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Question abou Patterson Algorithm

  1. Sep 8, 2011 #1
    Hi every one

    In the preliminaries section, the item c), there a proposition that say: "So by our choice of g we get "[tex]\theta/p | \psi/p[/tex]" whence "[tex]\theta | \psi[/tex]" ". I am not understanding this propositión, Please help me

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  3. Sep 17, 2011 #2

    micromass

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    That

    [tex]\theta/p~\vert~\psi/p[/tex]

    means that there exists a c such that

    [tex]c\theta/p=\psi/p[/tex]

    Now multiply both sides by p...
     
  4. Sep 18, 2011 #3
    thanks by your attention, but, a important part is: In the preliminaries section, the item c), there a proposition that say: "So by our choice of g we get "
    [tex]\theta/p | \psi/p[\tex], how i use "So by our choice of g we get", Why g havenot any degree?, why choice small as posible that contravening c)?
     
    Last edited: Sep 18, 2011
  5. Sep 18, 2011 #4

    HallsofIvy

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    juaninf, if you keep "bumping" your threads every few minutes, you are going to get banned.
     
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