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Math Help - Are these set assumptions correct?

  1. #1
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    Are these set assumptions correct?

    A = {n ∈ Z|2 ≤ n ≤ 10}, for positive integer Am = {n ∈ A| n is divisible by m}

    A = {2,3,4,5,6,7,8,9,10}

    A1 = {2,3,4,5,6,7,8,9,10}
    A2 = {2,4,6,8,10}
    A11 = {}

    A2nA3 (n = intersection)
    A2 = {2,4,6,8,10}
    A3 = {3,6,9}
    so A2nA3 = {6}

    Cartesian product A3A5A7
    A3 = {3,6,9}
    A5 = {5,10}
    A7 = {7}
    So Cartesian product of A3A5A7 = {(3,5,7),(3,5),(3,10,7),(3,10),(6,5,7),(6,5),(6,10 ,7),(6,10),(9,5,7),(9,5),(9,10,7),(9,10)}

    Powerset P(A3) = ({}, {3}, {6},{9}, {3,6},{3,9},{6,9},{3,6,9}

    Is {A2,A3,A5,A7} a partition of set A? no, they are not disjoint, A2,A3 share element 6.
    If B = (A-(A3 U A5 U A7)) are B, A3, A5,A7 a partition? yes, they are disjoint
    Is {A3,A5,A7} a partition of A? yes they are disjoint.

    Am I on the right track?
    Thanks for any help
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  2. #2
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    Quote Originally Posted by dunsta View Post
    A = {n ∈ Z|2 ≤ n ≤ 10}, for positive integer Am = {n ∈ A| n is divisible by m}

    A = {2,3,4,5,6,7,8,9,10}

    A1 = {2,3,4,5,6,7,8,9,10}
    A2 = {2,4,6,8,10}
    A11 = {}

    A2nA3 (n = intersection)
    A2 = {2,4,6,8,10}
    A3 = {3,6,9}
    so A2nA3 = {6}

    Cartesian product A3A5A7
    A3 = {3,6,9}
    A5 = {5,10}
    A7 = {7}
    So Cartesian product of A3A5A7 = {(3,5,7),(3,5),(3,10,7),(3,10),(6,5,7),(6,5),(6,10 ,7),(6,10),(9,5,7),(9,5),(9,10,7),(9,10)}

    Powerset P(A3) = ({}, {3}, {6},{9}, {3,6},{3,9},{6,9},{3,6,9}

    Is {A2,A3,A5,A7} a partition of set A? no, they are not disjoint, A2,A3 share element 6.
    If B = (A-(A3 U A5 U A7)) are B, A3, A5,A7 a partition? yes, they are disjoint
    Is {A3,A5,A7} a partition of A? yes they are disjoint.

    Am I on the right track?
    Thanks for any help
    I agree with everything except the Cartesian product and some of the partitioning logic.

    A partition must not only have pairwise disjoint sets, but also the union of those sets must equal the original set. Since 2 is not in A3, A5, or A7, therefore {A3,A5,A7} is not a partition of A.

    As for Cartesian product, one element must be chosen from each corresponding set; you will always have ordered triples, never ordered pairs, in the product you named. So A3A5A7 = {(3,5,7),(3,10,7),(6,5,7),(6,10,7),(9,5,7),(9,10,7 )}.

    Note that A2 n A3 is precisely the set of elements of A divisible by 6, and in general, A_m n A_n = A_(lcm(m,n)).
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