9 famous proofs of math: recent commentary

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Archimedes Plutonium

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Jun 29, 2014, 2:53:13 AM6/29/14

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Newsgroups: sci.math

Date: Sat, 28 Jun 2014 16:27:29 -0700 (PDT)

Subject: Big Small Tiling Conjecture that tells when a regular object/s can

tile Space #1865 Correcting Math

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Sat, 28 Jun 2014 23:27:30 +0000

Big Small Tiling Conjecture that tells when a regular object/s can tile Space #1865 Correcting Math

Alright, let make standard some definitions so that confusion is diminished.

Square when squashed forms Rhombus

Rectangle when squashed forms Parallelogram

Cube when squashed forms Rhomboid

Rectangular Solid when squashed forms Parallelepiped

Now, let me walk through the Logic of this Conjecture and its proof, even though I have rough conjecture and a rough proof in mind.

Rough Conjecture: given a regular solid/s joined into a cell (need to define cell here), that if we can divide the cell by a integer number and fit small cubes or rhomboids into that cell, means that the regular solid/s tile all of Space.

For example, suppose we did not know that a cube or rhomboid tiles all of Space, then we take our sample cube, say 5 by 5 by 5 whose volume is 125 and now we fetch 125 cubes of 1 by 1 by 1 and see if we can fill that larger cube. Obviously we can. Now we take the 1 by 1 by 1 and see if we can fill that with .5 by .5 by .5 cubes. Here we need 125 of .5 by .5 by .5 to fill the 1 by 1 by 1.

Now we need not stop with the cube figure or a rhomboid figure filled by smaller rhomboids. We can take the Rhombic Dodecahedron and divide it into a integer number of rhomboids that fills the Rhombic Dodecahedron.

Now we take the 2 Octahedron with 2 Tetrahedron and arrange them into a cell. Can we find a rhomboid of integer number to fill that cell? We cannot fill a single individual octahedron with small integer number of rhomboids, nor the individual tetrahedron. But in combination of 2 octahedron with 2 tetrahedron, we can find a integer number of small rhomboids that fills the cell.

So, this is the logic of the Conjecture and its proof. We know that cubes and rhomboids tile all of space. To find out if another figure tiles all of Space, we find out if a cell of that figure/s is able to be tiled by small cubes or small rhomboids.

So, the conjecture and the proof boils down to the idea that Space is already tiled by cubes and rhomboids. Any other Space tiler, must then be a smaller sized cube or rhomboid of a integer number.

END

Newsgroups: sci.math

Date: Sat, 28 Jun 2014 23:00:29 -0700 (PDT)

Subject: Kissing Point Faces to tile Space #1866 Correcting Math

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Sun, 29 Jun 2014 06:00:29 +0000

Kissing Point Faces to tile Space #1866 Correcting Math

On Saturday, June 28, 2014 6:27:29 PM UTC-5, Archimedes Plutonium wrote:

> Big Small Tiling Conjecture that tells when a regular object/s can tile Space #1865 Correcting Math

Alright, I am getting too many counterexamples such as a notch in a cube tiling. So although we can fill a notched cube by tiny small cubes except for the notch itself.

So that tells me I have some important concept missing.

The concept missing is face-filling. So if there was a notch in a face, another cube face cannot fill that face.

So I need Kissing Point Faces, where two faces join to form space filling.

Some progress is made though, in that I boiled down all space fillers as to either cubes or rhomboids (squashed cubes).

Now has anyone computed the ratio of how many small rhomboids fill a given rhombic dodecahedron?

Or how many fill a cutaway of the octahedron - tetrahedron stacking?

With rectangular space filling we can easily divide how many cubes would fill that space of rectangle solids.

So I need the Kissing Point Faces to get rid of counterexamples.

END

Newsgroups: sci.math

Date: Sat, 28 Jun 2014 23:47:25 -0700 (PDT)

Subject: Conjecture: all figures that tile space are a composition of cubes

and rhomboids #1867 Correcting Math

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Sun, 29 Jun 2014 06:47:25 +0000

Conjecture: all figures that tile space are a composition of cubes and rhomboids #1867 Correcting Math

I guess I need to backtrack here a bit with another conjecture, for I implied it within my earlier conjecture.

I implied that the only two building blocks of tiling are cubes and rhomboids (squashed cubes).

There is another candidate of triangular prisms but when put together, they are cubes or rhomboids. Here we have to muddy the water by asking which is more primitive, the triangular prism or the cubes and rhomboids?

Here I would ask if we think of all of Space, we usually think of it as a cube or squashed cube but hard pressed to think of Space as a gigantic triangular prism.

So, I have got to say that the triangular prism is a packing of 2 of them, just as the octrahedron with tetrahedron come in a ratio for tiling.

So, what this conjecture says that the tiling of Space all boils down to cubes and rhomboids. The proof maybe difficult.

Now my last two conjectures will eventually have to reconcile with this MathWorld page:

http://mathworld.wolfram.com/Space-FillingPolyhedron.html

And if we examine one of those figures on that page-- hexagonal prism.

So are my two conjectures-- all regular figures that tile space boil down to an integer number of cubes or rhomboids, and cubes and rhomboids are the only two building blocks, conforming to the hexagonal prism?

Well, if you make an incision forming two triangular prisms and join them leaving us with a rectangle and a rhomboid. So the hexagonal prism conforms.

Now the other figures on that MathWorld page, look to be somewhat difficult, but I am confident that they form a cell in which cubes and rhomboids fill all the space of that cell, and thus tile Space.

Archimedes Plutonium

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Jun 29, 2014, 4:42:12 PM6/29/14

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Date: Sun, 29 Jun 2014 12:47:05 -0700 (PDT)

Subject: Re: Stanford Univ professors endorsing Goldbach proof to arxiv

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Sun, 29 Jun 2014 19:47:05 +0000

So in mathematics, do we ever have a conjecture and following theorem about the odd numbers? We seem to only have them about even numbers such as No Odd Perfect or Goldbach. The action seems to lie in the even numbers.

Of course, the most important odd number is 1, but even there, no conjectures and theorems about 1 and the odd numbers. And the reason I am guessing of this lack of activity over odd numbers is that the odd numbers are asymmetrical and the even numbers as a set are symmetrical.

I am going to try to think up some important conjectures involving odd numbers.

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Date: Sun, 29 Jun 2014 13:00:25 -0700 (PDT)

Subject: Re: MIT professors endorse proof that "Maxwell Equations derive the

Peano axioms" to arxiv

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Sun, 29 Jun 2014 20:00:26 +0000

In this proof that the Maxwell Equations contain mathematical induction and thus Physics contains all of Number theory of mathematics, I should mention what Physics contains and Math does not contain.

Of course, I have to prove that the Maxwell Equations contains all of geometry of mathematics, not just Number theory. Here it is easy to show the Maxwell Equations contain the Pythagorean theorem for the theorem is equivalent to the Parallel Line Postulate of geometry.

But let me talk about what Physics contains in the Maxwell Equations which is above and superior to mathematics and which math can never deal with. In a sense, it is logic itself, of that of duality. The logic of Physics contains duality of a item having both properties of A and B wherein A is not equal to B.

In fact, we see duality in mathematics all the time in that we can have the Pythagorean theorem as a geometry picture or we can have it as Numbers of m^2 + n^2 = z^2. We do not say that the number 9 is a square, for they are two different things, one is a number, the other is a geometry figure.

The Logic of mathematics is only Aristotelian logic and cannot possess a duality logic along with Aristotelian logic.

The Logic of Physics is containing both Aristotelian and Duality Logic.

That is what the 4 Maxwell Equations are-- 2 of them about electricity and 2 of them about magnetism. Electricity is the dual of magnetism, neither equal to each other but contained in every physical thing and physical interaction.

In Mathematics, we cannot write any axiom set that is both Aristotelian and Duality Logic.

Archimedes Plutonium

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Jun 30, 2014, 7:12:36 PM6/30/14

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Date: Mon, 30 Jun 2014 13:33:26 -0700 (PDT)

Subject: rhomboid beats out over the MathWorld figures #1872 Correcting Math

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Mon, 30 Jun 2014 20:33:27 +0000

rhomboid beats out over the MathWorld figures #1872 Correcting Math

On Sunday, June 29, 2014 7:40:22 PM UTC-5, Archimedes Plutonium wrote:

(snipped)

> So take every one of those figures on MathWorld that alleges to be Space filling, and we know they are not Space filling when they reach the walls, floor and ceiling of the Space Cube. So the question is for the research student, what is that exact density for these:

> 1) unit rhombic dodecahedron

> 2) unit octahedra, truncated octahedron, with cubes in ratio 1:1:3

> 3) unit compound tetrahedra and truncated tetrahedra

> 4) unit triangular prisms not of a right angle

> 5) unit hexagonal prism

> 6) unit gyrobifastigium

> and many more shown on that page.

I do not recall seeing the rhomboid on that Mathworld page which would be erroneously thought of as a 100% tiler of Space. But when Space is a cube container the rhomboid-- squashed cube cannot fill that container 100%. However, the appropriate sized rhomboid would be, in my guess, the nearest to filling the Space but not able to be 100%. So that if we added the rhomboid figure to the MathWorld list that some research student will find the data, that the rhomboid excels those MathWorld figures.

We can sort of intuitively see this in that we make the rhomboid angles just a tiny tiny fraction off of 90 degrees and then packing them is almost like packing cubes.

Archimedes Plutonium

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Jul 2, 2014, 3:55:53 AM7/2/14

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Date: Tue, 1 Jul 2014 12:34:07 -0700 (PDT)

Subject: mathematicians need to go in and straighten out physicists, gemology,

chemistry on monoclinic crystal

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Tue, 01 Jul 2014 19:34:07 +0000

mathematicians need to go in and straighten out physicists, gemology, chemistry on monoclinic crystal

If one does a image search for monoclinic, one finds contradictions left and right on mathematics for they are claiming the Monoclinic crystal is a rectangular prism, when a rectangular prism has 6 faces of rectangles. They are claiming a monoclinic can have two 90 degree angles and be a parallelepiped.

So I think what has happened in the history of gemology, physics and chemistry, is that no-one in math came to sort these things out and tell them what is impossible and possible.

You can have a object whose alpha and gamma angles = 90 degrees but beta angle is not 90 degrees but you cannot call it a rectangular prism, nor even a parallelepiped, for such a object has no mathematical name. What it is, is a stacking of thin rectangular solids whose edges are therefor notched.

You cannot be drawing pictures of monoclinic crystals with straight line edges, but rather notched all along the edges.

You cannot use the terms rectangular prisms, parallelogram, parallelepiped nor draw pictures of them when describing Monoclinic Crystal Structure.

END

Newsgroups: sci.math

Date: Tue, 1 Jul 2014 12:52:20 -0700 (PDT)

Subject: packing rhomboids in cube container #1875 Correcting Math

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Tue, 01 Jul 2014 19:52:20 +0000

packing rhomboids in cube container #1875 Correcting Math

Alright, more work for the research student doing the tables of data on the MathWorld tiling figures.

I need to include rhomboids (squashed cubes) versus cubes and that of parallelepipeds (squashed rectangular solids) verse rectangular solids.

Here what I want is a data on using a cube container and how many rhomboids fit into it versus a rhomboid container and how many cubes fit into it.

For example, pretend infinity is 10 then a cube is 1000 cubic units and if we have 1 unit rhomboids, I can visualize that 100 of them (unless I made a mistake) will fail to tile the cube. Now reversing that situation where we have a rhomboid container whose volume is 1000 cubic units and how many unit cubes can fit into it? Is it 900, same as rhomboids into cube container?

I think I made the conjecture already that of the MathWorld figures that falsely claim to tile space, and where Mathworld omits the rhomboid and the parallelepiped, that I conjectured the closest to being 100 % tiling but missing is the rhomboid and parallelepiped.

In 3rd dimension, only the cube, rectangular solid, and right angled wedge can tile space 100%. I think the rhomboid and parallelepiped come closest to 100% but failing by a little bit compared to all those figures shown in MathWorld.

END

Newsgroups: sci.math

Date: Tue, 1 Jul 2014 15:51:47 -0700 (PDT)

Subject: mineralogy contradicts math with their monoclinic Re: mathematicians

need to go in and straighten out monoclinic crystal

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Tue, 01 Jul 2014 22:51:48 +0000

mineralogy contradicts math with their monoclinic Re: mathematicians need to go in and straighten out monoclinic crystal

--- quoting from ---

https://uwaterloo.ca/earth-sciences-museum/resources/crystal-shapes/monoclinic-crystal-system

Monoclinic crystal system

In the monoclinic system all the axes are different lengths. Two of them, the A and the C axes, meet at 90 degrees but the third one does not.

This crystal system has the following forms:

▪ 3 Pinacoids

▪ Prism and 2 Pinacoids

▪ Prism and 3 Pinacoids

▪ Prism and 5 Pinacoids

--- end quoting ---

What I am arguing here is that you cannot have a single solid figure composed of two 90 degree axes and a third axis not 90 degrees.

What you can have is say, razor blades as rectangles as rectangular solids and if you stack them at a oblique angle off 90 degrees, you can have what is called monoclinic. It would be a series of thin rectangular solids.

So do we want to have mineralogy science in utter contradiction to geometry of mathematics with their bogus monoclinic crystal, described as axes that are not 90 degrees?

We must have all sciences conforming to mathematics and not in contradiction to mathematics.

Archimedes Plutonium

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Jul 2, 2014, 4:00:00 AM7/2/14

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Date: Tue, 1 Jul 2014 22:00:36 -0700 (PDT)

Subject: continuity in math Re: Math Induction via multiplication #1876

Correcting Math

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Wed, 02 Jul 2014 05:00:36 +0000

continuity in math Re: Math Induction via multiplication #1876 Correcting Math

So, if continuity is never an issue in completeness or algebraic closure, then where is continuity to be found in math if at all? And what purpose does it have?

Well, the Infinite Numbers, those outside the boundaries of finite numbers and the irrationals form a set of numbers that is continuous between rational numbers and continuous beyond the macroinfinity borderline.

And what purpose does this continuity of Infinite Numbers provide? They form distance and length in geometry. In between any two infinite-numbers always exists another infinite-number. But we cannot say the same for rationals, because two rationals separated by 1*10^-603 have no more rationals between them.

END

Newsgroups: sci.math

Date: Tue, 1 Jul 2014 22:37:27 -0700 (PDT)

Subject: Alternative proof-- a necklace of cubes inside cuboid #1877

Correcting Math

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Wed, 02 Jul 2014 05:37:28 +0000

Alternative proof-- a necklace of cubes inside cuboid #1877 Correcting Math

Now it would be nice, really nice if the proof of no perfect cuboid could follow straightforward from the proof that no integer cube can be perfect-- all parameters as integers. For the proof that integer cubes cannot be perfect is that the diagonals have to be irrational.

So in some manner can we assemble a cuboid as a series of integer cubes and then utilize that fact that the diagonals of the cube are irrational? Well we can use Rational Cubes instead of integer cubes so we get smaller and smaller cubes.

But the trouble with this method is that the diagonals of the cuboid do not necessarily coincide with a series of diagonals of small cubes inside the cuboid that tiles the cuboid.

If somehow, I could relate the diagonals of the cubes tiling the cuboid with its diagonals, then I would have an alternative proof. Because we all know the sum of irrationals numbers remains an irrational number.

Now, maybe one way of doing this "relationship" of tying the diagonals of cubes with diagonals of cuboid is to ponder upon not tiling the entire cuboid but rather laying a taut string of cubes along the diagonal of the cuboid so that the diagonal of each cube is a part of the diagonal of the cuboid.

Now that would work, and then all I need to prove is that every cuboid diagonal can be exchanged for a string of cubes along the diagonal.

However, I would need to show that these cubes have rational sides, which should not be too difficult.

Let me sleep on this alternative proof.

END

Newsgroups: sci.math

Date: Tue, 1 Jul 2014 22:57:01 -0700 (PDT)

Subject: two concepts of density-- container density and kissing point lattice

density #1878 Correcting Math

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Wed, 02 Jul 2014 05:57:01 +0000

two concepts of density-- container density and kissing point lattice density #1878 Correcting Math

Alright, I really like how this proof of Kepler Packing has worked out. The Kepler conjecture is proven by lattice kissing point density. And then the more general conjecture given by Torquato and Jiao acknowledges that number of regular solids density is proven as a lattice kissing point density.

Finally, when we include Container Density, then the infinity borderline is called into play and that tiling of regular objects has only three Space tilers-- cubes, rectangular solids, and right angled wedges.

In most cases, the lattice density is far higher than the Container density. In sphere packing the lattice density is pi/sqrt18 = about 74%, but the container density of spheres is much smaller than 74%. The lattice density of Torquato and Jiao of octahedron is about 94% but as a container density is smaller.

The lattice density of rhombic dodecahedron in Old Math is 100%, but it does not tile Space because the container walls, floor, ceiling do not allow fractional rhombic dodecahedron and they cause the container density the Space density to be less than 100%.

So the issue of Packing in geometry is an issue of two types of density.

Archimedes Plutonium

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Jul 3, 2014, 4:36:13 AM7/3/14

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Date: Wed, 2 Jul 2014 01:33:28 -0700 (PDT)

Subject: powerful alternative proof of a string or chain of squares to build a

diagonal #1878 Correcting Math

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Wed, 02 Jul 2014 08:33:28 +0000

powerful alternative proof of a string or chain of squares to build a diagonal #1878 Correcting Math

Alright, I am pretty excited about this alternative method, only I have a suspicion it is too powerful and thus wrong, for which I have not found the error yet.

By too powerful I mean that we should be able to prove almost any rectangle diagonal is irrational. But what about a rectangle of sides 3 and 4, then the diagonal is not irrational.

The idea of the method of this proof is that in 2nd dimension given any integer sides of a rectangle we draw the diagonal and then we construct squares of tiny squares so that we have a chain or string of those squares touching vertices so that the diagonals of the squares when summed equals the total length of the diagonal of rectangle. The same applies to 3rd dimension with cuboids and a string of cubes on the diagonal.

So I am playing around with a graph paper and trying it out on a 3 by 2 rectangle whose diagonal would then be sqrt13. Now I need to formulate the size of square that will fit vertex to vertex and cover exactly the length of the diagonal of sqrt13. It must be a square with a rational number for its side, so that the tiny squares form a string or chain of those squares meeting the vertices of the 3 by 2 rectangle. So far I have not been able to manage how that would be necessarily a rational sided square using 3 and 2. So I have to sleep on this.

It sounds too powerful to be true, for it seems as though every rectangle of integer sides would have a irrational diagonal and we know that is not true.

So I need some rest and then perhaps can spot the error of this method. But if it works and there is no error in the method, then the method is so very powerful for many other proofs, not just "no perfect cuboid".

END

Newsgroups: sci.math

Date: Wed, 2 Jul 2014 03:19:34 -0700 (PDT)

Subject: Re: powerful alternative proof of a string or chain of squares to

build a diagonal #1879 Correcting Math

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Wed, 02 Jul 2014 10:19:35 +0000

Re: powerful alternative proof of a string or chain of squares to build a diagonal #1879 Correcting Math

On Wednesday, July 2, 2014 3:33:28 AM UTC-5, Archimedes Plutonium wrote:

> Alright, I am pretty excited about this alternative method, only I have a suspicion it is too powerful and thus wrong, for which I have not found the error yet.

Alright, the method kept me up and could not go to sleep.

> By too powerful I mean that we should be able to prove almost any rectangle diagonal is irrational. But what about a rectangle of sides 3 and 4, then the diagonal is not irrational.

I was very suspicious at first at how powerful the method is, and hence thought it to be wrong.

At very many attempts of creating a tiny square to cover the length of the diagonal of a integer rectangle, I failed until I thought of the idea of using at least two different sizes of squares, one size for the rectangle length and another for the rectangle width. And alternate them into the string that covers the diagonal.

> The idea of the method of this proof is that in 2nd dimension given any integer sides of a rectangle we draw the diagonal and then we construct squares of tiny squares so that we have a chain or string of those squares touching vertices so that the diagonals of the squares when summed equals the total length of the diagonal of rectangle. The same applies to 3rd dimension with cuboids and a string of cubes on the diagonal.

The idea is to cover the diagonal with a series of squares, where vertices touch and hence the length of the diagonal is equal to the summation of the tiny square hypoteni. Since all squares have irrational hypotenuse, the sum of irrationals is irrational.

> So I am playing around with a graph paper and trying it out on a 3 by 2 rectangle whose diagonal would then be sqrt13. Now I need to formulate the size of square that will fit vertex to vertex and cover exactly the length of the diagonal of sqrt13. It must be a square with a rational number for its side, so that the tiny squares form a string or chain of those squares meeting the vertices of the 3 by 2 rectangle. So far I have not been able to manage how that would be necessarily a rational sided square using 3 and 2. So I have to sleep on this.

If the rectangle were 3 by 4, then the diagonal is 5 and integer. And what saves this from the method is that a Pythagorean Triple multiplied by rational number k is still a P-triple with rational diagonal (hypotenuse). So I am in no trouble there.

But what numbers to assign to the small squares to cover the diagonal of a random rectangle which is not a P-triple?

For the 3 by 2 rectangle, I would need to divide by 2 to get 1.5 for 3, and then divide by 2 into 2 to get 1 for 2, but further divide by 2 to get 0.5. I want to reduce all rectangle sides such that they have a 0.5.

So the squares of 1.5 and 0.5 in alternation may cover the diagonal of the 3 by 2 rectangle. If not, if too large, then I go to the 0.15 square and 0.05 square until I reach the proper size that a string or chain of these alternating tiny squares exactly covers the diagonal of the 3 by 2 rectangle.

> It sounds too powerful to be true, for it seems as though every rectangle of integer sides would have a irrational diagonal and we know that is not true.

> So I need some rest and then perhaps can spot the error of this method. But if it works and there is no error in the method, then the method is so very powerful for many other proofs, not just "no perfect cuboid".

Let me sleep on that,, and see if it is good by tomorrow.

END

Newsgroups: sci.math

Date: Wed, 2 Jul 2014 13:31:06 -0700 (PDT)

Subject: what does the sequence .5, .05, .005, etc have to do with irrationals

#1880 Correcting Math

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Wed, 02 Jul 2014 20:31:07 +0000

what does the sequence .5, .05, .005, etc have to do with irrationals #1880 Correcting Math

Alright, the method is worth pursuing, although I think it is not a independent proof method from the two I already gave of No Perfect Cuboid. I think this method is part of those earlier methods. This method has some features worth pursuing.

For example, if the rectangle diagonal is a Pythagorean Triple say 3, 4, 5 then K time that of say .3, .4, .5 will also be triples. But if the rectangle diagonal is not a P-triple it must be irrational, not rational. And that is worth proving.

If the cuboid diagonal of any of its rectangles is not a P-triple, then the diagonal is irrational.

Now what I have been able to find out is somewhat remarkable and perhaps amazing as to why the numbers 0.5, 0.05, 0.005, etc seems to appear as a sort of "Irrational Number Building Block". Let me explain the idea.

So I have a rectangle of 3 by 2 and its diagonal is sqrt13. And what I want to do is build a string or chain of squares that covers the length of that sqrt13 where the squares are touching vertex to vertex.

So last night I was pondering if that was possible? Given any integer sided rectangle, can the diagonal length be covered by a string or chain of squares, tiny squares touching vertex to vertex (corner to corner) and encompassing exactly the distance or length of that diagonal? I was not sure and still amazed that is probably true.

Then I sought for some number value for these squares given any random rectangle, and the most amazing idea is that it involves the sequence of what I call 5's or "fives" or .5, .05, .005, etc etc.

So with my rectangle of 3 by 2 , what square can cover the length of sqrt13 where the squares are touching vertex to vertex? So I divide 3 by 2 and end up with 1.5 and divide 2 by 2 and end up with 1 but 1 is not a "fives" so I continue to divide 1 by 2 and end up with 0.5.

Now the sqrt13 = 3.60555.... and let me see if tiny squares of "fives" vertex to vertex recovers exactly that of 3.60555....

If I use 1.5 squares, they are too large so I shrink down to a .15 square and the hypotenuse would thus be .212132....

Likewise the .5 square is too large and I shrink down to a 0.05 square where the hypotenuse would be 0.0707103.... (and which this 7070 digit string appears often in these "fives").

Now, if I placed those squares, the .15 and the .05 as my chain of squares vertex touching vertex I find that dividing that I need 16.999... or in other words 17 of the .15 squares and 50.99.... of the 0.05 squares. So apparently I have a convergence to a integer by these "FIVES rule".

Now, if all the above is true, then it does not constitute an independent proof of No Perfect Cuboid. The impulse is to say that the interior diagonal of the Cuboid can be replaced by a chain of cubes of say 0.05 size and thus the diagonal is irrational number. That would be the compulsion, but we cannot do that because we do not know if the diagonal formed a P-triple.

We do know the diagonal could not form a P-triple because of the m and n's of 2mn, m^2-n^2, m^2+n^2.

So the chain of squares or cubes is not a independent proof of No Perfect Cuboid, but a additional fact that when you have a right triangle that does not form a P-triple, the hypotenuse is not going to be a rational number but an irrational number.

What is intriguing of the FIVES rule, is why the sequence of .5, .05, .005, etc is linked to irrational numbers. As though any irrational number can be built from that of squares of FIVES.

END

Newsgroups: sci.math

Date: Wed, 2 Jul 2014 16:11:24 -0700 (PDT)

Subject: Naturals reduce to sequence .5, .05, .005, . . related to irrationals

#1881 Correcting Math

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Wed, 02 Jul 2014 23:11:25 +0000

Naturals reduce to sequence .5, .05, .005, . . related to irrationals #1881 Correcting Math

Following up on my prior post about this remarkable sequence of 0.5, 0.05, 0.005, . . . that seems to capture all irrationals in a chain of squares linked by their vertex

/\/\/\/\

\/\/\/\/

Those are sad looking squares but meant to be squares linked by their vertices.

So what I am looking at in this new idea, is that if you get a Graph paper that has tiny squares inside of larger squares that a large square when the hypotenuse is drawn will be a length that is a connecting of tiny squares at their vertices. So the length of the hypotenuse of an integer square is irrational and the smaller squares that add up to that length are also irrational numbers added up to equal the larger irrational number.

So why not do this procedure on all integer rectangles? If we can find a small square that is the full length of its added up hypoteni to equal the rectangle hypotenuse.

And what I found is that we take any Natural larger than 0 and divide by 2. If the Natural is odd then we such as 7 then we end up with 3.5 and we have our FIVE digit. If the Natural is even such as 20 then we keep dividing by 2 until it ends up as 1 and then we finally divide by 2 to end up as 0.5 and we have our FIVE digit.

So, the question is, can this sequence of FIVES 0.5, 0.05, 0.005, 0.0005 , ... end up being a tiny square that covers the length of any and every irrational number, provided the rectangle has an even integer side (sic)?

If it can, it seems utterly remarkable as I showed earlier today that the sqrt13 with a square covering of 0.15 side is 16.999... or 17 of those squares and a square covering of 0.05 requires 50.99.... or 51 of those squares of 0.05 side to cover the length of 3.60555...

So, what is this magical relationship of FIVES with irrational numbers? Or have I made a error?

END

Newsgroups: sci.math

Date: Wed, 2 Jul 2014 16:28:21 -0700 (PDT)

Subject: explanation? Re: Naturals reduce to sequence .5, .05, .005, . .

related to irrationals #1882 Correcting Math

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Wed, 02 Jul 2014 23:28:21 +0000

explanation? Re: Naturals reduce to sequence .5, .05, .005, . . related to irrationals #1882 Correcting Math

On Wednesday, July 2, 2014 6:11:24 PM UTC-5, Archimedes Plutonium wrote:

> Following up on my prior post about this remarkable sequence of 0.5, 0.05, 0.005, . . . that seems to capture all irrationals in a chain of squares linked by their vertex

> /\/\/\/\

> \/\/\/\/

That is the best I can do of showing squares linked together at there vertices to cover the hypotenuse of a right triangle length.

> Those are sad looking squares but meant to be squares linked by their vertices.

> So what I am looking at in this new idea, is that if you get a Graph paper that has tiny squares inside of larger squares that a large square when the hypotenuse is drawn will be a length that is a connecting of tiny squares at their vertices. So the length of the hypotenuse of an integer square is irrational and the smaller squares that add up to that length are also irrational numbers added up to equal the larger irrational number.

> So why not do this procedure on all integer rectangles? If we can find a small square that is the full length of its added up hypoteni to equal the rectangle hypotenuse.

> And what I found is that we take any Natural larger than 0 and divide by 2. If the Natural is odd then we such as 7 then we end up with 3.5 and we have our FIVE digit. If the Natural is even such as 20 then we keep dividing by 2 until it ends up as 1 and then we finally divide by 2 to end up as 0.5 and we have our FIVE digit.

> So, the question is, can this sequence of FIVES 0.5, 0.05, 0.005, 0.0005 , ... end up being a tiny square that covers the length of any and every irrational number?

Sorry, that is a mistake of omission, for I should have said "any and every irrational number that comes from a rectangle in which one of its sides is an even number.

> If it can, it seems utterly remarkable as I showed earlier today that the sqrt13 with a square covering of 0.15 side is 16.999... or 17 of those squares and a square covering of 0.05 requires 50.99.... or 51 of those squares of 0.05 side to cover the length of 3.60555...

> So, what is this magical relationship of FIVES with irrational numbers? Or have I made a error?

Well I do not know if the above is true or I just found a coincidence in one example of a rectangle of 3 by 2 with sqrt13 hypotenuse. But if the above is true. Then we obviously need an explanation why a FIVE digit figure for a square, when summed, equals a irrational of a rectangle. I believe the explanation goes back to phi and the golden log spiral that the number phi involves the digit 5. Remember the rectangles of whirling squares to compose the golden ratio log spiral?

END

Newsgroups: sci.math

Date: Thu, 3 Jul 2014 01:30:36 -0700 (PDT)

Subject: utilizing this new method Re: Naturals reduce to sequence .5, .05,

.005, . . related to irrationals

From: Archimedes Plutonium <plutonium....@gmail.com>

Injection-Date: Thu, 03 Jul 2014 08:30:36 +0000

utilizing this new method Re: Naturals reduce to sequence .5, .05, .005, . . related to irrationals

I do not think I need this method for the proof of No Perfect Cuboid. I do not think this method makes a proof of No Perfect Cuboid.

I think what this method can do is focus on the structure of irrational numbers. We have little experience of adding irrational numbers.

The significance of the digit 5 for irrational numbers is made apparent and the importance of the square root of 0.5 = 0.7071067...

The idea of capturing the length of a irrational hypotenuse by the stringing together in a chain, tiny squares touching at vertices has never been attempted, at least I see no literature on this idea.

Can the method be used to prove a length is irrational in some other conjecture? None that I am aware of.

So here I have a method in search for work that can utilize the method.

Archimedes Plutonium

unread,

Jul 3, 2014, 4:37:03 AM7/3/14