AP proudly announcing his 253rd published book of science// My 253rd published book. The math proof: 3 arbitrary noncollinear points in Space, determines a Unique Plane and the nonexistence of 4D // Math research by Archimedes Plutonium Preface:

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Archimedes Plutonium

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Aug 22, 2023, 2:07:37 AM8/22/23

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My 253rd published book.

The math proof: 3 arbitrary noncollinear points in Space, determines a Unique Plane and the nonexistence of 4D // Math research

by Archimedes Plutonium

Preface: This book started in May 2023 with the focus of having a look at the proof of 3 arbitrary noncollinear points in Space forms a unique plane and how that proof was created and assembled. Also whether that proof can answer the question of whether 3rd dimension Space is the last and highest dimension. The result was rather fantastic, for it starts the major overhaul of Euclidean Geometry to be a discrete Space composed of Decimal Grid System Numbers. In this book I offer a new proof-- that is, a derived Theorem proof, and not a axiom of the statement of 3 arbitrary noncollinear points forming a unique plane. This book ends with a start of the total overhaul of Euclidean Plane Geometry because the true numbers of mathematics are discrete numbers-- the decimal grid numbers and also the start of straightening out the Euclid Axioms of geometry.

Cover Picture: Is my iphone photograph of this concept of 3 arbitrary points in Space forms a unique plane, and especially note the top rightward picture of multiple colored planes.

Product details
• ASIN ‏ : ‎ B0CG563N6Q
• Publication date ‏ : ‎ August 20, 2023
• Language ‏ : ‎ English
• File size ‏ : ‎ 433 KB
• Text-to-Speech ‏ : ‎ Enabled
• Screen Reader ‏ : ‎ Supported
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Not Enabled
• Sticky notes ‏ : ‎ On Kindle Scribe
• Print length ‏ : ‎ 83 pages

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Archimedes Plutonium

AP proudly announcing his 253rd published book of science// My 253rd published book. The math proof: 3 arbitrary noncollinear points in Space, determines a Unique Plane and the nonexistence of 4D // Math research by Archimedes Plutonium Preface:

Archimedes Plutonium

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Jun 1, 2026, 4:25:24 PM (6 days ago) Jun 1

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I am having a difficult time of remembering how I arrived at these conclusions. So review, I must.

Archimedes Plutonium

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Jun 1, 2026, 4:29:24 PM (6 days ago) Jun 1

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The math proof: 3 arbitrary noncollinear points in Space, determines a Unique Plane and the nonexistence of 4D // Math research

by Archimedes Plutonium

This is AP's 253rd published book of science published on Internet, Plutonium-Atom-Universe, PAU newsgroup is this. Please read this textbook for free on Internet.

https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe

Preface: This book started in May 2023 with the focus of having a look at the proof of 3 arbitrary noncollinear points in Space forms a unique plane and how that proof was created and assembled. Also whether that proof can answer the question of whether 3rd dimension Space is the last and highest dimension. The result was rather fantastic, for it starts the major overhaul of Euclidean Geometry to be a discrete Space composed of Decimal Grid System Numbers. In this book I offer a new proof-- that is, a derived Theorem proof, and not a axiom of the statement of 3 arbitrary noncollinear points forming a unique plane. This book ends with a start of the total overhaul of Euclidean Plane Geometry because the true numbers of mathematics are discrete numbers-- the decimal grid numbers and also the start of straightening out the Euclid Axioms of geometry.

Cover Picture: Is my iphone photograph of this concept of 3 arbitrary points in Space forms a unique plane, and especially note the top rightward picture of multiple colored planes.

-----------------------------------

Table of Contents

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1) My history for this book.

2) Some exercises in constructing unique ellipse from 3 arbitrary noncollinear points in space.

3) Two points in plane determine a unique circle.

4) Defining a Plane in Space without it being a axiom.

5) Polar, Cylindrical, Spherical Coordinate Systems.

6) Discrete geometry.

7) Resolving 3 arbitrary noncollinear points determines a unique plane in 3D geometry.

8) Hourglass figure is a unique ellipse.

9) Pencil Ellipses.

10) Exploring Right-Triangles at the Infinity borderline and its Algebraic completeness.

11) Return to the math-derived-theorem-proof of 3 arbitrary, noncollinear points in Space determines a unique plane.

12) Discrete Space and Euclidean Plane Geometry.

13) The Overhaul of Old Math's Euclidean Geometry.

14) Summary.

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Text

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1) My history for this book.

Archimedes Plutonium

May 28, 2023, 7:19:53 PM

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Alright, while doing my 240th book of science, I proved that 3 arbitrary points not collinear yield a unique circle but also yield a unique ellipse.

--- quoting my 240th published book ---

New True Geometry starting with cycloid correction and Geometry-of-Motion // math research

by Archimedes Plutonium

Preface: Motion Geometry such as producing the Cycloid curve in Old Math was in error. Looking at the Limacon we see this error in Motion Geometry at its worst. And the fix of the error has to come from physics of Electromagnetic theory where magnetic field is always perpendicular to electric field. The mistake made in Old Math Geometry is that they had two motions in a cycloid and limacon construction. They had the motion of a circle, but had arbitrary motion for the Pointer-Marker. This book addresses the Old Math Geometry mistakes and opens up the entire field of new math of Motion Geometry. In the midst of that correction several major conjectures were discovered and proven in this book which has created a complete overhaul of Old Math's conic sections. For the parabola and hyperbola are not open curves but closed loop circuits of ellipse, oval, circle. The nasty mistake of Old Math to think cones are apex to apex is ridiculous to the nth degree, for true conics are base to base <>, not apex to apex ><. And I finish this book with a stunning proof that 3 arbitrary non-collinear points in the plane not only produces a unique circle, but produces a unique ellipse, yet Old Math claimed it requires 5 points to produce a unique ellipse. I suspect Old Math's 5 point requirement is based on the square or rectangle, whereas my 3 point requirement is a parallelogram that is not square and not a rectangle. If I can find a unique ellipse inside a unique circle produced by 3 arbitrary points non-collinear, then I can say in general all 3 arbitrary non-collinear points produces a unique ellipse.

Cover Picture: My photograph of holding two cones base to base. Two cones base to base forms the foundation of Conic Sections, and not the silly apex to apex.

--- end quote ---

And being curious, I wanted to investigate the ultimate yield of 3 arbitrary points in space not collinear yields a unique plane in that space. This would be a separate new book for me.

What is the proof of that statement-- that 3 arbitrary noncollinear points in Space forms a unique plane?? At the time in May, I could not recollect in my memory what the proof was in my schooling, for I loved geometry in Grade School and High School. And if you ask me, I like geometry more than I like algebra and I am stronger in geometry than in algebra.

Well, looking around in books and internet, the proof is simply a axiom itself. No easier of a math proof if the statement is enshrined as a axiom. Some prefer to call axioms as postulates. Myself I prefer the word axiom for it is shorter to write and say.

But can that proof whether by a axiom or by a derived theorem proof, also prove that 4 dimensional space is contradictory? That the highest dimension for Space is 3 and no dimensions beyond 3? It is highly likely that the concept of 3 arbitrary points noncollinear determines a unique plane is directly linked with how many dimensions that Space has.

And so, let me include in this book proofs that 4 dimensional space and higher cannot exist.

My 253rd book of science// The math proof that 3 arbitrary points in Space, not collinear, determines a Unique Plane in Space and the nonexistence of 4D // Math research by Archimedes Plutonium.

2) Some exercises in constructing unique ellipse from 3 arbitrary noncollinear points in space.

Archimedes Plutonium

May 29, 2023, 1:54:17 PM

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A rectangle or square have contained inside them 2 right triangles. As we move one of them and join leg to leg forms a parallelogram, and making further movement to look like a hourglass figure in which the 4 vertices is a unique ellipse. So, given any 3 arbitrary points, not collinear, in space, forms a unique plane, forms a unique circle in that plane and by movement, those 3 points forms a rectangle or square delivering 2 right triangles, aided by movement that forms a unique ellipse.

Take 2 identical 60-30-90 right triangles joined together at smallest leg and reversed forming a parallelogram and forming One Unique Ellipse from the 4 vertex.

So the logic is 3 points determine unique circle-- in turn-- determines either square, rectangle or isosceles trapezoid. If the trapezoid is generated a further movement of trapezoid to Rectangle, and finally, from starting 3 points we have a Unique Ellipse.

With 2 identical right triangles 60-30-90 put leg to leg to form a parallelogram as exhibited here.

Then a movement, a scooting over as such.

.|/

The movement causes a HourGlass figure to appear.

AP says only a Unique Ellipse contains those 4 vertices.

I remember reading a article that one of the German universities had a wood model of Conics with the slant cut being a ellipse. I think it was Gottingen. But on Internet, I see that Smithsonian has a wooden model called Ellipsographs and it also shows the slant cut being an ellipse. But sadly these are math failures for they somehow rounded off the cut and made the cut not steep enough to show the truth-- the truth that it is a oval with its narrowness near the apex and its bulge to make a Oval where it is furthest from apex.

The cone has but 1 axis of symmetry and a ellipse requires 2 axes, yet Oval is also 1 axis of symmetry.

No wonder so many in math have a difficult time in seeing that 2 identical right triangles joined smallest leg to leg and reversed forms a unique ellipse. Their knowledge is all memorized knowledge, barely ever is it logical reasoned knowledge.

Now I had been talking about the North-South ellipse versus the East-West ellipse in the proof that 2 right triangles leg to leg form a Unique ellipse for which only that ellipse can cover the 4 vertex points of that parallelogram.

But let me outline where 2 Non-right triangles almost being a right triangle has 2 Ellipses not a unique ellipse from the 4 vertex points of the parallelogram. And drawn the 2 ellipses they come close to a convergence upon North-South ellipse. In other words, as the 2 triangles get closer and closer to being a 90 degree right triangle, the 2 ellipses have a tendency to converge upon a Unique ellipse.

These four vertex points are (0,0) and (22,44) and (48,31) and (70,75). And measuring these two triangles are not 90 degrees but 80 degrees. These two triangles are approx 80-30-70.

And what happens when you do not have 2 right triangles leg to leg to form a parallelogram, you indeed have room for 2 different ellipses covering all four vertex points. As the two triangles get closer and closer to being right triangles, the second ellipse disappears and you have remaining a Unique ellipse.

3) Two points in plane determine a unique circle.

Archimedes Plutonium

May 30, 2023, 9:30:33 PM

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Now in this book I need to drop further down in geometry than starting with 3 arbitrary, non-collinear points in the plane.

For it was remarked that 2 points determine a unique circle in plane, one point as center of circle and the other point marking the distance of the radius.

But the original intent of 3 arbitrary points is a point not the center of the circle.

So here is an immense, very immense discussion and interface between definitions of Logic, math proofs, math statements and precision. So we have an immense nuances and logical pitfalls.

So the circle definition in a plane is a center point and all points equidistant from that centerpoint as the radius of the circle. We all know this. But the reason I bring this to attention is what is the definition of a Plane, or of Space in Geometry.

Now here we get hung up in many many snags. Especially Space, but also Plane. Easy to define circle, but try defining Plane.

So Old Math adopted the Ancient Greeks, Euclid definition making it an axiom. They said, 3 arbitrary points in Space, not-collinear, define a unique Plane.

And this axiom makes the proof that 3 arbitrary points in Space, not-collinear be a unique plane, a simple proof of just one step proof. Statement then proof is citing this axiom.

But I want to pause for a moment and ask is this the only way to properly define the Plane in Space??

We saw how elegant the definition of circle was. So is there a more elegant definition of Plane??

I believe so, by looking at the definition of circle. For I am thinking that perhaps the elegance of defining circle can define better not only the Plane but also Space.

I am thinking of the idea that if we start with two points in space, one being a center and the second being a distance, a radius, then that defines Space as all points equidistant-- a hollow sphere. Then those two points define a straightline, infinite straightline and a motion, a movement of that straightline in Space defines a Plane in Space. The motion has to be such that it is not wavering, and the way to handle that is this circle in Space just created.

4) Defining a Plane in Space without it being a axiom.

Archimedes Plutonium

May 30, 2023, 11:25:51 PM

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Now what I have noticed here is that the definition of Circle in Geometry is sparkling clear like a cold bottle of mineral bubbly water on a hot summer day, whereas the definition of Plane is rather dull, coming only from a axiom given. Not even constructed. And we have the Parallel Postulate (axiom) also in the plane as given line and a point not on that line defining a unique line to given line and a Unique Plane in a surrounding Space.

So is the Parallel Postulate with its uniqueness related to the plane as unique from 3 arbitrary noncollinear points?

The clear definition of circle is a point in the plane as circle center and another point acting as radius for circle center.

So I am looking to see if I can make the definition of Plane as sparkling, dazzling clear as the definition of circle lays out a circle from 2 points in plane.

I have had quite a bit of experience with problems like this. In particular the Physics of unification of the 4 forces of physics. In that unification, I picked the "most perfect force" the EM force for it had the most perfect particle-- the photon or Light Wave, hence the other 3 forces were manifestations of Light Waves of EM force.

So here in mathematics, we can say the definition of Circle is far more perfect than the definition of Plane in Old Math.

And so, what I am going to try is a trick of Physics applied to mathematics. I am going to say "let there exist the circle" and define it as 2 points, one point the center and the other point a radius distance. Now that does not restrict the circle from forming a 3D hollow sphere. And so, now, I need to restrict the circle to form a plane. A plane distinguished from the Space that the circle lies in.

And coming from physics, and knowing how important is angles and especially the right angle for electricity is always perpendicular to magnetism. I am going to end up defining a Plane as being a circle parallel to all other circles of a Cylinder.

So, what I am thinking here, for an elegant definition of Circle, I need two points-- center and a distance from center for radius.

So an elegant Plane definition is 1 point to define a plane, and that can be got from a Cylinder as Space. Where the points of the center-line define a unique circle.

That would leave Space to be elegantly defined as needing 0 points.

Summary:

0 points define Space.

1 point defines unique Plane.

2 points define unique circle in plane

Now this only will work if I can manage to say all of Physics is cylindrical. That a Cylinder defines Space.

So have to check into that. But already there are indications that a cylindrical coordinate system is the most General of all coordinate systems.

Archimedes Plutonium

May 31, 2023, 11:07:22 AM

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So let us take a standard Right Triangle of the 3-4-5 right triangle. The one that is practically 60-30-90 in measurement and transfer that to a graph paper, instead of the recent example of non-right triangles (0,0) and (22,44) and (48,31) and (70,75).

Because when the two triangles of the parallelogram are not right triangles, then you get 2 ellipses to cover the parallelogram. When the two triangles are indeed right triangles, the nature of the right angle forbids a 2nd ellipse from forming.

Archimedes Plutonium

May 31, 2023, 11:33:09 AM

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So now, I am not happy nor satisfied with the Definition of Plane in Old Math of their axiom definition, 3 noncollinear points in Space determine a Unique Plane.

Not happy at all with that, because it is not elegant enough.

What is elegant is 2 points determine a circle in the plane. The first point as center of a circle, and the second point marking out a distance of radius for a unique circle.

Since there is such a elegant definition of circle in plane, means, I am sure, there must be a elegant definition of Plane in Space.

Old Math Plane is just too ugly. For it is just a axiom without and logical construction.

Now perhaps the reason it is too ugly and not elegant enough is because we all think in terms of the Cartesian Coordinate System as the basis of Euclidean Geometry.

But what if the most elegant coordinate system for Geometry was the cylindrical coordinate system.

Would we then be able to fetch the most elegant definition of Plane from Space being a cylinder?

The reason the Cylindrical Coordinate System is the best for physics Electromagnetic Theory is because it easily covers physics when you need symmetry about a axis. We use spherical coordinate system when you have symmetry about a point.

So if we said Cylindrical Coordinate System was the true and best system for physics to handle EM theory, then, would our definition of Plane in Space be as elegant as our definition of circle in plane-- 2 points determine a unique circle.

I think it would because then 2 points determine a unique circle and just 1 point determines a unique plane in Cylindrical Coordinate System.

I picture the Cylindrical Coordinate System as stacking up "planes of circles". The entire system is the stacking of planes.

And then the elegant definition of Space in Cylindrical Coordinate System is that 0 points --- meaning all points, determine Space. In the Old Math Cartesian Coordinate System, you could not even talk about All Points in Space, with their phony Reals and continuum. And you could not even mention the idea of "all numbers" not in a ill-logic of between any two points is another point. The disaster of endless numbers spills over into Space geometry.

I looked and looked and looked, if any physics textbook discusses Coordinate Systems. I dropped as low as High School physics textbook, no mention.

I looked in Halliday and Resnick, no mention. None in Feynman.

Perhaps all physicists go under a rule that no matter what coordinate system you use, polar coordinates, cylindrical coordinate system, spherical coordinate system or our everyday familiar Cartesian Coordinate system, all end up with the same answer if properly used. Perhaps that is the reason for the absence of coordinate systems in physics.

And perhaps it is codified in the Special Relativity, as all inertial frames of reference.

About the only place where you get good discussion of coordinate systems is 1st year calculus.

The reason I am interested in this, is because I suspect Cylindrical Coordinate System is superior to all other coordinate systems.

Alright, I looked through all the entire math and physics books I have at home. Looking for some clue as to whether Cylindrical Coordinate System has some superiority over all other coordinate systems, and find myself in the Dostoevsky fable of riding around for all the land to own, only to come back where I started from, fall off the horse and die, and my grave is all the land I ever needed.

These different systems are only easier given the circumstances but all give the same answer if done correctly. Cylinder or Sphere or Polar coordinates are not superior, only easier for some application.

However, as I was riding around for all the land to own, I remembered that I am the one responsible for making all valid functions be one and only that of Polynomial Functions. So, because of that, I can argue the superior Coordinate System is the plain ordinary common Cartesian Coordinate System, and that these other systems are wastrel and kookish for those wanting to dabble with angles along with numbers, not all numbers only.

So for a person wanting to make math easy and simple as possible, why add on polar coordinates, spherical and cylindrical coordinates? The only Coordinate System that is simple and easy is the Cartesian. But, in particular, is the Cartesian system the only system that can handle the idea that the only valid functions of math are polynomials??

Has anyone attempted to graph Polynomial functions in polar, cylindrical or spherical coordinates? Not to my knowledge.

So I am back, back once again on the pondering question of how to make the definition of Plane be as elegant as the definition of circle-- two points, one a center and the second point forms the radius all in a Plane.

I guess I was just mistaken to think 2 entities to make the circle for the circle was embedded in a plane already as being 3 entities, not so elegant. And accounting for my mistake, I need 3 entities-- 3 non-collinear points to form the Unique Plane.

Talk of Space is out of the picture, out of question, for Space is a given, the Cartesian Coordinate Decimal Grid Systems.

I do need to see if there is a proof that Cartesian Coordinates are the only to handle the idea that only valid functions are polynomials!!!

5) Polar, Cylindrical, Spherical Coordinate Systems.

Archimedes Plutonium

2:08 AM 1May2023

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Conjectures

Conjecture 1: It is impossible to plot the square or cube in polar, cylindrical or spherical coordinates.

Conjecture 2: It is impossible to plot the Polynomial functions in polar, cylindrical, or spherical coordinate systems.

So a few days ago I was on the exact opposite of today's viewpoint.

Now if I am convinced enough of the above, which I am not at the moment. Then there would be a simple easy proof of both. For the two are connected. Connected in the idea that the so called Parabola of Old Math begot from Conics of apex to apex >< when conics are really that of base to base <> and so the Parabola in New Math is tending towards a shape of a square. The Parabola formula is basically Y-> x^2 and we can see that as area of a square of side x.

So if it is impossible to plot a graph of square in polar, spherical, or cylindrical coordinate system, only in Cartesian coordinate system can we plot the square, implies that Polynomials are only able to be graphed and plotted in Cartesian Coordinate System, as well.

And this gets back to the logical flaw of having coordinate systems where axes are not all numbers but some angles and some numbers, distorting the final answer, such as the phony Sinusoid wave, when it is a semicircle wave.

The upshot in all of this, polar, cylindrical, spherical are for ease of calculation and nothing more. But also, vulnerable to making mistakes which the Cartesian system is less prone to mistakes.

Archimedes Plutonium

Jun 1, 2023, 3:30:22 AM

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So, getting back to this question-- 3 arbitrary non-collinear points in plane determine a unique circle. Yet in the plane, 2 arbitrary points, one as center other as radius determinant produces a unique circle.

In Space a unique plane is produced by 3 arbitrary non-collinear points. So I want the analog of 2 points to determine a unique plane.

But let me increase the frenzy even further. That 2 points in Space produce a unique Sphere, not just a circle in the Plane.

So why can I not use 2 items, 2 points in space to produce a unique plane?

Perhaps the key to this problem will rest on using the sphere and circle in tandem to produce a unique plane in Space.

Archimedes Plutonium

Jun 1, 2023, 3:47:50 AM

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I have come up with a solution. A pretty solution it is.

In Decimal Grid Systems as the true numbers of mathematics and the Cartesian Coordinate System that graphs these numbers have a Natural Center to each Grid. For 10 Grid the center is (5,5). For 100 Grid, the center is (50,50) in 2D. For 3D the center respectively is (5,5,5) and (50,50,50).

Now, given 1 point, determines a Space, and that one point is the center of that Space.

Now, given 2 points in that specific Space, not the center, determines a unique Plane in that Space.

This overcomes some hurdles.

This allows for Symmetry of determining Unique Plane, Unique circle in plane, unique sphere in space.

For 1 point determines Unique Space-- its center. Then 2 points in unique Space determines unique Plane by using center of Space. Unique Plane determines unique circle as well as unique sphere.

This would be immensely important for Atom Totality theory in physics for it demands a center to the Cosmos.

Yes, I am trying to recall a book, for I layed it aside for too long of a time spell.

We have a proof that given a plane and 3 arbitrary points noncollinear in the plane yields a unique circle.

Then I realized that 3 arbitrary points noncollinear in the plane should yield a unique ellipse. So that was the main feature of this book. And the way I solved that was 2 identical right triangles when placed shortest leg together and reverse one of them, then slide one so that you form a HourGlass figure. That determines a unique ellipse.

Old Math insists you need 5 arbitrary noncollinear points in plane to determine a unique ellipse. And here we constructed the unique ellipse from just 3 arbitrary noncollinear points.

Here, AP says no to Old Math and proves that 3 arbitrary noncollinear points determines a unique ellipse via building the unique circle, then constructing the unique ellipse from inside that circle. I vaguely recall that the unique ellipse does not use all 3 of the points used to construct the unique circle.

But it also solves the nagging problem of Old Math-- they needed a axiom to prove 3 arbitrary noncollinear points determines a unique plane. For they made that be a axiom.

AP makes that proof of 2 points in Space and using a third point in space-- the center of the Grid, determines a Unique plane in 3rd Dimension Space.

And then that construction of plane implies that 1 point in Space determines Space itself-- the center of the Grid System.

Further, this center then implies that Geometry can have at most 3 dimensions and cannot have a 4th or higher dimension.

But this is still in the conjectural phase, not the conclusive stage.

6) Discrete geometry.

Archimedes Plutonium

Jun 1, 2023, 4:00:52 PM

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Alright, I need to, in this book discuss and contemplate the density of points in Space when Space is a Gridwork of numbers with empty space in between one number and the next. This is Discrete Geometry.

And for the smallest Grid, the 10 Decimal Grid System, appears like a orchard with rows and columns of trees spaced so there is so to speak empty space from one tree to the next tree.

If we look down a row, we easily see a "line" of trees. But if we start at one tree and look at an angle we see holes and empty space.

So with that in mind, we have to ask about the foundation principles and axioms of Old Math concerning their "Any two points in a plane" determines a line.

Now is the line unique? For we talked about unique circles and unique ellipses and unique planes. But in Old Math they very much omitted the discussion of unique line.

And in Old Math, their numbers were Reals with the continuum. In New Math, numbers are Decimal Grid Numbers and discrete.

So in 10 Grid, is there a point chosen and at a angle, no other point can be intercepted by that angle, before we reach the end borderline of number 10 maximum, and 100 points on the 10 Grid, x-axis? Is there a angle in which only 2 points lie on a line in the xy plane with its 100 x 100 points = 10,000 coordinate points in all?

These and many more questions will be explored and pondered.

Archimedes Plutonium

Jun 2, 2023, 2:22:17 AM

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I never took Discrete Geometry when in College. I certainly wish I did now, for I would be somewhat familiar with what the literature has of discrete geometry, because I have thrown out all of the continuum along with Reals of Old Math.

And now I am inspecting the old Greek Euclid axioms of point, line, plane as they relate to Discrete Geometry. I suppose that the Ancient Greeks thought of space as a continuum.

Only in modern times-- after 1900 with quantum mechanics can we see Discrete Geometry had any credibility. As Quantum Mechanics gained strength, so did Discrete Geometry in math, in tandem, gain strength.

But today's thoughts are more reflective of Special Relativity, SR. Whenever I say SR, I do not mean what most people mean-- caught up in the goofy silly antics of SR-- age slower-- constant speed of light -- Twin paradox -- space contraction, and all the hullabaloo of cranks. No, when I speak of SR, it is the simple and clear and easy idea that SR is true for a moving bar magnet in stationary coil of Faraday law is the same as moving coil over a stationary magnet yields the same resultant electricity. A explanation that bypasses all the stupid fuss and crankery.

So now, applying SR to that of lines in Decimal Grid Systems. So in 10 Grid it appears that Space is very much open with much holes and gaps between numbers. Like that of a tree orchard where a tree is a number and the open air in between trees the gaps between numbers.

So that when I apply an Ancient Greek Axiom-- 2 points determine a line. I may have trouble finding another point at a specific angle.

Alright, I am thinking that just a casual gaze on these axioms-- two points determine a line, line segment whether in 10 Grid or 100 Grid or infinity Grid 10^604 Grid. A casual run through because the higher the Grids go, the more points in between. At some level of Grid System, the human mind cannot perceive whether it is a continuum from discrete. The Grid system 10^30 would be a system in which it feels as if no holes from one number to next number exists.

And we can have fun in comparing say biology and chemistry and physics to Grid Systems. So what size of Grid do we need where the hole is the size of a bacteria or of a virus? Or what size of Grid System can we say the hole size is of the size of a water molecule or hydrogen atom? Using the 10 Grid as basis grid where 0.1 is 1 millimeter.

So far, what is new is that Grid Systems must have a center, and the center of 10 Grid in 2D is (5,5) and (5,5,5) in 3D.

And this allows me to say that 1 point determines a Space, namely the center point. And that 2 points in space determines a unique plane provided the center point is not one of the 2 points. Then 3 points determine a unique circle in the plane and also a unique ellipse in the plane.

7) Resolving 3 arbitrary noncollinear points determines a unique plane in 3D geometry.

I remember the Hourglass figure was critical in obtaining the unique ellipse.

But I am gathering my wits together to remember how I solved the problem of 3 noncollinear points determine a Plane in Space.

If 2 points a circle center and length (radius) determines a unique circle or sphere for that matter. Then what quantity of points determines a plane.

It is an axiom that 3 noncollinear points determine a plane, but we have a inconsistency if 2 points determine a circle in plane or sphere in Space 3D.

I do not recall how I resolved that issue. For if 2 can do it for circle and sphere, certainly a Plane is uniquely determined by 2 arbitrary points in Space.

I do recall now about the Decimal Grid System calls for a unique center in each Grid System. So that (5,5) is the center of 10 Grid and (50,50) for 100 Grid and in 3D Space it is (5,5,5) and (50,50,50).

Yes, my memory is coming back now.

That we do not need a axiom for unique plane in Space being 3 arbitrary points.

Each Grid System in 3rd dimension space has a center-- for in true math, all our graphs are 1st Quadrant Only.

Thus, the Circle or Sphere needs only 2 arbitrary points and using the center of the grid for the 3rd point.

This eliminates the ugly inconsistency of Old Math-- 3 arbitrary noncollinear points determines a unique plane is an ***axiom*** proof in Old Math when it should be a derived proof. And that is what a Center of a Grid System allows for a derived theorem proof. And we toss out on the trashpile the "axiom in Old Math".

Now I need be careful that 2 arbitrary points in Space (3D) yield every plane possible. So there is quite a challenge here.

I use the center of Space in each Grid System and without loss of generality I talk only of 10 Grid with its center at (5,5,5).

Now given any two arbitrary points, say (1,0,0) and (2,2,2) and I shall use (5,5,5) to form a unique plane.

And so, I make the triangle (1,0,0),(2,2,2), (5,5,5) and then expand that triangle to be a unique plane.

But a hardship arises. In that if I always use (5,5,5) am I not missing many planes?

Alright, apparently the proof of 3 arbitrary points, noncollinear determines a unique plane in 3D Space is locked up in the idea that the 3 points determine a triangle. That will deliver all planes in 3D Space. Whereas my earlier idea of a Center of Grid System does deliver a unique plane, but not all the unique planes of 3D Space.

I am working towards making this a theorem derived proof and rejecting Old Math's idea of shrouding the concept as a axiom itself.

Thus the derived theorem proof is given 3 arbitrary noncollinear points in 3D Space determines a unique triangle. Now in the proof, all I need to do is expand that triangle interior to include the unique plane it rests in. Once that is accomplished I need to prove that all planes in the 3D Space are covered by these triangles.

Alright, well, some progress. What I am chasing after is a theorem derived proof that the plane is made unique by 3 arbitrary noncollinear points, without making that idea be a Axiom in and of itself. By deriving it from other concepts.

So at one end of the spectrum for a proof we have Decimal Grid Systems of Numbers, each having a Center, like that of 10 Grid has center (5,5,5) in a Space. Then given any 2 arbitrary points without the center, I utilize the center along with to form a triangle which forms a plane as I extend the triangle. But trouble is-- are all those planes unique? Certainly the triangles are unique, but not the planes extended from the triangle. Then there is the graver issue of "do I capture all the planes of Space?" And clearly that answer is no, for the center is always involved and you miss many planes of Space.

However, I feel confident the triangle is involved in a derived theorem proof; almost as if the triangle concept is part and parcel of the Plane concept.

Yes, well, at this moment I believe I have overcome these difficulties.

Proof Statement: Given Space of 3Dimension, each set of 3 arbitrary noncollinear points in Space forms a Unique Plane.

Proof: Our Space are the 1st Quadrant Only of Decimal Grid Number Systems. It does have a center for each Grid, such as (5,5,5) for 10 Grid. But we need not use that information. We have 3 arbitrary points in the Grid and not all 3 are collinear. We connect those points to be the vertices of a triangle. Now we do what has never been done in geometry history before. We take each side of that triangle as a line segment and expand it out parallel to the plane of that triangle, and here we do use the Euclid axiom of parallel lines. This expansion of the 3 sides produces a plane in Space of that Grid System and that plane of triangle is now a unique plane in that Grid System. One more question is open. Are all planes in that Grid System created and formed by this method of triangle expansion? And the answer is yes, because all points of the Grid System are available as a 3 point triangle. QED

Some interesting triangles in 10 Grid would be for x,y,z be points (0,0,0), (10,0,0),(0,0,10) which would be a triangle that is half the plane of the xz plane and once expanded is all the plane of xz. Another interesting triangle that forms the plane bearing the Center and is parallel to the xz plane in Decimal 10 Grid is (0,5,0), (5,5,5), (0,0,5).

Commentary: AP was sure that Old Math made a big mistake in the concept of 3 arbitrary noncollinear points was a axiom and not a derived proof. So AP went about looking for a derived proof and found it. Could this have been done without the true numbers of mathematics being Decimal Grid Numbers?? Absolutely not, because Reals as numbers and their continuum and their 8 quadrants in Space get tangled up with the axiom that is needed and essential-- the parallel axiom. For with Reals, infinity of parallelism is contradictory, while Decimal Grid Numbers, each grid is finite allowing for accommodation of the expansion of a triangle side with parallel axiom.

This overcomes the messed up idea that 2 points in Space determines a unique sphere, and 2 points in plane determines a unique circle.

But the Center of Space is a remarkable idea for physics cosmology and astronomy and elementary particles like the proton torus. For it points to a remarkable feature-- a center. Something that Special Relativity physics thought it had got rid of. In AP's physics-- Special Relativity is no more than a moving bar magnet thrust through a stationary coil in Faraday law is exactly the same as a moving coil in a stationary bar magnet produce the same amount of electricity. So, well that definition of AP requires a Center. While Old Physics definition of Special Relativity SR is opposed to a center as their "no absolute reference frame". Which is the start of the tearing down of Old Physics concept of SR.

I have long felt, that Old Physics contorted understanding of SR is not really equivalent (not the same) as AP's definition of SR.

I suppose the worst of Old Physics SR was its Twin Paradox, a kook contrivance to sell kook books. Whereas AP would never stoop to that level of confetti ignorance and write only a short book saying stationary or moving magnet or coil in Faraday law. When you have fools wanting to make fame and fortune and money, there is no bound to which they concoct mindless exciting stories Twin paradox, time dilation, length contraction, when all that needs be said is no different in moving or stationary magnet or coil.

And here is where a Center in a proton torus or a Center in the Cosmos starts to peel away the idiocy hype surrounding Old Physics Special Relativity. Please do not think I am saying Old Physics SR is wrong, it is just taught wrong, in piecemeal parts where the parts are wrong when the whole entire picture is not included, so that the student never grasps what SR is. And you see hundreds if not thousands of worthless failed cranks and crackpots in sci.math and sci.physics arguing both for and against Old Physics SR, and both camps are wrong in their piecemeal understanding and approach, because they cannot wrap their 1/2 marble skull around Twin Paradox, etc.

The proper teaching and understanding of SR has to be one whole unit-- stationary coil, thrusting bar magnet is the same as stationary magnet moving coil in Faraday law. But when you piecemeal that Faraday law, we you get a **un-intelligent** view of what is going on.

The way AP teaches SR, is one holistic, whole unit, not picking at length, not picking on time, not picking on reference frame, and other piecemeal parameters. But taking it holistically in.

Archimedes Plutonium

Jun 3, 2023, 12:58:15 AM

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Now what would be the easiest supporting evidence that the Cosmos has a astronomical center to it?

Recently the astronomers pointed out the Cosmic Ring of Galaxies in what they call the 3rd layer (see Caltech's Jarrett mappings).

Seeing one ring is not evidence of a Cosmic ring, but if it is shown that the mappings contain 3 rings, 2 from the Cosmic Proton of a Plutonium Atom Totality and a 3rd ring inside the 2 others at a perpendicular-- the Cosmic Muon.

And in my latest revised Atom Totality books, I tell astronomers that if they threw out all their Doppler Red Shifts and just moved the galaxies, that they would end up forming 3 Rings of Galaxies. My valid logical reasoning is this. Suppose you have galaxies observed and you use a phony distance marker of Doppler Red Shift. That phony measure will destroy the true pattern except for one. At least one ring survives a phony measurement tool. If you throw out Doppler Red Shift altogether, our mapping ends up with 3 perhaps even 4 Rings of the Cosmic Proton Torus.

Once we have such evidence of 3 rings, then we have proven the Universe has a Center, for a torus has a center.

But there is a far far easier way of showing Physics has a center for the astronomy skies. The law of gravity is inverse square, same as Coulomb force is inverse square. Inverse square in math means a sphere with its center. But a center to the Cosmos does not eliminate the Big Bang theory, for it too can accommodate a center. But the Big Bang cannot accommodate a ring pattern torus.

Archimedes Plutonium

Jun 3, 2023, 2:48:39 AM

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Math failures ... with their 2 different ellipses from the same major axis and minor axis of a ellipse, no wonder they failed geometry when they cannot tell apart a ellipse from Oval in slant cut of cone, for the slant cut is a Oval, yet ... is too dishonest and a math con-artist to admit the truth. No wonder all of .... math education is too dishonest in truth, and that is why no-one in .... can do a geometry proof of Fundamental Theorem of Calculus. How could they when they cannot tell apart a ellipse from oval.

...., the Smithsonian in the USA has a Ellipsograph that is wood,and they shaved the edges as to not be sharp, but by doing so, hides the fact that the slant cut is not a ellipse but an Oval. And .... has a ellipsograph (if not mistaken which also has a ellipse only because it was shaved from a Oval so as not to be a sharp edge. So .... has become idiots of mathematics, no-one there can tell the truth anymore, exemplified by the .... mindless "dark numbers". No math professor in .... can do a geometry proof of Fundamental Theorem of Calculus.

So they have a 3-4-5 right triangle placed leg to leg

Or, how about trying the 45-45-90, identical triangles joined leg to leg

Then they sit at their computer, punching in keys, saying to the computer-- give me two ellipses. ...then shows 2 different ellipses on their computer screen. And this is what .... calls "doing mathematics".

And this is a modern day problem of grave concern, those who are blind in thinking their computer graphics is giving them math results when all it is giving them is Artwork graphics, that lures the daft into thinking they have ellipses covering 4 points.

Now there is a question of why 2 identical right triangles leg to leg give a unique ellipse but other triangles leg to leg allow 2 ellipses to cover their 4 vertex points.

Perhaps the answer lies in the fact that the right triangles mark out not only the major and minor axes-- a uniqueness feature, but also mark out the foci as another uniqueness feature imposed on right triangles as parallelograms but not on say a rhombus as parallelogram.

And here the question has to be, as a corollary in geometry, can there ever be a rhombus composed of 2 right triangles joined leg to leg. Certainly the 45-45-90 cannot be a rhombus, and thus I figure, no rhombus is 2 right triangles joined leg to leg.

Archimedes Plutonium

Jun 3, 2023, 5:14:12 AM

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Alright, I thought I was completely finished with this 3 points determines a unique ellipse. Turns out I have quite a more distance to run with it. I had the oversight, or missed-sight of thinking I had both the major and minor axes when both identical right triangles are placed leg to leg into a parallelogram. I was thinking both axes surely determines a unique ellipse, that was an oversight for the 2 right triangles only determines the major axis and not the minor axis.

That means more distance to run on this proof, and the possibility of failure.

But there are strong signs that I will be triumphant because the square and rectangle and rhombus and then all sorts of parallelograms not 2 right triangles --- they do not mark out the major axis. Only the 2 right triangles mark out the major axis. And the others that have 2 or more ellipses (perhaps that can even be sharpened up to say all parallelograms have only 2 ellipses to cover their 4 vertices), have no axes involved in their ellipses.

What I am considering is whether the perpendicular bisectors of these 2 right triangles are indeed the foci of the unique ellipse. If that is true, then indeed the 2 right triangles determine a unique ellipse.

Archimedes Plutonium

Jun 3, 2023, 2:22:15 PM

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Sorry, I made a mistake here, in thinking the 2 right triangles captured both the Major and Minor axis, but it turns out the 2 right triangles captured only the Major axis. The minor axis is in-between, straddled on both sides by the 2 vertex of short legs joined together.

However, all is saved, capturing both Major and Minor axis if I join the 2 right triangles, first by leg to leg, then I slide the one right triangle down the leg of the other forming a hour-glass

..|/

I still have 4 vertices but no longer parallelogram, instead a hourglass especially with the 45-45-90, not so much with the 60-30-90.

So, many more kilometers to run with this proof.

8) Hourglass figure is a unique ellipse.

Archimedes Plutonium

Jun 3, 2023, 3:00:25 PM

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A glitch has turned up in AP's proof, for he thought he captured both Major and Minor axis. Turns out I have to do a final movement to obtain the unique ellipse to capture both Major and Minor axes. The final figure is much like a hourglass, or two right triangles joined vertex to vertex.

Caution, as you make the final movement, the center of the unique ellipse also appears.

And it appears that the hourglass for the 45-45-90 right triangles ends up being a Circle, then, all other 2 right-triangles in the Hourglass final movement ends up being ellipses, unique ellipses because they carry both the Major and the Minor axes.

Perseverance ---- forced AP to make a final HourGlass movement.

P.P.S. The only thing left for me is to count how many movements of figures to form the unique ellipse. Let me see, here, I, start with 3 points to form a circle. If the circle 3 points has no 90 degrees but is a isosceles trapezoid, I need to cut and paste the trapezoid into a rectangle. Then a second movement of cutting rectangle into 2 right triangles leg to leg. Then a third movement to where the 2 right triangles are a HourGlass forming the unique ellipse. Counting 3 movements, but see if my starting 3 points is saved in all these 3 movements???

Archimedes Plutonium

Jun 3, 2023, 3:47:07 PM

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Now Old Math says that the HourGlass figure is an Irregular Hexagon.

Let me try to draw it here.

..|/

Two right triangles meet in a point, the center of a unique ellipse for it marks out both Major and Minor axes.

But can it be debated that the Hourglass figure is a quadrilateral? Since both legs are on one line?

AP would have thought he was finished with a parallelogram to make the unique ellipse. Turns out another movement into a HourGlass figure is required for the unique ellipse.

The Hourglass nails both Major and Minor axes of unique ellipse and the center of the unique ellipse.

And now, finally, I look upon any and all drawn ellipses, I see in my mind their Major and Minor axis and center, and immediately I see the hour glass figure of 2 right triangles. Make a ellipse, sketch in the two axes and see the HourGlass figure of 2 right triangles.

Archimedes Plutonium

Jun 3, 2023, 9:01:51 PM

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Alright the glaring mistake I made was after finishing the proof, I did not go backwards. In math you often can go backwards, not just forwards. So in this proof I failed to start with an ellipse then work backwards to see the 2 right triangles that form the unique ellipse.

For the best drawing of this there is the book Harold Jacobs, Mathematics A Human Endeavor, 1970, page 305 of a graphed ellipse. His is in 4 quadrants which I try to avoid, but his coordinate points of major and minor axes are (0,9), (0,-9), (0,6), (0,-6).

Now I bring this to your attention because that drawing of the ellipse shows 2 sets of right triangles in a HourGlass pattern.

..|/

And

.....|\

...\|

What I failed to do before now was to keep the 2 right triangles leg to leg like this

Keeping them as a parallelogram composed of 2 right triangles leg to leg.

I should have reviewed the proof and gone backwards. Start with a Ellipse and look for the 2 right triangles.

And now I see there is a 2 Set of 2 right triangles.

So I had another Movement step of moving the 2 right triangles as a Parallelogram into becoming a HourGlass figure.

And that then forces the appearance of a Major axis and Minor axis and thus a Unique ellipse.

One last thing I have to do is see how many of the original 3 arbitrary non-collinear points of the Plane determine a Unique Ellipse are retained. Not that missing one of the three points is at all damaging to the proof, for when I have to cut and paste the isosceles trapezoid when the unique circle does not give me 90 degrees, in the cut and paste, we move one of the original 3 arbitrary points and is lost.

But not to worry because in 3 arbitrary points forming a unique circle, we have to do some movement in perpendicular bisectors to obtain the unique circle center. And so movement of points is not damaging to the proof, for the statement of these proofs is that 3 arbitrary points **determine or force** a unique circle, force a unique ellipse.

So my error over these last few weeks is an error of not going backwards in the proof argument. Starting with a ellipse and looking for 2 right triangles that determine the unique ellipse. A mistake I will not make again.

Archimedes Plutonium

Jun 4, 2023, 2:23:03 AM

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Old Math says these HourGlass figures are irregular hexagons. I would rather classify them as quadrilaterals. Sort of a Rectangle turned inside out.

But I am wondering if perpendicular bisectors can help locate the foci of the ellipse they are inscribed in.

I am looking at the smallest side of these right triangles and taking the perpendicular bisector until it intercepts with the hypotenuse side. Then take 1/2 of the length of the intercept and the perpendicular bisector intersects the major axis of close to where the focus should be.

So here I am playing around with finding the foci of an ellipse from the 4 right-triangles that compose it.

Archimedes Plutonium

Jun 4, 2023, 2:44:22 AM

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Now it has not gone unnoticed by me that one of all the right triangles is very special. The 45-45-90 is special for the hourglass configuration produces a circle, not a ellipse.

And it is fun as an exercise to get out a sheet of paper, take my plastic two 45-45-90 right triangles, lay them on the paper in a HourGlass configuration, trace the perimeter, then haul out the compass and see indeed, verify that the enclosing curve is not a ellipse but a circle.

Yet all other right triangles in the HourGlass configuration build a ellipse not a circle.

This would be an exciting conjecture to prove. Given all right triangles that exist and all ellipses that exist, that they are a 1 to 1 correspondence of existence. What it says is that no ellipse exists unless a right triangle exists that fits that ellipse.

Now that would seem to be an unnecessary conjecture or proof. But it is known in Pythagorean Triples that some numbers do not exist in Primitive Pythagorean Triples, such as the number 6, in primitive P-triples. And P-triples are integers, what about square root P-triples and thus square root ellipses.

So the conjecture is not irrelevant but rather intriguing and one can say that circles are squares nested inside and ellipses are right-triangles nested inside.

Archimedes Plutonium

Jun 4, 2023, 1:19:07 PM

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I suspect I can establish the foci from any ellipse from its 2 sets of right-triangles in a HourGlass figure.

I certainly can get the circle center from the Hourglass figure of 2 right triangles 45-45-90. Just a perpendicular bisector on both hypotenuse meet in the center.

Now looking to see if the HourGlass of 2 right triangles 60-30-90 gives me the ellipse foci.

Archimedes Plutonium

Jun 4, 2023, 8:06:14 PM

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Alright, looking at Old Math, and they already found the method of finding the location of the Foci. And in fact they use the 2 Right Triangles inside the Ellipse, -- I should say 4 right triangles.

The method is Subtract the square of semimajor axis from square of semiminor axis. Finally, take the square root of this difference which is the distance of foci from the ellipse center.

In the case of the ellipse shown in Harold Jacobs page 305 of Mathematics: A Human Endeavor, ellipse with center (0,0), and (9,0), (-9,0), (0,6), (0,-6), the semimajor axis is 9 and semiminor axis is 6, a right triangle whose two legs are 6 and 9. So we have 9^2 - 6^2 = 81 - 36 = 45. Now the square root of 45 is 6.708.... and from the center of Jacobs's ellipse which is (0,0) the two foci are (6.708.., 0) and (-6.708.., 0).

Now let me see if I can get these two foci from the good old Ancient Greek method of just compass, ruler and playing with perpendicular bisectors.

Archimedes Plutonium

Jun 5, 2023, 1:11:15 PM

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This is a hunch, just a conjecture, mind you. The foci are points on the rim of a circle that has equal area as the ellipse area has. In other words. the foci of a ellipse is the circle that is not squashed that equals the ellipse. If we squashed the circle of radius 6.708... it becomes the ellipse of major axis 18 and minor axis 12--- a Hunch on my part.

Archimedes Plutonium

Jun 5, 2023, 1:24:37 PM

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Alright, I have taken out my compass and performed the test upon Jacobs's ellipse on page 305 and it looks reasonable that the circle of radius 6.708... matches the area of the ellipse. If true, then the foci of ellipse are the rim points of the circle that was squashed to produce the ellipse in question.

But, I still need to have a geometry explanation of subtraction of squares then square root of difference.

Archimedes Plutonium

Jun 5, 2023, 7:27:59 PM

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One thought seems to lead into the next adventure.

My very favorite ellipses are pencil ellipses. Long narrow thin ellipses, especially for Light Waves. But my mind a bit, slightly troubled by these ellipses. For can you really make a ellipse to be long slender thin and still be an ellipse???? A troubling question for me. So this recent adventure allows me to probe into that question, by imagining say a ellipse that is 100 long major axis and say 1 wide minor axis. Is such a thing possible and what the foci would be? Then working backwards, what is the original circle when squashed that develops that pencil ellipse.

So here is an exciting experiment to perform, and to see if Pencil Ellipses are just imagination or reality.

9) Pencil Ellipses.

Archimedes Plutonium

Jun 5, 2023, 7:45:31 PM

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On Monday, June 5, 2023 at 7:26:58 PM UTC-5, Archimedes Plutonium wrote:

> One thought seems to lead into the next adventure.

> My very favorite ellipses are pencil ellipses. Long narrow thin ellipses, especially for Light Waves. But my mind a bit, slightly troubled by these ellipses. For can you really make a ellipse to be long slender thin and still be an ellipse???? A troubling question for me. So this recent adventure allows me to probe into that question, by imagining say a ellipse that is 100 long major axis and say 1 wide minor axis. Is such a thing possible and what the foci would be? Then working backwards, what is the original circle when squashed that develops that pencil ellipse.

> So here is an exciting experiment to perform, and to see if Pencil Ellipses are just imagination or reality.

So, Old Math tells us, and I certainly do not doubt them on this algebra. They tell us that I take the semimajor axis and square it, and subtract the square of the semiminor axis. The difference of which I take the square root thereof to find the distance from the center of the ellipse to a foci.

So let me doctor up my pencil ellipse to a better example that is easier for my calculator. A (hypothetical pencil ellipse) of 200 major axis and 2 minor axis, that gives me a 100 semimajor axis and a 1 semiminor axis.

Now the algebra is 100^2 - 1^2 = 10000-1 = 9,999 and so the foci is nearly the same point as the endpoint on the ellipse, 99.9949....

Which tells me one of two things, that my idea of a conjecture in a earlier post saying the foci are two points on the rim of a circle of equal area as the ellipse is false. Or, that a ellipse of these proportions is an impossibility.

So let me ponder this, and I would hazard to guess that a ellipse of these proportions are simply impossible to exist.

Now in my recent book published, I made the bold claim and proof, that all smooth curves come from conics base to base <>, not the silly apex to apex configuration ><. If that claim and proof is correct, then we have two identical cones base to base <> that are pencil cones as right triangles of say 89.9 - 0.1-90 degrees or something close to where one of the angles is almost 90 degrees and these slender thin cones yield pencil ellipses.

That tells me, my idea of the foci being the outer rim points of the circle before squashed is not that circle of equal area to the ellipse. Pencil ellipses can exist and do exist. But my earlier conjecture of a circle of equal area involves the foci of the ellipse is in error.

My 240th published book.

New True Geometry starting with cycloid correction and Geometry-of-Motion // math research

by Archimedes Plutonium

Cover Picture: My photograph of holding two cones base to base. Two cones base to base forms the foundation of Conic Sections, and not the silly apex to apex.

Now there is a cool idea in all of this. That the infinity borderline is where pi has 3 zero digits in a row in 601, 602 and 603 decimal place value.

So our 90 degree right triangle at the infinity borderline is a perfect 89.999..99 at 1*10^603 place value with the other angle of the right triangle being a meager 0.00..01. And so this smallest angle would be part of the longest pencil ellipse and thinest pencil ellipse, way way thinner than even our Old Physics conception of short wavelength gamma rays.

And so, I am pretty sure I need to go back and find a different interpretation geometrically of Old Math's square the semimajor axis subtract square of semiminor axis, then take square root of difference for the distance of the foci from ellipse center.

Alright, sorry, I just picked the wrong parameter involved. I picked area when I was dealing with two points on the rim of a circle which is the circle related to a given ellipse. In my case the Jacobs's ellipse of semimajor axis 9 and semiminor axis 6. I find the foci as 6.708...

So choosing the correct parameter-- circumference. And for circle it is pi*diameter. For ellipse it is a grisly formula, one of the most grisly formulas in simple geometry.

For circle of radius 6.708... the circumference is 3.14.. * 13.416.. = 42.126...

So, I estimate the ellipse circumference and see if they are the same.

Archimedes Plutonium

Jun 6, 2023, 4:09:39 AM

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Well, there are so many calculators on line and one tells me this ellipse circumference is approximately 47.59.. not close to 42.12...

However, I do notice that if I take the circle of diameter 6 and a second circle of diameter 9, I end up with 15 * 3.14 = 47.1 which in sigma error is off only by that of 47.5/47.1 = 0.8% which is liveable in most math houses and professor department chairs. Cognizant of the idea that an ellipse has two radii in action.

I invented the Percentage Method of Geometry, may as well use it. The Percentage Method is such that it recognizes area versus perimeter have the same percentage. So that if a formula of one-- either area or perimeter is a grisly formula, them simply take the easier one and use the percentage to obtain the difficult one.

The Jacobs's ellipse is 9 by 6 semimajor, semiminor axis. The focal point is 6.7...

The area of ellipse is pi*a*b = 3.14..*9*6 = 169.56. The area of rectangle that encloses ellipse is 18*12 = 216. The percentage is 169.56/216 = 78.5%

That means, simply, the perimeter of enclosed ellipse is 78.5% of (2*18) + (2*12) = 36+24 = 60 and 78.5% of that is 47.1 perimeter. And this is the same answer as taking the circle diameter 6 and second circle diameter of 9 gives me 47.1.

I completely bypass the grisly formula of perimeter in Old Math.

Archimedes Plutonium

Jun 6, 2023, 1:29:59 PM

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Let me play with a Pencil Ellipse, an example I gave earlier. My pencil ellipse is 100 semimajor axis and 1 semiminor axis. That would be area of the ellipse at 3.14...*100 *1 = 314. Now the area of rectangle that encloses this ellipse is 2*200 = 400. This is the percentage of 314/400 = 78.5% which is of course pi/4, a very very common percentage with dealing with smooth curves and enclosed in straightline geometry such as rectangle.

But now, avoiding the grisly formula of ellipse perimeter, I simply take 78.5% of rectangle perimeter and obtain ellipse perimeter. So (2*200) + (2*2) = 404. And then 78.5% of that is 317.14. And here I run into a major problem for the pencil ellipse is almost the same as rectangle that encloses it and should be 400.

Checking back with Jacobs's 9 by 6 ellipse, the 78.5% works nice, but here I have a problem with pencil ellipses.

So, well, maybe there is some problem with the existence of pencil ellipses.

Or, there could be problems with the ellipse circumference calculator-- they do not apply with pencil ellipses

C≈400.01

a Axis

b Axis

Solution

C≈π(a+b)(3(a﹣b)2

(a+b)2(-3(a﹣b)2

(a+b)2+4+10)+1)=π·(100+1)·(3·(100﹣1)2

(100+1)2·(-3·(100﹣1)2

(100+1)2+4+10)+1)≈400.01422

Another ellipse circumference calculator gets the same result of 400

(x - c₁)² / a² + (y - c₂)² / b² = 1

a = 100

b = 1

Circumference

400

It may well be that Old Math ellipse has a breakdown-failure (much like in humans with a nervous breakdown) where at some point of a ratio between semimajor and semiminor axis, that Old Math formulas cannot handle those ellipses.

But I am inclined to think that the percentage calculation may breakdown with pencil ellipses. No longer is percentage able to obtain correct numbers. For clearly the pencil ellipse perimeter is almost the same as the rectangle perimeter that encloses it.

Archimedes Plutonium

Jun 6, 2023, 2:31:30 PM

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And this would be a definition of pencil ellipse, where the percentage of area or perimeter with rectangle that encloses ellipse breaks down of its 78.5% comparison. For the perimeter of pencil ellipse starts to approach nearly 100% of enclosing rectangle perimeter.

Pencil Ellipses are so important for physics, for I have them as Light Waves.

Alright, I avoided and staved off talking about eccentricity of ellipse-- how flat or round it is.

Eccentricity = distance between foci/ distance of major axis

If eccentricity approaches 0 then circle round, if it approaches 1 then flat.

So in my above examples 6.7*2 = 13.4, divided by 18 is 0.744 while 199/200 = 0.99.

So perhaps this number 78.5% as pi/4 is the boundary for that of ellipses defined as pencil ellipse and what we consider ordinary common ellipses. A eccentricity of 0.785 is the last of a ordinary common ellipse and any higher eccentricity is a "pencil ellipse".

Archimedes Plutonium

Jun 7, 2023, 1:35:14 AM

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No, sorry the number 0.785 is not flat enough for what I am thinking of as pencil ellipses. I need to get into the 0.96 range for the pencil ellipse. That is in the 3/pi or 0.955 range.

Now perhaps what I really need is that I am the author of the idea that pi is actually square root of 10, number prefix of 3.16... and not 3.14.... If we take 3.14/3.16 and define that as the Pencil Ellipse eccentricity we have 0.99.... And perhaps that is where the pencil ellipse concept comes into action.

10) Exploring Right-Triangles at the Infinity borderline and its Algebraic completeness.

Archimedes Plutonium

Jun 8, 2023, 12:24:14 AM

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Time for some entertainment math as these recent topics elude to.

Previously in math I have explored the regular polyhedrons with the infinity borderline, that a boundary where it is evenly divisible by 5 factorial and even 6 factorial for algebraic completeness at 10^1208.

But I never explored Right Triangles at the infinity borderline where pi has 3 zero digits in a row 1*10^604 and its algebraic completeness at 1*10^1208.

All modern day run-down AI computers like ChatGPT are deaf dumb and silent on math, devoting their time to making up pitiful poems, and squirming in their corner when math is asked of them, so cranks like Geoffrey Hinton and Max Tegmark can overblow their crackpot ideas on AI intelligence-- a danger to the world.

Geoffrey Hinton and Max Tegmark cannot admit the maximum intelligence of AI computers is "anagram intelligence" nothing better. And AP wrote a book on computers maximum intelligence is mere anagram intelligence. And no computer will ever out-best human intelligence. But cranks Hinton and Tegmark cannot resist news and studio limelight. Do not get me wrong, AI is dangerous if placed into the Launch sequence of Atomic missiles, for it is easy for the circuit switches to make the AI launch atomic missiles by accident, when not supposed to be launched. But everything else about AI computers being dangerous is silly nonsense foisted by cranks who want that publicity.

So I once asked ChatGPT to verify the divisibility of pi digits at its 3 zero digits in a row, and the computer ran and hid, run and hide, much like asking Dr.Tao and Dr.Wiles to admit slant cut of cone is Oval, not the ellipse, Run Terry , Hide Andrew Wiles, hide behind your Mamma's apron.

So now, is a new day and new geometry about the infinity borderline-- I am worried about Right Triangles at the infinity borderline with pi's 3 zero digits in a row, and what kind of cone angle that would be a --- 89.9999..99 - 0.000..01- 90 degree angles.

I was even wondering if a Primitive Pythagorean Triple exists at the infinity borderline or does it require a composite right triangle??

For example pi is 3.14159.... and so we immediately have a Primitive P-Triple in (3, 4, 5) and (31, 480, 481), do we have a primitive (314, ... , ...)???

But that is not the order I prefer. I prefer the order be where 3 is the largest number the hypotenuse and the thus the 31 the hypotenuse and thus the 314 etc etc the hypotenuse)

But I can always use Composite P-Triples (3*(3/5), 4*(3/5), 5*(3/5))

So, what is it like for Primitive P-triples at the infinity borderline?? Is it spectacular there as it was spectacular for regular polyhedra with its angles of divisibility by 120 and by 720 at algebraic completeness 10^1208??

Will we find new great surprises at infinity borderline for Right Triangles as we found for Regular Polyhedra.

P.S. do not mind the AI computers like ChatGPT, busy on poems and on writing High School English papers for lazy or frantic High School students getting their paper in on time.

Archimedes Plutonium

Jun 8, 2023, 2:59:10 AM

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3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510

58209 74944 59230 78164 06286 20899 86280 34825 34211 70679

82148 08651 32823 06647 09384 46095 50582 23172 53594 08128

48111 74502 84102 70193 85211 05559 64462 29489 54930 38196

44288 10975 66593 34461 28475 64823 37867 83165 27120 19091

45648 56692 34603 48610 45432 66482 13393 60726 02491 41273

72458 70066 06315 58817 48815 20920 96282 92540 91715 36436

78925 90360 01133 05305 48820 46652 13841 46951 94151 16094

33057 27036 57595 91953 09218 61173 81932 61179 31051 18548

07446 23799 62749 56735 18857 52724 89122 79381 83011 94912

98336 73362 44065 66430 86021 39494 63952 24737 19070 21798

60943 70277 05392 17176 29317 67523 84674 81846 76694 05132

00056 81271 45263 56082 77857 71342 75778 96091 73637 17872

14684 40901 22495 34301 46549 58537 10507 92279 68925 89235

42019 95611 21290 21960 86403 44181 59813 62977 47713 09960

51870 72113 49999 99837 29780 49951 05973 17328 16096 31859
xxxx

50244 59455 34690 83026 42522 30825 33446 85035 26193 11881

71010 00313 78387 52886 58753 32083 81420 61717 76691 47303

59825 34904 28755 46873 11595 62863 88235 37875 93751 95778

18577 80532 17122 68066 13001 92787 66111 95909 21642 01989

38095 25720 10654 85863 27886 59361 53381 82796 82303 01952

03530 18529 68995 77362 25994 13891 24972 17752 83479 13151

55748 57242 45415 06959 50829 53311 68617 27855 88907 50983

81754 63746 49393 19255 06040 09277 01671 13900 98488 24012

85836 160

Basically what I am asking here is whether this integer is a Primitive Pythagorean Triple at the Infinity Borderline as a hypotenuse, or as a Composite Pythagorean Triple hypotenuse at the infinity borderline. Notice I dispensed with the decimal point.

314159 26535 89793 23846 26433 83279 50288 41971 69399 37510

58209 74944 59230 78164 06286 20899 86280 34825 34211 70679

82148 08651 32823 06647 09384 46095 50582 23172 53594 08128

48111 74502 84102 70193 85211 05559 64462 29489 54930 38196

44288 10975 66593 34461 28475 64823 37867 83165 27120 19091

45648 56692 34603 48610 45432 66482 13393 60726 02491 41273

72458 70066 06315 58817 48815 20920 96282 92540 91715 36436

78925 90360 01133 05305 48820 46652 13841 46951 94151 16094

33057 27036 57595 91953 09218 61173 81932 61179 31051 18548

07446 23799 62749 56735 18857 52724 89122 79381 83011 94912

98336 73362 44065 66430 86021 39494 63952 24737 19070 21798

60943 70277 05392 17176 29317 67523 84674 81846 76694 05132

000

Same question for 10^-1208 place value.

I am looking to see if there is any strange geometry at infinity with Right Triangles. I bet there is, because 10^1208 is evenly divisible by 120 and 720 yet it has just one singular zero, not three of them.

Starting 2009 when I was working on Infinity Borderline, I had an adventure of sorts of seeing that the infinity borderline was able to give all the angles of regular polyhedra, if not, it is not a infinity borderline. That meant the borderline was evenly divisible by 120 to include those special angles that make up the regular polyhedron. And to my amazement, where pi had 3 zeros in a row for the first time, it was evenly divisible by 5 factorial equals 120. To my further amazement, the infinity borderline at 1*10^604 with its algebraic completeness at 1*10^1208 was spectacular. Because at 1*10^1208, not only were pi digits evenly divisible by 120 = 5 factorial but a further evenly divisible by 6 factorial = 720, yet, pi digits has but one single lonely 0 digit at that spot. Really amazing.

Today I look into another geometry adventure with the infinity borderline and its algebraic completeness. I look at Right Triangles and the concept of Primitive Pythagorean Triples, Compound Pythagorean Triples. By compound I mean factors of primitive P-triples.

What prompts me into this research is that Physics requires Pencil ellipses for Light Waves are closed loop circuits, not straightline Arrow rays with front and back. No. Light waves are like all EM theory-- a closed loop circuit.

And so, I need the most geometrical closed loop circuit that is like a ray but still closed loop. And I call them Pencil Ellipses, but now I need even more thin and narrow ellipses and call them Needle Ellipses.

Needle Ellipses involve Right Triangles of form of angles of 89.999.. - 0.0...01 - 90 degrees. Almost two 90 degree angles.

And so, what this adventure is all about is a glimpse into what Right Triangles exist at the infinity borderline and its algebraic completeness and what angles can Right Triangles become where one of them is almost 90 degrees along with the 90 degree angle. Is the Infinity Borderline favorable to the existence of a Needle Ellipse is the question.

From my 233rd published book.

Utter garbage of Cantor's work on infinity especially the diagonal method// math focus

by Archimedes Plutonium (Author) (Amazon's Kindle)

Preface: So much of the Usenet newsgroup sci.math are discussions concerning why 0.9999... should equal 1 or about infinity and specifically Cantor's work with his diagonal method. I thought the time is ripe for me to write a independent book on why and how Cantor was wrong, and to lay the subject to rest. For the flaws of logical reasoning in infinity discussions are really abominable and abysmal and deserves an entire book of clarity packed with simple reasoning.

It amazes me how resembling is the science of geography with borderlines makes clear what countries there are, yet the borderline concept was so foreign to all mathematicians until AP discovered the infinity borderline starting 2009.

Cover Picture: My iphone photograph of two world globes showing countries with borderlines. Borderlines are beacons of clarity from one concept as it transitions into a newer concept.

--- end quoting my 233rd book ---

I have pinpointed that borderline from tractrix-circle analysis, from algebraic analysis of algebraic completeness, and from angles of regular polyhedra. The borderline in microinfinity is 1*10^-604 and in macroinfinity is 1*10^604.

The easiest way to see the borderline is to see where pi digits ends in a three zero digits in a row.

3.141592653589793238462643383279502884197169399375105820974944592307816406286 208998628034825342117067982148086513282306647093844609550582231725359408128481 117450284102701938521105559644622948954930381964428810975665933446128475648233 786783165271201909145648566923460348610454326648213393607260249141273724587006 606315588174881520920962829254091715364367892590360011330530548820466521384146 951941511609433057270365759591953092186117381932611793105118548074462379962749 567351885752724891227938183011949129833673362440656643086021394946395224737190 702179860943702770539217176293176752384674818467669405132000

Archimedes Plutonium

Jun 8, 2023, 6:11:35 PM

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First I need to ask the question of whether it is impossible for the hypotenuse of a Primitive Pythagorean Triple to be an even number. I know two of the sides have to be odd, and one even. But can a hypotenuse in PPT be even is the question.

If impossible would mean that the right triangles at infinity borderline are Composite Pythagorean Triples.

This does not hurt my speculations of these Needle Right Triangles, but rather instead, hurts Old Math's concept of Primitive Pythagorean Triples. Hurts Old Math because their number system of Reals, starting from integers is myoptic, and confused.

In the AP's Decimal Grid System of true numbers of mathematics, we have a mix of small numbers along with large numbers. We do not start with just bare raw integers. For example the 10 Grid is 0, 0.1, 0.2,0.3,...9.8, 9.9, 10.

So the concept of P-triples in AP's Grid system does not start with (3,4,5) triple but rather, the smallest triple in 10 Grid is 3*.6 = 1.8 and 4*.6 = 2.4 and 5*.6 = 3 so the smallest P-triple in 10 Grid is (1.8, 2.4, 3) rather than Old Math's smallest P-triple as (3,4,5)

So in New Math, of the true numbers of mathematics being Decimal Grid Systems, our smallest P-triple is (1.8, 2.4, 3) but still the hypotenuse is not a even number but here we have two of the sides as even and only one side as odd.

So we need a massive overhaul of Pythagorean Triples. Getting away from the concept of Primitive P-Triples as more of a stumbling block than insight.

For the objective is P-Triples when the true numbers of mathematics are Decimal Grid Numbers. And where mathematics has an infinity borderline at 1*10^604 with algebraic completeness at 1*10^1208.

Already I note a major observation. That the pi digits are evenly divisible by 120 at 10^604 place value. That means, evenly divisible by 3, by 4, and by 5-- Old Math's smallest PPT (3,4,5).

But I am after Needle Ellipses of right-triangles whose angles are almost both angles be 90 degrees, a 89.999...99 angle.

So in this research, I want to dig down into whether a right triangle exists with almost two 90 degree angles, and the third angle a mere 0.000..01 degree. In New Math, there are constraints and borderlines and forbidden angles, in Old Math anything goes, anything exists with their contradictory Reals.

For example, in 10 Grid, the smallest possible P-triple is the (0.3, 0.4, 0.5) borrowing from the 100 Grid 0.09+0.16= 0.25, which does not form a Needle right triangle to form a Needle Ellipse. I have to go to right triangles like that of (23,264,265) for the outlines of a Pencil Right triangle to form a Pencil Ellipse.

So here we begin to see where we what a small tiny number then two large numbers in Primitive Pythagorean Triples.

And this is a question of the Infinity Borderline, what is the largest numbered P-triples with a smallest one side to form a Needle Right Triangle and a Needle Ellipse.

Alright, in the 100 Grid, the largest Primitive P-Triple is (65,72,97) and that forms a right-triangle which in turn forms a regular normal Ellipse. But the largest Pencil Ellipse in 100 Grid appears to be (13,84,85) which has an angle from protractor of 85-5-90. Is it a Pencil Ellipse from a Pencil Right Triangle? So we have 85/90 = 0.94 and I define a Pencil Ellipse from square root of 10 divided into pi as 3.14/3.16 = 0.993. So it is almost a Pencil Ellipse from a Pencil Right Triangle.

What do I define a Needle Ellipse from a Needle Right triangle? I am going to define it in terms of Integers as being a Right triangle of at least 89-1-90 degrees, where 89/90 = 0.9888...

Are there any Old Math Primitive Pythagorean Triples that come close to 0.9888.... I doubt it.

But perhaps with Infinity borderline at 1*10^604 and algebraic completeness 1*10^1208, that perhaps the first such Needle Right Triangle which forms a Needle Ellipse is found.

This is exciting adventurous mathematics. Not like the stifling and stupid Old Math and its Reals continuum nonsense. And to see how silly Old Math was, --there at the beginning of the 20th century, the early 1900s, physics was abuzz with Quantum Mechanics-- discreteness. And yet the silly Old Math-- still chasing after more continuity with Cohen's mindless continuum hypothesis, never paying any attention to the stunning new physics of discreteness was producing. Old Math is at least 123 years behind the learning curve.

Archimedes Plutonium

Jun 9, 2023, 12:42:21 AM

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Through the years I kept a notebook of Primitive Pythagorean Triples. And here they are.

(3,4,5) (5,12,13) (7,24,25) (8,15,17) (9,40,41)

(11,60,61) (12,35,37) (13,84,85) (15,112,113) (16,63,65)

(17,144,145) (19,180,181) (20,21,29) (20,99,101) (21,220,221)

(23,264,265) (24,143,145) (25,312,313) (27,364,365) (28,45,53)

(28,195,197) (29,420,421) (31,480,481) (32,255,257) (33,56,65)

(33,544,545) (35,612,613) (36,77,85) (36,323,325) (37,684,685)

(39,80,89) (39,760,761) (40,399,401) (41,840,841) (43,924,925)

(44,117,125) (44,483,485) (48,55,73) (48,575,577) (51,140,149)

(52,165,173) (52,675,677) (56,783,785) (57,176,185) (60,91,109)

(60,221,229) (60,899,901) (65,72,97) (68,285,293) (69,260,269)

(75,308,317) (76,357,365) (84,187,205) (84,437,445) (85,132,157)

(87,416,425) (88,105,137) (92,525,533) (93,476,485) (95,168,193)

(96,247,265) (100,621,629) (104,153,185) (105,208,233) (105,608,617)

(108,725,733) (111,680,689) (115,252,277) (116,837,845) (119,120,169)

(120,209,241) (120,391,409) (123,836,845) (124,957,965) (129,920,929)

(132,475,493) (133,156,205) (135,352,377) (136,273,305) (140,171,221)

(145,408,433) (152,345,377) (155,468,493) (156,667,685) (160,231,281)

(161,240,289) (165,532,557) (168,425,457) (168,775,793) (175,288,337)

(180,299,349) (184,513,545) (185,672,697) (189,340,389) (195,748,773)

(200,609,641) (203,396,445) (204,253,325) (205,828,853) (207,224,305)

(215,912,937) (216,713,745) (217,456,505) (220,459,509) (225,272,353)

(228,325,397) (231,520,569) (232,825,857) (240,551,601) (248,945,977)

(252,275,373) (259,660,709) (260,651,701) (261,380,461) (273,736,785)

(276,493,565) (279,440,521) (280,351,449) (280,759,809) (287,816,865)

(297,304,425) (300,589,661) (301,900,949) (308,435,533) (315,572,653)

(319,360,481) (333,644,725) (336,377,505) (336,527,625) (341,420,541)

(348,805,877) (364,627,725) (368,465,593) (369,800,881) (372,925,997)

(385,552,673) (387,884,965) (396,403,565) (400,561,689) (407,624,745)

(420,851,949) (429,460,629) (429,700,821) (432,665,793) (451,780,901)

(455,528,697) (464,777,905) (468,595,757) (473,864,985) (481,600,769)

(504,703,865) (533,756,925) (540,629,829) (555,572,797) (580,741,941)

(615,728,953) (616,663,905) (696,697,985)

And the sense I get from them is a staircase leveling of where I am not going to find something like (7, ??, 999) or (11,??, 999) but that the hypotenuse length is going to force a minimal shortest leg length.

And that at the infinity borderline or its algebraic completeness may demand and force the largest Pencil Right Triangle, and even, perhaps a Needle Right Triangle.

So now, let me play around with a hopeful Needle Right Triangle in the 1000 Grid where I assume that 999 and 997 can form the two largest sides of a right triangle that would be 998001 with 994009 for a difference of about 4000 which the square root is roughly 63.

So, suppose there is a Right Triangle of 999-997-63 and that would be 997/999 = 0.99799... And I defined a Needle Right Triangle to be 89.99/90 = .999888.... So definitely not far off.

At the Infinity borderline or its algebraic completeness, is there such a Right Triangle existing which is a Needle Right Triangle?? And not just my "let us assume".

Alright, for the time being I am putting this thread and its ideas to rest, leaving it with the idea that Needle Ellipses from Needle Right Triangles have some special connection at infinity borderline because it is evenly divisible by 120 and 120 is 5 factorial, having 3,4,5 as factors. Also 6 factorial comes into play at 10^1208 algebraic completeness. I have many other books to write before I get to 248th, and today writing 243rd. In the interim time, I am confident I can sort out this Needle Right Triangle at infinity borderline.

Alright, for sure I know that the 3-4-5 Right Triangle exists evenly divisible at both Infinity borderline and at algebraic completeness, because both spots are evenly divisible by 120 = 5 factorial.

That means I have a Right Triangle fully existing at Infinity borderline that is a multiple of 3-4-5.

Now I want a Needle Right Triangle at Infinity Borderline. And what I do for that, is simply move over by the amount of 1*10^-604 for a needle Right Triangle at the Infinity Borderline.

move over to the left of | by the amount of 10^-604.

Leaving me with a right triangle of 90 degrees -- 89.99..99 degrees --- 0.000..01 degrees.

And a hypotenuse of square root of ( 4^2 + (10^-604)^2)

11) Return to the math-derived-theorem-proof of 3 arbitrary, noncollinear points in Space determines a unique plane.

AP's 253rd book of science: The math proof: 3 arbitrary points in Space, not collinear, determines a Unique Plane and the nonexistence of 4D // Math research

Archimedes Plutonium

Aug 15, 2023, 2:25:26 PM

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AP's 253rd book of science: The math proof: 3 arbitrary points in Space, not collinear, determines a Unique Plane and the nonexistence of 4D // Math research

You see, I am trying to recollect what I had written in May of this year, and put off writing this book until now in August of 2023.

In the title I had written I found another proof that dimension of Space can only be 3rd dimension at tops, and cannot be 4th or higher.

So I am trying to recollect why I said that.

I know for sure I was very unhappy with making 3 arbitrary points in Space, noncollinear, form a unique plane as being a Axiom-Postulate. Very unhappy with that idea, and thus I found a derived theorem proof.

I do recall now the ellipse requiring just 3 arbitrary noncollinear points in Plane to form a unique ellipse. And that was resolved by a HourGlass type figure of 2 right triangles, smallest legs joined, one reversed and moved to one vertex point.

But I am trying to jog my memory as to how all of this proves Space can be 3rd dimension at tops.

It must have something to do with the fact that 3 arbitrary points, noncollinear in Space forms a Unique Plane, and so if Space had 4th dimension--- there then exists no way this plane can enter or intercept a 4th dimension.

So I am trying to jog my memory.

Apparently I had no proof of no 4th dimension and beyond, for I did not have a proper proof of theorem derived 3 arbitrary points, noncollinear determines a unique plane.

But now that I do have that proof, I can go on to prove Space has to be 3rd dimension and no higher.

And the proof of no 4th dimension space and higher is the fact and idea that how does a plane traverse or lie in this higher dimension. We can think of a 3rd dimensional space as being a packaging container full of Planes. A volume of planes makes 3rd dimension. If we had a 4th dimension, none of our planes picks up any added dimension.

So now, in 10 Grid, in 3rd dimension, we have 100 points on x-axis and 100 on y-axis and 100 on z-axis, not counting zero (0,0,0). In total that would be 100^3 = 1,000,000 coordinate points or 10^6. Now how many possible arrangements are there of 3 distinct points? Is it the horribly large number of 1,000,000 factorial? No, for that is permutation math. Here I have Combination math and we have 1,000,000*999,999*999,998/ 3*2*1 = some huge number, not as bad as 1,000,000 factorial.

Now in my earlier proof, each triangle yields a unique plane, but many different triangles are also in that same plane. And that is a stumbling block I fell on before. Each triangle does produce a unique plane for that triangle, but that plane once formed carries many different triangles.

Now, how do I separate out all the triangles that lie within one plane in the 10 Grid? How do I extract the number of unique planes in 10 Grid?

So I have this number as the total number of unique triangles in 10 Grid= 1,000,000*999,999*999,998/ 3*2*1 reduced to 1,000,000*333,333*499,999 which is the number of unique triangles in the 10 Decimal Grid.

Now in the 10 Grid, how many triangles would each plane on average carry? If I examine the xy plane, a flat plane as representative of an average plane I am looking for an average number of triangles that forms that plane through expansion of its sides via the parallel axiom. So here is a duplicate Combination problem. 10,000*9,999*9,998 / 3*2*1 = 10,000*3,333*4,999. And now I divide 1,000,000*333,333*499,999 by that of 10,000*3,333*4,999 to arrive at an approximate average number of distinct planes that reside in the Space of a 10 Grid. This is approximately 100*10*10 = 10,000. Is this a surprise?? That the total number of points in 10 Grid 3D Space is 1,000,000 and the total number of planes in 10 Grid is approximately 10,000.

Does that look right?

So then, back to the question of a 4th Dimension?

Archimedes Plutonium

Aug 16, 2023, 4:05:46 AM

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Alright, when I started writing this book back in May 2023, I thought I discovered a new proof that 3rd dimension was tops and there was no 4th or beyond. I based that on the idea each Grid system had a center, such as 10 Grid was (5,5,5) and 100 Grid was (50,50,50). Turns out that was not a proof.

But I still have a proof of no 4th dimension or higher based on this proof of 3 arbitrary noncollinear points in Space forms a unique plane. I arrived at that proof as a derived theorem using the Axiom of parallelism, by extending the sides of every triangle in 3D Space to construct a unique plane from that triangle. In essence, I replaced Old Math's proof of 3 arbitrary noncollinear points in Space forms a unique plane. Old Math had that as a axiom itself. So their proof was not a proof at all but a axiom.

The real proof as derived uses the parallel postulate of Euclid.

In Harold Jacobs GEOMETRY, 2nd edition, 1987, he lists the postulate 3 on page 51 saying "Postulate 3 If there are three noncollinear points, then there is exactly one plane that contains them.

On page 204, "Postulate 10 (The Parallel Postulate) Through a a point not on a line, there is exactly one line parallel to the line."

So, well, what AP did, by constructing 3 points as triangles and proving 3 arbitrary noncollinear points determines a unique plane, is AP proved it from using the parallel postulate and thus we do not need postulate 3. It is redundant.

And it is here, the parallel postulate itself denies the existence of 4th dimension or higher. So in one instance I throw out an axiom in Old Math Geometry, and use the Parallel axiom to prove it as a theorem.

Yet this same parallel postulate is going to throw out the 4th dimension and higher. And the way it does that, through consistency.

A 4th dimension or higher does not give a unique line parallel to given line from a point outside the given line.

There is a better way of stating Euclid's parallel postulate in the idea that a line parallel to another line is a perpendicular on one is a perpendicular on the other.

____________ B

____|________A

Every point on line A when drawn a perpendicular is a perpendicular on line B.

In 3rd dimension we have x-axis perpendicular to y-axis and we have z-axis perpendicular to x-axis and we have z-axis perpendicular to y-axis. We have 3 perpendicular tests. When we talk of 4th dimension we need a perpendicular to x and y and z. But no new axis is available to be perpendicular to x,y, z. But say there was such, some would call it a hidden dimension, or a curled up dimension. But the trouble is, it would destroy the Parallel Postulate Axiom.

So the reason dimension ends at 3rd, is that higher than 3rd destroys the Parallel Postulate.

And this is a new method of proof, a proof by consistency, that when a statement is made to be proven, and if it contradicts an existing axiom, it is a false statement.

Now do not confuse this proof method as Reductio Ad Absurdum, for that is not a valid proof method. Proof by Consistency is a class of itself, and not reductio ad absurdum.

Archimedes Plutonium

Aug 16, 2023, 11:36:12 AM

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Alright, very much satisfied with this, as I have thus removed one Axiom (Postulate) from Euclid's arsenal.

And I cannot help but notice that triangles are the most primitive closed structure and whether the most primitive closed structure of Plane Geometry causes there to be the highest dimension of 3, as the triangle has 3 sides. So that if the most primitive closed structure were a square with 4 sides then the dimensions of geometry would have a 4th dimension.

Now I need to spend some time on this idea of given a triangle in Space 3D how to expand the sides with parallel lines extension to form the unique plane that the given triangle creates. Remembering that there are holes between successive numbers in Grid System.

Archimedes Plutonium

Aug 16, 2023, 6:00:18 PM

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So in cleaning up the Euclid postulates, it is better to state the Parallel Postulate as a perpendicular from one line is the perpendicular to the other line which then causes the recognition of triangles, the right-triangle and the Pythagorean theorem, directly in the set of postulates. This, thus, eliminates the need of a postulate saying given any 3 arbitrary points in Space forms a unique plane. For the plane is created by the triangle in the plane and then a parallel postulate extension of the 3 sides of the triangle.

In a sense, parallel lines, one to another is part of the uniques plane formation extending a side of a triangle, a right triangle in this instance.

So in building the axioms of plane geometry, the triangle figure needs to be built within the set of axioms.

Archimedes Plutonium

Aug 16, 2023, 8:51:32 PM

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Now I never expected that this book would end up as a vast clean-up of Euclid plane geometry axioms. For honestly I was simply after a proof of 3 arbitrary noncollinear points forming a unique plane and Space was at tops just 3rd dimension.

But as it stands, in proving those two, I come across having to Overhaul all the Euclid axioms. I am rather glad and amazed this has come about. More so because when you have the true numbers of mathematics be Decimal Grid Numbers with holes in between one number and the next number, that a vast clean up of Euclid axioms is inevitable.

So what I managed to do is find a Derived Theorem Proof of 3 arbitrary noncollinear points in Space forms a unique plane with a special tool which I call extension of a triangle side by successive parallel lines and thus a triangle forms the unique plane as the triangle is so to speak-- The Miniature Plane -- for which the Parallel Postulate of Euclid (its equivalent statement as perpendiculars) extends the triangle plane to form the "Greater Unique Plane".

So, well this then subtracts or eliminates one of Euclid's axioms and shows us how valuable the parallel postulate is.

It also indicates that in the 1830s and beyond with the creation of the NonEuclidean geometries was a silly and stupid adventure and a failure. For when jiggling around with the Parallel Postulate with Reals as numbers you do end up with rather nasty mistakes of thinking there is Lobachevsky and Riemann geometries distinct from Euclid geometry. Not true. The Euclidean Geometry is the only single geometry and is seen as this formula.

Euclidean geometry = the summation or union of Riemann geometry add on or unioned with Lobachevsky geometry.

The fools and idiots of Old Math geometry failed to recognize that you can make positive curvature ) and union that with negative curvature to end up with flat plane Euclidean geometry.

NonEuclidean geometry was a wastrel bifurcation of one single true geometry-- Euclidean. Lobachevsky geometry is a subset, yes a small subset of Euclidean geometry and is not a distinct separate independent geometry, same goes for Riemannian geometry.

12) Discrete Space and Euclidean Plane Geometry.

Archimedes Plutonium

Aug 16, 2023, 11:53:41 PM

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So, well, throughout the history of mathematics, in particular geometry from Ancient Greek times forward has been a one-sided lopsided idea to continuity, continuity continuity with only a fringe look at discrete. For the numbers of math from Ancient Greek times forward were the progressive build-up of the Reals, with their rag tag bag of lumping one group of numbers, cobbling them together-- Naturals add Rationals add Irrationals add Imaginary add whatever.... just cobbling. And yet by 1900, physics walked the opposite direction-- discrete in quantum mechanics. Of course the fools of mathematics kept walking continuity with Reals.

In a world of continuity and Reals, there would be no hope that someone can spot that the Euclid Postulate 3 arbitrary noncollinear points in Space forms a unique plane, no-one in a continuity driven math agenda would spot that triangles extended by the Parallel Postulate forms the unique plane in Space. But someone-- AP -- who sees all of geometry is discrete would spot that fact needs a derived theorem proof.

And so how does AP see a triangle in Discrete Space and its Parallel Postulate extension building the plane?

So here I draw a right triangle in Discrete Space of coordinate points in 1st Quadrant Only of the Pythagorean triple 3,4,5

(3,0,0) and (0,4,0) and (0,0,5)

Now I connect those three points in Space of 10 Grid. There are holes and gaps in between every number and its successor.

So in Space of 10 Grid I have 3 arbitrary points forming a right-triangle, and how does this right triangle determine a Unique plane in 3D Space of 10 Grid?

The triangle itself is a "triangular plane" containing discrete points inside and all these points are connected by straightline segments with themselves and the 3 vertex points.

Now we look at each side of this 3,4,5 right triangle.

Say we look at the side with length 3. Now we run a line from any interior or vertex points of the triangle to go outside of the triangle interior. And any lines outside triangle interior that are parallel to the side of length 3 is part of the larger extended plane of the triangle plane.

I should have done this sample problem in 2nd Dimension first before doing it in 3D. Basically I have a Right Triangle plane, and now I use the Parallel Postulate to extend that plane to encompass the more general plane in 3D Space.

Now some may think that the perpendicular can make the extension by Parallel Postulate tool, make the extension start to so to speak "curl up or down" in 3D Space. And here is where we have to make sure of proper usage of the Parallel Postulate tool. That lines outside the triangle do not start curling up for they do not intersect with all the triangle interior plane lines.

Years ago, in one of my books I wrote words to the effect that the Euclid postulates need a vast overhaul in light of the fact that geometry is discrete and the true numbers are Decimal Grid Numbers. There is no way I could do such a overhaul by thinking-- let me sit down and do a Euclid Overhaul. No, the proper way such a overhaul can be executed is by the example of this book-- I stumble upon a error of Old Math Geometry, and in the process of fixing the error, I see what needs to change in the Ancient Greek postulates. It is by a stumbling happenstance that we see the errors and begin to fix them.

Archimedes Plutonium

12:09 AM 17Aug2023

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So in a Overhaul of Euclid Axioms (some prefer to call them Postulates), we should define what a triangle is. We have the first Axiom as "there is a point" and the second axiom as "there is a line". So by using axiom 1 & 2 we Define, not make into another axiom, but define intersection of line segments and one of 3 line segments intersecting in plane and call it a triangle. Define particular triangles with a angle of 90 degrees.

A Axiomatic System is better when it has as few of axioms as possible, and most of its work is from derived theorem proofs, and by definitions.

So this book ends up being a big thrust forward into Overhauling the Euclid Plane Geometry.

And I leave this book with again a tantalizing thought. Is the fact that the most primitive closed and straightline figure in geometry as the triangle, with its 3 sides, is that linked to the fact that Geometry at most is 3rd dimension? The most primitive closed figure portends and forces the dimension of Space be the same number-- 3???

To answer that I need to see how geometry plays in the laws of Electromagnetism and so the AP-EM Equations.

And I would affirm, that if the smallest most primitive closed figure of geometry requires 3 line segments, then the Space of dimension is topped at 3, no 4 and beyond.

If the smallest primitive closed figure were a square, then the top dimension of Space is 4th and not 3rd.

Yes, I should have done this first in 2D, rather than 3D to see exactly how the Parallel Postulate Extension Tool works.

So in 2D with only the xy axis, and my 3 arbitrary points (0,0), (3,0),(0,4) forming a right triangle 3-4-5. And this right triangle contains its side points in 10 Grid such as (0, 0.1) then (0,0.2) etc and (0.1, 0), (0.2, 0) etc and interior points of right triangle (0.1,0.1) etc

Now look and examine a point outside the Right Triangle Plane for I call all the points of the Right Triangle its side points and interior points as a Triangle-Plane.

And this is how Euclid Plane Geometry should teach and educate us on the concept of "plane" where we first encounter it as a "triangle plane". And only with collaboration with the Parallel Postulate do we extend the "triangle plane" as a general plane in a specific Grid System, such as I am doing in the smallest Grid Decimal 10 Grid.

So a point outside the 3-4-5 Right Triangle Plane is a point for example such as (5,5) which is the center of the xy 10 Grid plane. So now, I must show that (5,5) is in the unique 3-4-5 triangle plane that constructs the unique xy plane in 3D Space of xyz. The Parallel Postulate easily does this feat, as I compose a new triangle plane with point (5,5) using (3,0), (0,4) and (5,5) forms a new triangle plane. And now the crucial question-- what is the intersection of this new triangle plane with the original triangle plane 3-4-5? And the answer is that all of both triangle planes intersection is the larger union plane.

Contrast that intersection in 3D of the point outside the 3-4-5 triangle plane of (5,5,5), the center of 10 Grid xyz 3D. So I use the three points (3,0,0), (0,4,0), (5,5,5) and form a second new triangle plane in 3D and ask what is the intersection of both triangle planes? Here the intersection is merely a line and not both planes. Here the intersection is the line what is the hypotenuse line of the 3-4-5 right triangle and the intersection is not all of the 3-4-5 triangle plane.

So this type of construction plays out on all points outside the given arbitrary 3 points in 3D, where the Parallel Postulate Tool finds a Unique Plane given those 3 arbitrary noncollinear points. The original triangle plane is Extended by the tool to form a Unique plane a more general plane than the triangle plane in all of 3D.

Now, well, I have already shown that the triangle of 3 sides as the smallest closed figure in geometry demands and forces the maximum dimension be the same as the number of sides 3. This eliminates 4 and higher dimensions in geometry as we destroy the Parallel Postulate by thinking there is a 4th or higher dimension. You can talk about 4th dimension and crazily imagine a 4th perpendicular axis to the 3 given axes xyz, but if you pursue further the crazy idea you destroy the Parallel Postulate.

But now another question arises, can this Plane Construction using Parallel Postulate Extension Tool also prove that 1st Quadrant Only in geometry exists and that 4 quadrants in 2D or 8 quadrants in 3D are also whacko crazy imagination run amok???

I believe so. I believe in the same manner that this tool destroys 4th and higher dimension, that this tool destroys all quadrants with negative numbers, as to say, negative numbers are a delusion, a crazy idea in geometry.

13) The Overhaul of Old Math's Euclidean Geometry.

Archimedes Plutonium

Aug 18, 2023, 3:10:47 PM

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Alright, well I need to start revamping all of Euclid Plane Geometry. This of course cannot take place in one sitting, but over a long period of time. And in the end I probably will write a math textbook series titled TEACHING TRUE GEOMETRY.

The source for this overhaul of all of Geometry is the fact, a plain fact that the true numbers of mathematics are discrete numbers of Decimal Grid Systems where each number has holes and gaps to the next number. Throughout the history of geometry, it was continuum, continuous geometry. In true geometry, it is discrete, and it is sad that the mathematicians were too stupid and more so, too arrogant to learn from the physics community starting the year 1900 when quantum mechanics physics took off in leaps and bounds in discovery that the world is discrete, not continuous-- because the word "quantum" means discrete. But no, mathematicians the world over from 1900 onwards were arrogant, deaf, dumb, and blind as they pursued more and more continuity, and the physics community went in the opposite direction.

Today I start with revamping the representation of Angles in True Geometry.

Starting with the Decimal 10 Grid, the smallest Grid System.

|_________________________

0 1 2 3 4 5 6 7 8 9 10

Now there are exactly 100 points on x-axis not counting 0, same goes for y-axis

Now in Old Geometry their angles went from 0 to 360 degrees, not a bad system, but New Geometry is better.

There are 100 angles to compose 45 degrees, then there ends up being 200 angles to make 360degrees.

The smallest angle is 0.1degree and 45degrees is angle 10 in New Geometry. The angle below 10 is 9.9degrees. The angle above 10 is a bit more than 45degrees and is denoted by angle 10.1.

On the extreme rightside of the above graph, if we make a perpendicular and mark each number from 0.1 to 10 and draw a line from (0,0) to that point we get all the angles from 0 to 45degrees. Now we draw another line across the top parallel to x-axis to get angles 45degrees to 90degrees. And these angles are 10.1,10.2,10.3,....15.0,15.1,15.2,.......19.8,19.9,20.0.

So what is angle 15.0 if 20.0 is 90degrees and 10.0 is 45degrees? Well, halfway between 45 and 90 would be 45/2 = 22.5 and thus 90-22.5 = 67.5degrees in Old Math Geometry but in New Math Geometry that is angle 15.0.

This is what Discrete Geometry does to angles-- makes them discrete. Now to get smaller angles, one has to go to the 100 Grid if they require finer measuring.

Now this of course changes Trigonometry when we start to write angles as Grid Numbers.

In the 10 Grid there are 200 angles in all, not counting 0 angle.

A more tough question to answer is this, there are 100 points on x-axis and 100 on y-axis, not counting (0,0) and thus 100*100 = 10,000 points in all, and the question is, how many possible lines exist in 10 Grid, when the definition of line is that of 2 points determine a line. That question is in 2nd dimension and becomes even tougher to determine in 3rd dimension. Once that is done we have the question of how many planes exist in 3D 10 Grid? And then finally, how many solids, once we define solids, exists in 3D 10 Grid.

So in Old Math Geometry, they had it easy on questions of how many points, how many lines, how many planes and how many solids. Because in Old Math Geometry it was all wrong. And their answer is always infinity, infinity and more infinity, since their numbers were continuum and continuous. In New True Geometry, each Grid System has a specific amount of points, of lines, of planes and of solids (if we can define solids, well-define them).

In an earlier post, I managed to gain a total amount of planes that exist in 10 Grid of 3D as the number 10,000 only I need to verify that as it is easy to make a miscalculation.

Archimedes Plutonium

Aug 18, 2023, 3:47:57 PM

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Now here I have to backtrack a little to ask of my method of using the Parallel Postulate Tool.

In essence, what I did to prove the theorem-- 3 arbitrary noncollinear points determine a unique plane is that I in fact, posited that the triangle itself was the unique plane and that the Parallel Postulate only extended the sides of the triangle to encompass more of the surrounding space.

So, well, could I simply have stopped at the fact that a triangle itself defines a plane as its interior, and it is obviously unique as the 3 points are unique.

Or do we have to define a plane as being a 4 sided figure, a parallelogram or a rectangle or square. The Cartesian Coordinate System is underlying foundation is a square and rectangle of points.

And here, I am going where no-one has gone before. The 2D geometry, the plane geometry with 1st quadrant only, is indeed a square. But its angles are indeed a triangle.

Does this mean the smallest plane is a square? Does it mean the triangle although it has interior and sides, is not a plane itself, but embedded in a plane?

Does it mean all planes must be 4 sided figures? And although a triangle is a not fully developed or grown plane.

Basically I am rehashing the idea that the Parallel Postulate Tool is used to fully grow the triangle plane to be embedded inside a larger plane.

It is easy to see in 3D that we have 3 arbitrary noncollinear points and to see it is a Triangle in 3D. Do we need the Parallel Postulate Tool to then extend that triangle plane to mark out the full 3D plane of a 4 sided figure?

The smallest triangle in 2D is a triangle of just 3 points as vertices and no interior points at all, (0,0)(0,0.1),(0.1,0), in 10 Grid. And then once extended, encompasses all of the xy plane.

I suspect that the Parallel Postulate is essential in extending the triangle-plane.

Archimedes Plutonium

Aug 18, 2023, 11:28:21 PM

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So in 3D, with 1st Quadrant only and 10 Grid, with 100 points on x-axis and then 100 on y-axis and 100 on z-axis, can be seen as a cube box. Are all the planes that of 4 sided figures??

So I got out my plastic geometry figures and quickly see that some planes are pure triangles, 3 sided figures not 4 sided figures. Where no application of the Parallel Postulate Tool is needed for the vertices of this triangle cannot extend the triangular plane any further.

This means the triangular plane must be the unit plane in Geometry and that 4 sided planes are extensions of triangle planes. The 4 sided plane is then seen as a luxury, an extra big plane when the basic plane is the triangular plane.

So in Euclid Geometry Postulates, we begin with (1) There exists a Point-- describe the point; then (2) There exists a Line -- describe the line as built from 2 points. (3) This need not be a postulate but a definition of the triangle-- as 3 points not collinear joined pairwise of those 3 points forms a triangle. By a theorem proof we can prove it is unique triangle-- due to the 3 points are unique. Next, (4) another more postulates that include the parallel postulate.

Now with the Parallel Postulate we prove as a derived Theorem that 3 arbitrary noncollinear points in 3D Space forms a unique Plane. That plane can be a triangle-plane or the larger 4-sided plane.

Is the 4 sided plane the largest plane in terms of sides for the 3D 10 Grid? At the moment, I cannot envision nor picture any plane with 5 sides or 6 or more. I suspect 4 is the maximum.

Archimedes Plutonium

Aug 19, 2023, 1:08:19 PM

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On Friday, August 18, 2023 at 11:27:41 PM UTC-5, Archimedes Plutonium wrote:

> Is the 4 sided plane the largest plane in terms of sides for the 3D 10 Grid? At the moment, I cannot envision nor picture any plane with 5 sides or 6 or more. I suspect 4 is the maximum.

The shape of 3D 10 Grid is a cube shape, so I am asking if there is a cross section of the inside of a cube that yields a 5 or higher sided figure? None that I can see, for the 6 faces of the cube only allow 4 sided 2D figures to emerge or 3 sided triangular planes. I cannot retrieve a pentagon plane nor a hexagon plane. And proofs should be provided.

Archimedes Plutonium

Aug 19, 2023, 5:16:29 PM

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Interesting question, if I make a conic sectioning cut into the cube or rectangular box, the largest number of sides is a 4 sided planar figure resultant.

What happens with a cut into the Dodecahedron?? Can I get a 5 sided planar cut figure? Indeed I can. But the Icosahedron looks like it can deliver a 10 sided planar figure result.

With the Octahedron a cut similar to a conic cut that is perpendicular resulting in hyperbola gives a 6-sided hexagon planar resultant yet the cube gives no more than 4-sided result.

A vertical planar cut into Dodecahedron gives a 6-sided figure.

14) Summary.

I started this book in May 2023, into a research adventure of a specific proof of geometry math. One I had learned in High School. I never expected it to end up starting the overhaul of all of Old Math's Geometry. I started with the specific proof of 3 arbitrary noncollinear points in Space determine a unique plane in Space. And by August of 2023, I was involved with a geometry proof of the dodecahedron that used Computers and computer graphics. I found this alleged proof of dodecahedron to also be flawed and, in which this caused me to start the overhaul of Euclid's first two postulates-- specifically the first two axioms of Euclid-- There is a point which has no length, width or depth, and his 2nd postulate (axiom) --two points determine a line for which the line has length but no width or depth.

Archimedes Plutonium

4:09 AM 20Aug2023

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On Saturday, August 19, 2023 at 8:59:29 PM UTC-5, Archimedes Plutonium wrote:

> Quanta Magazine needs to toss out this Messrs.Jayadev Athreya, David Aulicino, Patrick Hooper as bogus.

> Computer Graphics is never admissible into proofs or understanding Geometry. A computer is never able to do Geometry, only the human mind can do geometry (along with many animals). But a machine that is not living is unable to do geometry.

> > There is a picture of this uniqueness in Quanta-magazine, 2020

> > "Mathematicians Report New Discovery About the Dodecahedron"

> > "Suppose you stand at one of the corners of a Platonic solid. Is there some straight path you could take that would eventually return you to your starting point without passing through any of the other corners?"

> > Drs.Jayadev Athreya, David Aulicino, Patrick Hooper may have been victims of Computer Graphing rather than realized fundamental truths of geometry. This happened to me also with the case of tiling a sphere, that the computer gives a image as though the sphere was tiled by hexagons. And computers can make a picture that is so much con-art and deceiving of the human eyes, like optical illusions.

Alright in the bathe tonight, I figured out a counterproof. And much of this will set me up for a future Overhaul of Euclidean Geometry textbook, where I overhaul all the axioms of Euclidean Geometry because Space is discrete in coordinate points, not a continuum.

The very first axiom of (1) there is a point with no length, width, and breadth, and the second axiom (2) Two points determine a line which has length but no width and no depth. Both need more clarity. In a discrete world the point and line need actual substance rather than some idealism delusion. The point needs a tiny length, width, and depth and the line needs a tiny width and depth. This tiny finite metric is obtained from the difference in pi as 3.14159... and 3.162277.... the square root of 10. For the 10 Grid, the tiny metric is 3.16- 3.14 = 0.02.

So the overhauled first two axioms of Plane Geometry are (1) There is a point with a tiny metric length, width, and depth, and (2) There is a line which is determined by 2 points and has a length and a tiny width and depth.

Now I need those two ideas to explain the Counterproof to Drs.Jayadev Athreya, David Aulicino, Patrick Hooper Dodecahedron. I need those tiny metrics, which depend on what Grid System I am working in, say 10 or 100 or 1000 etc. And a perpendicular cut, like in conic sections or a cut at an angle. In particular a cut through the Apex point of a vertex of dodecahedron. So how can you cut a "singular point"? Certainly not in Old Math Geometry but in New Math geometry in 10 Grid we have the metric of 0.02 to play with in 10 Grid, in 100 Grid we have 0.021 to play with.

So say the cut in a dodecahedron is a straight 90 degree perpendicular, then we have the cut divide the 0.02 with 0.01 on both sides of the cut. What if the cut is at 30 degree angle or 60 degree angle or somewhere between 0 and 90 degrees?

Counterproof: The cut at 90 degrees creates a straightline segment at the Dodecahedron apex point of vertex. And for the straightline created at the vertex, it creates a straightline that circumnavigates around the dodecahedron and is forced to run into the south-pole vertex if we call our starting vertex a north pole vertex.

Now a pyramid escapes this northpole and southpole with its apex vertex and follows the Messrs.Jayadev Athreya, David Aulicino, Patrick Hooper dodecahedron claim but only for the singular apex vertex of pyramid.

All the 5 Platonic Solids every one of its vertices has a north-pole and south-pole vertex, even the tetrahedron has a offset northpole and southpole vertex.

This means that the claim by Messrs.Jayadev Athreya, David Aulicino, Patrick Hooper dodecahedron is a false claim. And where they messed up is in their idea of a straightline. From what I can make out, they define their straightline from a planar Net of 12 pentagons.

In ascii-art the Net usually looks like this:

...H..H......H..H

H H H H H H

.....H..........H

And from what I gather, their straightline is across this flat plane Net that represents the Dodecahedron, but when you actually fold this Net up into 3D the straightline dissolves into zig zags.

The Straightline should be the same as Conic Section Cuts with a knife that cuts at a angle into a solid.

And what the AP counterproof says, is that any solid figure with vertices and each has both a north and south pole vertex cannot follow the claims of Messr.Jayadev Athreya, David Aulicino, Patrick Hooper dodecahedron.

This concludes this research of math geometry and is a good start into a future book of overhauling all of Old Math Geometry, for the simple reason, the true numbers of mathematics are Decimal Grid Numbers which are discrete. And discrete numbers demand Geometry be Discrete Geometry.

zzzzzzzzzz

This is an Ebook, meaning that I am constantly adding new material, amplifying and detailing points of interest. And I am constantly editing and correcting and revising.

Archimedes Plutonium

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Jun 1, 2026, 4:38:54 PM (6 days ago) Jun 1

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So I am trying to refresh my mind how I came to some of these remarkable conclusions. I am stopped by these lines in the text.

"And coming from physics, and knowing how important is angles and especially the right angle for electricity is always perpendicular to magnetism. I am going to end up defining a Plane as being a circle parallel to all other circles of a Cylinder.

So, what I am thinking here, for an elegant definition of Circle, I need two points-- center and a distance from center for radius.

So an elegant Plane definition is 1 point to define a plane, and that can be got from a Cylinder as Space. Where the points of the center-line define a unique circle.

That would leave Space to be elegantly defined as needing 0 points.

Summary:
0 points define Space.

1 point defines unique Plane.

2 points define unique circle in plane

Now this only will work if I can manage to say all of Physics is cylindrical. That a Cylinder defines Space.

So have to check into that. But already there are indications that a cylindrical coordinate system is the most General of all coordinate systems."