Mathematics
If A = {1, 2, 3, 4, 5}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6} ; verify :
(i) A - (B ∪ C) = (A - B) ∩ (A - C)
(ii) A - (B ∩ C) = (A - B) ∪ (A - C).
Answer
A = {1, 2, 3, 4, 5}
B = {2, 4, 6, 8}
C = {3, 4, 5, 6}
(i) A - (B ∪ C) = (A - B) ∩ (A - C)
Taking LHS : A - (B ∪ C)
B ∪ C - contains all the elements in set B and C.
B ∪ C = {2, 4, 6, 8} ∪ {3, 4, 5, 6}
B ∪ C = {2, 3, 4, 5, 6, 8}
A - (B ∪ C) - contains all the elements which are in set A but not in B ∪ C.
A - (B ∪ C) = {1, 2, 3, 4, 5} - {2, 3, 4, 5, 6, 8}
A - (B ∪ C) = {1} ……………(1)
Taking RHS : (A - B) ∩ (A - C)
A - B - contains all the elements which are in set A but not in B.
A - B = {1, 2, 3, 4, 5} - {2, 4, 6, 8}
A - B = {1, 3, 5}
A - C - contains all the elements which are in set A but not in C.
A - C = {1, 2, 3, 4, 5} - {3, 4, 5, 6}
A - C = {1, 2}
(A - B) ∩ (A - C) - contains all the common elements in set (A - B) and (A - C).
(A - B) ∩ (A - C) = {1, 3, 5} ∩ {1, 2}
(A - B) ∩ (A - C) = {1} ……………(2)
From (1) and (2), we can see
LHS = RHS
Hence, A - (B ∪ C) = (A - B) ∩ (A - C)
(ii) A - (B ∩ C) = (A - B) ∪ (A - C).
Taking LHS : A - (B ∩ C)
B ∩ C - contains all the common elements in set B and C.
B ∩ C = {2, 4, 6, 8} ∩ {3, 4, 5, 6}
B ∩ C = {4, 6}
A - (B ∩ C) - contains all the elements which are in set A but not in B ∩ C.
A - (B ∩ C) = {1, 2, 3, 4, 5} - {4, 6}
A - (B ∩ C) = {1, 2, 3, 5} ……………(3)
Taking RHS : (A - B) ∪ (A - C)
A - B - contains all the elements which are in set A but not in B.
A - B = {1, 2, 3, 4, 5} - {2, 4, 6, 8}
A - B = {1, 3, 5}
A - C - contains all the elements which are in set A but not in C.
A - C = {1, 2, 3, 4, 5} - {3, 4, 5, 6}
A - C = {1, 2}
(A - B) ∪ (A - C) - contains all the elements in set (A - B) and (A - C).
(A - B) ∪ (A - C) = {1, 3, 5} ∪ {1, 2}
(A - B) ∪ (A - C) = {1, 2, 3, 5} ……………(4)
From (3) and (4), we can see
LHS = RHS
Hence, A - (B ∩ C) = (A - B) ∪ (A - C)
Related Questions
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