
when pi is sqrt10 then algebraic closure is exactly 1*10^1207 with infinity border as Floorpi*10^603; Perfect inverse number #1584 Correcting Math
Posted:
Feb 23, 2014 3:47 PM


when pi is sqrt10 then algebraic closure is exactly 1*10^1207 with infinity border as Floorpi*10^603; Perfect inverse number #1584 Correcting Math
Now whenever anyone starts a conversation in mathematics saying how unique 10 is and that it is the Algebra inverse number since 1/10 is 0.1 or any multiple of 10 such as 10^3 has a "perfect inverse" of 10^3. So 10 is the number in mathematics that is the PERFECT INVERSE number. The trouble one usually finds in mathematics is that most mathematicians are filled up with garbage on this topic for they immediately lash into contorted notions that 10 is only special because of biology evolution has given humanity 10 fingers to count with, as if humanity had 8 fingers that these same mathematicians could foster their silly and daffy notions that 10 is not special.
Well, 10 is special, no matter how many fingers humanity has. And 10 is base independent and what I mean by that is that you have to arrive at base 10 first in mathematics in order to work and translate into the minorbases of any base other than the base 10. In other words, base 10 is the superior base and the reason for this is because base 10 is the only base in which the square root of 10 is close to and actually replaces pi. In binary base, 10 is of course the number 2 but the square root of 10 in base 2 is 1.414.. whereas in base 10, square root of 10 is 3.1622.. which is the replacement of pi.
So why is 10 probably the single most extraordinary number in mathematics? Because it is "true pi" and where the pi of old times as 3.14159.. is the number that is the preliminary or prelude pi. Old pi is much like scaffolding in building an architectural building where the scaffolding is necessary, but once the building is up, the scaffolding is removed and what the building in math of pi is, is that of square root of 10.
So we use Old pi of 3.14159.. to use in finding the borderline of finite with infinite as to where the Tractrix area crosses over that of the Circle area since both are equal at infinity. We find this spot to be Floor pi*10^603 where old pi has three zero digits in a row for the tractrix to catch up and briefly surpass the area of corresponding circle. But Floor pi*10^603 has no clean inverse but a messy inverse. So we take infinity to be 1*10^603 and its microinfinity to be 1*10^603. But notice that now we throw out Old pi and use sqrt10 as pi and notice that squaring sqrt10*10^603 delivers to us not a messy number but that of precisely 1*10^1207 for the squaring of square root of 10 is 10.
So, this leads us into the question, do I have evidence or proof of what I say above? Well, there is good evidence and proof in the fact that in order to square the polygon, that a semicircle is required or necessary and that we cannot square the polygon unless we have existing the OVER ARCHING semicircle to turn a rectangle into a square.
Now, why is this so? Well the best answer I have is that if you take any righttriangle and instead of the Pythagorean theorem being three squares on the two legs and hypotenuse where a^2 + b^2 = c^2, that three semicircles on that righttriangle obeys the same formula of a^2 + b^2 = c^2, so to transform a rectangle which is made of two right triangles, we require the Pythagorean theorem of Semicircles.
Because we require out of necessity the Pythagorean theorem of Semicircles, means that pi at infinity is not 3.14159.. but is rather instead square root of 10. And we can prove this by substituting the Over Arching semicircle in the proof below with that of a 10^603 polygon replacing the circle.
So in New Math, with true pi being square root of 10, what happens is that all curved objects are thrown out as nonexistent, as fakes, and that the true object of a curve or circle are tiny straightline segments so small that the human eye sees them as curves or circles when in fact they are simply tiny straightlines.
Now here is another website that transforms a polygon into that of a square. And I was looking to see if they could do it without involving a circle.
 quoting from  http://www.qc.edu.hk/math/Certificate%20Level/Polygon%20area.htm
Finally we have to construct a square of area equal to the area of a given rectangle. We use the same rectangle produced in Step 2, that is, AGHI.
Given rectangle AGHI,
1. Produce AG to J such that GJ = GH. 2. Find the mid point O, of AJ. 3. Use O as center, OA as radius, draw a semicircle. 4. Produce GH to meet the semicircle at K. 5. Use GK as side, construct a square GKLM. 6. Area of AGHI = Area of GKLM
 end quoting from website 

Recently I reopened the old newsgroup of 1990s and there one can read my recent posts without the hassle of mockers and hatemongers.
https://groups.google.com/forum/?hl=en#!forum/plutoniumatomuniverse Archimedes Plutonium

