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Topic: Squaring the polygon is as common as squaring the circle #1582
Correcting Math

Replies: 3   Last Post: Feb 23, 2014 12:22 AM

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 plutonium.archimedes@gmail.com Posts: 9,428 Registered: 3/31/08
Squaring the polygon is as common as squaring the circle #1582
Correcting Math

Posted: Feb 22, 2014 4:04 PM

Squaring the polygon is as common as squaring the circle #1582 Correcting Math

In New Math, since circles do not exist because they are just compilations of tiny straightline segments, means all circles are polygons, regular polygons that have the smallest unit of distance as a side. Such as 1*10^-603 or 1*10^-1206. That is why the alternate pi is square root of 10 so that spherical coordinates are that of 1*10^-1206 while planar coordinates are 1*10^-603.

Now, does anyone have the quickest means of converting a regular polygon into a square? Is it cut up into triangles and pieced back together to form a square?

You see, it is sad that the history of math took the route of thinking pi was silly transcendental so as to escape squaring the circle, when the route math should have taken to be truthful, is that no circles exist but that a polygon with tiny straightlines is the circle and hence, squaring the polygon is always true.

Yesterday I wrote:
Now a lot of people are going to instantly misunderstand me as to what I am doing with pi. I am finding out what the real pi is, not the old fake pi used in current mathematics. How can there be a fake pi? Well, if the world has no curves, no circles but rather all of these items are tiny straightline segments so small that the mind imagines them to be a curve or a smooth curve or a circle, when in fact they are compilations of tiny straightlines.

Your computer itself is proof of what I speak because the circles you see on the computer screen are actually tiny straight lines so small that your brain imagines them to be circular smooth curve.

Did you ever wonder why pi was thought of as transcendental-- never a root to a polynomial? Well the reason is that pi never existed in the first place other than a Rational number of 3.14159..132000 that ended at 603 digits rightward of the decimal point, which is a number no more significant than say 3.3333... which has 603 digits rightward of the decimal point.

So if smooth curves never exist other than being Compilations of tiny straightline segments, then we need a Formula for the perimeter of a many sided polygon that is imagined as a circle. What is the formula for the perimeter or circumference of a polygon that looks like a circle?

Well, that perimeter or circumference must be a formula similar to C = pi * diameter, only where pi is not 3.14159.. but rather where it is a number 3.14...
6000 having 603 digits rightward of decimal point and evenly divisible by 120^3?

Why 120^3? because if you take the 5 regular polyhedrons and placed them inside of what we imagine a sphere and examine the vertices, the 50 vertices that intersect with the "imagined sphere" surface, that those vertices form a pattern that requires the True Pi to be evenly divisible by 120^3.

Now, in New True Math, squaring the circle exists, because no circles themselves exist. The polygon that imitates the circle does exist and it is easily rendered into a square.

Earlier today I wrote:
Why pi has to be evenly divisible by 120^3 at infinity border; A Synthethic pi #1580 Correcting Math

Alright, I am presuming one fact for which I have not investigated. I am assuming the reason the Computer when I had access delivered the fact that pi digits are evenly divisible by 120 at that of 10^-245 and then at 10^-601, 10^-602, 10^-603. So I am assuming that the even divisibility of pi at 601, 602, 603 is because pi has three zero digits in a row, and not the fact that pi is evenly divisible by 120^3 = 1728000 at 10^-603. I am presuming that fact and which is suggested by the fact that the digits leading into 32000 are digits that look to be not divisible by 120^3. as that of

3.14159.. 84674818467669405132000 (pi to the 603rd digits rightward of decimal point)

At least I do not think those digits indicate even divisibility by that of 120^3.

But when you command a computer for even divisibility by 120, the computer is going to give you a answer of 120 and not tell you it was 120^3.

So, assuming the above is only 120 even divisibility, I then seek to find a synthetic manufactured pi that has 603 digits rightward of decimal point and which is evenly divisible by 120^3 and has ending digits of just three zeroes
as looking like this 3.14..6000.

For example, if we take 314159000 and divide by 1728000 we get about 181.8. Now we take 1728000 and multiply by 182 and get 314496000. Now that newly synthesized pi is almost the same as pi only slightly larger and having the characteristic feature of even divisibility by 120^3.

Now, why pursue this manufacturing of a new pi? Well, the theory behind it is that mathematics has no curves at all but are totally tiny straightline segments and when they are so tiny, we only visualize them as a smooth curve. So the pi of Old Math when circles do not exist at all, means the circle is really some immense large number regular polygon. Now Nature provides us with Regular Polyhedron, of which there are 5 in total. Now, if we enclosed all 5 of those regular polyhedron inside a sphere and studied their intersection with the surface of the sphere by the vertices of the regular polyhedron, they form a pattern on that surface. If I remember correctly, there are 50 vertices in all of those 5 existing regular polyhedron. Now how are those vertices spaced relative to one another as the intersection on the sphere?

Well, the spacing would be a even divisibility of 120 by 120 by 120. What I mean by that is consider the xy plane as a cross section and then the yz plane as a cross section and then the xz plane as a cross section of a sphere in 3rd dimension. You need the xy, the yz, and the xz planes to fully describe the sphere and the enclosed regular polyhedron. So this means that pi has to be evenly divisible at infinity by 120^3 in order to preserve the even divisibility of regular polyhedron.

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Recently I re-opened the old newsgroup of 1990s and there one can read my recent posts without the hassle of mockers and hatemongers.