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January 2nd, 2017, 12:31 AM  #1 
Senior Member Joined: Dec 2012 Posts: 979 Thanks: 24  Fermat's Last Theorem in Set Theory
Fermat's Last Theorem in Set Theory. Knwoing that: $C^n=A^n+B^n$ Is equal to state: $C^n=2A^n+\Delta$ or $C^n=2B^n\Delta$ Remembering that $C$ has to be an Odd Value. Than Fermat the Last can be shown on the cartesian plane as: Or in Set theory as (where we consider the function $Y=2X^n$, that is the analitic extension of our problem, as the main Set, and our FLT Integer or Rational "operator" the SubSet): (where I hope it's clear we are using the Inverse function from $ P \to X\in \mathbb{R}$ Maschke you agree on this? We can stop here as a proof ? I don't like this very much because I think someone still have some doubt, so I go ahead using ordinal to armor the proof (I very hope !): As shown Rational Powers obey to my Ordinal: $M_{n,X}=(X^n(X1)^n)$ And more in general in the Rationals: \[ M_{n,K,x }= {n \choose 1}x^{n1}/K + {n \choose 2}x^{n2}/K^2 + {n \choose 3}x^{n3}/K^3 +... +/ \frac{1}{K^n} \] that means that any Rational Height of the Gnomons $M_{n,k,i}$, so the ordinal $M_{n,k}$ computated for the ith value happen FROM $n>2$ (derivate = curve) JUST at Non Symmetric distance between two following value $(i  (i1))$: both in the Domain (X) and in the CoDomain (Y), so: $M_{n,k,i} \neq \frac {Y_{i}  Y_{i1}}{2}$ This is clear since bellow a monotone rising curve if we join two following points $ (X_{i}, Y_{i})  (X_{i1}, Y_{i1})$ with a segment, it's clear that the area bellow the curve by this two points is littlest than the one bellow the segment for the same points. So for the Tricotomic Law, can't be "always different", and "equal": So Fermat's equation $C^n=A^n+B^n$ Lead to: $C^n=2A^n+\Delta^{}$ or $C^n=2B^n\Delta^{+}$ But this means in $Y$, and for all the following derivates, so for all Rational value we can have from $C^n/K$, they must be all symetric, so obey to the same ordinal (differs just in value): $\delta^{}=\delta^{+}$ this is the contraddiction we are looking for several hundred years... (I hope) ...was a 8 years long way to get here... Last edited by complicatemodulus; January 2nd, 2017 at 12:53 AM. 
January 7th, 2017, 11:34 PM  #2 
Senior Member Joined: Dec 2012 Posts: 979 Thanks: 24 
To make this I really risk the divorce... But it's the really Very END (some check on the computations is required, but the story is right from the beginning..). If Fermat was wrong, we can make a new ordinal (different from my Complicate Modulus Columns) that has to fit Power property, so that can square the Derivate $Y'=2X^{n1}$ with rectangles of Integer (or rational) Area. Regardless for the numbers the new Ordinal has to respect the fact that:  the Exceeding Area (the one between the the rectangular Gnomon's Roof, and the derivate, curve, that it's lower)  and the Missing Area (the one between the derivate, curve, higher, and the Roof of the Rectangular Gnomon's, bellow) MUST BE EQUAL, but as you can see, and imagine, if we hold the same symetric Ratio once we mve on a curve the Exceeding Area with its Convex lower border makes more area than the Missing one that has a Concave upper border: So dividing our initial condition by $K$ so dividing $P=?C^3/K$ we have infinite new Exceeding/Missing areas that are NOT equal, due to the difference of curvature of the Derivate, and we can imagine (or see in the Table) that this Ratio is not 1, and not constant. So we have now a more interesting Math point of view of the 2th Derivate... Here the Table A=5, B=6 P=341 ; $C=341^{(1/3)}= 6,98...$ ; Delta= 91 I know this in terms of ordinal from long time (before I study what ordinals are), but to let this be clear on the graph takes me a very, very long work... and rigors that I've to force me to apply... My health has suffered a lot, as well as my marriage, but someone had to do ... Ciao Stefano Here what happen on the Balancing Point. I've a 200 pages paper draft on that... will be a long work to be finished and I don't know if I will see it completed in life... Last edited by complicatemodulus; January 7th, 2017 at 11:39 PM. 
January 8th, 2017, 01:28 AM  #3 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,232 Thanks: 427 Math Focus: Yet to find out.  I really don't think that's necessary. Devoting your life to any field will surely come with sacrifices and compromises that interfere with your personal life, i think that's totally healthy, and it defines ones character. But there is always a balance, not too much, not too little.

January 8th, 2017, 09:13 AM  #4  
Senior Member Joined: Dec 2012 Posts: 979 Thanks: 24  Quote:
Balancing ourself it's not soo difficoult... the problem it's to appear balanced... to the others... And when you see time it's going and thinks are still so intersting and much deep then where you are... Another invariant appear on the red carpet: $r_k/q_k = 1,19820791643669...$ or there is some computational/typing or else error in my table or again it's time to put the hands on the shovel... Here the first graph to show that the balancing point $BP$ cannot be in the middle In Abscissa, the previous also prove cannot be in the middle in the Ordinate. $BP$ has to be the point where Exceeding Area $A^+$ it's equal to the Missing Area $A^$ To prove that the Area $T1<T2$ we take as reference the Triangular Missing /exceeding areas that we know has ratio 1:1. The same to prove that the Area $T3>T4$, so BP must be (Always if the derivate is a curve) lower/right than MP. The condition to square the area bellow a curve with a rectangular gnomon is: Exceeding Area $A^+$ it's equal to the Missing Area $A^$, than we must have: Last edited by complicatemodulus; January 8th, 2017 at 09:53 AM.  
January 13th, 2017, 08:06 AM  #5  
Senior Member Joined: Dec 2012 Posts: 979 Thanks: 24  Quote:
I check also $n=5$ and the difference it's more evident. So the good invariant for Power's derivate (from $n>=2$, from the first to all the significative, so non flat, following ones ) rest the equal areas $A^+=A^$ respect to the Roof of the Integer or Rational Gnomons (the Rectangular column follows from the Telescoping Sum Property).  
January 14th, 2017, 11:49 PM  #6 
Senior Member Joined: Dec 2012 Posts: 979 Thanks: 24 
A draft of the first part of my work it's available here: https://www.academia.edu/30658894/Th..._Vol.1_2_draft Under revision, weekly updated. Vol.2 Will need more time... 
January 24th, 2017, 06:26 AM  #7 
Senior Member Joined: Dec 2012 Posts: 979 Thanks: 24 
To avoid readers loss of time I hope next week the revised final version of The TwoHandClock will be available on the same link. On the draft there are all the main cocepts but it's full of errors of various nature. I'm still in doubt if wait some more because there is a new entry regarding the Trapezoidal Gnomons versus the Rectangular ones... but this open a new branch of study for a pletora of Diophantine equations of nth degree with rational results... 
January 25th, 2017, 09:41 AM  #8 
Senior Member Joined: Dec 2012 Posts: 979 Thanks: 24 
Now the draft is more clean (not jet the final version...): https://www.academia.edu/30658894/Th...ock_Vol.1_of_2 
February 3rd, 2017, 07:02 AM  #9 
Senior Member Joined: Dec 2012 Posts: 979 Thanks: 24 
Here the final picture explain the difference betweens $n=2$ where the area below the linear derivate is a triangle, so squarable in the integers in case the irrational part vanish in the product, while for $n>2$ the only way to square an area that has an irrational right limit is via integration. Last edited by complicatemodulus; February 3rd, 2017 at 07:10 AM. 

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