Abstract: We will prove by means of Cantor's mapping between natural numbers and positive fractions that his approach to actual infinity implies the existence of numbers which cannot be applied as defined individuals. We will call them dark numbers.

1. Outline of the proof

(1) We assume that all natural numbers are existing and are indexing all integer fractions in a matrix of all positive fractions.
(2) Then we distribute, according to Cantor's prescription, these indices over the whole matrix. We observe that in every step prescribed by Cantor the set of indices does not increase and the set of not indexed fractions does not decrease.
(3) Therefore it is impossible to index all fractions in a definable way. Indexing many fractions together "in the limit" would be undefined and can be excluded according to section 2 below. Reducing the discrepancy step by step would imply a first event after finitely many steps.
(4) In case of a complete mapping of  into the matrix, i.e., when every index has entered its final position, only indexed fractions are visible in the matrix.
(5) We conclude from the invisible but doubtless present not indexed fractions that they are attached to invisible positions of the matrix.
(6) By symmetry considerations also the first column of the matrix and therefore also  contains invisible, so-called dark elements.
(7) Hence also the initial mapping of natural numbers and integer fractions cannot have been complete. Bijections, i.e., complete mappings, of actually infinite sets and are impossible.

2. Rejecting the limit idea

When dealing with Cantor's mappings between infinite sets, it is argued usually that these mappings require a "limit" to be completed or that they cannot be completed. Such arguing has to be rejected flatly. For this reason some of Cantor's statements are quoted below.

"If we think the numbers p/q in such an order [...] then every number p/q comes at an absolutely fixed position of a simple infinite sequence" [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 126]

"The infinite sequence thus defined has the peculiar property to contain the positive rational numbers completely, and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
"thus we get the epitome (ω) of all real algebraic numbers [...] and with respect to this order we can talk about the th algebraic number where not a single one of this epitome () has been forgotten." [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 116]

"such that every element of the set stands at a definite position of this sequence" [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 152]

The clarity of these expressions is noteworthy: all and every, completely, at an absolutely fixed position, th number, where not a single one has been forgotten.

"In fact, according to the above definition of cardinality, the cardinal number M remains unchanged if in place of an element or of each of some elements, or even of each of all elements m of M another thing is substituted." [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 283]

This opportunity will be utilized to replace the pairs of the bijection by matrices or to attach a matrix to every pair of the bijection, respectively..

3. The proof

If all positive fractions m/n are existing, then they all are contained in the matrix

1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ...
     .

If all natural numbers k are existing, then they can be used as indices to index the integer fractions m/1 of the first column. Denoting indexed fractions by X and not indexed fractions by O, we obtain the matrix

XOOO...
XOOO...
XOOO...
XOOO...
XOOO...
     .

Cantor claimed that all natural numbers k are existing and can be applied to index all positive fractions m/n. They are distributed according to

k = (m + n - 1)(m + n - 2)/2 + m .
The result is a sequence of fractions

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

This sequence is modelled here in the language of matrices. The indices are taken from their initial positions in the first column and are distributed in the given order.

Index 1 remains at fraction 1/1, the first term of the sequence. The next term, 1/2, is indexed with 2 which is taken from its initial position 2/1

XXOO...
OOOO...
XOOO...
XOOO...
XOOO...
     .

Then index 3 is taken from its initial position 3/1 and is attached to 2/1

XXOO...
XOOO...
OOOO...
XOOO...
XOOO...
     .

Then index 4 is taken from its initial position 4/1 and is attached to 1/3

XXXO...
XOOO...
OOOO...
OOOO...
XOOO...
     .

Then index 5 is taken from its initial position 5/1 and is attached to 2/2

XXXO...
XXOO...
OOOO...
OOOO...
OOOO...
     .

And so on. When finally all exchanges of X and O have been carried out and, according to Cantor, all indices have been issued, it turns out that no fraction without index is visible any longer

XXXX...
XXXX...
XXXX...
XXXX...
XXXX...
    

but by the process of lossless exchange of X and O no O can have left the matrix as long as finite natural numbers are issued as indices. Therefore there are not less fractions without index than at the beginning.

We know that all O and as many fractions without index are remaining, but we cannot find any one. Where are they? The only possible explanation is that they are attached to dark positions.

By means of symmetry considerations we can conclude that every column including the integer fractions and therefore also the natural numbers contain dark elements. Cantor's indexing covers only the potentially infinite collection of visible fractions, not the actually infinite set of all fractions. This concerns also every other attempt to index the fractions and even the identical mapping. Bijections, i.e., complete mappings, of actually infinite sets and are impossible.

4. Counterarguments

Now and then it is argued, in spite of the preconditions explicitly quoted in section 2, that a set-theoretical or analytical limit should be applied.. This however would imply that all the O remain present in all definable matrices until "in the limit" these infinitely many O have to leave in an undefinable way; hence infinitely many fractions have to become indexed "in the limit" such that none of them can be checked  contrary to the proper meaning of indexing.

Some set theorists reject it as inadmissible to "limit" the indices by starting in the first column. But that means only to check that the set of natural numbers has the same size as the set of integer fractions. In contrast to Cantor's procedure the origin of the natural numbers is remembered. But this  the only difference to Cantor's approach  does not interfere with the indexing prescription and would not destroy the bijection if it really existed.

Finally, the counter argument that in spite of lossless exchange of X and O a loss of O could be tolerated suffers from deliberately contradicting basic logic.

Beweis der Existenz dunkler Zahlen

Abstract: Im Folgenden wird am Beispiel von Cantors Abbildung zwischen natürlichen Zahlen und positiven Brüchen bewiesen, dass seine Sichtweise aktual unendlicher Mengen die Existenz von nicht individuell verwendbaren Zahlen impliziert. Wir wollen sie dunkle Zahlen nennen.

1. Grundzüge des Beweises

(1) Wir nehmen an, dass alle natürlichen Zahlen existieren und alle ganzzahligen Brüche in einer Matrix aller positiven Brüche indizieren.
(2) Dann verteilen wir diese Indizes nach Cantors Vorschrift über die gesamte Matrix. Dabei ergibt sich, dass in jedem der von Cantor vorgegebenen Schritte die Menge der Indizes nicht zu- und die Menge der nicht indizierten Brüche nicht abnimmt.
(3) Also ist es nicht möglich, alle Brüche definierbar zu indizieren. Eine gemeinsame Indizierung vieler Brüche "im Grenzfalle" wäre undefinierbar und kann nach dem folgenden Abschnitt 2 ausgeschlossen werden. Eine schrittweise Verminderung der Diskrepanz im Verlauf der Folge würde hingegen eine erste Verminderung nach endlich vielen Schritten erfordern.
(4) Bei vollständiger Abbildung von  in die Brüche, d.h. wenn jeder Index seinen endgültigen Platz bezogen hat, sind in der Matrix aber nur indizierte Brüche erkennbar.
(5) Nun schließen wir aus dem Fehlen erkennbarer nicht indizierter Brüche, dass sie sich auf nicht erkennbaren Positionen innerhalb der Matrix befinden.
(6) Aus Symmetrieüberlegungen ergibt sich, dass auch die erste Spalte der Matrix und damit auch die Menge der natürlichen Zahlen nicht erkennbare, sogenannte dunkle Elemente enthält.
(7) Bijektionen zwischen aktual unendlichen Mengen und  können daher nicht existieren.

2. Ausschluss der Limes-Idee

Wenn man Cantors Abbildungen unendlicher Mengen behandelt, so werden stets nur allzubald Argumente angeführt, wonach diese Abbildungen nur im Grenzfalle gelten oder überhaupt niemals vollendet sind. Solche Argumente müssen mit Entschiedenheit zurückgewiesen werden. Es geht im Folgenden ausschließlich um Cantors Behauptungen, die hier zu diesem Zwecke eingefügt werden.

Insane WM, and why can no-one at Gottingen Univ or Germany admit slant cut of cone is Oval, never ellipse.

We do not want to hear another insane post and thread by WM who needs a psychiatrist not a sci.math account.

Metin Tolan,Laura Covi, Anja Karliczek head of Germany Federal Ministry of Education & Research, why does Germany keep flooding sci.math with WM math bullshit of dark numbers that fill up the front pages of sci.math with his insane bullshit going on for 3 decades now????
>

THIS Play act spam to hog front page of sci.math

Sergi_o 58 "Three proofs..." but Sergi_o and WM cannot even admit the truth of slant cut of cone is Oval, never ellipse
Chris Thomasson 57 "Three proofs...." yet Chris, WM the jackarses of math cannot admit slant cut of cone is Oval, never ellipse
WM 40 "Three proofs..."
Jim Burns 993 "Three proofs...." spam failure Burns cannot even admit slant cut of cone is Oval, never ellipse
Ross Finlayson 976 "Three proofs of dark numbers..." yet arsewipe WM & Ross cannot even admit slant cut of cone is Oval, never ellipse.
Sergi_o 915 "Three proofs of dark numbers..." 3:28PM
Fritz Feldhase a geometry moron with his ellipse a conic section but everyday spamming sci.math 881 "Three proofs of dark numbers...." 9:09PM
Jim Burns 871 "Three proofs of dark numbers...." 4:52PM


> Chris Thomasson 870 "Three proofs of dark numbers...."
>
>
> Brigitta Wolff, is this some type of German comedy show, WM daily spamming sci.math and have a team of comedians-- Chris Thomasson, Jim Burns, TheRafters, Sergi_o, day after day after day keep this WM imbecile with dark numbers as the first 5 post threads on the top of the leader board.
>
> Is there a German word for this Spamming behavior-- Shitligkeitzitter
>
>
> > >
> > > WM is a insane poster who cannot even admit slant cut of cone is Oval, never ellipse, so insane is WM that he floods sci.math with his mindless dark numbers bullshit.
> > >
> > > He cannot admit the truth of math, so he should not be posting in sci..math.
> > >
> > > Why, Anja, WM is so insane, the fruitcake cannot even understand Boole logic is wrong with 2 OR 1 = 3 and WM has AND as subtraction.
> > >
> > > So mindless is WM, he cannot even ask the question,-- is the muon the true electron of atoms and the 0.5MeV particle the Dirac magnetic monopole..
> > >
> > > Yet this German shitturd WM floods the front page of sci.math every day for 3 decades with his never ending insane bullshit of dark numbers.
> 
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>
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>Thorsten Theobald,Yury Person, Wolfgang Mueckenheim ever admit slant cut of cone is oval, never the ellipse, ever do a geometry proof of FTC, no all failures of mathematics, and idling away.
> 
>
> Germany's Muck the Puke taking up oxygen out of sci.math with his endless and mindless dark numbers, ellipse a conic when that is an oval, and the failure of all of Germany-- never a geometry proof of Fundamental Theorem of Calculus
>
> siren Sergio,Laura Covi,Andreas Dillmann,Gottingen no one in Gottingen Germany can admit slant cut of cone is Oval, never ellipse, all they seem to do is play with WM dark numbers KuhscheiBe. Nor can anyone do a geometry proof of Fundamental Theorem of Calculus, because Germany spends its time on Muck the Puck "dark numbers" please, need your help to get WM moved over to sci.logic and out of sci.math with his crazy postings-- dark numbers. He is a fool, not a scientist.
>
> > > > > >
> > > > > > B. Schmidt,Metin Tolin, please, need you help to get WM moved over to sci.logic and out of sci.math with his crazy postings-- dark numbers.. He is a fool, not a scientist.
> > > > > > 
> > > > > > > stir-crazy-W.Mueckenheim-the fool still thinks slant cut of cone is ellipse when in truth it is an oval, and the failure of logic WM believes in Boole's AND as TFFF when in truth it is TTTF to avoid what the nitwit WM has as 2 OR 1 = 3 with AND as subtraction. WM is a math failure for the idiot never knew calculus was geometry and therefore never sought a geometry proof of Fundamental Theorem of Calculus.
> > > > > > > Universitat Augsburg, Germany, rector Sabine Doering-Manteuffel
> > > > > > > Math dept Ronald H.W.Hoppe, B. Schmidt, Sarah Friedrich, Stefan Grosskinsky, Friedrich Pukelsheim, Mirjam Dur, Ralf Werner.
> > > > > > >
> > > > > > > Hochschule Augsburg, Wolfgang Mueckenheim
> > > > > > > Augsburg- Friedrich Pukelsheim-Gottingen,Metin Tolin,Ariane Frey, Wolfgang Glatzel why does Wolfgang Mueckenheim the idiot with "dark numbers" & Dr. Tao fail geometry so so badly,
> > > > > > > >
> > > > > > > > The idiots of math never knew calculus was geometry, erst, they would provide a geometry proof of Fundamental Theorem of Calculus, why they are so banal stupid in math, they still believe slant cut in cone is an ellipse, when in reality it is a oval.
> > > > > > > > They are not mathematicians but mindless fuckdogs of math.
> > > > > >
> > > > > > > > Maybe they wear glasses and cannot see properly. Maybe WM & Tao were never good in math, for they cannot even tell apart a ellipse from oval. They cannot even ask the question which is the atom's true electron-- muon or 0.5MeV particle?
> > > > > > > >
> > > > > > > > Univ Augsburg Germany math-- Hello-- Wolfgang Mueckenheim the fool of math wasting everybodies time -- for WM is a math failure with his slant cut in cone as ellipse when in truth it is a Oval. And now, that fool of math with his "dark numbers". Can you whisk him off to a "shrink in Germany and put him on medications"??
> > > > > > > >
> > > > > > > >
> > > > > > > > Stefan Grobkinsky, Sarah Friedrich, Mirjam Dur, Ralf Werner, Friedrich Pukelsheim, Kai Cieliebak, Urs Frauenfelder, Jennifer Gruber, Yannis Bahni, Zhen Gao, Sungho Kim, Shuaipeng Liu, Julius Natrup, Marian Poppr, Kevin Ruck, Evgeny Volkov, Frederic Wagner
> > > > > >
> > > > > > Julius Natrup, Marian Poppr,B. Schmidt,Metin Tolin, please, need you help to get WM moved over to sci.logic and out of sci.math with his crazy postings-- dark numbers. He is a fool, not a scientist.
> > > > > > Sarah Friedrich please help shuffle insane WM to sci.logic with his mindless dark-numbers.
> > > > > > 
> > > > > > >
> > > > > > > Stefan Grobkinsky, Sarah Friedrich, of Augsburg Germany, why does the USA have to have piped in turds from Wolfgang Mueckenheim of his crazy "dark numbers". So Russia cuts gas to Germany, yet Germany pipes the turds of WM into the USA. That is not fair.
> > > > > > >
> > > > > > >
> > > > > > > Wolfgang Mueckenheim math-mindless-fuckdog with his mindless "dark numbers"
> > > > > > >
> > > > > > > Scoot him over to sci.logic-- for he fails math.
> > > > > > > Germany's insane WM fails math
> > > > > > > 1) he cannot accept slant cut of cone is oval, never ellipse
> > > > > > > 2) he accepts Boole logic of AND truth table is TFFF which leads to 2 OR 1 = 3 with AND as subtraction.
> > > > > > > 3) arsehole WM, never understood calculus is geometry and needs a geometry proof of Fundamental Theorem of Calculus
> > > > > > > 4) WM cannot even read a proof of math-- for the 7 Circle Theorem of 1974 is false and invalid
> > > > > > >
> > > > > > Metin Tolin,Julius Natrup, Marian Poppr,B. Schmidt, please, need you help to get WM and his mindless "dark numbers bullshit" moved over to sci.logic and out of sci.math with his crazy postings-- dark numbers. He is a fool, not a scientist.
> > > > > > > >
> > > > > > > >
> > > > > > > > Wolfgang Mueckenheim math-mindless-fuckdog with his mindless "dark numbers"
> > > > > > > >
> > > > > > > > Scoot him over to sci.logic-- for he fails math.
> > > > > > > > Germany's insane WM fails math
> > > > > > > > 1) he cannot accept slant cut of cone is oval, never ellipse
> 
> 
> > > > > > > > 2) he accepts Boole logic of AND truth table is TFFF which leads to 2 OR 1 = 3 with AND as subtraction.
> > > > > > > > 3) fool WM, never understood calculus is geometry and needs a geometry proof of Fundamental Theorem of Calculus
> > > > > > > > 4) WM cannot even read a proof of math-- for the 7 Circle Theorem of 1974 is false and invalid
> > > > > > > >
> > > > > > > > Yet, every day the fool WM pollutes sci.math with his dark numbers bullshit.
> > > > > > > Sarah Friedrich please help shuffle insane WM to sci.logic with his mindless dark-numbers.
> > 
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> > > > > > > > Stefan Grobkinsky, Sarah Friedrich, of Augsburg Germany, why does the USA have to have piped in turds from Wolfgang Mueckenheim of his crazy "dark numbers". So Russia cuts gas to Germany, yet Germany pipes the turds of WM into the USA. That is not fair.
> > > > Universitat Augsburg, Germany, rector Sabine Doering-Manteuffel
> > > > Math dept Ronald H.W.Hoppe, B. Schmidt, Sarah Friedrich, Stefan Grosskinsky, Friedrich Pukelsheim, Mirjam Dur, Ralf Werner.
> > > >
> > > > Hochschule Augsburg, Wolfgang Mueckenheim
> > > > Gottingen Univ math
> > > >
> > > > Dorothea Bahns, Laurent Bartholdi, Valentin Blomer, Jorg Brüdern, Stefan Halverscheid, Harald Andres Helfgott, Madeleine Jotz Lean, Ralf Meyer, Preda Mihailescu, Walther Dietrich Paravicini, Viktor Pidstrygach, Thomas Schick, Evelina Viada, Ingo Frank Witt, Chenchang Zhu
> > > >
> > > > Eternal-September.org
> > > > Wolfgang M. Weyand
> > > > Berliner Strasse
> > > > Bad Homburg
> > > >
> > > > Goethe Universitat Physics dept
> > > >
> > > > Brigitta Wolff president
> > > >
> > > > Jurgen Habermass
> > > > Horst Stocker
> > > > Gerd Binnig
> > > > Horst Ludwig Stormer
> > > > Peter Grunberg
> > > >
> > > > math
> > > > Alex Kuronya
> > > > Martin Moller
> > > > Jakob Stix
> > > > Annette Werner
> > > > Andreas Bernig
> > > > Esther Cabezas-Rivas
> > > > Hans Crauel
> > > > Thomas Gerstner
> > > > Bastian von Harrach
> > > > Thomas Mettler
> > > > Tobias Weth
> > > > Amin Coja-Oghlan
> > > > Raman Sanyal
> > > > Thorsten Theobald
> > > > Yury Person
> > > >
> > > >
> > > > Gottingen Univ physics
> > > > Prof. Dr. Karsten Bahr
> > > > Prof. Dr. Peter Bloechl
> > > > Prof. Dr. Eberhard Bodenschatz
> > > > Prof. Laura Covi, PhD
> > > > Prof. Dr. Andreas Dillmann
> > > > Prof. Dr. Stefan Dreizler
> > > > Prof. Dr. Jörg Enderlein
> > > > Prof. Dr. Laurent Gizon
> > > > Prof. Dr. Ariane Frey
> > > > apl. Prof. Dr. Wolfgang Glatzel
> > > > Prof. Dr. Fabian Heidrich-Meisner
> > > > Prof. Dr. Hans Christian Hofsäss
> > > > Prof. Dr. Andreas Janshoff
> > > > Prof. Dr. Christian Jooß
> > > > Prof. Dr. Stefan Kehrein
> > > > Prof. Dr. Stefan Klumpp
> > > > Prof. Dr. Sarah Köster
> > > > Prof. Dr. Reiner Kree
> > > > Prof. Dr. Matthias Krüger
> > > > Prof. Dr. Stanley Lai
> > > > Prof. Dr. Stefan Mathias
> > > > apl. Prof. Dr. Vasile Mosneaga
> > > > Prof. Dr. Marcus Müller
> > > > Prof. Dr. Jens Niemeyer
> > > > apl. Prof. Dr. Astrid Pundt
> > > > Prof. Dr. Arnulf Quadt
> > > > apl. Prof. Dr. Karl-Henning Rehren
> > > > Prof. Dr. Ansgar Reiners
> > > > Prof. Dr. Angela Rizzi
> > > > Prof. Dr. Claus Ropers
> > > > Prof. Dr. Tim Salditt
> > > > Prof. Dr. Konrad Samwer
> > > > Prof. Dr. Christoph Schmidt
> > > > apl. Prof. Dr. Susanne Schneider
> > > > Prof. Dr. Steffen Schumann
> > > > Prof. Dr. Simone Techert
> > > > apl. Prof. Dr. Michael Seibt
> > > > Prof. Dr. Peter Sollich
> > > > Prof. Dr. Andreas Tilgner
> > > > Prof. Cynthia A. Volkert
> > > > Prof. Dr. Florentin Wörgötter
> > > > Prof. Dr. Annette Zippelius
>
> 
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> 3rd published book
>
> AP's Proof-Ellipse was never a Conic Section // Math proof series, book 1 Kindle Edition
> by Archimedes Plutonium (Author)
>
> Ever since Ancient Greek Times it was thought the slant cut into a cone is the ellipse. That was false. For the slant cut in every cone is a Oval, never an Ellipse. This book is a proof that the slant cut is a oval, never the ellipse. A slant cut into the Cylinder is in fact a ellipse, but never in a cone.
>
> Product details
> • ASIN ‏ : ‎ B07PLSDQWC
> • Publication date ‏ : ‎ March 11, 2019
> • Language ‏ : ‎ English
> • File size ‏ : ‎ 1621 KB
> • Text-to-Speech ‏ : ‎ Enabled
> • Enhanced typesetting ‏ : ‎ Enabled
> • X-Ray ‏ : ‎ Not Enabled
> • Word Wise ‏ : ‎ Not Enabled
> • Print length ‏ : ‎ 20 pages
> • Lending ‏ : ‎ Enabled
> •
> •
>
> Proofs Ellipse is never a Conic section, always a Cylinder section and a Well Defined Oval definition//Student teaches professor series, book 5 Kindle Edition
> by Archimedes Plutonium (Author)
> 
>
> Last revision was 14May2022. This is AP's 68th published book of science.
>
> Preface: A similar book on single cone cut is a oval, never a ellipse was published in 11Mar2019 as AP's 3rd published book, but Amazon Kindle converted it to pdf file, and since then, I was never able to edit this pdf file, and decided rather than struggle and waste time, decided to leave it frozen as is in pdf format. Any new news or edition of ellipse is never a conic in single cone is now done in this book. The last thing a scientist wants to do is wade and waddle through format, when all a scientist ever wants to do is science itself. So all my new news and thoughts of Conic Sections is carried out in this 68th book of AP. And believe you me, I have plenty of new news.
>
> In the course of 2019 through 2022, I have had to explain this proof often on Usenet, sci.math and sci.physics. And one thing that constant explaining does for a mind of science, is reduce the proof to its stripped down minimum format, to bare bones skeleton proof. I can prove the slant cut in single cone is a Oval, never the ellipse in just a one sentence proof. Proof-- A single cone and oval have just one axis of symmetry, while a ellipse requires 2 axes of symmetry, hence slant cut is always a oval, never the ellipse.
>
> Product details
> • ASIN ‏ : ‎ B081TWQ1G6
> • Publication date ‏ : ‎ November 21, 2019
> • Language ‏ : ‎ English
> • File size ‏ : ‎ 827 KB
> • Simultaneous device usage ‏ : ‎ Unlimited
> • Text-to-Speech ‏ : ‎ Enabled
> • Screen Reader ‏ : ‎ Supported
> • Enhanced typesetting ‏ : ‎ Enabled
> • X-Ray ‏ : ‎ Not Enabled
> • Word Wise ‏ : ‎ Not Enabled
> • Print length ‏ : ‎ 51 pages
> • Lending ‏ : ‎ Enabled
>
> #12-2, 11th published book
>
> World's First Geometry Proof of Fundamental Theorem of Calculus// Math proof series, book 2 Kindle Edition
> by Archimedes Plutonium (Author)
>
> Last revision was 15Dec2021. This is AP's 11th published book of science.
> Preface:
> Actually my title is too modest, for the proof that lies within this book makes it the World's First Valid Proof of Fundamental Theorem of Calculus, for in my modesty, I just wanted to emphasis that calculus was geometry and needed a geometry proof. Not being modest, there has never been a valid proof of FTC until AP's 2015 proof. This also implies that only a geometry proof of FTC constitutes a valid proof of FTC.
>
> Calculus needs a geometry proof of Fundamental Theorem of Calculus. But none could ever be obtained in Old Math so long as they had a huge mass of mistakes, errors, fakes and con-artist trickery such as the "limit analysis".. And very surprising that most math professors cannot tell the difference between a "proving something" and that of "analyzing something". As if an analysis is the same as a proof. We often analyze various things each and every day, but few if none of us consider a analysis as a proof. Yet that is what happened in the science of mathematics where they took an analysis and elevated it to the stature of being a proof, when it was never a proof.
>
> To give a Geometry Proof of Fundamental Theorem of Calculus requires math be cleaned-up and cleaned-out of most of math's mistakes and errors. So in a sense, a Geometry FTC proof is a exercise in Consistency of all of Mathematics. In order to prove a FTC geometry proof, requires throwing out the error filled mess of Old Math. Can the Reals be the true numbers of mathematics if the Reals cannot deliver a Geometry proof of FTC? Can the functions that are not polynomial functions allow us to give a Geometry proof of FTC? Can a Coordinate System in 2D have 4 quadrants and still give a Geometry proof of FTC? Can a equation of mathematics with a number that is _not a positive decimal Grid Number_ all alone on the right side of the equation, at all times, allow us to give a Geometry proof of the FTC?
>
> Cover Picture: Is my hand written, one page geometry proof of the Fundamental Theorem of Calculus, the world's first geometry proof of FTC, 2013-2015, by AP.
>
>
> Product details
> ASIN ‏ : ‎ B07PQTNHMY
> Publication date ‏ : ‎ March 14, 2019
> Language ‏ : ‎ English
> File size ‏ : ‎ 1309 KB
> Text-to-Speech ‏ : ‎ Enabled
> Screen Reader ‏ : ‎ Supported
> Enhanced typesetting ‏ : ‎ Enabled
> X-Ray ‏ : ‎ Not Enabled
> Word Wise ‏ : ‎ Not Enabled
> Print length ‏ : ‎ 154 pages
> Lending ‏ : ‎ Enabled
> Amazon Best Sellers Rank: #128,729 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
> #2 in 45-Minute Science & Math Short Reads
> #134 in Calculus (Books)
> #20 in Calculus (Kindle Store)

lördag 19 november 2022 kl. 23:20:49 UTC+1 skrev WM:
> Abstract: We will prove by means of Cantor's mapping between natural numbers and positive fractions that his approach to actual infinity implies the existence of numbers which cannot be applied as defined individuals. We will call them dark numbers.

Good luck! You haven't even defined what any of that means

>
>
> 1. Outline of the proof
>
> (1) We assume that all natural numbers are existing and are indexing all integer fractions in a matrix of all positive fractions.
> (2) Then we distribute, according to Cantor's prescription, these indices over the whole matrix. We observe that in every step prescribed by Cantor the set of indices does not increase and the set of not indexed fractions does not decrease.

And here you go wrong instantly, it is not a step by step process!

> (3) Therefore it is impossible to index all fractions in a definable way. Indexing many fractions together "in the limit" would be undefined and can be excluded according to section 2 below. Reducing the discrepancy step by step would imply a first event after finitely many steps.
> (4) In case of a complete mapping of  into the matrix, i.e., when every index has entered its final position, only indexed fractions are visible in the matrix.
> (5) We conclude from the invisible but doubtless present not indexed fractions that they are attached to invisible positions of the matrix.
> (6) By symmetry considerations also the first column of the matrix and therefore also  contains invisible, so-called dark elements.
> (7) Hence also the initial mapping of natural numbers and integer fractions cannot have been complete. Bijections, i.e., complete mappings, of actually infinite sets and are impossible.
>
>
> 2. Rejecting the limit idea
>
> When dealing with Cantor's mappings between infinite sets, it is argued usually that these mappings require a "limit" to be completed or that they cannot be completed. Such arguing has to be rejected flatly. For this reason some of Cantor's statements are quoted below.
>
> "If we think the numbers p/q in such an order [...] then every number p/q comes at an absolutely fixed position of a simple infinite sequence" [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 126]
>
> "The infinite sequence thus defined has the peculiar property to contain the positive rational numbers completely, and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
> "thus we get the epitome (ω) of all real algebraic numbers [...] and with respect to this order we can talk about the th algebraic number where not a single one of this epitome () has been forgotten." [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 116]
>
> "such that every element of the set stands at a definite position of this sequence" [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 152]
>
> The clarity of these expressions is noteworthy: all and every, completely, at an absolutely fixed position, th number, where not a single one has been forgotten.
>
> "In fact, according to the above definition of cardinality, the cardinal number M remains unchanged if in place of an element or of each of some elements, or even of each of all elements m of M another thing is substituted." [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 283]
>
> This opportunity will be utilized to replace the pairs of the bijection by matrices or to attach a matrix to every pair of the bijection, respectively.
>
>
> 3. The proof
>
> If all positive fractions m/n are existing, then they all are contained in the matrix
>
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> 5/1, 5/2, 5/3, 5/4, ...
>      .
>
> If all natural numbers k are existing, then they can be used as indices to index the integer fractions m/1 of the first column. Denoting indexed fractions by X and not indexed fractions by O, we obtain the matrix
>
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> XOOO...
>      .
>
> Cantor claimed that all natural numbers k are existing and can be applied to index all positive fractions m/n. They are distributed according to
>
> k = (m + n - 1)(m + n - 2)/2 + m .
> The result is a sequence of fractions
>
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... .
>
> This sequence is modelled here in the language of matrices. The indices are taken from their initial positions in the first column and are distributed in the given order.
>
> Index 1 remains at fraction 1/1, the first term of the sequence. The next term, 1/2, is indexed with 2 which is taken from its initial position 2/1
>
> XXOO...
> OOOO...
> XOOO...
> XOOO...
> XOOO...
>      .
>
> Then index 3 is taken from its initial position 3/1 and is attached to 2/1
>
> XXOO...
> XOOO...
> OOOO...
> XOOO...
> XOOO...
>      .
>
> Then index 4 is taken from its initial position 4/1 and is attached to 1/3
>
> XXXO...
> XOOO...
> OOOO...
> OOOO...
> XOOO...
>      .
>
> Then index 5 is taken from its initial position 5/1 and is attached to 2/2
>
> XXXO...
> XXOO...
> OOOO...
> OOOO...
> OOOO...
>      .
>
> And so on. When finally all exchanges of X and O have been carried out and, according to Cantor, all indices have been issued, it turns out that no fraction without index is visible any longer
>
> XXXX...
> XXXX...
> XXXX...
> XXXX...
> XXXX...
>     
>
> but by the process of lossless exchange of X and O no O can have left the matrix as long as finite natural numbers are issued as indices. Therefore there are not less fractions without index than at the beginning.
>
> We know that all O and as many fractions without index are remaining, but we cannot find any one. Where are they? The only possible explanation is that they are attached to dark positions.
>
> By means of symmetry considerations we can conclude that every column including the integer fractions and therefore also the natural numbers contain dark elements. Cantor's indexing covers only the potentially infinite collection of visible fractions, not the actually infinite set of all fractions.. This concerns also every other attempt to index the fractions and even the identical mapping. Bijections, i.e., complete mappings, of actually infinite sets and are impossible.
>
>
> 4. Counterarguments
>
> Now and then it is argued, in spite of the preconditions explicitly quoted in section 2, that a set-theoretical or analytical limit should be applied. This however would imply that all the O remain present in all definable matrices until "in the limit" these infinitely many O have to leave in an undefinable way; hence infinitely many fractions have to become indexed "in the limit" such that none of them can be checked  contrary to the proper meaning of indexing.
>
> Some set theorists reject it as inadmissible to "limit" the indices by starting in the first column. But that means only to check that the set of natural numbers has the same size as the set of integer fractions. In contrast to Cantor's procedure the origin of the natural numbers is remembered. But this  the only difference to Cantor's approach  does not interfere with the indexing prescription and would not destroy the bijection if it really existed.
>
> Finally, the counter argument that in spite of lossless exchange of X and O a loss of O could be tolerated suffers from deliberately contradicting basic logic.
>
> Regards, WM

You are still trying to do a step by step thing, which is invalid

zelos...@gmail.com schrieb am Montag, 21. November 2022 um 07:36:10 UTC+1:
> lördag 19 november 2022 kl. 23:20:49 UTC+1 skrev WM:
> > Abstract: We will prove by means of Cantor's mapping between natural numbers and positive fractions that his approach to actual infinity implies the existence of numbers which cannot be applied as defined individuals. We will call them dark numbers.
> Good luck! You haven't even defined what any of that means

Dark numbers are numbers which cannot be defined individuallyx.

> > Finally, the counter argument that in spite of lossless exchange of X and O a loss of O could be tolerated suffers from deliberately contradicting basic logic.
>
> You are still trying to do a step by step thing, which is invalid

which is valid for every step which can be defined individually. For all those steps the difference between O's and X's remains constant because no O leaves the matrix.

But a much simpler example is this one. Consider the sequence of strings
21111111111111111111111111...
12111111111111111111111111...
11211111111111111111111111...
11121111111111111111111111...
and so on, exchanging always the 2 and the 1 following next upon the 2.
Will the 2 leave the string? Why should it? There are obviously enough places.
Nevertheless, in the final state only the string
11111111111111111111111111...
is visible. The 2, if present yet, occupies a dark place.

Regards, WM

måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 07:36:10 UTC+1:
> > lördag 19 november 2022 kl. 23:20:49 UTC+1 skrev WM:
> > > Abstract: We will prove by means of Cantor's mapping between natural numbers and positive fractions that his approach to actual infinity implies the existence of numbers which cannot be applied as defined individuals. We will call them dark numbers.
> > Good luck! You haven't even defined what any of that means
> Dark numbers are numbers which cannot be defined individuallyx.

What the fuckd oes that even mean? You haven't defined what any of that means!

> > > Finally, the counter argument that in spite of lossless exchange of X and O a loss of O could be tolerated suffers from deliberately contradicting basic logic.
> >
> > You are still trying to do a step by step thing, which is invalid
> which is valid for every step which can be defined individually. For all those steps the difference between O's and X's remains constant because no O leaves the matrix.

That still means fucking nothing!

And your matrix is still a red herring!

>
> But a much simpler example is this one. Consider the sequence of strings
> 21111111111111111111111111...
> 12111111111111111111111111...
> 11211111111111111111111111...
> 11121111111111111111111111...
> and so on, exchanging always the 2 and the 1 following next upon the 2.
> Will the 2 leave the string? Why should it? There are obviously enough places.
> Nevertheless, in the final state only the string
> 11111111111111111111111111...
> is visible. The 2, if present yet, occupies a dark place.
>
> Regards, WM

Still haven't defined what "dark" even means.

zelos...@gmail.com schrieb am Montag, 21. November 2022 um 11:08:59 UTC+1:
> måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:

> > But a much simpler example is this one. Consider the sequence of strings
> > 21111111111111111111111111...
> > 12111111111111111111111111...
> > 11211111111111111111111111...
> > 11121111111111111111111111...
> > and so on, exchanging always the 2 and the 1 following next upon the 2.
> > Will the 2 leave the string? Why should it? There are obviously enough places.
> > Nevertheless, in the final state only the string
> > 11111111111111111111111111...
> > is visible. The 2, if present yet, occupies a dark place.
> >
> Still haven't defined what "dark" even means.

Can you see the 2 in 11111111111111111111111111... ?

Regards, WM

On Monday, 21 November 2022 at 05:26:55 UTC-4, WM wrote:
[...]
> But a much simpler example is this one. Consider the sequence of strings
> 21111111111111111111111111...
> 12111111111111111111111111...
> 11211111111111111111111111...
> 11121111111111111111111111...
> and so on, exchanging always the 2 and the 1 following next upon the 2.
> Will the 2 leave the string? Why should it? There are obviously enough places.
> Nevertheless, in the final state only the string
> 11111111111111111111111111...
> is visible. The 2, if present yet, occupies a dark place.

There is no "final state" to a step-wise infinite process, you fucking idiot. All you have is a limit state, which is *NOT* subject to the rules that govern the composition of the strings in the sequence. You have many times been given the hint that the limit of the sequence 1, 1/2, 1/3, 1/4, ... is zero, even though every term of the sequence is greater than zero. How hard can that be to comprehend? And after twenty or so years you *STILL* don't get it? The only dark thing is your mind.

måndag 21 november 2022 kl. 13:39:12 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 11:08:59 UTC+1:
> > måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:
>
> > > But a much simpler example is this one. Consider the sequence of strings
> > > 21111111111111111111111111...
> > > 12111111111111111111111111...
> > > 11211111111111111111111111...
> > > 11121111111111111111111111...
> > > and so on, exchanging always the 2 and the 1 following next upon the 2.
> > > Will the 2 leave the string? Why should it? There are obviously enough places.
> > > Nevertheless, in the final state only the string
> > > 11111111111111111111111111...
> > > is visible. The 2, if present yet, occupies a dark place.
> > >
> > Still haven't defined what "dark" even means.
> Can you see the 2 in 11111111111111111111111111... ?
>
> Regards, WM
Can you see the 9 in 11?

Gus Gassmann schrieb am Montag, 21. November 2022 um 14:06:15 UTC+1:
> On Monday, 21 November 2022 at 05:26:55 UTC-4, WM wrote:
> [...]
> > But a much simpler example is this one. Consider the sequence of strings
> > 21111111111111111111111111...
> > 12111111111111111111111111...
> > 11211111111111111111111111...
> > 11121111111111111111111111...
> > and so on, exchanging always the 2 and the 1 following next upon the 2.
> > Will the 2 leave the string? Why should it? There are obviously enough places.
> > Nevertheless, in the final state only the string
> > 11111111111111111111111111...
> > is visible. The 2, if present yet, occupies a dark place.
> There is no "final state" to a step-wise infinite process,

Then there is no state where indexing is complete and has been finished. But that is our starting point.

> All you have is a limit state, which is *NOT* subject to the rules that govern the composition of the strings in the sequence.

and which is not a term of this sequence. But here we consider only terms of the sequence.

> You have many times been given the hint that the limit of the sequence 1, 1/2, 1/3, 1/4, ... is zero, even though every term of the sequence is greater than zero.

The limit is not a term of the sequence. But all visible terms of the sequence have almost all terms as successors.

Regards, WM

zelos...@gmail.com schrieb am Montag, 21. November 2022 um 14:43:03 UTC+1:
> måndag 21 november 2022 kl. 13:39:12 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Montag, 21. November 2022 um 11:08:59 UTC+1:
> > > måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:
> >
> > > > But a much simpler example is this one. Consider the sequence of strings
> > > > 21111111111111111111111111...
> > > > 12111111111111111111111111...
> > > > 11211111111111111111111111...
> > > > 11121111111111111111111111...
> > > > and so on, exchanging always the 2 and the 1 following next upon the 2.
> > > > Will the 2 leave the string? Why should it? There are obviously enough places.
> > > > Nevertheless, in the final state only the string
> > > > 11111111111111111111111111...
> > > > is visible. The 2, if present yet, occupies a dark place.
> > > >
> > > Still haven't defined what "dark" even means.
> > Can you see the 2 in 11111111111111111111111111... ?

> Can you see the 9 in 11?

9 is not involved in this example but 2 is.

Regards, WM

On 11/21/2022 6:39 AM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 11:08:59 UTC+1:
>> måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:
>
>>> But a much simpler example is this one. Consider the sequence of strings
>>> 21111111111111111111111111...
>>> 12111111111111111111111111...
>>> 11211111111111111111111111...
>>> 11121111111111111111111111...
>>> and so on, exchanging always the 2 and the 1 following next upon the 2.
>>> Will the 2 leave the string? Why should it? There are obviously enough places.
>>> Nevertheless, in the final state only the string
>>> 11111111111111111111111111...
>>> is visible. The 2, if present yet, occupies a dark place.
>>>
>> Still haven't defined what "dark" even means.
>
> Can you see the 2 in 11111111111111111111111111... ?

notion says it is all 1's

if you want a 2 in it, put it in there.

>
> Regards, WM

again you twattle about in minor math notation

your one of the "0.999... not equal to 1.000..." trolls, minor silly stuff only found in boring places.

On 11/21/2022 9:35 AM, WM wrote:
> Gus Gassmann schrieb am Montag, 21. November 2022 um 14:06:15 UTC+1:
>> On Monday, 21 November 2022 at 05:26:55 UTC-4, WM wrote:
>> [...]
>>> But a much simpler example is this one. Consider the sequence of strings
>>> 21111111111111111111111111...
>>> 12111111111111111111111111...
>>> 11211111111111111111111111...
>>> 11121111111111111111111111...
>>> and so on, exchanging always the 2 and the 1 following next upon the 2.
>>> Will the 2 leave the string? Why should it? There are obviously enough places.
>>> Nevertheless, in the final state only the string
>>> 11111111111111111111111111...
>>> is visible. The 2, if present yet, occupies a dark place.
>> There is no "final state" to a step-wise infinite process,
>
> Then there is no state where indexing is complete and has been finished.

so you agree you are misleading people.

>
>> All you have is a limit state, which is *NOT* subject to the rules that govern the composition of the strings in the sequence.
>
> and which is not a term of this sequence. But here we consider only terms of the sequence.

all above sequences have a term 2 in them, there is no last or final state/step

>
>> You have many times been given the hint that the limit of the sequence 1, 1/2, 1/3, 1/4, ... is zero, even though every term of the sequence is greater than zero.
>
> The limit is not a term of the sequence. But all visible terms of the sequence have almost all terms as successors.

irrelevant.

>
> Regards, WM

On Monday, 21 November 2022 at 11:36:00 UTC-4, WM wrote:
> Gus Gassmann schrieb am Montag, 21. November 2022 um 14:06:15 UTC+1:
> > On Monday, 21 November 2022 at 05:26:55 UTC-4, WM wrote:
> > [...]
> > > But a much simpler example is this one. Consider the sequence of strings
> > > 21111111111111111111111111...
> > > 12111111111111111111111111...
> > > 11211111111111111111111111...
> > > 11121111111111111111111111...
> > > and so on, exchanging always the 2 and the 1 following next upon the 2.
> > > Will the 2 leave the string? Why should it? There are obviously enough places.
> > > Nevertheless, in the final state only the string
> > > 11111111111111111111111111...
> > > is visible. The 2, if present yet, occupies a dark place.
> > There is no "final state" to a step-wise infinite process,
> Then there is no state where indexing is complete and has been finished. But that is our starting point.

Of course there is no (finite) state at which indexing is complete. Why the *FUCK* do you insist that there has to be? Try to get it through that last remaining neuron of yours that this is an infinite process, which has a *LIMIT*. Ben Bacarisse took great pains to explain that to you, although it is obvious that nothing sticks anymore, Teflon-brain.

[snip remaining bullshit]

måndag 21 november 2022 kl. 16:36:51 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 14:43:03 UTC+1:
> > måndag 21 november 2022 kl. 13:39:12 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Montag, 21. November 2022 um 11:08:59 UTC+1:
> > > > måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:
> > >
> > > > > But a much simpler example is this one. Consider the sequence of strings
> > > > > 21111111111111111111111111...
> > > > > 12111111111111111111111111...
> > > > > 11211111111111111111111111...
> > > > > 11121111111111111111111111...
> > > > > and so on, exchanging always the 2 and the 1 following next upon the 2.
> > > > > Will the 2 leave the string? Why should it? There are obviously enough places.
> > > > > Nevertheless, in the final state only the string
> > > > > 11111111111111111111111111...
> > > > > is visible. The 2, if present yet, occupies a dark place.
> > > > >
> > > > Still haven't defined what "dark" even means.
> > > Can you see the 2 in 11111111111111111111111111... ?
> > Can you see the 9 in 11?
> 9 is not involved in this example but 2 is.
>
> Regards, WM
if it is ...111 there is no 2 in it.

On 11/21/2022 9:36 AM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 14:43:03 UTC+1:
>> måndag 21 november 2022 kl. 13:39:12 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 11:08:59 UTC+1:
>>>> måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:
>>>
>>>>> But a much simpler example is this one. Consider the sequence of strings
>>>>> 21111111111111111111111111...
>>>>> 12111111111111111111111111...
>>>>> 11211111111111111111111111...
>>>>> 11121111111111111111111111...
>>>>> and so on, exchanging always the 2 and the 1 following next upon the 2.
>>>>> Will the 2 leave the string? Why should it? There are obviously enough places.
>>>>> Nevertheless, in the final state only the string
>>>>> 11111111111111111111111111...
>>>>> is visible. The 2, if present yet, occupies a dark place.
>>>>>
>>>> Still haven't defined what "dark" even means.
>>> Can you see the 2 in 11111111111111111111111111... ?
>
>> Can you see the 9 in 11?
>
> 9 is not involved in this example but 2 is.
>
> Regards, WM

Wrong, 9 is the generator function of all 1's

1/9 = .1111111111111111111111111111...

On 11/21/2022 3:26 AM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 07:36:10 UTC+1:
>> lördag 19 november 2022 kl. 23:20:49 UTC+1 skrev WM:
>>> Abstract: We will prove by means of Cantor's mapping between natural numbers and positive fractions that his approach to actual infinity implies the existence of numbers which cannot be applied as defined individuals. We will call them dark numbers.
>> Good luck! You haven't even defined what any of that means
>
> Dark numbers are numbers which cannot be defined individuallyx.
>
>>> Finally, the counter argument that in spite of lossless exchange of X and O a loss of O could be tolerated suffers from deliberately contradicting basic logic.
>>
>> You are still trying to do a step by step thing, which is invalid
>
> which is valid for every step which can be defined individually. For all those steps the difference between O's and X's remains constant because no O leaves the matrix.
>
> But a much simpler example is this one. Consider the sequence of strings
> 21111111111111111111111111...
> 12111111111111111111111111...
> 11211111111111111111111111...
> 11121111111111111111111111...
> and so on, exchanging always the 2 and the 1 following next upon the 2.
> Will the 2 leave the string? Why should it? There are obviously enough places.
> Nevertheless, in the final state only the string
> 11111111111111111111111111...
> is visible. The 2, if present yet, occupies a dark place.

.....it is collected and supposited into the ever expanding dark regions of your mind.

>
> Regards, WM

does 1/2 = 0.5 yet ?

Sergi o schrieb am Montag, 21. November 2022 um 18:30:31 UTC+1:
> On 11/21/2022 9:36 AM, WM wrote:
> > zelos...@gmail.com schrieb am Montag, 21. November 2022 um 14:43:03 UTC+1:
> >> måndag 21 november 2022 kl. 13:39:12 UTC+1 skrev WM:
> >>> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 11:08:59 UTC+1:
> >>>> måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:
> >>>
> >>>>> But a much simpler example is this one. Consider the sequence of strings
> >>>>> 21111111111111111111111111...
> >>>>> 12111111111111111111111111...
> >>>>> 11211111111111111111111111...
> >>>>> 11121111111111111111111111...
> >>>>> and so on, exchanging always the 2 and the 1 following next upon the 2.
> >>>>> Will the 2 leave the string? Why should it? There are obviously enough places.
> >>>>> Nevertheless, in the final state only the string
> >>>>> 11111111111111111111111111...
> >>>>> is visible. The 2, if present yet, occupies a dark place.
> >>>>>
> >>>> Still haven't defined what "dark" even means.
> >>> Can you see the 2 in 11111111111111111111111111... ?
> >
> >> Can you see the 9 in 11?
> >
> > 9 is not involved in this example but 2 is.
> >
> Wrong, 9 is the generator function of all 1's
>
> 1/9 = .1111111111111111111111111111...

Read if you can. I wrote 1111111111111111111111111111... .

Regards, WM

On 11/21/2022 12:28 PM, WM wrote:
> Sergi o schrieb am Montag, 21. November 2022 um 18:30:31 UTC+1:
>> On 11/21/2022 9:36 AM, WM wrote:
>>> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 14:43:03 UTC+1:
>>>> måndag 21 november 2022 kl. 13:39:12 UTC+1 skrev WM:
>>>>> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 11:08:59 UTC+1:
>>>>>> måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:
>>>>>
>>>>>>> But a much simpler example is this one. Consider the sequence of strings
>>>>>>> 21111111111111111111111111...
>>>>>>> 12111111111111111111111111...
>>>>>>> 11211111111111111111111111...
>>>>>>> 11121111111111111111111111...
>>>>>>> and so on, exchanging always the 2 and the 1 following next upon the 2.
>>>>>>> Will the 2 leave the string? Why should it? There are obviously enough places.
>>>>>>> Nevertheless, in the final state only the string
>>>>>>> 11111111111111111111111111...
>>>>>>> is visible. The 2, if present yet, occupies a dark place.
>>>>>>>
>>>>>> Still haven't defined what "dark" even means.
>>>>> Can you see the 2 in 11111111111111111111111111... ?
>>>
>>>> Can you see the 9 in 11?
>>>
>>> 9 is not involved in this example but 2 is.
>>>
>> Wrong, 9 is the generator function of all 1's
>>
>> 1/9 = .1111111111111111111111111111...
>
> Read if you can. I wrote 1111111111111111111111111111... .
>
> Regards, WM

Irrelevant.

Gus Gassmann schrieb am Montag, 21. November 2022 um 17:36:03 UTC+1:
> On Monday, 21 November 2022 at 11:36:00 UTC-4, WM wrote:
> > Gus Gassmann schrieb am Montag, 21. November 2022 um 14:06:15 UTC+1:
> > > On Monday, 21 November 2022 at 05:26:55 UTC-4, WM wrote:
> > > [...]
> > > > But a much simpler example is this one. Consider the sequence of strings
> > > > 21111111111111111111111111...
> > > > 12111111111111111111111111...
> > > > 11211111111111111111111111...
> > > > 11121111111111111111111111...
> > > > and so on, exchanging always the 2 and the 1 following next upon the 2.
> > > > Will the 2 leave the string? Why should it? There are obviously enough places.
> > > > Nevertheless, in the final state only the string
> > > > 11111111111111111111111111...
> > > > is visible. The 2, if present yet, occupies a dark place.
> > > There is no "final state" to a step-wise infinite process,
> > Then there is no state where indexing is complete and has been finished.. But that is our starting point.

> this is an infinite process, which has a *LIMIT*. Ben Bacarisse took great pains to explain that to you, although it is obvious that nothing sticks anymore, Teflon-brain.
>
The limit is not involved when "every number p/q comes at an absolutely fixed position of a simple infinite sequence". But that is not so obvious to see as in my matrices.

If in every step all O remain which are existing from the beginning and when in the limit they suddenly disappear, then it is impossible to know their indices. Then they do not have indices. Therefore the limit cannot be applied if Cantor's claim shall be satisfied: "The infinite sequence thus defined has the peculiar property to contain the positive rational numbers completely, and each of them only once at a determined place." Note that every "determined place" means an individual index.

Regards, WM

On 11/21/2022 12:36 PM, WM wrote:
> Gus Gassmann schrieb am Montag, 21. November 2022 um 17:36:03 UTC+1:
>> On Monday, 21 November 2022 at 11:36:00 UTC-4, WM wrote:
>>> Gus Gassmann schrieb am Montag, 21. November 2022 um 14:06:15 UTC+1:
>>>> On Monday, 21 November 2022 at 05:26:55 UTC-4, WM wrote:
>>>> [...]
>>>>> But a much simpler example is this one. Consider the sequence of strings
>>>>> 21111111111111111111111111...
>>>>> 12111111111111111111111111...
>>>>> 11211111111111111111111111...
>>>>> 11121111111111111111111111...
>>>>> and so on, exchanging always the 2 and the 1 following next upon the 2.
>>>>> Will the 2 leave the string? Why should it? There are obviously enough places.
>>>>> Nevertheless, in the final state only the string
>>>>> 11111111111111111111111111...
>>>>> is visible. The 2, if present yet, occupies a dark place.
>>>> There is no "final state" to a step-wise infinite process,
>>> Then there is no state where indexing is complete and has been finished. But that is our starting point.
>
>> this is an infinite process, which has a *LIMIT*. Ben Bacarisse took great pains to explain that to you, although it is obvious that nothing sticks anymore, Teflon-brain.
>>
> The limit is not involved when "every number p/q comes at an absolutely fixed position of a simple infinite sequence". But that is not so obvious to see as in my matrices.

your matrices are no good at all, because you lose indexed positions of then fractions in your SwapadoodleFestival. Fail.

>
> If in every step all O remain which are existing from the beginning and when in the limit they suddenly disappear, then it is impossible to know their indices.

That is why Your matrices Fail. They do not work. You cannot explain how the fraction got its index removed!

> Then they do not have indices.

You intentionally stripped them off, admit it! you took off the stickies!!

> Therefore the limit

there is no limit.

> cannot be applied if Cantor's claim shall be satisfied: "The infinite sequence thus defined has the peculiar property to contain the positive rational numbers completely, and each of them only once at a determined place." Note that every "determined place" means an individual index.

no, it means the fractions have been indexed to the natural numbers.

>
> Regards, WM

does 1/9 = .1111111111111111111111111... yet ?

zelos...@gmail.com schrieb am Montag, 21. November 2022 um 17:46:24 UTC+1:
> måndag 21 november 2022 kl. 16:36:51 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Montag, 21. November 2022 um 14:43:03 UTC+1:
> > > måndag 21 november 2022 kl. 13:39:12 UTC+1 skrev WM:
> > > > zelos...@gmail.com schrieb am Montag, 21. November 2022 um 11:08:59 UTC+1:
> > > > > måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:
> > > >
> > > > > > But a much simpler example is this one. Consider the sequence of strings
> > > > > > 21111111111111111111111111...
> > > > > > 12111111111111111111111111...
> > > > > > 11211111111111111111111111...
> > > > > > 11121111111111111111111111...
> > > > > > and so on, exchanging always the 2 and the 1 following next upon the 2.
> > > > > > Will the 2 leave the string? Why should it? There are obviously enough places.
> > > > > > Nevertheless, in the final state only the string
> > > > > > 11111111111111111111111111...
> > > > > > is visible. The 2, if present yet, occupies a dark place.
> > > > > >
> > > > > Still haven't defined what "dark" even means.
> > > > Can you see the 2 in 11111111111111111111111111... ?
> > > Can you see the 9 in 11?
> > 9 is not involved in this example but 2 is.
> >
> if it is ...111 there is no 2 in it.

It is 11111111111111111111111111... But 2 has not left. The number of digits is the same in all terms of the sequence.

Regards, WM

On Monday, 21 November 2022 at 14:36:16 UTC-4, WM wrote:
> Gus Gassmann schrieb am Montag, 21. November 2022 um 17:36:03 UTC+1:
> > On Monday, 21 November 2022 at 11:36:00 UTC-4, WM wrote:
> > > Gus Gassmann schrieb am Montag, 21. November 2022 um 14:06:15 UTC+1:
> > > > On Monday, 21 November 2022 at 05:26:55 UTC-4, WM wrote:
> > > > [...]
> > > > > But a much simpler example is this one. Consider the sequence of strings
> > > > > 21111111111111111111111111...
> > > > > 12111111111111111111111111...
> > > > > 11211111111111111111111111...
> > > > > 11121111111111111111111111...
> > > > > and so on, exchanging always the 2 and the 1 following next upon the 2.
> > > > > Will the 2 leave the string? Why should it? There are obviously enough places.
> > > > > Nevertheless, in the final state only the string
> > > > > 11111111111111111111111111...
> > > > > is visible. The 2, if present yet, occupies a dark place.
> > > > There is no "final state" to a step-wise infinite process,
> > > Then there is no state where indexing is complete and has been finished. But that is our starting point.
> > this is an infinite process, which has a *LIMIT*. Ben Bacarisse took great pains to explain that to you, although it is obvious that nothing sticks anymore, Teflon-brain.
> >
> The limit is not involved when "every number p/q comes at an absolutely fixed position of a simple infinite sequence". But that is not so obvious to see as in my matrices.

And again you have no clue what you are talking about. "[E]very number p/q...." does not describe a (potentially infinite) sequence (with a limiting behavior), but an actually infinite mapping (which contains the limit). Cantor does things one way (the easy and proper way), and you insist on doing it the other way, which is more difficult and requires a limit (although you have never had even an inkling of what is going on). You are a fucking moron, always have been, and always will be. Please piss off.

On 11/21/2022 1:13 PM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 17:46:24 UTC+1:
>> måndag 21 november 2022 kl. 16:36:51 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 14:43:03 UTC+1:
>>>> måndag 21 november 2022 kl. 13:39:12 UTC+1 skrev WM:
>>>>> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 11:08:59 UTC+1:
>>>>>> måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:
>>>>>
>>>>>>> But a much simpler example is this one. Consider the sequence of strings
>>>>>>> 21111111111111111111111111...
>>>>>>> 12111111111111111111111111...
>>>>>>> 11211111111111111111111111...
>>>>>>> 11121111111111111111111111...
>>>>>>> and so on, exchanging always the 2 and the 1 following next upon the 2.
>>>>>>> Will the 2 leave the string? Why should it? There are obviously enough places.
>>>>>>> Nevertheless, in the final state only the string
>>>>>>> 11111111111111111111111111...
>>>>>>> is visible. The 2, if present yet, occupies a dark place.
>>>>>>>
>>>>>> Still haven't defined what "dark" even means.
>>>>> Can you see the 2 in 11111111111111111111111111... ?
>>>> Can you see the 9 in 11?
>>> 9 is not involved in this example but 2 is.
>>>
>> if it is ...111 there is no 2 in it.
>
> It is 11111111111111111111111111... But 2 has not left. The number of digits is the same in all terms of the sequence.

no. 11111111111111111111111111... means all 1's only and no 2.

try again.

>
> Regards, WM

On 11/21/2022 9:34 AM, Sergi o wrote:
> On 11/21/2022 3:26 AM, WM wrote:
>> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 07:36:10
>> UTC+1:
>>> lördag 19 november 2022 kl. 23:20:49 UTC+1 skrev WM:
>>>> Abstract: We will prove by means of Cantor's mapping between natural
>>>> numbers and positive fractions that his approach to actual infinity
>>>> implies the existence of numbers which cannot be applied as defined
>>>> individuals. We will call them dark numbers.
>>> Good luck! You haven't even defined what any of that means
>>
>> Dark numbers are numbers which cannot be defined individuallyx.
>>
>>>> Finally, the counter argument that in spite of lossless exchange of
>>>> X and O a loss of O could be tolerated suffers from deliberately
>>>> contradicting basic logic.
>>>
>>> You are still trying to do a step by step thing, which is invalid
>>
>> which is valid for every step which can be defined individually. For
>> all those steps the difference between O's and X's remains constant
>> because no O leaves the matrix.
>>
>> But a much simpler example is this one. Consider the sequence of strings
>> 21111111111111111111111111...
>> 12111111111111111111111111...
>> 11211111111111111111111111...
>> 11121111111111111111111111...
>> and so on, exchanging always the 2 and the 1 following next upon the 2.
>> Will the 2 leave the string? Why should it? There are obviously enough
>> places.
>> Nevertheless, in the final state only the string
>> 11111111111111111111111111...
>> is visible. The 2, if present yet, occupies a dark place.
>
> ....it is collected and supposited into the ever expanding dark regions
> of your mind.
>
>>
>> Regards, WM
>
>
>  does 1/2 = 0.5  yet ?

Why does 1/2 = .5 seem to baffle WM?

On 11/21/2022 4:39 AM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 11:08:59 UTC+1:
>> måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:
>
>>> But a much simpler example is this one. Consider the sequence of strings
>>> 21111111111111111111111111...
>>> 12111111111111111111111111...
>>> 11211111111111111111111111...
>>> 11121111111111111111111111...
>>> and so on, exchanging always the 2 and the 1 following next upon the 2.
>>> Will the 2 leave the string? Why should it? There are obviously enough places.
>>> Nevertheless, in the final state only the string
>>> 11111111111111111111111111...
>>> is visible. The 2, if present yet, occupies a dark place.
>>>
>> Still haven't defined what "dark" even means.
>
> Can you see the 2 in 11111111111111111111111111... ?

Wow! You are dense. 2 must be dark. lol!

Can you see a 2 in

12
112
1112
11112
111112
1111112
11111112
[...]

1...2

This number can be constructed via direct function using only a single
natural number (including zero) at its input, so its not dark.