My Math Forum Supersymmetric Complicate Numbers

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 August 21st, 2017, 10:16 PM #1 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Supersymmetric Complicate Numbers Looking at the Class of he Rest of my Complicate Numbers, I discover what I've called Supersymmetric Complicate Numbers "P". The Integers $P$ defined as: $P =A^n+B^n$ ; $n>1$ $A,B \in \mathbb{N^+}$ will be writable as Complicate Numbers in a "Supersymmetric" way (I don't have any idea if this has some connection to the known Supersymmetric property in physics): Remembering we wrote a Number as Complicate Number writing it as the Sum of its $Integer$ $n-th$ $Root$ plus a $Rest$ we can see that: $P$ as a Result of this Sum, written as a Complicate Number, is characterized by the Commutative Property between the $Integer Root$ part and the $Rest$: $P = (A M_n + 0 ) + (B M_n + 0) = (A M_n + B^n ) = ( B M_n + A^n)$ For example: $P= 341 = 5^3 + 6^3 = (5 M_3 + 6^3) = (6 M_3 + 5^3)$ And it can also be written as Two complicate Numbers with The Same Absolute Value of Class of Rest: since: $$A^n = \sum_{1}^{A} M_n$$ $$B^n = \sum_{1}^{B} M_n$$ with $M_n = (x^n-(x-1)^n)$ than: $$P= A^n+B^n =\sum_{1}^{A} M_n + \sum_{1}^{B} M_n = 2* A^n + \Delta =2* B^n -\Delta$$ with $$\Delta = \sum_{A+1}^{B} M_n$$ $P= 341 = 2* 5^3 + 91 = 2* 6^3 - 91 = (5 [2M_3] + 91) = (6 [2M_3] - 91)$ ... I'm alone in a wonderfull swiming pool... Last edited by complicatemodulus; August 21st, 2017 at 10:28 PM.
 August 22nd, 2017, 04:27 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,821 Thanks: 1047 Math Focus: Elementary mathematics and beyond You still haven't explained what "complicate numbers" are.
 August 22nd, 2017, 08:04 AM #3 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Sorry, I'm boring this forum from several years with them. But here once again: Are numbers I invented as a Bijection with all $P\in \mathbb{N^+}$. Are made by a Pure Power, I call Integer n-th Root of $P$, plus a $Rest$: For example taking of $n=2$: $1 = 1^2+0 = 1M_2+0$ $2= 1^2+1 = 1M_2+1$ $3= 1^2+2 = 1M_2+2$ $4= 2^2+0 = 2M_2+0$ $5= 2^2+0 = 2M_2+1$ Where more in general for the n-th power the Complicate Modulo is: $M_n=(x^n-(x-1)^n)$ And you can find the Integer Root as a recursive difference using this function as Modulo: For example Square root of 5 is equal to make it modulo $M_2$: 1 . 2x-1 = 1 . 5-1 = 4 2 . 2x-1 = 3 . 4-3 = 1 so 2 is the Integer Square Root of 5, so 5= 2^2+1 in fact if you can't go over because: 3 . 2x-1 = 5 . 1-5 = -4 And the same using a special trick to work with Rationals. Here an example of how my Complicate Numbers can be shown on a Two-Hand-Clock: So is a way to cut a number with a "modulo" that is a function and not just a number. Than you've back rising slices, plus a Rest, and you always know exactly which integer you're talking of, plus they becomes very useful in case you are working with problems involving Powers. I'm preparing a book on that... Last edited by skipjack; August 22nd, 2017 at 09:26 AM.
 August 22nd, 2017, 08:50 AM #4 Senior Member   Joined: May 2016 From: USA Posts: 1,047 Thanks: 430 People say they do not understand what you are talking about, and you repeat yourself. $P_1 = 1^{any} + 1^{any} = 2.$ $P_2 = 1^{any} + 2^2 = 5.$ $P_3 = 2^2 + 2^2 = 8.$ $P_4 = 1^{any} + 2^3 = 9.$ $P_5 = 1^{any} + 3^2 = 10.$ So I thought I understood what your set P was. Then you give as your first three examples: $1^2 + 0^2 = 1.$ Zero is not a positive integer. $1^2 + 1^2 = 2.$ OK. $1^2 + 2^1 = 3.$ But 2 is not raised to a power greater than 1. So far you have failed to define P, let alone anything based on P. Incompetent is way too generous an assessment.
August 22nd, 2017, 09:31 PM   #5
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Quote:
 Originally Posted by JeffM1 ... So far you have failed to define P, let alone anything based on P. Incompetent is way too generous an assessment.
The definition of a Complicate Modulus Number is now 4-5 page long since it start as in the example (and as you probably missunderstand) and than I realize it can be generalized.

The first Complicate number I invent was made as:

$$P= \left \lfloor{\sqrt{P}}\right \rfloor ^2 + \left( P - \left \lfloor{\sqrt{P}}\right \rfloor ^2 \right)$$

Than I go up to:

$$P \in \mathbb{R}| P= a M_n + Rest = \left( \lfloor{{P^{(1/n)}}}\rfloor \right)^n + Rest = \sum _{x=1}^{\left( \lfloor{{P^{(1/n)}}} \rfloor \right)} M_n + Rest$$

Than I call this "Reduced Complicate Modulus Numbers" since there can be a more general "Non Reduced Complicate Modulus Numbers"

$$P \in \mathbb{R} : P= a [k M_n] + Rest$$

Is the General Complicate Modulus Number that can also fit (so returns a Zero as Rest) for non perfect powers, but for example for numbers like $k*A^n$ with $k \neq A$, or $k=A$ too if you like / need.

Where $[k M_n] = k* (x^n-(x-1)^n)$

So the title of the post isreferring to some special $P$ that can be currently written in the known algebra as $P= A^n+B^n$, while once it can be written as a Complicate Number becomes... etc.. (pls return to the original post here).

...I'm fully alone here, so I'm also fully free to swim where I prefer / or I'm able to

Last edited by complicatemodulus; August 22nd, 2017 at 09:45 PM.

 August 22nd, 2017, 11:33 PM #6 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,600 Thanks: 546 Math Focus: Yet to find out. I feel like it's a "if you can't explain it simply, you don't understand it well enough" sort of situation... Thanks from cjem
August 23rd, 2017, 12:05 AM   #7
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Quote:
 Originally Posted by Joppy I feel like it's a "if you can't explain it simply, you don't understand it well enough" sort of situation...
I think now is clear enough. Of course on my book will be better...

 August 23rd, 2017, 12:08 AM #8 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,600 Thanks: 546 Math Focus: Yet to find out. But I don't understand what you are trying to say. And here's the thing. Perhaps I shouldn't understand, since I don't know anything about number theory. However, my point is that, you should be able to explain your idea in the simplest way possible, such that someone like me (with little training) can understand! But every time I look at your threads, I just see the same graphs, clocks, and weird summations.. Can you explain your idea without the use of symbols? Sent from my iPhone using Tapatalk
 August 23rd, 2017, 12:28 AM #9 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Sorry I think is clear enough in math. Here the last chance: And in any case you've to buy my book on amazon (probably end of september) to read all what I discover starting from this very basic trick.
 August 23rd, 2017, 01:28 AM #10 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,600 Thanks: 546 Math Focus: Yet to find out. i find this incomprehensible... labels seem to be scattered all over the place. It is stated that the case of 'n=2' is being considered, but n doesn't appear anywhere. How am I to know what it signifies? What does rest mean? What values can X take? What is k and what does it mean to multiply k by 'modulus'? In the case of 'Classic modular arithmetic', what are the points a) and b) are referring to?

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