Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Two proofs of FLT, one with Generalized-FLT plus a lemma, and the
second a pure geometry proof of FLT

Replies: 2   Last Post: Oct 2, 2017 8:57 AM

 Messages: [ Previous | Next ]
 plutonium.archimedes@gmail.com Posts: 18,572 Registered: 3/31/08
Two proofs of FLT, one with Generalized-FLT plus a lemma, and the
second a pure geometry proof of FLT

Posted: Oct 2, 2017 12:50 AM

Newsgroups: sci.math
Date: Sun, 1 Oct 2017 15:22:38 -0700 (PDT)

Subject: yes, convinced I have two proofs of FLT Re: my Second proof of FLT, a
pure geometry proof
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Sun, 01 Oct 2017 22:22:39 +0000

yes, convinced I have two proofs of FLT Re: my Second proof of FLT, a pure geometry proof

Alright, I am convinced I have two different proofs of FLT. The first was a Generalized FLT with Lemma, and the second was a direct pure geometry proof. The first involved some geometry of the Condensed Rectangle, but the Lemma was a pure arithmetic law of exponents. The second involved all geometry, except the -- steps of a impossible barrier of arithmetic law of exponents.

So both proofs involve a violation of arithmetic laws of exponents.

Characteristics and traits of Generalized FLT with Lemma-- here we examine a set of numbers all pooled together by conglomerating all exponents into one huge set::

exp3 ?{1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, . .}
exp4 ?{1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, . .}
exp5 ?{1, 32, 243, 1024, 3125, 7776, 16807, . . }

Generalized FLT: asks-- if we conglomerate all of those exponents from 3 onwards into just one big set of numbers

conglomerated exponents {1, 8, 16, 27, 32, 64, 81, 125, 216, 243, 256, . .}

And asks if there is a A+B =C and we immediately have solutions

8 + 8 = 16

27 + 216 = 243

343 + 2401 = 2744

So Generalized FLT has solutions, no problem about that, but what is the characteristics of those solutions? And the character trait is that they have a factor in common because we are allowed to pick numbers with different exponents. So, now, having proven Generalized FLT and wanting to prove FLT has no solutions, we find that if we are restricted to a constant exponent, no solutions are available, for it violates the Arithmetic of law of exponents-- the Lemma.

Now the Features, Characteristics, Traits of the Pure Geometry proof of FLT. Here, we simply look at a set of constant exponents such as exp3

exp3 ?{1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, . .}

And by observation just cannot see a A,B,C such that A+B=C, unlike the Generalized where we easily spot solutions.

So, we cannot visually see any solutions to exp3 and wonder whether any exists. So we do a geometry proof. We say each of those numbers is a Perfect Cube. Now by logic we take each of those perfect cubes, except 1, and notch them, by removing a smaller perfect cube such as 8-1 = 7 , or, 27-8 = 19,  or, 64-27 = 37. And, we notch every one of those perfect cubes. Now we ask, by logic, if a Notched Cube is ever equal to a Perfect UnNotched Cube. If the answer is no, means FLT is proven to never have any solutions.

Now we ask, why can no Notched Cube be equal to a Perfect Cube? And much like the Lemma in Generalized FLT, is a Arithmetic Law of Exponents Barrier, that same Arithmetic law of exponents as barrier comes into play in this geometry proof.

If we examine the Algebra of a Notched Cube we see it follows along these lines;; (x+1)^3 - x^3 = 3x^2 +3x +1

So, in the geometry proof, we ask for a m^3 = 3x^2 + 3x +1 where m,x are counting numbers. And that is a impossible Arithmetic of Law of Exponents.

Now, for exp4 and higher, this arithmetic impossibility is met with again in that the notched figure is a -1 exponent short of the figure.

So, the two proofs match each other in terms of why FLT cannot have any solutions-- breaks the Arithmetic law of exponents.

Now, the great lesson of FLT. The great lesson it teaches us, is that dimensions beyond 3rd are nonexistent. There is no 4th dimension or 5th or higher cube. Cubes stop at 3rd dimension. So that A^4 or A^5 are A*A^3 and A*A*A^3. Where we take the extra A's as scalars of a Perfect Cube A^3.

So that A^4 is a Rectangular Solid and that A^5 is another Rectangular Solid. And we can have Notched Rectangular Solids or UnNotched Rectangular Solids. This is the great lesson learned from FLT. The proof was difficult. But the application of FLT, that is important, is that mathematics must now always treat 4th dimension and higher, as merely extensions of 3rd dimension. So if presented with a A^6 + B^6 = C^6, we realize those are j*A^3 + k*B^3 = h*C^3.

Old Math was hideous math when it came to dimension, for it is sheer hideous nonsense and stupidity to think of A^6 + B^6 = C^6 as involving 6th dimensional cubes. Leave it to professors of mathematics, sitting around in ivory towers, wanting to become famous, to dream up the idea of a 4th dimension cube or 5th or 6th etc.

AP

Newsgroups: sci.math
Date: Sun, 1 Oct 2017 21:32:12 -0700 (PDT)

Subject: preparing Pure Geometry proof of FLT for array Re: my Second proof of
FLT, a pure geometry proof
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Mon, 02 Oct 2017 04:32:13 +0000

preparing Pure Geometry proof of FLT for array Re: my Second proof of FLT, a pure geometry proof

Now, let me prepare the pure Geometry proof of FLT for the Array. The array if you do not remember is where math is distilled to a minimum, no gushing oozing extras, just the pure engine of how the proof works.

Theorem Statement of pure geometry proof of Fermat's Last Theorem:: Given A,B,C, n as Counting numbers, 1,2,3,4,?.. A^n + B^n = C^n where n is 3 or larger, there are no solutions to be found. The pure geometry proof is based on cubes of geometry and notched cubes, where we subtract the volume of one cube from another cube. And in the proof we define all further exponents higher than 3 as extended cubes which are rectangular solids, so that A^5, for example is the scalar A*A multiplied by the cube A^3. This is a pure geometry proof for it delves into the features, characteristics and traits of the cube. And the proof logic rests on the idea that if no Notched Cube is equal to a Perfect Cube itself, exists, then, hence no A,B,C exists to deliver a solution for A^n + B^n = C^n. For exponent 4 and higher, we generalize notched cubes to notched rectangular solids.

Proof Statement::

Successor Notched Cubes
1^3
2^3   and 8-1 = 7
3^3   and 27-8 = 19
4^3   and 64-27 = 37
5^3   and 125-64 = 61
6^3   and 216-125 = 91
7^3   and 343-216 = 127
8^3   and 512-343 = 169
9^3   and 729-512 = 217
10^3   and 1000-729 = 271
11^3   and 1331-1000 = 331
12^3   and 1728-1331 = 397

Some Examples of Staggered Notched Cubes

3^3   and 27-1 = 26
4^3   and 64-8 = 56
5^3   and 125-27 = 98
6^3   and 216-64 = 152
7^3   and 343-125 = 218
8^3   and 512-216 = 296
9^3   and 729-343 = 386
10^3   and 1000-512 = 488
11^3   and 1331-729 = 602
12^3   and 1728-1000 = 728

We conglomerate all notched Cubes into a sequence such as 7, 19, 26, 37, 56, 61, ?..

And ask, is there a Perfect Cube that can equal the volume of any of these numbers of the list of all Notched Cubes?

If we examine the Algebra of a Notched Cube we see it follows along these lines;; (x+1)^3 - x^3 = 3x^2 +3x +1

So, in the geometry proof, we ask for a m^3 = 3x^2 + 3x +1 where m,x are counting numbers. And that is a impossible Arithmetic of Law of Exponents.

Now, for exp4 and higher, this arithmetic impossibility is met with again in that the notched figure is a exponent of -1 short of the figure. Hence, never any solutions to A^n + B^n = C^n.
QED

Now, the great lesson of FLT. The great lesson it teaches us, is that dimensions beyond 3rd are nonexistent. There is no 4th dimension or 5th or higher cube. Cubes stop at 3rd dimension. So that A^4 or A^5 are A*A^3 and A*A*A^3. Where we take the extra A's as scalars of a Perfect Cube A^3.

So that A^4 is a Rectangular Solid and that A^5 is another Rectangular Solid. And we can have Notched Rectangular Solids or UnNotched Rectangular Solids. This is the great lesson learned from FLT. The proof was difficult. But the application of FLT, that is important, is that mathematics must now always treat 4th dimension and higher, as merely extensions of 3rd dimension. So if presented with a A^6 + B^6 = C^6, we realize those are j*A^3 + k*B^3 = h*C^3.

AP