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why sqrt10 is the new pi in New Math, since circles do not exist; Squaring the Polygon #1583 Correcting Math
Posted:
Feb 23, 2014 1:47 AM


why sqrt10 is the new pi in New Math, since circles do not exist; Squaring the Polygon #1583 Correcting Math
Now in New Math, we find the infinity borderline by the unit Tractrix area crosses over the corresponding unit circle area for the first time at Floorpi*10^603 and the reason it does so is because pi has three zero digits in a row for the first time that allows the Tractrix area to increase while the circle area lies stagnant in zeroes.
So in New Math, there are no curves in geometry as there is no continuum but rather points with holes in between successor points. And since geometry is points with holes in between means that every joining of points that exist is a joining by a tiny straightline segment. So no smooth curve exists and a circle is not really a smooth curve but is a polygon if our eyes could microscope down to the tiny straightline segments. This is proven by modern day computers because every circle you see, you are really seeing tiny straightlines compiled to give the illusion of a smooth curve.
So that pi as the measure of pi times diameter as the circumference of a circle has to be reshaped, to mean pi times "new pi" is the perimeter or circumference of a many sided regular polygon. How many sides? Well if infinity is 10^603 then it is a 10^603 regular polygon.
Now for years I have been saying that infinity border for planar coordinates is 1*10^603 for it gives a easy and simple inverse of 1*10^603, whereas 3.1415..*10^603 gives no simple inverse for a microinfinity. And I called Floor pi * 10^603 as spherical coordinates, but never giving justification other than to say, the algebra is made that much easier to use 1*10^603 rather than Floor pi. Now I seem to have learned how I can reconcile 1*10^603 as macroinfinity border rather than Floorpi*10^603.
If we take real pi to be sqrt10 = 3.162277.. and then we take the Algebraic Completeness of 1*10^603 to be that of 1*10^1206, we see that we can take sqrt10 as the 1*10^1206, for if you take the square root of 1*101206 you end up with 1*10^603. So here I found justification in making square root of 10 to be the real pi of the world, rather than the 3.14159.. version. That version was begot from thinking the world is a continuum and has no border between finite and infinity.
This also allows pi as sqrt10 to be irrational for it is irrational since it has at least two different numbers to compose its conjugates such as 3.1622 and 3.1624 as conjugates in abbreviated form since multiply those two together gives us 10.000 in abbreviated form (actually there must be at least 604 or more digits rightward of the decimal point).
So, now, I went to check on an old proof. I do not know who first proved this proof that given any polygon, it can be squared. I went to check because I sort of remember that the proof requires the use of a semicircle. Now can we do the proof by never using a circle? I am a little hesitant to say yes. And if circles do not exist, then the circle if required would be a regular polygon itself. So here is a means of cross checking a proof for we would need a regular polygon to prove the area of a polygon is turned into a square.
Anyone know who first proved the transformation of polygon area into square? I would guess Archimedes of ancient Greek. But it may have come later.
I thought it wise to quote as much of the proof so that in future, if the website is removed we still have the data.
 quoting from http://www.proofwiki.org/wiki/Construction_of_Square_Equal_to_Given_Polygon 
Let A be the given polygon. Construct the rectangle BCDE equal to the given polygon. ?If it so happens that BE=ED , then BCDE is a square, and the construction is complete. ?Suppose BE?ED . Then WLOG suppose BE>ED . Produce BE from E and construct on it EF=ED . Bisect BF at G . Construct the semicircle BHF with center G and radius GF (see diagram). ?Produce DE from D to H . ?From Difference of Two Squares, the rectangle contained by BE and EF together with the square on EG is equal to the square on GF . But GF=GH . So the rectangle contained by BE and EF together with the square on EG is equal to the square on GH . From Pythagoras's Theorem, the square on GH equals the squares on GE and EH . Then the rectangle contained by BE and EF together with the square on EG is equal to the squares on GE and EH . Subtract the square on GE from each. Then the rectangle contained by BE and EF is equal to the square on EH . So the square on EH is equal to the rectangle BCDE . ?So the square on EH is equal to the given polygon, as required.
 end quoting 

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