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ETH Zurich looking (perhaps publish) AP's World's first valid proof of Fundamental Theorem of Algebra
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Archimedes Plutonium
2017-06-14 18:40:29 UTC
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On Wednesday, June 14, 2017 at 10:27:11 AM UTC-5, ***@gmail.com wrote:


77.58.43.158 jan burse

ETH Zurich

Paul Biran, Marc Burger, Patrick Cheridito, Manfred Einsiedler, Paul Embrechts, Giovanni Felder, Alessio Figalli, Norbert Hungerbuhler, Tom Ilmanen, Horst Knorrer, Emmanuel Kowalski, Urs Lang, Rahul Pandharipande, Richard Pink, Tristan Riviere, Dietmar Salamon, Martin Schweizer, Mete Soner, Michael Struwe, Benjamin Sudakov, Alain Sznitman, Josef Teichmann, Wendelin Werner, Thomas Willwacher


1st VALID PROOF OF FUNDAMENTAL THEOREM OF ALGEBRA //Correcting Math pre-6th ed

by Archimedes Plutonium

...,


Old Math statement of FTA:: The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number A is a complex number with an imaginary part equal to zero.

....,
Comment:: New Math statement theorem of FTA need not be so long, because in New Math we recognize that sqrt-1 is simply A/0 where A is a real-number (New Real Number).

Theorem-Statement of FTA:: given any polynomial with New Real Numbers, that there always exists at least one New Real Number, call it A, as a solution.

Proof of FTA: x^2 +1 = 0, goes to 1/x^2 = -1, with solution x=0 as 1/0 = sqrt-1, thus any polynomial is 1/polynomial = +-k has at least, one solution for A.

Comment:: Notice the beauty of the Statement of Theorem then Proof are almost identical in length of words. This is what happens when you have a true proof of a statement in mathematics.

Comment:: there is a huge fallout of this proof, in the fact that "i" an imaginary number is no longer needed because the Reals in 0 and any Real A where we have A/0 is a imaginary number and is a complex number. So, all of a sudden the Complex Plane and imaginary number dissolves out of math history into a gutter of shame. Whenever see "i" what that really is, is A/0 where A is any Real number.
....,


AP
b***@gmail.com
2017-06-14 21:23:33 UTC
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repaste alarm, brainless spammer AP strikes again.
Post by Archimedes Plutonium
77.58.43.158 jan burse
ETH Zurich
Paul Biran, Marc Burger, Patrick Cheridito, Manfred Einsiedler, Paul Embrechts, Giovanni Felder, Alessio Figalli, Norbert Hungerbuhler, Tom Ilmanen, Horst Knorrer, Emmanuel Kowalski, Urs Lang, Rahul Pandharipande, Richard Pink, Tristan Riviere, Dietmar Salamon, Martin Schweizer, Mete Soner, Michael Struwe, Benjamin Sudakov, Alain Sznitman, Josef Teichmann, Wendelin Werner, Thomas Willwacher
1st VALID PROOF OF FUNDAMENTAL THEOREM OF ALGEBRA //Correcting Math pre-6th ed
by Archimedes Plutonium
...,
Old Math statement of FTA:: The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number A is a complex number with an imaginary part equal to zero.
....,
Comment:: New Math statement theorem of FTA need not be so long, because in New Math we recognize that sqrt-1 is simply A/0 where A is a real-number (New Real Number).
Theorem-Statement of FTA:: given any polynomial with New Real Numbers, that there always exists at least one New Real Number, call it A, as a solution.
Proof of FTA: x^2 +1 = 0, goes to 1/x^2 = -1, with solution x=0 as 1/0 = sqrt-1, thus any polynomial is 1/polynomial = +-k has at least, one solution for A.
Comment:: Notice the beauty of the Statement of Theorem then Proof are almost identical in length of words. This is what happens when you have a true proof of a statement in mathematics.
Comment:: there is a huge fallout of this proof, in the fact that "i" an imaginary number is no longer needed because the Reals in 0 and any Real A where we have A/0 is a imaginary number and is a complex number. So, all of a sudden the Complex Plane and imaginary number dissolves out of math history into a gutter of shame. Whenever see "i" what that really is, is A/0 where A is any Real number.
....,
AP
Archimedes Plutonium
2017-06-15 04:47:55 UTC
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Post by Archimedes Plutonium
77.58.43.158 jan burse
ETH Zurich
Paul Biran, Marc Burger, Patrick Cheridito, Manfred Einsiedler, Paul Embrechts, Giovanni Felder, Alessio Figalli, Norbert Hungerbuhler, Tom Ilmanen, Horst Knorrer, Emmanuel Kowalski, Urs Lang, Rahul Pandharipande, Richard Pink, Tristan Riviere, Dietmar Salamon, Martin Schweizer, Mete Soner, Michael Struwe, Benjamin Sudakov, Alain Sznitman, Josef Teichmann, Wendelin Werner, Thomas Willwacher
World's first valid Prime Number Theorem proof done via Math-Induction//Correcting Math pre-6th ed


MATHEMATICAL-INDUCTION PROOF OF PRIME NUMBER THEOREM PNT; first valid proof of PNT


Re: beefing up my Math Induction proof Page84, 12-1, World's first valid Prime Number Theorem proof done via Math-Induction

On Wednesday, April 12, 2017 at 3:37:37 PM UTC-5, Don Redmond wrote:
(snipped)
Post by Archimedes Plutonium
Firstly, I don't see where the induction argument is. It looks more like you're trying to show that the sum of the first n positive integers is 27 and readjusting at each step. That's not a proof, that's the arithmetic equivalence of doodling.
Secondly, why does the sum of the reciprocals of the powers of 2, from 0 onward equaling 2 seem so strange. After N steps you have 2 - 2^(-N). Even on your grid system this will look like 2 for N large enough. Similarly for the sum of the reciprocals of the triangular numbers we get, after N steps we get 2 - (2/(N + 1), which also look like 2 on your grid system for N large enough. We have nothing to doctor here.
Don
Hi Don, yes I can make that much more clear as to what I was Inducting.

Since no computer on Earth can tell us what the primes counting is at 1*10^604, I thought Math Induction is the only means of establishing the proof.

So I assembled a precision definition of when two series are equal. You need a case 1 equality of 100/4 = 25 is our case 1. Then, jumping ahead to the 10^604 Grid, I need to have an equality there. So, given case 1 and the ending case as equal, and everything in between can be Math Induction doctored. I prove PNT true.

What I Induct, is the entire process of doctoring a Grid.

I have case 1, now I inspect Grid 10^4, and doctor it so we have equality.

Now I take Grid 10^5 and do the same thing.

Then I say given case 1 and given the process of Grid 10^4 then Grid 10^5, that we can doctor the end Grid to form a equality, means that I Math Inducted the entire process to have beginning equality and have ending equality.

Now, you may complain that you can prove every series is equal to another series, but not true. For example, the Riemann Hypothesis with the Euler series, there is no beginning equality, meaning that the Riemann Hypothesis is a fakery to even start. Both the Euler series and Riemann series have no beginning equality, hence, the RH is a nonstarter.

Another example. The Even numbers series 2, 4, 6, 8, 10, .... versus Odd numbers 1, 3, 5, 7, 9, .... Are they equal? Old Math says yes. Precision Math says no.

Give me a starting equality.

The 2 does not equal 1

The second sum 2+4 does not equal 1+3.

Never, along those two Series can you get a Starting Block Equality.

With Prime Number Theorem, you easily have a Starting Equality 100/4 = 25 primes.

Don, question, why are mathematicians ever so derelict in defining equality between two series? Why are they so lazy and ignorant in well defining Series Equality? Even a High School kid recognizes that when you say -- "these two series are equal" recognizes that at some specific term we should have the number. But, no, Old Math says two series are equal because at Infinity puff the magic dragon says they are equal.

So, Don, are you not sick and tired of telling young kids, series A is equal to series B because puff the magic dragon at infinity makes them equal. I sure am sick and tired of hearing puff the magic dragon makes them equal.

AP

Re: beefing up my Math Induction proof Page84, 12-1, World's first valid Prime Number Theorem proof done via Math-Induction

- show quoted text -
Speaking of doctoring up, I need to doctor up the above sentence to read this::

"these two series are equal" recognizes that at some specific term we should have the numbers the same for both series.

When I first learned about Series, in College, I was awestruck by the lack of coordination of series concept, the lack of a definition of when one series equals another series, and what really aggravated me, was that no-one in math seems to understand that if you say series A equals series B, that there has to be a STARTING block equality

1+3+5+7+.....

is never equal to

2+ 4+6+8+... because you can never get a nth term that is equal in both

Now, the Series

2+2+4+6+8+.... can equal 1+3+5+7+.... because we have a starting block equality of the 2nd terms

1+3 = 2+2

And thus, with that Starting Block Equality we can build into the Series a doctored ending equality.

So, Old Math is very very pitifully poor on Series equality and the reason is easy to see-- they have a moron definition of infinity-- no borderline.

AP

Re: beefing up my Math Induction proof Page84, 12-1, World's first valid Prime Number Theorem proof done via Math-Induction

- show quoted text -
Speaking of doctoring up, I need to doctor up the above sentence to read this::

"these two series are equal" recognizes that at some specific term we should have the numbers the same for both series.

When I first learned about Series, in College, I was awestruck by the lack of coordination of series concept, the lack of a definition of when one series equals another series, and what really aggravated me, was that no-one in math seems to understand that if you say series A equals series B, that there has to be a STARTING block equality

1+3+5+7+.....

is never equal to

2+ 4+6+8+... because you can never get a nth term that is equal in both

Now, the Series

2+2+4+6+8+.... can equal 1+3+5+7+.... because we have a starting block equality of the 2nd terms

1+3 = 2+2

And thus, with that Starting Block Equality we can build into the Series a doctored ending equality.

So, Old Math is very very pitifully poor on Series equality and the reason is easy to see-- they have a moron definition of infinity-- no borderline.

AP



In New Math we have Grid Systems:

10 Grid
100 Grid
1000 Grid
10^4 Grid
10^5 Grid
10^6 Grid
etc

Primes, the actual count of Primes in those Grids listed above follow this progression


This is the Grid PROGRESSION

10   4 actual primes, 10/2 = 5 predicted when using the formula of base sqrt10 is exponent 2

10^2         25 actual       25 SPOT ON EXACT with sqrt10 base 100/4 which Here is 4

10^3         168 actual       lower bound 1000/7 = 142,  
upper bound 1000/6 = 166   HERE 6

10^4         1,229  actual   lower bound 10,000/9 = 1111 ,  
upper bound 10,000/8 = 1250   HERE 8

10^5         9,592 actual   lower bound 100,000/11 =  9090   ,  
upper bound 100,000/10 = 10,000      HERE 10

10^6         78,498 actual   lower bound 1,000,000/13 = 76,923 ,   HERE 13
upper bound 1,000,000/12 = 83,333  

10^7          664,579 actual     lower bound 10^7/16 = 625,000   ,  
upper bound 10^7/15 = 666,666    HERE 15

10^8          5,761,455 actual      lower bound 10^8/18 = 5,555,555    
 , upper bound 10^8/17 = 5,882,352  HERE 17

10^9         50,847,534 actual     lower 10^9/20 = 50,000,000   ,     HERE 20
upper 10^9/19 = 52,631,578

10^10        455,052,511 actual   lower 10^10/22 = 454,545,454   ,    HERE 22
 upper 10^10/21 = 476,190,476

10^11        4,118,054,813 actual     lower 10^11/24 = 4,116,666,667   ,    HERE 24
upper 10^11/23 = 4,347,826,087

10^12        37,607,912,018 actual   lower 10^12/26 = 38,461,538,46-     HERE 26
upper 10^12/25 = 40,000,000,000


SERIES representation of Actual Primes rather than Progression representation

4 + (25-4) + (168-25) + (1229-168) + (9592-1229) + . .

Progression of Predicted primes from formula using base sqrt10

10/2 =5, 100/4 =25, 1000/6 =166.666.. , 10000/8 = 1250, . .

Series representation of Predicted Primes using Formula sqrt10 as base with its exponent

5 + (25-5) + (166.666..- 25) + (1250 - 166.666..) + . .


ACTUAL INFINITY BORDERLINE IS 1*10^604 but since that is cumbersome to work with and since no computer has ever calculated the actual-primes from 0 to 1*10^604 we PRETEND infinity border is 10^4 and use that via MATH-INDUCTION to find the Doctored Formula.

MATHEMATICAL INDUCTION PROOFS to tell if two different Series are equal to each other at infinity requires these items:
1) Starting Equality
2) Mid Section of N to N+1 provided by Grid System
3) Ending Equality at the infinity border

Here we pretend 10^4 is the infinity border and look to find a DOCTORED formula of base sqrt10 exponent that delivers and Ending Equality.

10^4 Grid has actual 1229 primes yet predicts 10000/8 = 1250

So we Doctor our Formula of base sqrt10 exponent  to use pretend Grid 10^4 where N is 4 exponent on 10000/8 and subtract actual primes of Grid N-2 (that is 10^2 Grid) and add actual primes of Grid N-3 ( which is 10 Grid). So we have arithmetic wise, we have 1250 - 25 + 4 = 1229.

So we have in the end here, we have a Starting Equality of 100/4 = 25 primes in 100 Grid and we have a Ending Equality of 10000/8 subtract 25 add 4 = 1229.

Now, we apply that SCHEME or Pattern to the true infinity borderline of 1*10^604 and so we have something like this 10^604/1208 subtract Grid N-2 add Grid N-3 or add Grid N-1 add Grid N-2.

We do whatever it takes to Doctor the formula so that it makes the End result equal to actual primes at infinity border.

Now, we have a starting equality 100/4 and a Doctored End Equality. Now we do not worry about applying the Doctored Formula on the Mid Section terms, for we certainly have to apply that doctoring. Just so long as we have the Start and End terms agree with actual Prime Count.

Now, let me show you this method on the Zeta Series of the Riemann Hypothesis and show you why they are never equal for no starting-equality is possible at infinity. Show you on the Harmonic series (Oresme) and why it never diverges. And show you on the reciprocal two series of 2-doubled compared to triangular numbers alleged to converge to 2, why they are not equal at infinity border since it is impossible to doctor them at infinity border for a Ending equality. However, on the reciprocals we can find a different series for each that does converge at infinity border and hence are equal series.

What I am developing here is the first time in math history that we clean up the concept of two different Series equaling each other. This has never been done before for the simple reason that when you have Infinity as a notion, mere notion and opinion as to what infinity is, you cannot have a concept of Series equality at infinity. Only when you have a precise infinity with a borderline, can you have precision Series understanding.


Very crude dot picture of 5f6, 94TH
ELECTRON DOT CLOUD of 231Pu


::\ ::|:: /::
::\::|::/::
_ _
(:Y:)
- -
::/::|::\::
::/ ::|:: \::
One of those dots is the Milky Way galaxy. And each dot represents another galaxy.
            . \ .  . | .   /.
           . . \. . .|. . /. .
              ..\....|.../...
               ::\:::|::/::
---------------      -------------
--------------- (Y) -------------
---------------      --------------
               ::/:::|::\::
              ../....|...\...
           . . /. . .|. . \. .
            . / .  . | .   \ .


http://www.iw.net/~a_plutonium/ 
whole entire Universe is just one big atom 
where dots of the electron-dot-cloud are galaxies

I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of spammers, off-topic-misfits, front-page-hogs, stalking mockers, suppression-bullies, and demonizers.     

https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe        
Archimedes Plutonium
Archimedes Plutonium
2017-06-15 12:07:34 UTC
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On Wednesday, June 14, 2017 at 5:33:43 AM UTC-5, ***@gmail.com wrote:

Jan Burse
77.58.43.158

ETH Zurich

Paul Biran, Marc Burger, Patrick Cheridito, Manfred Einsiedler, Paul Embrechts, Giovanni Felder, Alessio Figalli, Norbert Hungerbuhler, Tom Ilmanen, Horst Knorrer, Emmanuel Kowalski, Urs Lang, Rahul Pandharipande, Richard Pink, Tristan Riviere, Dietmar Salamon, Martin Schweizer, Mete Soner,
Michael Struwe, Benjamin Sudakov, Alain Sznitman, Josef Teichmann, Wendelin Werner, Thomas Willwacher

Page85, 12-2, many major proofs
AP's Proof of Riemann Hypothesis, part 1 of 2

PART 1 of 2: LOGICAL FLAWS & DISPROOF OF THE RIEMANN HYPOTHESIS //Correcting Math 5th ed

MAJOR LOGICAL FLAWS & DISPROOF OF THE RIEMANN HYPOTHESIS RH

by Archimedes Plutonium

Introduction:

the connection with PNT and RH Re: Page86, Part 2 of 2: LOGICAL FLAWS & DISPROOF OF THE RIEMANN HYPOTHESIS //Correcting Math 5th ed

Now these first two proofs-- Prime Number Theorem PNT and Riemann Hypothesis RH, follow a pattern, a obvious bold and strong pattern.

The pattern is that Old Math has a crazy quilt definition of Series equality. Since Old Math has no borderline with infinity, they have no good definition of what it means for two series to be equal. And so sloppy are they, that they imagine the Series 1 + 1/2 + 1/3 + 1/4 +, . . . the harmonic series is equal to the Series 1+1+1+.... at infinity, Old Math believes those two Series are equal. That is how crazy Old Math was.

And then, of course, when something like PNT comes or RH comes, that having a crazy notion of what is equality for Series, it is little wonder that the nutters of Old Math have troubles with PNT and RH.

AP

Comparing RH to the Prime Number Theorem PNT, in that proof Mathematical Induction on Series, I had a Starting Equality of 100/sqrt10^4 and using sqrt10 base rather than "e" base I have a Starting Equality of 100/4 = 25, and at infinity I Doctor the Ending Equality.

Now, we compare Riemann Hypothesis with its zetas and we also compare the Harmonic Series. The Zetas and the Harmonic Series are never able to give us a Starting Equality to apply a Math Induction proof of RH. What that means is RH is unprovable. It is a false conjecture.

The reason that Riemann Hypothesis is always a failure, is because the two series of Zetas are never equal to each other term per term, because they lack a Starting Equality to form a proof by Mathematical Induction. The Prime Number Theorem is provable because it has a Starting Equality of 100/sqrt10^4 where we have 100/4 by replacing "e base" with sqrt10, and although the prediction count by using sqrt10 is far more sloppy than other formulas based on "e", there can never be a Starting Equality with "e".

The way to prove PNT or RH is via Math Induction, with a Starting Equality and let the Grid Systems be the Math Induction format of "if N then N+1". and at the Infinity borderline wash away the imperfections for a Ending Equality by Doctoring or by tacking on more terms to Equilabrate each series at infinity. This can be seen in the Proof of the Prime Number Theorem. So long as we have a Starting Equality, we can prove PNT.

Why do the Zetas never equal each other but rather-- asymptotically approach one another? It is because they both have BadFractions such as 1/3 = .333... which the 3 digits go beyond the infinity borderline and never able to be tamed into a Starting Equality. Unlike what are GoodFractions 1/8 = 0.125000...

THE LOGIC of the Mechanics of RH proof: the logic here is that RH should parallel the proof of the PNT. Not that Prime Number Theorem is equivalent to RH, but parallel in proof structure.

I have never compared PNT to RH and how RH should be proved using PNT in parallel concert.

Both PNT and RH have huge problems with Series, but it did not stop PNT from having a proof.

CONVERGENCY THEORY precisely defined:

Convergency theory, and what we have is a less strict form of equality. In math we have equality but also we have convergency to make two concepts, each distinct from one another converge to equality at infinity via Math-Induction on series. Not asymptotic approach of two different series.

Convergency boils down to five items:

i) starting equality terms of two items in comparison
ii) Middle terms close together by a lower and upper bounds factor, a "if N then N+1" which is demonstrated by the Grid System.
iii) the terms near the infinity borderline and are either Doctored of formula or are tacked on terms to Equalibrated both series.
iv) the final terms of the two items in comparison are equal "at infinity"
v) convergency has a very close similarity to how Mathematical Induction works in that a starting equality, a ending equality and we say all the terms in the two items under comparison are "convergent equal" in the Middle section

However, I do see a huge problem in a ** starting equality ** for the Zetas. I managed to find starting equalities in PNT such as the primes from 0 to 100 are 100/4 where the 4 is begot from using (sqrt10)^4 rather than using "e log".

So, I see huge problem in getting a STARTING EQUALITY for the Riemann and Euler Zetas.

So, what I suspect is going to happen is that a proof of PNT is possible because these conditions are able to be satisfied but not satisfied for the zetas of RH.

The RH was invented in a time period of 1800s where infinity was ill-defined and never well-defined and so was the theory of Series. So that the RH was solid Old Math with foggy notion of infinity and a infinity without a border between finite and infinite, so that a Oresme series of 1 + 1/2 + 1/3 + 1/4 + . . in Old Math was considered Divergent, of course because infinity was a screwy notion of "forever without any borders".  In New Math, the borderline is found to be 1*10^604 and so Oresme's Series is a finite number Convergent series as should be all Fractional Term series converge. We see from Grid Systems where we pretend that 100 is infinity border that 100 terms in the Harmonic Series 1+1/2 + 1/3 + 1/4 + . . + 1/100 would converge to approx 5 or thereabouts so that 5% of 100 as infinity border indicates that the larger powers of 10 as infinity border would have a decreasing percent for convergence. So, in Oresme to Euler and Riemann with their ill-defined infinity they would have thought the Harmonic Series diverges when in fact it converges. That effectively puts an end to a proof for RH.

The subject of Logic was never really all that big during the time of both Euler and Riemann and they had errors of logic in their thinking and published work which today's modern day mathematicians have never come face to face with.

So Equality of Series is similar to Mathematical Induction, in that you need a starting case, and then you assume "n" and if "n+1" is true, then the set is equal to the Counting Numbers. For the Zetas, we never have a Starting Equality.

Similar to the Prime Number Theorem of the accounting of the abundance of primes. The formula Li(x) is terribly close to equaling the amount of primes, but it too suffers from never having a starting equality, because the logarithmic function Ln can never give equality. So when I replace Li(x) with sqrt10 as basis, we see that for 100/4 where the 4 is (sqrt(10))^4 gives exactly 25 primes from 1 to 100. So we have a Starting Equality and at the Infinity borderband we can engineer a ending equality and so we have equality of amount of primes with the formula Sqrt10 basis.


Page86, 12-3, many major proofs
AP's Proof of Riemann Hypothesis, part 2 of 2

PART 2 of 2: LOGICAL FLAWS & DISPROOF OF THE RIEMANN HYPOTHESIS //Correcting Math 5th ed.

by Archimedes Plutonium

Subtraction Fallacy of Zetas
___________________________


Alright, now for the subtraction fallacy, as if the "never equality for the zetas" was not punishing enough.

Let me now walk through the exact error of logic, the second fatal flaw of RH and why the zetas are never equal due to the subtraction fallacy, and why the Riemann Hypothesis is phony.

Missing Dollar Fallacy:

I do not know how old is the Missing Dollar Paradox, whether it is just a recent 20th century knowledge, and I suspect so. The important thing about that Paradox is that mathematics easily goes astray when you have the infinite series, and you have subtraction involved. Because it is very difficult to reckon the "basis for any subtraction with infinite terms". But if anyone is in doubt that Euler and others committed this fallacy, simply needs to look at the Euler zeta compared to the Riemann zeta and realize, that although they approach one another asymptotically, they never actually become equal, just as the tangent function never actually meets or intersects or equals at x=pi/2 for division by zero is undefined. The zetas never equal, not because of division by zero, but simply due to the fact that the Euler zeta is always larger.

Three soldiers go to a hotel in World War 2 before deployed to Europe. They sign in and each pays $10 for their room. The owner comes in and sees he has 3 soldiers going to Europe so sends the bellman up with $5 in singles and says give them back this money. The bellman is rather corrupt and also cannot divide 5 by 3 evenly so decides to give each soldier back just $1 and pocket the $2. So, what is our accounting? Well each soldier payed $9 for his room for a total of $27 and the bellman pocketed $2 so where is the missing $1?

If we look at page 102 to 103 of Derbyshire's Prime Obsession book of 2003 we see a subtraction involved in infinite sum series. On page 103, Derbyshire says this: "When subtracting the left-hand sides, treat (1-(1/2^s)Z(s) as a blob, a single number (which of course it is, for any given s). I have one of this blob, and I have 1/3^s of it."


In the below old post of mine earlier this year I discussed the fallacy of what I call "Misplaced Subtraction". In that dollar fallacy, if you subtract 1 from 10 and then have 9x3= 27 and with the 2 you end up with 29, so where is the missing dollar? If you properly do the arithmetic, 25 was paid and so 25+3 = 28 with the bellman pocketing 2. It is this fallacy, that Euler committed and which Riemann and later followers would not correct, but rather use the flawed math.

So, on page 102 we see Derbyshire with the Riemann zeta and for my purpose I will use the terms of just Z = 1 + 1/2 + 1/3 for those three are sufficient to reveal the Logical Error of Euler. When I add 1+1/2 +1/3, I get 1.83333...
And when I multiply by 1/2, I get 1/2Z = 1/2 (1.83333...) which is nothing wrong yet. However, it is the subtraction that the fallacy catches those unaware, unsuspecting. It is the Misplaced Subtraction that makes the Riemann zeta never equal to the Euler zeta.

So, on page 102 where Derbyshire has this, only reduced to my example at hand:

(1 - 1/2) Z = 1 + 1/3 + 1/5

And adding 1+1/3 + 1/5 is not going to be 1/2(1.8333...) but rather is that of 1.5333... or, if I included the next terms of 1/7 then 1/9 etc etc, nowhere am I going to get 1/2(1.8333...). And the reason being, well, is quite simple, in that the Fallacy of Subtraction of Series, just as the Missing Dollar Fallacy (Paradox).

When you do subtraction in mathematics, you have to be very careful of what "base" you subtract from, so that if you subtract 1 from 10 to get 9 and 9x3=27 when you should have subtracted 5 from 30 in the Missing Dollar Paradox below.

Well, in the Hotel Fallacy, the error is to never acknowledge the 25 platform of what the soldiers paid, but rather to think that the only platform is the 30 platform, so that when subtracting or adding with only 30 in mind, the fallacy creeps into the accounting. Same thing with the Euler & Derbyshire accounting that as you treat these "blobs" you treat the Zetas as only a infinity platform with no other platform to consider. And that causes the error.

Euler and all his future followers made that same fallacy with the zetas. It is an easy mistake to make, especially when you have two infinite series. And especially when you have division coupled with subtraction as in the Euler zeta. Now below is an old post of April 2011 where I talk about this Euler mistake and point to the page 102 starting in the Derbyshire book. When you subtract that second series from the first series, you no longer are guaranteed equality. For example of the Missing Dollar:

3 x $9 was paid and the bellman pocketed $2 = 27 + 2 does not equal 30

$10 + $10 + $10 = $30

So the error of Euler is that he needed to first establish an infinity border, and say he finds it to be 10^604. Then, his Riemann zeta of additive terms is smaller than 10^604 and is a specific finite rational number. Now when Euler multiplies by 1/2^s the equations are still equal, but, when Euler (via Derbyshire) subtracts the two equations, there are two different and separated ACCOUNTING BASIS and so they are no longer equal in the subtraction.

For example in the Missing Dollar Fallacy:

3x9 = 27 add on 2 is 29 and not 30

First accounting basis 25 in the till + 1 + 1 + 1 for each soldier + 2 for bellman = 30

Second accounting basis 25 in till + 2 for bellmann = 3 x 9 for soldiers

In the first accounting our infinity border is 30 and in the second accounting our infinity border is 27

Yet Euler in his proof had 30 = 27 or had 30 = 29.

Very crude dot picture of 5f6, 94TH
ELECTRON DOT CLOUD of 231Pu


::\ ::|:: /::
::\::|::/::
_ _
(:Y:)
- -
::/::|::\::
::/ ::|:: \::
One of those dots is the Milky Way galaxy. And each dot represents another galaxy.
            . \ .  . | .   /.
           . . \. . .|. . /. .
              ..\....|.../...
               ::\:::|::/::
---------------      -------------
--------------- (Y) -------------
---------------      --------------
               ::/:::|::\::
              ../....|...\...
           . . /. . .|. . \. .
            . / .  . | .   \ .


http://www.iw.net/~a_plutonium/ 
whole entire Universe is just one big atom 
where dots of the electron-dot-cloud are galaxies

I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of spammers, off-topic-misfits, front-page-hogs, stalking mockers, suppression-bullies, and demonizers.     

https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe        
Archimedes Plutonium
Archimedes Plutonium
2017-06-15 21:08:09 UTC
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Post by Archimedes Plutonium
Jan Burse
77.58.43.158
ETH Zurich
2> Paul Biran, Marc Burger, Patrick Cheridito, Manfred Einsiedler, Paul Embrechts, Giovanni Felder, Alessio Figalli, Norbert Hungerbuhler, Tom Ilmanen, Horst Knorrer, Emmanuel Kowalski, Urs Lang, Rahul Pandharipande, Richard Pink, Tristan Riviere, Dietmar Salamon, Martin Schweizer, Mete Soner,
2> Michael Struwe, Benjamin Sudakov, Alain Sznitman, Josef Teichmann, Wendelin Werner, Thomas Willwacher
Array of Math using the Conservation Principle for proofs

Theorem-statement:: the sqrt2 is irrational. By irrational is meant sqrt2 is never the ratio of two integers A and B, as A/B, where A and B are reduced in lowest form-- they share no common factor.

Proving Statement:: the existence of A/B = sqrt2 implies A^2/B^2 = 2. This means A^2 = 2B^2 means both are even and share a common factor, hence no A/B.

Comment:: AP would show that the fundamental meaning or irrational number is that it is two numbers involved and playing as one. So the number sqrt2 is the two numbers of 1.414 along with 1.415 in 1000 Grid.

Comment:: a useful contrast of proving No odd perfect, except 1, exists, alongside the proof of sqrt2 is irrational, is instructive since both use the same method of proof.

Theorem Statement:: Only 1 can be Odd Perfect and no odd number larger than 1 can be perfect. By perfect is meant all factors of the number in question except the number itself is added up and if equal to the number is thus perfect. A special note is that 1 is the only singleton factor while all other factors are cofactors. For example, 9 is 1+3+3, not 1+3. And we group the factors of given odd number k. We group the factors into two groups of a m/k and p/k where k is the odd number. So that if m+p = k then k is odd perfect. A deficient odd number is one in which m+p falls short of being k. The first odd deficient is 3 with 1/3 + 0/3, the later comes 9 with 3/9 + 4/9, later comes 15 with 5/15 + 4/15.

Proof Statement:: We ask what is preventing odd deficient numbers from becoming perfect, from reaching perfect condition. We look to see if there is a pattern that prevents a odd number from attaining perfect, so we set about constructing an odd perfect number. We take any odd deficient and sum its factors and place the sum fully in the m grouping, leaving the p grouping 0, so that m falls short of k and m is odd because of singleton 1. That means for every odd number which is short of being perfect must have a p value of a even number and hence k an odd number has a even number factor. So the barrier in constructing a odd perfect number is the fact that a even number is a factor in dividing a odd number-- impossible construction.

Comment:: notice how the proof that sqrt2 is irrational has a back and forth tussle with even and odd and the same goes for odd perfect proof.

....,

.....,

Statement theorem::
Euclid Proof of Infinitude of Primes: Statement, the set of all primes { 2, 3, 5, 7, 11, . . . } is an infinite set.


Statement Proof Euclid Infinitude of Primes: given any finite set of primes, multiply the lot and add 1, and you easily find a new prime not on the original list.

Comment:: By year 2000, we recognize that in mathematics, especially with the infinitude of primes proof, that the Ancient Greeks had no definition of what it means to be finite or means to be infinite. A infinity borderline between finite and infinite only began to dawn on mathematicians by the time of AP. Once you well define infinity with a borderline, the proof of Infinitude of Primes takes on an entirely new criterion-- that of density, which to any logical person, being finite and being infinite is a question of density-- is it rare or is it common and frequent?



Statement Fact: For a set of Naturals to be infinite means there are at least 1*10^604 such numbers to be found from 1 to 1*10^1208.

Statement Fact: In New Math, the algebraic closure of 1*10^604 is its square 1*10^1208.

Comment:: Anticipating the proof of the infinitude of Twin Primes, in that one can find 1^10^604 Twin Primes between 1 and 1*10^1208, and in fact, you only need go to about 10^610 to find 10^604 twin primes. For regular primes we need go to about 10^607 and for twin primes, we need go to 10^610.


Comment:: Infinity, as our commonsense always dictated before math professors muddied the works, commonsense says that Infinity is a density measurement. What is the density of Fibonacci primes, can we get 10^604 of them between 1 and 10^1208 and be fully finished with the question, because density tells us they are finite.

Facts:: The Primes in the interval 1 to 100 are these: 
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 
67, 71, 73, 79, 83, 89, 97

1 to 10^2         25 primes and 10 perfect-squares
1 to 10^3         168 primes and 31 perfect-squares  
1 to 10^4         1,229 primes and 100 perfect-squares 
1 to 10^5         9,592  primes and 316 perfect-squares  
1 to 10^6         78,498 primes and 1000 perfect-squares 
1 to 10^7         664,579 primes and 3162 perfect-squares    

So, what we see is a constancy of perfect squares alternating between 100-- and 316-- patterns that are far smaller in number than the primes of that interval. From this we conclude that Primes are an infinite set and is a far far easier proof method than any proof of the infinitude of primes ever given before.


Comment:: Now we prove Twin Primes is an infinite set, using the Perfect Square Test of comparison, since Perfect Squares is the smallest infinite set in terms of density.

Facts::

There are 15 twin primes in that interval 1 to 100 and only 10 perfect-squares: 
3, 5, 7, 11, 13, 17, 19, 29, 31, 41, 43, 59, 61, 71, 73

1 to 100 has 15 Twin Primes and 10 Perfect Squares
1 to 10^3 has 35 Twin Primes and 31 Perfect Squares
1 to 10^4 has 205 Twin Primes and 100 Perfect Squares
1 to 10^5 has 1,224 Twin Primes and 316 Perfect Squares
1 to 10^6 has 8,169 Twin Primes and 1000 Perfect Squares
1 to 10^7 has 58,980 Twin Primes and 3162 Perfect Squares

Statement:: So here again we see that the Twin Primes far outnumber the Perfect Squares in density and thus the Twin Primes are an infinite set. We contrast how the Fibonacci Numbers and Fibonacci Primes would be finite sets yet the primes and twin-primes are infinite sets because we measured them against the Perfect-Squares set.
....,

....,

S_k

S_k+1

....,

....,

Comments: The Fundamental theorems were entered in earlier in the Array in historical timeline. But here we repeat them as bunched together for it may spur new ideas.

Fundamental Theorem of Arithmetic
______________________________

Theorem Statement of FTArith.:: Also called the Unique Prime Factorization Theorem, for it is about Natural numbers, the set 1, 2, 3, 4, etc etc and that they can be expressed each, beyond the number 1, as a unique product of prime numbers. For example 7 = 7, and 10 = 2*5 and 24 = 2*2*2*3. We leave out 1 for it is unit, and leave it out for then we have no uniqueness since 1 = 1*1*..*1 any number of 1s.

Proof-Statement of FTArith::  If N is prime end of job. If N is composite divide until only primes remain. Are these remaining primes unique? Primes p1, p2, ..p_k if not unique and another collection of primes q1,q2, . .q_j different from p's also equals N would violate the prior proved theorem that if prime r divides uv then r divides u or v.



....,

Comment:: Here we have the fine example of where the proof statement is smaller in length than the theorem-statement. But it only goes to show, that if you have a valid proof, you cut away the fat, and that is ironic since FTArithmetic is UPFAT, unique prime factorization theorem.

....,

Very crude dot picture of 5f6, 94TH
ELECTRON DOT CLOUD of 231Pu


::\ ::|:: /::
::\::|::/::
_ _
(:Y:)
- -
::/::|::\::
::/ ::|:: \::
One of those dots is the Milky Way galaxy. And each dot represents another galaxy.
            . \ .  . | .   /.
           . . \. . .|. . /. .
              ..\....|.../...
               ::\:::|::/::
---------------      -------------
--------------- (Y) -------------
---------------      --------------
               ::/:::|::\::
              ../....|...\...
           . . /. . .|. . \. .
            . / .  . | .   \ .


http://www.iw.net/~a_plutonium/ 
whole entire Universe is just one big atom 
where dots of the electron-dot-cloud are galaxies

I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of spammers, off-topic-misfits, front-page-hogs, stalking mockers, suppression-bullies, and demonizers.     

https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe        
Archimedes Plutonium
Archimedes Plutonium
2017-06-16 20:54:34 UTC
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Post by Archimedes Plutonium
77.58.43.158 jan burse
3> ETH Zurich
4> Paul Biran, Marc Burger, Patrick Cheridito, Manfred Einsiedler, Paul Embrechts, Giovanni Felder, Alessio Figalli, Norbert Hungerbuhler, Tom Ilmanen, Horst Knorrer, Emmanuel Kowalski, Urs Lang, Rahul Pandharipande, Richard Pink, Tristan Riviere, Dietmar Salamon, Martin Schweizer, Mete Soner, Michael Struwe, Benjamin Sudakov, Alain Sznitman, Josef Teichmann, Wendelin Werner, Thomas Willwacher
5>
On Friday, June 16, 2017 at 3:46:04 PM UTC-5, Archimedes Plutonium wrote:
b***@gmail.com
2017-06-16 21:05:10 UTC
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Most stupid crank AP fell off the chair while writing the below.
His tesla headphones evaporated the last of his brain cells.
Post by Archimedes Plutonium
Post by Archimedes Plutonium
77.58.43.158 jan burse
3> ETH Zurich
4> Paul Biran, Marc Burger, Patrick Cheridito, Manfred Einsiedler, Paul Embrechts, Giovanni Felder, Alessio Figalli, Norbert Hungerbuhler, Tom Ilmanen, Horst Knorrer, Emmanuel Kowalski, Urs Lang, Rahul Pandharipande, Richard Pink, Tristan Riviere, Dietmar Salamon, Martin Schweizer, Mete Soner, Michael Struwe, Benjamin Sudakov, Alain Sznitman, Josef Teichmann, Wendelin Werner, Thomas Willwacher
5>
Archimedes Plutonium
2017-06-22 06:20:04 UTC
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On Friday, June 16, 2017 at 4:05:22 PM UTC-5, ***@gmail.com wrote:

Burse loves AP's Riemann Hypothesis proof and wants to inscribe it into the library::




Page85, 12-2, many major proofs
AP's Proof of Riemann Hypothesis, part 1 of 2

PART 1 of 2: LOGICAL FLAWS & DISPROOF OF THE RIEMANN HYPOTHESIS //Correcting Math 5th ed

MAJOR LOGICAL FLAWS & DISPROOF OF THE RIEMANN HYPOTHESIS RH

by Archimedes Plutonium

Introduction:

the connection with PNT and RH Re: Page86, Part 2 of 2: LOGICAL FLAWS & DISPROOF OF THE RIEMANN HYPOTHESIS //Correcting Math 5th ed

Now these first two proofs-- Prime Number Theorem PNT and Riemann Hypothesis RH, follow a pattern, a obvious bold and strong pattern.

The pattern is that Old Math has a crazy quilt definition of Series equality. Since Old Math has no borderline with infinity, they have no good definition of what it means for two series to be equal. And so sloppy are they, that they imagine the Series 1 + 1/2 + 1/3 + 1/4 +, . . . the harmonic series is equal to the Series 1+1+1+.... at infinity, Old Math believes those two Series are equal. That is how crazy Old Math was.

And then, of course, when something like PNT comes or RH comes, that having a crazy notion of what is equality for Series, it is little wonder that the nutters of Old Math have troubles with PNT and RH.

AP

Comparing RH to the Prime Number Theorem PNT, in that proof Mathematical Induction on Series, I had a Starting Equality of 100/sqrt10^4 and using sqrt10 base rather than "e" base I have a Starting Equality of 100/4 = 25, and at infinity I Doctor the Ending Equality.

Now, we compare Riemann Hypothesis with its zetas and we also compare the Harmonic Series. The Zetas and the Harmonic Series are never able to give us a Starting Equality to apply a Math Induction proof of RH. What that means is RH is unprovable. It is a false conjecture.

The reason that Riemann Hypothesis is always a failure, is because the two series of Zetas are never equal to each other term per term, because they lack a Starting Equality to form a proof by Mathematical Induction. The Prime Number Theorem is provable because it has a Starting Equality of 100/sqrt10^4 where we have 100/4 by replacing "e base" with sqrt10, and although the prediction count by using sqrt10 is far more sloppy than other formulas based on "e", there can never be a Starting Equality with "e".

The way to prove PNT or RH is via Math Induction, with a Starting Equality and let the Grid Systems be the Math Induction format of "if N then N+1". and at the Infinity borderline wash away the imperfections for a Ending Equality by Doctoring or by tacking on more terms to Equilabrate each series at infinity. This can be seen in the Proof of the Prime Number Theorem. So long as we have a Starting Equality, we can prove PNT.

Why do the Zetas never equal each other but rather-- asymptotically approach one another? It is because they both have BadFractions such as 1/3 = .333... which the 3 digits go beyond the infinity borderline and never able to be tamed into a Starting Equality. Unlike what are GoodFractions 1/8 = 0.125000...

THE LOGIC of the Mechanics of RH proof: the logic here is that RH should parallel the proof of the PNT. Not that Prime Number Theorem is equivalent to RH, but parallel in proof structure.

I have never compared PNT to RH and how RH should be proved using PNT in parallel concert.

Both PNT and RH have huge problems with Series, but it did not stop PNT from having a proof.

CONVERGENCY THEORY precisely defined:

Convergency theory, and what we have is a less strict form of equality. In math we have equality but also we have convergency to make two concepts, each distinct from one another converge to equality at infinity via Math-Induction on series. Not asymptotic approach of two different series.

Convergency boils down to five items:

i) starting equality terms of two items in comparison
ii) Middle terms close together by a lower and upper bounds factor, a "if N then N+1" which is demonstrated by the Grid System.
iii) the terms near the infinity borderline and are either Doctored of formula or are tacked on terms to Equalibrated both series.
iv) the final terms of the two items in comparison are equal "at infinity"
v) convergency has a very close similarity to how Mathematical Induction works in that a starting equality, a ending equality and we say all the terms in the two items under comparison are "convergent equal" in the Middle section

However, I do see a huge problem in a ** starting equality ** for the Zetas. I managed to find starting equalities in PNT such as the primes from 0 to 100 are 100/4 where the 4 is begot from using (sqrt10)^4 rather than using "e log".

So, I see huge problem in getting a STARTING EQUALITY for the Riemann and Euler Zetas.

So, what I suspect is going to happen is that a proof of PNT is possible because these conditions are able to be satisfied but not satisfied for the zetas of RH.

The RH was invented in a time period of 1800s where infinity was ill-defined and never well-defined and so was the theory of Series. So that the RH was solid Old Math with foggy notion of infinity and a infinity without a border between finite and infinite, so that a Oresme series of 1 + 1/2 + 1/3 + 1/4 + . . in Old Math was considered Divergent, of course because infinity was a screwy notion of "forever without any borders".  In New Math, the borderline is found to be 1*10^604 and so Oresme's Series is a finite number Convergent series as should be all Fractional Term series converge. We see from Grid Systems where we pretend that 100 is infinity border that 100 terms in the Harmonic Series 1+1/2 + 1/3 + 1/4 + . . + 1/100 would converge to approx 5 or thereabouts so that 5% of 100 as infinity border indicates that the larger powers of 10 as infinity border would have a decreasing percent for convergence. So, in Oresme to Euler and Riemann with their ill-defined infinity they would have thought the Harmonic Series diverges when in fact it converges. That effectively puts an end to a proof for RH.

The subject of Logic was never really all that big during the time of both Euler and Riemann and they had errors of logic in their thinking and published work which today's modern day mathematicians have never come face to face with.

So Equality of Series is similar to Mathematical Induction, in that you need a starting case, and then you assume "n" and if "n+1" is true, then the set is equal to the Counting Numbers. For the Zetas, we never have a Starting Equality.

Similar to the Prime Number Theorem of the accounting of the abundance of primes. The formula Li(x) is terribly close to equaling the amount of primes, but it too suffers from never having a starting equality, because the logarithmic function Ln can never give equality. So when I replace Li(x) with sqrt10 as basis, we see that for 100/4 where the 4 is (sqrt(10))^4 gives exactly 25 primes from 1 to 100. So we have a Starting Equality and at the Infinity borderband we can engineer a ending equality and so we have equality of amount of primes with the formula Sqrt10 basis.

Very crude dot picture of 5f6, 94TH
ELECTRON DOT CLOUD of 231Pu


::\ ::|:: /::
::\::|::/::
_ _
(:Y:)
- -
::/::|::\::
::/ ::|:: \::
One of those dots is the Milky Way galaxy. And each dot represents another galaxy.
            . \ .  . | .   /.
           . . \. . .|. . /. .
              ..\....|.../...
               ::\:::|::/::
---------------      -------------
--------------- (Y) -------------
---------------      --------------
               ::/:::|::\::
              ../....|...\...
           . . /. . .|. . \. .
            . / .  . | .   \ .


http://www.iw.net/~a_plutonium/ 
whole entire Universe is just one big atom 
where dots of the electron-dot-cloud are galaxies

I re-opened the old newsgroup PAU of 1990s and there one can read my recent posts without the hassle of spammers, off-topic-misfits, front-page-hogs, stalking mockers, suppression-bullies, and demonizers.     

https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe        
Archimedes Plutonium
bassam king karzeddin
2017-06-22 08:46:19 UTC
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Post by Archimedes Plutonium
77.58.43.158 jan burse
ETH Zurich
Paul Biran, Marc Burger, Patrick Cheridito, Manfred Einsiedler, Paul Embrechts, Giovanni Felder, Alessio Figalli, Norbert Hungerbuhler, Tom Ilmanen, Horst Knorrer, Emmanuel Kowalski, Urs Lang, Rahul Pandharipande, Richard Pink, Tristan Riviere, Dietmar Salamon, Martin Schweizer, Mete Soner, Michael Struwe, Benjamin Sudakov, Alain Sznitman, Josef Teichmann, Wendelin Werner, Thomas Willwacher
1st VALID PROOF OF FUNDAMENTAL THEOREM OF ALGEBRA //Correcting Math pre-6th ed
by Archimedes Plutonium
...,
Old Math statement of FTA:: The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number A is a complex number with an imaginary part equal to zero.
....,
Comment:: New Math statement theorem of FTA need not be so long, because in New Math we recognize that sqrt-1 is simply A/0 where A is a real-number (New Real Number).
Theorem-Statement of FTA:: given any polynomial with New Real Numbers, that there always exists at least one New Real Number, call it A, as a solution.
Proof of FTA: x^2 +1 = 0, goes to 1/x^2 = -1, with solution x=0 as 1/0 = sqrt-1, thus any polynomial is 1/polynomial = +-k has at least, one solution for A.
Comment:: Notice the beauty of the Statement of Theorem then Proof are almost identical in length of words. This is what happens when you have a true proof of a statement in mathematics.
Comment:: there is a huge fallout of this proof, in the fact that "i" an imaginary number is no longer needed because the Reals in 0 and any Real A where we have A/0 is a imaginary number and is a complex number. So, all of a sudden the Complex Plane and imaginary number dissolves out of math history into a gutter of shame. Whenever see "i" what that really is, is A/0 where A is any Real number.
....,
AP
Actually, the equation (x^2 + 1 = 0), was the most famous unsolved puzzle for all the old ancient genius mathematicians, starting from Summaries, Baby lions, Eygeptions, Arabs, Muslims, Chinese, Greeks, Indeans, ...etc

Until a drunk genius had aroused a few centuries back and solved it in just a few seconds, wonder!

"Let us decide that there is a solution but of course unreal, and call it imaginary (i) unity, such that when multiplied by itself must give you back your original number unity but also in mirror, so wonderful

And since then, the Doors were opened so widely for all those non-mathematicians to be considered as genius people also, too shameful indeed and only forever
And the vast majority of so ignorant people nowadays are also considered scientist, thinkers, ..., etc, wonder!

And you (reader) would not like that for sure, but for so silly obvious known reasons for sure, where this doesn't count at all

And please do not think always foolishly that this is a kind of insult for Top Professional Mathematicians since they had already painted themselves with its endless shame for quite many centuries now, for sure

And the poor minds mathematicians do not understand yet why people do not generally digest what they usually eat

BKK

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