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Pairwise disjoint mathematical proof

  1. Oct 19, 2012 #1
    Set 1 {{1, 2}, {2, 3}, {1, 2, 3}, {2, 3, 4}, {2, 3, 5}, . . . , {2, 3, n}};
    Set 2 {{3, 5}, {5, 4}, {3, 5, 4}, {4, 5, 1}, {4, 5, 2},{3, 5, 6}, {3, 5, 7}, . . . , {3, 5, n}};
    Set 3 {{3, 1}, {1, 4}, {3, 1, 4}, {1, 4, 2}, {1, 3, 5}, {1, 3, 6}, . . . , {1, 3, n}};
    Set 4 {{1, 5}, {5, 2}, {1, 5, 2}, {2, 4}, {4, 3}, {5, 6}, {5, 7}, . . . , {5, n}};
    Set 5 {{3, 6}, {6, 4}, {3, 6, 4}, {4, 6, 1}, {4, 6, 2}, {4, 6, 5},{5, 6, 7}, {5, 6, 8}, . . . , {5, 6, n}};
    Set 6 for n = 8,
    {{3, 7}, {7, 4}, {3, 7, 4}, {4, 7, 1}, {4, 7, 2}, {4, 7, 5},{4, 7, 6}, {5, 7, 8}},
    and for n ≥ 9, {{3, 7}, {7, 4}, {3, 7, 4}, {4, 7, 1}, {4, 7, 2}, {4, 7, 5}, {4, 7, 6},{5, 7, 8}, {5, 7, 9}, . . . , {5, 7, n}};
    for n ≥ 11 and for k ∈ {7, . . . , n − 4},
    set K: {{3, k + 1}, {k + 1, 4}, {3, k + 1, 4},{4, k + 1, 1}, {4, k + 1, 2},{4, k + 1, 5}, {4, k + 1, 6},
    . . . , {4, k + 1, k},{5, k + 1, k + 2}, {5, k + 1, k + 3}, . . . , {5, k + 1, n}};
    for n ≥ 10,
    set n-3: {{3, n − 2}, {n − 2, 4}, {3, n − 2, 4}, {4, n − 2, 1}, {4, n − 2, 2},
    {4, n − 2, 5}, {4, n − 2, 6}, . . . , {4, n − 2, n − 3},
    {5, n − 2, n − 1}, {5, n − 2, n}};
    for n ≥ 9,
    set n-2 : {{3, n − 1}, {n − 1, 4}, {3, n − 1, 4}, {4, n − 1, 1}, {4, n − 1, 2},
    {4, n − 1, 5}, {4, n − 1, 6}, . . . , {4, n − 1, n − 2}, {5, n − 1, n}};
    Set n-1 : {{3, n}, {n, 4}, {3, n, 4}, {4, n, 1}, {4, n, 2},
    {4, n, 5}, {4, n, 6}, . . . , {4, n, n − 1}}.

    How to prove the n−1 sets are pairwise disjoint in general

    I could see it by placing a value for n greater than 8.
     
  2. jcsd
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