timro21
Posts:
55
Registered:
1/31/07


Re: New Method in proof that No Perfect Cuboid is constructible #1499 Correcting Math
Posted:
Feb 2, 2014 9:45 PM


On Saturday, 1 February 2014 19:32:53 UTC+10, Archimedes Plutonium wrote: > New Method in proof that No Perfect Cuboid is constructible #1499 Correcting Math > > > > Sorry, I cannot go to bed when I have a math proof on my mind. So I took a hot shower, which works wonders for me in clear thinking. > > > > I could not find anyway of relating a Space Diagonal to Face Diagonal in one triangle of a cuboid; and to separate that triangle from any other common triangle. So, clearly I need a different figure of geometry. This reminds me of the FLT proof garnered after I looked at Beal and realized I do not need triangles or angles, but rather rectangles. > > > > So, here is the proof that No Perfect Cuboid can be constructed > > > > Take any cuboid that has integer edges and integer face diagonals. > > > > 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. > > > > Now, place an identical cuboid on top of the one in the picture. Now we see the rectangle formed from the facediagonals as that of 6 x 8sqrt10. > > > > Now we see the figure formed by the Spacediagonals is not a rectangle but rather a parallelogram with angle theta. > > > > The area formula for parallelogram is Area = B*C*sin(theta). > > > > In other words, the sin(theta) will deliver to me the fact that if one diagonal is integer, the other diagonal must be irrational due to the sin(theta). > > > > Now, also notice that the parallelogram formed by the Space diagonals of the two cuboids stacked one on top the other is embedded inside the larger rectangle of 12 x 8sqrt10. > > > > So, I have the two diagonals embedded in one another, one in a rectangle, the other in a parallelogram. It is here, where I can show that if the one is integer, the other, due to the sin(theta) must be irrational. I am too tired at the moment and will continue tomorrow. > > > > QED > > > > > >  > > > > 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
sine(theta) can take ALL values between 1 and 1, including ALL rational values in thus range, eg 2/3, 37/52, 129/200, etc

