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6.5-HISTORY of AP's four proofs-- Pure Geometry Proof of FLT//AP's 4 Proofs of Fermat's Last Theorem

Started byArchimedes Plutonium <plutonium.archimedes@gmail.com>
First post2018-09-30 19:21 -0700
Last post2018-10-01 02:27 +0000
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  6.5-HISTORY of AP's four proofs-- Pure Geometry Proof of FLT//AP's 4   Proofs of Fermat's Last Theorem Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-09-30 19:21 -0700
    Re: Archimedes "Village Idiot" Plutonium flunked the Math Test of a lifetime-generation test moroney@world.std.spaamtrap.com (Michael Moroney) - 2018-10-01 02:27 +0000

#424641 — 6.5-HISTORY of AP's four proofs-- Pure Geometry Proof of FLT//AP's 4 Proofs of Fermat's Last Theorem

FromArchimedes Plutonium <plutonium.archimedes@gmail.com>
Date2018-09-30 19:21 -0700
Subject6.5-HISTORY of AP's four proofs-- Pure Geometry Proof of FLT//AP's 4 Proofs of Fermat's Last Theorem
Message-ID<948bda6a-4d2c-41d8-9a32-36840a167f6b@googlegroups.com>
6.5Newsgroups: sci.math 
Date: Sun, 30 Sep 2018 00:24:50 -0700 (PDT) 

Subject: 6.5-HISTORY of AP's four proofs-- Pure Geometry Proof of FLT//AP's 4 
 Proofs of Fermat's Last Theorem 
From: Archimedes Plutonium <plutonium....@gmail.com> 
Injection-Date: Sun, 30 Sep 2018 07:24:50 +0000 

6-HISTORY of AP's four proofs-- Pure Geometry Proof of FLT//AP's 4 Proofs of Fermat's Last Theorem 

6-HISTORY of AP's four proofs-- Pure Geometry Proof of FLT//AP's 4 Proofs of Fermat's Last Theorem 

On Saturday, September 29, 2018 at 9:32:23 PM UTC-5, Archimedes Plutonium wrote: 
(snipped except for Pure Geometry Proof) 

> 
> d) The Pure Algebra Proof of FLT works in a Pure Geometry Proof of FLT although the final proving mechanism is the same as above. In the Geometry proof, we rely upon a perpendicular dissection of a given arbitrary cube in exp3 or a "ghost cube" in exp4 or higher. The beauty of the Pure Geometry proof of FLT is that it requires Ghost Figures in geometry, since there is no dimension beyond 3rd dimension. 
> 
> 

Alright, I am in year 2018, telling the story of how Archimedes Plutonium found 4 proofs of Fermat's Last Theorem. I am too tired to do this Pure Geometry proof tonight, for it is the largest write up of all the 4 proofs. Large because it has so much new Geometry never before seen in mathematics. I am talking about Ghost figures, such as when I made a cube out of two paper clips of two squares separated by a side length. When we normally think of a cube, we think of a solid object as a cube. But that need not be the case of most cubes, for a cube that consists of two squares is also a cube. 

Now most people in mathematics have believed that FLT proof is a useless item of math for it does not speak to much of any other math except proving it. Sort of the question, what use is it. And many have thought that FLT is as useless as they come. But I differ in opinion because of the fact that FLT pure Geometry forces mathematics to have Ghost Figure Geometry. Because in True Mathematics and more importantly True Physics, the 3rd dimension is the last and final dimension. There is no 4th and higher dimension and these are poppycock dreams of kooks of science and math. Dimension ends with 3rd. 

Why all the geometry fuss? Well, FLT is not only about exp3 as 3rd dimension but then exp4 , exp5, exp6 etc etc. So, how can we ever have a Geometry proof of FLT at all, if we have no geometry to represent exp4 and beyond? So here is where mathematics Geometry is introduced to the Geometry of Ghost figures. Two paper clip squares and a side can represent a exp5 ghost cube. Or, 5 faces of a cube of 5 squares can represent exp5. When we get to exponent 13 we run out of attributes of a cube, so then we start by taking 1/2 of a side as a ghost cube. 

Do this tomorrow,,,,, 


Newsgroups: sci.math 
Date: Sun, 30 Sep 2018 11:22:11 -0700 (PDT) 

Subject: pure geometry proof of FLT with its Ghost figures Re: 6-HISTORY of 
 AP's four proofs-- Pure Geometry Proof of FLT//AP's 4 Proofs of Fermat's Last Theorem 
From: Archimedes Plutonium <plutonium....@gmail.com> 
Injection-Date: Sun, 30 Sep 2018 18:22:11 +0000 

pure geometry proof of FLT with its Ghost figures Re: 6-HISTORY of AP's four proofs-- Pure Geometry Proof of FLT//AP's 4 Proofs of Fermat's Last Theorem 

- show quoted text - 
Alright, let me outline this for it carries much and a lot of brand new mathematics. 

I do a lot of physics especially electromagnetism, Maxwell theory about electricity and magnetism and we need box figures made into coils. And so one coil is a bunch of copper wire wrapped around a rectangular box or a cube. Now one complete square circuit of wire can be considered to be a "square of geometry" yet all it has is a bare wire shaped in the form of a square. It has 4 sides and 4 vertex. It has no interior, unless you want to think of the interior as Space between the 4 sides. Now if I had two of these identical square wires space a distance of the side length of one wire square apart, you can see that these two square wires form a 3rd dimension cube. To all purposes, those two square wires form a cube, call it a ghost cube if you want to, but a cube just as legitimate as a cube wooden block solid is a cube. 

So I need Ghost Figures of Geometry, because of FLT, for exp4 and higher. 

Dimensions in physics and in mathematics ends at 3rd dimension. The 3rd dimension is the last and final dimension. Do not let the fool and kook physicist of mathematician snake oil persuade you that 4th dimension and higher exist, for they do not exist. 

So, in FLT pure geometry proof. I need to deal with exp4 and higher with some sort of geometry figure, and those figures are Ghost Cubes, all Ghost Cubes. 

In exp4, I can just simple take 4 faces of a cube to represent 2^4 or 3^4 or 4^4 etc. In exp5, I can take 5 faces of a ghost cube to represent 3^5 or 4^5 etc. Or I can take the vertex, the 8 vertices of a cube. At exp13 I have to be more imaginative, because a cube has 8 vertex, 6 faces, 12 sides and I have run out of numbers for attributes of a cube. So at 13 and higher, I start using 1/2 of a side, that gets me to exp24, at exp25 I start using 1/2 of 1/2 of a side, etc etc. Always keeping and maintaining the Cube. 

So now I have Geometry figures to do a FLT pure geometry, and the way this works is Perpendicular Dissection of a Cube, ghost cube if not exp3. 

You are presented a cube of exp3, say 10^3 or say a cube of exp4 such as 10^4 or exp5 like 10^5. 

You run your fingers along the top face of these cubes and at one point you stop and bring in the Perpendicular Slicer, perpendicular to the axis of the cube. And there you make a perpendicular dissection. Leaving the original cube to now become two smaller rectangular boxes. Put the two boxes back together and you have your original cube back. 

Alright so far, and now we haul in all those Substitutions of Primitive Pythagorean Triples, PPT. Each of those two rectangular boxes is a A^3 + B^3 and the original cube is C^3, or if in exp4 A^4 + B^4 = C^4 etc. 

Now the pure geometry proof becomes almost identical to the pure algebra proof of FLT. For the two sliced boxes that we cut the original cube into. We inspect those two boxes and they are rectangular boxes not cubes. And what we want to know is -- once we turn those boxes into smaller cubes than the original cube whence they came from, is the A^3 and B^3 both Whole Number smaller cubes from C^3, or is the A^3 and B^3 fractional cubes (what Old Math called irrational number side cubes) 

And here, the Pure Geometry proof uses the identical same algebra mechanism to answer why all those two boxes can never be whole number cubes. 

A major algebra math lesson to learn here is that of solving a list of formulas simultaneously 

So we have nine formulas: 

(1) a,b not equal to 0 
(2) b>a 
(3) 2ab 
(4) b^2-a^2 
(5) b^2 + a^2 
(6) A^3 + B^3 = C^3 if working in exp3 (specific to the exponent) 
(7) cube root A if working in exp3 
(8) cube root B if working in exp3 
(9) cube root C if working in exp3 

And asked to solve for Whole Number solutions, of the a,b,A,B,C, and for exponent3 or higher, the only whole numbers that can solve all nine formulas is the whole number 0 and no others, but 0,0,0 is not a Triple solution as a,b cannot be 0. 

Now I need explain that Pure Geometry proof is only a name, for you can see there is much algebra involved, but the pure geometry involved are these cubes and dissection and then conversion of boxes into cubes. 

AP

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#424642 — Re: Archimedes "Village Idiot" Plutonium flunked the Math Test of a lifetime-generation test

Frommoroney@world.std.spaamtrap.com (Michael Moroney)
Date2018-10-01 02:27 +0000
SubjectRe: Archimedes "Village Idiot" Plutonium flunked the Math Test of a lifetime-generation test
Message-ID<pos0lk$td1$1@pcls7.std.com>
In reply to#424641
Math Failure Archimedes Plutonium <plutonium.archimedes@gmail.com> fails:

>Newsgroups: sci.math
>Subject: Re: 6.5-HISTORY of AP's four proofs-- Pure Geometry Proof of FLT//AP's 4 Proofs of Fermat's Last Theorem

You have been sufficiently punished for today's insubordination. You may 
now go off to continue to fail at math and physics.

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