Answer:
a) = {0, 2, 3, 4, 5, 6, 8}, b = {0} empty set. c = {0, 1, 6, 7, 8, 9} d = {0, 1, 3, 5, 6, 7, 8, 9} e = {0, 1, 5, 6, 7, 8, 9} f = {2,4} .
Step-by-step explanation:
Given:
S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {0, 2, 4, 6, 8}
B = {1, 3, 5, 7, 9}
C = {2, 3, 4, 5}
D = {1, 6, 7}
Solution:
a) A∪C = A + C - A∩C
= {0, 2, 4, 6, 8} + {2, 3, 4, 5} - {2, 4}
= {0, 2, 2, 3, 4, 4, 5, 6, 8} - {2, 4}
= {0, 2, 3, 4, 5, 6, 8}
b) A∩B = ?
A∪B = A + B - A∩B
A∪B + A∩B = A + B
A∩B = A + B - A∪B
= {0, 2, 4, 6, 8} + {1, 3, 5, 7, 9} - {0, 1, 2, 3, 4, 5, 6, 8, 9}
= {0,1, 2, 3, 4, 5, 6,7, 8, 9} - {0, 1, 2, 3, 4, 5, 6, 8, 9}
A∩B = {0} empty set.
c) C′ = S - C
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 4, 5}
= {0, 1, 6, 7, 8, 9}
d) (C′∩D)∪B = ?
(C′∩D) = {0, 1, 6, 7, 8, 9} ∩ {1, 6, 7}
= {1, 6, 7}
B = {1, 3, 5, 7, 9}
(C′∩D)∪B = (C′∩D) + B - (C′∩D)∩B
= {0, 1, 6, 7, 8, 9} + {1, 3, 5, 7, 9} - {1, 7, 9,}
= {0, 1, 1, 3, 5, 6, 7, 7, 8, 9, 9} - {1, 7, 9,}
= {0, 1, 3, 5, 6, 7, 8, 9}
e) (S∩C)′ = S - (S∩C)′
(S∩C) = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} ∩ {2, 3, 4, 5}
= {2, 3, 4, 5}
(S∩C)′ = S - (S∩C)
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 4, 5}
= {0, 1, 5, 6, 7, 8, 9}
f) A∩C∩D′ = ?
D′ = S - D
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 6, 7}
= {0, 2, 3, 4, 5, 8, 9}
A∩C∩D′ = {0, 2, 4, 6, 8} ∩ {2, 3, 4, 5} ∩ {0, 2, 3, 4, 5, 8, 9}
= {2,4}
