On Jun 18, 1:02 pm, Archimedes Plutonium <plutonium.archime...@gmail.com> wrote: > On Jun 18, 5:20 am, Archimedes Plutonium > > <plutonium.archime...@gmail.com> wrote: > > Let me work backwards and design the number I want. It must be evenly > > divisible by > > 2,3,4,5, thus 120 to obey Euler regular polyhedra formula. It must be > > a perfect > > cube. > > > Now floor pi 10^603 obeys Euler formula, but is not a perfect cube. So > > let us try > > 1x10^603 for it is a perfect cube but does not obey division by 3. > > > So, is there any number between 10^603 and floor pi 10^603 that obeys > > those > > requirements? Let me sleep on it. > > Alright I slept on the question and problem and like this outcome. > > I can take the number 120 as the baseline number with repeated self > multiplication > to arrive at a number between 10^603 and 10^604. Since 120 is evenly > divisible by > 2,3,4,5 or 5!, then I am guaranteed of even divisibility to satisfy > Euler regular > polyhedra formula for geometry. Our borderline must satisfy geometry > and essentially > the Euler formula is one of the most important. There are others, such > as the formula > that pseudosphere and respective sphere have equal area at infinity. > But I think with > all those zeroes the pseudosphere will have equal area. > > Now the problem I face here, in engineering this borderline, is > whether it exists > between 10^603 and floor pi 10^603, or whether I have to be more > lenient and require > it to exist between 10^603 and 10^604? Or, whether it even exists is a > question mark? > > It is obvious such a engineered number must satisfy Euler formula and > pseudosphere area > and other geometry results, but it is not at all obvious that it must > be a perfect cube. > I have no justification for making this engineered borderline infinity > number a perfect cube, other than to say that Physics is 3rd > dimensional and so volume of Space, finite Space would end at the > borderline and there is no reason for space to prefer the x or y > or z axis over one another. This is the Maxwell Equations in that > space is symmetrical of > 3rd dimension. So for that reason, I seek to have the borderline a > perfect cube. >
Well, the Maxwell Equations are Elliptic geometry where cubes and squares do not exist, but rather circles and spheres. This is the Coulomb law or inverse square law. And that law, or the Maxwell Equations in toto are symmetrical. When we translate that into Euclidean geometry, the sphere becomes a cube. So I need the borderline of finite with infinite to be a perfect cube. So if I keep multiplying by only the number 120, do I end up with a number that falls between 10^603 and 10^604 and is a perfect cube?
> So, does there exist a number that is between 10^603 and 10^604 that > is a root multiple of > 120, that is 120^n, and for which it is evenly divisible by 2,3,4,5 > and is a perfect cube? > Is there such a number that fits that description and is it a unique > number between 10^603 > and 10^604? >