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Topic: New Method in proof that No Perfect Cuboid is constructible #1499
Correcting Math

Replies: 30   Last Post: Feb 11, 2014 2:20 PM

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 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
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> 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.
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> 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.
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> So, here is the proof that No Perfect Cuboid can be constructed
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> Take any cuboid that has integer edges and integer face diagonals.
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> We can look at the diagram on this website:
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> http://www.algebra.com/algebra/homework/Rectangles.faq.question.674558.html
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> Now that is a picture of a cuboid that is 6, 8, 24, with face-diagonal of 8sqrt10 and space-diagonal of 26.
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> Now, place an identical cuboid on top of the one in the picture. Now we see the rectangle formed from the face-diagonals as that of 6 x 8sqrt10.
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> Now we see the figure formed by the Space-diagonals is not a rectangle but rather a parallelogram with angle theta.
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> The area formula for parallelogram is Area = B*C*sin(theta).
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> 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).
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> 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.
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> 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.
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> QED
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> --
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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.
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> 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