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19972NO absolute-perfect cuboid exists

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  • leavemsg1
    Mar 31 5:04 PM
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      it has prime numbers in it, so...

      i. an absolute-perfect cuboid cannot exist.

      ii. it would have to happen before PT's exceed 2^3 < 3^2.

      iii. only primitive Pythagorean Triples have minimum expanse.

      iv. their standard derivation:

      m,n; m^2-n^2; 2mn; m^2+n^2

      2,1; 3= 2^2-1^2 4=2*2*1 5= 2^2+1^2

      3,2; 5= 3^2-2^2 12=2*3*2 13= 3^2+2^2

      4,3; 7= 2^2-1^2 24=2*4*3 25= 4^2+3^2

      v. a desciption of their minimum expanse property:

      2,1 3= 2+1 ... 5= 3*1+2

      3,2 5= 3+2 ... 13= 5*2+3

      4,3 7= 4+3 ... 25= 7*3+4

      m,n = m+n = (m+n)n +m

      no other PT's are of minimum expanse after 2^3 < 3^2,
      so the idiosyncracy of finding an absolute perfect cube
      becomes more absurd as the search is continued further.

      logically, only a semi-perfect cuboid could exist with
      edges (3), (4), (12), and main diagonal (13); if it were
      an absolute-perfect cuboid, then the other face diagonal
      would have to be integral, as in (3), (4) & (5)'s case.

      vi. an absolute-perfect cuboid cannot exist!
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