
math professors Princeton Univ. endorsing Beal & FLT proofs to arxiv
Posted:
May 15, 2014 3:33 PM


I decided I needed to stiffen up the proof of Beal below. To a logical person, they often streamline their proofs and do not bother with adding some information which people not used to logic have a difficult time in seeing. For me, it is redundant to include the distributive law into the proof of Beal, but to those who do not do math but only on occasion, they would not see the logic unless I explain those very details. So below is a more stiffened up proof of Beal.
Proof of the Beal conjecture that proves Fermat's Last Theorem
I proved Beal and FLT (Fermat's Last Theorem) before I discovered that Logical Material Implication Table of T,F,T,T is incorrect and should be that of T,F,undefined, undefined. What this does is remove reductio ad absurdum as a mathematics proof technique as only a probability technique and no longer a deduction. It means the only valid proof method for mathematics, is construction proofs. My Beal and FLT are construction proofs. Another feature of the true Logic Implication operator, is that corollaries of mathematics cannot be proven as "stand alone conjectures". And that a corollary requires the overarching theorem attending the corollary be proven beforehand. This implies that Wiles's FLT is not true for in order to truly prove FLT, a proof of Beal had to come first.
We see the relationship between a Theorem and its Corollary very easily from Beal and FLT, in that we have all these cases to worry about. The history of FLT was a case study, prove it in the case of exponent this and that, but never a proof of all of FLT, and that is because corollaries are never proven as stand alone theorems and must have their theorem (Beal) proven first. This is because Logic Implication is not T,F,T,T where false proofs hide behind those last two T, T when they should be undefined, undefined. The reason Implication must be T,F,undefined, undefined is because mathematics has 2/0 and 0/0 where division by 0 is undefined, and that Old Logic with its T,F,T,T does not allow for division by zero.
Detailed Proofs Beal's conjecture with its FLT corollary
Both proofs of Beal and FLT are based on a fact of geometry, that you can represent a number with its cofactors as the sides of a rectangle. And to prove either Beals or FLT is a simple matter of stacking two rectangles that have equal sides, A and B to produce a third rectangle C which has a side equalling the _shared side_ of A and B. ________________________ DETAILED PROOF OF BEAL ________________________
It is a constructive proof.
We make the table of all the numbers possible in the Beal Conjecture as the conglomeration of exponents of 3 or larger as this set:
{1, 8, 16, 27, 32, 64, 81, 125, 128, 216, 243, 256, . .}
Here we have conglomerated exp3 and exp4 and exp5 etc etc into one set.
We know Beal has solutions of A+B=C in that set for here are three examples:
2^3 + 2^3 = 2^4 with prime divisor 2 3^3 + 6^3 = 3^5 with prime divisor 3 7^3 + 7^4 = 14^3 with prime divisor 7
What we need to prove is that all solutions have a prime divisor in common, ie all three rectangles of A, B, C so that A+B=C, have one shared side equal to one another.
Definition of CondensedRectangle: given any number in the set of conglomerated exponents, we construct rectangles of that number from its unit squares whose sides are cofactors of the number. For instance, rectangle of 216 units as either 12x18 units, or 9x24 units, or 6x36 units or 3x72 units, or 2x108, but never a 1x216 units. We exclude 1 times the number as a condensed rectangle. So a condensedrectangle is one in which it is composed of cofactors of the number in question, except for 1, and the number itself for 1x216 units is not a condensedrectangle.
Now for the constructive proof that Beal solutions must have a common prime divisor.
We stack CondensedRectangles of the numberspace that Beal's conjecture applies:
Solution Number Space for Beal now becomes these condensedrectangles:
{ 2*4, 2*8, 4*4, 3*9, 2*16, 4*8, 2*32, 4*16, 8*8,. . . }
We convert each of those numbers into CondensedRectangle, except 1 of course, and where many numbers have several condensed rectangles so the Solution Space of Numbers increases by a large amount. If an A and B as condensedrectangles have the same side such as 3x9 units and 9x24units wherein you stack them on their shared side of 9 and which matches another number of its condensedrectangle such as 9x27 units, then you have a Beal solution of A+B=C. For if we were to take the 9 by 27 condensed rectangle it decomposes into 3x9 and 9x24.
Now, the question is, are all A+B equal to a C, form stacked condensed rectangles that share a common side?
All stackable condensedrectangles must have one side the same for the two rectangles to stack, in the case above it is the side 9 with its common divisor of the prime 3.
If any other solution to Beal had A stacked upon B without a common side between them, then the figure formed cannot be a overall new rectangle but something that looks like this:
HHHHHH HHHHHH HHHHHHHHHH HHHHHHHHHH
That is not a condensed rectangle and all the numbers of the Solution Space, except the number 1, are condensed rectangles. That above figure is 6sided figure. That is a 6sided figure, yet a rectangle is only a 4 sided figure. That is not a Condensed Rectangle, represented by its cofactors. So that equality can not be achieved by any other stacking than condensed rectangles equalling condensed rectangles.
Then the question is, can we have an A and B with a shared common side equal to a C that is a condensed rectangle with no common side shared with A and B?
The question is, how can I be sure that all the A, B, C such that A + B = C have a common prime divisor?
As if the question is asking whether the Condensed Rectangles covered the question by forbidding any equality unless there is a common shared side in A in B and in C.
Well, the answer is easily enough covered for the Condensed Rectangles eliminates any possibility of A,B,C where A+B = C and not have a shared side by all three of A, B, C. It does this by the Distributive law of integers.
All the numbers in the solution space of Beal have condensed rectangles, except the number 1.
All the A, B, C are written as condensed rectangles of (s*t) + (r*p) = (u*v).
For example (9*3) + (9*24) = (9*27) which is 3^3 + 6^3 = 3^5 which is 27 + 216 = 243.
In order for (s*t) + (r*p) = (u*v) then the s, r, u have to be equal sides. Have to be (9*t) + (9*p) = (9*v) in the example of 27 + 216 = 243.
Distributive Law (9*3)/9 + (9*24)/9 = (9*27)/9 where 3 + 24 = 27
So, give me three numbers A, B, C, chosen at random from the Solution Space of Beal {1, 8, 16, 27, 32, 64, 81, 125, 128, 216, 243, 256, . .} except of course 1, and can those three chosen random A, B, C obey A + B = C? Only if they obey the Distributive Law can the A + B equal to the C when we write the A, B, C as condensed rectangles.
QED
_________________________________________ Detailed Proof of FLT using condensedrectangles _________________________________________
It is a construction proof method for we show that it is impossible to construct A+B = C inside of a specific exponent.
Fermat's Last Theorem FLT conjecture says there are no solutions to the equation a^y + b^y = c^y where a,b,c,y are positive integers and y is greater than 2.
The number Space that governs FLT is this:
exp3 {1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, . .}
exp4 {1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, . .}
exp5 {1, 32, 243, 1024, 3125, 7776, 16807, 32,768, 59,049, 100,000, 161,051, 248,832, 371,293, . .}
exp6 ..... . . . .
So in FLT we ask whether there are any triples, A,B,C in any one of those _specific exponents_ such that A+B=C. In FLT, our solution space is only one particular exponent such as 3 or 4, or 5 to hunt down and find a A,B,C to satisfy A+B=C.
In the proof we use CondensedRectangles which is defined as a rectangle composed of unit squares of the cofactors of a number, except for 1 x number itself. So the number 27 in exp3 has Condensed Rectangles of 3x9 only. The number 125 in exp3 has condensed rectangles of 5x25 only, and the number 81 in exp4 has condensed rectangles of 3x27 and 9x9.
Now in the proof of Beal, solutions of A + B = C require a common divisor.
2^3 + 2^3 = 2^4 with prime divisor 2 3^3 + 6^3 = 3^5 with prime divisor 3 7^3 + 7^4 = 14^3 with prime divisor 7
Here in FLT, we need not even worry about exponents because all solutions to Beal encompass exponents so that if these were FLT solutions:
2^? + 2^? = 2^? with prime divisor 2 3^? + 6^? = 3^? with prime divisor 3 7^? + 7^? = 14^? with prime divisor 7
So in that construction of a solution to FLT there is a common divisor and so now we divide the equation by the common divisor, and we get this:
1^y + 1^y = 1^y 1^z + 2^z = 1^z 1^w + 1^w = 2^w
And those are impossible constructions. So the proof of FLT requires Beal be proven first in order to display that a Beal proof cannot coexist unless FLT has no solutions.
QED

Recently I reopened the old newsgroup of 1990s and there one can read my recent posts without the hassle of mockers and hatemongers.
https://groups.google.com/forum/?hl=en#!forum/plutoniumatomuniverse Archimedes Plutonium

