1. Joined
    29 Dec '08
    Moves
    6788
    18 Jun '12 19:09
    No...wait...! (cue sound of crickets)
  2. Standard memberkaroly aczel
    the Devil himself
    Brisbane,QLD
    Joined
    11 Apr '09
    Moves
    91587
    18 Jun '12 19:24
    Lol
  3. Standard memberAgerg
    The 'edit'or
    converging to it
    Joined
    21 Aug '06
    Moves
    11458
    18 Jun '12 21:281 edit
    Let B be the boycott operator applied to a poster p_i, with B^2(p_i) = B(B(p_i)) denoting the boycott operator applied to the boycotter of p_i. Let n be a positive integer, and 0 be the empty boycott. Then there exists k such that for all n > k, B^n(p_i) = 0.


    Proof left as an exercise for the reader.
  4. Standard memberSwissGambit
    Caninus Interruptus
    2014.05.01
    Joined
    11 Apr '07
    Moves
    92274
    19 Jun '12 04:57
    Originally posted by Agerg
    Let B be the boycott operator applied to a poster p_i, with B^2(p_i) = B(B(p_i)) denoting the boycott operator applied to the boycotter of p_i. Let n be a positive integer, and 0 be the empty boycott. Then there exists k such that for all n > k, B^n(p_i) = 0.


    Proof left as an exercise for the reader.
    Take it to Posers and Puzzles, Spanky.
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