Mathematics

Vortex Based Mathematics

[latexpage]

Note: I have no idea whether this is a load of crap or not, but at the very least its a fun mathematical adventure for nerds. Just remember to take some of the more philosophical stuff with a grain of salt and do your own research before coming to any major conclusions.

Background

Marko Rodin = original founder of vortex-based mathematics

Randy Powell = continues the work of Rodin today

Relevant Operations

Modular arithmetic notation ( defined for the purposes of this article as follows)

  •  \oplus _k   ⇒ addition modulo k
    • Example: 6 \oplus _5 5 = (6+5)\mod{5} = 11 \mod{5} = 1
  • \ominus _k   ⇒  subtraction modulo k
    • Example: 4 \ominus _7 8 = (4-8) \mod{7}= -4 \mod{7} = 3
  • \otimes _k   ⇒  multiplication modulo k
    • Example: 3 \otimes _9 7 = (3*7) \mod{9}= 21 \mod{9} = 3
  • \oslash _k   ⇒  division modulo k
    • This operation is a little more complicated than the first three
    • Importantly, care must be taken when using modulo division – it isn’t necesarilly a unique or sufficient operation
      • e.g., 9 \oslash _9 x = 9     \forall       x
    • Steps used for calculating a \oslash _k b
      • Determine the lowest possible value m that results in a + (k \cdot m) obtaining a value which is a multiple of b
      • Plug in m into the following expression: a \oslash b = (a + (k \cdot m)) / b
    • Example: find 8 \oslash _6 5
      • Givens: a = 8, b = 5,  k = 6
      • 8 + 6(2) = 20 is the lowest multiple of 5 we can obtain using a + (k \cdot m))
        • Thus, m = 2
      • 8 \oslash _6 5 = (8 + 6(2)) / 5 = 20 / 5 = 4
  • For the purposes of this article, let’s assume we are working in modulo 9 (i.e., our normal 0,1,…,9 digit system)

Digital sum (or root) = an operation taken on a positive rational real number x \in \mathbb{Q} ^+ that returns an integer n = 1 ... k (we are assuming k = 9) which corresponds to repeatedly taking the sum of all digits in x repeatedly until obtaining a single digit

  • Examples:

x = 25679 \Rightarrow 2 + 5 + 6 + 7 + 9 = 29 \Rightarrow 2 + 9 = 11 \Rightarrow 1 + 1 = 2 \Rightarrow n = 2

x = 3.6793 \Rightarrow 3 + 6 + 7 + 9 + 3 = 28 \Rightarrow 2 + 8 = 10 \Rightarrow 1 + 0 = 1 \Rightarrow n = 1

Doubling = pick a starting number x_0, multiply it by two, take the digital sum of the result (x_1) and repeat the process until you get x_i = x_0

  • As it turns out, in mod 9, and starting at x_0 = 1, we get x_6 = x_0
    • Proof:
      • x_0 = 1 \Rightarrow 2 x_0 = 2(1) = 2 = x_1
      • x_1= 2 \Rightarrow 2 x_1 = 2(2) = 4 = x_2
      • x_2 = 4 \Rightarrow x_2 = 2(4) = 8 = x_3
      • x_3 = 8 \Rightarrow 2 x_3 = 2(8) = 16 \Rightarrow 1 + 6 = 7 = x_4
      • x_4 = 7 \Rightarrow 2 x_4 = 2(7) = 14 \Rightarrow 1 + 4 = 5 = x_5
      • x_5 = 5 \Rightarrow 2 x_4 = 2(5) = 10 \Rightarrow 1 + 0 = 1 = x_6 = x_0
    • Importantly, if we ignored the step where we take the digital sum, each x_{i} value will have a digital sum corresponding to the the value it would have obtained if we did include the digital sum step
    • Morover, this \begin{matrix} 1 & 2 & 4 & 8 & 7 & 5 & 1 & ... \end{matrix}    cycle appears as long as we choose an x_0 with a digital sum of either 1, 2, 4, 5, 7, or 8
      • Example: x_0 = 7
        • x_1 = 2(7) = 14 \Rightarrow 1 + 4 = 5
        • x_2 = 2(5) = 10 \Rightarrow 1 + 0 = 1
        • x_3 = 2(1) = 2
        • x_4 = 2(2) = 4
        • x_5 = 2(4) = 8
        • x_6 = 2(8) = 16 \Rightarrow 1 + 6 = 7
      • Here is an image depicting this pattern as a sequence of transitions between nodes/states: blue path
      • Note: we also get this pattern from taking the digital sum of sequential powers of two
        • Proof:
          • 2^0 = 1
          • 2^1 = 2
          • 2^2 = 4
          • 2^3 = 8
          • 2^4 = 16 \Rightarrow 1+6 = 7
          • 2^5 = 32 \Rightarrow 3 + 2 = 5
          • 2^6  = 64 \Rightarrow 6 + 4 = 10 \Rightarrow 1+0=1
  • However, in mod 9, and starting at x_0 = 3, 6, \mathrm{or} 9 or rational numbers whose digital sum is divisible by one of these numbers, we get different patterns
    • If we let x_0 = 3, we get…
      • x_1 = 2(3) = 6
      • x_2 = 2(6) = 12 \Rightarrow 1+2 = 3 = x_0
  • Thus, for any given x_0 \in \mathbb{Q}, every x_i = 2x_{i-1} can be mapped to a sigle digit using doubling

Diagram

vortex_mathematics

  • Blue nodes corresponds with “physical” energy states
  • Red nodes corresponds with “metaphysical” energy states

Sources

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