On Thursday, December 14, 2023 at 6:08:40 AM UTC-4, WM wrote:
On 13.12.2023 20:41, William wrote:
need dark numbers.
Only producing the putative A shows that "dark" numbers are needed.No. Producing B
does not involve a limit
On 14.12.2023 15:23, William wrote:
On Thursday, December 14, 2023 at 6:08:40 AM UTC-4, WM wrote:
On 13.12.2023 20:41, William wrote:
need dark numbers.
Only producing the putative A shows that "dark" numbers are needed.No. Producing B
does not involve a limit
It involves the realization of all natural numbers
k = (m + n - 1)(m + n - 2)/2 + m
at matrix places
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
This requires a limit or a last step.
On 2023-12-14 15:22:50 +0000, WM said:What proves completeness in your opinion?
This requires a limit or a last step.
You say but don't show.
On 15.12.2023 10:58, Mikko wrote:
On 2023-12-14 15:22:50 +0000, WM said:
This requires a limit or a last step.
You say but don't show.
What proves completeness in your opinion?
On 12/15/2023 7:12 AM, WM wrote:
On 15.12.2023 10:58, Mikko wrote:
On 2023-12-14 15:22:50 +0000, WM said:
This requires a limit or a last step.
You say but don't show.
What proves completeness in your opinion?
Write something which is true of
each one of multiple objects.
For example,
|
| x is the point between the fore- and hindpart
| of a non-empty split F,H of the rationals ℚ
What's hidden in plain sight is:
when those are the points we're discussing,
|
| x is the point between the fore- and hindpart
| of a non-empty split F,H of the rationals ℚ
|
is true for each of what we're discussing.
Complete.
In instances involving infinity,
we do not _accomplish_ completeness.
On 15.12.2023 16:57, Jim Burns wrote:
On 12/15/2023 7:12 AM, WM wrote:
On 15.12.2023 10:58, Mikko wrote:
On 2023-12-14 15:22:50 +0000, WM said:
This requires a limit or a last step.
You say but don't show.
What proves completeness in your opinion?
Write something which is true of
each one of multiple objects.
That does not make it true.
For example,
|
| x is the point between the fore- and hindpart
| of a non-empty split F,H of the rationals ℚ
In instances involving infinity,
we do not _accomplish_ completeness.
That's it!
Fritz Feldhase schrieb am Freitag, 15. Dezember 2023 um 14:36:48 UTC+1:uncountable because eps cannot be made small enough to leave only finitely many uncounted. Therefore every definition of completeness in this respect is nonsense.
On Friday, December 15, 2023 at 1:12:39 PM UTC+1, WM wrote:
What proves completeness [...]?
A DEFINITION of /completeness/ and a PROOF (of a statement referring to completeness).
Completeness in counting the natural numbers cannot be proven because the contrary can be proven. And that is simple: For every eps > 0 there are only finitely many unit fractions in (eps, 1] counted, but infinitely many in (0, eps] uncounted and
Regards, WMSo NOTHING in your logic system can completely talk about the whole of
On 15.12.2023 10:58, Mikko wrote:
On 2023-12-14 15:22:50 +0000, WM said:
What proves completeness in your opinion?This requires a limit or a last step.
You say but don't show.
On 2023-12-15 12:12:36 +0000, WM said:
On 15.12.2023 10:58, Mikko wrote:
On 2023-12-14 15:22:50 +0000, WM said:What proves completeness in your opinion?
This requires a limit or a last step.
You say but don't show.
A proof of completeness. Details depend on what kind of completeness
you mean.
On 12/15/23 11:51 AM, WM wrote:
Fritz Feldhase schrieb am Freitag, 15. Dezember 2023 um 14:36:48 UTC+1:So NOTHING in your logic system can completely talk about the whole of Natural Numbers, so it just can't handle them.
On Friday, December 15, 2023 at 1:12:39 PM UTC+1, WM wrote:
What proves completeness [...]?
A DEFINITION of /completeness/ and a PROOF (of a statement referring to
completeness).
Completeness in counting the natural numbers cannot be proven because the
contrary can be proven. And that is simple: For every eps > 0 there are only >> finitely many unit fractions in (eps, 1] counted, but infinitely many in (0, eps]
uncounted and uncountable because eps cannot be made small enough to leave only
finitely many uncounted. Therefore every definition of completeness in this respect
is nonsense.
Le 16/12/2023 à 02:05, Richard Damon a écrit :
On 12/15/23 11:51 AM, WM wrote:
Fritz Feldhase schrieb am Freitag, 15. Dezember 2023 um 14:36:48 UTC+1: >>>> On Friday, December 15, 2023 at 1:12:39 PM UTC+1, WM wrote:So NOTHING in your logic system can completely talk about the whole of
What proves completeness [...]?
A DEFINITION of /completeness/ and a PROOF (of a statement referring
to completeness).
Completeness in counting the natural numbers cannot be proven because
the contrary can be proven. And that is simple: For every eps > 0
there are only finitely many unit fractions in (eps, 1] counted, but
infinitely many in (0, eps] uncounted and uncountable because eps
cannot be made small enough to leave only finitely many uncounted.
Therefore every definition of completeness in this respect is nonsense.
Natural Numbers, so it just can't handle them.
Neither can yours. XOOO...
XOOO...
XOOO...
XOOO...
..
The indeX transferred from matrix position (k, j) to matrix position (m,
n) implies that, in exchange, O is transferred from matrix position (m,
n) to matrix position (k, j). An O is indicating a not indexed position.
No O will ever get off of the matrix. Hence not all positions will be indeXed.
But you believe that. So you are provably wrong.
Regards, WM
On 12/16/23 7:26 AM, WM wrote:
Le 16/12/2023 à 02:05, Richard Damon a écrit :But that arguement is irrelevent.
On 12/15/23 11:51 AM, WM wrote:
Fritz Feldhase schrieb am Freitag, 15. Dezember 2023 um 14:36:48 UTC+1: >>>>> On Friday, December 15, 2023 at 1:12:39 PM UTC+1, WM wrote:So NOTHING in your logic system can completely talk about the whole of
What proves completeness [...]?
A DEFINITION of /completeness/ and a PROOF (of a statement referring >>>>> to completeness).
Completeness in counting the natural numbers cannot be proven because
the contrary can be proven. And that is simple: For every eps > 0
there are only finitely many unit fractions in (eps, 1] counted, but
infinitely many in (0, eps] uncounted and uncountable because eps
cannot be made small enough to leave only finitely many uncounted.
Therefore every definition of completeness in this respect is nonsense. >>>>
Natural Numbers, so it just can't handle them.
Neither can yours. XOOO...
XOOO...
XOOO...
XOOO...
..
The indeX transferred from matrix position (k, j) to matrix position (m,
n) implies that, in exchange, O is transferred from matrix position (m,
n) to matrix position (k, j). An O is indicating a not indexed position.
No O will ever get off of the matrix. Hence not all positions will be
indeXed.
But you believe that. So you are provably wrong.
There are many ways that fail to index
the set,
but the key is that they are shown equal if there exist An
indexing that matches them.
On 12/15/2023 12:09 PM, WM wrote:
Something which is true of
each one of multiple objects
is
something which is true of
each one of multiple objects.
I'm familiar with contexts in which
it is useful to say of a function that
it doesn't jump.
This is not a fantasy.example,
not Scrooge McDuck's vaults.
How we say _what we intend to say_
which is that a function doesn't jump,
is to say something once which is
true in infinitely.many ways.
In instances involving infinity,
we do not _accomplish_ completeness.
That's it!
Our methods take that into account.
We begin at completeness and
maintain completeness as we learn
by not-first-false augmenting.
We don't accomplish completeness and
we don't need to.
Le 16/12/2023 à 13:48, Richard Damon a écrit :
On 12/16/23 7:26 AM, WM wrote:
Le 16/12/2023 à 02:05, Richard Damon a écrit :But that arguement is irrelevent.
On 12/15/23 11:51 AM, WM wrote:
Fritz Feldhase schrieb am Freitag, 15. Dezember 2023 um 14:36:48So NOTHING in your logic system can completely talk about the whole
UTC+1:
On Friday, December 15, 2023 at 1:12:39 PM UTC+1, WM wrote:
What proves completeness [...]?
A DEFINITION of /completeness/ and a PROOF (of a statement
referring to completeness).
Completeness in counting the natural numbers cannot be proven
because the contrary can be proven. And that is simple: For every
eps > 0 there are only finitely many unit fractions in (eps, 1]
counted, but infinitely many in (0, eps] uncounted and uncountable
because eps cannot be made small enough to leave only finitely many
uncounted. Therefore every definition of completeness in this
respect is nonsense.
of Natural Numbers, so it just can't handle them.
Neither can yours. XOOO...
XOOO...
XOOO...
XOOO...
..
The indeX transferred from matrix position (k, j) to matrix position
(m, n) implies that, in exchange, O is transferred from matrix
position (m, n) to matrix position (k, j). An O is indicating a not
indexed position. No O will ever get off of the matrix. Hence not all
positions will be indeXed.
But you believe that. So you are provably wrong.
No,it is exactly Cantor's way. Only one step is inserted: The first
column is put in bojection with the natural numbers: 1/n <--> n. But if
that is not posible, then all is impossible.
There are many ways that fail to index the set,
There are only ways that fail. Otherwise every injection would be a bijection.
but the key is that they are shown equal if there exist An indexing
that matches them.
It is hard to understand how Cantor could convince the matheologians of
this silly idea. But in order to help you I did use Cantor's formula k =
(m + n - 1)(m + n - 2)/2 + m with the resulting sequence 1/1, 1/2, 2/1,
1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5,
3/4, 4/3, 5/2, 6/1, ...
Regards, WM
Le 15/12/2023 à 19:31, Jim Burns a écrit :
On 12/15/2023 12:09 PM, WM wrote:
In instances involving infinity,
we do not _accomplish_ completeness.
That's it!
Our methods take that into account.
We begin at completeness and
maintain completeness as we learn
by not-first-false augmenting.
The set of O remains completely
in the matrix.
We don't accomplish completeness and
we don't need to.
A bijection [would] require it.
Something which is true of
each one of multiple objects
is
something which is true of
each one of multiple objects.
Never an O leaves the matrix.
Le 16/12/2023 à 09:07, Mikko a écrit :
On 2023-12-15 12:12:36 +0000, WM said:
On 15.12.2023 10:58, Mikko wrote:
On 2023-12-14 15:22:50 +0000, WM said:What proves completeness in your opinion?
This requires a limit or a last step.
You say but don't show.
A proof of completeness. Details depend on what kind of completeness
you mean.
You "prove" that Cantor enumerates all fractions.
That is wrong. The matrix
XOOO...
XOOO...
XOOO...
XOOO...
..will never be covered by indeXes.
The indeX transferred from matrix position (k, j) to matrix position (m, n)
implies that,
On 2023-12-16 12:18:21 +0000, WM said:
That is wrong. The matrix
XOOO...
XOOO...
XOOO...
XOOO...
..will never be covered by indeXes.
Yes it is. Every position in the matrix has a row index and a column
indes. Positions not covered are not part of the matrix.
The indeX transferred from matrix position (k, j) to matrix position (m, n)
What does "transferred" mean?
On 12/16/2023 8:27 AM, WM wrote:
Never an O leaves the matrix.
The O's are overwritten by X's from ⟨k,1⟩
Yes, you can biject n <--> 1/n, you can also biject k <--> n/m with a different mapping, This is NOT a "Contradiction"
The fact that the first mapping missed an infinite number of terms of
the second is irrelvent.
Le 16/12/2023 à 15:25, Richard Damon a écrit :
Yes, you can biject n <--> 1/n, you can also biject k <--> n/m with a
different mapping, This is NOT a "Contradiction"
The fact that the first mapping missed an infinite number of terms of
the second is irrelvent.
Not in mathematics!
But the fact that the second mapping misses an infinite number of terms, proved by the eternal presence of O, proves that Cantor's way fails.
Regards, WM
Yes, in mathematics of infinite sets, the fact that one mapping misses
even an infinite number of members means nothing,
Le 17/12/2023 à 21:23, Richard Damon a écrit :
Yes, in mathematics of infinite sets, the fact that one mapping misses
even an infinite number of members means nothing,
But here Cantor's mapping misses an infinite number of fractions.
Regards, WM
Le 17/12/2023 à 09:53, Mikko a écrit :
On 2023-12-16 12:18:21 +0000, WM said:
That is wrong. The matrix
XOOO...
XOOO...
XOOO...
XOOO...
..will never be covered by indeXes.
Yes it is. Every position in the matrix has a row index and a column
indes. Positions not covered are not part of the matrix.
There are not enough indices.
The indeX transferred from matrix position (k, j) to matrix position (m, n) >>What does "transferred" mean?
If (m, n) is to be indexed according to Cantors formula, then the indeX
is taken from its position (k, j).
On 12/17/23 4:18 PM, WM wrote:
Le 17/12/2023 à 21:23, Richard Damon a écrit :
Yes, in mathematics of infinite sets, the fact that one mapping misses
even an infinite number of members means nothing,
But here Cantor's mapping misses an infinite number of fractions.
You aren't using the right mapping, showing your ignorance.
On 2023-12-17 19:36:59 +0000, WM said:
The indeX transferred from matrix position (k, j) to matrix position (m, n)
What does "transferred" mean?
If (m, n) is to be indexed according to Cantors formula, then the indeX
is taken from its position (k, j).
How does that imply anything?
Le 18/12/2023 à 01:09, Richard Damon a écrit :
On 12/17/23 4:18 PM, WM wrote:
Le 17/12/2023 à 21:23, Richard Damon a écrit :
Yes, in mathematics of infinite sets, the fact that one mapping
misses even an infinite number of members means nothing,
But here Cantor's mapping misses an infinite number of fractions.
You aren't using the right mapping, showing your ignorance.
Where does my mapping deviate from Cantor's
k = (m + n - 1)(m + n - 2)/2 + m
with the result
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
for the first time?
Regards, WM
Le 18/12/2023 à 10:44, Mikko a écrit :
On 2023-12-17 19:36:59 +0000, WM said:
The indices are distributet according to Cantor's formulaThe indeX transferred from matrix position (k, j) to matrix
position (m, n)
What does "transferred" mean?
If (m, n) is to be indexed according to Cantors formula, then the
indeX is taken from its position (k, j).
How does that imply anything?
k = (m + n - 1)(m + n - 2)/2 + m
resulting in
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
but where the indices are removed, there are no indices.
Index 1 remains at fraction 1/1, the first term of the sequence.
1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
..
The next term, 1/2, is indexed with 2 which is taken from its initial position 2/1
XXOO...
OOOO...
XOOO...
XOOO...
XOOO...
..
Then index 3 is taken from its initial position 3/1 and is attached to 2/1
XXOO...
XOOO...
OOOO...
XOOO...
XOOO...
..
Then index 4 is taken from its initial position 4/1 and is attached to 1/3
XXXO...
XOOO...
OOOO...
OOOO...
XOOO...
..
Then index 5 is taken from its initial position 5/1 and is attached to 2/2
XXXO...
XXOO...
OOOO...
OOOO...
OOOO...
..
The O will remain forever, indicating not indexed positions.
Regards, WM
Le 16/12/2023 à 20:27, Jim Burns a écrit :
On 12/16/2023 8:27 AM, WM wrote:
Never an O leaves the matrix.
The O's are overwritten by X's from ⟨k,1⟩
But they remain in the matrix.
The set of positione covered by O
minus
the set of positions covered by X
is infinite and will never decrease.
Le 18/12/2023 à 10:44, Mikko a écrit :
On 2023-12-17 19:36:59 +0000, WM said:
The indices are distributet according to Cantor's formulaThe indeX transferred from matrix position (k, j) to matrix position (m, n)
What does "transferred" mean?
If (m, n) is to be indexed according to Cantors formula, then the indeX
is taken from its position (k, j).
How does that imply anything?
k = (m + n - 1)(m + n - 2)/2 + m
resulting in
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
but where the indices are removed, there are no indices.
Infinity minus infinity
is not an operation.
An operation needs
no less than one result and
no more than one result.
On 2023-12-18 10:52:59 +0000, WM said:
Le 18/12/2023 à 10:44, Mikko a écrit :
On 2023-12-17 19:36:59 +0000, WM said:
The indeX transferred from matrix position (k, j) to matrix position (m, n)
What does "transferred" mean?
If (m, n) is to be indexed according to Cantors formula, then the indeX >>>> is taken from its position (k, j).
How does that imply anything?
The indices are distributet according to Cantor's formula
k = (m + n - 1)(m + n - 2)/2 + m
resulting in
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
but where the indices are removed, there are no indices.
If any of the indices is removed you no longer have a matrix.
On 12/18/23 5:47 AM, WM wrote:
Le 18/12/2023 à 01:09, Richard Damon a écrit :
On 12/17/23 4:18 PM, WM wrote:
Le 17/12/2023 à 21:23, Richard Damon a écrit :
Yes, in mathematics of infinite sets, the fact that one mapping
misses even an infinite number of members means nothing,
But here Cantor's mapping misses an infinite number of fractions.
You aren't using the right mapping, showing your ignorance.
Where does my mapping deviate from Cantor's
k = (m + n - 1)(m + n - 2)/2 + m
with the result
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
for the first time?
Which shows that your matrix B having values like
1/1 2/1 3/1 ...
1/2 2/2 3/2 ...
1/3 2/3 3/3 ...
.
.
.
is exactly indexed by a matrix form like:
1 3 6 10
2 5 9
4 8 13
7 12
11
Where the number at the m,n location of your fractions is indexed by the number at the m,n location of the Natural Numbers organized in that fashion.
You seem to think that just because you can make a pattern that doesn't
cover it (like trying to go all the way down first),
it means you can't
cover it.
On 12/18/23 5:52 AM, WM wrote:
The O will remain forever, indicating not indexed positions.And the problem is you aren't supposed to be taking the indexes out of
B,
Le 18/12/2023 à 17:06, Richard Damon a écrit :
On 12/18/23 5:52 AM, WM wrote:
The O will remain forever, indicating not indexed positions.And the problem is you aren't supposed to be taking the indexes out of B,
What forbids to index the first column first? Or what forbids to use the integer fractions n/1 for indexing?
Regards, WM
On 12/19/23 1:54 PM, WM wrote:
What forbids to index the first column first? Or what forbids to use the
integer fractions n/1 for indexing?
Nothing "forbids" it, but the rule is that if ANY indexing works, then
we have established that they are equal sizes.
The fact that some attempts doesn't cover is irrelevant.
You can't disprove that a mapping exists by showing that one mapping fails.
Le 19/12/2023 à 10:58, Mikko a écrit :
On 2023-12-18 10:52:59 +0000, WM said:
If any of the indices is removed you no longer have a matrix.
The matrix remains what it is.
Le 19/12/2023 à 21:07, Richard Damon a écrit :
On 12/19/23 1:54 PM, WM wrote:
What forbids to index the first column first? Or what forbids to use
the integer fractions n/1 for indexing?
Nothing "forbids" it, but the rule is that if ANY indexing works, then
we have established that they are equal sizes.
First, this is a stupid rule. But second, my proof concerns the most
famous and praised indexing and shows that this very mapping does not work.
The fact that some attempts doesn't cover is irrelevant.
You can't disprove that a mapping exists by showing that one mapping
fails.
I have shown that Cantor's mapping fails. And I can show that every
other mapping will fail too.
Regards, WM
On 2023-12-19 12:00:46 +0000, WM said:
If any of the indices is removed you no longer have a matrix.
The matrix remains what it is.
You can't change the matrix so that it remanis as it was.
If it remains as it was it is not changed.
On 12/20/23 5:51 AM, WM wrote:
I have shown that Cantor's mapping fails. And I can show that every
other mapping will fail too.
Nope, you have shown that a mapping that has some surface simulatity to Cantor's fails.
Cantor showed a mapping of k to m/n not of m/1 to m/n.
Cantor maps the NATURAL NUMBERS (not the intergral rationals) to the rationals.
Cantor maps the NATURAL NUMBERS (not the intergral rationals) to the rationals.
Le 20/12/2023 à 14:17, Richard Damon a écrit :
On 12/20/23 5:51 AM, WM wrote:
I have shown that Cantor's mapping fails. And I can show that every
other mapping will fail too.
Nope, you have shown that a mapping that has some surface simulatity
to Cantor's fails.
What is the difference?
Cantor showed a mapping of k to m/n not of m/1 to m/n.
I first use the bijection between integers and integer fractions k <-->
k/1. If that fails, then every mapping between infinite sets fails.
Cantor maps the NATURAL NUMBERS (not the intergral rationals) to the
rationals.
So do I.
Regards, WM
On 12/20/23 14:17, Richard Damon wrote:
Cantor maps the NATURAL NUMBERS (not the intergral rationals) to the
rationals.
I am missing the start of this conversation. Aren't integral rationals exactly the integers? Are there any integers which are not rational?
Le 20/12/2023 à 13:05, Mikko a écrit :
On 2023-12-19 12:00:46 +0000, WM said:
If any of the indices is removed you no longer have a matrix.
The matrix remains what it is.
You can't change the matrix so that it remanis as it was.
If it remains as it was it is not changed.
The matrix remains, only the covering by X changes.
On 12/20/23 14:17, Richard Damon wrote:
Cantor maps the NATURAL NUMBERS (not the intergral rationals) to the
rationals.
I am missing the start of this conversation. Aren't integral rationals exactly the integers?
On 2023-12-20 16:28:09 +0000, WM said:
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex.
When it "changes" it does not really change, there just is another "covering", and after successive changes a sequence of such "coverings".
Le 21/12/2023 à 10:41, Mikko a écrit :
On 2023-12-20 16:28:09 +0000, WM said:
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex.
It has the same (m, n), but not all positions have been counted by k
k = (m + n - 1)(m + n - 2)/2 + m.
Cantor described a function from the SET of natural numbers and the SET of rational numbers,
whereas OBP 'constructs' a few 'mappings',
each between some particular natural number and some rational number.
On 2023-12-21 10:57:10 +0000, WM said:
Le 21/12/2023 à 10:41, Mikko a écrit :
On 2023-12-20 16:28:09 +0000, WM said:
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex.
It has the same (m, n), but not all positions have been counted by k
k = (m + n - 1)(m + n - 2)/2 + m.
Positions are not covered by. Some of them are covered by X.
Each position as m and n so each position has k.
Le 21/12/2023 à 10:41, Mikko a écrit :
On 2023-12-20 16:28:09 +0000, WM said:
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex.
It has the same (m, n), but not all positions have been counted by k
k = (m + n - 1)(m + n - 2)/2 + m.
XOOO...
XOOO...
XOOO...
XOOO...
..
When it "changes" it does not really change, there just is another
"covering", and after successive changes a sequence of such "coverings".
And none covers the whole matrix by indeXes.
Regards, WM
Le 20/12/2023 à 19:04, immibis a écrit :
On 12/20/23 14:17, Richard Damon wrote:
Cantor maps the NATURAL NUMBERS (not the intergral rationals) to the
rationals.
I am missing the start of this conversation. Aren't integral rationals
exactly the integers?
Of course. Therefore it is irrelevant whether we start with the first
column 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, ...
5/1, 5/2, 5/3, 5/4, ...
..
or first biject the integer fractions with the natural numbers
1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
..
In no case it is possible to cover the whole matrix by indeXes
XOOO...
XOOO...
XOOO...
XOOO...
..
Regards, WM
Le 21/12/2023 à 12:48, Mikko a écrit :
On 2023-12-21 10:57:10 +0000, WM said:
Le 21/12/2023 à 10:41, Mikko a écrit :
On 2023-12-20 16:28:09 +0000, WM said:
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex.
It has the same (m, n), but not all positions have been counted by k
k = (m + n - 1)(m + n - 2)/2 + m.
Positions are not covered by. Some of them are covered by X.
Each position as m and n so each position has k.
That is an error. Each O remains in the matrix at a not indexed place.
Le 19/12/2023 à 21:07, Richard Damon a écrit :
On 12/19/23 1:54 PM, WM wrote:
What forbids
to index the first column first?
Or what forbids
to use the integer fractions n/1 for indexing?
Nothing "forbids" it,
but the rule is that
if ANY indexing works,
then we have established that
they are equal sizes.
First, this is a stupid rule.
On 12/21/23 6:05 AM, WM wrote:
Le 20/12/2023 à 19:04, immibis a écrit :
On 12/20/23 14:17, Richard Damon wrote:
Cantor maps the NATURAL NUMBERS (not the intergral rationals) to the
rationals.
I am missing the start of this conversation. Aren't integral rationals
exactly the integers?
Of course. Therefore it is irrelevant whether we start with the first
column 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, ...
5/1, 5/2, 5/3, 5/4, ...
..
or first biject the integer fractions with the natural numbers
1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
..
In no case it is possible to cover the whole matrix by indeXes
XOOO...
XOOO...
XOOO...
XOOO...
..
IF you are trying to say that there isn't a mapping for the values in
the first column to the whole set, you are just wrong and the mapping
has been given.
If you are arguing that the shape of the first column doesn't ever cover
the matrix, that is just an irrelevent stupidity, as no one says it should.
We cover the matrix:
1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
...
with values mappes as following:
1 2 4 7 11 16 22
3 5 8 12 17 23
6 9 13 18 24
10 14 19 25
15 20 26
21 26
27
Note, all of these *VALUES* can be found in the first column of the rationals, but that is irrelevant, as we are showing a mapping of
indexes from the Natural Numbers to rational matrix.
On 2023-12-21 11:56:10 +0000, WM said:
Le 21/12/2023 à 12:48, Mikko a écrit :
On 2023-12-21 10:57:10 +0000, WM said:
Le 21/12/2023 à 10:41, Mikko a écrit :
On 2023-12-20 16:28:09 +0000, WM said:
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex.
It has the same (m, n), but not all positions have been counted by k
k = (m + n - 1)(m + n - 2)/2 + m.
Positions are not covered by. Some of them are covered by X.
Each position as m and n so each position has k.
That is an error. Each O remains in the matrix at a not indexed place.
By the definition of "matrix" there are only indexed places.
On 12/21/23 5:57 AM, WM wrote:
Le 21/12/2023 à 10:41, Mikko a écrit :So, you just don't understand what the bijection is doing.
On 2023-12-20 16:28:09 +0000, WM said:
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex.
It has the same (m, n), but not all positions have been counted by k
k = (m + n - 1)(m + n - 2)/2 + m.
XOOO...
XOOO...
XOOO...
XOOO...
..
When it "changes" it does not really change, there just is another
"covering", and after successive changes a sequence of such "coverings".
And none covers the whole matrix by indeXes.
On 12/20/2023 5:51 AM, WM wrote:
Le 19/12/2023 à 21:07, Richard Damon a écrit :
On 12/19/23 1:54 PM, WM wrote:
What forbids
to index the first column first?
Or what forbids
to use the integer fractions n/1 for indexing?
Nothing "forbids" it,
but the rule is that
if ANY indexing works,
then we have established that
they are equal sizes.
First, this is a stupid rule.
The rule is the same for finite and infinite sets.
What you (WM) are probably calling stupid
isn't so much a _rule_ rule
as it is the _observation_ that
the flock-of-sheep rule isn't correct
for all sets.
Le 21/12/2023 à 16:16, Richard Damon a écrit :
On 12/21/23 6:05 AM, WM wrote:
Le 20/12/2023 à 19:04, immibis a écrit :
On 12/20/23 14:17, Richard Damon wrote:
Cantor maps the NATURAL NUMBERS (not the intergral rationals) to
the rationals.
I am missing the start of this conversation. Aren't integral
rationals exactly the integers?
Of course. Therefore it is irrelevant whether we start with the first
column 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, ...
5/1, 5/2, 5/3, 5/4, ...
..
or first biject the integer fractions with the natural numbers
1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
..
In no case it is possible to cover the whole matrix by indeXes
XOOO...
XOOO...
XOOO...
XOOO...
..
IF you are trying to say that there isn't a mapping for the values in
the first column to the whole set, you are just wrong and the mapping
has been given.
It appears so, but the O indicate not indeXed fractions. This has to be
taken into account.
If you are arguing that the shape of the first column doesn't ever
cover the matrix, that is just an irrelevent stupidity, as no one says
it should.
That is Cantor's claim. All indexes have to be appßlied. They can be
taken from the first column.
We cover the matrix:
1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
...
with values mappes as following:
1 2 4 7 11 16 22
3 5 8 12 17 23
6 9 13 18 24
10 14 19 25
15 20 26
21 26
27
Note, all of these *VALUES* can be found in the first column of the
rationals, but that is irrelevant, as we are showing a mapping of
indexes from the Natural Numbers to rational matrix.
You are mistaken. All your indeXes up to every indeX applied in your
mapping belong to a finite set {1, 2, 3, ..., X} which has ℵo
successors. My proof shows that the O, indicating not indexed fractions,
will remain forever within the matrix, being even the majority.
Regards, WM
Le 21/12/2023 à 16:18, Richard Damon a écrit :
On 12/21/23 5:57 AM, WM wrote:
Le 21/12/2023 à 10:41, Mikko a écrit :So, you just don't understand what the bijection is doing.
On 2023-12-20 16:28:09 +0000, WM said:
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex.
It has the same (m, n), but not all positions have been counted by k
k = (m + n - 1)(m + n - 2)/2 + m.
XOOO...
XOOO...
XOOO...
XOOO...
..
When it "changes" it does not really change, there just is another
"covering", and after successive changes a sequence of such
"coverings".
And none covers the whole matrix by indeXes.
A bijection should index every element of the matrix, but it does not, because the O remain. Therefore it is not a bijection.
Regards, WM
Le 21/12/2023 à 17:45, Mikko a écrit :
On 2023-12-21 11:56:10 +0000, WM said:
Le 21/12/2023 à 12:48, Mikko a écrit :
On 2023-12-21 10:57:10 +0000, WM said:
Le 21/12/2023 à 10:41, Mikko a écrit :
On 2023-12-20 16:28:09 +0000, WM said:
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex.
It has the same (m, n), but not all positions have been counted by k >>>>> k = (m + n - 1)(m + n - 2)/2 + m.
Positions are not covered by. Some of them are covered by X.
Each position as m and n so each position has k.
That is an error. Each O remains in the matrix at a not indexed place.
By the definition of "matrix" there are only indexed places.
But not indexed by Cantor's k = (m + n - 1)(m + n - 2)/2 + m.
Le 21/12/2023 à 17:45, Mikko a écrit :
On 2023-12-21 11:56:10 +0000, WM said:
Le 21/12/2023 à 12:48, Mikko a écrit :
On 2023-12-21 10:57:10 +0000, WM said:
Le 21/12/2023 à 10:41, Mikko a écrit :
On 2023-12-20 16:28:09 +0000, WM said:
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex.
It has the same (m, n), but not all positions have been counted by k >>>>> k = (m + n - 1)(m + n - 2)/2 + m.
Positions are not covered by. Some of them are covered by X.
Each position as m and n so each position has k.
That is an error. Each O remains in the matrix at a not indexed place.
By the definition of "matrix" there are only indexed places.
But not indexed by Cantor's k = (m + n - 1)(m + n - 2)/2 + m.
Regards, WM
On 12/22/23 09:13, WM wrote:
Le 21/12/2023 à 19:18, Jim Burns a écrit :
On 12/20/2023 5:51 AM, WM wrote:
Le 19/12/2023 à 21:07, Richard Damon a écrit :
On 12/19/23 1:54 PM, WM wrote:
What forbids
to index the first column first?
Or what forbids
to use the integer fractions n/1 for indexing?
Nothing "forbids" it,
but the rule is that
if ANY indexing works,
then we have established that
they are equal sizes.
First, this is a stupid rule.
The rule is the same for finite and infinite sets.
No. For finite sets of equal numerosity any injection is a surjection
and any surjection is an injection.
Wrong rule. If any injection between two sets exists, they have equal sizes.
Le 21/12/2023 à 19:18, Jim Burns a écrit :
On 12/20/2023 5:51 AM, WM wrote:
Le 19/12/2023 à 21:07, Richard Damon a écrit :
On 12/19/23 1:54 PM, WM wrote:
What forbids
to index the first column first?
Or what forbids
to use the integer fractions n/1 for indexing?
Nothing "forbids" it,
but the rule is that
if ANY indexing works,
then we have established that
they are equal sizes.
First, this is a stupid rule.
The rule is the same for finite and infinite sets.
No. For finite sets of equal numerosity any injection is a surjection
and any surjection is an injection.
On 12/22/23 09:09, WM wrote:
Le 21/12/2023 à 17:45, Mikko a écrit :All of them are indexed by Cantor
On 2023-12-21 11:56:10 +0000, WM said:
Le 21/12/2023 à 12:48, Mikko a écrit :
On 2023-12-21 10:57:10 +0000, WM said:
Le 21/12/2023 à 10:41, Mikko a écrit :
On 2023-12-20 16:28:09 +0000, WM said:It has the same (m, n), but not all positions have been counted by k >>>>>> k = (m + n - 1)(m + n - 2)/2 + m.
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex. >>>>>>
Positions are not covered by. Some of them are covered by X.
Each position as m and n so each position has k.
That is an error. Each O remains in the matrix at a not indexed place.
By the definition of "matrix" there are only indexed places.
But not indexed by Cantor's k = (m + n - 1)(m + n - 2)/2 + m.
No, you are showing you can create a non-covering mapping,
but the rule
says that if at least ONE covering mapping exists, we meet the requirement.
On 2023-12-22 08:09:14 +0000, WM said:
Le 21/12/2023 à 17:45, Mikko a écrit :
On 2023-12-21 11:56:10 +0000, WM said:
Le 21/12/2023 à 12:48, Mikko a écrit :
On 2023-12-21 10:57:10 +0000, WM said:
Le 21/12/2023 à 10:41, Mikko a écrit :
On 2023-12-20 16:28:09 +0000, WM said:It has the same (m, n), but not all positions have been counted by k >>>>>> k = (m + n - 1)(m + n - 2)/2 + m.
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex. >>>>>>
Positions are not covered by. Some of them are covered by X.
Each position as m and n so each position has k.
That is an error. Each O remains in the matrix at a not indexed place.
By the definition of "matrix" there are only indexed places.
But not indexed by Cantor's k = (m + n - 1)(m + n - 2)/2 + m.
It is indexed by two natural numbers because the matrices meintioned
above and earlier in this discussion are matrices that are indexed by
two natural numbers.
You have called these two numbers m and n.
For each natural number k there is only one pair of natural numbers
m and n so that k = (m + n - 1)(m + n - 2)/2 + m. Therefore k is
sufficient to identify a position in any of the matrices.
On 12/22/23 3:07 AM, WM wrote:
Le 21/12/2023 à 16:18, Richard Damon a écrit :So, you are showning that the bijecton for k to k/1 doesn't cover all of
On 12/21/23 5:57 AM, WM wrote:
Le 21/12/2023 à 10:41, Mikko a écrit :So, you just don't understand what the bijection is doing.
On 2023-12-20 16:28:09 +0000, WM said:
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex.
It has the same (m, n), but not all positions have been counted by k
k = (m + n - 1)(m + n - 2)/2 + m.
XOOO...
XOOO...
XOOO...
XOOO...
..
When it "changes" it does not really change, there just is another
"covering", and after successive changes a sequence of such
"coverings".
And none covers the whole matrix by indeXes.
A bijection should index every element of the matrix, but it does not,
because the O remain. Therefore it is not a bijection.
m/n, but that doesn't show that the bijecton for k to m/n where
k = (m + n - 1)(m + n - 2)/2 + m doesn't work.
Le 22/12/2023 à 15:05, Richard Damon a écrit :
No, you are showing you can create a non-covering mapping,
I show that Cantor's mapping is not covering the matrix.
but the rule says that if at least ONE covering mapping exists, we
meet the requirement.
Who made that rule?
Regards, WM
Le 22/12/2023 à 16:15, immibis a écrit :
On 12/22/23 09:13, WM wrote:
Le 21/12/2023 à 19:18, Jim Burns a écrit :
On 12/20/2023 5:51 AM, WM wrote:
Le 19/12/2023 à 21:07, Richard Damon a écrit :
On 12/19/23 1:54 PM, WM wrote:
What forbids
to index the first column first?
Or what forbids
to use the integer fractions n/1 for indexing?
Nothing "forbids" it,
but the rule is that
if ANY indexing works,
then we have established that
they are equal sizes.
First, this is a stupid rule.
The rule is the same for finite and infinite sets.
No. For finite sets of equal numerosity any injection is a surjection
and any surjection is an injection.
Wrong rule. If any injection between two sets exists, they have equal
sizes.
Yes, that is awrong rule. See the injection of natnumbers into the reals.
Regards, WM
Le 22/12/2023 à 15:19, Mikko a écrit :
On 2023-12-22 08:09:14 +0000, WM said:
Le 21/12/2023 à 17:45, Mikko a écrit :
On 2023-12-21 11:56:10 +0000, WM said:
Le 21/12/2023 à 12:48, Mikko a écrit :By the definition of "matrix" there are only indexed places.
On 2023-12-21 10:57:10 +0000, WM said:
Le 21/12/2023 à 10:41, Mikko a écrit :
On 2023-12-20 16:28:09 +0000, WM said:It has the same (m, n), but not all positions have been counted by k >>>>>>> k = (m + n - 1)(m + n - 2)/2 + m.
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex. >>>>>>>
Positions are not covered by. Some of them are covered by X.
Each position as m and n so each position has k.
That is an error. Each O remains in the matrix at a not indexed place. >>>>
But not indexed by Cantor's k = (m + n - 1)(m + n - 2)/2 + m.
It is indexed by two natural numbers because the matrices meintioned
above and earlier in this discussion are matrices that are indexed by
two natural numbers.
As the O, not indexed by Cantor prove, most of the matrix' (m, n) are dark. Only the upper left corner contains visible m and n.
You have called these two numbers m and n.
For each natural number k there is only one pair of natural numbers
m and n so that k = (m + n - 1)(m + n - 2)/2 + m. Therefore k is
sufficient to identify a position in any of the matrices.
The remaining O prove the contrary.
Regards, WM
Le 22/12/2023 à 15:08, Richard Damon a écrit :
On 12/22/23 3:07 AM, WM wrote:
Le 21/12/2023 à 16:18, Richard Damon a écrit :So, you are showning that the bijecton for k to k/1 doesn't cover all
On 12/21/23 5:57 AM, WM wrote:
Le 21/12/2023 à 10:41, Mikko a écrit :So, you just don't understand what the bijection is doing.
On 2023-12-20 16:28:09 +0000, WM said:
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex.
It has the same (m, n), but not all positions have been counted by k >>>>> k = (m + n - 1)(m + n - 2)/2 + m.
XOOO...
XOOO...
XOOO...
XOOO...
..
When it "changes" it does not really change, there just is another >>>>>> "covering", and after successive changes a sequence of such
"coverings".
And none covers the whole matrix by indeXes.
A bijection should index every element of the matrix, but it does
not, because the O remain. Therefore it is not a bijection.
of m/n, but that doesn't show that the bijecton for k to m/n where
k = (m + n - 1)(m + n - 2)/2 + m doesn't work.
If the first bijection k to k/1 is a bijection, then I apply in the
second stepCantor's procedure. If not, then there are no infinite
bijections at all.
Regards, WM
Le 21/12/2023 à 19:18, Jim Burns a écrit :
On 12/20/2023 5:51 AM, WM wrote:
First, this is a stupid rule.
What you (WM) are probably calling stupid
isn't so much a _rule_ rule
as it is the _observation_ that
the flock-of-sheep rule isn't correct
for all sets.
That is not an observation
That is not an observation
but an assumption
of people who deny or are unable to
understand actual infinity.
Le 22/12/2023 à 15:19, Mikko a écrit :
On 2023-12-22 08:09:14 +0000, WM said:
Le 21/12/2023 à 17:45, Mikko a écrit :
On 2023-12-21 11:56:10 +0000, WM said:
Le 21/12/2023 à 12:48, Mikko a écrit :By the definition of "matrix" there are only indexed places.
On 2023-12-21 10:57:10 +0000, WM said:
Le 21/12/2023 à 10:41, Mikko a écrit :
On 2023-12-20 16:28:09 +0000, WM said:It has the same (m, n), but not all positions have been counted by k >>>>>>> k = (m + n - 1)(m + n - 2)/2 + m.
The matrix remains, only the covering by X changes.
Your first "covering" is just another matrix with the same idicex. >>>>>>>
Positions are not covered by. Some of them are covered by X.
Each position as m and n so each position has k.
That is an error. Each O remains in the matrix at a not indexed place. >>>>
But not indexed by Cantor's k = (m + n - 1)(m + n - 2)/2 + m.
It is indexed by two natural numbers because the matrices meintioned
above and earlier in this discussion are matrices that are indexed by
two natural numbers.
As the O, not indexed by Cantor prove, most of the matrix' (m, n) are dark. Only the upper left corner contains visible m and n.
On 12/22/23 11:23 AM, WM wrote:
but the rule says that if at least ONE covering mapping exists, we
meet the requirement.
Who made that rule?
It wasn't "Made" but "Discovered".
It is a natural consequence of the
need for a consistent set of logical rules to describe infinite set.
𝕍 is the least.upper.bound of FISONs ⟨1,…,n⟩
Your explanation is that A contains elements
which "cannot be said" to self-equal.
Le 22/12/2023 à 17:59, Richard Damon a écrit :
On 12/22/23 11:23 AM, WM wrote:
but the rule says that if at least ONE covering mapping exists, we
meet the requirement.
Who made that rule?
It wasn't "Made" but "Discovered".
It was "discovered" that there are as many fractions as prime numbers.
No it is believed by matheologians who cannot think from 1 to 3. (They
would discover that between every pair of natural numebers there are infinitely,many fraction.
It is a natural consequence of the need for a consistent set of
logical rules to describe infinite set.
That is not consistent but foolish.
Regards, WM
Le 23/12/2023 à 00:06, Jim Burns a écrit :
On 12/22/2023 3:13 AM, WM wrote:
Le 21/12/2023 à 19:18, Jim Burns a écrit :
On 12/20/2023 5:51 AM, WM wrote:
First, this is a stupid rule.
What you (WM) are probably calling stupid
isn't so much a _rule_ rule
as it is the _observation_ that
the flock-of-sheep rule isn't correct
for all sets.
That is not an observation
𝕍 is the least.upper.bound of FISONs ⟨1,…,n⟩
Does not exist.
The collection of FISONs is
potentially infinite.
First, this is a stupid rule.
What you (WM) are probably calling stupid
isn't so much a _rule_ rule
as it is the _observation_ that
the flock-of-sheep rule isn't correct
for all sets.
That is not an observation
𝕍 is the least.upper.bound of FISONs ⟨1,…,n⟩
Does not exist.
𝕍 is the least.upper.bound of FISONs ⟨1,…,n⟩
Does not exist.
Your explanation is that A contains elements
which "cannot be said" to self-equal.
Which cannot be *proved* to be anything
because they cannot be chosen as individuals.
But I assume that they all are natural numbers,
all have unique prime decompositions and
all are equal to themselves.
You don't seem to understand basic set theory.
Each claim in such a sequence is true.
Le 23/12/2023 à 14:44, Richard Damon a écrit :
You don't seem to understand basic set theory.
Teach me. Does NUF(x) increase from NUF(0) = 0 to NUF(x>0) > 0 in steps
of height 1 or not?
Regards, WM
0, as your logic of NUF(x) being 1 at the "lowest" Unit Fraction is aflawed assumption, YOU need to prove the existance of such a point, you
Le 23/12/2023 à 18:47, Jim Burns a écrit :
Each claim in such a sequence is true.
Does NUF(x) increase
from NUF(0) = 0 to NUF(x>0) > 0
in steps [of] height 1 or not?
Does NUF(x) increase
from NUF(0) = 0 to NUF(x>0) > 0
in steps [of] height 1 or not?
On 12/24/23 3:31 AM, WM wrote:
Le 23/12/2023 à 14:44, Richard Damon a écrit :
You don't seem to understand basic set theory.
Teach me. Does NUF(x) increase from NUF(0) = 0 to NUF(x>0) > 0 in steps
of height 1 or not?
Not at finite numbers x.
The "points" where NUF(x) might increase in units of height of 1, would
be at the infintesimals, with a proper definition of NUF to handle them.
Otherwise, it is just discontinuous,
because it doesn't take into
account unbounded nature of the unit fractions.
your logic of NUF(x) being 1 at the "lowest" Unit Fraction is a
flawed assumption,
You are just showing your ignorance of how mathematics actually works,
Le 24/12/2023 à 18:11, Richard Damon a écrit :
On 12/24/23 3:31 AM, WM wrote:
Le 23/12/2023 à 14:44, Richard Damon a écrit :
You don't seem to understand basic set theory.
Teach me. Does NUF(x) increase from NUF(0) = 0 to NUF(x>0) > 0 in
steps of height 1 or not?
Not at finite numbers x.
It can increase only at unit fractions. They all are finite numbers.
The "points" where NUF(x) might increase in units of height of 1,
would be at the infintesimals, with a proper definition of NUF to
handle them.
Unit fractions are not infinitesimals because natnumbers are not infinite.
Otherwise, it is just discontinuous,
Not ain agreement with mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
because it doesn't take into account unbounded nature of the unit
fractions.
Mathjematics shows that the unit fractions appear unbounded only because
the dark unit fractions cannot ne seen.
your logic of NUF(x) being 1 at the "lowest" Unit Fraction is a flawed
assumption,
It is simple and basic mathematics, valid for all unit fractions: n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
You are just showing your ignorance of how mathematics actually works,
Not like this? n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
Regards, WM
Le 24/12/2023 à 18:11, Richard Damon a écrit :
[...]
It can increase only at unit fractions.
On 12/24/2023 3:30 AM, WM wrote:
Le 23/12/2023 à 18:47, Jim Burns a écrit :
Each claim in such a sequence is true.
Does NUF(x) increase
from NUF(0) = 0 to NUF(x>0) > 0
in steps [of] height 1 or not?
You (WM) _want_ to only be told a claim
without an argument for it.
You are _demanding_ no.argument.
The following isn't what you're demanding:
----
This true claim:
| for each unit.fraction ⅟j
| there is a unit fraction ⅟kⱼ such that
| ⅟kⱼ < ⅟j
|
| Example: Consider kⱼ = j+1
| It's a true claim.
Does NUF(x) increase
from NUF(0) = 0 to NUF(x>0) > 0
in steps [of] height 1 or not?
No.
That particular quantifier shift,
On 12/25/2023 2:29 PM, WM wrote:
Le 24/12/2023 à 18:11, Richard Damon a écrit :
[...]
It can increase only at unit fractions.
It = |{⅟n|⅟n<x}|
It can increase only _near_ unit fractions.
|{⅟n|⅟n<x}| does not increase at _point_ x
|{⅟n|⅟n<x}| increases between _points_ x x′
|{⅟n|⅟n<x}| increases _near_ 0
For any positive x
|{⅟n|⅟n<0}| < |{⅟n|⅟n<x}|
0 is is not a unit fraction.
However, 0 is near unit fractions.
There is no positive point x near 0
which does not have two unit fractions
between it and 0
0 < ⅟mₓ⁺¹ < ⅟mₓ < x
Thus,
|{⅟n|⅟n<x}| increases by more than 1 _near_ 0
On 12/25/23 2:29 PM, WM wrote:
Le 24/12/2023 à 18:11, Richard Damon a écrit :
On 12/24/23 3:31 AM, WM wrote:
Le 23/12/2023 à 14:44, Richard Damon a écrit :
You don't seem to understand basic set theory.
Teach me. Does NUF(x) increase from NUF(0) = 0 to NUF(x>0) > 0 in
steps of height 1 or not?
Not at finite numbers x.
It can increase only at unit fractions. They all are finite numbers.
And, since there is no minimum unit fraction, it can never have a finite value that is greater than 0.
The "points" where NUF(x) might increase in units of height of 1,
would be at the infintesimals, with a proper definition of NUF to
handle them.
Unit fractions are not infinitesimals because natnumbers are not infinite.
Right, but that is the only points that you could try to define a NUF
that has a finite value (other than 0)
Otherwise, it is just discontinuous,
Not in agreement with mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
Where does that say your NUF can't be discontinous at 0?
because it doesn't take into account unbounded nature of the unit
fractions.
Mathjematics shows that the unit fractions appear unbounded only because
the dark unit fractions cannot ne seen.
Nope. They ARE Unbounded as there is no bound to how small a Unit
Fraction can get.
your logic of NUF(x) being 1 at the "lowest" Unit Fraction is a flawed
assumption,
It is simple and basic mathematics, valid for all unit fractions: n ∈ ℕ: >> 1/n - 1/(n+1) = d_n > 0.
Which does imply what you are trying to claim it does.
THat says from one unit fraction to another, it can only change by 1.
Since 0 isn't a unit fration, is says nothing about how it increase from there to any unit fraction,
Your argument is just based on
there needing to be a smallest Unit
fraction, and thus a highest Natural Number, but such a thing doesn't actually exist.
Le 25/12/2023 à 20:47, Richard Damon a écrit :
On 12/25/23 2:29 PM, WM wrote:
Le 24/12/2023 à 18:11, Richard Damon a écrit :
On 12/24/23 3:31 AM, WM wrote:
Le 23/12/2023 à 14:44, Richard Damon a écrit :
You don't seem to understand basic set theory.
Teach me. Does NUF(x) increase from NUF(0) = 0 to NUF(x>0) > 0 in
steps of height 1 or not?
Not at finite numbers x.
It can increase only at unit fractions. They all are finite numbers.
And, since there is no minimum unit fraction, it can never have a
finite value that is greater than 0.
The solution is dark unit fractions.
The "points" where NUF(x) might increase in units of height of 1,
would be at the infintesimals, with a proper definition of NUF to
handle them.
Unit fractions are not infinitesimals because natnumbers are not
infinite.
Right, but that is the only points that you could try to define a NUF
that has a finite value (other than 0)
The solution is dark unit fractions.
Where does that say your NUF can't be discontinous at 0?
Otherwise, it is just discontinuous,
Not in agreement with mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0. >>
After every unit fractions there is a gap. So NUF cannot increase by
more than 1 at every point.
because it doesn't take into account unbounded nature of the unit
fractions.
Mathjematics shows that the unit fractions appear unbounded only
because the dark unit fractions cannot ne seen.
Nope. They ARE Unbounded as there is no bound to how small a Unit
Fraction can get.
After every unit fractions there is a gap. So NUF cannot increase by
more than 1 at every point.
your logic of NUF(x) being 1 at the "lowest" Unit Fraction is a
flawed assumption,
It is simple and basic mathematics, valid for all unit fractions: n ∈
ℕ: 1/n - 1/(n+1) = d_n > 0.
Which does imply what you are trying to claim it does.
Of course.
THat says from one unit fraction to another, it can only change by 1.
Since 0 isn't a unit fration, is says nothing about how it increase
from there to any unit fraction,
but from one unit fraction to another, it can only change by 1.
Your argument is just based on
from one unit fraction to another, it can only change by 1.
there needing to be a smallest Unit fraction, and thus a highest
Natural Number, but such a thing doesn't actually exist.
Nevertheless, the increase by more than one from 0 to ℵo cannot be discerned. These unit fractions are not available as individuals.
Regards, WM
Le 24/12/2023 à 21:44, Jim Burns a écrit :
On 12/24/2023 3:30 AM, WM wrote:
Le 23/12/2023 à 18:47, Jim Burns a écrit :
Each claim in such a sequence is true.
Does NUF(x) increase
from NUF(0) = 0 to NUF(x>0) > 0
in steps [of] height 1 or not?
You (WM) _want_ to only be told a claim
without an argument for it.
You are _demanding_ no.argument.
I am interested only whether you agree with mathematics or not.
The following isn't what you're demanding:
----
This true claim:
| for each unit.fraction ⅟j
| there is a unit fraction ⅟kⱼ such that
| ⅟kⱼ < ⅟j
|
| Example: Consider kⱼ = j+1
| It's a true claim.
That is true if for each j there is j+1.
But you seem to be convinced that ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 should
be disregarded. Why?
Does NUF(x) increase
from NUF(0) = 0 to NUF(x>0) > 0
in steps [of] height 1 or not?
No.
When going over points with gaps in linear order you will encounter a
first one.
That particular quantifier shift,
Nonsense. If every point of (0, 1] has the property to have ℵ₀ unit fractions at its left-hand side, then all points of (0. 1] have the
property to have ℵ₀ unit fractions at their left-hand side, then the interval (0, 1] has the property to have ℵ₀ unit fractions at its left-hand side. That is simplest logic and mathematics. Because if the
latter was wrong, then this wrongness had to be proved. This could be
done only by showing a point og (0, 1] for which it was wrong.
Regards, WM
Le 25/12/2023 à 22:59, Jim Burns a écrit :
On 12/25/2023 2:29 PM, WM wrote:
Le 24/12/2023 à 18:11, Richard Damon a écrit :
[...]
It can increase only at unit fractions.
It = |{⅟n|⅟n<x}|
It can increase only _near_ unit fractions.
|{⅟n|⅟n<x}| does not increase at _point_ x
|{⅟n|⅟n<x}| increases between _points_ x x′
|{⅟n|⅟n<x}| increases _near_ 0
For any positive x
|{⅟n|⅟n<0}| < |{⅟n|⅟n<x}|
0 is is not a unit fraction.
However, 0 is near unit fractions.
There is no positive point x near 0
which does not have two unit fractions
between it and 0
0 < ⅟mₓ⁺¹ < ⅟mₓ < x
No such point can be found. But if no such point existed, then never the complete set could be applied for enumerating purposes.
Thus,
|{⅟n|⅟n<x}| increases by more than 1 _near_ 0
That contradicts that gaps between all unit fractions exist. If all unit fractions have distances, then the distance after each one causes a halt
in the increase of NUF(x). That is true for all and therefore also for
the first unit fraction.
Regards, WM
On 12/25/23 5:47 PM, WM wrote:
Your argument is just based on
from one unit fraction to another, it can only change by 1.
And 0 isn't a unit fraction, so it isn't constrained.
Nevertheless, the increase by more than one from 0 to ℵo cannot be
discerned. These unit fractions are not available as individuals.
YOU may not discern it, but it does happen.
On 12/25/23 5:26 PM, WM wrote:
When going over points with gaps in linear order you will encounter a
first one.
Nope, not when the gaps keep on shrinking as fast as they do.
At the point 1/n, the size of the gap is such that you would expect to
be able to put at least another n gaps between you and the origin, so
that gaps shrink faster than they approach 0.
And what is WRONG with the concept that ALL finite number x > 0 (or unit fractions) have an infinite number between them and zero?
Le 26/12/2023 à 01:38, Richard Damon a écrit :
On 12/25/23 5:47 PM, WM wrote:
Your argument is just based on
from one unit fraction to another, it can only change by 1.
And 0 isn't a unit fraction, so it isn't constrained.
But as soon as a unit fraction is encountered, it is followed by a gap.
Nevertheless, the increase by more than one from 0 to ℵo cannot be
discerned. These unit fractions are not available as individuals.
YOU may not discern it, but it does happen.
Can you discern it?
Regards, WM
Le 26/12/2023 à 04:59, Richard Damon a écrit :
On 12/25/23 5:26 PM, WM wrote:
When going over points with gaps in linear order you will encounter a
first one.
Nope, not when the gaps keep on shrinking as fast as they do.
Non-empty gaps remain non-empty.
At the point 1/n, the size of the gap is such that you would expect to
be able to put at least another n gaps between you and the origin, so
that gaps shrink faster than they approach 0.
All gaps are containing more than one point.
And what is WRONG with the concept that ALL finite number x > 0 (or
unit fractions) have an infinite number between them and zero?
It clashes with ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 where d_n consists of uncountably many points.
Regards, WM
On 12/26/23 5:21 AM, WM wrote:
But as soon as a unit fraction is encountered, it is followed by a gap.
But before that unit fraction, there are more unit fractions,
Nevertheless, the increase by more than one from 0 to ℵo cannot be
discerned. These unit fractions are not available as individuals.
YOU may not discern it, but it does happen.
Can you discern it?
Yes, I can discern that the numbers get unboundedly dense towards 0,
How? That says that BETWEEN any TWO numbers there is a finite distance,
but that distance can get unboundedly small, so we can fit an unbounded number of them in the finite space.
On 12/26/2023 5:26 AM, WM wrote:
What makes a gap a gap
Instead of "points with gaps",
consider "points next to each other"
Two points are _next_ to each other
if there is no point between them.
Between them is empty.
⅟1 is next to ⅟2
⅟n is next to ⅟n⁺¹ and next to ⅟n⁻¹
If the first unit fraction existed,
it would be next to 0
No unit fraction is next to 0
No unit fraction is the first unit fraction.
When going over points with
gaps in linear order
you will encounter a first one.
No.
The first unit fraction not.exists.
Le 26/12/2023 à 04:59, Richard Damon a écrit :
On 12/25/23 5:26 PM, WM wrote:
When going over points with
gaps in linear order
you will encounter a first one.
Nope,
not when the gaps keep on shrinking
as fast as they do.
Non-empty gaps remain non-empty.
When going over points with
gaps in linear order
you will encounter a first one.
Le 26/12/2023 à 19:26, Jim Burns a écrit :
On 12/26/2023 5:26 AM, WM wrote:
Non-empty gaps remain non-empty.
What makes a gap a gap
Non-empty means that the aps contain points,
even uncountably many,
none of which is a unit fraction.
Instead of "points with gaps",
consider "points next to each other"
There are no
unit fractions next to each other.
Two points are _next_ to each other
if there is no point between them.
Between them is empty.
Such points exist only in the form that
one of them is dark.
⅟1 is next to ⅟2
⅟n is next to ⅟n⁺¹ and next to ⅟n⁻¹
No.
There are uncountably many points
between them.
If the first unit fraction existed,
it would be next to 0
Not necessarily.
But we will never know.
Le 26/12/2023 à 13:53, Richard Damon a écrit :
How? That says that BETWEEN any TWO numbers there is a finite
distance, but that distance can get unboundedly small, so we can fit
an unbounded number of them in the finite space.
Each gap contains an uncountably many points.
Regards, WM
We begin our discussion of unit fractions
with claims true of each unit fraction,
seenᵂᴹ or unseenᵂᴹ
If the first unit fraction existed,
it would be next to 0
Not necessarily.
Necessarily.
Because of what "first" means.
On 12/26/23 1:17 PM, WM wrote:
Yes, I can discern that the numbers get unboundedly dense towards 0,
But since you cannot distinguish these dense unit fractions, they are dark.
Who says we can not distinguish them.
What we can't do is to try to name them all individually at once, since
the set has an infinite membership.
On 12/26/23 1:20 PM, WM wrote:
Le 26/12/2023 à 13:53, Richard Damon a écrit :
How? That says that BETWEEN any TWO numbers there is a finite
distance, but that distance can get unboundedly small, so we can fit
an unbounded number of them in the finite space.
Each gap contains an uncountably many points.
This doesn't say anything about the need for there to be a "first" Unit Fraction for NUF(x) to be 1 at.
Le 27/12/2023 à 03:29, Richard Damon a écrit :
On 12/26/23 1:17 PM, WM wrote:
Yes, I can discern that the numbers get unboundedly dense towards 0,
But since you cannot distinguish these dense unit fractions, they are
dark.
Who says we can not distinguish them.
You. All you can discern is not "unboundedly dense".
What we can't do is to try to name them all individually at once,
since the set has an infinite membership.
All you can name, at onece or later, belong to a finite initial segment
much smaller than the whole set. Almost all cannot be named.
Regards, WM
Le 26/12/2023 à 21:23, Jim Burns a écrit :
We begin our discussion of unit fractions
with claims true of each unit fraction,
seenᵂᴹ or unseenᵂᴹ
Yes, in particular this one:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0makes no claim for 0
If the first unit fraction existed,
it would be next to 0
Not necessarily.
Necessarily.
Because of what "first" means.
"Next to" means that
there is no point in between.
First means there is
no unit fraction in between.
"Next to" means that
there is no point in between.
"Next to" in the unit.fractions means that
there is no unit.fraction between.
Le 27/12/2023 à 17:45, Jim Burns a écrit :
"Next to" means that
there is no point in between.
"Next to" in the unit.fractions means that
there is no unit.fraction between.
One could use that meaning but it may be misleading. Therefore I don't
use it.
Regards, WM
On 12/27/23 2:24 PM, WM wrote:
Le 27/12/2023 à 17:45, Jim Burns a écrit :
"Next to" means that
there is no point in between.
"Next to" in the unit.fractions means that
there is no unit.fraction between.
One could use that meaning but it may be misleading. Therefore I don't
use it.
But, when our domain of discussion in the "Unit Fractions" then 1/2 is
"next to" 1/3, as there is nothing IN THE SET between them.
On 12/27/23 4:35 AM, WM wrote:
Le 27/12/2023 à 03:29, Richard Damon a écrit :
On 12/26/23 1:17 PM, WM wrote:
Yes, I can discern that the numbers get unboundedly dense towards 0,
But since you cannot distinguish these dense unit fractions, they are
dark.
Who says we can not distinguish them.
You. All you can discern is not "unboundedly dense".
You can discern ANY of the number in that unboundedly dense set.
And all Natural Numbers are finite, and thus can be use to define a
finite initial segment, so ALL can used.
Le 27/12/2023 à 21:06, Richard Damon a écrit :
On 12/27/23 2:24 PM, WM wrote:
Le 27/12/2023 à 17:45, Jim Burns a écrit :
"Next to" means that
there is no point in between.
"Next to" in the unit.fractions means that
there is no unit.fraction between.
One could use that meaning but it may be misleading. Therefore I
don't use it.
But, when our domain of discussion in the "Unit Fractions" then 1/2 is
"next to" 1/3, as there is nothing IN THE SET between them.
But that use encourages you to forget that uncountably many points lie between any two unit fractions. Therefore the function NUF(x) has always
a level between any two unit fractions. Therefoe never infinitely many
can be at the start together.
Regards, WM
Le 27/12/2023 à 14:37, Richard Damon a écrit :
On 12/27/23 4:35 AM, WM wrote:
Le 27/12/2023 à 03:29, Richard Damon a écrit :
On 12/26/23 1:17 PM, WM wrote:
But since you cannot distinguish these dense unit fractions, they
Yes, I can discern that the numbers get unboundedly dense towards 0, >>>>>
are dark.
Who says we can not distinguish them.
You. All you can discern is not "unboundedly dense".
You can discern ANY of the number in that unboundedly dense set.
Each discerned unit fraction has a distance from its next unit fractions
and hence is not unboundedly dense.
And all Natural Numbers are finite, and thus can be use to define a
finite initial segment, so ALL can used.
All have distances and therefore NUF(x) can never grow from 0 to
infinity without passing a first unit fraction and being constant
afterwards for a while.
Regards, WM
On 12/27/23 5:15 PM, WM wrote:
Therefore the function NUF(x) has always
a level between any two unit fractions. Therefore never infinitely many
can be at the start together.
But there can be infinitely many before any unit fraction,
Le 27/12/2023 à 17:45, Jim Burns a écrit :
On 12/27/2023 3:37 AM, WM wrote:
Le 26/12/2023 à 21:23, Jim Burns a écrit :
If the first unit fraction existed,
it would be next to 0
Not necessarily.
Necessarily.
Because of what "first" means.
"Next to" means that
there is no point in between.
"Next to" in the unit.fractions means that
there is no unit.fraction between.
One could use that meaning
but it may be misleading.
Therefore I don't use it.
On 12/27/23 5:47 PM, WM wrote:
Each discerned unit fraction has a distance from its next unit fractions
and hence is not unboundedly dense.
Then what is the BOUND?
Remember, that distance reduces without any limit to its size, so you
claim of not unboundedly dense is proved to be untrue.
All have distances and therefore NUF(x) can never grow from 0 to
infinity without passing a first unit fraction and being constant
afterwards for a while.
Which only applies *IF* there IS a "first" Unit Fraction, which there isn't.
| ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
makes no claim for 0
because for no n ∈ ℕ is 1/n = 0
Le 28/12/2023 à 03:57, Richard Damon a écrit :
On 12/27/23 5:15 PM, WM wrote:
Therefore the function NUF(x) has always a level between any two unit
fractions. Therefore never infinitely many can be at the start together.
But there can be infinitely many before any unit fraction,
My logic does not allow that.
Regards, WM
Le 28/12/2023 à 03:58, Richard Damon a écrit :
On 12/27/23 5:47 PM, WM wrote:
Each discerned unit fraction has a distance from its next unit
fractions and hence is not unboundedly dense.
Then what is the BOUND?
The bound is zero. The accumulation point contains dark unit fractions.
Remember, that distance reduces without any limit to its size, so you
claim of not unboundedly dense is proved to be untrue.
Every chosen unit fraction and its neighbours are not unboundedly dense.
The dark unit fractions appear unboundedly dense.
All have distances and therefore NUF(x) can never grow from 0 to
infinity without passing a first unit fraction and being constant
afterwards for a while.
Which only applies *IF* there IS a "first" Unit Fraction, which there
isn't.
If there are any unit fractions in separated oder, then there is a first
one.
Regards, WM
Le 28/12/2023 à 11:39, Jim Burns a écrit :
| ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
makes no claim for 0
because for no n ∈ ℕ is 1/n = 0
That is true. But as soon as one unit fraction is encountered, it is
followed by a gap consisting of points which are not unit fractions.
Hence NUF(x) is a step function.
According to ZFC the number of unit fractions grows between 0 and (0, 1]
by more than 2.
This is impossible because between any two unit fractions there are "uncountably" many points which do not fit between 0 and (0, 1]. This
proves that ZFC does not deliver correct mathematics in this case, like
in several others: https://www.researchgate.net/publication/376587391_The_seven_deadly_sins_of_set_theory
Regards, WM
Le 28/12/2023 à 11:39, Jim Burns a écrit :
[...]
According to ZFC
the number of unit fractions grows
between 0 and (0, 1]
by more than 2.
| ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
makes no claim for 0
because for no n ∈ ℕ is 1/n = 0
That is true.
But as soon as one unit fraction is encountered,
it is followed by a gap consisting of
points which are not unit fractions.
Hence NUF(x) is a step function.
| ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
makes no claim for 0
because for no n ∈ ℕ is 1/n = 0
That is true.
But as soon as one unit fraction is encountered,
it is followed by a gap consisting of
points which are not unit fractions.
Hence NUF(x) is a step function.
According to ZFC
the number of unit fractions grows
between 0 and (0, 1]
by more than 2.
This is impossible because
between any two unit fractions there are
"uncountably" many points which do not fit
between 0 and (0, 1].
This proves that ZFC does not deliver
correct mathematics in this case,
like in several others:
Le 28/12/2023 à 11:39, Jim Burns a écrit :
[...]
Hence NUF(x) is a step function.
Hence NUF(x) is a step function.
On 12/28/2023 6:02 AM, WM wrote:
No, ZFC doesn't say this:
the number of unit fractions grows
between 0 and (0, 1]
by more than 2.
According to ZFC (meaning "standardly")
nothing visibleᵂᴹ and nothing darkᵂᴹ
is between 0 and (0,1]
Something between 0 and (0,1] is
below (0,1] and also
between 0 and 1, thus in (0,1]
Below and in (0,1] is contradictory.
On 12/28/23 5:44 AM, WM wrote:
Le 28/12/2023 à 03:57, Richard Damon a écrit :It could be that you are just stuck on ZFC, and don't understand the
On 12/27/23 5:15 PM, WM wrote:
Therefore the function NUF(x) has always a level between any two unitBut there can be infinitely many before any unit fraction,
fractions. Therefore never infinitely many can be at the start together. >>
My logic does not allow that.
limits of the number system it generates.
Le 28/12/2023 à 14:17, Richard Damon a écrit :
On 12/28/23 5:44 AM, WM wrote:
Le 28/12/2023 à 03:57, Richard Damon a écrit :It could be that you are just stuck on ZFC, and don't understand the
On 12/27/23 5:15 PM, WM wrote:
Therefore the function NUF(x) has always a level between any two
unit fractions. Therefore never infinitely many can be at the start
together.
But there can be infinitely many before any unit fraction,
My logic does not allow that.
limits of the number system it generates.
I have set out to prove that ZFC can't handle this problem.
Regards, WM
On 12/18/23 5:47 AM, WM wrote:
You aren't using the right mapping, showing your ignorance.
Where does my mapping deviate from Cantor's
k = (m + n - 1)(m + n - 2)/2 + m
with the result
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
for the first time?
Which shows that your matrix B having values like
1/1 2/1 3/1 ...
1/2 2/2 3/2 ...
1/3 2/3 3/3 ...
.
.
.
is exactly indexed by a matrix form like:
1 3 6 10
2 5 9
4 8 13
7 12
11
Le 18/12/2023 à 16:00, Richard Damon a écrit :
On 12/18/23 5:47 AM, WM wrote:
You aren't using the right mapping, showing your ignorance.
Where does my mapping deviate from Cantor's
k = (m + n - 1)(m + n - 2)/2 + m
with the result
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
for the first time?
Which shows that your matrix B having values like
1/1 2/1 3/1 ...
1/2 2/2 3/2 ...
1/3 2/3 3/3 ...
.
.
.
is exactly indexed by a matrix form like:
1 3 6 10
2 5 9
4 8 13
7 12
11
This is a triangle, never a matrix.
Regards, WM
On 12/30/23 12:00 PM, WM wrote:
Le 18/12/2023 à 16:00, Richard Damon a écrit :
On 12/18/23 5:47 AM, WM wrote:
You aren't using the right mapping, showing your ignorance.
Where does my mapping deviate from Cantor's
k = (m + n - 1)(m + n - 2)/2 + m
with the result
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
for the first time?
Which shows that your matrix B having values like
1/1 2/1 3/1 ...
1/2 2/2 3/2 ...
1/3 2/3 3/3 ...
.
.
.
is exactly indexed by a matrix form like:
1 3 6 10
2 5 9
4 8 13
7 12
11
This is a triangle, never a matrix.
WHat point in the matrix does it never cover?
Le 30/12/2023 à 18:33, Richard Damon a écrit :
On 12/30/23 12:00 PM, WM wrote:
Le 18/12/2023 à 16:00, Richard Damon a écrit :
On 12/18/23 5:47 AM, WM wrote:
You aren't using the right mapping, showing your ignorance.
Where does my mapping deviate from Cantor's
k = (m + n - 1)(m + n - 2)/2 + m
with the result
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3,
4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
for the first time?
Which shows that your matrix B having values like
1/1 2/1 3/1 ...
1/2 2/2 3/2 ...
1/3 2/3 3/3 ...
.
.
.
is exactly indexed by a matrix form like:
1 3 6 10
2 5 9
4 8 13
7 12
11
This is a triangle, never a matrix.
WHat point in the matrix does it never cover?
Almost all.
XOOO...
XOOO...
XOOO...
XOOO...
..
Regards, WM
Le 28/12/2023 à 19:17, Jim Burns a écrit :
On 12/28/2023 6:02 AM, WM wrote:
[...]
No, ZFC doesn't say this:
the number of unit fractions grows
between 0 and (0, 1]
by more than 2.
According to ZFC (meaning "standardly")
nothing visibleᵂᴹ and nothing darkᵂᴹ
is between 0 and (0,1]
But NUF has collected ℵo unit fractions
before any positive point.
They do not come from Santa Claus.
They do not come from Santa Claus.
But NUF has collected ℵo unit fractions
before any positive point.
ℵ₀.many unit.fractions do not
come from Santa Claus.
They do not come from anyone or anywhere.
They have already been here.
On 12/30/23 12:45 PM, WM wrote:
Le 30/12/2023 à 18:33, Richard Damon a écrit :But your x's aren't the triangle,
On 12/30/23 12:00 PM, WM wrote:
Le 18/12/2023 à 16:00, Richard Damon a écrit :
On 12/18/23 5:47 AM, WM wrote:
You aren't using the right mapping, showing your ignorance.
Where does my mapping deviate from Cantor's
k = (m + n - 1)(m + n - 2)/2 + m
with the result
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3,
4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
for the first time?
Which shows that your matrix B having values like
1/1 2/1 3/1 ...
1/2 2/2 3/2 ...
1/3 2/3 3/3 ...
.
.
.
is exactly indexed by a matrix form like:
1 3 6 10
2 5 9
4 8 13
7 12
11
This is a triangle, never a matrix.
WHat point in the matrix does it never cover?
Almost all.
XOOO...
XOOO...
XOOO...
XOOO...
...
Le 31/12/2023 à 05:42, Jim Burns a écrit :
ℵ₀.many unit.fractions do not
come from Santa Claus.
They do not come from anyone or anywhere.
They have already been here.
They are
before any positive x but
after 0
without any internal distances?
Le 30/12/2023 à 19:07, Richard Damon a écrit :
On 12/30/23 12:45 PM, WM wrote:
Le 30/12/2023 à 18:33, Richard Damon a écrit :But your x's aren't the triangle,
On 12/30/23 12:00 PM, WM wrote:
Le 18/12/2023 à 16:00, Richard Damon a écrit :
On 12/18/23 5:47 AM, WM wrote:
You aren't using the right mapping, showing your ignorance.
Where does my mapping deviate from Cantor's
k = (m + n - 1)(m + n - 2)/2 + m
with the result
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, >>>>>>> 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
for the first time?
Which shows that your matrix B having values like
1/1 2/1 3/1 ...
1/2 2/2 3/2 ...
1/3 2/3 3/3 ...
.
.
.
is exactly indexed by a matrix form like:
1 3 6 10
2 5 9
4 8 13
7 12
11
This is a triangle, never a matrix.
WHat point in the matrix does it never cover?
Almost all.
XOOO...
XOOO...
XOOO...
XOOO...
...
They become the triangle, never more.
Regards, WM
Le 31/12/2023 à 05:42, Jim Burns a écrit :
ℵ₀.many unit.fractions do not
come from Santa Claus.
They do not come from anyone or anywhere.
They have already been here.
They are before any positive x but after 0 without any internal distances? Goodbye.
Regards, WM
On 12/31/2023 4:51 AM, WM wrote:
Le 31/12/2023 à 05:42, Jim Burns a écrit :
ℵ₀.many unit.fractions do not
come from Santa Claus.
They do not come from anyone or anywhere.
They have already been here.
They are
before any positive x but
after 0
without any internal distances?
No.
None is before each positive x.
Le 31/12/2023 à 13:04, Jim Burns a écrit :
On 12/31/2023 4:51 AM, WM wrote:
Le 31/12/2023 à 05:42, Jim Burns a écrit :
ℵ₀.many unit.fractions do not
come from Santa Claus.
They do not come from anyone or anywhere.
They have already been here.
They are
before any positive x but
after 0
without any internal distances?
No.
None is before each positive x.
Fine.
The first unit fraction is a positive x itself.
But you cannot find
an x with less than ℵ₀ smaller unit fractions.
They are dark.
On 12/31/2023 4:51 AM, WM wrote:
Le 31/12/2023 à 05:42, Jim Burns a écrit :
ℵ₀.many unit.fractions do not
come from Santa Claus.
They do not come from anyone or anywhere.
They have alreadcy been here.
They are
before any positive x but
after 0
without any internal distances?
They are
before any positive x but
after 0
without any internal distances?
No.
None is before each positive x.
Each two have a positive distance.
Your (WM's) quantifier magic is bad.
For each cardinal n which.can.grow.by.1
more than n.many unit.fractions are in (0,x]
mₓ < ⅟x ≤ mₓ⁺¹ < ... < mₓ⁺ⁿ⁺¹
0 < ⅟mₓ⁺ⁿ⁺¹ < ... < ⅟mₓ⁺¹ ≤ x
For each cardinal n which.can.grow.by.1
NUF(x) = |(-∞,x]∩⅟ℕ₁| is not n
NUF(x) is
not a cardinal which.can.grow.by.1
On 12/31/23 4:35 AM, WM wrote:
Le 30/12/2023 à 19:07, Richard Damon a écrit :
On 12/30/23 12:45 PM, WM wrote:
Le 30/12/2023 à 18:33, Richard Damon a écrit :But your x's aren't the triangle,
On 12/30/23 12:00 PM, WM wrote:
Le 18/12/2023 à 16:00, Richard Damon a écrit :
On 12/18/23 5:47 AM, WM wrote:
You aren't using the right mapping, showing your ignorance.
Where does my mapping deviate from Cantor's
k = (m + n - 1)(m + n - 2)/2 + m
with the result
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, >>>>>>>> 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
for the first time?
Which shows that your matrix B having values like
1/1 2/1 3/1 ...
1/2 2/2 3/2 ...
1/3 2/3 3/3 ...
.
.
.
is exactly indexed by a matrix form like:
1 3 6 10
2 5 9
4 8 13
7 12
11
This is a triangle, never a matrix.
WHat point in the matrix does it never cover?
Almost all.
XOOO...
XOOO...
XOOO...
XOOO...
...
They become the triangle, never more.
WHat point in the matrix can't they reach?
You seem to like making claims you can't actually back up,
Le 31/12/2023 à 13:46, Richard Damon a écrit :
On 12/31/23 4:35 AM, WM wrote:
Le 30/12/2023 à 19:07, Richard Damon a écrit :
On 12/30/23 12:45 PM, WM wrote:
Le 30/12/2023 à 18:33, Richard Damon a écrit :But your x's aren't the triangle,
On 12/30/23 12:00 PM, WM wrote:
Le 18/12/2023 à 16:00, Richard Damon a écrit :
On 12/18/23 5:47 AM, WM wrote:
You aren't using the right mapping, showing your ignorance. >>>>>>>>>Where does my mapping deviate from Cantor's
k = (m + n - 1)(m + n - 2)/2 + m
with the result
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4,
3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
for the first time?
Which shows that your matrix B having values like
1/1 2/1 3/1 ...
1/2 2/2 3/2 ...
1/3 2/3 3/3 ...
.
.
.
is exactly indexed by a matrix form like:
1 3 6 10
2 5 9
4 8 13
7 12
11
This is a triangle, never a matrix.
WHat point in the matrix does it never cover?
Almost all.
XOOO...
XOOO...
XOOO...
XOOO...
...
They become the triangle, never more.
WHat point in the matrix can't they reach?
Such a point cannot be given. However, as you recognized yourself, the X
from the first column cannot cover the whole matrix. The O will remain.
You seem to like making claims you can't actually back up,
I can prove them using mathematics: The o will remain. Therefore their positions exist. But they cannot be given. Conclusion: They are at dark positions.
Regards, WM
On 1/1/24 12:21 PM, WM wrote:
WHat point in the matrix can't they reach?
Such a point cannot be given. However, as you recognized yourself, the X
from the first column cannot cover the whole matrix. The O will remain.
No O will remain, as some x from below will map to it
On 1/1/2024 10:34 AM, WM wrote:
The first unit fraction is a positive x itself.
The first unit fraction not.exists.
Le 01/01/2024 à 18:11, Jim Burns a écrit :
On 1/1/2024 10:34 AM, WM wrote:
The first unit fraction is a positive x itself.
The first unit fraction not.exists.
The [unit] fractions are placed in linear order
with distances and none is smaller than zero.
That enforces a first one.
Le 01/01/2024 à 19:19, Richard Damon a écrit :
On 1/1/24 12:21 PM, WM wrote:
WHat point in the matrix can't they reach?
Such a point cannot be given. However, as you recognized yourself,
the X from the first column cannot cover the whole matrix. The O will
remain.
No O will remain, as some x from below will map to it
By definition of exchange of O and X never an O will leave the matrix.
Regards, WM
On 1/2/2024 2:30 PM, WM wrote:
The [unit] fractions are placed in linear order
with distances and none is smaller than zero.
That enforces a first one.
For each cardinal n which.can.change.by.1
On 1/2/24 2:35 PM, WM wrote:
Le 01/01/2024 à 19:19, Richard Damon a écrit :
On 1/1/24 12:21 PM, WM wrote:
WHat point in the matrix can't they reach?
Such a point cannot be given. However, as you recognized yourself,
the X from the first column cannot cover the whole matrix. The O will
remain.
No O will remain, as some x from below will map to it
By definition of exchange of O and X never an O will leave the matrix.
But we are not "swapping" them but mapping the X to its new position
Note also, every O that you move to the column with the X, will later be covered by an even later x, until the o's all disappear into infinity.
Le 03/01/2024 à 00:59, Richard Damon a écrit :
On 1/2/24 2:35 PM, WM wrote:
Le 01/01/2024 à 19:19, Richard Damon a écrit :
On 1/1/24 12:21 PM, WM wrote:
WHat point in the matrix can't they reach?
Such a point cannot be given. However, as you recognized yourself,
the X from the first column cannot cover the whole matrix. The O
will remain.
No O will remain, as some x from below will map to it
By definition of exchange of O and X never an O will leave the matrix.
But we are not "swapping" them but mapping the X to its new position
and mapping the O to the position which the X has left.
Note also, every O that you move to the column with the X, will later
be covered by an even later x, until the o's all disappear into infinity.
They do not disappear because by exchanging with an X from the matrix
the O remains in the matrix.
Regards, WM
Le 02/01/2024 à 22:15, Jim Burns a écrit :
On 1/2/2024 2:30 PM, WM wrote:
The [unit] fractions are placed in linear order
with distances and none is smaller than zero.
That enforces a first one.
For each cardinal n which.can.change.by.1
For every unit fraction.
On 1/3/24 5:27 AM, WM wrote:
Le 03/01/2024 à 00:59, Richard Damon a écrit :
On 1/2/24 2:35 PM, WM wrote:
Le 01/01/2024 à 19:19, Richard Damon a écrit :
On 1/1/24 12:21 PM, WM wrote:
WHat point in the matrix can't they reach?
Such a point cannot be given. However, as you recognized yourself, >>>>>> the X from the first column cannot cover the whole matrix. The O
will remain.
No O will remain, as some x from below will map to it
By definition of exchange of O and X never an O will leave the matrix.
But we are not "swapping" them but mapping the X to its new position
and mapping the O to the position which the X has left.
Why?
And "exchanging" isn't the operation being performed,
On 1/3/2024 5:22 AM, WM wrote:
Le 02/01/2024 à 22:15, Jim Burns a écrit :
On 1/2/2024 2:30 PM, WM wrote:
The [unit] fractions are placed in linear order
with distances and none is smaller than zero.
That enforces a first one.
For each cardinal n which.can.change.by.1
For every unit fraction.
For each cardinal
|⅟mₓ⁺ⁿ,…,⅟mₓ⁺¹| which.can.change.by.1
Le 03/01/2024 à 12:28, Jim Burns a écrit :
On 1/3/2024 5:22 AM, WM wrote:
Le 02/01/2024 à 22:15, Jim Burns a écrit :
On 1/2/2024 2:30 PM, WM wrote:
The [unit] fractions are placed in linear order
with distances and none is smaller than zero.
That enforces a first one.
For each cardinal n which.can.change.by.1
For every unit fraction.
For each cardinal
|⅟mₓ⁺ⁿ,…,⅟mₓ⁺¹| which.can.change.by.1
For every unit fraction!
The [unit] fractions are placed in linear order
with distances and none is smaller than zero.
That enforces a first one.
Le 03/01/2024 à 13:08, Richard Damon a écrit :
On 1/3/24 5:27 AM, WM wrote:
Le 03/01/2024 à 00:59, Richard Damon a écrit :
On 1/2/24 2:35 PM, WM wrote:
Le 01/01/2024 à 19:19, Richard Damon a écrit :But we are not "swapping" them but mapping the X to its new position
On 1/1/24 12:21 PM, WM wrote:
WHat point in the matrix can't they reach?
Such a point cannot be given. However, as you recognized
yourself, the X from the first column cannot cover the whole
matrix. The O will remain.
No O will remain, as some x from below will map to it
By definition of exchange of O and X never an O will leave the matrix. >>>
and mapping the O to the position which the X has left.
Why?
Because there is no index remaining. This is indicated by O aka withOut indeX.
And "exchanging" isn't the operation being performed,
The game is defined so. You are not allowed to change it.
Regards, WM
On 1/3/24 10:11 AM, WM wrote:
Le 03/01/2024 à 13:08, Richard Damon a écrit :
But we are not "swapping" them but mapping the X to its new position
By definition of exchange of O and X never an O will leave the matrix. >>>>
and mapping the O to the position which the X has left.
Why?
Because there is no index remaining. This is indicated by O aka withOut
indeX.
Why isn't the index remaining?
What used it up?
Remember, we are using the k's to mark off the m,n, that doesn't make
the k go away.
And "exchanging" isn't the operation being performed,
The game is defined so. You are not allowed to change it.
But you are playing the wrong game.
Cantor played a different game, which it seems you don't even know how
to play.
Le 04/01/2024 à 01:35, Richard Damon a écrit :
On 1/3/24 10:11 AM, WM wrote:
Le 03/01/2024 à 13:08, Richard Damon a écrit :
and mapping the O to the position which the X has left.
By definition of exchange of O and X never an O will leave the
matrix.
But we are not "swapping" them but mapping the X to its new position >>>>>
Why?
Because there is no index remaining. This is indicated by O aka
withOut indeX.
Why isn't the index remaining?
The index is placed according to Cant6or's formula. It must leave its
initial position which is now without index.
What used it up?
Remember, we are using the k's to mark off the m,n, that doesn't make
the k go away.
They go away from their initial positions in the first column.
And "exchanging" isn't the operation being performed,
The game is defined so. You are not allowed to change it.
But you are playing the wrong game.
No, you are not understanding it.
Cantor played a different game, which it seems you don't even know how
to play.
I repeat exactly his game:
k = (m + n - 1)(m + n - 2)/2 + m
The k come from the first column
1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
..
and go to
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX... ..........................................
Regards, WM
A cardinal larger than
each cardinal which.can.grow.by.1
is not
any cardinal which.can.grow.by.1.
and is
a cardinal which.CANNOT.grow.by.1.
On 1/4/24 5:39 AM, WM wrote:
Le 04/01/2024 à 01:35, Richard Damon a écrit :
On 1/3/24 10:11 AM, WM wrote:
Le 03/01/2024 à 13:08, Richard Damon a écrit :
and mapping the O to the position which the X has left.
By definition of exchange of O and X never an O will leave the >>>>>>>> matrix.
But we are not "swapping" them but mapping the X to its new position >>>>>>
Why?
Because there is no index remaining. This is indicated by O aka
withOut indeX.
Why isn't the index remaining?
The index is placed according to Cant6or's formula. It must leave its
initial position which is now without index.
WHY?
You don't understand what the operation defined is.
What used it up?
Remember, we are using the k's to mark off the m,n, that doesn't make
the k go away.
They go away from their initial positions in the first column.
Nope.
And "exchanging" isn't the operation being performed,
The game is defined so. You are not allowed to change it.
But you are playing the wrong game.
No, you are not understanding it.
You can't win the game if you aren't playing the right game.
Cantor played a different game, which it seems you don't even know how
to play.
I repeat exactly his game:
k = (m + n - 1)(m + n - 2)/2 + m
The k come from the first column
1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
..
and go to
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX...
..........................................
And we show that every element of the "matrix" has a matching point in
the list of Natural Numbers,
Le 03/01/2024 à 18:59, Jim Burns a écrit :
A cardinal larger than
each cardinal which.can.grow.by.1
is not
any cardinal which.can.grow.by.1.
and is
a cardinal which.CANNOT.grow.by.1.
Every unit fraction obeys maths.
Le 04/01/2024 à 13:26, Richard Damon a écrit :
On 1/4/24 5:39 AM, WM wrote:
Le 04/01/2024 à 01:35, Richard Damon a écrit :
On 1/3/24 10:11 AM, WM wrote:
Le 03/01/2024 à 13:08, Richard Damon a écrit :
By definition of exchange of O and X never an O will leave the >>>>>>>>> matrix.
But we are not "swapping" them but mapping the X to its new
position
and mapping the O to the position which the X has left.
Why?
Because there is no index remaining. This is indicated by O aka
withOut indeX.
Why isn't the index remaining?
The index is placed according to Cant6or's formula. It must leave its
initial position which is now without index.
WHY?
You don't understand what the operation defined is.
What used it up?
Remember, we are using the k's to mark off the m,n, that doesn't
make the k go away.
They go away from their initial positions in the first column.
Nope.
Yes, they do.
And "exchanging" isn't the operation being performed,
The game is defined so. You are not allowed to change it.
But you are playing the wrong game.
No, you are not understanding it.
You can't win the game if you aren't playing the right game.
I do the the game exactly according to the rules given by Cantor with
one exception: First I use a bijection between ℕ and the first column 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, ...
5/1, 5/2, 5/3, 5/4, ...
..
to get
1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
..
If however that destroys Cantor's game, then it has never worked.
Cantor played a different game, which it seems you don't even know
how to play.
I repeat exactly his game:
k = (m + n - 1)(m + n - 2)/2 + m
The k come from the first column
1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
..
and go to
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2,
5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX...
..........................................
And we show that every element of the "matrix" has a matching point in
the list of Natural Numbers,
No, You show that all known natural numbers match all known unit
fractions. Therefore they appear to be in bijection. Only a very, very
stupid mind can believe that there are as many natural numbers divisible
by 10^10^1000 as fractions or even as algebraic numbers.
But since you do not even understand the basics of this discussion, it
is useless to continue. EOD
Regards, WM
| Sysop: | Keyop |
|---|---|
| Location: | Huddersfield, West Yorkshire, UK |
| Users: | 715 |
| Nodes: | 16 (2 / 14) |
| Uptime: | 34:29:08 |
| Calls: | 12,109 |
| Files: | 15,006 |
| Messages: | 6,518,342 |