Solve the following equation.
1 answer:
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Answer:
a. {1, 2, 3, 4, 5, 6, 7}
b. {1, 2, 3, 4, 5, 6, 8}
c. {4, 5, 7}
d. {1, 4, 5}
Step-by-step explanation:
The union of two sets is the list of elements in either set. The intersection of two sets is the list of elements in both sets. The complement of a set is the list of elements in the universal set that are not in the set being complemented. The complement of a set can be indicated with an apostrophe: A' is the complement of set A, for example.
<h3>a. AU(BUC)</h3>B∪C = {2, 5, 6, 7} ∪ {3, 4, 6, 7} = {2, 3, 4, 5, 6, 7}
A∪(B∪C) = {1, 4, 5, 7} ∪ {2, 3, 4, 5, 6, 7} = {1, 2, 3, 4, 5, 6, 7}
<h3>b. (AN(BNC))'</h3>B∩C = {2, 5, 6, 7} ∩ {3, 4, 6, 7} = {6, 7}
A ∩ (B∩C) = {1, 4, 5, 7} ∩ {6, 7} = {7}
(A∩(B∩C))' = {7}' = {1, 2, 3, 4, 5, 6, 8}
<h3>c. (ANB)U(ANC)</h3>A∩Β = {1, 4, 5, 7} ∩ {2, 5, 6, 7} = {5, 7}
Α∩C = {1, 4, 5, 7} ∩ {3, 4, 6, 7} = {4, 7}
(A∩B)∪(A∩C) = {5, 7} ∪ {4, 7} = {4, 5, 7}
<h3>d. (ANB')U(ANC')</h3>(A∩B')∪(A∩C') = A∩(B'∪C') = A∩(B∩C)'
(B∩C)' = {6, 7}' = {1, 2, 3, 4, 5, 8}
A∩(B∩C)' = {1, 4, 5, 7} ∩ {1, 2, 3, 4, 5, 8} = {1, 4, 5}
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<em>Additional comment</em>
In part (d) we made use of De Morgan's law for sets:
B'∪C' = (B∩C)'
We also made use of the distributive property for sets:
A∩(B∪C) = (A∩B)∪(A∩C)
