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Re: [PrimeNumbers] Dimostrazione geometrica

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  • Adriano Palma
    ** Proprietary ** ** High Priority ** Sir, your so-called proof is utterly incomprehensible. In order to avoid further linguistic chaos you are kindly
    Message 1 of 4 , Oct 23, 2009
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      ** Proprietary **
      ** High Priority **

      Sir, your so-called proof is utterly incomprehensible.
      In order to avoid further linguistic "chaos" you are kindly invited to
      exhibit the ration of the form x/y, where x and y are integers such
      that
      x/y times x/y equals 2. That is the only way you have to prove your
      claim. If you achieve that geometrically, set theoretically or
      algebraically is an interesting but, to be sure, secondary issue, given
      the magnitude of what you claim to have discovered.
      regards



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      >>> "Edgar J. Delpero." <edgar_james2002@...> 10/22/2009 9:19 PM
      >>>


      La somma dei triangoli rettangoli Aurii costruito nei cateti, divisi
      per due è uguale all'area del triagolo rettangolo costruito
      sull'ipotenusa o diagonali.

      Esempio: n=2 geometrica dimostrazione.
      h=1 e b=1=1^2
      d=1.414213562..., allora i due rettangoli Aureii addizionati sono
      uguali alla diagonale, ovvero (h+b)^3.
      h=1.414213562..., b=1.41421356
      Allora, [(h+b)^3]/2 = Area=d sostanzialmente, il valore dell'area è
      uguale al valore della diagonale.
      Conclussione: esistono due numeri m>n, tale che m/n siano uguali
      all'area ed all'ipotenusa o diagonale:m>n, tale che m/n=A=d.
      Ruotine :
      per m ed n =m/n.
      m=3
      n=2
      m=[(1+RADQ3)/2]^2*4*4*4*4*4*4*4*4*4*4*4*4*4.=125226845
      n=(125226845)/RADQ2 =88548751.3

      m/n=A=d=(125226845)/88548751.3

      importante collocare al numeratore qualsiasi numero "primo" o della
      forma 2n-1, e al denominatore RADQn cercata.
      La dimostrazione si può scrivere infinite volte !!!..,
      n=2
      m= 127 important "primenumber o 2n-1".
      n=2
      m=[(1+RADQ3)/127]^2*4*4*4*4*4*4*4*4*4*4*4*4.=2525622892
      n=(2525622892)/RADQ2=1785885074
      m/n=A=d=(2525622892)/1785885074=RADQ2
      Posso fare anche una dimostrazione biunivoca del tipo Cantoriana.
      D'altro canto per ogni numero irrazionale vi sono due numeri m/n=RADQx

      CVD

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