# Thread: Are these set assumptions correct?

1. ## 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 A3×A5×A7
A3 = {3,6,9}
A5 = {5,10}
A7 = {7}
So Cartesian product of A3×A5×A7 = {(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

2. Originally Posted by dunsta
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 A3×A5×A7
A3 = {3,6,9}
A5 = {5,10}
A7 = {7}
So Cartesian product of A3×A5×A7 = {(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 A3×A5×A7 = {(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)).