On Saturday, December 9, 2023 at 7:43:53 AM UTC-4, WM wrote:
William schrieb am Freitag, 8. Dezember 2023 um 22:02:44 UTC+1:
B(1) } shrinks by one element to the set of elements of (|N x |N)The set of elements of (|N x |N) { Positions of A(1)\Positions ofNope. The set "positions of A(n)\positions of B(n)" shrinks.Please show where the first event happens.
{ Positions of A(2)\Positions of B(2) }
Which element is that?The element of the set |Nx|N, (1,0) .
On 09.12.2023 17:43, William wrote:
Which element is that?
The element of the set |Nx|N, (1,0) .
No, all elements of all A(n) remain.
All elements of all B(n) remain.
All elements of the difference remain.
On 12/10/2023 3:48 AM, WM wrote:
On 09.12.2023 17:43, William wrote:
Which element is that?
The element of the set |Nx|N, (1,0) .
No, all elements of all A(n) remain.
All elements of all B(n) remain. All elements of the difference remain.
Initially,
A\B is unpopulated by B.elements.
You (WM) deny that and claim that
after all and only non-populating swaps,
A\B has become populated.
s
(I guess) your explanationᵂᴹ is that
it's darkᵂᴹ
On Sunday, December 10, 2023 at 4:57:50 AM UTC-4, WM wrote:
On 09.12.2023 17:43, William wrote:
No, all elements of all A(n) remain. All elements of all B(n) remain.Which element is that?The element of the set |Nx|N, (1,0) .
All elements of the difference remain.
let S be the positions of A(1) (also the positions of A(2))
(1,0) is an element of S
P(1) = the positions of B(1) = {(0,0),(0,1)}
P(2) = the positions of B(2) = {(0,0),(0,1),(1,0)|
(1,0) is an element of S\P(1)
(1,0) is not an element of S\P(2)
On Monday, December 11, 2023 at 6:21:09 AM UTC-4, WM wrote:not mean replace)
On 10.12.2023 14:07, William wrote:So what? The elements of A(0) do not remain in A(n)\B(n) ("\" does
(1,0) is an element of S\P(1)The elements of A(0) replaced by B(n) remain in the matrix.
(1,0) is not an element of S\P(2)
On Tuesday, December 12, 2023 at 6:02:41 AM UTC-4, WM wrote:
William schrieb am Montag, 11. Dezember 2023 um 12:38:27 UTC+1:
On Monday, December 11, 2023 at 6:21:09 AM UTC-4, WM wrote:not mean replace)
On 10.12.2023 14:07, William wrote:So what? The elements of A(0) do not remain in A(n)\B(n) ("\" does
(1,0) is an element of S\P(1)The elements of A(0) replaced by B(n) remain in the matrix.
(1,0) is not an element of S\P(2)
B(0) = B(1) is the first column of A(0) = A(1):
Nope B(0)=B(1) has one element
B(0)=B(1) =
1
A(0)=A(1)=
1, 1/2, 1/3, 1/4, ...Nope. B(2) has two elements
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, ...
...
A(2) =
1, 2, 1/3, 1/4, ...
1/2, 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, ...
...
B(2) =
1, 2
3
4
5
...
B(2) =
1, 2
Nope. A(2) and A(0) have the same elements
A(2) \ B(2) has precisely the same elements as A(0) \ B(0).
B(0) and B(2) do not have the same elements.
On 10.12.2023 21:05, Jim Burns wrote:
You (WM) deny that and claim that
after all and only non-populating swaps,
A\B has become populated.
The elements of A\B are never changed.
First B is only a column B(0).
Later it becomes a triangle (never a square).
(I guess) your explanationᵂᴹ is that
it's darkᵂᴹ
Fact is that B never disappears.
On 12/11/2023 5:28 AM, WM wrote:
On 10.12.2023 21:05, Jim Burns wrote:
You (WM) deny that and claim that
after all and only non-populating swaps,
A\B has become populated.
The elements of A\B are never changed.
Yes.
After all B.swaps ⟨i,j⟩↔⟨kᵢⱼ,1⟩ inside B
for all B.locations ⟨i,j⟩ in B
all fractions i/j (j≠1) are not anywhere in B
Fact is that B never disappears.
After all B.swaps ⟨i,j⟩↔⟨kᵢⱼ,1⟩ inside B
for all B.locations ⟨i,j⟩ in B
all fractions i/j (j≠1) are not anywhere in B
I'm willing to call that disappearing
The problem with explaining
that so-called "disappearance" as
not-a-disappearance
but as darkᵂᴹ events happening unseenᵂᴹ
is that
we don't seeᵂᴹ or not-seeᵂᴹ any of the
fractions or B.locations.
We see
descriptive and not-first-false claims
about fractions and B.locations.
We can see (by seeing claims)
that B matches its first column.
On Tuesday, December 12, 2023 at 1:18:03 PM UTC-4, WM wrote:
On 12.12.2023 15:29, William wrote:
Please adhere to my notation
Whatever notation you use, you only need dark numbers to get a location for a "fraction" in your putative matrix A.
On Tuesday, December 12, 2023 at 3:19:10 PM UTC-4, WM wrote:
On 12.12.2023 18:56, William wrote:
On Tuesday, December 12, 2023 at 1:18:03 PM UTC-4, WM wrote:I need locations for all fractions p/q with q > 1 because I know that
On 12.12.2023 15:29, William wrote:
Please adhere to my notation
Whatever notation you use, you only need dark numbers to get a location for a "fraction" in your putative matrix A.
they never can leave the matrix.
Indeed, the "fractions" never leave the A(n) which are the only matrices produced by your billiards game.
Since your billiards game never produces the putative matrix A, it does not show the existence of "dark numbers".
On Tuesday, December 12, 2023 at 5:46:27 PM UTC-4, WM wrote:By the formula used in the OP.
If A is not produced by usingNope. B is defined.
Cantor's formula, then also B is not produced by Cantor's formula.
Your putative A is not defined.
On Wednesday, December 13, 2023 at 7:40:43 AM UTC-4, WM wrote:
William schrieb am Dienstag, 12. Dezember 2023 um 22:54:13 UTC+1:
Your putative A is not defined.All matrices A(n) are defined
Correct. So what?. Your putative A is not defined.
On Wednesday, December 13, 2023 at 12:40:05 PM UTC-4, WM wrote:
On 13.12.2023 13:12, William wrote:
On Wednesday, December 13, 2023 at 7:40:43 AM UTC-4, WM wrote:So what? It is irrelevant. There is no matrix which has lost any
William schrieb am Dienstag, 12. Dezember 2023 um 22:54:13 UTC+1:
Your putative A is not defined.All matrices A(n) are defined
Correct. So what?. Your putative A is not defined.
fraction. Hence all are there and occupy not indexed positions.
Nope, In each A(n) only a finite number of elements have been moved, and each of them has only been moved a finite number of positions.
However, your putative A is not defined.
On 12.12.2023 18:55, Jim Burns wrote:
On 12/11/2023 5:28 AM, WM wrote:
On 10.12.2023 21:05, Jim Burns wrote:
You (WM) deny that and claim that
after all and only non-populating swaps,
A\B has become populated.
The elements of A\B are never changed.
Yes.
They are all positive fractions
with exception of unit fractions.
You (WM) deny that and claim that
after all and only non-populating swaps,
A\B has become populated.
The elements of A\B are never changed.
Yes.
They are all positive fractions
with exception of unit fractions.
After all B.swaps ⟨i,j⟩↔⟨kᵢⱼ,1⟩ inside B
for all B.locations ⟨i,j⟩ in B
all fractions i/j (j≠1) are not anywhere in B
No, they are not in B but also
not outside of A
(which exists after all swaps as well as B).
Fact is that B never disappears.
After all B.swaps ⟨i,j⟩↔⟨kᵢⱼ,1⟩ inside B
for all B.locations ⟨i,j⟩ in B
all fractions i/j (j≠1) are not anywhere in B
I'm willing to call that disappearing
They are in the darkness.
Fact is that B never disappears.
After all B.swaps ⟨i,j⟩↔⟨kᵢⱼ,1⟩ inside B
for all B.locations ⟨i,j⟩ in B
all fractions i/j (j≠1) are not anywhere in B
I'm willing to call that disappearing
They are in the darkness.
The problem with explaining
that so-called "disappearance" as
not-a-disappearance
but as darkᵂᴹ events happening unseenᵂᴹ
is that
we don't seeᵂᴹ or not-seeᵂᴹ any of the
fractions or B.locations.
We see
descriptive and not-first-false claims
about fractions and B.locations.
In fact most of all matrices, including B
is dark.
All that can be defined of any infinite matrix,
including B, is
a triangle in the upper left corner.
We can see (by seeing claims)
that B matches its first column.
The first column can be seen until
some defined n. But almost all is dark:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
On Wednesday, December 13, 2023 at 2:58:14 PM UTC-4, WM wrote:
On 13.12.2023 19:36, William wrote:
On Wednesday, December 13, 2023 at 12:40:05 PM UTC-4, WM wrote:All that can be moved has been moved,
On 13.12.2023 13:12, William wrote:
On Wednesday, December 13, 2023 at 7:40:43 AM UTC-4, WM wrote:So what? It is irrelevant. There is no matrix which has lost any
William schrieb am Dienstag, 12. Dezember 2023 um 22:54:13 UTC+1:
Your putative A is not defined.All matrices A(n) are defined
Correct. So what?. Your putative A is not defined.
fraction. Hence all are there and occupy not indexed positions.
Nope, In each A(n) only a finite number of elements have been moved, and each of them has only been moved a finite number of positions.
This does not change the fact that in each A(n) only a finite number of elements have been moved and each of them has only been moved a finite number of times.
Hence producing any A(n) does not show that "dark" numbers are needed.
Only producing the putative A shows that "dark" numbers are needed.
die Matrix B = (b_n,m) wird nicht "produziert", sondern ist (wiefolgt) _definiert_.
Note: The matrix B = (b_n,m) is defined by b_n,m = (m + n - 1)(m + n- 2)/2 + m for all n,m e IN.
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 Thursday, December 14, 2023 at 11:32:42 AM UTC-4, WM wrote:
Profilbild von WM
WM
ungelesen,
16:22 (vor 9 Minuten)
an
On 14.12.2023 15:23, William wrote:
On Thursday, December 14, 2023 at 6:08:40 AM UTC-4, WM wrote:It involves the realization of all natural numbers
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
Nope it does not involve "realization"
k = (m + n - 1)(m + n - 2)/2 + mAn algorithm which defines B. A matrix is defined if there is an algorithm which gives the value, given an element of |Nx|N. There is no algorithm for your putative A.
On 13.12.2023 23:14, Jim Burns wrote:
nothing that
could disprove the fact that
by exchanging X and O in
XOOO...
XOOO...
XOOO...
XOOO...
...
the matrix will never be covered with X.
Simple geometry.
On Thursday, December 14, 2023 at 12:23:01 PM UTC-4, WM wrote:
On 14.12.2023 16:45, William wrote:Nope, your billiards game defines A(n) for each n. Your billiards game does not define A,
A matrix is defined if there is an algorithm which gives the value, given an element of |Nx|N. There is no algorithm for your putative A.You are wrong. The algorithm is given in the OP.
On 12/14/2023 5:16 AM, WM wrote:
On 13.12.2023 23:14, Jim Burns wrote:
nothing that
could disprove the fact that
by exchanging X and O in
XOOO...
XOOO...
XOOO...
XOOO...
...
the matrix will never be covered with X.
Simple geometry.
"Geometry", he says.
Geometry is thronging with
sets you deny exist.
Sets which are the same "size" as
proper subsets is the explanation.
It is Cantor's explanation.
Profilbild von WM
WM
ungelesen,
16:22 (vor 9 Minuten)
an
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. Your belief alone is not sufficient.
Regards, WM
On 14.12.2023 20:00, Jim Burns wrote:
On 12/14/2023 5:16 AM, WM wrote:
nothing that
could disprove the fact that
by exchanging X and O in
XOOO...
XOOO...
XOOO...
XOOO...
...
the matrix will never be covered with X.
Simple geometry.
"Geometry", he says.
Geometry is thronging with
sets you deny exist.
Sets exist and
their elements remain existing when
redistributing them.
Sets which are the same "size" as
proper subsets is the explanation.
It is Cantor's explanation.
It is wrong.
Consider the unit fractions.
For every x > 0 that you choose
you know that
infinitely many unit fractions are smaller.
What can you do to distinguish them
in order to enumerate them? Nothing!
No chosen x will separate them.
On Thursday, December 14, 2023 at 3:24:58 PM UTC-4, WM wrote:
On 14.12.2023 17:34, William wrote:A. I.e. given (n,m) an element of |Nx|N how can you determine the value A[m,n]. Your game of billiards does not give a method.
On Thursday, December 14, 2023 at 12:23:01 PM UTC-4, WM wrote:What defines
On 14.12.2023 16:45, William wrote:Nope, your billiards game defines A(n) for each n. Your billiards game does not define A,
A matrix is defined if there is an algorithm which gives the value, given an element of |Nx|N. There is no algorithm for your putative A.You are wrong. The algorithm is given in the OP.
On 12/14/23 10:32 AM, WM wrote:
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. Your belief alone is not
sufficient.
So, what IS the "last Step" that is required?
You don't seem to understand that Natural Numbers are Unbounded, and
thus there isn't a "last" one.
On 12/14/2023 2:31 PM, WM wrote:
Because we cannot perform a supertask,
there are only finitely.many claims we've made,
Irony:
It is because we are finite that we know that
our claims are correct about infinity.
Sets which are the same "size" as
proper subsets is the explanation.
It is Cantor's explanation.
It is wrong.
Consider the unit fractions.
For every x > 0 that you choose
you know that
infinitely many unit fractions are smaller.
Our choosing and our not-choosing
does not change our description of reals,
rationals, integers, naturals, or
unit.fractions to a different description.
Our choosing and our not-choosing
does not change the visibly.not.first.false
to anything less visibly.not.first.false.
Perhaps you're depending on an unstated
| since we aren't infinite beings,
| we can't choose infinitely-many things.
I agree that we finite beings can't choose
infinitely.many things.
But that's not how we know these facts.
What can you do to distinguish them
in order to enumerate them? Nothing!
No chosen x will separate them.
I agree that we finite beings can't distinguish
infinitely.many things.
I agree that we finite beings can't enumerate
infinitely.many things.
But that's not how we know these facts.
on 12/7/2023, WM supposed :
My game of billiards defines
and places every natural number that is defined by
k = (m + n - 1)(m + n - 2)/2 + m.
Are you saying that the pairing formula 'defines' the natural numbers?
On Friday, December 15, 2023 at 6:10:03 AM UTC-4, WM wrote:
Dark elements
Are not elements of the set of natural numbers.
Your game of billiards does not provide an algorithm that defines A.
On 07.12.2023 13:20, FromTheRafters wrote:
on 12/7/2023, WM supposed :
My game of billiards defines and places every natural number that is
defined by
k = (m + n - 1)(m + n - 2)/2 + m.
Are you saying that the pairing formula 'defines' the natural numbers?
The index k of the position (m, n) is defined by the choice of m and n.
For dark positions (that cannot b chosen) no index can be defined.
Therefore the formula does ot define the whole matrix B.
Regards, WM
On 15.12.2023 02:00, Richard Damon wrote:
On 12/14/23 10:32 AM, WM wrote:
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. Your belief alone is not
sufficient.
So, what IS the "last Step" that is required?
You don't seem to understand that Natural Numbers are Unbounded, and
thus there isn't a "last" one.
Therefore the complete indexing can only happen in the limit or never.
In fact it will never happen.
Simple example: For all x > 0 that can be chosen there exist infinitely
many smaller fractions. They cannot be distinguished. They cannot be
indexed.
Regards, WM
On 15.12.2023 05:49, Jim Burns wrote:
On 12/14/2023 2:31 PM, WM wrote:
What can you do to distinguish them
in order to enumerate them? Nothing!
No chosen x will separate them.
I agree that we finite beings can't distinguish
infinitely.many things.
That is not the problem!
Simply try to distinguish
any unit fraction which has
not infinitely many smaller ones.
Such unit fractions must exist,
because
they occupy separate points in a linear order.
Never infinitely many sit at the same point.
WM has brought this to us :
On 15.12.2023 05:49, Jim Burns wrote:
On 12/14/2023 2:31 PM, WM wrote:
Because we cannot perform a supertask,
there are only finitely.many claims we've made,
How do you enumerate the algebraic numbers?
All at once.
On 12/15/2023 5:35 AM, WM wrote:
On 15.12.2023 05:49, Jim Burns wrote:
On 12/14/2023 2:31 PM, WM wrote:
What can you do to distinguish them
in order to enumerate them? Nothing!
No chosen x will separate them.
I agree that we finite beings can't distinguish
infinitely.many things.
That is not the problem!
Simply try to distinguish
any unit fraction which has
not infinitely many smaller ones.
Such unit fractions must exist,
because
they occupy separate points in a linear order.
No.
Le 15/12/2023 à 18:27, FromTheRafters a écrit :
WM has brought this to us :
On 15.12.2023 05:49, Jim Burns wrote:
On 12/14/2023 2:31 PM, WM wrote:
Because we cannot perform a supertask,
there are only finitely.many claims we've made,
How do you enumerate the algebraic numbers?
All at once.
Not true, impossible. All at once is an oximoronic statement.
On 12/15/23 5:17 AM, WM wrote:
The index k of the position (m, n) is defined by the choice of m and n.The problem is that there are no positions that can not be chosen.
For dark positions (that cannot b chosen) no index can be defined.
Therefore the formula does ot define the whole matrix B.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That is your false assumption.
On 12/15/23 5:22 AM, WM wrote:
Therefore the complete indexing can only happen in the limit or never.
In fact it will never happen.
No, it happens "at the limit" since to set is countably infinite.
Simple example: For all x > 0 that can be chosen there exist infinitely
many smaller fractions. They cannot be distinguished. They cannot be
indexed.
If x is your unit fractions or the rationals, then they CAN be indexed,
just not from lowest to highest.
Le 15/12/2023 à 18:58, Jim Burns a écrit :
On 12/15/2023 5:35 AM, WM wrote:
On 15.12.2023 05:49, Jim Burns wrote:
On 12/14/2023 2:31 PM, WM wrote:
What can you do to distinguish them
in order to enumerate them? Nothing!
No chosen x will separate them.
I agree that we finite beings can't distinguish
infinitely.many things.
That is not the problem!
Simply try to distinguish
any unit fraction which has
not infinitely many smaller ones.
Such unit fractions must exist,
because
they occupy separate points in a linear order.
No.
Yes. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n.
∀x ∈ (0, 1]: NUF(x) = ℵo is wrong.
Regards, WM
Le 15/12/2023 à 13:53, Richard Damon a écrit :
On 12/15/23 5:22 AM, WM wrote:
Therefore the complete indexing can only happen in the limit or never.
In fact it will never happen.
No, it happens "at the limit" since to set is countably infinite.
Nothing happens at the limit i.e., after all finite terms.
Simple example: For all x > 0 that can be chosen there exist
infinitely many smaller fractions. They cannot be distinguished. They
cannot be indexed.
If x is your unit fractions or the rationals, then they CAN be
indexed, just not from lowest to highest.
Show us how you do that for the ℵ₀ unit fractions which are smaller than every chosen x.
Regards, WM
Given n such that 1/n+1 < x <= 1/n, (or n == 0 if 1 < x) then the ℵ₀unit fractions smaller than that x are the set of 1/(n+k) for k = 1, 2,
On 15.12.2023 09:25, William wrote:
On Thursday, December 14, 2023 at 3:24:58 PM UTC-4, WM wrote:
On 14.12.2023 17:34, William wrote:A. I.e. given (n,m) an element of |Nx|N how can you determine the
On Thursday, December 14, 2023 at 12:23:01 PM UTC-4, WM wrote:What defines
On 14.12.2023 16:45, William wrote:Nope, your billiards game defines A(n) for each n. Your billiards
A matrix is defined if there is an algorithm which gives theYou are wrong. The algorithm is given in the OP.
value, given an element of |Nx|N. There is no algorithm for your
putative A.
game does not define A,
value A[m,n]. Your game of billiards does not give a method.
Dark elements cannot be determined as individuals.
The same problem is inherent in enumerating the algebraic numbers.
The same problem is inherent in enumerating the fractions.
For every eps > 0 there are infinitely many fractions between 0 and eps.
How should they be indexed?
All those tasks cannot be executed.
Regards, WM
Le 15/12/2023 à 13:55, Richard Damon a écrit :
On 12/15/23 5:17 AM, WM wrote:
The index k of the position (m, n) is defined by the choice of m and n.The problem is that there are no positions that can not be chosen.
For dark positions (that cannot b chosen) no index can be defined.
Therefore the formula does ot define the whole matrix B.
You don't see them. But they are there.
Try to identify those which I cannot identify because they are larger
than every identified natural number:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That is your false assumption.
No that is mathematics, contrary to the belief of matheologial cranks.
Regards, WM
On 12/15/23 5:43 PM, WM wrote:
Show us how you do that for the ℵ₀ unit fractions which are smaller than >> every chosen x.unit fractions smaller than that x are the set of 1/(n+k) for k = 1, 2,
Given n such that 1/n+1 < x <= 1/n, (or n == 0 if 1 < x) then the ℵ₀
3, ...
That IS an unbounded enumeration that visits EVERY unit fraction smaller
than x. It enumerates them from highest to lowest, since the unit
fractions form a decreasing series.
On 12/15/23 5:48 PM, WM wrote:
I said:
The problem is that there are no positions that can not be chosen.
And that means that there is no element of ℕ that can't become a member
of ℕ_def,
and thus, there are no member of ℕ that are always left
behind, and thus no elements left to be "dark".
You say ℕ_def is "potentially" infinite, which means "Dark Numbers" are "potentially non-existent",
On 12/15/23 5:32 PM, WM wrote:
That is not the problem!
Simply try to distinguish
any unit fraction which has
not infinitely many smaller ones.
Such unit fractions must exist,
because
they occupy separate points in a linear order.
No.
Yes. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n.
∀x ∈ (0, 1]: NUF(x) = ℵo is wrong.
Then name a value that contradicts it!
Your concept of the function says it only has "finite values>0" for
input values that can not be used as individual numbers, and thus,
nowhere is your NUF finite and not 0.
Le 16/12/2023 à 02:29, Richard Damon a écrit :
On 12/15/23 5:48 PM, WM wrote:
I said:
The problem is that there are no positions that can not be chosen.
And that means that there is no element of ℕ that can't become a member
of ℕ_def,
That is wrong.
and thus, there are no member of ℕ that are always left
behind, and thus no elements left to be "dark".
You despise mathematics which proves the contrary:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
You say ℕ_def is "potentially" infinite, which means "Dark Numbers" are
"potentially non-existent",
Dark numbers are the matter that is required to make numbers visible.
Le 16/12/2023 à 02:29, Richard Damon a écrit :
On 12/15/23 5:48 PM, WM wrote:
I said:
The problem is that there are no positions that can not be chosen.
And that means that there is no element of ℕ that can't become a
member of ℕ_def,
That is wrong.
and thus, there are no member of ℕ that are always left behind, and
thus no elements left to be "dark".
You despise mathematics which proves the contrary:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
You say ℕ_def is "potentially" infinite, which means "Dark Numbers"
are "potentially non-existent",
Dark numbers are the matter that is required to make numbers visible.
But almost all dark numbers cannot be made visible because otherwise
nothing would be left between n and ω.
Regards, WM
On 12/16/23 3:37 AM, WM wrote:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
But you can't define the actual set ℕ_def, so that doesn't mean anything.
WM was thinking very hard :
Le 15/12/2023 à 21:52, Jim Reburns a écrit :
Not true, impossible. All at once is an oximoronic statement.
Enumerating is a stepwise process.
No it isn't. This is just a bijective function being applied to the
whole set. Nothing about steps at all.
For algebraic numbers the polynomials of
height n have to be searched one after the other.
Have they found them all yet?
On 12/16/23 3:32 AM, WM wrote:
Le 16/12/2023 à 02:21, Richard Damon a écrit :
On 12/15/23 5:43 PM, WM wrote:
Show us how you do that for the ℵ₀ unit fractions which are smaller >>>> than every chosen x.2, 3, ...
Given n such that 1/n+1 < x <= 1/n, (or n == 0 if 1 < x) then the ℵ₀ >>> unit fractions smaller than that x are the set of 1/(n+k) for k = 1,
That IS an unbounded enumeration that visits EVERY unit fraction
smaller than x. It enumerates them from highest to lowest, since the
unit fractions form a decreasing series.
In every case ℵ₀ unit fractions are left. Show us a unit fraction having >> less than ℵ₀ smaller unit fractions. Fail.
Note that by ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 the increase from 0 happens
in steps of height 1. If you deny that then you are outside of
mathematics and logic in the realm of glowing faith but irrelevant for
mathematical work.
I don't need to, that is bounded logic,
You are the one that is going "outside" of mathematics,
Le 16/12/2023 à 13:13, FromTheRafters a écrit :
WM was thinking very hard :
Le 15/12/2023 à 21:52, Jim Reburns a écrit :
Not true, impossible. All at once is an oximoronic statement.
Enumerating is a stepwise process.
No it isn't. This is just a bijective function being applied to the
whole set. Nothing about steps at all.
This answer shows that unrealistic fools have seized power and try to
shut down mathematics. Counting is done step by step.
For algebraic numbers the polynomials of height n have to be searched
one after the other.
Have they found them all yet?
No infinite set will ever be counted.
Regards, WM
Le 16/12/2023 à 12:56, Richard Damon a écrit :
On 12/16/23 3:37 AM, WM wrote:
There is no set but only a potentially infinite collection. Just like∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
But you can't define the actual set ℕ_def, so that doesn't mean anything. >>
the collection of known prime numbers.
Regards, WM
Le 16/12/2023 à 12:56, Richard Damon a écrit :
On 12/16/23 3:32 AM, WM wrote:
Le 16/12/2023 à 02:21, Richard Damon a écrit :
On 12/15/23 5:43 PM, WM wrote:
Show us how you do that for the ℵ₀ unit fractions which are smaller >>>>> than every chosen x.unit fractions smaller than that x are the set of 1/(n+k) for k = 1,
Given n such that 1/n+1 < x <= 1/n, (or n == 0 if 1 < x) then the ℵ₀
2, 3, ...
That IS an unbounded enumeration that visits EVERY unit fraction
smaller than x. It enumerates them from highest to lowest, since the
unit fractions form a decreasing series.
In every case ℵ₀ unit fractions are left. Show us a unit fraction
having less than ℵ₀ smaller unit fractions. Fail.
Note that by ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 the increase from 0
happens in steps of height 1. If you deny that then you are outside
of mathematics and logic in the realm of glowing faith but irrelevant
for mathematical work.
I don't need to, that is bounded logic,
It is mathematics.
You are the one that is going "outside" of mathematics,
No. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 is basic.
Regards, WM
Le 16/12/2023 à 12:56, Richard Damon a écrit :
On 12/16/23 3:37 AM, WM wrote:
There is no set but only a potentially infinite collection. Just∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
But you can't define the actual set ℕ_def, so that doesn't mean anything. >>
like the collection of known prime numbers.
On 12/16/23 12:47 PM, WM wrote:
Le 16/12/2023 à 13:13, FromTheRafters a écrit :
WM was thinking very hard :
Le 15/12/2023 à 21:52, Jim Reburns a écrit :
Not true, impossible. All at once is an oximoronic statement.
Enumerating is a stepwise process.
No it isn't. This is just a bijective function being applied to the
whole set. Nothing about steps at all.
This answer shows that unrealistic fools have seized power and try to
shut down mathematics. Counting is done step by step.
Then you can't have the full "Counting Numbers" as you never get to the end.
For algebraic numbers the polynomials of height n have to be searched
one after the other.
Have they found them all yet?
No infinite set will ever be counted.
Which is why you can't insist on "counting" step by step sets that are infinite.
On 12/16/23 12:53 PM, WM wrote:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 is basic.
Which doesn't say what you want it to say,
On 12/16/23 12:50 PM, WM wrote:
Le 16/12/2023 à 12:56, Richard Damon a écrit :
On 12/16/23 3:37 AM, WM wrote:There is no set but only a potentially infinite collection. Just like
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
But you can't define the actual set ℕ_def, so that doesn't mean anything. >>>
the collection of known prime numbers.
Regards, WM
Which makes it NOT a "Set of Numbers" and thus can't be removed from a
"Set of Numbers"
This makes your "Dark Numbers" not numbers that CAN'T be used
individually, just those that haven't yet been, but still could be in
the future.
Le 16/12/2023 à 19:02, Richard Damon a écrit :
On 12/16/23 12:47 PM, WM wrote:
Le 16/12/2023 à 13:13, FromTheRafters a écrit :
WM was thinking very hard :
Le 15/12/2023 à 21:52, Jim Reburns a écrit :
Not true, impossible. All at once is an oximoronic statement.
Enumerating is a stepwise process.
No it isn't. This is just a bijective function being applied to the
whole set. Nothing about steps at all.
This answer shows that unrealistic fools have seized power and try to
shut down mathematics. Counting is done step by step.
Then you can't have the full "Counting Numbers" as you never get to
the end.
Bingo!
For algebraic numbers the polynomials of height n have to be
searched one after the other.
Have they found them all yet?
No infinite set will ever be counted.
Which is why you can't insist on "counting" step by step sets that are
infinite.
The desire to count infinite sets does not relieve the rules of counting.
Regards, WM
Le 16/12/2023 à 19:03, Richard Damon a écrit :
On 12/16/23 12:53 PM, WM wrote:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 is basic.
Which doesn't say what you want it to say,
It says that NUF(x) cannot grow to 2 without having paused at 1.
Regards, WM
Le 16/12/2023 à 19:05, Richard Damon a écrit :
On 12/16/23 12:50 PM, WM wrote:
Le 16/12/2023 à 12:56, Richard Damon a écrit :
On 12/16/23 3:37 AM, WM wrote:There is no set but only a potentially infinite collection. Just like
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
But you can't define the actual set ℕ_def, so that doesn't mean
anything.
the collection of known prime numbers.
Regards, WM
Which makes it NOT a "Set of Numbers" and thus can't be removed from a
"Set of Numbers"
Yes, it is hard to delimit ℕ_def.
This makes your "Dark Numbers" not numbers that CAN'T be used
individually, just those that haven't yet been, but still could be in
the future.
Many can be used, but almost all cannot. Because you cannot use a unit fraction that has only few smaller unit fractions.
Regards, WM
On 15.12.2023 05:49, Jim Burns wrote:
Irony:
It is because we are finite that we know that
our claims are correct about infinity.
No,
because your are stupid, you believe that.
That is one of the strange properties of unboundely small values, while
any distance is finite, you can have an infinite number of them before
any finite distance.
On 12/16/23 12:53 PM, WM wrote:
No. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 is basic.Which doesn't say what you want it to say,
The LOGIC you apply to that statement
can't handle unbounded values
On 16.12.2023 21:51, Richard Damon wrote:
That is one of the strange properties of unboundely small values,
while any distance is finite, you can have an infinite number of them
before any finite distance.
That is not a strange property but the belief of a crank.
Regards, WM
On 16.12.2023 19:03, Richard Damon wrote:
On 12/16/23 12:53 PM, WM wrote:
No. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 is basic.Which doesn't say what you want it to say,
The LOGIC you apply to that statement
says that even one non-vanishing distance is > 0.
can't handle unbounded values
shows the nonsense entertained by matheologians in brightest light.
Regards, WM
That your brain is unable to comprehend how this might work is a problem
with your brain, not mathematics.
On 16.12.2023 21:51, Richard Damon wrote:
That is one of the strange properties of
unboundely small values,
while any distance is finite,
you can have an infinite number of them
before any finite distance.
That is not a strange property
but the belief of a crank.
On 12/17/2023 7:38 AM, WM wrote:
On 16.12.2023 21:51, Richard Damon wrote:
That is one of the strange properties of
unboundely small values,
while any distance is finite,
you can have an infinite number of them
before any finite distance.
That is not a strange property
but the belief of a crank.
Le 17/12/2023 à 13:55, Richard Damon a écrit :
That your brain is unable to comprehend how this might work is aYou said: while any distance is finite, you can have an infinite number
problem with your brain, not mathematics.
of them before any finite distance.
How do you distinguish then some of these infinitely many unit fractions which lie before any finite distance but are not dark?
Regards, WM
On 12/17/23 8:27 AM, WM wrote:
Le 17/12/2023 à 13:55, Richard Damon a écrit :
That your brain is unable to comprehend how this might work is aYou said: while any distance is finite, you can have an infinite
problem with your brain, not mathematics.
number of them before any finite distance.
How do you distinguish then some of these infinitely many unit
fractions which lie before any finite distance but are not dark?
Regards, WM
We can give every one of them a unique name, as they are one over some natural number, which all have names.
On 12/17/23 8:27 AM, WM wrote:
Le 17/12/2023 à 13:55, Richard Damon a écrit :We can give every one of them a unique name, as they are one over some natural number, which all have names.
That your brain is unable to comprehend how this might work is aYou said: while any distance is finite, you can have an infinite number
problem with your brain, not mathematics.
of them before any finite distance.
How do you distinguish then some of these infinitely many unit fractions
which lie before any finite distance but are not dark?
On 12/17/23 12:31 PM, Richard Damon wrote:
On 12/17/23 8:27 AM, WM wrote:
You said: while any distance is finite, you can have an infiniteWe can give every one of them a unique name, as they are one over some
number of them before any finite distance.
How do you distinguish then some of these infinitely many unit
fractions which lie before any finite distance but are not dark?
natural number, which all have names.
In fact, I will ask why you think there is any problem giving them a name.
After all at the point 1/n the length of the flat point before it is of lenght 1/n - 1/(n+1) = 1/(n*(n+1)), so even if all the proceeding lines
were the same length, we would have at least n+1 more gaps, and the fact
that they get smaller we get even more.
So even without actually traveling to that smaller space, we can see
that given we are at 1/n, we can get to at least 1/(2n) in the space available, so there is always room for more.
Le 17/12/2023 à 18:31, Richard Damon a écrit :
On 12/17/23 8:27 AM, WM wrote:
Le 17/12/2023 à 13:55, Richard Damon a écrit :We can give every one of them a unique name, as they are one over some
That your brain is unable to comprehend how this might work is aYou said: while any distance is finite, you can have an infinite
problem with your brain, not mathematics.
number of them before any finite distance.
How do you distinguish then some of these infinitely many unit
fractions which lie before any finite distance but are not dark?
natural number, which all have names.
Then give one of those unit fractions which lie before any finite
distance a name.
Regards, WM
Le 17/12/2023 à 18:53, Richard Damon a écrit :
On 12/17/23 12:31 PM, Richard Damon wrote:
On 12/17/23 8:27 AM, WM wrote:
You said: while any distance is finite, you can have an infiniteWe can give every one of them a unique name, as they are one over
number of them before any finite distance.
How do you distinguish then some of these infinitely many unit
fractions which lie before any finite distance but are not dark?
some natural number, which all have names.
In fact, I will ask why you think there is any problem giving them a
name.
That's easy. Every name you give belongs to a unit fraction which dos
not lie before any finite distance.
After all at the point 1/n the length of the flat point before it is
of lenght 1/n - 1/(n+1) = 1/(n*(n+1)), so even if all the proceeding
lines were the same length, we would have at least n+1 more gaps, and
the fact that they get smaller we get even more.
So even without actually traveling to that smaller space, we can see
that given we are at 1/n, we can get to at least 1/(2n) in the space
available, so there is always room for more.
All named unit fractions have finite distances from 0. I am asking for
those which "lie before any finite distance" but are not dark.
Regards, WM
Le 17/12/2023 à 18:31, Richard Damon a écrit :
On 12/17/23 8:27 AM, WM wrote:
Le 17/12/2023 à 13:55, Richard Damon a écrit :We can give every one of them a unique name, as they are one over some
That your brain is unable to comprehend how this might work is aYou said: while any distance is finite, you can have an infinite
problem with your brain, not mathematics.
number of them before any finite distance.
How do you distinguish then some of these infinitely many unit
fractions which lie before any finite distance but are not dark?
natural number, which all have names.
Then give one of those unit fractions which lie before any finite
distance a name.
Regards, WM
Le 17/12/2023 à 16:24, Jim Burns a écrit :
On 12/17/2023 7:38 AM, WM wrote:
On 16.12.2023 21:51, Richard Damon wrote:
That is one of the strange properties of
unboundely small values,
while any distance is finite,
you can have an infinite number of them
before any finite distance.
That is not a strange property
but the belief of a crank.
Here you have found a congenial fellow.
Let an infinite number of bObs disappear
Here you have found a congenial fellow.
and let an infinite number of
finite distances disappear.
But oppose darkness vehemently.
On 12/17/2023 12:20 PM, WM wrote:
Perhaps
you consider not-asserting contradictions
to be vehemence, and
Define
super.visibleᵂᴹ n to be
visibleᵂᴹ n with
each number before n visibleᵂᴹ
On 12/17/23 2:17 PM, WM wrote:
Note, "Any" means I can choose one.
You mean give a unit fraction that is before EVERY unit fraction, but
that is logically impossible, how can there be a member of a set that
isn't part of the set it is supposed to be part of.
On 12/17/23 2:17 PM, WM wrote:
We can give every one of them a unique name, as they are one over some
natural number, which all have names.
Then give one of those unit fractions which lie before any finite
distance a name.
Non-existant.
On 12/17/23 2:22 PM, WM wrote:
Le 17/12/2023 à 18:53, Richard Damon a écrit :
On 12/17/23 12:31 PM, Richard Damon wrote:
On 12/17/23 8:27 AM, WM wrote:
You said: while any distance is finite, you can have an infiniteWe can give every one of them a unique name, as they are one over
number of them before any finite distance.
How do you distinguish then some of these infinitely many unit
fractions which lie before any finite distance but are not dark?
some natural number, which all have names.
In fact, I will ask why you think there is any problem giving them a
name.
That's easy. Every name you give belongs to a unit fraction which dos
not lie before any finite distance.
Why does it need to lay before "ANY" finite distance, as that presumes
that there IS a "first" finite difference
After all at the point 1/n the length of the flat point before it is
of lenght 1/n - 1/(n+1) = 1/(n*(n+1)), so even if all the proceeding
lines were the same length, we would have at least n+1 more gaps, and
the fact that they get smaller we get even more.
So even without actually traveling to that smaller space, we can see
that given we are at 1/n, we can get to at least 1/(2n) in the space
available, so there is always room for more.
All named unit fractions have finite distances from 0. I am asking for
those which "lie before any finite distance" but are not dark.
Why?
That would be asking me to find your non-existant "Dark" numbers.
Le 17/12/2023 à 21:18, Richard Damon a écrit :
On 12/17/23 2:22 PM, WM wrote:
Le 17/12/2023 à 18:53, Richard Damon a écrit :
On 12/17/23 12:31 PM, Richard Damon wrote:
On 12/17/23 8:27 AM, WM wrote:
You said: while any distance is finite, you can have an infiniteWe can give every one of them a unique name, as they are one over
number of them before any finite distance.
How do you distinguish then some of these infinitely many unit
fractions which lie before any finite distance but are not dark?
some natural number, which all have names.
In fact, I will ask why you think there is any problem giving them a
name.
That's easy. Every name you give belongs to a unit fraction which dos
not lie before any finite distance.
Why does it need to lay before "ANY" finite distance, as that presumes
that there IS a "first" finite difference
That is one of the strange properties of unboundely small values, while
any distance is finite, you can have an infinite number of them before
any finite distance.
After all at the point 1/n the length of the flat point before it is
of lenght 1/n - 1/(n+1) = 1/(n*(n+1)), so even if all the proceeding
lines were the same length, we would have at least n+1 more gaps,
and the fact that they get smaller we get even more.
So even without actually traveling to that smaller space, we can see
that given we are at 1/n, we can get to at least 1/(2n) in the space
available, so there is always room for more.
All named unit fractions have finite distances from 0. I am asking
for those which "lie before any finite distance" but are not dark.
Why?
That would be asking me to find your non-existant "Dark" numbers.
So it is. That is one of the strange properties of unboundely small
values, while any distance is finite, you can have an infinite number of
them before any finite distance. They are dark.
Regards, WM
Le 17/12/2023 à 21:21, Richard Damon a écrit :
On 12/17/23 2:17 PM, WM wrote:
Note, "Any" means I can choose one.
No, any means there is no other. Any includes the first distance between
two unit fractions.
You mean give a unit fraction that is before EVERY unit fraction, but
that is logically impossible, how can there be a member of a set that
isn't part of the set it is supposed to be part of.
So you agree that the unit fractions and their distances are all there
and that NUF can not increase by more than 1 without being constant for
a finite period,
As it starts with 0, the first increase is by 1.
Regards, WM
Before any given finite point, there are infinitely many smaller points.
So, your "darkness" property is based on the exisnce of things that do
not exist.
On 12/18/23 6:19 AM, WM wrote:
Right, so NUF(x) is infinite for ANY finite number, and thus for ANY
unit fraction, so it is NEVER 1 at any, so there is no smallest unit fraction.
That is one of the strange properties of unboundely small
values, while any distance is finite, you can have an infinite number of
them before any finite distance. They are dark.
They are not "Dark", they are just more of the finite unit fractions.
no need to make some of them "dark", as all of them are "definable"
Nope, NUF goes from 0 to 1 in the range of infintesimal numbers which
are NOT "Unit Fractions" so outside the actual domain of the numbers described
You need to clarify what number system you are working in. If we are
talking about the Natural Number (which can be built on some simple set theory) then these infintesimals, (or the transfinite numbers) don't exist.
Le 18/12/2023 à 15:21, Richard Damon a écrit :
Before any given finite point, there are infinitely many smaller points.
That means these are points, ℵ₀ unit fractions, which cannot be distinguished from each other by defining a point between them.
So, your "darkness" property is based on the exisnce of things that do
not exist.
The darkness property is based upon the non-existence of separators
between the ℵ₀ unit fractions.
Regards, WM
Le 18/12/2023 à 15:21, Richard Damon a écrit :
On 12/18/23 6:19 AM, WM wrote:
Right, so NUF(x) is infinite for ANY finite number, and thus for ANY
unit fraction, so it is NEVER 1 at any, so there is no smallest unit
fraction.
This is inconsistent with the property, accepted by you: NUF can not
increase by more than 1 without being constant for a finite period.
NUF(0) = 0 implies NUF(x) = 1.
That is one of the strange properties of unboundely small values,
while any distance is finite, you can have an infinite number of them
before any finite distance. They are dark.
They are not "Dark", they are just more of the finite unit fractions.
They cannot be separated by a finite distance.
no need to make some of them "dark", as all of them are "definable"
That contradicts what you said above: you can have an infinite number of
them before any finite distance. Please decide what you want to believe.
Regards, WM
Le 18/12/2023 à 15:27, Richard Damon a écrit :
Nope, NUF goes from 0 to 1 in the range of infintesimal numbers which
are NOT "Unit Fractions" so outside the actual domain of the numbers
described
NUF counts only unit fractions. It cannot increase to 1 without a unit fraction.
You need to clarify what number system you are working in. If we are
talking about the Natural Number (which can be built on some simple
set theory) then these infintesimals, (or the transfinite numbers)
don't exist.
They are not relevant. If NUF(x) = 1, then (0, x] has covered one unit fraction.
Regards, WM
Le 18/12/2023 à 02:35, Jim Burns a écrit :
On 12/17/2023 12:20 PM, WM wrote:
Perhaps
you consider not-asserting contradictions
to be vehemence, and
I consider Bob leaving the matrix as impossible.
Define
super.visibleᵂᴹ n to be
visibleᵂᴹ n with
each number before n visibleᵂᴹ
Each number before a visible number
is visible.
This is the definitions of visible n:
A FISON {1, 2, 3, ..., n} exists.
Regards, WM
Le 18/12/2023 à 02:35, Jim Burns a écrit :
Define
super.visibleᵂᴹ n to be
visibleᵂᴹ n with
each number before n visibleᵂᴹ
Each number before a visible number
is visible.
This is the definitions of visible n:
A FISON {1, 2, 3, ..., n} exists.
I consider Bob leaving the matrix as
impossible.
I consider Bob leaving the matrix as
impossible.
I consider Bob leaving the matrix as
impossible.
Nope, because every unit fraction is distinguishable individually, and
there are even rational values between them.
On 12/18/2023 6:36 AM, WM wrote:
Perhaps
you consider not-asserting contradictions
to be vehemence, and
I consider Bob leaving the matrix as impossible.
Each number before a visible number
is visible.
This is the definitions of visible n:
A FISON {1, 2, 3, ..., n} exists.
Okay.
On 12/18/23 11:00 AM, WM wrote:
Le 18/12/2023 à 15:27, Richard Damon a écrit :
You need to clarify what number system you are working in. If we are
talking about the Natural Number (which can be built on some simple
set theory) then these infintesimals, (or the transfinite numbers)
don't exist.
They are not relevant. If NUF(x) = 1, then (0, x] has covered one unit
fraction.
Which means that x is the smallest unit fraction, which doesn't exist.
So, NUF is NOT actually defined over the finite values, and is based on
bad logic.
On 12/18/23 10:56 AM, WM wrote:
Le 18/12/2023 à 15:21, Richard Damon a écrit :
On 12/18/23 6:19 AM, WM wrote:
Right, so NUF(x) is infinite for ANY finite number, and thus for ANY
unit fraction, so it is NEVER 1 at any, so there is no smallest unit
fraction.
This is inconsistent with the property, accepted by you: NUF can not
increase by more than 1 without being constant for a finite period.
NUF(0) = 0 implies NUF(x) = 1.
Yes, at an infintesimally small value of x.
NUF has a discontinuity at 0+, as the incremental property only applies
for FINITE values of x > 0.
That is one of the strange properties of unboundely small values,
while any distance is finite, you can have an infinite number of them
before any finite distance. They are dark.
They are not "Dark", they are just more of the finite unit fractions.
They cannot be separated by a finite distance.
But they are,
you don't seem to understand that "finite" can get
vanishingly small.
If "any" means, chose a specific one, then yes, for "Any" finite value,
there are an infinite number of finite values below it.
If "Any" means, "All", then the statement is just nonsense, it is asking
for an element of the set that isn't an element of the set.
Thus, for "any" definable x, there is an infinite number of definable
unit fractions below it, so for all definable x, the value of NUF(x) is infinite,
Since "dark" x's can't be used individually, and thus we can't evaluate NUF(x) at dark values, we have no points where NUF(x) is 1
there being
a discontinuiry of NUF(x) at 0+ at the gap between the definite number
0, and all the finite x > 0, caused by the fact that numbers are "dense".
Le 18/12/2023 à 17:27, Richard Damon a écrit :
Nope, because every unit fraction is distinguishable individually, and
there are even rational values between them.
Then not ℵ₀ unit fractions would remain below every eps which can be
used to distinguish them.
Regards, WM
Le 18/12/2023 à 17:31, Richard Damon a écrit :
On 12/18/23 11:00 AM, WM wrote:
Le 18/12/2023 à 15:27, Richard Damon a écrit :
You need to clarify what number system you are working in. If we are
talking about the Natural Number (which can be built on some simple
set theory) then these infintesimals, (or the transfinite numbers)
don't exist.
We are using only natural numbers and their reciprocals. No infinitesimals.
They are not relevant. If NUF(x) = 1, then (0, x] has covered one
unit fraction.
Which means that x is the smallest unit fraction, which doesn't exist.
It does exist, if this is correct for all natural numbers:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
So, NUF is NOT actually defined over the finite values, and is based
on bad logic.
NUF is defined everywhere.
Regards, WM
Le 18/12/2023 à 17:48, Jim Burns a écrit :
On 12/18/2023 6:36 AM, WM wrote:
I consider Bob leaving the matrix
as impossible.
Exchanging between two matrix-position
will never decrease
the exchanged elements.
Each number before a visible number
is visible.
This is the definitions of visible n:
A FISON {1, 2, 3, ..., n} exists.
Okay.
There is no sharp border between
visible and dark
On 12/19/2023 2:16 AM, WM wrote:
Le 18/12/2023 à 17:21, Richard Damon a écrit :
On 12/18/23 10:56 AM, WM wrote:
Le 18/12/2023 à 15:21, Richard Damon a écrit :
On 12/18/23 6:19 AM, WM wrote:
Right, so NUF(x) is infinite for ANY finite number, and thus for
ANY unit fraction, so it is NEVER 1 at any, so there is no smallest
unit fraction.
This is inconsistent with the property, accepted by you: NUF can not
increase by more than 1 without being constant for a finite period.
NUF(0) = 0 implies NUF(x) = 1.
Yes, at an infintesimally small value of x.
No unit fraction is infinitesimally small. But perhaps you call dark
values infinitesimally small.
Huh? 1/1, 1/2, 1/3, ...
They do get infinitely small, yet never equal zero...
On 12/19/2023 6:07 PM, Richard Damon wrote:
On 12/19/23 7:58 PM, Chris M. Thomasson wrote:
On 12/19/2023 2:16 AM, WM wrote:
Le 18/12/2023 à 17:21, Richard Damon a écrit :
On 12/18/23 10:56 AM, WM wrote:
Le 18/12/2023 à 15:21, Richard Damon a écrit :
On 12/18/23 6:19 AM, WM wrote:
Right, so NUF(x) is infinite for ANY finite number, and thus for >>>>>>> ANY unit fraction, so it is NEVER 1 at any, so there is no
smallest unit fraction.
This is inconsistent with the property, accepted by you: NUF can
not increase by more than 1 without being constant for a finite
period. NUF(0) = 0 implies NUF(x) = 1.
Yes, at an infintesimally small value of x.
No unit fraction is infinitesimally small. But perhaps you call dark
values infinitesimally small.
Huh? 1/1, 1/2, 1/3, ...
They do get infinitely small, yet never equal zero...
No, they get unboundedly small, but every one is finite, so it isn't
"infinitely" small. The key is unboundedly small means there is no
smallest item, but is just ever getting smaller, but stays a finitely
describable value.
NUF(x) = 1 only at an INFINITELY small value, which is smaller than
all UNBOUNDLY small values. The "Infinitely small values" are a
different type of number, NOT part of the standard definition of unit
fractions or the unbounded numeber systems like rational or reals.
These are similar classes of numbers as that which deals with the
mathematics of Omega and its friends. In other words, the
Transfinites, which WM has admitted are out of this domain of discussion.
In one sense, his "dark" numbers could be these transfinites, except
he claims that his "dark numbers" are specifically part of the Natural
Numbers / Unit Fractions (which the Transfinite are not). Also, the
Transfinite numbers are "nameable" and "usable individually" (just not
with sets limited to the finite numbers).
Humm... How about, humm, their ability to go to infinity is unbounded?
Is that okay? They do get smaller...
NUF actually does those finite steps at infinitesimals,
your NUF fails to be defined for x > 0.
On 12/19/23 5:16 AM, WM wrote:
Right, so NUF(x) is infinite for ANY finite number, and thus for ANY >>>>> unit fraction, so it is NEVER 1 at any, so there is no smallest unit >>>>> fraction.
This is inconsistent with the property, accepted by you: NUF can not
increase by more than 1 without being constant for a finite period.
NUF(0) = 0 implies NUF(x) = 1.
Yes, at an infintesimally small value of x.
No unit fraction is infinitesimally small. But perhaps you call dark
values infinitesimally small.
But they are UNBOUNDEDLY small, which means there is no smallest.
On 12/19/2023 5:42 AM, WM wrote:
Le 18/12/2023 à 17:48, Jim Burns a écrit :
On 12/18/2023 6:36 AM, WM wrote:
I consider Bob leaving the matrix
as impossible.
Exchanging between two matrix-position
will never decrease
the exchanged elements.
I think it can be a good and useful thing
to say
| 2¹ᐟ² is impossible
in a discussion about rationals.
However, that type of impossibility
has no power outside that discussion.
2¹ᐟ² is possible in other discussions.
There are matrices ⟨1,…,i⟩×⟨1,…,j⟩ for which
Bob leaving ⟨1,…,i⟩×⟨1,…,j⟩ is impossible.
On 12/19/23 5:25 AM, WM wrote:
Le 18/12/2023 à 17:27, Richard Damon a écrit :
Nope, because every unit fraction is distinguishable individually, and
there are even rational values between them.
Then not ℵ₀ unit fractions would remain below every eps which can be
used to distinguish them.
Why?
Remember, you are talking about an UNBOUNDED set
so it is inexhaustible.
No, they get unboundedly small, but every one is finite, so it isn't "infinitely" small.
NUF(x) = 1 only at an INFINITELY small value, which is smaller than all UNBOUNDLY small values.
Le 19/12/2023 à 15:07, Richard Damon a écrit :
On 12/19/23 5:16 AM, WM wrote:
Right, so NUF(x) is infinite for ANY finite number, and thus for
ANY unit fraction, so it is NEVER 1 at any, so there is no
smallest unit fraction.
This is inconsistent with the property, accepted by you: NUF can
not increase by more than 1 without being constant for a finite
period. NUF(0) = 0 implies NUF(x) = 1.
Yes, at an infintesimally small value of x.
No unit fraction is infinitesimally small. But perhaps you call dark
values infinitesimally small.
But they are UNBOUNDEDLY small, which means there is no smallest.
Each dunit fraction is finite and larger than zero. There is a smallest.
An increase of NUF(x) before every x > 0 is impossible. There is a first increase after 0, but not to more than 1, because after every unit
fraction there is a distance d_n > 0.
Regards, WM
Le 19/12/2023 à 15:07, Richard Damon a écrit :
On 12/19/23 5:25 AM, WM wrote:
Le 18/12/2023 à 17:27, Richard Damon a écrit :
Nope, because every unit fraction is distinguishable individually,
and there are even rational values between them.
Then not ℵ₀ unit fractions would remain below every eps which can be >>> used to distinguish them.
Why?
Remember, you are talking about an UNBOUNDED set
Wrong. The set of unit fractions is bounded by 0 and 1.
so it is inexhaustible.
NUF(x) exists for every point.
Regards, WM
Le 20/12/2023 à 03:07, Richard Damon a écrit :
No, they get unboundedly small, but every one is finite, so it isn't
"infinitely" small.
Correct.
NUF(x) = 1 only at an INFINITELY small value, which is smaller than
all UNBOUNDLY small values.
Smaller than all eps > 0 that can be chosen. So it is.
But you cannot oppose to this result: NUF(0) = 0 ad NUF(eps) = ℵo. That means the increase from 0 to ℵo cannot be seen and discerned. The axiom
of choice is invalid here.
Regards, WM
Le 19/12/2023 à 15:14, Richard Damon a écrit :
NUF actually does those finite steps at infinitesimals,
Whatever you call them, these steps cannot be investigated as
individuals. They are dark.
your NUF fails to be defined for x > 0.
The function is defined but its values are dark for small x.
Regards, WM
On 12/20/23 5:00 AM, WM wrote:
Each unit fraction is finite and larger than zero. There is a smallest.No, there is no smallest finite.
An increase of NUF(x) before every x > 0 is impossible. There is a first
increase after 0, but not to more than 1, because after every unit
fraction there is a distance d_n > 0.
If you disagree, name it.
There is no positive point with less than ℵo unit fractionsat its left-hand side.
All positive points with no exception have ℵo unit fractions attheir left-hand
The interval (0, 1] has aleph_0 unit fractions at its left-hand side. Contradiction.
Probably you support the "logic" of matheologians like Feldhase.
Le 20/12/2023 à 13:42, Richard Damon a écrit :
On 12/20/23 5:00 AM, WM wrote:
Each unit fraction is finite and larger than zero. There is aNo, there is no smallest finite.
smallest. An increase of NUF(x) before every x > 0 is impossible.
There is a first increase after 0, but not to more than 1, because
after every unit fraction there is a distance d_n > 0.
If you disagree, name it.
It is dark. ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 holds for all natnumbers.
You disagree with mathematics and with my logic:
Every positive point has ℵo unit fractions at its left-hand side.
There is no positive point with less than ℵo unit fractionsat its left-hand side.
All positive points with no exception have ℵo unit fractions attheir left-hand
side.
The interval (0, 1] has aleph_0 unit fractions at its left-hand side. Contradiction.
Probably you support the "logic" of matheologians like Feldhase.
There is no positive point with less than ℵo unit fractions at its left-hand side.
But that does not imply that ℵo unit fractions lie left of the interval.
But that is nonsense - if all points can be selected as the axiom of
choice proposes.
Regards, WM
On 12/20/23 5:36 AM, WM wrote:
Smaller than all eps > 0 that can be chosen. So it is.
Which isn't a finite number.
There can not be a "finite number" that is smaller than ALL finite
numbers
But you cannot oppose to this result: NUF(0) = 0 ad NUF(eps) = ℵo. That
means the increase from 0 to ℵo cannot be seen and discerned. The axiom
of choice is invalid here.
Right, and the points it increases are NOT "finite" numbers, and thus
NOT the "unit fractions" that you talk about.
Le 20/12/2023 à 13:54, Richard Damon a écrit :
On 12/20/23 5:36 AM, WM wrote:
Smaller than all eps > 0 that can be chosen. So it is.
Which isn't a finite number.
There are many finite numbers smaller than every esp > 0 that can be
chosen.
There can not be a "finite number" that is smaller than ALL finite
numbers
Choose an eps > 0 with less than almost all unit fractions in (0, eps].
Fail.
But you cannot oppose to this result: NUF(0) = 0 ad NUF(eps) = ℵo.
That means the increase from 0 to ℵo cannot be seen and discerned.
The axiom of choice is invalid here.
Right, and the points it increases are NOT "finite" numbers, and thus
NOT the "unit fractions" that you talk about.
How should the number of unit fractions increase before all unit fractions? Finest logic!
Regards, WM
On 12/20/23 11:18 AM, WM wrote:
Le 20/12/2023 à 13:54, Richard Damon a écrit :
On 12/20/23 5:36 AM, WM wrote:
Smaller than all eps > 0 that can be chosen. So it is.
Which isn't a finite number.
There are many finite numbers smaller than every esp > 0 that can be
chosen.
SO you agree that there is no "smallest" eps that can be used.
There can not be a "finite number" that is smaller than ALL finite
numbers
Choose an eps > 0 with less than almost all unit fractions in (0, eps].
Fail.
Why should you be able to?
Choose a number greater than 1/2 of infinity? That is a nonsense question,
On 12/20/23 11:14 AM, WM wrote:
Every positive point has ℵo unit fractions at its left-hand side.
There is no positive point with less than ℵo unit fractionsat its left-hand side.
All positive points with no exception have ℵo unit fractions attheir left-hand
side.
The interval (0, 1] has aleph_0 unit fractions at its left-hand side.
Contradiction.
What contradiction?
But that does not imply that ℵo unit fractions lie left of the interval.
Why not?
But that is nonsense - if all points can be selected as the axiom of
choice proposes.
Why?
What in the "Axiom of Choice" says we can choose the "End" of an
Unbounded set?
Le 19/12/2023 à 19:15, Jim Burns a écrit :
On 12/19/2023 5:42 AM, WM wrote:
Le 18/12/2023 à 17:48, Jim Burns a écrit :
On 12/18/2023 6:36 AM, WM wrote:
I consider Bob leaving the matrix
as impossible.
Exchanging between two matrix-position
will never decrease
the exchanged elements.
I think it can be a good and useful thing
to say
| 2¹ᐟ² is impossible
in a discussion about rationals.
However, that type of impossibility
has no power outside that discussion.
2¹ᐟ² is possible in other discussions.
There are matrices ⟨1,…,i⟩×⟨1,…,j⟩ for which
Bob leaving ⟨1,…,i⟩×⟨1,…,j⟩ is impossible.
Only those matrices exist,
because otherwise logic is violated.
By definition only
exchanging X and O without loss
is allowed.
Le 20/12/2023 à 17:38, Richard Damon a écrit :
On 12/20/23 11:14 AM, WM wrote:
Every positive point has ℵo unit fractions at its left-hand side.
There is no positive point with less than ℵo unit fractionsat its left-hand side.
All positive points with no exception have ℵo unit fractions attheir left-hand
side.
The interval (0, 1] has aleph_0 unit fractions at its left-handside.
Contradiction.
What contradiction?
There is no unit fraction left-hand of the interval.
But that does not imply that ℵo unit fractions lie left of the interval. >>Why not?
Because they all are positive.
But that is nonsense - if all points can be selected as the axiom of
choice proposes.
Why?
What in the "Axiom of Choice" says we can choose the "End" of an
Unbounded set?
It says that we can choose every element. But here almost all remain
below every choice.
Regards, WM
On 12/19/2023 6:50 PM, Richard Damon wrote:
On 12/19/23 9:27 PM, Chris M. Thomasson wrote:
On 12/19/2023 6:07 PM, Richard Damon wrote:
On 12/19/23 7:58 PM, Chris M. Thomasson wrote:
On 12/19/2023 2:16 AM, WM wrote:
Le 18/12/2023 à 17:21, Richard Damon a écrit :
On 12/18/23 10:56 AM, WM wrote:
Le 18/12/2023 à 15:21, Richard Damon a écrit :
On 12/18/23 6:19 AM, WM wrote:
Right, so NUF(x) is infinite for ANY finite number, and thus >>>>>>>>> for ANY unit fraction, so it is NEVER 1 at any, so there is no >>>>>>>>> smallest unit fraction.
This is inconsistent with the property, accepted by you: NUF can >>>>>>>> not increase by more than 1 without being constant for a finite >>>>>>>> period. NUF(0) = 0 implies NUF(x) = 1.
Yes, at an infintesimally small value of x.
No unit fraction is infinitesimally small. But perhaps you call
dark values infinitesimally small.
Huh? 1/1, 1/2, 1/3, ...
They do get infinitely small, yet never equal zero...
No, they get unboundedly small, but every one is finite, so it isn't
"infinitely" small. The key is unboundedly small means there is no
smallest item, but is just ever getting smaller, but stays a
finitely describable value.
NUF(x) = 1 only at an INFINITELY small value, which is smaller than
all UNBOUNDLY small values. The "Infinitely small values" are a
different type of number, NOT part of the standard definition of
unit fractions or the unbounded numeber systems like rational or reals. >>>>
These are similar classes of numbers as that which deals with the
mathematics of Omega and its friends. In other words, the
Transfinites, which WM has admitted are out of this domain of
discussion.
In one sense, his "dark" numbers could be these transfinites, except
he claims that his "dark numbers" are specifically part of the
Natural Numbers / Unit Fractions (which the Transfinite are not).
Also, the Transfinite numbers are "nameable" and "usable
individually" (just not with sets limited to the finite numbers).
Humm... How about, humm, their ability to go to infinity is
unbounded? Is that okay? They do get smaller...
Yes, that could be an ok wording (in my opinion). The key is that the
unit fractions, like the Natural Numbers, never REACH the
infintesimal/infinite limit, but approach it without a limiting bound.
Agreed! Thanks Richard. :^)
But all ARE "accessible", just unbounded.
Le 20/12/2023 à 23:16, Richard Damon a écrit :
But all ARE "accessible", just unbounded.
Between NUF(0) = 0 and for every x > 0: NUF(x) = ℵ₀ there are ℵ₀ unit fractions, in a non-vanishing part of the real axis, which cannot be
chosen by any x. How would you choose one of them? Don't claim that it
could be done. Show how it can be done!
Regards, WM
Le 21/12/2023 à 16:35, Richard Damon a écrit :
On 12/21/23 5:45 AM, WM wrote:
Le 20/12/2023 à 23:16, Richard Damon a écrit :
But all ARE "accessible", just unbounded.
Between NUF(0) = 0 and for every x > 0: NUF(x) = ℵ₀ there are ℵ₀ unit
fractions, in a non-vanishing part of the real axis, which cannot be
chosen by any x. How would you choose one of them? Don't claim that
it could be done. Show how it can be done!
What unit fraction can't be chosen?
Those which cannot be chosen, almost all.
Given ANY x>0 that you choose, we can find an n such that 1/(n+1) < x,
that n being a natural number greater than 1/x-1
and almost all unit fractions, namely ℵo being smaller. You cannot
discern them, separate them (although they are separated by uncountably
many points, none of which you can access), choose them.
Given that n, the unit fractions 1/(n+k), for k = 1, 2, 3, 4, 5, ...
are all smaller, and individually definable.
Note, this can be done for ANY x you choose.
Of course, but ℵo unit fractions, almost all unit fractions, remain unchosen.
They exist inside of the interval (0, 1] and have uncountably many
points between each other. Hence, ther is a first one.
Regards, WM
On 12/21/23 5:45 AM, WM wrote:
Le 20/12/2023 à 23:16, Richard Damon a écrit :
But all ARE "accessible", just unbounded.
Between NUF(0) = 0 and for every x > 0: NUF(x) = ℵ₀ there are ℵ₀ unit
fractions, in a non-vanishing part of the real axis, which cannot be
chosen by any x. How would you choose one of them? Don't claim that it
could be done. Show how it can be done!
What unit fraction can't be chosen?
Given ANY x>0 that you choose, we can find an n such that 1/(n+1) < x,
that n being a natural number greater than 1/x-1
Given that n, the unit fractions 1/(n+k), for k = 1, 2, 3, 4, 5, ... are
all smaller, and individually definable.
Note, this can be done for ANY x you choose.
Le 20/12/2023 à 23:16, Richard Damon a écrit :
But all ARE "accessible", just unbounded.
Between NUF(0) = 0 and
for every x > 0: NUF(x) = ℵ₀
there are ℵ₀ unit fractions,
in a non-vanishing part of the real axis,
[...]
Between NUF(0) = 0 and
for every x > 0: NUF(x) = ℵ₀
there are ℵ₀ unit fractions,
in a non-vanishing part of the real axis,
which cannot be chosen by any x.
On 12/21/2023 5:45 AM, WM wrote:
Between NUF(0) = 0 and
for every x > 0: NUF(x) = ℵ₀
there are ℵ₀ unit fractions,
in a non-vanishing part of the real axis,
which cannot be chosen by any x.
No.
For each x > 0
for each unit.fraction u of ℵ₀.many in ⅟ℕₓ
there is an xᵤ > 0
by which u "can be chosen"
meaning
such that u ∈ (xᵤ,1]
On 12/21/23 2:35 PM, WM wrote:
But all ARE "accessible", just unbounded.
Between NUF(0) = 0 and for every x > 0: NUF(x) = ℵ₀ there are ℵ₀ unit
fractions, in a non-vanishing part of the real axis, which cannot be
chosen by any x. How would you choose one of them? Don't claim that
it could be done. Show how it can be done!
What unit fraction can't be chosen?
Those which cannot be chosen, almost all.
If you can't show them, you can't say they exist.
Your "proof" is you can't see them, so they must be "dark", but they
don't actually need to exist.
All your arguement shows is that they are not bounded, but that was
already known.
Your logic system just can't handle unbounded sets,
Of course, but ℵo unit fractions, almost all unit fractions, remain
unchosen.
Please describe which one can't be?
Remember, ALL Natural Numbers ARE describable,
But if there was, then there is also a smalller one (since x/2 will also
be a unit fraction, and smaller than it)
Le 21/12/2023 à 21:08, Richard Damon a écrit :
On 12/21/23 2:35 PM, WM wrote:
But all ARE "accessible", just unbounded.
Between NUF(0) = 0 and for every x > 0: NUF(x) = ℵ₀ there are ℵ₀ >>>>> unit fractions, in a non-vanishing part of the real axis, which
cannot be chosen by any x. How would you choose one of them? Don't
claim that it could be done. Show how it can be done!
What unit fraction can't be chosen?
Those which cannot be chosen, almost all.
If you can't show them, you can't say they exist.
The proof of existence is taken from the assumption of actual infinity
and of the mathematical formula here:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
Your "proof" is you can't see them, so they must be "dark", but they
don't actually need to exist.
All your arguement shows is that they are not bounded, but that was
already known.
The fractions have a greatest lower bound, namely zero. It is called
their infimum.
Your logic system just can't handle unbounded sets,
The unit fractions, like all positive fractions have an infimum, that is
a bound.
Of course, but ℵo unit fractions, almost all unit fractions, remain
unchosen.
Please describe which one can't be?
Remember, ALL Natural Numbers ARE describable,
You are wrong. It can be seen by the fact that
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
The set of undefined natnumbers will never be emptied.
But if there was, then there is also a smalller one (since x/2 will
also be a unit fraction, and smaller than it)
That is provable for visible unit fractions only.
Regards, WM
Le 22/12/2023 à 05:19, Jim Burns a écrit :<> Consider xᵤ = u/2
On 12/21/2023 5:45 AM, WM wrote:
Between NUF(0) = 0 and
for every x > 0: NUF(x) = ℵ₀
there are ℵ₀ unit fractions,
in a non-vanishing part of the real axis,
which cannot be chosen by any x.
No.
For each x > 0
for each unit.fraction u of ℵ₀.many in ⅟ℕₓ
there is an xᵤ > 0
by which u "can be chosen"
meaning
such that u ∈ (xᵤ,1]
Proof.
ℵ₀ unit fractions populate
at least ℵ₀ points
which cannot be left-hand of all points,
because they are just these points.
"There does not exist
a point of (0, 1] with [fewer] than
ℵ₀ smaller unit fractions"
is so simple to prove wrong
(like disappearing Bob),
that it should no longer be claimed.
Therefore all your arguing in favour of
this nonsense will no longer be read by me.
Your simple and wrong proof
uses an unjustified quantifier shift.
You doing the equivalent of:
On 12/22/23 3:23 AM, WM wrote:
What unit fraction can't be chosen?
Those which cannot be chosen, almost all.
If you can't show them, you can't say they exist.
The proof of existence is taken from the assumption of actual infinity
and of the mathematical formula here:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
Please define your assumption of "actual infinity".
If you mean that some n ∈ ℕ is actually infinite
Since Unit Fractions are UNBOUNDED, that lower bound does not exist as a member of the set,
The set of undefined natnumbers will never be emptied.
So? That is just a property of using Bounded operations on Unbounded sets.
[...]
with less than ℵo smaller unit fractions,
[...]
Le 22/12/2023 à 14:56, Richard Damon a écrit :
On 12/22/23 3:23 AM, WM wrote:
What unit fraction can't be chosen?
Those which cannot be chosen, almost all.
If you can't show them, you can't say they exist.
The proof of existence is taken from the assumption of actual
infinity and of the mathematical formula here:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
Please define your assumption of "actual infinity".
The elements of a set exist such that none can be added.
If you mean that some n ∈ ℕ is actually infinite
No n is infinite and no 1/n is zero. Therefore NUF(x) increases only
after 0 for the first time.
Since Unit Fractions are UNBOUNDED, that lower bound does not exist as
a member of the set,
That would be a minimum. The infimum is explained here: https://en.wikipedia.org/wiki/Infimum_and_supremum
The set of unit fractions has the infimum 0.
The set of undefined natnumbers will never be emptied.
So? That is just a property of using Bounded operations on Unbounded
sets.
No, the set of unit fractions is bounded.
Regards, WM
you are assuming that NUF is definable by this set of words.
Since
at ANY 1/n, there are an infinite number of unit fractions with values
below it, it never has a finite value.
Since Unit Fractions are UNBOUNDED, that lower bound does not exist as
a member of the set,
That would be a minimum. The infimum is explained here:
https://en.wikipedia.org/wiki/Infimum_and_supremum
The set of unit fractions has the infimum 0.
It CAN'T be because the infimum must be a member of the set,
Le 23/12/2023 à 14:29, Richard Damon a écrit :
you are assuming that NUF is definable by this set of words.
Number of Unit Fractions between 0 and x.
Since at ANY 1/n, there are an infinite number of unit fractions with
values below it, it never has a finite value.
NUF(0) = 0, with increasing x it grows. The question is whether it can
grow by more than one unit fractions at one and the same x. The answer
is no, given by mathematics
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
Since Unit Fractions are UNBOUNDED, that lower bound does not exist
as a member of the set,
That would be a minimum. The infimum is explained here:
https://en.wikipedia.org/wiki/Infimum_and_supremum
The set of unit fractions has the infimum 0.
It CAN'T be because the infimum must be a member of the set,
Try to correct Wikipedia. You are obliged to correct the lie.
Regards, WM
On 12/23/23 9:53 AM, WM wrote:
NUF(0) = 0, with increasing x it grows. The question is whether it can
grow by more than one unit fractions at one and the same x. The answer
is no, given by mathematics
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
It can only grow by one at any finite number, but since there is a discontuity in the fationals between 0 and the rationals x > 0, it can
jump at that point.
The problem is that you definition of NUF just isn't actually realizable
Since Unit Fractions are UNBOUNDED, that lower bound does not exist
as a member of the set,
That would be a minimum. The infimum is explained here:
https://en.wikipedia.org/wiki/Infimum_and_supremum
The set of unit fractions has the infimum 0.
It CAN'T be because the infimum must be a member of the set,
Try to correct Wikipedia. You are obliged to correct the lie.
What "Lie"?
That bounds might not exsit?
or that bounds must be a member of the set in question?
Your game of billiards does not provide an algorithm that defines A.
On 25.12.2023 19:04, William wrote:
Your game of billiards does not provide an algorithm that defines A.
There is only one algorithm, namely k = (m + n - 1)(m + n - 2)/2 + m.
It stems from Cantor and yields
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, ... .
Mine is the same 1/k = (m + n - 1)(m + n - 2)/2 + m.
But whether or not A is defined, the matrix
XOOO...
XOOO...
XOOO...
XOOO...
...
will never be covered by X. This shows that never all matrix positions
will be indeXed.
Regards, WM
The algorithm is DEFINED as drawing k, from the LINE of the Natural
Numbers, and showing that this covers the MATRIX indexed by m,n.
On Tuesday, December 26, 2023 at 5:52:43 AM UTC-4, WM wrote:
On 25.12.2023 19:04, William wrote:
There is only one algorithm, namely k = (m + n - 1)(m + n - 2)/2 + m.
Your game of billiards does not provide an algorithm that defines A.
Correct, This defines B. Your game of billiards does not define A.
Le 26/12/2023 à 13:35, Richard Damon a écrit :
The algorithm is DEFINED as drawing k, from the LINE of the Natural
Numbers, and showing that this covers the MATRIX indexed by m,n.
Precisely this happens after the integer fractions have been replaced by
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, ...
..
Should this simple replacement destroy the process, then the whole
process is impossible.
Regards, WM
On 12/27/23 3:20 AM, WM wrote:
Le 26/12/2023 à 13:35, Richard Damon a écrit :
The algorithm is DEFINED as drawing k, from the LINE of the Natural
Numbers, and showing that this covers the MATRIX indexed by m,n.
Precisely this happens after the integer fractions have been replaced by
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, ...
..
Should this simple replacement destroy the process, then the whole
process is impossible.
The k terms, of your first column, are the INPUTS,
Le 27/12/2023 à 14:12, Richard Damon a écrit :
On 12/27/23 3:20 AM, WM wrote:
Le 26/12/2023 à 13:35, Richard Damon a écrit :
The algorithm is DEFINED as drawing k, from the LINE of the Natural
Numbers, and showing that this covers the MATRIX indexed by m,n.
Precisely this happens after the integer fractions have been replaced
by 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, ...
..
Should this simple replacement destroy the process, then the whole
process is impossible.
The k terms, of your first column, are the INPUTS,
Of course. And everybody sees that they will never cover the whole matrix.
Regards, WM
On 12/27/23 11:54 AM, WM wrote:
Le 27/12/2023 à 14:12, Richard Damon a écrit :
On 12/27/23 3:20 AM, WM wrote:
Le 26/12/2023 à 13:35, Richard Damon a écrit :
The algorithm is DEFINED as drawing k, from the LINE of the Natural
Numbers, and showing that this covers the MATRIX indexed by m,n.
Precisely this happens after the integer fractions have been replaced
by 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, ...
..
Should this simple replacement destroy the process, then the whole
process is impossible.
The k terms, of your first column, are the INPUTS,
Of course. And everybody sees that they will never cover the whole matrix.
WHo needs the INPUT to cover, the test was if the OUTPUT cover.
On 12/27/23 12:39 PM, WM wrote:
Fritz Feldhase schrieb am Mittwoch, 27. Dezember 2023 um 16:22:14 UTC+1:
Nonsense about quantifiers.
My proof has nothing to do with quantifier shift.
The proof is this: If before and after every unit fraction 1/n there is
a distance (containing uncountably many points) up to the next unit
fraction, then over this distance the Number of Unit Fractions between 0
and x, NUF(x), is constant. Unless any unit fraction evades this law,
NUF(x) must increase in steps of height 1, starting from NUF(0) = 0.
NUF(0) = 0.
NUF(x) for any finite x > 0 is infinite,
because at ANY finite point,
due to density, there are an infinite number of unit fractions before it.
The constant gap before the point x, where 1/(n+1) < x < 1/n is of
length < x/n so there is clearly room for at least n more items before
it, so there is plenty of space for more values.
There is not finite point where NUF(x) = 1, as such a point would, by necessity, be a "First" Unit Fraction, meaning that was a "highest"
Natural Number, but, BY DEFINITION, such a thing does not exist, as ALL Natural numbers have a successor.
So, unless YOU can find a value that evades THAT law, you arguement is
based on the assumption of something tha just doesn't exist.
On Wednesday, December 27, 2023 at 4:16:47 PM UTC+1, Fritz Feldhasewrote:
wrote:On Wednesday, December 27, 2023 at 3:52:32 PM UTC+1, Fritz Feldhase
smaller unit fractions"On Friday, December 22, 2023 at 9:30:58 AM UTC+1, WM wrote:
"There does not exist a point of (0, 1] with less than ℵ₀
is so simple to prove wrong
Oh please, Mr. Mückenheim, prove it wrong for us
its left-hand side.There is no positive point with less than ℵo unit fractions at
fractions at their left-hand side.All positive points with no exception have ℵo unit
left-hand side.Using symbolic language: A x e IR+: E^ℵo u e UF: u < x.
So far so good.
The interval (0, 1] has aleph_0 unit fractions at its
logical fallacy, since it can "lead" from a true statement to a false one).Nope. Here you are performing a /quantifier shift/ (which is a
Hint: Using symbolic language you latter statement reads:
E^ℵo u e UF: A x e (0, 1]: u < x.
But to get E^ℵo u e UF: A x e (0, 1]: u < x
Le 27/12/2023 à 19:13, Richard Damon a écrit :
On 12/27/23 11:54 AM, WM wrote:
Le 27/12/2023 à 14:12, Richard Damon a écrit :
On 12/27/23 3:20 AM, WM wrote:
Le 26/12/2023 à 13:35, Richard Damon a écrit :
The algorithm is DEFINED as drawing k, from the LINE of the
Natural Numbers, and showing that this covers the MATRIX indexed
by m,n.
Precisely this happens after the integer fractions have been
replaced by 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, ...
..
Should this simple replacement destroy the process, then the whole
process is impossible.
The k terms, of your first column, are the INPUTS,
Of course. And everybody sees that they will never cover the whole
matrix.
WHo needs the INPUT to cover, the test was if the OUTPUT cover.
Whatever you wish to express. Fact is that the elements of the first
column cannot cover the whole matrix. But that is what Cantor claims.
Regards, WM
On 12/27/23 2:01 PM, WM wrote:
Whatever you wish to express. Fact is that the elements of the first
column cannot cover the whole matrix. But that is what Cantor claims.
No, he never mapped "Matrix B to Matrix B", but the Natural Numbers to
the pairs m,n.
On 12/27/23 2:05 PM, WM wrote:
NUF(0) = 0.
NUF(x) for any finite x > 0 is infinite,
Your claim is disproved by the fact that every single unit fraction has
a level of constant NUF(x) behind it.
How does that disprove that it is infinite for all finite x. Since we
can consider "infinite" to be constant over that invertal.
Le 27/12/2023 à 20:34, Richard Damon a écrit :
On 12/27/23 2:01 PM, WM wrote:
Whatever you wish to express. Fact is that the elements of the first
column cannot cover the whole matrix. But that is what Cantor claims.
No, he never mapped "Matrix B to Matrix B", but the Natural Numbers to
the pairs m,n.
But there are to few natural numbers. There are to few indices X in
XOOO...
XOOO...
XOOO...
XOOO...
..
cover the whole matrix.
Regards, WM
But there isn't a "gap" between 0 and the unbounded set of unit
fractions, so you "set theory" can't actually say anything about what
happens there.
Is can't increase infinitely anywhere BETWEEN UNIT FRACTIONS,
but
between 0 and the positive numbers, it isn't defined
junp, in fact it MUST because the Unit Fractions get unboundly small so
there can't be a "first" unit fraction for NUF to be 1 at.
On 12/27/23 4:50 PM, WM wrote:
But there are too few natural numbers. There are too few indices X in
XOOO...
XOOO...
XOOO...
XOOO...
..
cover the whole matrix.
But there is not, when you transform k -> m,n where
k = = (m + n - 1)(m + n - 2)/2 + m
you find there is exactly one combination of m,n for every k, so there
are just as many k's as m,n combinations.
Thus, we have found a valid bijection.
On Wednesday, December 27, 2023 at 7:30:52 PM UTC+1, WM wrote:
A x e (0, 1]: E^ℵo u: u < x
is same as
A (x, 1]: E^ℵo u: u < (x, 1]
Huh?! "A (x, 1]: E^ℵo u: u < (x, 1]" ist noch nicht mal eine wff,
Le 28/12/2023 à 03:35, Richard Damon a écrit :
But there isn't a "gap" between 0 and the unbounded set of unit
fractions, so you "set theory" can't actually say anything about what
happens there.
Is can't increase infinitely anywhere BETWEEN UNIT FRACTIONS,
Correct. Therefore the set cannot exist without gap. At most one unit fraction can.
but between 0 and the positive numbers, it isn't defined
incorrect.
, so can arbitrarily
junp, in fact it MUST because the Unit Fractions get unboundly small
so there can't be a "first" unit fraction for NUF to be 1 at.
ZFC supplies for the function Number of Unit Fractions between 0 and x
the result
NUF(x) = 0 for x =< 0 and NUF(x) = ℵo for x > 0.
So, 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
On 12/28/23 3:44 AM, WM wrote:
Le 28/12/2023 à 03:35, Richard Damon a écrit :
On 12/27/23 4:50 PM, WM wrote:
But there are too few natural numbers. There are too few indices X in
XOOO...
XOOO...
XOOO...
XOOO...
..
cover the whole matrix.
But there is not, when you transform k -> m,n where
k = = (m + n - 1)(m + n - 2)/2 + m
you find there is exactly one combination of m,n for every k, so there
are just as many k's as m,n combinations.
Hence there is an internal contradiction.
WHAT is in contradiction?
He talks NOTHING of "taking from a column", that is YOUR invention.
On 12/28/23 3:42 AM, WM wrote:
ZFC supplies for the function Number of Unit Fractions between 0 and xYou do understand that ZFC does not provide a full definition of the
the result
NUF(x) = 0 for x =< 0 and NUF(x) = ℵo for x > 0.
So, 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
"Natural Numbers",
and never claims to?
Thus, trying to use it in a system that DOES have a full definition of
the Natural Numbers leads you to inconsistencies.
Le 28/12/2023 à 14:07, Richard Damon a écrit :
On 12/28/23 3:44 AM, WM wrote:
Le 28/12/2023 à 03:35, Richard Damon a écrit :
On 12/27/23 4:50 PM, WM wrote:
But there are too few natural numbers. There are too few indices X in >>>>>
XOOO...
XOOO...
XOOO...
XOOO...
..
cover the whole matrix.
But there is not, when you transform k -> m,n where
k = = (m + n - 1)(m + n - 2)/2 + m
you find there is exactly one combination of m,n for every k, so
there are just as many k's as m,n combinations.
Hence there is an internal contradiction.
WHAT is in contradiction?
XOOO...
XOOO...
XOOO...
XOOO...
..
He talks NOTHING of "taking from a column", that is YOUR invention.
He does not even talk about a matrix. But he talks about a series (older
form of sequence) an that applies all natural numbers as indices of the terms. They are the integer fractions, i.e., the elements of the first column. You need not even much intelligence to understand that - but a bit.
Regards, WM
On 12/28/23 11:11 AM, WM wrote:
WHAT is in contradiction?
XOOO...
XOOO...
XOOO...
XOOO...
..
And what is the contradiction about that?
Who expects that one column of a matrix will ever fill the full area of
the SAME matrix?
If you do, your logic is just broken.
Right, he shows a mapping of the "1 dimensional" sequence of k, to the
"2 dimensional" sequence set of m,n. We can display the 2-dimensional sequence in a matrix
On 12/28/23 11:28 AM, WM wrote:
Can you give a reference for this opinion? Is there a "professional"
mathematician who confessed this?
Just read Godel. He shows that any system that creates the full (or
actually, just enough of the) mathematics of the Natural Numbers must be incomplete, but also points out the ZFC can be complete but not be able
to prove it consistency. This is because ZFC doesn't generate the full mathematics.
This is because ZFC just uses First Order Logic, which isn't enough to
create the full mathematics, (but is enough to generate the set of numbers).
The Mathematics requries adding at least induction, which is a second
order logical form, and thus beyond basic ZFC.
As to formal references, I will admit this isn't my field of expertise,
and thus I don't know the formal source material that lays this out. I
am really an "Engineer" that uses this sort of Theory to make things
work, and thus have an understanding of the basic principles.
Le 28/12/2023 à 19:34, Richard Damon a écrit :
On 12/28/23 11:28 AM, WM wrote:
Can you give a reference for this opinion? Is there a "professional"
mathematician who confessed this?
Just read Godel. He shows that any system that creates the full (or
actually, just enough of the) mathematics of the Natural Numbers must
be incomplete, but also points out the ZFC can be complete but not be
able to prove it consistency. This is because ZFC doesn't generate the
full mathematics.
And this should prevent the function NUF(x) to be observed completely?
This is because ZFC just uses First Order Logic, which isn't enough to
create the full mathematics, (but is enough to generate the set of
numbers).
This distinction is simply nonsense.There is logic and there is unlogic. That's all. But anyhow NUF(x) is first order logic.
The Mathematics requries adding at least induction, which is a second
order logical form, and thus beyond basic ZFC.
To my knowledge, no set theorist does agree with you.
As to formal references, I will admit this isn't my field of
expertise, and thus I don't know the formal source material that lays
this out. I am really an "Engineer" that uses this sort of Theory to
make things work, and thus have an understanding of the basic principles.
An engineer who really would apply set theory and mathematical "logic"
is doomed to failure.
Regards, WM
On Wednesday, December 27, 2023 at 4:24:30 AM UTC-4, WM wrote:larger than B and containing not indexed fractions.
My game shows that the matrix A is
Your game does not provide a definition for the putative A
On Saturday, December 30, 2023 at 12:44:39 PM UTC-4, WM wrote:
I.e the putative A is not defined. Yet you continue to use the properties of this non existent thingYour game does not provide a definition for the putative AThat is because
On 04.01.2024 17:19, William wrote:
On Saturday, December 30, 2023 at 12:44:39 PM UTC-4, WM wrote:
I.e the putative A is not defined. Yet you continue to use theYour game does not provide a definition for the putative AThat is because
properties of this non existent thing
This matrix is of same size:
XOOO...
XOOO...
XOOO...
XOOO...
...
Its size does not change when reordering the X. The not indexed places, indicated by O, will be reordered too, but will never be indexed. Only
that is of interest.
Regards, WM
On 1/5/24 5:14 AM, WM wrote:
On 04.01.2024 17:19, William wrote:
On Saturday, December 30, 2023 at 12:44:39 PM UTC-4, WM wrote:
I.e the putative A is not defined. Yet you continue to use theYour game does not provide a definition for the putative AThat is because
properties of this non existent thing
This matrix is of same size:
XOOO...
XOOO...
XOOO...
XOOO...
...
Its size does not change when reordering the X. The not indexed
places, indicated by O, will be reordered too, but will never be
indexed. Only that is of interest.
Regards, WM
In other words, you don't understand Cantor's proof and are just trying
to use "common sense" on things that are actually known to defy "common sense" because infinity introduces some "strangeness"
Note, if you do it right, you can actually make the O's "disappear",
because you can swap them out to an unbounded distance from the start,
and pull in the unbounded number of x's on that first column into the
part of the matrix that we can see.
Yes, the Os are "still there" but have disappered.
Remember, one property of infinite sets is that a proper subset of an
infinte set may still have exactly the same size as the original set
itself.
Thus, the original set of all cells with X or O, can be (and in this
case is) the exact same size as both the set of Xs and the set of Os and those sets are also exactly the same size.
If your mind and logic can't handle that fact, then it is just
insufficient to handle this.
A simple proof of this fact is look at the Natural Numbers and the Even Numbers. Clearly the Even Numbers are a Proper Subset of the Natural
Numbers, as one way to make that set is to remove all the odd numbers,
which should leave us with a set that is one-half the size.
But another way to build the set of even numbers is to transform the
elements of the set of Natural Numbers by doubling each element. A set
built by transforming every element to a new unique element can't be of different size as the original set.
Thus, we have shown that for sets of infinite size, the Natural Numbers,
the subset of it , The Even Natural Numbers, is both a proper subset,
and of the same size.
*Does the directly executed D(D) halt?*
On Friday, January 5, 2024 at 6:14:17 AM UTC-4, WM wrote:
On 04.01.2024 17:19, William wrote:
On Saturday, December 30, 2023 at 12:44:39 PM UTC-4, WM wrote:
this non existent thingI.e the putative A is not defined. Yet you continue to use the properties ofYour game does not provide a definition for the putative AThat is because
This matrixA does not exist, There is no "the matrix".
On 1/5/24 5:14 AM, WM wrote:
On 04.01.2024 17:19, William wrote:
On Saturday, December 30, 2023 at 12:44:39 PM UTC-4, WM wrote:
I.e the putative A is not defined. Yet you continue to use theYour game does not provide a definition for the putative AThat is because
properties of this non existent thing
This matrix is of same size:
XOOO...
XOOO...
XOOO...
XOOO...
...
Its size does not change when reordering the X. The not indexed places,
indicated by O, will be reordered too, but will never be indexed. Only
that is of interest.
In other words, you don't
and are just trying
to use "common sense" on things that are actually known to defy "common sense" because infinity introduces some "strangeness"
Note, if you do it right, you can actually make the O's "disappear",
because you can swap them out to an unbounded distance from the start,
Yes, the Os are "still there" but have disappered.
Remember, one property of infinite sets is that a proper subset of an
infinte set may still have exactly the same size as the original set itself.
Le 05/01/2024 à 16:04, Richard Damon a écrit :
On 1/5/24 5:14 AM, WM wrote:
On 04.01.2024 17:19, William wrote:
On Saturday, December 30, 2023 at 12:44:39 PM UTC-4, WM wrote:
I.e the putative A is not defined. Yet you continue to use theYour game does not provide a definition for the putative AThat is because
properties of this non existent thing
This matrix is of same size:
XOOO...
XOOO...
XOOO...
XOOO...
...
Its size does not change when reordering the X. The not indexed
places, indicated by O, will be reordered too, but will never be
indexed. Only that is of interest.
In other words, you don't
believe in nonsense
and are just trying to use "common sense" on things that are actually
known to defy "common sense" because infinity introduces some
"strangeness"
Only when the religion of matheology is concerned.
Note, if you do it right, you can actually make the O's "disappear",
because you can swap them out to an unbounded distance from the start,
Yes, they disappear but they cannot leave the matrix by exchange with x. Simplest logic.
Yes, the Os are "still there" but have disappered.
That means, they have settled at dark places.
Remember, one property of infinite sets is that a proper subset of an
infinte set may still have exactly the same size as the original set
itself.
That is shown to be the belief of cranks by just the present argument.
It cannot be used as an argument to corrupt logic.
Regards, WM
On 1/8/24 5:15 AM, WM wrote:
If actual infinity exists, then the matrix A(0) =Right, "exchanging" can't get rid of anything,
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, ...
..
exists, and every reordering of the X by exchanging X and O does not
remove any O. Hence not all fractions can get indexed.
so something based on
exchanging can't show two things are of the same length as you always
will have xs and os.
You are just too stupid to understand where you are wrong.
Le 08/01/2024 à 13:21, Richard Damon a écrit :
On 1/8/24 5:15 AM, WM wrote:
If actual infinity exists, then the matrix A(0) =Right, "exchanging" can't get rid of anything,
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, ...
..
exists, and every reordering of the X by exchanging X and O does not
remove any O. Hence not all fractions can get indexed.
Please inform Jim Burns of this invaluable recognition.
so something based on exchanging can't show two things are of the sameBut if the indexing should be done by integer fractions, or if the replacement of integer fractions by integers is allowed, then Cantor's "bijection" is "completed" by exchanging. Today I have taught my
length as you always will have xs and os.
students of this argument. No-one has contradicted.
You are just too stupid to understand where you are wrong.
But you know it? Where? Is the bijection n <--> 1/n impossible?
Everything else happens just as Cantor did it.
Regards, WM
On 1/9/24 11:57 AM, WM wrote:
But you know it? Where? Is the bijection n <--> 1/n impossible?
Everything else happens just as Cantor did it.
Regards, WM
No, we can biject n <--> 1/n, just as we can biject n <--> m/n with a different pattern.
On Monday, January 8, 2024 at 6:15:16 AM UTC-4, WM wrote:
the matrix A(0)
Is not the putative matrix A. There is no matrix A
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