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October 23rd, 2016, 10:31 PM   #1
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Find the Rule for Complicate Numbers


Complicate Numbers

is a "complex like" number, given by the

Integer n-th root (of the given number) + his Integer Rest.

To identify the degree of the Integer Root, insthead of "i" will be used the letter "M" plus the degree of the Root, and to avoid missunderstanding, the Integer Root of a Given Number A ( here with latex as: $\lfloor(A^{1/n})\rfloor$) come first:

$\displaystyle A^{1/2} = \lfloor(A^{1/2})\rfloor M2 + Rest $
$\displaystyle A^{1/3} = \lfloor(A^{1/3})\rfloor M2 + Rest $
$\displaystyle A^{1/4} = \lfloor(A^{1/4})\rfloor M2 + Rest $


$\displaystyle Rest = A - \lfloor(A^{1/n})\rfloor $

For example:

$\displaystyle A= \sqrt 2 = 1 M2 + 1$

$\displaystyle B= \sqrt 3 = 1 M2 + 2$

$\displaystyle C= \sqrt 4 = 2 M2 + 0$


To have the numbers without using the, integer of "" function, just make the recoursive subtraction of M2=2x-1 from the given number A we are trying to make the square root. (I already wrote about this several times in this forum...)

The play is:

Find the rule (if any) for the product AxB.

hint: AxA is the first trick you've to solve.

...I can't really understand why nobody is interested to play in this uncontaminated new free math land... be clear... don't search for my help... I'm playing (free!) severals light years ahead...

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October 24th, 2016, 11:03 PM   #2
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More simple...

Find the Rule for: A+B

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October 26th, 2016, 09:17 PM   #3
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$\displaystyle A=(aM2+b)$

$\displaystyle B=(cM2+d)$

The rule for A*B is, in base Decimal:

$\displaystyle A*B = (a^2*c^2+a^2*d+bc^2+bd) $

than to have it back the result as a Complicate Modulus Number:

$\displaystyle A*B = |(a^2*c^2+a^2*d+bc^2+bd)|M2 $


$\displaystyle A*B = \Delta_{1}^{\lfloor A*B \rfloor} |(a^2*c^2+a^2*d+bc^2+bd)|_{(2x-1)}$

that is the recoursive difference from the Decimal Number AxB, to the Complicate Modulus Number modulus M2=(2x-1) or: AxB|M2


$\displaystyle A=(1M2+1) = \sqrt {2}$

$\displaystyle B=(3M2+1) = \sqrt {10}$

The rule for A*B is: we need to go base Decimal to solve:

$\displaystyle A*B = (1*3^2+1^2*1+3^2+1) = 20 $

$\displaystyle A*B = \Delta_{1}^{\lfloor A*B \rfloor} |20|_{(2x-1)} = 4M2+4 $


1-> 20-1 = 19
2-> 19-3= 16
3-> 16-5= 11
4-> 11-7 = 4 Rest

So $A*B = 4M2+4 $

Fast ? Not at all...

Usefull ?

We will see when will be necessary to works with MQbit: a special Qbit I invent that can be 1,0 or a "well oriented" Vector...

Last edited by complicatemodulus; October 26th, 2016 at 09:31 PM.
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