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joi, 2 octombrie 2025

A Paradox of Infinite Long Division // Un paradosso della divisione infinita

 Everything started from the aspects provided in the Post "An easy...".(I still haven't found a simple explanation.)

         I will extract from there the equation 

$26^3-24^3-12^3+1^3=(2^3+1)\cdot 225$

which expresses that  $2025=9\cdot 225$.

(under construction)

longdivision

\begin{array}{ccccccc}\;\;\;26^3&-24^3&\;&-12^3&+1&\vdash&2^3+1\\-26^3&\;&-13^3&&&&13^3-12^3-\left(\frac{13}{2}\right)^3+\left(\frac{13}{4}\right )^3-\left (\frac{13}{8}\right)^3+\left (\frac{13}{16}\right)^3-\left (\frac{13}{32}\right )^3+\cdots\\\hline \setminus&-24^3&-13^3&-12^3&+1\\\;&24^3&\;&12^3\\\hline \;&\setminus&-13^3&\setminus&+1\\\;&\;&13^3&+\left(\frac{13}{2}\right)^3\\\hline \;&\;&\setminus&\left(\frac{13}{2}\right)^3&\;&+1\\\;&\;&\;&-\left(\frac{13}{2}\right)^3&-\left(\frac{13}{4}\right)^3\\\hline \;&\;&\;&\setminus&-\left(\frac{13}{4}\right)^3&+1\\\;&\;&\;&\;&\left(\frac{13}{4}\right)^3+\left (\frac{13}{8}\right)^3\\\hline \;&\;&\;&\;&\left (\frac{13}{8}\right)^3&+1\\\;&\;&\;&\;&-\left(\frac{13}{8}\right)^3-\left(\frac{13}{16}\right)^3\\\hline \;&\;&\;&\;&-\left(\frac{13}{16}\right)^3&+1\\\;&\;&\;&\;&\left(\frac{13}{16}\right)^3+\left(\frac{13}{32}\right)^3\\\hline \;&\;&\;&\;&\left(\frac{13}{32}\right)^3&+1\\\:&\;&\;&\;&\dots&\dots\end{array}

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