O at 1/2 moves to 2/1
> because (1+2-1)(1+2-2)/2 + 1 = 2

So it is.
>
> The O at 2/1 moves to 3/1
> because (2+1-1)*(2+1-2)/2 + 2 = 3
>
> The O at 3/1 moves to 6/1
> because (3+1-1)*(3+1-2)/2 + 3 = 6

After the move k = 8 we have:

XXXXOOOOO...
XXXOOOOOO...
XOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
....
>
> No O remains fixed.

Yes, you are right. My mistake. The X, when moved will remain fixed

> Other than 1/1 n/d moves to a location i/1 with an
> index later than n/d. That later-indexed location
> moves to a still later-indexed location, ad infinitum.

Of course. But since every X stems from a finite place, every transposed O willland at a finite place.
>
> No definable O moves to a dark location.
>
> i = (n+d-1)*(n+d-2)/2 + n
>
> definable n, definable d
> definable n+d-1, definable n+d-2
> either 2 divides n+d-1 or 2 divides n+d-2
> 2 divides (n+d-1)*(n+d-2)
> definable (n+d-1)*(n+d-2)/2 + n
> i = (n+d-1)*(n+2-2)/2 + n
> definable i

Right. Every move ends at a definable place.

Therefore the number of O's in the matrix does not change. Agreed?

Regards, WM

Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 19:08:11 UTC+1:

> n/d ---> i/1
> moves the X's and O's within the defined 𝔽⁺

Yes. Every transposition where the X_i1 moves to O_nd and vice versa will preserve the numbers of X's and O's.
>
> You want the reason to be existence of dark numbers.
> But i = (n+d-1)*(n+d-2)/2 + n
> Whether or not dark numbers exist, no O moves to
> a dark number.

Right.
>
> Dark numbers don't explain how all of
> 𝔽⁺ = { n/d | n,d ∈ ⋃{{1,...,k}} }
> fits in only its first column.

They don't fit in the first column.
>
> The reason that all of 𝔽⁺ can move to only its first column
> is that 𝔽⁺ is NOT the first kind of collection,
> the sheep-kind, the pebble-kind.

Whatever. Even if all O's fit in the first column, they stay within the matrix. Therefore their number is preserved.
>
> You:
> Look! 𝔽⁺ does not behave like the first kind of collection.

I am not interested in the kind of collection. I am only interested in the fact that no O can leave, and hence not all fractions can be indeXed.

Regards, WM

On 1/19/2022 12:23 PM, WM wrote:
> FromTheRafters schrieb am Mittwoch, 19. Januar 2022 um 13:23:08 UTC+1:
>> WM expressed precisely :
>
>>> Look, I did nothing else but ask where Cantor's indexes come from.
>>> Matheologians never did so and thus remained in ignorance. It is clear that
>>> the elements of one column will never flood the whole matrix because for
>>> every taken X (necessarily in the finite) there opens a gap receiving an O
>>> (necessarily in the finite). It is simply unbelievable that people could
>>> believe that this could change!!!
>> If it doesn't work the wrong way, try doing it the right way.
>
> I did. My simple precondition is: Every Index is taken from a finite place.

stealing #'s off the sides of sheeps, shame!

> Hence every O lands at a finite place.

you don't need a "place" to take numbers from or land in in Math.

> No magic, no limits, simply application of
> k = (m + n - 1)(m + n - 2)/2 + m.

wrong. Cantor does it clearly. Yours is boogered.

> The only violation of Cantor's way is to ask where the indexes come from.

wrong. you just do not know how to do math, so quit trying.

>
> If you don't like my result, simply don't ask, and you may stay happy with set theory.

set theory is just fine without your permanent confusion.

>
> Regards, WM

On 1/19/2022 12:27 PM, WM wrote:
> sergio schrieb am Mittwoch, 19. Januar 2022 um 14:39:49 UTC+1:
>> On 1/19/2022 3:38 AM, WM wrote:
>
>>> Every O moves only once and then remains fixed --- in the first column --- above the never empty rest of X's.
>> Using a boogered Tic-Tac-Toe as a substitute for Cantor's Enumeration is silly.
>
> It is allowed to use diagrams in order to get the overview, in particular to know where the indxes come from.

no, you use it to fool people. Indexes are NOT in the matrix. Fail.

>>>
>>>> the replacement has ended
>>>> before step i = (n+d-1)*(n+d-2)/2 + n
>>>
>>> No. We can agree that the replacement will never end and therefore never all fractions will be indexed, but the fact remains: before the replacement has ended, no O will leave the matrixboard.
>> because you are doing it wrong.
>
> What is done wrong?

you are manipulating the Matrix of fractions.

there is absolutely no need to do any of what you are doing except to intentionally arrive at a false conclusion.

>
>> Just wiki Cantor's Enumeration and use that,
>
> I do: k = (m + n - 1)(m + n - 2)/2 + m

on the wrong matrix.

>
>> and see where your mistakes are.
>
> My "mistake" is to show where the indexes come from. That's the only deviation from Cantor-

intentional mistakes, that fail. There is no need to show where indexes come from.

>
> Regards, WM

Your shell game is kaput.

On 1/19/2022 12:56 PM, WM wrote:
> Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 19:08:11 UTC+1:
>
<snip bad math>

> Regards, WM

Fail.

On 1/19/2022 12:17 PM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 19. Januar 2022 um 12:31:53 UTC+1:
>> onsdag 19 januari 2022 kl. 10:38:59 UTC+1 skrev WM:
>>> Jim Burns schrieb am Mittwoch, 19. Januar 2022 um 02:03:36 UTC+1:
>>>> On 1/18/2022 2:26 PM, WM wrote:
>>>>> Jim Burns schrieb
>>>>> am Dienstag, 18. Januar 2022 um 15:58:27 UTC+1:
>>>>>> Without ending somewhere, they're not the first kind.
>>>>>
>>>>> But without ending all O's will remain.
>>>> If the O on n/d remains,
>>> Every O moves only once and then remains fixed --- in the first column --- above the never empty rest of X's.
>>>> the replacement has ended
>>>> before step i = (n+d-1)*(n+d-2)/2 + n
>>> No. We can agree that the replacement will never end and therefore never all fractions will be indexed, but the fact remains: before the replacement has ended, no O will leave the matrixboard.
>>>
>>> Look, I did nothing else but ask where Cantor's indexes come from. Matheologians never did so and thus remained in ignorance. It is clear that the elements of one column will never flood the whole matrix because for every taken X (necessarily in the finite) there opens a gap receiving an O (necessarily in the finite). It is simply unbelievable that people could believe that this could change!!!
>>>
>> Again, your re-arranging is entirely fuckign IRRELEVANT
>
> Why do you think so?

it is obvious it is irrelevant.

>
> Of course nobody need be interested in my model. But for those beein interested I can state that my model applies Cantor's first diagonal arguments meticulously.

Liar.

> Every fraction is indexed as he did. The only difference is that I ask where the indexes come from.

only because you do not know how to use numbers, and you failed.

>
> Since only transpositions are applied, always exchanging two items, an index and a void card at *finite places*, but never creating indices, it is clear that not all fractions can be indexed. And there is absolutely no mistake in my model because it is so simple. Everybody not being prejudiced by inertia and othodox belief, should easily be able to understand this most important correction of mathematics since Hippasos.
>
> Regards, WM

yoru shell game is uncovered, you manipulated the matrix, to deceive others.

You are a Deceiver.

On Wednesday, January 19, 2022 at 2:32:18 PM UTC-4, WM wrote:
> William schrieb am Mittwoch, 19. Januar 2022 um 16:06:05 UTC+1:
> > On Wednesday, January 19, 2022 at 5:25:44 AM UTC-4, WM wrote:
> > > William schrieb am Mittwoch, 19. Januar 2022 um 00:39:06 UTC+1:
> > > > On Tuesday, January 18, 2022 at 6:08:45 PM UTC-4, WM wrote:
> > >
> > > > > > > But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's
> > > > > > >
> > > > > > > o
> > > > > > > oo
> > > > > > > ooo
> > > > > > > ...
> > > > > > >
> > > > > > > than are o's avaible.
> > > > > > Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available.
> > > > > FISONs have less than aleph_0 o's.
> > > > Nothing you can say will change the fact that here are aleph_0 lines. Thus there are aleph_0 o's available.
> > > OK, if you believe in infinite FISONs,
> > Nope. I do not believe in an infinite FISON. Each line is finite.
> Each line is the union of all preceding lines + o.
Correct. Does not, nothing you say can, change the fact that there are an infinite number of lines.
---
William Hughes

William schrieb am Mittwoch, 19. Januar 2022 um 21:24:37 UTC+1:
> On Wednesday, January 19, 2022 at 2:32:18 PM UTC-4, WM wrote:
> > William schrieb am Mittwoch, 19. Januar 2022 um 16:06:05 UTC+1:
> > > On Wednesday, January 19, 2022 at 5:25:44 AM UTC-4, WM wrote:
> > > > William schrieb am Mittwoch, 19. Januar 2022 um 00:39:06 UTC+1:
> > > > > On Tuesday, January 18, 2022 at 6:08:45 PM UTC-4, WM wrote:
> > > >
> > > > > > > > But ℵo FISONs are not available because of the pigeon hole principle. You cannot distinguish more strings of o's
> > > > > > > >
> > > > > > > > o
> > > > > > > > oo
> > > > > > > > ooo
> > > > > > > > ...
> > > > > > > >
> > > > > > > > than are o's avaible.
> > > > > > > Agreed. There are aleph_0 lines, Thus there are aleph_0 o's available.
> > > > > > FISONs have less than aleph_0 o's.
> > > > > Nothing you can say will change the fact that here are aleph_0 lines. Thus there are aleph_0 o's available.
> > > > OK, if you believe in infinite FISONs,
> > > Nope. I do not believe in an infinite FISON. Each line is finite.
> > Each line is the union of all preceding lines + o.
> Correct. Does not, nothing you say can, change the fact that there are an infinite number of lines.

An infinite number of finite lines. That implies that the infinity is potential.

Regards, WM