On 02.06.2025 03:56, Ben Bacarisse wrote:
WM <wolfgang.mueckenheim@tha.de> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte
des Unendlichen" and "Kleine Geschichte der Mathematik" at Technische
Hochschule Augsburg.)
On 31.05.2025 02:20, Ben Bacarisse wrote:
WM <wolfgang.mueckenheim@tha.de> writes:
>
It has been shown to the student by many arguments that the bijection
fails.
Which of the conditions of being a bijection (as presented in your book)
does b fail to meet? (I'm betting you won't say.)
>
The condition to be definable.
There are two conditions (as you know perfectly well) and it meets both
(as you also know perfectly well). Here are the conditions:
bijektiv (oder eineindeutig), wenn f injektiv und surjektiv ist
Is b not injective? Is b not surjektiv? Here's b again so you can
check for yourself that it is both:
Not all natural numbers of Cantor's set can be individually defined:
Since all natural numbers can be reduced to the empty set by subtracting them collectively,
ℕ \ {1, 2, 3, ...} = { }
they could also be reduced to the empty set by subtracting them individually - if this was possible. But then the well-order would force the existence of a last one. Contradiction.
Therefore the bijection is only possible for the potentially infinite Peano-set ℕ_def. They cannot all be manipulated because every subtracted subset has successors.
You called me a liar for saying this. I /would/ actually have to accept
the nonsense you teach (or at least copy it out) to get full marks.
>
You could also disprove it. But you carefully evade.
Eh? You just told me that what I wrote would not get full marks. I
would have to parrot at least some of your nonsense to get more. That's
all I was saying before when you said I was lying.
You could disprove it, if you could.
...
Do you not have to write marking schemes for your exams? And if in
fact you do, what do yours say about alternative answers?
>
If a student had ever disproved my proofs he would have got additional
points.
Do you always refuse to answer simple questions? Do you have to write
marking schemes for your exams? I'm just trying to find out if there is
documentary evidence of what a student at your college has to write to
get full marks.
You really don't want to say, do you?
Here are two:
https://www.hs-augsburg.de/~mueckenh/GU/Pruefung%20GU1001.pdfhttps://www.hs-augsburg.de/~mueckenh/GU/Pruefung%20GU1007.pdfBy the way, I can see why you don't want to show any marking scheme. It
would have to include the junk you expect the students to say,
Why do you call it junk? You cannot contradict even one of many proofs. If you could, you would not talk about nonsense like whether 1 is a definable natural number.
1. Dark Numbers
Not all numbers can be chosen, expressed, and communicated as individuals such that the receiver knows what the sender has meant. We call those numbers dark numbers. Much evidence has been collected and discussed [1]. But in the following we will present the shortest proof of their existence. Of course the facilities to express numbers depend on the environment and the power of the applied system. But this proof shows that, independent of the system, infinitely many natural numbers will remain dark forever.
A simple example is provided already by the denominators of the harmonic series (1/n). Whatever attempts are made to express denominators m as large as possible, the sum from 1/1 to 1/m is finite while the remaining part of the series diverges.
2. Kempner Series
The harmonic series diverges. But as Kempner [2] has shown in 1914, deleting all terms containing the digit 9 turns it into a converging series, the Kempner series, here abbreviated as K(9). That means that the complement C(9) of removed terms
C(9) = 1/9 + 1/19 + 1/29 + ...
all containing the digit 9, carries the divergence alone. All other terms can be removed. Same is true when all terms containing the digit 8 are removed. That means that the complement C(8)
C(8) = 1/8 + 1/18 +1/28 + ...
of the Kempner series K(8) carries the divergence alone. Since here those terms containing the digits 9 without digits 8 belong to the converging series K(8) we can conclude that the divergence is caused by the intersection only, i.e., by all terms containing the digits 8 and 9 simultaneously:
C(8) ∩ C(9) = 1/89 + 1/98 + 1/189 + ...
3. Proof
But not all terms containing 8 and 9 are needed. We can continue and remove all terms containing 1, 2, 3, 4, 5, 6, 7, or 0 in the denominator without changing this result because the ten corresponding Kempner series K(0), K(1), ..., K(9) are converging and their complements C(0), C(1), ..., C(9) are diverging. But only the intersection of all complements carries the divergence. That means that only the terms containing all the digits 0 to 9 simultaneously constitute the diverging series.
But that is not the end! We can remove any natural number k, like 2025, and the remaining Kempner series will converge. For proof use base 2026 where 2025 is a digit. This extends to every defined number, i.e., every number k that can be defined, chosen, and communicated such that the receiver knows what the sender has meant. When the terms containing k are deleted, then the remaining series converges.
4. Result
The diverging part of the harmonic series is constituted only by intersection of all complements C(k) of Kempner series K(k) of defined natural numbers k, i.e., by all the terms containing the digit sequences of all defined natural numbers. No defined natural number exists which must be left out. Terms which, although being larger than every defined number, do not contain all defined digit sequences, for instance not Ramsey's number, belong to converging Kempner series and not to the diverging series of the intersection of all complements. All infinitely many terms containing not the digit 1 or not the digit sequence 2025 or not the digit sequence of Ramsey's number can be deleted without violating the divergence.
All Kempner series K(k) of defined, i.e., finite numbers k split off in this way are converging and therefore the sum of their always finite sums is finite too although it can be very large [3]. The divergence however remains. It is carried only by terms which are dark and greater than all digit sequences of all defined numbers --- we can even say greater than all digit sequences of all definable numbers because, when larger numbers will be defined in future, they will behave in the same way. It is impossible to choose a natural number such that the intersection of the complements of all Kempner series of larger numbers is finite.
This is a proof of the huge set of undefinable or dark numbers.
Literature
[1] W. Mückenheim: "Evidence for Dark Numbers", ELIVA Press, Chisinau, 2024, pp. 1-36.
[2] A. J. Kempner: "A Curious Convergent Series", American Mathematical Monthly 21 (2), 1914, pp. 48–50.
[3] T. Schmelzer, R. Baillie: "Summing a Curious, Slowly Convergent Series", American Mathematical Monthly 115 (6), 2008, pp. 525–540.
Regards, WM
Simple enough for every reader?
Re: Simple enough for every reader?
