Discussion:
A dark number for WM...
(too old to reply)
Chris M. Thomasson
2023-12-18 03:08:51 UTC
Permalink
I am thinking of a number. A Hint:



lol! ;^)
WM
2023-12-28 08:38:36 UTC
Permalink
Post by Chris M. Thomasson
I am thinking of a number.
Better think of maths.

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
Richard Damon
2023-12-28 13:07:50 UTC
Permalink
Post by WM
Post by Chris M. Thomasson
I am thinking of a number.
Better think of maths.
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
https://www.researchgate.net/publication/376587391_The_seven_deadly_sins_of_set_theory
Regards, WM
And between 0 and (0, 1] is NOT a pair of unit fractions, so the logic
doesn't fit, and there isn't a first unit fraction to work your argueent
that it goes from 0 to first unit fraction to something.

Of course ZFC doesn't deliver the corrct mathematics in this case, as
you are working in a mathematical system above ZFC. You don't seem to
understand what ZFC IS, it isn't a theory about all mathematics, but a
theory of a simple mathematics (Not even reaching to the full power of
the Natural Numbers) that can have some useful properties to explore.

IT is well known that pushing ZFC too far, extending it beyond its
design, will tend to. make it "blow up", and you can't determine in the
system, if that has happened yet.
WM
2023-12-30 11:27:06 UTC
Permalink
Post by Richard Damon
Of course ZFC doesn't deliver the corrct mathematics in this case,
It claims to cover and correctly describe all unit fractions.
Post by Richard Damon
IT is well known that pushing ZFC too far, extending it beyond its
design, will tend to. make it "blow up", and you can't determine in the
system, if that has happened yet.
In mathematics, I can determine that after every unit fraction NUF(x) is
constant. Hence there is a first step of height 1.

Regards, WM
Fritz Feldhase
2023-12-30 14:25:49 UTC
Permalink
In mathematics, I can determine that after every unit fraction NUF(x) is constant.
Ja, Mückenheim. NUF(x) ist KONSTANT auf (0, 1]. Also auch "after every unit fraction", Dumbo.
Hence there is a first step of height 1.
Dummschwatz.

Also nochmal: NUF(x) ist KONSTANT auf (0, 1], und zwar gilt NUF(x) = aleph_0 für alle x in (0, 1].

Es gibt keinen "first step of height 1". Für x = 0 ist NUF(x) = 0 und für alle x e (0, 1] ist NUF(x) = aleph_0.
Richard Damon
2023-12-30 16:12:54 UTC
Permalink
Post by WM
Post by Richard Damon
Of course ZFC doesn't deliver the corrct mathematics in this case,
It claims to cover and correctly describe all unit fractions.
But never says that there is a smallest of them.

There is a "first" but the order is based on the order of the Natural
Number, so the first is 1/1. There is no Smallest.

That is the problem with your logic, you don't understand how unbounded
sets work.
Post by WM
Post by Richard Damon
IT is well known that pushing ZFC too far, extending it beyond its
design, will tend to. make it "blow up", and you can't determine in
the system, if that has happened yet.
In mathematics, I can determine that after every unit fraction NUF(x) is
constant. Hence there is a first step of height 1.
Only by FALSELY assuming there IS a first step. Since there actually is
no first step, you can't get it down to one.

If there WAS a value of x that NUF(x) = 1, then that x must be the
smallest unit fraction, but we know that if x is a unit fraction, so is
x/2, which is smaller, so no smallest x exists for NUF(x) to be 1.

The resolution would need to be expanding your number set to include the
infinitesimals, where there does exist a "smallest value x", but you
have explicitly stated that you are not using them.

Note also, ZFC can gernerate sets of other Number Systems, even
"exotics" ones, if you inject other axioms into the system, like your
assertion that there exist a smallest Unit Fraction. Since you have
shown a lack of understanding of exactly what you are doing, this is a
likely case for you, and your "dark numbers" may just be some exotics
that are not actually Natural Numbers, but numbers resulting from your
"extra" axioms, like there is a smallest unit fraction.
Post by WM
Regards, WM
WM
2023-12-30 16:40:34 UTC
Permalink
Post by Richard Damon
Post by WM
In mathematics, I can determine that after every unit fraction NUF(x) is
constant. Hence there is a first step of height 1.
Only by FALSELY assuming there IS a first step.
Wrong. This being constant proves that there is a first step.
Post by Richard Damon
If there WAS a value of x that NUF(x) = 1, then that x must be the
smallest unit fraction, but we know that if x is a unit fraction, so is
x/2, which is smaller, so no smallest x exists for NUF(x) to be 1.
One of these two claims is wrong:
1) After every unit fraction NUF(x) is constant.
2) For every n there is n+1.

The solution is dark numbers, which also appear in other contexts like the
intersection of endsegments and the x-O-problem.
Post by Richard Damon
The resolution would need to be expanding your number set to include the
infinitesimals, where there does exist a "smallest value x", but you
have explicitly stated that you are not using them.
There are only natural numbers and fractions concerned with these
problems.

Regards, WM
Richard Damon
2023-12-30 16:47:03 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
In mathematics, I can determine that after every unit fraction NUF(x)
is constant. Hence there is a first step of height 1.
Only by FALSELY assuming there IS a first step.
Wrong. This being constant proves that there is a first step.
But it is constant for less than the distance x, so it can't be the
first step.
Post by WM
Post by Richard Damon
If there WAS a value of x that NUF(x) = 1, then that x must be the
smallest unit fraction, but we know that if x is a unit fraction, so
is x/2, which is smaller, so no smallest x exists for NUF(x) to be 1.
1) After every unit fraction NUF(x) is constant.
2) For every n there is n+1.
Why do one of these need to be wrong?
Post by WM
The solution is dark numbers, which also appear in other contexts like
the intersection of endsegments and the x-O-problem.
No, there seem to be a solution for a non-problem cause by bad thinking
Post by WM
Post by Richard Damon
The resolution would need to be expanding your number set to include
the infinitesimals, where there does exist a "smallest value x", but
you have explicitly stated that you are not using them.
There are only natural numbers and fractions concerned with these problems.
And with just Unit Fractions, NUF(x) is never 1, and you logic that says
so based on the FALSE assumption that there exists a Naturl Number n
that doesn't have a successor, so its unit fraction can be first.

In other words, your theory is a theory of Unicorns.
Post by WM
Regards, WM
WM
2023-12-30 17:23:31 UTC
Permalink
Post by Fritz Feldhase
NUF(x) ist KONSTANT auf (0, 1], und zwar gilt NUF(x) = aleph_0 für alle x in
(0, 1].
That means you cannot distinguish ℵo unit fractions by choosing an x > 0
or by any other means.

Regards, WM
Richard Damon
2023-12-30 17:39:06 UTC
Permalink
Post by WM
Post by Fritz Feldhase
NUF(x) ist KONSTANT auf (0, 1], und zwar gilt NUF(x) = aleph_0 für
alle x in (0, 1].
That means you cannot distinguish ℵo unit fractions by choosing an x > 0
or by any other means.
Regards, WM
What are you smoking?

The fact that NUF(x) has the value of "aleph_0", means that NUF(x) isn't
a function on the finites (as aleph_0 isn't a finite number).

Note also, that it is a fact that aleph_0 + 1 is aleph_0, means that the
concept "Constant" as used for finite numbers doesn't apply to it.

Thus we can still have aleph_0 unit fractions in the range of (0,1] that
are all distinguishable.

Your logic just doesn't work, because you don't understand the domain of
requard for it and use it outside of its applicability.

It is a known fact that when you introduce infinites into math, you need
to change the rules, as finite math can not handle them.
WM
2023-12-30 19:16:15 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Fritz Feldhase
NUF(x) ist KONSTANT auf (0, 1], und zwar gilt NUF(x) = aleph_0 für
alle x in (0, 1].
That means you cannot distinguish ℵo unit fractions by choosing an x > 0
or by any other means.
Thus we can still have aleph_0 unit fractions in the range of (0,1] that
are all distinguishable.
Try it. Fail. Feldhase correctly has recognized that it is impossible by
the choice of an x to distinguish ℵo unit fractions which are at the
left-hand side of every x > 0 that can be choosen.

Regards, WM
Richard Damon
2023-12-30 19:29:35 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
Post by Fritz Feldhase
NUF(x) ist KONSTANT auf (0, 1], und zwar gilt NUF(x) = aleph_0 für
alle x in (0, 1].
That means you cannot distinguish ℵo unit fractions by choosing an x
Post by Fritz Feldhase
0 or by any other means.
Thus we can still have aleph_0 unit fractions in the range of (0,1]
that are all distinguishable.
Try it. Fail. Feldhase correctly has recognized that it is impossible by
the choice of an x to distinguish ℵo unit fractions which are at the
left-hand side of every x > 0 that can be choosen.
Regards, WM
Why not?

Note, is isn't for EVERY, but for ANY.

You are using the wrong qualificaiton.
WM
2023-12-31 08:57:12 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
Thus we can still have aleph_0 unit fractions in the range of (0,1]
that are all distinguishable.
Try it. Fail. Feldhase correctly has recognized that it is impossible by
the choice of an x to distinguish ℵo unit fractions which are at the
left-hand side of every x > 0 that can be choosen.
Why not?
Try it. Then you will see it.
Post by Richard Damon
Note, is isn't for EVERY, but for ANY.
Any and every chosen unit fraction leaves ℵo smaller unit fractions
undistinguished.
Post by Richard Damon
You are using the wrong qualificaiton.
You are lacking some qualification.

Regards, WM
Richard Damon
2023-12-31 12:38:40 UTC
Permalink
Post by WM
Post by WM
Post by Richard Damon
Thus we can still have aleph_0 unit fractions in the range of (0,1]
that are all distinguishable. >> Try it. Fail. Feldhase correctly has recognized that it is
impossible by>> the choice of an x to
Post by WM
Post by WM
distinguish ℵo unit fractions which are at the
left-hand side of
every x > 0 that can be choosen.
Why not?
Try it. Then you will see it.
I did. The fact you don't understand is your fault.

IF you want to insist that I write out the distinguishing for all, I
will reply do it for the full N_vis

"Distinguishing" must be for an individual, not a whole set, or your
logic is just incapable of handling infinite sets.
Post by WM
Post by WM
Note, is isn't for EVERY, but for ANY.
Any and every chosen unit fraction leaves ℵo smaller unit fractions
undistinguished.
It leaves ℵo unit fractions that CAN BE distinguished.
Post by WM
Post by WM
You are using the wrong qualificaiton.
You are lacking some qualification.
Which one? or is it just "dark"
Post by WM
Regards, WM
Your inability to understand basic theory isn't a reason to presume
extraordinary theory is needed.

Your logic is just bad.
WM
2024-01-01 14:20:28 UTC
Permalink
Post by Richard Damon
Post by WM
Post by WM
Post by Richard Damon
Thus we can still have aleph_0 unit fractions in the range of (0,1]
that are all distinguishable. >> Try it. Fail. Feldhase correctly has
recognized that it is
impossible by>> the choice of an x to
Post by WM
Post by WM
distinguish ℵo unit fractions which are at the
left-hand side of
every x > 0 that can be choosen.
Why not?
Try it. Then you will see it.
I did.
Which one is next to zero?
Post by Richard Damon
IF you want to insist that I write out the distinguishing for all, I
will reply do it for the full N_vis
N_vis is potentially infinite. I did never claim that I could nam
Post by Richard Damon
Post by WM
Any and every chosen unit fraction leaves ℵo smaller unit fractions
undistinguished.
It leaves ℵo unit fractions that CAN BE distinguished.
It leaves ℵo unit fractions which are left undistinguished.

Regards, WM
Richard Damon
2024-01-01 18:19:18 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Thus we can still have aleph_0 unit fractions in the range of
(0,1] >> that are all distinguishable. >> Try it. Fail. Feldhase
correctly has recognized that it is
impossible by>> the choice of an x to
Post by WM
distinguish ℵo unit fractions which are at the left-hand side of
every x > 0 that can be choosen.
Why not?
Try it. Then you will see it.
I did.
Which one is next to zero?
None of them. Why does there need to be one?
Post by WM
Post by Richard Damon
IF you want to insist that I write out the distinguishing for all, I
will reply do it for the full N_vis
N_vis is potentially infinite. I did never claim that I could nam
No, N_vis IS infinite (as a set) because ALL Natural Numbers are infinite.
Post by WM
Post by Richard Damon
Post by WM
Any and every chosen unit fraction leaves ℵo smaller unit fractions
undistinguished.
It leaves ℵo unit fractions that CAN BE distinguished.
It leaves ℵo unit fractions which are left undistinguished.
So?

Just because we haven't written out a name for something doesn't mean we
can't.

It seems your "dark" is not a property of the number it self, but a
transit property of what we have done with them. A "dark number" loses
its darkness when we look at it and write out its name.

Thus, it isn't an interesting property for the number itself.
Post by WM
Regards, WM
WM
2024-01-02 19:17:11 UTC
Permalink
Post by Richard Damon
Post by WM
Which one is next to zero?
None of them. Why does there need to be one?
Because we have a linear set with gaps between the members. That enforces
a first one.
Post by Richard Damon
Post by WM
It leaves ℵo unit fractions which are left undistinguished.
So?
Just because we haven't written out a name for something doesn't mean we
can't.
But ℵo will remain unnamed forever.
Post by Richard Damon
It seems your "dark" is not a property of the number it self, but a
transit property of what we have done with them. A "dark number" loses
its darkness when we look at it and write out its name.
Yes. But for ℵo numbers the property cannot be changed.
Post by Richard Damon
Thus, it isn't an interesting property for the number itself.
It is interesting enough to disprove Cantor'stheory.

Regards, WM
FromTheRafters
2024-01-02 20:42:27 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
Which one is next to zero?
None of them. Why does there need to be one?
Because we have a linear set with gaps between the members.
There are no gaps between the members, those are commas.
Post by WM
That enforces a first one.
No, it simply inverts the last failed attempt with the naturals because
they had no 'last one'.
Ross Finlayson
2024-01-02 21:27:56 UTC
Permalink
Post by FromTheRafters
Post by WM
Post by Richard Damon
Post by WM
Which one is next to zero?
None of them. Why does there need to be one?
Because we have a linear set with gaps between the members.
There are no gaps between the members, those are commas.
Post by WM
That enforces a first one.
No, it simply inverts the last failed attempt with the naturals because
they had no 'last one'.
There are various models of large numbers and infinite nubmers,
about fragments and parts, that there are models of numbers with
none infinite and models of integers with many.

This is often called Archimedean and non-Archimedean,
and illustrates in the wider dialectic, the very notions.

The "almost all" and "almost none" the elements of Cantor space,
by their density of 0's and 1's, sits next to a particular construction,
the "square Cantor space", that bridge the inductive impasse and
make one half.

Otherwise usual infinite limits and continuum limits speak to
the usual notions of completions and compactifications of things,
and the most usual "the law of large numbers".

How many primes are at infinity? There's a model for each.

"Powers and Roots"

...
A design engineer makes airplanes out of paper
... I have here a copy of van der Waerden's Moderne Algebra
WM
2024-01-03 10:54:51 UTC
Permalink
Post by Fritz Feldhase
Post by WM
the X cannot cover the matrix.
1. Instead "the X" we should use the infinitely many elements in IN,
namely: 1, 2, 3, 4, ...
Post by Fritz Feldhase
==> THIS WAY we may express (or prescribe) which element occupies
which position in the considered matrices.

This can also be done with the X:

XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX...
...........................................
Post by Fritz Feldhase
2. Now the elements 1, 2, 3, 4, ... c a n (since they do) "cover" the
matrix B = (b_n,m)_(n,m e IN) defined with b_n,m = (m + n - 1)(m + n -
2)/2 + m for all n,m e IN.
Post by Fritz Feldhase
3. On the other hand they do NOT "cover" the matrix A = (a_n,m)
defined with a_n,1 = n for all n e IN and a_n,m = 0 for all n e IN\{1},
m e IN.

Not even after having been arranged as matrix B. Therefore B, the matrix
of defined indices is much smaller than the matrix A of all positive
fractions.

By the way that is clear because ∀n ∈ ℕ: between n and n+1 there are
more than two fractions on the real line.

Regards, WM
Fritz Feldhase
2024-01-03 15:52:12 UTC
Permalink
The matrix B is much smaller than the matrix A
Mückenheim, Du bist für jede Art von Mathematik zu doof und zu blöde.

Beweis: Wir betrachten 2 Matrizen A = (a_n.m)_(n,m e IN) und B = (b_n.m)_(n,m e IN). Diese haben - wie man sieht - das selbe "Format". Insbesondere gilt also A = B genau dann wenn für alle n,m e IN: a_n,m = b_n,m. Die Matrx A ist also weder "kleiner" noch "größer" als die Matrix B, sondern "gleichgroß".

In einem zweiten Schritt spezifizieren wir die Elemente der beiden Matrizen: Für die Elemente a_n,m der Matrize A gelte: a_n,1 = n for all n e IN and a_n,m = 0 for all n e IN\{1},
m e IN; für die Elemente b_n,m der Matrize B gelte: b_n,m = (m + n - 1)(m + n - 2)/2 + m for all n,m e IN.

Hinweis: Die Matrizen ändern nicht ihre Größe in Abhängigkeit von den konkreten Elementen auf ihren jeweiligen Positionen (a_n,m bzw. b_n,m).

Und jetzt geh scheißen, Mückenheim!
Fritz Feldhase
2024-01-03 15:57:09 UTC
Permalink
∀n ∈ ℕ: between n and n+1 there are more than two fractions on the real line.
WOW, das ist beinahe abelpreiswürdig, Mückenheim!

How would you know?!
WM
2024-01-03 08:20:36 UTC
Permalink
Post by FromTheRafters
Post by WM
Post by Richard Damon
Post by WM
Which one is next to zero?
None of them. Why does there need to be one?
Because we have a linear set with gaps between the members.
There are no gaps between the members, those are commas.
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
d_n is not a comma.

Regards, WM
Richard Damon
2024-01-02 23:55:48 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
Which one is next to zero?
None of them. Why does there need to be one?
Because we have a linear set with gaps between the members. That
enforces a first one.
Nope. You are using BOUNDED logic rules on an UNBOUNDED set.
Post by WM
Post by Richard Damon
Post by WM
It leaves ℵo unit fractions which are left undistinguished.
So?
Just because we haven't written out a name for something doesn't mean
we can't.
But ℵo will remain unnamed forever.
But all are nameable.
Post by WM
Post by Richard Damon
It seems your "dark" is not a property of the number it self, but a
transit property of what we have done with them. A "dark number" loses
its darkness when we look at it and write out its name.
Yes. But for ℵo numbers the property cannot be changed.
We can for any of them.
Post by WM
Post by Richard Damon
Thus, it isn't an interesting property for the number itself.
It is interesting enough to disprove Cantor'stheory.
Regards, WM
Nope. Just proves you don't understand Cantor.
WM
2024-01-03 08:28:53 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Which one is next to zero?
None of them. Why does there need to be one?
Because we have a linear set with gaps between the members. That
enforces a first one.
Nope. You are using BOUNDED logic rules on an UNBOUNDED set.
I am using logic. You are using matheological belief.
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
It leaves ℵo unit fractions which are left undistinguished.
So?
Just because we haven't written out a name for something doesn't mean
we can't.
But ℵo will remain unnamed forever.
But all are nameable.
Not the last ℵo.
Post by Richard Damon
Post by WM
Post by Richard Damon
It seems your "dark" is not a property of the number it self, but a
transit property of what we have done with them. A "dark number" loses
its darkness when we look at it and write out its name.
Yes. But for ℵo numbers the property cannot be changed.
We can for any of them.
Not for the last ℵo.

Regards, WM
Richard Damon
2024-01-03 11:58:44 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Which one is next to zero?
None of them. Why does there need to be one?
Because we have a linear set with gaps between the members. That
enforces a first one.
Nope. You are using BOUNDED logic rules on an UNBOUNDED set.
I am using logic. You are using matheological belief.
No, YOU ard using invalid logic, not understanding the limitations of
the logic you are trying to use.

I am using proven math principles.

If your logic disagrees with proven facts, it is just wrong.
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
It leaves ℵo unit fractions which are left undistinguished.
So?
Just because we haven't written out a name for something doesn't
mean we can't.
But ℵo will remain unnamed forever.
But all are nameable.
Not the last ℵo.
Yes they are.
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
It seems your "dark" is not a property of the number it self, but a
transit property of what we have done with them. A "dark number"
loses its darkness when we look at it and write out its name.
Yes. But for ℵo numbers the property cannot be changed.
We can for any of them.
Not for the last ℵo.
Regards, WM
Yes, all of them.

Just saying they aren't doesn't make it so.

You are just proving your stupidity.
Fritz Feldhase
2024-01-03 16:19:52 UTC
Permalink
Post by WM
Post by Richard Damon
But ℵo [unit fractions] will remain unnamed forever.
But all are nameable.
Not the last ℵo.
What does this even mean?
Yes they [all --FF] are.
***@Mückenheim:

If this is the sequence of unit fractions

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

I will consider the terms in the sequence

"I/I", "I/II", "I/III", ...

(in the given order) and use them as names for the unit fractions (in the given order).

You see, this way each and every unit fraction has a name.

Hint: If u is a unit fraction, then there is exactly one natural number, let's call it n, such that u = 1/n. Then the n-th term in the sequence ("I/I", "I/II", "I/III", ...) is u's name.

The other way round, given the name "I/I...I", where the "|...|" part after the symbol "/" conists of, say, n "I", the unit fraction this name denotes is the n-th term in the sequence (1/1, 1/2, 1/3, ...).
Post by WM
Post by Richard Damon
Post by Richard Damon
It seems your "dark" is not a property of the number it self, but a
transit property of what we have done with them. A "dark number"
loses its darkness when we look at it and write out its name.
Yes. But for ℵo numbers the property cannot be changed.
Nonsense.

If n is a natural number, we can (in principle) write out its name. Well, _at least a Turing Machine can_.
Post by WM
Post by Richard Damon
We can for any of them.
Not for the last ℵo.
What does this even mean?!

It seems that WM considers "the last ℵo [natural numbers]" to be a fixed set of natural numbers. (What an idiot.)
Yes, all of them.
Just saying they aren't doesn't make it so.
You are just proving your stupidity.
That's WM's usual MO.
Chris M. Thomasson
2024-01-03 21:37:11 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Which one is next to zero?
None of them. Why does there need to be one?
Because we have a linear set with gaps between the members. That
enforces a first one.
Nope. You are using BOUNDED logic rules on an UNBOUNDED set.
I am using logic. You are using matheological belief.
No, YOU ard using invalid logic, not understanding the limitations of
the logic you are trying to use.
I am using proven math principles.
If your logic disagrees with proven facts, it is just wrong.
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
It leaves ℵo unit fractions which are left undistinguished.
So?
Just because we haven't written out a name for something doesn't
mean we can't.
But ℵo will remain unnamed forever.
But all are nameable.
Not the last ℵo.
Yes they are.
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
It seems your "dark" is not a property of the number it self, but a
transit property of what we have done with them. A "dark number"
loses its darkness when we look at it and write out its name.
Yes. But for ℵo numbers the property cannot be changed.
We can for any of them.
Not for the last ℵo.
Regards, WM
Yes, all of them.
Just saying they aren't doesn't make it so.
You are just proving your stupidity.
Arguing with WM reminds me of the following scene:



Its pointless. His stupidity shows no bounds. Yet he does not even know
what unbounded means...
WM
2024-01-04 10:13:23 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
It seems your "dark" is not a property of the number it self, but a
transit property of what we have done with them. A "dark number"
loses its darkness when we look at it and write out its name.
Yes. But for ℵo numbers the property cannot be changed.
We can for any of them.
Not for the last ℵo.
Yes, all of them.
"All" which have ℵo successors. But that are not all.
Post by Richard Damon
Just saying they aren't doesn't make it so.
Just saying they can be named does not make it so. After every named
number ℵo remain. That cannoit be changed.

Regards, WM
Richard Damon
2024-01-04 12:13:41 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
It seems your "dark" is not a property of the number it self, but
a transit property of what we have done with them. A "dark number"
loses its darkness when we look at it and write out its name.
Yes. But for ℵo numbers the property cannot be changed.
We can for any of them.
Not for the last ℵo.
Yes, all of them.
"All" which have ℵo successors. But that are not all.
Which ones don't?

Answer, Only the ones that are not themselves Natural Numbers!

Remember, a basic property of ALL Natural Numbers (by their definitions)
is that every Natural Number has a successor Natural Number, and every
Natural Number has a finite construction, and that finite construction
can be turned into a name.
Post by WM
Post by Richard Damon
Just saying they aren't doesn't make it so.
Just saying they can be named does not make it so. After every named
number ℵo remain. That cannoit be  changed.
But that doesn't mean they all cannot be named?

I have shown why they all can be named, and the fact that you don't
beleive it just shows you don't understand the actual mathematics of the
set.

Your problem seems to be that you are stuck in a logic system that just
can't handled unbounded sets.
Post by WM
Regards, WM
WM
2024-01-05 10:32:05 UTC
Permalink
Post by Richard Damon
Post by WM
Just saying they can be named does not make it so. After every named
number ℵo remain. That cannot be  changed.
I have shown why they all can be named,
If it is possible, then do it or let those do it whom you expect they
could do it. If it is impossible, then you are a liar.

Regards, WM
Richard Damon
2024-01-05 15:04:34 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
Just saying they can be named does not make it so. After every named
number ℵo remain. That cannot be  changed.
I have shown why they all can be named,
If it is possible, then do it or let those do it whom you expect they
could do it. If it is impossible, then you are a liar.
Regards, WM
So, you don't understand that the Natural Numbers are exactly the
OPPOSITE of your "dark" concept. They can all be used individually, but
they can not be "individually named" as the whole set.

May claim is that I can name "any" number I choose. Say, Ten Billion,
three hundred and seventy five Million, six hundred and thirty seven
thousand, four hundred and twenty two.

I can do that to any individual number I want.

I admit I can't individually name ALL of them at once.


Now, they can be formulaically named at once, in fact, the notation

1, 2, 3, 4, ... effectively does that.

But your logic can't handle that, as it can't handle infinite sets.
WM
2024-01-08 10:28:08 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Just saying they can be named does not make it so. After every named
number ℵo remain. That cannot be  changed.
I have shown why they all can be named,
If it is possible, then do it or let those do it whom you expect they
could do it. If it is impossible, then you are a liar.
May claim is that I can name "any" number I choose. Say, Ten Billion,
three hundred and seventy five Million, six hundred and thirty seven
thousand, four hundred and twenty two.
I can do that to any individual number I want.
But you cannot want it for every natural number because after every that
you want, ℵ are following.
Post by Richard Damon
I admit I can't individually name ALL of them at once.
You can name and subtract all at once collectively, such that no
successors remain. You cannot do so with in dividual numbers. There always
ℵ successors remain. This is a big difference.
Post by Richard Damon
Now, they can be formulaically named at once, in fact, the notation
1, 2, 3, 4, ... effectively does that.
Yes, but the "..." show that this means handling them collectively.
Post by Richard Damon
But your logic can't handle that, as it can't handle infinite sets.
I can handle them collectively only - just like you. But I have recognized
the difference. You not.

Regards, WM
Richard Damon
2024-01-08 12:21:05 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Just saying they can be named does not make it so. After every
named number ℵo remain. That cannot be  changed.
I have shown why they all can be named,
If it is possible, then do it or let those do it whom you expect they
could do it. If it is impossible, then you are a liar.
May claim is that I can name "any" number I choose. Say, Ten Billion,
three hundred and seventy five Million, six hundred and thirty seven
thousand, four hundred and twenty two.
I can do that to any individual number I want.
But you cannot want it for every natural number because after every that
you want, ℵ are following.
Why can't i "want" it for every natural number. I can want for a lot.

Your logic is just "dark".
Post by WM
Post by Richard Damon
I admit I can't individually name ALL of them at once.
You can name and subtract all at once collectively, such that no
successors remain. You cannot do so with in dividual numbers. There
always ℵ successors remain. This is a big difference.
Right, each number can be used individually.

All numbers must be used collectively.

That is the nature of unbounded sets,
Post by WM
Post by Richard Damon
Now, they can be formulaically named at once, in fact, the notation
1, 2, 3, 4, ... effectively does that.
Yes, but the "..." show that this means handling them collectively.
but each can be done individually by the pattern
Post by WM
Post by Richard Damon
But your logic can't handle that, as it can't handle infinite sets.
I can handle them collectively only - just like you. But I have
recognized the difference. You not.
Regards, WM
So, what is the collective set of just dark numbers.

If they can be used collectively, you should be able to show such a set.
WM
2024-01-09 17:10:00 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Just saying they can be named does not make it so. After every
named number ℵo remain. That cannot be  changed.
I have shown why they all can be named,
If it is possible, then do it or let those do it whom you expect they
could do it. If it is impossible, then you are a liar.
May claim is that I can name "any" number I choose. Say, Ten Billion,
three hundred and seventy five Million, six hundred and thirty seven
thousand, four hundred and twenty two.
I can do that to any individual number I want.
But you cannot want it for every natural number because after every that
you want, ℵ are following.
Why can't i "want" it for every natural number.
Because every wanted number has ℵ successors. You cannot remove them
individually but only collectively. The latter shows that all can be
removed.
Post by Richard Damon
Your logic is just "dark".
The blind sees only darkness.
Post by Richard Damon
Post by WM
Post by Richard Damon
I admit I can't individually name ALL of them at once.
You can name and subtract all at once collectively, such that no
successors remain. You cannot do so with in dividual numbers. There
always ℵ successors remain. This is a big difference.
Right, each number can be used individually.
All numbers must be used collectively.
That is the nature of unbounded sets,
Yes. That shows that not all numbers can be addressed individually. That
is what I call dark.

Regards, WM
Richard Damon
2024-01-10 02:58:59 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Just saying they can be named does not make it so. After every
named number ℵo remain. That cannot be  changed.
I have shown why they all can be named,
If it is possible, then do it or let those do it whom you expect
they could do it. If it is impossible, then you are a liar.
May claim is that I can name "any" number I choose. Say, Ten
Billion, three hundred and seventy five Million, six hundred and
thirty seven thousand, four hundred and twenty two.
I can do that to any individual number I want.
But you cannot want it for every natural number because after every
that you want, ℵ are following.
Why can't i "want" it for every natural number.
Because every wanted number has ℵ successors. You cannot remove them
individually but only collectively. The latter shows that all can be
removed.
So?

Your logic is just flawed if you expect to be able to create an infinite
set with a finite set.
Post by WM
Post by Richard Damon
Your logic is just "dark".
The blind sees only darkness.
Post by Richard Damon
Post by WM
Post by Richard Damon
I admit I can't individually name ALL of them at once.
You can name and subtract all at once collectively, such that no
successors remain. You cannot do so with in dividual numbers. There
always ℵ successors remain. This is a big difference.
Right, each number can be used individually.
All numbers must be used collectively.
That is the nature of unbounded sets,
Yes. That shows that not all numbers can be addressed individually. That
is what I call dark.
Regards, WM
Nope, where are the numbers that can't be addressed individually. The
fact that there are always more, but also that there are more addresses
available, doesn't show that you can't address them all.

There can not be an actual Natural Number that can not be indexed, in
fact, that is essentially a tautology, as the Natural Numbers ARE the
indexes, so every Natural Number can be indexed by itself.

Now, your "dark" numbers could be those numbers past the finite numbers,
what are called the "transfinite numbers", except you specifically say
they aren't but that they are parts of the finite numbers that just
can't be "named" (even though the definitions of the numbers provide a
way to name any of them).

Also, the Transfinite numbers CAN be "named" individually, so they can't
be your "dark" numbers.

All your "dark" numbers are, is an artifact of trying to use logic that
can't handle infinite sets on infinite sets.
Ross Finlayson
2024-01-10 05:45:28 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Just saying they can be named does not make it so. After every
named number ℵo remain. That cannot be changed.
I have shown why they all can be named,
If it is possible, then do it or let those do it whom you expect
they could do it. If it is impossible, then you are a liar.
May claim is that I can name "any" number I choose. Say, Ten
Billion, three hundred and seventy five Million, six hundred and
thirty seven thousand, four hundred and twenty two.
I can do that to any individual number I want.
But you cannot want it for every natural number because after every
that you want, ℵ are following.
Why can't i "want" it for every natural number.
Because every wanted number has ℵ successors. You cannot remove them
individually but only collectively. The latter shows that all can be
removed.
So?
Your logic is just flawed if you expect to be able to create an infinite
set with a finite set.
Post by WM
Post by Richard Damon
Your logic is just "dark".
The blind sees only darkness.
Post by Richard Damon
Post by WM
Post by Richard Damon
I admit I can't individually name ALL of them at once.
You can name and subtract all at once collectively, such that no
successors remain. You cannot do so with in dividual numbers. There
always ℵ successors remain. This is a big difference.
Right, each number can be used individually.
All numbers must be used collectively.
That is the nature of unbounded sets,
Yes. That shows that not all numbers can be addressed individually. That
is what I call dark.
Regards, WM
Nope, where are the numbers that can't be addressed individually. The
fact that there are always more, but also that there are more addresses
available, doesn't show that you can't address them all.
There can not be an actual Natural Number that can not be indexed, in
fact, that is essentially a tautology, as the Natural Numbers ARE the
indexes, so every Natural Number can be indexed by itself.
Now, your "dark" numbers could be those numbers past the finite numbers,
what are called the "transfinite numbers", except you specifically say
they aren't but that they are parts of the finite numbers that just
can't be "named" (even though the definitions of the numbers provide a
way to name any of them).
Also, the Transfinite numbers CAN be "named" individually, so they can't
be your "dark" numbers.
All your "dark" numbers are, is an artifact of trying to use logic that
can't handle infinite sets on infinite sets.
The "brick-batting Wolfgang Muckenheim" has been going on
around here forever.

I.e. the only goal anymore is to make him see the light.

A few times I about got him to adopt some more defensible
theories, but, perhaps it's cultural but there's some things
he can't, ..., understand. I think though that those are old things
and that time heals wounds. Last I heard, ....

So, what you do, is queue up the Blues Brothers, there's a scene,
where, Jake sees the light.

I don't care for musicals but there are exceptions when it's performances.
(Opera can be pretty good, though.)

Also I look about as down on you as you on him.
It's also about same, I figure you can learn better.

Luckily most of it is _old_ learning.
Chris M. Thomasson
2024-01-10 07:35:27 UTC
Permalink
Post by Ross Finlayson
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Just saying they can be named does not make it so. After every
named number ℵo remain. That cannot be changed.
I have shown why they all can be named,
If it is possible, then do it or let those do it whom you expect
they could do it. If it is impossible, then you are a liar.
May claim is that I can name "any" number I choose. Say, Ten
Billion, three hundred and seventy five Million, six hundred and
thirty seven thousand, four hundred and twenty two.
I can do that to any individual number I want.
But you cannot want it for every natural number because after every
that you want, ℵ are following.
Why can't i "want" it for every natural number.
Because every wanted number has ℵ successors. You cannot remove them
individually but only collectively. The latter shows that all can be
removed.
So?
Your logic is just flawed if you expect to be able to create an infinite
set with a finite set.
Post by WM
Post by Richard Damon
Your logic is just "dark".
The blind sees only darkness.
Post by Richard Damon
Post by WM
Post by Richard Damon
I admit I can't individually name ALL of them at once.
You can name and subtract all at once collectively, such that no
successors remain. You cannot do so with in dividual numbers. There
always ℵ successors remain. This is a big difference.
Right, each number can be used individually.
All numbers must be used collectively.
That is the nature of unbounded sets,
Yes. That shows that not all numbers can be addressed individually. That
is what I call dark.
Regards, WM
Nope, where are the numbers that can't be addressed individually. The
fact that there are always more, but also that there are more addresses
available, doesn't show that you can't address them all.
There can not be an actual Natural Number that can not be indexed, in
fact, that is essentially a tautology, as the Natural Numbers ARE the
indexes, so every Natural Number can be indexed by itself.
Now, your "dark" numbers could be those numbers past the finite numbers,
what are called the "transfinite numbers", except you specifically say
they aren't but that they are parts of the finite numbers that just
can't be "named" (even though the definitions of the numbers provide a
way to name any of them).
Also, the Transfinite numbers CAN be "named" individually, so they can't
be your "dark" numbers.
All your "dark" numbers are, is an artifact of trying to use logic that
can't handle infinite sets on infinite sets.
The "brick-batting Wolfgang Muckenheim" has been going on
around here forever.
I.e. the only goal anymore is to make him see the light.
A few times I about got him to adopt some more defensible
theories, but, perhaps it's cultural but there's some things
he can't, ..., understand. I think though that those are old things
and that time heals wounds. Last I heard, ....
So, what you do, is queue up the Blues Brothers, there's a scene,
where, Jake sees the light.
I don't care for musicals but there are exceptions when it's performances.
(Opera can be pretty good, though.)
Also I look about as down on you as you on him.
It's also about same, I figure you can learn better.
Luckily most of it is _old_ learning.
When WM students get an F, they might want to listen to this song. Think
about WM:

http://youtu.be/mRD0-GxqHVo
WM
2024-01-10 18:49:45 UTC
Permalink
Post by Richard Damon
There can not be an actual Natural Number that can not be indexed, in
fact, that is essentially a tautology, as the Natural Numbers ARE the
indexes, so every Natural Number can be indexed by itself.
But we cannot use these indices. Proof by the unit fractions 1/n. Since
all have distances, there is a first or smallest one, but we cannot
discern it.
Post by Richard Damon
Also, the Transfinite numbers CAN be "named" individually, so they can't
be your "dark" numbers.
Dark numbers like ω have no finite initial segments.

Regards, WM
Richard Damon
2024-01-11 00:49:23 UTC
Permalink
Post by WM
Post by Richard Damon
There can not be an actual Natural Number that can not be indexed, in
fact, that is essentially a tautology, as the Natural Numbers ARE the
indexes, so every Natural Number can be indexed by itself.
But we cannot use these indices. Proof by the unit fractions 1/n. Since
all have distances, there is a first or smallest one, but we cannot
discern it.
You are just indexing from the wrong end.

1st is 1/1
2nd is 1/2
Post by WM
Post by Richard Damon
Also, the Transfinite numbers CAN be "named" individually, so they
can't be your "dark" numbers.
Dark numbers like ω have no finite initial segments.
But ω isn't a "Natural Number", it is never actually reached by the
construction formula, only approached.

And ω is individually nameable, so isn' "dark" by your defiition.

Seems you have a bit of a definition problem.
Post by WM
Regards, WM
Chris M. Thomasson
2024-01-11 03:45:04 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
There can not be an actual Natural Number that can not be indexed, in
fact, that is essentially a tautology, as the Natural Numbers ARE the
indexes, so every Natural Number can be indexed by itself.
But we cannot use these indices. Proof by the unit fractions 1/n.
Since all have distances, there is a first or smallest one, but we
cannot discern it.
You are just indexing from the wrong end.
1st is 1/1
2nd is 1/2
Hyper dyslexic, and an ass to boot?
Post by Richard Damon
Post by WM
Post by Richard Damon
Also, the Transfinite numbers CAN be "named" individually, so they
can't be your "dark" numbers.
Dark numbers like ω have no finite initial segments.
But ω isn't a "Natural Number", it is never actually reached by the
construction formula, only approached.
And ω is individually nameable, so isn' "dark" by your defiition.
Seems you have a bit of a definition problem.
Post by WM
Regards, WM
WM
2024-01-11 21:31:53 UTC
Permalink
Post by Richard Damon
You are just indexing from the wrong end.
Points which are existing can be indexed from what we like.
Post by Richard Damon
But ω isn't a "Natural Number", it is never actually reached by the
construction formula, only approached.
0 is reached by the cursor moving from 1 to 0.
Post by Richard Damon
And ω is individually nameable, so isn' "dark" by your defiition.
Seems you have a bit of a definition problem.
No, you are not able to read maths texts.

Definition: A natural number is "identified" or (individually) "defined"
or "instantiated" if it can be communicated such that sender and receiver
understand the same and can link it by a finite initial segment to the
origin 0.

Regards, WM
Richard Damon
2024-01-12 03:07:05 UTC
Permalink
Post by WM
Post by Richard Damon
You are just indexing from the wrong end.
Points which are existing can be indexed from what we like.
So you say, and then you can't find the index.
Post by WM
Post by Richard Damon
But ω isn't a "Natural Number", it is never actually reached by the
construction formula, only approached.
0 is reached by the cursor moving from 1 to 0.
But 0 isn't ω. You seem to be having problems with definitions.
Post by WM
Post by Richard Damon
And ω is individually nameable, so isn' "dark" by your defiition.
Seems you have a bit of a definition problem.
No, you are not able to read maths texts.
Really, we could name the numbers you said couldn't be named.
Post by WM
Definition: A natural number is "identified" or (individually) "defined"
or "instantiated" if it can be communicated such that sender and
receiver understand the same and can link it by a finite initial segment
to the origin 0.
Regards, WM
And ω isn't a "Natural Number" as we can't reach it but the finite (but
unbounded) application of the successor operator, which is what defines
the set of Natural Numbers.

So, ω not having a finite initial sequence doesn't make it "Dark", it
makes it not a Natural Number.

You don't seem to understand the difference.
WM
2024-01-12 13:54:37 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
You are just indexing from the wrong end.
Points which are existing can be indexed from what we like.
So you say, and then you can't find the index.
This proves the existence of dark points.
Post by Richard Damon
Post by WM
Post by Richard Damon
But ω isn't a "Natural Number", it is never actually reached by the
construction formula, only approached.
0 is reached by the cursor moving from 1 to 0.
But 0 isn't ω.
Both have equivalent properties in that they lie at the ends of the
sequences of natural numbers and of unit fractions.

Regards, WM
Richard Damon
2024-01-12 23:46:28 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
You are just indexing from the wrong end.
Points which are existing can be indexed from what we like.
So you say, and then you can't find the index.
This proves the existence of dark points.
Nope, because all those numbers ARE indexed from the other end,
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
But ω isn't a "Natural Number", it is never actually reached by the
construction formula, only approached.
0 is reached by the cursor moving from 1 to 0.
But 0 isn't ω.
Both have equivalent properties in that they lie at the ends of the
sequences of natural numbers and of unit fractions.
And both of them are OUTSIDE the ends of the sequences, so not part of
the sets.
Post by WM
Regards, WM
WM
2024-01-13 10:14:18 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
You are just indexing from the wrong end.
Points which are existing can be indexed from what we like.
So you say, and then you can't find the index.
This proves the existence of dark points.
Nope, because all those numbers ARE indexed from the other end,
No. ℵ are missing.
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
But ω isn't a "Natural Number", it is never actually reached by the
construction formula, only approached.
0 is reached by the cursor moving from 1 to 0.
But 0 isn't ω.
Both have equivalent properties in that they lie at the ends of the
sequences of natural numbers and of unit fractions.
And both of them are OUTSIDE the ends of the sequences, so not part of
the sets.
Correct. But between all definable positive real numbers and these ends
there are ℵ dark natnumbers and ℵ dark unit fractions, respectively.

Regards, WM
Richard Damon
2024-01-13 13:22:45 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
You are just indexing from the wrong end.
Points which are existing can be indexed from what we like.
So you say, and then you can't find the index.
This proves the existence of dark points.
Nope, because all those numbers ARE indexed from the other end,
No. ℵ are missing.
Which ones?

It seem your "Dark" numbers are just numbers that are not existing. We
can't "see" them because they don't exist.
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
But ω isn't a "Natural Number", it is never actually reached by
the construction formula, only approached.
0 is reached by the cursor moving from 1 to 0.
But 0 isn't ω.
Both have equivalent properties in that they lie at the ends of the
sequences of natural numbers and of unit fractions.
And both of them are OUTSIDE the ends of the sequences, so not part of
the sets.
Correct. But between all definable positive real numbers and these ends
there are ℵ dark natnumbers and ℵ dark unit fractions, respectively.
That is clearly false unless your "natnumbers" are not the Natural
Numbers and your unit fractions are not the actual Unit Fractions.

Between the Real Numbers 0.4 and 0.45 are NO Natural Numbers or Unit
Fractions, so you claim is just proved to be invalid.

If your "natnumbers" are something else, please define what that set of
numbers is, and how they are generated.

It almosts sounds like an intentional deception to confuse people.

As I have said, there ARE "transfinite" numbers, (that are NOT part of
the finite number sets) that have some of the properties you claim
(except for the not being namable or usable individually) but you have
seemed to try to make it clear that these are not what you are talking
about.
Post by WM
Regards, WM
Ross Finlayson
2024-01-14 01:06:12 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
You are just indexing from the wrong end.
Points which are existing can be indexed from what we like.
So you say, and then you can't find the index.
This proves the existence of dark points.
Nope, because all those numbers ARE indexed from the other end,
No. ℵ are missing.
Which ones?
It seem your "Dark" numbers are just numbers that are not existing. We
can't "see" them because they don't exist.
[...]
Or a dark number to a child can be 42 simply because the child has not
thought of it yet?
Actually most children know any number's somewhere betwen zero and infinity.
With zero inclusive.
Infinity
FromTheRafters
2024-01-14 11:56:39 UTC
Permalink
Post by Ross Finlayson
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
You are just indexing from the wrong end.
Points which are existing can be indexed from what we like.
So you say, and then you can't find the index.
This proves the existence of dark points.
Nope, because all those numbers ARE indexed from the other end,
No. ℵ are missing.
Which ones?
It seem your "Dark" numbers are just numbers that are not existing. We
can't "see" them because they don't exist.
[...]
Or a dark number to a child can be 42 simply because the child has not
thought of it yet?
Actually most children know any number's somewhere betwen zero and infinity.
With zero inclusive.
Infinity
Infinity is usually NaN but zero usually is.
Volney
2024-01-15 17:08:08 UTC
Permalink
Some of my teachers wanted me to join GATE when, iirc, was in the 4'th
grade, but I did not want to. They even got to my parents.
What is GATE?
FromTheRafters
2024-01-15 17:19:24 UTC
Permalink
Post by Volney
Some of my teachers wanted me to join GATE when, iirc, was in the 4'th
grade, but I did not want to. They even got to my parents.
What is GATE?
Education for gifted children.
WM
2024-01-14 12:27:56 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
Nope, because all those numbers ARE indexed from the other end,
No. ℵ are missing.
Which ones?
The dark numbers.
Post by Richard Damon
It seem your "Dark" numbers are just numbers that are not existing. We
can't "see" them because they don't exist.
But we know that there are ℵ unit fractions between zero and the
smallest named unit fraction.
Post by Richard Damon
Post by WM
between all definable positive real numbers and these ends
there are ℵ dark natnumbers and ℵ dark unit fractions, respectively.
Between the Real Numbers 0.4 and 0.45 are NO Natural Numbers or Unit
Fractions, so you claim is just proved to be invalid.
Between the smallest visible unit fraction and zero there are ℵ dark
unit fractions.
Post by Richard Damon
As I have said, there ARE "transfinite" numbers, (that are NOT part of
the finite number sets) that have some of the properties you claim
(except for the not being namable or usable individually) but you have
seemed to try to make it clear that these are not what you are talking
about.
So it is. I am talking about natural numbers and the reciprocals.

Regards, WM
Richard Damon
2024-01-14 13:04:57 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Nope, because all those numbers ARE indexed from the other end,
No. ℵ are missing.
Which ones?
The dark numbers.
So you say, but cant show.

You are just chasing phantoms in your own mind.
Post by WM
Post by Richard Damon
It seem your "Dark" numbers are just numbers that are not existing. We
can't "see" them because they don't exist.
But we know that there are ℵ unit fractions between zero and the
smallest named unit fraction.
But there isn't a "smallest namable unit fraction".

"Named", as in someone has written the name, isn't a property of the
number, and thus not applicable here.
Post by WM
Post by Richard Damon
Post by WM
between all definable positive real numbers and these ends there are
ℵ dark natnumbers and ℵ dark unit fractions, respectively.
Between the Real Numbers 0.4 and 0.45 are NO Natural Numbers or Unit
Fractions, so you claim is just proved to be invalid.
Between the smallest visible unit fraction and zero there are ℵ dark
unit fractions.
But there is no "Smallest Visible Unit Fraction". If you think there is,
name it.

After all, it IS "visible" so namable.
Post by WM
Post by Richard Damon
As I have said, there ARE "transfinite" numbers, (that are NOT part of
the finite number sets) that have some of the properties you claim
(except for the not being namable or usable individually) but you have
seemed to try to make it clear that these are not what you are talking
about.
So it is. I am talking about natural numbers and the reciprocals.
And the numbers you are trying to talk about can't exist in those sets,
as all of the numbers in that set are "visible" and "namable" and Usable
individually.

So, you are stuck in a contradiction.

You seem there is no actual boundary past which the visible numbers
can't reach, so all the numbers you want to call "dark" are visible, so
they aren't dark.

You are just lost in bad logic.
Post by WM
Regards, WM
Chris M. Thomasson
2024-01-15 19:36:46 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Nope, because all those numbers ARE indexed from the other end,
No. ℵ are missing.
Which ones?
The dark numbers.
So you say, but cant show.
You are just chasing phantoms in your own mind.
Post by WM
Post by Richard Damon
It seem your "Dark" numbers are just numbers that are not
existing. We can't "see" them because they don't exist.
But we know that there are ℵ unit fractions between zero and the
smallest named unit fraction.
But there isn't a "smallest namable unit fraction".
"Named", as in someone has written the name, isn't a property of the
number, and thus not applicable here.
Post by WM
Post by Richard Damon
Post by WM
between all definable positive real numbers and these ends there
are ℵ dark natnumbers and ℵ dark unit fractions, respectively.
Between the Real Numbers 0.4 and 0.45 are NO Natural Numbers or
Unit Fractions, so you claim is just proved to be invalid.
Between the smallest visible unit fraction and zero there are ℵ
dark unit fractions.
But there is no "Smallest Visible Unit Fraction". If you think there
is, name it.
After all, it IS "visible" so namable.
WM is on some sort of moronic quest to name the smallest visible unit
fraction? Wow! ;^) lol.
Post by Richard Damon
Post by WM
Post by Richard Damon
As I have said, there ARE "transfinite" numbers, (that are NOT
part of the finite number sets) that have some of the properties
you claim (except for the not being namable or usable
individually) but you have seemed to try to make it clear that
these are not what you are talking about.
So it is. I am talking about natural numbers and the reciprocals.
And the numbers you are trying to talk about can't exist in those
sets, as all of the numbers in that set are "visible" and "namable"
and Usable individually.
So, you are stuck in a contradiction.
You seem there is no actual boundary past which the visible numbers
can't reach, so all the numbers you want to call "dark" are visible,
so they aren't dark.
You are just lost in bad logic.
Big time.
The funny part is that once WM tries to name the smallest unit
fraction, we can all name another one that is smaller, closer to zero,
so to speak...
;^D
He tried and failed with the reduced naturals in Q+ which are
essentially {1/1, 2/1, 3/1,...} and seems to believe that inverting them
to {1/1, 1/2, 1/3,} will make the whole set a different size and make
his point about Cantor's diagonal method(s) clear.
To the rest of us it is obvious that one set can be indexed by its
denominators and the other by its numerators.
Well, that pretty much sums up WM's main folly? I think so. Thanks. :^)

Strange that it is a teacher... Humm....

Fritz Feldhase
2024-01-11 03:49:20 UTC
Permalink
the unit fractions 1/n. [...] there is a first or smallest one
Nein, Du hirnloser Affe: Wenn s ein Stammbruch ist, dann ist 1/(1/s + 1) ein Stammbruch, der kleiner als s ist.

Also gibt es keinen kleisten Stammbruch, Du Spinner.

Erzählst Du Deinen Studenten s o einen Schwachsinn? Und d i e fressen das? Oder anders: So etwas läßt man Dich "unterrichten"? Die Erde ist eine Scheibe. Der Mond besteht aus grünem Käse. 1 ist keine natürliche Zahl, sondern ein Korkenzieher. usw .
WM
2024-01-01 14:31:53 UTC
Permalink
the matrix
X O O O...
X O O O...
X O O O...
X O O O...
...
will never be covered by X.
Das hat vor allem etwas damit zu tun, dass sich eine "gegebene" Matrix nicht
verändert,
But the covering X change their positions.
Nun kann man aber auch eine Matrix A = (a_n,m)_(n,m e IN) definieren, die statt
Xen z. B. die natürlichen Zahlen 1, 2, 3, ... in der ersten Spalte enthält (und
1 0 0 ...
2 0 0 ...
3 0 0 ...
...
ODER aber auch eine Matrix B = (b_n,m)_(n,m e IN) , die die Zahlen 1, 2, 3, 4,
... VERTEILT über sämtliche "Positionen" der Matrix enthält, und zwar gemäß
der Definition b_n.m = <1000-mal gepostet>.
1 2 4 ...
3 5 8 ...
6 9 13 ...
...
If both matrices were of same size, then the covering would be possible.
What should hinder it? However it is impossible. That falsifies your
claim.

Regards, WM
Fritz Feldhase
2024-01-01 15:17:23 UTC
Permalink
Post by WM
Nun kann man aber eine Matrix A = (a_n,m)_(n,m e IN) definieren, die statt
Xen z. B. die natürlichen Zahlen 1, 2, 3, ... in der ersten Spalte enthält (und
1 0 0 ...
2 0 0 ...
3 0 0 ...
...
ODER aber auch eine Matrix B = (b_n,m)_(n,m e IN) , die die Zahlen 1, 2, 3, 4,
... VERTEILT über sämtliche "Positionen" der Matrix enthält, und zwar gemäß
der Definition b_n.m = <1000-mal gepostet>.
1 2 4 ...
3 5 8 ...
6 9 13 ...
...
If both matrices were of same size, then the covering would be possible.
Exactly, and both matrices ARE "of the same size", since they have the same DOMAIN for their indices, dumbo (if you didn't notice).

Hint: A = (a_n,m)_(n,m e IN) and B = (b_n,m)_(n,m e IN).

This means that A = B iff for all n,m e IN: a_n.m = b_n,m as well as A =/= B iff for some n,m e IN: a_n.m =/= b_n,m.

Du bist wirklich für jede Art von Mathematik zu dumm und zu blöde.
Richard Damon
2024-01-01 15:33:29 UTC
Permalink
Post by WM
the matrix > > X O O O... > X O O O... > X O O O... > X O O O...
... > will never be covered by X.
Das hat vor allem etwas damit zu tun, dass sich eine "gegebene" Matrix
nicht verändert,
But the covering X change their positions.
And thus your arguement breaks.
Post by WM
Nun kann man aber auch eine Matrix A = (a_n,m)_(n,m e IN) definieren,
die statt Xen z. B. die natürlichen Zahlen 1, 2, 3, ... in der ersten
1 0 0 ... 2 0 0 ... 3 0 0 ... ...
ODER aber auch eine Matrix B = (b_n,m)_(n,m e IN) , die die Zahlen 1,
2, 3, 4, ... VERTEILT über sämtliche "Positionen" der Matrix enthält,
und zwar gemäß der Definition b_n.m = <1000-mal gepostet>.
1 2 4 ... 3 5 8 ... 6 9 13 ... ...
If both matrices were of same size, then the covering would be possible.
What should hinder it? However it is impossible. That falsifies your claim.
But they ARE of the sames size, both have alpha_0 rows and columns and
elements.
Post by WM
Regards, WM
Also, you seem to have a very broken Usenet reader (or are jus being
intentionally deceptive), as your references don't match the message you
are quoting. And the messages you are quoting don't seem to be from the
group you are posting in.

This doesn't say much about why people should trust what you say, since
you seem to naturally do "broken" methods.
WM
2024-01-01 19:18:24 UTC
Permalink
Post by Richard Damon
Post by WM
the matrix > > X O O O... > X O O O... > X O O O... > X O O O...
... > will never be covered by X.
Das hat vor allem etwas damit zu tun, dass sich eine "gegebene" Matrix
nicht verändert,
But the covering X change their positions.
And thus your arguement breaks.
No.
Post by Richard Damon
Post by WM
If both matrices were of same size, then the covering would be possible.
What should hinder it? However it is impossible. That falsifies your claim.
But they ARE of the sames size, both have alpha_0 rows and columns and
elements.
As my proof shows, ℵo is not a valid measure. (Note the words you are
trying to use are aleph_0 and quantifier.)
Post by Richard Damon
Also, you seem to have a very broken Usenet reader (or are jus being
intentionally deceptive), as your references don't match the message you
are quoting. And the messages you are quoting don't seem to be from the
group you are posting in.
The reason is that Franz Fritsche aka Fritz Feldhase is so ill-mannered
that his writing are rejected by news readers. But as long as Google is
active, I can see them. Alassince Google requires many Captchas (probably
triggered by the Korean spammers in sci.logic), I answer his contributions
here. It may happen, that by manual addressing mistakes occur.

Regards, WM
Richard Damon
2024-01-01 19:52:22 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
the matrix > > X O O O... > X O O O... > X O O O... > X O O O...
... > will never be covered by X.
Das hat vor allem etwas damit zu tun, dass sich eine "gegebene"
Matrix nicht verändert,
But the covering X change their positions.
And thus your arguement breaks.
No.
Yes, it does. Your stupidity to not understand it, doesn't change that
your logic is just broken.
Post by WM
Post by Richard Damon
Post by WM
If both matrices were of same size, then the covering would be
possible. What should hinder it? However it is impossible. That
falsifies your claim.
But they ARE of the sames size, both have alpha_0 rows and columns and
elements.
As my proof shows, ℵo is not a valid measure. (Note the words you are
trying to use are aleph_0 and quantifier.)
WHy not? You don't seem to understand the math needed for "infinite
measures"
Post by WM
Post by Richard Damon
Also, you seem to have a very broken Usenet reader (or are jus being
intentionally deceptive), as your references don't match the message
you are quoting. And the messages you are quoting don't seem to be
from the group you are posting in.
The reason is that Franz Fritsche aka Fritz Feldhase is so ill-mannered
that his writing are rejected by news readers. But as long as Google is
active, I can see them. Alassince Google requires many Captchas
(probably triggered by the Korean spammers in sci.logic), I answer his
contributions here. It may happen, that by manual addressing mistakes
occur.
If someone is so ill-mannered as to have there messages being blocked at
the news-server leverl (which doesn't really sound to be a thing),
intenntionally doing things to get around that is just also ill-mannered.

It seems you have the same sort of problem with usenet as you do with
mathematics, not understanding the rules and thus just ignoring them.

The reference headers of a message are supposed to point to the message
you are responding to. Erasing the message and putting a different
message in the message, just shows you are ignorant of what you are
doing, just like using the wrong rules of mathematics says you don't
know what you are doing in it.

Perhaps his problem is that he is posting in German, in a group that is
designated as an "English" language newsgroup (as most of sci.* is.)
Post by WM
Regards, WM
WM
2024-01-02 19:11:18 UTC
Permalink
Post by Richard Damon
Post by WM
As my proof shows, ℵo is not a valid measure. (Note the words you are
trying to use are aleph_0 and quantifier.)
WHy not?
Because the X cannot cover the matrix.

Regards, WM
Richard Damon
2024-01-02 23:55:46 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
As my proof shows, ℵo is not a valid measure. (Note the words you are
trying to use are aleph_0 and quantifier.)
WHy not?
Because the X cannot cover the matrix.
Regards, WM
But the remapped ones do.

You are just pointing the gun the wrong way and shooting yourself.
WM
2024-01-03 08:25:57 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
As my proof shows, ℵo is not a valid measure. (Note the words you are
trying to use are aleph_0 and quantifier.)
WHy not?
Because the X cannot cover the matrix.
But the remapped ones do.
It appears so:
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX...
.............................................................................

But since never an O leaves, the apparent picture is not the whole
picture.

Le 28/12/2023 à 19:34, Richard Damon a écrit: Who expects that one
column of a matrix will ever fill the full area of the SAME matrix?


Regards, WM
Richard Damon
2024-01-03 11:58:46 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
As my proof shows, ℵo is not a valid measure. (Note the words you
are trying to use are aleph_0 and quantifier.)
WHy not?
Because the X cannot cover the matrix.
But the remapped ones do.
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX...
.............................................................................
But since never an O leaves, the apparent picture is not the whole picture.
Le 28/12/2023 à 19:34, Richard Damon a écrit: Who expects that one
column of a matrix will ever fill the full area of the SAME matrix?
Regards, WM
In other words, you don't understand what you are talking about.

That seems normal for you.

Using illogic, you can make anything seems "reasonable" until you
understand how things actually work.
Ross Finlayson
2024-01-03 17:34:15 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
As my proof shows, ℵo is not a valid measure. (Note the words you
are trying to use are aleph_0 and quantifier.)
WHy not?
Because the X cannot cover the matrix.
But the remapped ones do.
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX...
.............................................................................
But since never an O leaves, the apparent picture is not the whole picture.
Le 28/12/2023 à 19:34, Richard Damon a écrit: Who expects that one
column of a matrix will ever fill the full area of the SAME matrix?
Regards, WM
In other words, you don't understand what you are talking about.
That seems normal for you.
Using illogic, you can make anything seems "reasonable" until you
understand how things actually work.
Like ex falso quodlibet?

It seems like Virgil used to say, "reke your own rede".

...
WM
2024-01-04 10:16:03 UTC
Permalink
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
As my proof shows, ℵo is not a valid measure. (Note the words you
are trying to use are aleph_0 and quantifier.)
WHy not?
Because the X cannot cover the matrix.
But the remapped ones do.
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX...
.............................................................................
But since never an O leaves, the apparent picture is not the whole picture.
Le 28/12/2023 à 19:34, Richard Damon a écrit: Who expects that one
column of a matrix will ever fill the full area of the SAME matrix?
In other words, you don't understand what you are talking about.
The sentence above was yours.

Regards, WM
Richard Damon
2024-01-04 12:15:10 UTC
Permalink
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
As my proof shows, ℵo is not a valid measure. (Note the words you
are trying to use are aleph_0 and quantifier.)
WHy not?
Because the X cannot cover the matrix.
But the remapped ones do.
XOOO... XXOO... XXOO... XXXO... ... XXXX...
XOOO... OOOO... XOOO... XOOO... ... XXXX...
XOOO... XOOO... OOOO... OOOO... ... XXXX...
XOOO... XOOO... XOOO... OOOO... ... XXXX...
.............................................................................
But since never an O leaves, the apparent picture is not the whole picture.
Le 28/12/2023 à 19:34, Richard Damon a écrit: Who expects that one
column of a matrix will ever fill the full area of the SAME matrix?
In other words, you don't understand what you are talking about.
The sentence above was yours.
Right YOU DON'T UNDERERSTAND THE PROBLEM.

You are just proving yourself too stupid to discuss these things with.
Post by WM
Regards, WM
WM
2024-01-04 10:20:07 UTC
Permalink
Post by Fritz Feldhase
∀n ∈ ℕ: between n and n+1 there are more than two fractions on the real
line.
WOW, das ist beinahe abelpreiswürdig,
It shows in a simple manner that for all intervals (n, n+1] there are more
fractions than natural numbers. In mathematics this proves that all
indexing can only happen "in the limit", i.e., nowhere.

Regards, WM
Richard Damon
2024-01-04 12:21:41 UTC
Permalink
Post by WM
Post by WM
∀n ∈ ℕ: between n and n+1 there are more than two fractions on the
real line.
WOW, das ist beinahe abelpreiswürdig,
It shows in a simple manner that for all intervals (n, n+1] there are
more fractions than natural numbers. In mathematics this proves that all
indexing can only happen "in the limit", i.e., nowhere.
Regards, WM
But that isn't the question/claim.

The claim is that in any interval, there are a countable infinite number
of rational numbers.

That means we can either form a bijection between them, or a pair of
injections (one each way) to show that neither set is fundamentally
bigger. (some attempts are allowed to fail, the question is does a
successful one exist).

THAT can be done, so we can show the sets (the rationals in an interval,
and the Natural Numbers) are the same size.

Your failure to understand that just shows your ignorance.

In particular, your inability to understand that there CAN be "failed"
mappings that don't work, shows your lack of understanding.

Yes, when you reject that idea, your mathematics just becomes contradictory.
Ross Finlayson
2024-01-04 18:47:27 UTC
Permalink
Post by Richard Damon
Post by WM
Post by WM
∀n ∈ ℕ: between n and n+1 there are more than two fractions on the
real line.
WOW, das ist beinahe abelpreiswürdig,
It shows in a simple manner that for all intervals (n, n+1] there are
more fractions than natural numbers. In mathematics this proves that all
indexing can only happen "in the limit", i.e., nowhere.
Regards, WM
But that isn't the question/claim.
The claim is that in any interval, there are a countable infinite number
of rational numbers.
That means we can either form a bijection between them, or a pair of
injections (one each way) to show that neither set is fundamentally
bigger. (some attempts are allowed to fail, the question is does a
successful one exist).
THAT can be done, so we can show the sets (the rationals in an interval,
and the Natural Numbers) are the same size.
Your failure to understand that just shows your ignorance.
In particular, your inability to understand that there CAN be "failed"
mappings that don't work, shows your lack of understanding.
Yes, when you reject that idea, your mathematics just becomes contradictory.
..., heh, ....
WM
2024-01-04 10:24:20 UTC
Permalink
Post by Fritz Feldhase
Post by WM
Post by Richard Damon
We can for any of them.
Not for the last ℵo.
What does this even mean?!
It means that "all" natural numbers which have ℵo natural numbers
following are not all natural numbers.
Post by Fritz Feldhase
It seems that WM considers "the last ℵo [natural numbers]" to be a fixed set
of natural numbers.
No, the visible numbers are potentially infinite. But ℵo numbers remain
dark forever. Otherwise only finitely many numbers would follow upon some
visible number.

Regards, WM
Richard Damon
2024-01-04 12:23:16 UTC
Permalink
Post by WM
On Wednesday, January 3, 2024 at 12:58:56 PM UTC+1, Richard Damon
wrote: > On 1/3/24 3:28 AM, WM wrote: > > Le 03/01/2024 à 00:55,
Richard Damon a écrit : > >> On 1/2/24 2:17 PM, WM wrote: > > > >
We can for any of them. > > > > > Not for the last ℵo.
What does this even mean?!
It means that "all" natural numbers which have ℵo natural numbers
following are not all natural numbers.
It seems that WM considers "the last ℵo [natural numbers]" to be a
fixed set of natural numbers.
No, the visible numbers are potentially infinite. But ℵo numbers remain
dark forever. Otherwise only finitely many numbers would follow upon
some visible number.
Regards, WM
No, the "visible" numbers are ACTUALLY INFINITE in number (but always
finite in value). There is no "Potentially" here by the normal definition.

You logic is just based on bad expectations.
Chris M. Thomasson
2024-01-04 22:40:00 UTC
Permalink
Post by WM
On Wednesday, January 3, 2024 at 12:58:56 PM UTC+1, Richard Damon
wrote: > On 1/3/24 3:28 AM, WM wrote: > > Le 03/01/2024 à 00:55,
Richard Damon a écrit : > >> On 1/2/24 2:17 PM, WM wrote: > > > >
We can for any of them. > > > > > Not for the last ℵo.
What does this even mean?!
It means that "all" natural numbers which have ℵo natural numbers
following are not all natural numbers.
It seems that WM considers "the last ℵo [natural numbers]" to be a
fixed set of natural numbers.
No, the visible numbers are potentially infinite. But ℵo numbers remain
dark forever. Otherwise only finitely many numbers would follow upon
some visible number.
Moron.
Fritz Feldhase
2024-01-02 23:58:30 UTC
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Post by WM
the X cannot cover the matrix.
1. Instead "the X" we should use the infinitely many elements in IN, namely: 1, 2, 3, 4, ...
==> THIS WAY we may express (or prescribe) which element occupies which position in the considered matrices.

2. Now the elements 1, 2, 3, 4, ... c a n (since they do) "cover" the matrix B = (b_n,m)_(n,m e IN) defined with b_n,m = (m + n - 1)(m + n - 2)/2 + m for all n,m e IN.

3. On the other hand they do NOT "cover" the matrix A = (a_n,m) defined with a_n,1 = n for all n e IN and a_n,m = 0 for all n e IN\{1}, m e IN.
Chris M. Thomasson
2023-12-28 23:42:09 UTC
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Post by WM
Post by Chris M. Thomasson
I am thinking of a number.
Better think of maths.
LOL!

I am thinking of a number, its not dark to me, but dark to you? Is that
what you mean by dark? ;^o
Post by WM
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
https://www.researchgate.net/publication/376587391_The_seven_deadly_sins_of_set_theory
Regards, WM
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