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Phi physics help

  1. Jul 14, 2008 #1
    [tex]\foralln\inN\varphi(n)/mid/n[/tex]
     
  2. jcsd
  3. Jul 14, 2008 #2
    Re: Phi

    i made a mistake in writing
     
  4. Jul 14, 2008 #3

    CRGreathouse

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    Re: Phi

    I imagine you meant
    [tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex] (which is false; [itex]\varphi(3)\!\not\,\,\mid3[/itex])
    but I'm not sure what the question is.
     
  5. Jul 14, 2008 #4
    Re: Phi

    how we prove
    [tex]
    \forall n\in\mathbb{N}\;\varphi(n)\mid n
    [/tex]
     
    Last edited by a moderator: Jul 15, 2008
  6. Jul 14, 2008 #5
    Re: Phi

    How we prove that?
    \forall n\in\mathbb{N}\;\varphi(n)\mid n
     
  7. Jul 14, 2008 #6
    Re: Phi

    how we prove the statement in post 3
     
  8. Jul 15, 2008 #7

    CRGreathouse

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    Re: Phi

    [tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex]

    You can't, it's false. It only holds for 1, 2, 4, 6, 8, 12, 16, ... = A007694.
     
  9. Jul 15, 2008 #8
    Re: Phi

    Why can’t we derive a contradiction in order to show that it’s false?
     
  10. Jul 15, 2008 #9

    CRGreathouse

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    Re: Phi

    I gave a contradiction, 3, in my first post.
     
  11. Jul 16, 2008 #10

    HallsofIvy

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    Re: Phi

    CRGreathouse, he asked how you prove the contradiction you gave in post 3 and you answered "You can't, it's false. "! You were, of course, referring to his original post, not the post you quoted.

    roam, you prove the contradiction by doing the arithmetic. What is [itex]\phi(3)[/itex]?
     
  12. Jul 16, 2008 #11

    CRGreathouse

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    Re: Phi

    Ah. I took that to mean 'How do we prove the statement "[tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex]" from post #3', rather than 'How do we prove the statement "[tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex] [...] is false" from post #3'. To me, "[tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex]" was the only mathematical statement in post #3; "(which is false[...])" is a nonrestrictive clause. '"[tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex]" is false' would have been a mathematical statement, but one I only implied. That's why I was so confused by post #6.

    Of course a contradiction is an easy way to show that [tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex] fails.
     
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