1 as Division Through Infinite Change

What is the basis for reason? And mathematics?

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Eodnhoj7
Posts: 8595
Joined: Mon Mar 13, 2017 3:18 am

1 as Division Through Infinite Change

Post by Eodnhoj7 »

1) One exists if and only if there is 1.

2) The existence of 1 occurs if and only if there is 1 and 1 occurs if and only if there exists 1. “Existence” is inherent in 1 and vice versa as both form and function, where “existence” can be viewed as either form or function depending upon the relativistic point of measurement, One exists through itself, hence is directed towards itself.

3) One directed towards itself divides itself as 1 and 1. This self direction results into a division of 1.

4) This division of 1, through one tending towards 1, results in 2 through the fraction of 1. 1 however maintains itself as 1 considering all division through 1 and extensions of 1 requires 1 in itself to maintain itself as the same. 1 dividing itself results in 2 as it folds through itself. “∢" observes the inherent “number of “1’s”” which compose the fraction as a form of self folding.

(1/1 ∢ 2) → ((1/1)/1 ∢ 3) → (((1/1)/1)/1 ∢ 4)→ ((((1/1)/1)/1)/1 ∢ 5)


5) 1 progresses to 2 through self division, with 2 (as 1 dividing itself) dividing 1 as 3, etc. All acts of 1 dividing itself through fractals results in whole numbers.

6) 1 as self-dividing, maintains this division ad-infinitum, hence 1 tends towards infinity as 1.

7) 1 as self dividing, forms 1 as its own boundary through division and in these respects 1 exists as infinite division.

8 ) 1 exists as infinite change through itself with 1 existing as division in itself. This division is a result of 1 tending towards itself as its own predicate to maintain its existence.






((∃1 ↔1) ∧ (1 ↔∃1))→(((1 →1) →1/1)∢ 2 ) →(((1 →1→1) →(1/1)/1)∢ 3 ) → ∞






1→ ((1 →1) →1/1)∢ 2 → (1/((1/1∢ 2) )) ∢ 3 → (1/((1/((1/1∢ 2) )) ∢ 3)) ∢ 4 → ∞






1/((1/1∢ 2)→((1/1)/1 ∢ 3)→(((1/1)/1)/1 ∢ 4) →((((1/1)/1)/1)/1∢ 5) →(((((1/1)/1)/1)/1)/1 ∢ 6)→((((((1/1)/1)/1)/1)/1)/1∢ 7) →(((((((1/1)/1)/1)/1)/1)/1)/1 ∢ 8)→ ∞ )






1/(1→ ((1 →1) →1/1)∢ 2 → (1/((1/1∢ 2) )) ∢ 3 → (1/((1/((1/1∢ 2) )) ∢ 3)) ∢ 4 → ∞)





(1/(1→2→3→4→5→6→7→8→ ∞))∢ (2→3→4→5→6→7→8→9 → ∞)




((1/(1→2→3→4→5→6→7→8→ ∞))∢ (2→3→4→5→6→7→8→9 → ∞))
= (1 ∢ 1/∆(1→∞) )


Thread with paper attached at bottom to give clarity to equations:

http://arktos.boards.net/thread/2595/1- ... llTo=28710
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