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| Started by | Archimedes Plutonium <plutonium.archimedes@gmail.com> |
|---|---|
| First post | 2018-09-13 18:49 -0700 |
| Last post | 2018-09-16 02:08 +0000 |
| Articles | 4 — 2 participants |
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Donald Trump has more moral fiber than does Andrew Beal, and AMS with Charles Fefferman, Ron Graham, Daniel Mauldin Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-13 18:49 -0700
publish the AP ellipse proof in AMS Re: Donald Trump has more moral fiber than does Andrew Beal, and AMS with Charles Fefferman, Ron Graham, Daniel Mauldin Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-14 14:42 -0700
Re: publish the AP ellipse proof in AMS Re: Donald Trump has more moral fiber than does Andrew Beal, and AMS with Charles Fefferman, Ron Graham, Daniel Mauldin Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-15 03:58 -0700
Re: publish the AP ellipse proof in AMS Re: Donald Trump has more moral fiber than does Andrew Beal, and AMS with Charles Fefferman, Ron Graham, Daniel Mauldin moroney@world.std.spaamtrap.com (Michael Moroney) - 2018-09-16 02:08 +0000
| From | Archimedes Plutonium <plutonium.archimedes@gmail.com> |
|---|---|
| Date | 2018-09-13 18:49 -0700 |
| Subject | Donald Trump has more moral fiber than does Andrew Beal, and AMS with Charles Fefferman, Ron Graham, Daniel Mauldin |
| Message-ID | <dbedf25c-c290-450b-a5b9-55c312799208@googlegroups.com> |
Donald Trump at least has an excuse to liar-- to shore up his base supporters, but there is no way in this world that Andrew Beal, AMS, Charles Fefferman, Ron Graham, Daniel Mauldin, Ken Ribet have any moral ground to stand on, as they continue to believe and preach a ellipse is a conic, when it was proven by AP some 5 years ago that the ellipse is never a conic, but a cylinder section. So why are these so called math professors, banker still preaching false junk, and liaring to students? Why? Do they want to compete with Trump in liaring?
See AP proof of ellipse --
Proofs ellipse is never a conic, always a cylinder section by
Archimedes Plutonium
--------------------
Conics = oval, 2 proofs, synthetic, analytic
Synthetic Geometry & Analytical Geometry Proofs that Conic section =
Oval, never an ellipse-- World's first proofs thereof
by Archimedes Plutonium
_Synthetic Geometry proofs that Cylinder section= Ellipse// Conic section= Oval
First Synthetic Geometry proofs, later the Analytic Geometry proofs.
Alright I need to get this prepared for the MATH ARRAY of proofs, that
the Ellipse is a Cylinder section, and that the Conic section is an
oval, never an ellipse
PROOF that Cylinder Section is an Ellipse, never a Oval::
I would have proven it by Symmetry. Where I indulge the reader to
place a circle inside the cylinder and have it mounted on a swivel, a
tiny rod fastened to the circle so that you can pivot and rotate the
circle. Then my proof argument would be to say--when the circle plate
is parallel with base, it is a circle but rotate it slightly in the
cylinder and determine what figure is produced. When rotated at the
diameter, the extra area added to the upper portion equals the extra
area added to bottom portion in cylinder, symmetrical area added,
hence a ellipse. QED
Now for proof that the Conic section cannot be an ellipse but an oval,
I again would apply the same proof argument by symmetry.
Proof:: Take a cone in general, and build a circle that rotates on a
axis. Rotate the circle just a tiny bit for it is bound to get stuck
or impeded by the upward slanted walls of the cone. Rotate as far as
you possibly can. Now filling in the area upwards is far smaller than
filling in the area downwards. Hence, only 1 axis of symmetry, not 2
axes of symmetry. Define Oval as having 1 axis of symmetry. Thus a
oval, never an ellipse. QED
The above two proofs are Synthetic Geometry proofs, which means they
need no numbers, just some concepts and axioms to make the proof work.
A Synthetic geometry proof is where you need no numbers, no coordinate
points, no arithmetic, but just using concepts and axioms. A Analytic
Geometry proof is where numbers are involved, if only just coordinate
points.
Array:: Analytic Geometry proof that Cylinder section= Ellipse//Conic
section = Oval, never ellipse
Now I did 3 Experiments and 3 models of the problem, but it turns out
that one model is superior over all the other models. One model is the
best of all.
That model is where you construct a cone and a cylinder and then
implant a circle inside the cone and cylinder attached to a handle so
that you can rotate the circle inside. Mine uses a long nail that I
poked holes into the side of a cylinder and another one inside a cone
made from heavy wax paper of magazine covers. And I used a Mason or
Kerr used lid and I attached them to the nail by drilling two holes
into each lid and running a wire as fastener. All of this done so I
can rotate or pivot the circle inside the cylinder and cone. You need
a long nail, for if you make the models too small or too skinny, you
lose clarity.
ARRAY, Analytic Geometry Proof, Cylinder Section is a Ellipse::
E
__
.-' `-.
.' `.
/ \
; ;
| G c | H
; ;
\ /
`. .'
`-. _____ .-'
F
The above is a view of a ellipse with center c and is produced by the
Sectioning of a Cylinder as long as the cut is not perpendicular to
the base, and as long as the cut involves two points not larger than
the height of the cylinder walls. What we want to prove is that the
cut is always a ellipse, which is a plane figure of two axes of
symmetry with a Major Axis and Minor Axis and center at c.
Side view of Cylinder EGFH above with entry point cut at E and exit
point cut at F and where c denotes the central axis of the cylinder
and where x denotes a circle at c parallel with the base-circle of
cylinder
| |
| | E
| |
| |
|x c |x
| |
| |
| |
|F |
| |
| |
| |
So, what is the proof that figure EGFH is always an ellipse in the
cylinder section? The line segment GH is the diameter of the circle
base of cylinder and the cylinder axis cuts this diameter in half such
that Gc = cH. Now we only need to show that Fc = cE. This is done from
the right triangles cxF and cxE, for we note that by Angle-Side-Angle
these two right triangles are congruent and hence Fc = cE, our second
axis of symmetry and thus figure EGFH is always an ellipse. QED
Array proof:: Analytic Geometry proof that Conic section= Oval// never ellipse
ARRAY, Analytic Geometry Proof, Conic Section is a Oval, never an ellipse::
A
,'" "`.
/ \
C | c | D
\ /
` . ___ .'
B
The above is a view of a figure formed from the cut of a conic with
center c as the axis of the cone and is produced by the Sectioning of
a Cone as long as the cut is not perpendicular to the base, and as
long as the cut is not a hyperbola, parabola or circle (nor line).
What we want to prove is that this cut is always a oval, never an
ellipse. An oval is defined as a plane figure of just one axis of
symmetry and possessing a center, c, with a Major Diameter as the axis
of symmetry and a Minor Diameter. In our diagram above, the major
diameter is AB and minor diameter is CD.
Alright, almost the same as with Cylinder section where we proved the
center was half way between Major Axis and Minor Axis of cylinder,
only in the case of the Conic, we find that the center is half way
between CD the Minor Diameter, but the center is not halfway in
between the Major Diameter, and all of that because of the reason the
slanted walls of the cone cause the distance cA to be far smaller than
the distance cB. In the diagram below we have the circle of x centered
at c and parallel to base. The angle at cx is not 90 degrees as in
cylinder. The angle of cAx is not the same as the angle cBx, as in the
case of the cylinder, because the walls of the cone-for line segments-
are slanted versus parallel in the cylinder. Triangles cAx and cBx are
not congruent, and thus, the distance of cA is not equal to cB,
leaving only one axis of symmetry AB, not CD.
/ \A
x/ c \x
B/ \
Hence, every cut in the Cone, not a hyperbola, not a parabola, not a
circle (not a line) is a Oval, never an ellipse.
QED
--Archimedes Plutonium
Very crude dot picture of 5f6, 94TH
ELECTRON=muon 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.
Read my recent posts in peace and quiet.
https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe
Archimedes Plutonium
On Wednesday, August 1, 2018 at 4:20:27 PM UTC-5, burs...@gmail.com wrote:
> WIth 1 million $ you can buy a lot of wheelchairs:
>
> |_ |_ |_ |_ |_ |_ |_
> |_| |_| |_| |_| |_| |_| |_|
> * * * * * * * * * * * * * *
>
> Am Samstag, 28. Juli 2018 15:18:32 UTC+2 schrieb Archimedes Plutonium:
> > Unless of course, well, if Andrew Beal with proxy American Mathematical Society of Charles Fefferman, Ron Graham, Daniel Mauldin, have stipulated that the winner of Beal's donation has to be Jewish. I did not read all the terms of the prize payout. And there is no way anyone in the world can steal AP's proof, for you have to mention at least once-- the word "rectangle". Because a proof of Generalized FLT is all about a rectangle replacing a number.
> >
> > Besides, AP is throwing in another proof for free-- That the Ellipse is never a conic section, always a cylinder section. And it gives the AMS really -- some math to do at their place, other than what they do at this moment in time-- nothing mathematical, but only corruption of mathematics.
> >
> > The Generalized FLT was proven by AP as long ago as 2014
> >
> > AP proves Generalized FLT in 2014, and today is 28JUL2018, over 4 years for anyone at AMS to digest this proof.
> >
> >
> > Newsgroups: sci.math
> > Date: Wed, 8 Jan 2014 02:32:14 -0800 (PST)
> > Complaints-To: groups...@google.com
> > Injection-Info: glegroupsg2000goo.googlegroups.com; posting-host=166.137.110.8;
> > posting-account=fsC03QkAAAAwkSNcSEKmlcR-W_HNitEd
> > NNTP-Posting-Host: 166.137.110.8
> > User-Agent: G2/1.0
> > MIME-Version: 1.0
> > Message-ID: <0930d831-785f-4440-bf56-132cc17c5707@googlegroups.com>
> > Subject: taking on both FLT and Beal in one proof; using Condensed Rectangles
> > #1388 Correcting Math
> > From: Archimedes Plutonium <plutonium....@gmail.com>
> > Injection-Date: Wed, 08 Jan 2014 10:32:14 +0000
> >
> >
> > taking on both FLT and Beal in one proof; using Condensed Rectangles #1388 Correcting Math
> >
> > Alright, I was curious about something which lead to the Beal conjecture which lead to the idea that if I can prove FLT, the same method should prove Beal, sounds reasonable to a logical person.
> >
> > So I looked up the smallest triple that solves Beal and it is 2^3 + 2^3 = 2^4 and the next smallest is 3^3 + 6^3 = 3^5. So, unlike FLT where I am looking for 1+2B=C in conjunction with A+B=C, in Beal we are looking for 2B=C in conjunction with A+B=C.
> >
> > So, I thought and thought about it, and realized I was looking at the wrong geometric figure to base the FLT proof upon. I cannot do the Beal on angles and triangles. What I can do the Beal on is what I call, _condensed rectangles_.
> >
> > A condensed rectangle is that of given a number, how to compact the number as unit squares into the most smallest sides of a rectangle. For instance, given 27,216,243 the most condensed rectangle for each of those three numbers is 3x9, then 9x24, then 9 by 27. In other words, given a number you take the unit squares of that number and try to make those unit squares into a square as much as possible even though most times it will be a rectangle. So given the number 36, we could make a 2 by 18 rectangle or a 3 by 12 rectangle or a 4 by 9 rectangle or a 6 by 6 rectangle and the most condense of those is a 6x6 or a square. Now most numbers of their condensed rectangle will not be a square. Many numbers will be just a 1 by something rectangle such as 9 is 1x9.
> >
> > So, what I am going to do is drop the angle and triangle and apply _condensed rectangle_ to both FLT and Beal.
> >
> > Now I will use these tables for FLT:
> >
> > exp3 {1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, . .}
> >
> > exp4 {1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, . .}
> >
> > exp5 {1, 32, 243, 1024, 3125, 7776, 16807, 32,768, 59,049, 100,000, 161,051, 248,832, 371,293, . .}
> >
> > And for Beal, I will use a conglomerated table composed of all those exponents into one set starting with exp3
> >
> > {1, 8, 16, 27, 32, 64, 81, 125, 243, 256, . .}
> > Notice in Beal we immediately get the primitive triple of 2B=C, whereas in FLT the primitive triple of 1+2B=C occurs in exponent2 and nowhere else. In both Beal and FLT we want to know of A+B=C.
> >
> > In Beal we have solutions of A+B=C but want to know if any solutions without a prime factor for A,B,C. It is this requirement that made me realize I was chasing after the wrong geometry figure of a angle and triangle and must chase after just _condensed rectangles_.
> >
> > In FLT, no A,B,C exists for A+B=C because no 1+2B=C exists in exp3 or higher. In Beal, 2B=C exists and then A+B=C exists but with that requirement of a prime factor.
> > --
> >
> > Recently I re-opened the old newsgroup of 1990s and there one can read my recent posts without the hassle of mockers and hatemongers.
> >
> > https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe
> >
> > Archimedes Plutonium
> >
> > Here is the latest update on Generalized FLT, for I sort of held back for some years now, because it troubled me to go from proof of Generalized to proof of FLT itself, but now, I accept those final few lines of proof-- see Karl Runge website.
> >
> >
> > Newsgroups: sci.math
> > Date: Tue, 24 Jul 2018 21:47:12 -0700 (PDT)
> >
> > Subject: Review of my Generalized FLT proof and its corollary FLT
> > From: Archimedes Plutonium <plutonium....@gmail.com>
> > Injection-Date: Wed, 25 Jul 2018 04:47:13 +0000
> >
> > Review of my Generalized FLT proof and its corollary FLT
> > I need a complete review of this proof-- especially the part where it is said-- once you have generalized FLT, you easily have FLT. I do not see the fallout as some on the Internet claim.
> >
> >
> > Well the sci.math has a complete history of my adventures with Fermat's Last Theorem starting with August 1993, and developing many twists and turns. Several times I had to throw out my prior attempts.
> >
> > What emerged was decades later, a final true proof of FLT. It has one huge starting surprise to it. Hugely Startling Surprise. The surprise is that a class of math proofs cannot be proven until a "more general theorem that overlays the "lesser theorem tucked away inside" the more general theorem is proven FIRST"
> >
> > So that-- there never exists a proof of FLT until you first prove true the Generalized FLT
> >
> > What is Generalized FLT? Glad you asked, for it is:
> >
> > A^x + B^y = C^z where A,B,C,x,y,z are positive integers with x,y,z greater than 2
> >
> > FLT is of course
> >
> > A^n + B^n = C^n where A,B,C, n are positive integers and n is greater than 2
> >
> > Newsgroups: sci.math
> > Date: Thu, 9 Jun 2016 13:38:48 -0700 (PDT)
> >
> > Subject: Proof of Generalized FLT & Fermat's Last Theorem #178 Correcting Math 4th ed
> > From: Archimedes Plutonium <plutonium....@gmail.com>
> > Injection-Date: Thu, 09 Jun 2016 20:38:49 +0000
> >
> >
> > Proof of Generalized FLT & Fermat's Last Theorem #178 Correcting Math 4th ed
> >
> > Proof of Generalized FLT & Fermat's Last Theorem #178 Correcting Math 4th ed
> >
> > PROOFS of Generalized FLT & FERMAT's LAST THEOREM
> >
> >
> > PROOFS of Generalized FLT & FERMAT's LAST THEOREM as corollary
> >
> > Editing and commentary corner: Once Generalized FLT and FLT are turned into a geometrical perspective-- condensed rectangles, then the proof is as easy as making homemade cherry pie. And the importance of proving Generalized FLT first is that FLT never has a proof until Generalized FLT is proven first.
> >
> > Summary of text: Both proofs of Generalized FLT and FLT are based on a fact of geometry, that you can represent a number with its cofactors as the sides of a rectangle. And to prove either Generalized FLT or FLT is a simple matter of stacking two rectangles that have equal sides, A and B to produce a third rectangle C which has a side equalling the _shared side_ of A and B.
> >
> > TEXT:
> >
> > I decided I needed to stiffen up the proof of Generalized FLT below. To a logical person, they often streamline their proofs and do not bother with adding some information which people not used to logic have a difficult time in seeing. For me, it is redundant to include the distributive law into the proof of Generalized FLT, but to those who do not do math but only on occasion, they would not see the logic unless I explain those very details. So below is a more stiffened up proof of Generalized FLT.
> >
> > Proof of the Generalized FLT conjecture that proves Fermat's Last Theorem
> >
> > What is Generalized FLT? Glad you asked, for it is:
> >
> > A^x + B^y = C^z where A,B,C,x,y,z are positive integers with x,y,z greater than 2
> >
> > FLT is of course
> >
> > A^n + B^n = C^n where A,B,C, n are positive integers and n is greater than 2
> >
> > I proved Generalized FLT and FLT (Fermat's Last Theorem) before I discovered that Logical Material Implication Table of T,F,T,T is incorrect and should be that of T,F,undefined, undefined. What this does is remove reductio ad absurdum as a mathematics proof technique as only a probability technique and no longer a deduction. It means the only valid proof method for mathematics, is construction proofs. My Generalized FLT and FLT are construction proofs. Another feature of the true Logic Implication operator, is that corollaries of mathematics cannot be proven as "stand alone conjectures". And that a corollary requires the over-arching- theorem attending the corollary be proven beforehand. This implies that Wiles's FLT is not true for in order to truly prove FLT, a proof of Generalized FLT had to come first.
> >
> > We see the relationship between a Theorem and its Corollary very easily from Generalized FLT and FLT, in that we have all these cases to worry about. The history of FLT was a case study, prove it in the case of exponent this and that, but never a proof of all of FLT, and that is because corollaries are never proven as stand alone theorems and must have their theorem (Generalized FLT) proven first. This is because Logic Implication is not T,F,T,T where false proofs hide behind those last two T, T when they should be undefined, undefined. The reason Implication must be T,F,undefined, undefined is because mathematics has 2/0 and 0/0 where division by 0 is undefined, and that Old Logic with its T,F,T,T does not allow for division by zero.
> >
> > Detailed Proofs Generalized FLT conjecture with its FLT corollary
> >
> >
> > Both proofs of Generalized FLT and FLT are based on a fact of geometry, that you can represent a number with its cofactors as the sides of a rectangle. And to prove either Generalized FLT or FLT is a simple matter of stacking two rectangles that have equal sides, A and B to produce a third rectangle C which has a side equalling the _shared side_ of A and B.
> > ________________________
> > DETAILED PROOF OF Generalized FLT
> > ________________________
> >
> > It is a constructive proof.
> >
> > We make the table of all the numbers possible in the Generalized FLT Conjecture as the conglomeration of exponents of 3 or larger as this set:
> >
> > {1, 8, 16, 27, 32, 64, 81, 125, 128, 216, 243, 256, . .}
> >
> > Here we have conglomerated exp3 and exp4 and exp5 etc etc into one set.
> >
> > We know Generalized FLT has solutions of A+B=C in that set for here are three examples:
> >
> > 2^3 + 2^3 = 2^4 with prime divisor 2
> > 3^3 + 6^3 = 3^5 with prime divisor 3
> > 7^3 + 7^4 = 14^3 with prime divisor 7
> >
> > What we need to prove is that all solutions have a prime divisor in common, ie all three rectangles of A, B, C so that A+B=C, have one shared side equal to one another.
> >
> > Definition of Condensed-Rectangle: given any number in the set of conglomerated exponents, we construct rectangles of that number from its unit squares whose sides are cofactors of the number. For instance, rectangle of 216 units as either 12x18 units, or 9x24 units, or 6x36 units or 3x72 units, or 2x108, but never a 1x216 units. We exclude 1 times the number as a condensed rectangle. So a condensed-rectangle is one in which it is composed of cofactors of the number in question, except for 1, and the number itself for 1x216 units is not a condensed-rectangle.
> >
> > Now for the constructive proof that Generalized FLT solutions must have a common prime divisor.
> >
> > We stack Condensed-Rectangles of the number-space that Generalized FLT conjecture applies:
> >
> > Solution Number Space for Generalized FLT now becomes these condensed-rectangles:
> >
> > { 2*4, 2*8, 4*4, 3*9, 2*16, 4*8, 2*32, 4*16, 8*8,. . . }
> >
> > We convert each of those numbers into Condensed-Rectangle, except 1 of course, and where many numbers have several condensed rectangles so the Solution Space of Numbers increases by a large amount. If an A and B as condensed-rectangles have the same side such as 3x9 units and 9x24units
> > wherein you stack them on their shared side of 9 and which matches another number of its condensed-rectangle such as 9x27 units, then you have a Generalized FLT solution of A+B=C. For if we were to take the 9 by 27 condensed rectangle it decomposes into 3x9 and 9x24.
> >
> > Now, the question is, are all A+B equal to a C, form stacked condensed rectangles that share a common side?
> >
> > All stackable condensed-rectangles must have one side the same for the two rectangles to stack, in the case above it is the side 9 with its common divisor of the prime 3.
> >
> > If any other solution to Generalized FLT had A stacked upon B without a common side between them, then the figure formed cannot be a overall new rectangle but something that looks like this:
> >
> > HHHHHH
> > HHHHHH
> > HHHHHHHHHH
> > HHHHHHHHHH
> >
> > That is not a condensed rectangle and all the numbers of the Solution Space, except the number 1, are condensed rectangles. That above figure is 6-sided figure. That is a 6-sided figure, yet a rectangle is only a 4 sided figure. That is not a Condensed Rectangle, represented by its cofactors. So that equality can not be achieved by any other stacking than condensed rectangles equalling condensed rectangles.
> >
> > Then the question is, can we have an A and B with a shared common side equal to a C that is a condensed rectangle with no common side shared with A and B?
> >
> > The question is, how can I be sure that all the A, B, C such that A + B = C have a common prime divisor?
> >
> > As if the question is asking whether the Condensed Rectangles covered the question by forbidding any equality unless there is a common shared side in A in B and in C.
> >
> > Well, the answer is easily enough covered for the Condensed Rectangles eliminates any possibility of A,B,C where A+B = C and not have a shared side by all three of A, B, C. It does this by the Distributive law of integers.
> >
> > All the numbers in the solution space of Generalized FLT have condensed rectangles, except the number 1.
> >
> > All the A, B, C are written as condensed rectangles of (s*t) + (r*p) = (u*v).
> >
> > For example (9*3) + (9*24) = (9*27) which is 3^3 + 6^3 = 3^5 which is 27 + 216 = 243.
> >
> > In order for (s*t) + (r*p) = (u*v) then the s, r, u have to be equal sides. Have to be (9*t) + (9*p) = (9*v) in the example of 27 + 216 = 243.
> >
> > Distributive Law (9*3)/9 + (9*24)/9 = (9*27)/9 where 3 + 24 = 27
> >
> > So, give me three numbers A, B, C, chosen arbitrarily from the Solution Space of Generalized FLT {1, 8, 16, 27, 32, 64, 81, 125, 128, 216, 243, 256, . .} except of course 1, and can those three chosen arbitrarily A, B, C obey A + B = C? Only if they obey the Distributive Law can the A + B equal to the C when we write the A, B, C as condensed rectangles.
> >
> > QED
> >
> >
> > _________________________________________
> > Detailed Proof of FLT using condensed-rectangles
> > _________________________________________
> >
> > It is a construction proof method for we show that it is impossible to construct A+B = C inside of a specific exponent.
> >
> > Fermat's Last Theorem FLT conjecture says there are no solutions to the equation a^y + b^y = c^y where a,b,c,y are positive integers and y is greater than 2.
> >
> > The number Space that governs FLT is this:
> >
> > exp3 {1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, . .}
> >
> > exp4 {1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, . .}
> >
> > exp5 {1, 32, 243, 1024, 3125, 7776, 16807, 32,768, 59,049, 100,000, 161,051, 248,832, 371,293, . .}
> >
> > exp6 .....
> > .
> > .
> > .
> > .
> >
> > So in FLT we ask whether there are any triples, A,B,C in any one of those _specific exponents_ such that A+B=C. In FLT, our solution space is only one particular exponent such as 3 or 4, or 5 to hunt down and find a A,B,C to satisfy A+B=C.
> >
> > In the proof we use Condensed-Rectangles which is defined as a rectangle composed of unit squares of the cofactors of a number, except for 1 x number itself. So the number 27 in exp3 has Condensed Rectangles of 3x9 only. The number 125 in exp3 has condensed rectangles of 5x25 only, and the number 81 in exp4 has condensed rectangles of 3x27 and 9x9.
> >
> > Now in the proof of Generalized FLT solutions of A + B = C require a common divisor.
> >
> > 2^3 + 2^3 = 2^4 with prime divisor 2
> > 3^3 + 6^3 = 3^5 with prime divisor 3
> > 7^3 + 7^4 = 14^3 with prime divisor 7
> >
> > Here in FLT, we need not even worry about exponents because all solutions to Generalized FLT encompass exponents so that if these were FLT solutions:
> >
> > 2^? + 2^? = 2^? with prime divisor 2
> > 3^? + 6^? = 3^? with prime divisor 3
> > 7^? + 7^? = 14^? with prime divisor 7
> >
> > So in that construction of a solution to FLT there is a common divisor and so now we divide the equation by the common divisor, and we get this:
> >
> > 1^y + 1^y = 1^y
> > 1^z + 2^z = 1^z
> > 1^w + 1^w = 2^w
> >
> > And those are impossible constructions. So the proof of FLT requires Generalized FLT be proven first in order to display that a Generalized FLT proof cannot co-exist unless FLT has no solutions.
> >
> > QED
> >
> > AP
> >
> >
> > Newsgroups: sci.math
> > Date: Tue, 24 Jul 2018 22:48:48 -0700 (PDT)
> >
> > Subject: Re: Review of my Generalized FLT proof and its corollary FLT
> > From: Archimedes Plutonium <plutonium....@gmail.com>
> > Injection-Date: Wed, 25 Jul 2018 05:48:49 +0000
> >
> >
> > The condensed rectangles is good, for it allows us to recognize 2^3 as 8, in 8+8 = 16, where 16 is 2^4
> >
> > So you have
> > ___________
> > | |2
> > |__________|
> > 4
> > ___________
> > | |
> > |__________|
> >
> > Numbers as Rectangles
> >
> > and thus 8 rectangle stacked on 8 rectangle
> >
> > is a larger rectangle of 4 by 4
> >
> > So numbers as rectangles is totally feasible
> >
> > And makes the chore of finding triples as solutions in the generalized set
> >
> > TABLE of numbers in Generalized FLT
> > the table of all the numbers possible in the Generalized FLT Conjecture as the conglomeration of exponents of 3 or larger as this set:
> > {1, 8, 16, 27, 32, 64, 81, 125, 128, 216, 243, 256, . .}
> >
> > So, we browse through that set looking for Triples A,B, C that is A+B= C, we did one above of 8+8= 16
> >
> > Now we ask ourselves, when will we find solutions and do all solutions have a common feature, a common characteristic? And the answer is that in order to be equal, for A+B to equal C, we have to have a common shared side to stack. In the case above, the 4 is common side to each rectangle.
> >
> > In essence, that is the proof, and nothing more needs to be said about Generalized FLT. If two rectangles do not share a common side (a common factor) then they add up to not being a rectangle of 4 sides but a
> > figure with 6 sides.
> >
> >
> > HHHHHH
> > HHHHHH
> > HHHHHHHHHH
> > HHHHHHHHHH
> >
> > When you turn all the numbers in the Generalized FLT space into rectangles, only those two rectangles that add up to a new larger rectangle are solutions to a triple.
> >
> > So Generalized FLT was simple and easy.
> >
> > Now going from Generalized to the Corollary (in prior posts I called it a lemma by mistake, it is a corollary), of FLT, and here is where I have some difficulty.
> >
> > According to a number of people on Internet, once you proven Generalized FLT, they claim you automatically without fuss have a proof of FLT itself.
> >
> > I find that hard to swallow.
> >
> > Because, say that for exp3, that D,E,F were solutions, and those on Internet would say that D,E,F carry a common factor, due to Generalized FLT. But, I would retort-- so what, from the distributive law, if DEF was a solution, it has an infinite number of solutions due to distributive law where kD, kE, kF are infinitely many solutions as k can be any constant. So why fuss and bother about one factor solution from Generalized FLT.
> >
> > So, either the Internet folks have some missing logic, or, AP is dense about this movement from Generalized FLT to FLT itself. And I rather think the former.
> >
> > AP
> >
> > Newsgroups: sci.math
> > Date: Wed, 25 Jul 2018 00:17:48 -0700 (PDT)
> >
> > Subject: Re: Review of my Generalized FLT proof and its corollary FLT
> > From: Archimedes Plutonium <plutonium....@gmail.com>
> > Injection-Date: Wed, 25 Jul 2018 07:17:48 +0000
> >
> > On Wednesday, July 25, 2018 at 2:07:29 AM UTC-5, Archimedes Plutonium wrote:
> > > Now here is one of several websites discussing how a proof of Generalized FLT would immediately prove FLT
> > >
> > > —- quoting KarlRunge.com
> > >
> >
> > Now I am going to replace a name with the words "Generalized FLT" from this quote simply because I cringe at the idea of money hucksterism advertisements in mathematics
> >
> > > It is easy to show that if Generalized FLT(sic) is true then it implies Fermat's Last Theorem: since for x = y = z > 2 , assume a solution A, B, C and then divide the equation the common factor= [gcd(A,B,C)]x, and you get a new solution with gcd(A',B',C') = 1, in contradiction of Generalized FLT(sic). Thus there can be no solutions for x = y = z > 2. This was conjectured by Fermat in the 1600's....
> >
> >
> > Alright, I am beginning to see what "other mathematicians" feel is a proof of FLT once Generalized is begot. They think that a solution of FLT contradicts Generalized FLT.
> >
> > I think what they think, that reduction of the solution, by factoring out, leaves the new triple without any factors and hence violates the Generalized.
> >
> > So, I accept their assessment. That Generalized is really fully in charge and can easily conquer FLT.
> >
> > AP
> >
> > Newsgroups: sci.math
> > Date: Wed, 25 Jul 2018 09:54:09 -0700 (PDT)
> >
> > Subject: why it bothered me for years Re: Review of my Generalized FLT proof
> > and its corollary FLT
> > From: Archimedes Plutonium <plutonium....@gmail.com>
> > Injection-Date: Wed, 25 Jul 2018 16:54:09 +0000
> >
> > why it bothered me for years Re: Review of my Generalized FLT proof and its corollary FLT
> >
> > - show quoted text -
> > I am going to "give in" and accept the above as the final proof of FLT as a corollary of Generalized FLT, although I feel a bit what we used to call "hornswaggled". I get a sort of feeling of the proof is "rigged", for it does not seem to flow free and clear.
> >
> > Suppose A,B,C has a A^3 + B^3 = C^3, then because of Generalized factor out its common factor gives D^3 + E^3 = F^3, now, because it has no common factor it violates Generalized FLT.
> >
> > You see, that bothers me-- you use Generalized one moment, next moment you use it again to say it is violated.
> >
> > Troubling, it bothers me, and the reason I have hesitated for years in accepting it.
> >
> > AP
> >
> > On Wednesday, July 25, 2018 at 4:22:24 PM UTC-5, Archimedes Plutonium wrote:
> > FLT needed a superstructure to solve it Re: Review of my Generalized FLT proof and its corollary FLT
> >
> > - show quoted text -
> > Bother me as it did, I accept it now fully, especially the GCD greatest common divisor part.
> >
> > And the importance of FLT, is that it appears to be the first theorem in all of mathematics that needs a overarching theorem to prove it as a corollary. In other words, some proofs are "tiny proofs" that need a larger structure to get hands around. I am not aware of any other mathematical proof that needed a superstructure to penetrate it.
> >
> > AP
> >
> > ==========
> >
> > AP is aware of the Mickey Mouse game the AMS is playing with this prize-- with their obnoxious demands of having one of their own idiots publish a proof, then have themselves reveiw the published proof-- all self serving corruption. For what AP says-- those fools cannot even realize the Ellipse was never a conic section which AP proved by 2016 and yet, here it is 2018 and the American Mathematical Society under leadership of Charles Fefferman, Ron Graham, Daniel Mauldin, have not opened that discussion to the world, but rather, has ignored it, even though sci.math is full of the discussion-- ellipse is not a conic-- and that due to corruption Charles Fefferman, Ron Graham, Daniel Mauldin, are wanting to teach to everyone in schools in books, the texts, that the ellipse is a conic-- When, when even a bright High School Student can experiment with a cylinder can and a circle lid, that the lid will be an oval inside the cone and be a ellipse inside the cylinder. One can even imagine the experiment in the "mind eye" and know it is true.
> >
> > So, Andrew Beal, and AMS with Charles Fefferman, Ron Graham, Daniel Mauldin, kindly pay Archimedes Plutonium what he is rightly due.
> >
> > AP
On Thursday, September 13, 2018 at 7:20:15 PM UTC-5, Earle JOnes wrote:
>
> I want to better understand your work.
>
> Consider this equation:
>
> x^2 / a^2 + y^2 / b^2 = 1
>
> If a = b, this is the equation of a circle.
> If a not= b, this is the equation of an ellipse.
>
> What is the equation of an oval?
>
> Thanks,
>
> earle
> *
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| From | Archimedes Plutonium <plutonium.archimedes@gmail.com> |
|---|---|
| Date | 2018-09-14 14:42 -0700 |
| Subject | publish the AP ellipse proof in AMS Re: Donald Trump has more moral fiber than does Andrew Beal, and AMS with Charles Fefferman, Ron Graham, Daniel Mauldin |
| Message-ID | <a52a4176-2e77-4a43-8bf1-22953a5171c5@googlegroups.com> |
| In reply to | #423534 |
On Friday, September 14, 2018 at 6:56:41 AM UTC-5, Michael Moroney wrote: > > hi! Could you give a pointer to a paper which explains these new bonds? > This is very interesting news if true, so I would like to see the > science behind this. Yes, have Andrew Beal in colloboration with AMS, Charles Fefferman, Ron Graham, Ken Ribet, Daniel Mauldin, John Horton Conway, Thomas Hales publish in the next AMS that the Ellipse is never a conic as per AP's proof, ___and then go from there_____, a solid recognition by academic math professors that they made a mistake with the ellipse, and are now willing to correct it. For many textbooks have to redact that out of their textbook. Truth always takes priority over hassles of money-- and probably one of the reasons math professors are so slow to act on this ellipse truth-- inconvenient for them. And, even though you-- professors of math who never want to fix and clean up error filled mathematics, your students deserve better, and you should quit teaching if you cannot come around to admitting a ellipse is a cylinder section, never a conic section. For it is embarrassing to have students smarter than the math professor who by taking a cone, cylinder can and a round lid, can prove on the spot, in the classroom that the cone yields an oval, never an ellipse. Proofs ellipse is never a conic, always a cylinder section by Archimedes Plutonium -------------------- Conics = oval, 2 proofs, synthetic, analytic Synthetic Geometry & Analytical Geometry Proofs that Conic section = Oval, never an ellipse-- World's first proofs thereof by Archimedes Plutonium _Synthetic Geometry proofs that Cylinder section= Ellipse// Conic section= Oval First Synthetic Geometry proofs, later the Analytic Geometry proofs. Alright I need to get this prepared for the MATH ARRAY of proofs, that the Ellipse is a Cylinder section, and that the Conic section is an oval, never an ellipse PROOF that Cylinder Section is an Ellipse, never a Oval:: I would have proven it by Symmetry. Where I indulge the reader to place a circle inside the cylinder and have it mounted on a swivel, a tiny rod fastened to the circle so that you can pivot and rotate the circle. Then my proof argument would be to say--when the circle plate is parallel with base, it is a circle but rotate it slightly in the cylinder and determine what figure is produced. When rotated at the diameter, the extra area added to the upper portion equals the extra area added to bottom portion in cylinder, symmetrical area added, hence a ellipse. QED Now for proof that the Conic section cannot be an ellipse but an oval, I again would apply the same proof argument by symmetry. Proof:: Take a cone in general, and build a circle that rotates on a axis. Rotate the circle just a tiny bit for it is bound to get stuck or impeded by the upward slanted walls of the cone. Rotate as far as you possibly can. Now filling in the area upwards is far smaller than filling in the area downwards. Hence, only 1 axis of symmetry, not 2 axes of symmetry. Define Oval as having 1 axis of symmetry. Thus a oval, never an ellipse. QED The above two proofs are Synthetic Geometry proofs, which means they need no numbers, just some concepts and axioms to make the proof work. A Synthetic geometry proof is where you need no numbers, no coordinate points, no arithmetic, but just using concepts and axioms. A Analytic Geometry proof is where numbers are involved, if only just coordinate points. Array:: Analytic Geometry proof that Cylinder section= Ellipse//Conic section = Oval, never ellipse Now I did 3 Experiments and 3 models of the problem, but it turns out that one model is superior over all the other models. One model is the best of all. That model is where you construct a cone and a cylinder and then implant a circle inside the cone and cylinder attached to a handle so that you can rotate the circle inside. Mine uses a long nail that I poked holes into the side of a cylinder and another one inside a cone made from heavy wax paper of magazine covers. And I used a Mason or Kerr used lid and I attached them to the nail by drilling two holes into each lid and running a wire as fastener. All of this done so I can rotate or pivot the circle inside the cylinder and cone. You need a long nail, for if you make the models too small or too skinny, you lose clarity. ARRAY, Analytic Geometry Proof, Cylinder Section is a Ellipse:: E __ .-' `-. .' `. / \ ; ; | G c | H ; ; \ / `. .' `-. _____ .-' F The above is a view of a ellipse with center c and is produced by the Sectioning of a Cylinder as long as the cut is not perpendicular to the base, and as long as the cut involves two points not larger than the height of the cylinder walls. What we want to prove is that the cut is always a ellipse, which is a plane figure of two axes of symmetry with a Major Axis and Minor Axis and center at c. Side view of Cylinder EGFH above with entry point cut at E and exit point cut at F and where c denotes the central axis of the cylinder and where x denotes a circle at c parallel with the base-circle of cylinder | | | | E | | | | |x c |x | | | | | | |F | | | | | | | So, what is the proof that figure EGFH is always an ellipse in the cylinder section? The line segment GH is the diameter of the circle base of cylinder and the cylinder axis cuts this diameter in half such that Gc = cH. Now we only need to show that Fc = cE. This is done from the right triangles cxF and cxE, for we note that by Angle-Side-Angle these two right triangles are congruent and hence Fc = cE, our second axis of symmetry and thus figure EGFH is always an ellipse. QED Array proof:: Analytic Geometry proof that Conic section= Oval// never ellipse ARRAY, Analytic Geometry Proof, Conic Section is a Oval, never an ellipse:: A ,'" "`. / \ C | c | D \ / ` . ___ .' B The above is a view of a figure formed from the cut of a conic with center c as the axis of the cone and is produced by the Sectioning of a Cone as long as the cut is not perpendicular to the base, and as long as the cut is not a hyperbola, parabola or circle (nor line). What we want to prove is that this cut is always a oval, never an ellipse. An oval is defined as a plane figure of just one axis of symmetry and possessing a center, c, with a Major Diameter as the axis of symmetry and a Minor Diameter. In our diagram above, the major diameter is AB and minor diameter is CD. Alright, almost the same as with Cylinder section where we proved the center was half way between Major Axis and Minor Axis of cylinder, only in the case of the Conic, we find that the center is half way between CD the Minor Diameter, but the center is not halfway in between the Major Diameter, and all of that because of the reason the slanted walls of the cone cause the distance cA to be far smaller than the distance cB. In the diagram below we have the circle of x centered at c and parallel to base. The angle at cx is not 90 degrees as in cylinder. The angle of cAx is not the same as the angle cBx, as in the case of the cylinder, because the walls of the cone-for line segments- are slanted versus parallel in the cylinder. Triangles cAx and cBx are not congruent, and thus, the distance of cA is not equal to cB, leaving only one axis of symmetry AB, not CD. / \A x/ c \x B/ \ Hence, every cut in the Cone, not a hyperbola, not a parabola, not a circle (not a line) is a Oval, never an ellipse. QED --Archimedes Plutonium Very crude dot picture of 5f6, 94TH ELECTRON=muon 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. Read my recent posts in peace and quiet. https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe Archimedes Plutonium
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| From | Archimedes Plutonium <plutonium.archimedes@gmail.com> |
|---|---|
| Date | 2018-09-15 03:58 -0700 |
| Subject | Re: publish the AP ellipse proof in AMS Re: Donald Trump has more moral fiber than does Andrew Beal, and AMS with Charles Fefferman, Ron Graham, Daniel Mauldin |
| Message-ID | <5b0b684c-7993-4db9-b4af-47e159e7f5c9@googlegroups.com> |
| In reply to | #423605 |
On Saturday, September 15, 2018 at 4:08:56 AM UTC-5, burs...@gmail.com wrote: > A minimal wheelchair: > > |_ > |_| > * * > > This one is customized, its has AP AP writes:: Burse and his Swiss wheelchair fantasy-- maybe he thinks a wheelchair can make Andrew Beal, and AMS with Charles Fefferman, Ron Graham, Daniel Mauldin, Ken Ribet understand the below proof that a ellipse is never a conic. AP's proof the ellipse is never a Conic Section, always a Cylinder section, and how the proof works Let us analyze AP's Proof On Friday, September 14, 2018 at 6:57:36 PM UTC-5, Archimedes Plutonium wrote: > > Array:: Analytic Geometry proof that Cylinder section= Ellipse//Conic > section = Oval, never ellipse > > Now I did 3 Experiments and 3 models of the problem, but it turns out > that one model is superior over all the other models. One model is the > best of all. > > That model is where you construct a cone and a cylinder and then > implant a circle inside the cone and cylinder attached to a handle so > that you can rotate the circle inside. Mine uses a long nail that I > poked holes into the side of a cylinder and another one inside a cone > made from heavy wax paper of magazine covers. And I used a Mason or > Kerr used lid and I attached them to the nail by drilling two holes > into each lid and running a wire as fastener. All of this done so I > can rotate or pivot the circle inside the cylinder and cone. You need > a long nail, for if you make the models too small or too skinny, you > lose clarity. > > ARRAY, Analytic Geometry Proof, Cylinder Section is a Ellipse:: > > > E > __ > .-' `-. > .' `. > / \ > ; ; > | G c | H > ; ; > \ / > `. .' > `-. _____ .-' > F > Alright, focus on the distance from c to F in the cone-cut compared to the distance from c to E In a Cylinder cut, those two distances are the same because a cylinder has two axes of symmetry. The side view of a cylinder is this | | | | | | That allows cE to be the same distance as cF But the side view of the cone is /\E /c \ F / \ The distance c to E is shorter because the slant of the side walls of the cone are in the direction of shortening cE, whereas the slant opposite c in cF make that distance larger than cE > The above is a view of a ellipse with center c and is produced by the > Sectioning of a Cylinder as long as the cut is not perpendicular to > the base, and as long as the cut involves two points not larger than > the height of the cylinder walls. What we want to prove is that the > cut is always a ellipse, which is a plane figure of two axes of > symmetry with a Major Axis and Minor Axis and center at c. > > Side view of Cylinder EGFH above with entry point cut at E and exit > point cut at F and where c denotes the central axis of the cylinder > and where x denotes a circle at c parallel with the base-circle of > cylinder > > | | > | | E > | | > | | > |x c |x > | | > | | > | | > |F | > | | > | | > | | > > So we can see that the distance cE = cF in cylinder for the walls are Parallel to one another, giving distance symmetry But in the Cone, the walls are not parallel, shortening the distance cE compared to cF. Leaving only one axis of symmetry that of cx. The oval is the conic section of a cut at a slant, while the cylinder cut at a slant is a ellipse. The Oval has just one axis of symmetry. > So, what is the proof that figure EGFH is always an ellipse in the > cylinder section? The line segment GH is the diameter of the circle > base of cylinder and the cylinder axis cuts this diameter in half such > that Gc = cH. Now we only need to show that Fc = cE. This is done from > the right triangles cxF and cxE, for we note that by Angle-Side-Angle > these two right triangles are congruent and hence Fc = cE, our second > axis of symmetry and thus figure EGFH is always an ellipse. QED > > > > Array proof:: Analytic Geometry proof that Conic section= Oval// never ellipse > > ARRAY, Analytic Geometry Proof, Conic Section is a Oval, never an ellipse:: > > > A > ,'" "`. > / \ > C | c | D > \ / > ` . ___ .' > B > > The above is a view of a figure formed from the cut of a conic with > center c as the axis of the cone and is produced by the Sectioning of > a Cone as long as the cut is not perpendicular to the base, and as > long as the cut is not a hyperbola, parabola or circle (nor line). > What we want to prove is that this cut is always a oval, never an > ellipse. An oval is defined as a plane figure of just one axis of > symmetry and possessing a center, c, with a Major Diameter as the axis > of symmetry and a Minor Diameter. In our diagram above, the major > diameter is AB and minor diameter is CD. > > Alright, almost the same as with Cylinder section where we proved the > center was half way between Major Axis and Minor Axis of cylinder, > only in the case of the Conic, we find that the center is half way > between CD the Minor Diameter, but the center is not halfway in > between the Major Diameter, and all of that because of the reason the > slanted walls of the cone cause the distance cA to be far smaller than > the distance cB. In the diagram below we have the circle of x centered > at c and parallel to base. The angle at cx is not 90 degrees as in > cylinder. The angle of cAx is not the same as the angle cBx, as in the > case of the cylinder, because the walls of the cone-for line segments- > are slanted versus parallel in the cylinder. Triangles cAx and cBx are > not congruent, and thus, the distance of cA is not equal to cB, > leaving only one axis of symmetry AB, not CD. > > / \A > x/ c \x > B/ \ > > Hence, every cut in the Cone, not a hyperbola, not a parabola, not a > circle (not a line) is a Oval, never an ellipse. > > QED > > --Archimedes Plutonium > > Very crude dot picture of 5f6, 94TH > ELECTRON=muon 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. > > Read my recent posts in peace and quiet. > > https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe > Archimedes Plutonium
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| From | moroney@world.std.spaamtrap.com (Michael Moroney) |
|---|---|
| Date | 2018-09-16 02:08 +0000 |
| Subject | Re: publish the AP ellipse proof in AMS Re: Donald Trump has more moral fiber than does Andrew Beal, and AMS with Charles Fefferman, Ron Graham, Daniel Mauldin |
| Message-ID | <pnkdvo$f2c$3@pcls7.std.com> |
| In reply to | #423605 |
Archimedes Plutonium <plutonium.archimedes@gmail.com> writes: >On Friday, September 14, 2018 at 6:56:41 AM UTC-5, Michael Moroney wrote: >> hi! Could you give a pointer to a paper which explains these new bonds? >> This is very interesting news if true, so I would like to see the >> science behind this. >Yes, have Andrew Beal in colloboration with AMS, Charles Fefferman, Ron >Graham, Ken Ribet, Daniel Mauldin, John Horton Conway, Thomas Hales publish in the next AMS that the Ellipse is never a conic as per AP's OK, so the answer is 'no'. Gotcha.
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