So the mystery of pi-- what Old Math called -- transcendental is no longer really a mystery at all, for there are TWO numbers that act like pi, one of which is very much algebraic-- the square root of 10.
And why does mathematics require for one number to be in fact TWO numbers? It is because physics has quantum duality, meaning that something is both particle and wave simultaneously. The difference between 3.162... and 3.141... is a miniscule difference
of 0.021... In fact there are many many prefix numbers digits of important physics constants-- especially Magnetic Flux Quantum, and weak mixing angle (although we ignore the Standard Model b.s.), Electron g-factor (really the Dirac magnetic monopole g-
factor) and the Muon-g-factor, the Rydberg unit of energy.
Why does mathematics need at least One number that is represented by TWO Numbers? It is because of physics duality demands one number of mathematics define the Sigma Error contained within mathematics. Pi as both 3.16... and 3.14... sets the standard for
equality in mathematics as sigma error of 3.16/3.14 = 0.6%. If in physics experiments, the outcome is 0.6% or less you have equality of concepts. If more, then you do not have equality of concepts.
For the layperson in all of this, can be summarized as that geometry needs a sigma error, for a line is of length and no width and no depth. But, that means a contradiction for a line has to have some width some depth to exist. In steps our 0.021 factor
in that a Line in Geometry has at least a 0.021 factor of width and depth in order to exist.
Now, I worked on Infinity Borderline for pi as 3.14.... some decades in the past from this date 12Aug2023, ignoring square root of 10. So now, since pi is actually TWO numbers acting as one, I need to see what the infinity borderline is for 3.16... For 3.
14... where pi has its first three zero digits in a row at 10^-604 place value is the infinity borderline and its algebraic completeness at 10^-1208. Both of these spots is evenly divisible by 5! and in fact 10^-1208 is evenly divisible by 6! as one
computer read-out claimed (needs further verification)
So for 3.16... I need to see if 10^-604 and 10^-1208 are evenly divisible by both 5!= 120 and 6! = 720. Why 5! you ask? Because at infinity we need the existence of smooth polyhedra and polygons in geometry.
Now immediately I recognize there is a "9" digit for 3.16... in the 10^-604 spot so I know it is not evenly divisible there by 5! or 6!. However in the Algebraic Completeness of Infinity borderline there is a "0" digit and a good chance it is evenly
divisible by both 120 and 720.
Archimedes Plutonium<
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3:58 AM 12Aug2023
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Archimedes Plutonium<
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