
professors of UC Berkeley endorsing No Perfect Cuboid to arxiv #1887 Correcting Math
Posted:
Jul 4, 2014 4:34 PM


Alright, below are two proofs of No Perfect Cuboid in which the algebra proof provides proofs that No Perfect Right Triangular Prism exists. Of course a test of that proof is that a Perfect Equilateral Triangular Prism exists whose sides are 5, and 12. In the regular polygon conjecture I need to fill in some holes of how all the even regular polygons beyond the 4gon contain an irrational length.
And, overall, I suppose I need to shorten, or condense these proofs to bare bones. I did not do this the last time around so will try better this time.
XXXXXXX 2nd proof of No Perfect Cuboid is Constructible using Algebra XXXXXXXXX
In this 2nd proof of No Perfect Cuboid I use Ptriple algebra of this:
(2mn, m^2  n^2, m^2 + n^2)
Alright, I do have a 2nd proof of No Perfect Cuboid which carries with it a proof that No Perfect Right Triangular Prism.
I wanted to do better than my first proof of No Perfect Cuboid so that it brings in the parallelepipeds, tetrahedron, and right triangular prism. So I ventured for some time looking for a naturalsquare that was embedded so that I could instantly say, a irrational number has to be present in a cuboid. So, I was looking mostly for a geometry proof, not a number crunching algebra proof. As it turns out, the proof of No Perfect Regular Polygons beyond the 3gon (described next in a conjecture) is a geometry proof, but the proof of No Perfect Cuboid or Right Triangular Prism is a algebra proof. What I mean is that a Pythagorean Triple is based upon the algebra of this:
2mn, m^2n^2, m^2 + n^2 where m and n are positive integers
Those m and n's form all Ptriples of primitive and if we include the factor of "k" encompass composite Ptriples.
So, how does the proof of No Perfect Cuboid and No Perfect Right Triangular Prism (call it the wedge) come about from those m and n's?
Now below I have 3 examples of a "almost perfect cuboid" and we can also use them for the wedge for the wedge is a cuboid cut in half along the spacediagonal where the space diagonal becomes a face diagonal of the wedge.
Imperfect cuboid that misses by one diagonal:
(153,672,3sqrt52777) where the irrational number is close to 689.19 (104,153,185) (104,672, 680) (104, 3sqrt52777, 697)
(a, c, d) (b, a, e) (b, c, f) (b, d, g)
Imperfect Cuboid that misses by one side:
(7800, 18720, 20280) (7800, sqrt211773121, 16511) where the sqrt number is near 14552.4 (sqrt211773121, 18720, 23711) (7800, 23711, 24961)
(a, c, e) (a, b, f) (b, c, d) (a, d, g)
My own set of a defective perfect cuboid was given earlier as this set:
(12,16,20) (15, 20, 25) (12,15,sqrt369) where the sqrt is close to 19.2 (15,16,sqrt481) where the sqrt is close to 21.9
(a, c, d) (b, d, e) (a, b, f) (b, c, g)
Now, every cuboid or wedge is going to have those 4 sets which are the distinct right triangles from the 3 distinct faces and a 4th right triangle from the spacediagonal (in wedge also).
So the diophantine equations are truly these 4 sets of right triangles composed in the m and n's, and not the commonly given set shown in Wikipedia or MathWorld.
If you inspect closely and carefully the above near misses and their 4 sets, you will see that there is a special right triangle triple that is composed of the longest leg of two previous right triangle triples.
For example in the first set notice that 153 and 672 are the longest leg of the following right triangles.
(153,672,3sqrt52777) (104,153,185) (104,672, 680)
So, why no perfect cuboid, why no perfect right triangular wedge?
Because one of the right triangles in the set of 4 is a right triangle composed of the longest leg of two previous right triangles.
So that means the m and n's of their positive integer solutions that make up all Ptriples (if we include a factor of k), implies that a m and n for one Ptriple along with the m and n of a different Ptriple can have their longest leg form a third Ptriple with the m and n's fully intact and no changes.
That is why a irrational number must occur from this Ptriple formed of the longest legs of 2 other Ptriples.
Now here is a good reference website in case anyone wants more information on the 2mn, m^2  n^2, and m^2 + n^2 as the backbone algebra of Ptriples.
http://www.maths.surrey.ac.uk/hostedsites/R.Knott/Pythag/pythag.html#moretriples
It is this algebra that I use to prove No Perfect Cuboid and no Perfect Right Triangular Prism (wedge).
> > (153,672,3sqrt52777) where the irrational number is close to 689.19 > > (104,153,185) > > (104,672, 680) > > (104, 3sqrt52777, 697) > > > > (a, c, d) > > (b, a, e) > > (b, c, f) > > (b, d, g) > >
Now I feel almost like a auto mechanic disassembling a engine and showing how it all works.
Now the 2mn and m^2n^2 can shift around as to being either the short leg or long leg of right triangle
In the case of the near miss above we have m=13, n=4 where m^2n^2 = 153 and 2*13*4 = 104 and m^2 + n^2 = 169 + 16 = 185 for (104,153,185).
Now for (104,672, 680) that factors down to 13, 84, 85 and the m=7 and n= 6, so where the 84 is 2mn = 2*7*6 = 84. The 13 is 7^2  6^2 = 49  36 = 13 as m^2 n^2.
So we focus on the (153,672,3sqrt52777) and ask why the hypotenuse must be irrational? Is it because the triple is composed of two previous Ptriples where we stole the longest leg, one in the form of 2mn of 2*7*6 and the other in the form of m^2n^2 = 13^2  4^2 = 153. So in this Triple (not a Ptriple for it has the irrational number) we have one leg that uses m=7, n=6 whilst the other leg used m=13, n=4.
So, in the process of having a Triple composed of two longest legs of two prior Triples, means we can not reconcile either of these two longest legs causing the hypotenuse to be irrational.
So what we have is a triple of this form:
(longest leg from other Ptriple, longest leg from second Ptriple, irrational number)
So, the key to the proof of No Perfect Cuboid and No Perfect Triangular Prism is all due to the fact that whenever you pull two longest legs from other Ptriples, your triple must have a irrational number.
> > > > Imperfect Cuboid that misses by one side: > > > > (7800, 18720, 20280) > > (7800, sqrt211773121, 16511) where the sqrt number is near 14552.4 > > (sqrt211773121, 18720, 23711) > > (7800, 23711, 24961) > > > > (a, c, e) > > (a, b, f) > > (b, c, d) > > (a, d, g) > >
Now in this case study, it was a side of the cuboid with an irrational number, but if you notice the same algebra is present in this case study where you have three Triples:
(a, c, e) (a, b, f) where the longest leg of each of those c and b is pulled out and used to form (b, c, d)
What that does is cause irreconcilable m and n's of b and c and forces a irrational number to appear.
> > My own set of a defective perfect cuboid was given earlier as this set: > > > > (12,16,20) > > (15, 20, 25) > > (12,15,sqrt369) where the sqrt is close to 19.2 > > (15,16,sqrt481) where the sqrt is close to 21.9 > > > > (a, c, d) > > (b, d, e) > > (a, b, f) > > (b, c, g) > >
In this case, again, the longest legs of two Triples are pulled out and used to form a new Triple. And here again, the m and n's cannot be reconciled and forces a irrational number for hypotenuse.
> > Now, every cuboid or wedge is going to have those 4 sets which are the distinct right triangles from the 3 distinct faces and a 4th right triangle from the spacediagonal (in wedge also). > >
So the Cuboid and Wedge will always have those 4 sets of Triples, with adjoining sides of 4 right triangles.
> > So the diophantine equations are truly these 4 sets of right triangles, and not the commonly given set shown in Wikipedia or MathWorld. >
Those Diophantine Equations are much much too general. They are not incorrect, just hopelessly too general to ever provide a solution. What is needed is a Diophantine Equations based on the 4 set of Triples and with their m and n's of each of those 4 Triples. It is a set of Equations of that, which shows no allinteger solution.
QED
In the pursuit of No Perfect Cuboid I ran into a conjecture of regular polygons.
Now I wanted a easy geometry proof of No Perfect Cuboid involving natural squares embedded in the regular solid, because if a integer side square was embedded then its diagonal is irrational.
No Perfect Octahedron: proof: it contains a square of four of its vertices and that causes a diagonal to be an irrational number. QED
No Perfect Cube: proof: it contains squares and hence those squares form irrational diagonal lengths. QED
No Perfect Dodecahedron: Proof: it contains a cube of 8 of its vertices and thus contains a square whose diagonal is an irrational number. QED
No Perfect Icosahedron: Proof: I noticed on my icosahedron dice that 4 vertices form a rectangle inside the icosahedron, so I had to find out more about that rectangle. Turns out the rectangle is a golden ratio rectangle of 1::1.618.. and so it has a diagonal that is an irrational number length, hence no perfect icosahedron. QED
The search for embedded natural squares lead me to conjure a conjecture over regular polygons in 2nd dimension that none of them could be "perfect" sides and all diagonals as integer lengths.
XXXXXXXXX Proof of CONJECTURE POLYGONS XXXXXXX
Conjecture: every regular polygon beyond 3gon must have at least a side or one of its diagonals as irrational. In other words, the side and all diagonals must have at least one which is an irrational length.
Now this is a dazzling new conjecture, because it is not obvious or apparent that regular polygons would have a irrational length.
Now the Robbins pentagon can be perfect, but it is not a regularpolygon.
I have not seen anything remotely like this conjecture in the math literature. It appears that noone ever had a mind to extending the theory of "perfect" to include regular polygons. Everyone was probably so focused on proving no perfect cuboid, that they never bothered to extend the reach of perfect to 2nd dimension and the regular polygons.
Now let me state the conjecture differently. What it says is basically that given any regular polygon beyond 3gon, that it must contain at least 1 side and 1 diagonal, distinct diagonal that is. So the 4gon the square has 1 side and 1 distinct diagonal. Same as the 5gon which has 1 side and 1 distinct diagonal. What the conjecture says is that the collection of side and distinct diagonals of a regular polygon can not all be integer or rational numbers, that at least one of those has to be an irrational number length. So in the case of the square or pentagon, either the side is irrational or the diagonal is irrational.
Now that is a pretty powerful conjecture if proven true. For it says that in Nature, if you want to be a Regular Polygon, that at least one of the side or diagonals is irrational starting with the 4gon. Now the 3gon has only a side and no diagonals so it falls outside the purview of the conjecture.
Now in the proof of the Conjecture I am going to use the fact that a right triangle has inscribed squares as seen from this website:
http://jwilson.coe.uga.edu/emt725/class/trabue/SquareinRtTng/SQinRT.html
Now it looks as though the proof of the Regular Polygons beyond 3gon all of which must have at least one of its diagonals or its side as irrational is split into two different camps of proof, where the odd number ngons 5, 7, 9, 11, 13, etc etc are proven by the above square inscribed in right triangle. Whereas the even numbered ngons, 4, 6, 8, 10, 12, etc etc are proven by a different inscribed square of a right triangle. Some of the even ngons have already a Natural square such as 4, 8, 12 and so immediately we can dispel them since they satisfy the conjecture for they have a diagonal that is irrational. But that leaves many of the even numbered ngons such as 6, 10, 14, 18 with a different proof than the odd numbered ngons.
Alright, let me review as to why the pentagon side or diagonal must be irrational where both cannot be integer or rational.
Looking at that diagram by Wilson we see the length CP must be irrational for it is the diagonal of the square. But it is also the leg of the right triangle CPA. In the pentagon figure CP is part of the diagonal while the square side Cb is part of the side of the pentagon, as shown in the complement diagram of Wikipedia reference below.
> > From that website we see two sides of square lie on sides of the right triangle. > > > > From this website of Wikipedia > > > >  quoting the caption of picture  > > http://en.wikipedia.org/wiki/Ptolemy_theorem#Pentagon > > > > Complement of the pentagon chord > >  end quote 
Alright, I had better try to defend these statements. The reader is required to have both figures to compare and the best way is to have two computers, one with the Wilson picture and one with the Wikipedia complement of pentagon picture.
> > > > Alright, let me review as to why the pentagon side or diagonal must be irrational where both cannot be integer or rational. > > > > > > > > http://jwilson.coe.uga.edu/emt725/class/trabue/SquareinRtTng/SQinRT.html >
> > http://en.wikipedia.org/wiki/Ptolemy_theorem#Pentagon > > > > Complement of the pentagon chord
> > Looking at that diagram by Wilson we see the length CP must be irrational for it is the diagonal of the square. But it is also the leg of the right triangle CPA. In the pentagon figure CP is part of the diagonal while the square side Cb is part of the side of the pentagon, as shown in the complement diagram of Wikipedia reference below. >
Alright, the diagonal in Wikipedia is EB with half its length as b. The b becomes the diagonal of the inscribed square as ABF in Wikipedia.
So, comparing the two pictures, the CBA of Wilson is the ABF of Wikipedia. CBA of Wilson where C is the right angle and BFA of Wikipedia where B is the right angle.
CP length in Wilson, is the "b" length in Wikipedia. And this length is the hypotenuse of the square, so it is irrational in length.
The square sides are along AB and BF in Wikipedia. So the side of the pentagon "a" is integer or rational but the "b" length in Wikipedia is 1/2 of the diagonal of the pentagon and must be irrational.
I hope this covers it all and explains why the pentagon side and diagonal, one of which must always be an irrational number.
XXXXXXXXX 1st Proof that NO PERFECT CUBOID is constructible XXXXXXXXXXXX
Proof No Perfect Cuboid is constructible; parallelogram A = bc(sin(t))
Now a good picture found on the internet for the reader to follow concerning this proof is this picture: We can look at the diagram on this website:
http://www.algebra.com/algebra/homework/Rectangles.faq.question.674558.html
Now that is a picture of a cuboid that is 6, 8, 24, with facediagonal of 8sqrt10 and spacediagonal of 26.
If you take a cuboid, an Euler brick such as the (44, 117, 240) and stack an identical brick on top then a special rectangle forms with the 2(44) by 267.
If we take the smallest known Euler brick discovered by Halcke in 1719 as (44, 117, 240) with face diagonals of 125, 244, 267. That we have a rectangle that is 44 by 267 with hypotenuse of sqrt73225 = 5sqrt2929 = 270.601...
And we have a special parallelogram enclosed inside that 88 by 267 rectangle of a parallelogram of sides 5sqrt2929 by 44.
In the Parallelogram method I retrieve a special rectangle which has embedded inside a parallelogram that intersects at two vertices and intersects at the midpoint of 88 at two spots. The picture on the webpage and the diagram below can help the reader to see this special rectangle with parallelogram inside.
http://www.algebra.com/algebra/homework/Rectangles.faq.question.674558.html
_____________ ___________ / / / / / /
___________ ____________
So, in the above picture, that parallelogram intersects two vertices and two of the sides at its midpoint.
So, what makes the Parallelogram method work is the sin(theta) involved in area. If the outer rectangle was all integers with an integer overall area, then, due to the sin(theta) the interior area of the parallelogram is irrational and its sides cannot be all integers, and vice versa. So the proof of the No Perfect Cuboid exists pivots totally on the fact that the area of a parallelogram has a sin(theta) that causes either one of the overall rectangle or enclosed parallelogram to not have all integer sides.
Alright I need to fill in the steps of the end of the proof, showing that the Space diagonal must be irrational if the Face diagonal is integer with attendant edge as integer. So I need to fill in those final steps.
Now here is the picture of the end of that proof: __________ \   \ space diagonal  \  _________\ \   \ space diagonal  \  _________\ face diagonal
Now that is a picture of two cuboids stacked one on top the other. And we see the parallelogram formed by the two space diagonals whose area is a(b)(sin t).
So we know from Perfect Cuboid properties that the face diagonal and all the edges are integers. What we do not know is whether we can have the space diagonal an integer also.
By using the parallelogram formula of area we have sin(t) as irrational, for we have ruled out sin30 and sin150 because cosine would make the cuboid edge irrational.
So, if the space diagonal is irrational, we have no further to go for we end the proof. The other possibility is that the space diagonal is integer. So now we know the edge involved is integer and the space diagonal is integer would mean that Integer times Integer times sin(t) is irrational and so the parallelogram area is irrational.
Now we remove the upper triangle in the above diagram:
\  \  \ _________\ \   \ space diagonal  \  _________\ face diagonal
And what we have remaining is a trapezoid. Now the area of a trapezoid is 1/2height(a +b) where a and b are bases.
Now we have concluded so far that the area of the parallelogram is irrational and the area of the lower triangle is rational or integer for it is 1/2 base x height and the base and height are integers.
So, the addition of irrational plus rational ends up as irrational overall.
Now look at the area of the trapezoid where its two bases are the integer edges of the cuboid, for one base is twice the other but both are integer, and the height of that trapezoid is integer for it is the face diagonal. So the overall trapezoid is rational or integer.
But the area overall of the parallelogram plus lower triangle is irrational. So we have a impossible situation here in the construction. It is impossible for the Space diagonal to be integer. If the Space diagonal were irrational, then the area of the parallelogram can be rational if need be, for we then have irrational times integer times sin(t) = Area.
When the space diagonal is integer, then we violate the construction of both parallelogram, trapezoid and the triangles involved.
So in the construction of a Perfect Cuboid, we reach the impossible construction phase when both face diagonal and space diagonal are integers.
QED

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

