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August 17th, 2016, 12:03 AM   #41
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Ok end of the joke...

I try to explain where there is the "BUG" that do not allow us to pass from the Step Sum (where all is correct !) to the limit, AFTER THE EXTRACTION OF THE KNOWN CUBE/SQUARE:

The trick fails since once we re-EXTRACT THE "KNOWN" CUBE/SQUARE, we FIX $C\in Q$ so from this moment we broke the possibility to pass to the limit via "continum".

I already said that till we work in the rational we are working with gears, while at the limit we work with smooth weels.

So the error is that when we again transfrom the Step Sum in "Known" Cube/Square we are coupling a gear with a finite number of teeth with what at the limit becomes a smooth weel... asking to this strange couple to be able to move $dx$...

Unfortunately for us the trip is, for so, wrong and say us nothing about the nature of $C$, just that we have to work with it without the restriction $C\in Q$...


Last edited by complicatemodulus; August 17th, 2016 at 12:40 AM.
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August 17th, 2016, 11:27 PM   #42
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For n=2 will be easy to see where the error is:

So starting from the well known:

\[ \left(\frac{B}{A} -\frac{1}{KA} \right) ^2 \neq \left( \left(\frac{C}{A} -\frac{1}{KA} \right) ^2- 1 \right) \]

\[ \sum_{1/KA}^{B/A-1/KA} \frac{2x}{(KA)}-\frac{1}{(KA)^2} \neq \sum_{1/KA}^{C/A-1/KA} \frac{2x}{(KA)}-\frac{1}{(KA)^2} - 1 \]

\[ \sum_{B/A}^{C/A-1/KA} \frac{2x}{(KA)}-\frac{1}{(KA)^2} \neq 1 \]

Till here all is correct since if we again pass to the limit for $K\to\infty$ we have again:

\[ \lim_{K\to\infty} \sum_{B/A}^{C/A-1/KA} \frac{2x}{(KA)}-\frac{1}{(KA)^2} = \int_{B/A}^{C/A} 2x dx= \frac{C^2}{B^2} -\frac{B^2}{A^2} = 1 \]

While if we rest in Q and we try to translate and "dismount" just part of the Step Sum we make an error since we break the continuum property, infact we wil not return to the original equation:

\[ \sum_{1/KA}^{(C-B)/A} \frac{2(x+B/A-1/KA)}{(KA)}-\frac{1}{(KA)^2} \neq 1 \]

\[ \sum_{1/KA}^{(C-B)/A} \frac{2x}{(KA)}-\frac{1}{(KA)^2} + ((C-B)/A) (2B/KA^2-2/(KA)^2)\neq 1 \]

\[ ((C-B)/A)^2 + ((C-B)/A) (2B/(KA^2)-2/(KA)^2) \neq 1 \]

$-2BC/A^2+B^2/A^2+C^2/A^2+2BC/(KA^3)-2B^2/(KA^3)+2B/(K^2A^3)-2C/(K^2A^3) \neq 1$

That at the limit for $K\to\infty$ becomes a new equation:

$-2BC/A^2+B^2/A^2+C^2/A^2 = 1$


$\left( \frac{C-B}{A}\right)^2 = 1$

The same will happen on the presented trick for $n=3$, so in general we just have a confirmation that we are not allowed to "Rational" translate a Rational/Integer AREA and dismount BELLOW A CONTINUOS CURVE, but this will not exclude that is not possible in some case, we know is $n=2$ where the linear dipendence of y' to x allow to find some A,B,C limits for what the FLT trick works...

...this is what I hope to receive back from some expert reader... if someone was following me...

I again leave for some months...


Last edited by complicatemodulus; August 17th, 2016 at 11:34 PM.
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