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Is One Equivalent To Infinity?

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#1 eodnhoj7



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Posted 19 October 2017 - 10:24 AM

Presented Argument:

1)  "1" exists as "unity" and "wholeness".

2)   As unity and wholeness "1" must maintain a self-reflective symmetry in order to exist.  We observe this in common life where a physical or abstract structure maintains its stability through the reflection
     of points, with these points in turn forming a symmetry between eachother which allows the structure to exist.

3)   "1" reflects upon itself to maintain itself.  In doing so it reflects "2" (as: 1 ≡ 1 ≅ 2), "3" (as: 1 ≡ 1 ≡ 1 ≅ 3), "4"              (as: 1 ≡  1 ≡ 1 ≡ 1 ≅ 4), etc. unto infinity. In this respect all numbers are strictly structural extensions of 1.

4)   As structural extensions of "1" all numbers reflect both "1", themselves "1x", and eachother "1y" unto infinity.  In this respect 1 reflects infinity.

5)   All number, including "1", continually manifests through self-reflective symmetry unto infinity.  Infinity, as "totality", is synonymous with both "unity" and "stability" for there is no deficiency in it.  In this respect both 1 and infinity are equal.

6)   The self-reflective nature of 1, 1xy, and infinity observes a circular reflective symmetry, and in this respect observes self-reflection as the "maintenance of structure through the maintenance of center(s)".

7)   The nature of "1" as infinite through reflective symmetry, observes all number as structural extensions of "1" as mere approximates of "1".  In this respect, all approximates observe a form of "deficiency in unity."  This deficiency is unity is not a thing in and of itself, as all number is composed of "1" and exists as "1" reflecting upon itself, therefore it is equivalent to 0.  0 is strictly the limit of infinity, as an observation that only infinity exists as 1.

Agree, disagree, don't know?  Explain why.