Let A = {a, b, c, d}, B = {b, c, e} and C = {a, b, e}. Find A ∪ B.
Answer
A = {a, b, c, d}
B = {b, c, e}
A ∪ B - contains all the elements in set A and B.
A ∪ B = {a, b, c, d} ∪ {b, c, e}
Hence, A ∪ B = {a, b, c, d, e}.
Let A = {a, b, c, d}, B = {b, c, e} and C = {a, b, e}. Find B ∪ C.
Answer
B = {b, c, e}
C = {a, b, e}
B ∪ C - contains all the elements in set B and C.
B ∪ C = {b, c, e} ∪ {a, b, e}
Hence, B ∪ C = {a, b, c, e}.
Let A = {a, b, c, d}, B = {b, c, e} and C = {a, b, e}. Find A ∪ C.
Answer
A = {a, b, c, d}
C = {a, b, e}
A ∪ C - contains all the elements in set A and C.
A ∪ C = {a, b, c, d} ∪ {a, b, e}
Hence, A ∪ C = {a, b, c, d, e}.
Let A = {a, b, c, d}, B = {b, c, e} and C = {a, b, e}. Find A ∩ B.
Answer
A = {a, b, c, d}
B = {b, c, e}
A ∩ B - contains all the common elements in set A and B.
A ∩ B = {a, b, c, d} ∩ {b, c, e}
Hence, A ∩ B = {b, c}.
Let A = {a, b, c, d}, B = {b, c, e} and C = {a, b, e}. Find B ∩ C.
Answer
B = {b, c, e}
C = {a, b, e}
B ∩ C - contains all the common elements in set B and C.
B ∩ C = {b, c, e} ∩ {a, b, e}
Hence, B ∩ C = {b, e}.
Let A = {a, b, c, d}, B = {b, c, e} and C = {a, b, e}. Find A ∩ C.
Answer
A = {a, b, c, d}
C = {a, b, e}
A ∩ C - contains all the common elements in set A and C.
A ∩ C = {a, b, c, d} ∩ {a, b, e}
Hence, A ∩ C = {a, b}.
Let A = {2, 3, 4, 6}, B = {5, 7, 8} and C = {2, 7, 8, 9}. Find A ∪ B.
Answer
A = {2, 3, 4, 6}
B = {5, 7, 8}
A ∪ B - contains all the elements in set A and B.
A ∪ B = {2, 3, 4, 6} ∪ {5, 7, 8}
Hence, A ∪ B = {2, 3, 4, 5, 6, 7, 8}.
Let A = {2, 3, 4, 6}, B = {5, 7, 8} and C = {2, 7, 8, 9}. Find B ∪ C.
Answer
B = {5, 7, 8}
C = {2, 7, 8, 9}
B ∪ C - contains all the elements in set B and C.
B ∪ C = {5, 7, 8} ∪ {2, 7, 8, 9}
Hence, B ∪ C = {2, 5, 7, 8, 9}.
Let A = {2, 3, 4, 6}, B = {5, 7, 8} and C = {2, 7, 8, 9}. Find A ∪ C.
Answer
A = {2, 3, 4, 6}
C = {2, 7, 8, 9}
A ∪ C - contains all the elements in set A and C.
A ∪ C = {2, 3, 4, 6} ∪ {2, 7, 8, 9}
Hence, A ∪ C = {2, 3, 4, 6, 7, 8, 9}.
Let A = {2, 3, 4, 6}, B = {5, 7, 8} and C = {2, 7, 8, 9}. Find A ∩ B.
Answer
A = {2, 3, 4, 6}
B = {5, 7, 8}
A ∩ B - contains all the common elements in set A and B.
A ∩ B = {2, 3, 4, 6} ∩ {5, 7, 8}
As there is no element common to A and B,
Hence, A ∩ B = φ.
Let A = {2, 3, 4, 6}, B = {5, 7, 8} and C = {2, 7, 8, 9}. Find A ∩ C.
Answer
A = {2, 3, 4, 6}
C = {2, 7, 8, 9}
A ∩ C - contains all the common elements in set A and C.
A ∩ C = {2, 3, 4, 6} ∩ {2, 7, 8, 9}
Hence, A ∩ C = {2}.
Let A = {2, 3, 4, 6}, B = {5, 7, 8} and C = {2, 7, 8, 9}. Find B ∩ C.
Answer
B = {5, 7, 8}
C = {2, 7, 8, 9}
B ∩ C - contains all the common elements in set B and C.
B ∩ C = {5, 7, 8} ∩ {2, 7, 8, 9}
Hence, B ∩ C = {7, 8}.
Let A = {1, 4, 7, 8} and B = {4, 6, 8, 9}. Find A - B.
Answer
A = {1, 4, 7, 8}
B = {4, 6, 8, 9}
A - B - contains all the elements in set A but not in B.
A - B = {1, 4, 7, 8} - {4, 6, 8, 9}
Hence, A - B = {1, 7}.
Let A = {1, 4, 7, 8} and B = {4, 6, 8, 9}. Find B - A.
Answer
A = {1, 4, 7, 8}
B = {4, 6, 8, 9}
B - A - contains all the elements in set B but not in A.
B - A = {4, 6, 8, 9} - {1, 4, 7, 8}
Hence, B - A = {6, 9}.
Let ξ = {13, 14, 15, 16, 17, 18, 19, 20, 21}, A = {13, 17, 19} and B = {14, 16, 18, 20}. Find A′.
Answer
ξ = {13, 14, 15, 16, 17, 18, 19, 20, 21}
A = {13, 17, 19}
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {13, 14, 15, 16, 17, 18, 19, 20, 21} - {13, 17, 19}
Hence, A′ = {14, 15, 16, 18, 20, 21}.
Let ξ = {13, 14, 15, 16, 17, 18, 19, 20, 21}, A = {13, 17, 19} and B = {14, 16, 18, 20}. Find B′.
Answer
ξ = {13, 14, 15, 16, 17, 18, 19, 20, 21}
B = {14, 16, 18, 20}
B′ - contains all the elements in universal set which are not in B.
B′ = ξ - B
B′ = {13, 14, 15, 16, 17, 18, 19, 20, 21} - {14, 16, 18, 20}
Hence, B′ = {13, 15, 17, 19, 21}.
Let ξ = {x | x ∈ Z, -4 ≤ x ≤ 4}, A = {x | x ∈ W, x < 4} and B = {x | x ∈ N, 2 < x ≤ 4}. Find A′.
Answer
ξ = {x | x ∈ Z, -4 ≤ x ≤ 4}
ξ = {-4, -3, -2, -1, 0, 1, 2, 3, 4}
A = {x | x ∈ W, x < 4}
A = {0, 1, 2, 3}
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {-4, -3, -2, -1, 0, 1, 2, 3, 4} - {0, 1, 2, 3}
Hence, A′ = {-4, -3, -2, -1, 4}.
Let ξ = {x | x ∈ Z, -4 ≤ x ≤ 4}, A = {x | x ∈ W, x < 4} and B = {x | x ∈ N, 2 < x ≤ 4}. Find B′.
Answer
ξ = {x | x ∈ Z, -4 ≤ x ≤ 4}
ξ = {-4, -3, -2, -1, 0, 1, 2, 3, 4}
B = {x | x ∈ N, 2 < x ≤ 4}
B = {3, 4}
B′ - contains all the elements in universal set which are not in B.
B′ = ξ - B
B′ = {-4, -3, -2, -1, 0, 1, 2, 3, 4} - {3, 4}
Hence, B′ = {-4, -3, -2, -1, 0, 1, 2}.
Let ξ = {x | x ∈ N, x is a factor of 144}, A = {x | x ∈ N, x is a factor of 24}, B = {x | x ∈ N, x is a factor of 36}, C = {x | x ∈ N, x is a factor of 48}. Find A′.
Answer
ξ = {x | x ∈ N, x is a factor of 144}
144 = 1 × 144 = 2 × 72 = 3 × 48 = 4 × 36 = 6 × 24 = 8 × 18 = 9 × 16 = 12 × 12
ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}
A = {x | x ∈ N, x is a factor of 24}
24 = 1 × 24 = 2 × 12 = 3 × 8 = 4 × 6
A = {1, 2, 3, 4, 6, 8, 12, 24}
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144} - {1, 2, 3, 4, 6, 8, 12, 24}
Hence, A′ = {9, 16, 18, 36, 48, 72, 144}.
Let ξ = {x | x ∈ N, x is a factor of 144}, A = {x | x ∈ N, x is a factor of 24}, B = {x | x ∈ N, x is a factor of 36}, C = {x | x ∈ N, x is a factor of 48}. Find B′.
Answer
ξ = {x | x ∈ N, x is a factor of 144}
144 = 1 × 144 = 2 × 72 = 3 × 48 = 4 × 36 = 6 × 24 = 8 × 18 = 9 × 16 = 12 × 12
ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}
B = {x | x ∈ N, x is a factor of 36}
36 = 1 × 36 = 2 × 18 = 3 × 12 = 4 × 9 = 6 × 6
B = {1, 2, 3, 4, 6, 9, 12, 18, 36}
B′ - contains all the elements in universal set which are not in B.
B′ = ξ - B
B′ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144} - {1, 2, 3, 4, 6, 9, 12, 18, 36}
Hence, B′ = {8, 16, 24, 48, 72, 144}.
Let ξ = {x | x ∈ N, x is a factor of 144}, A = {x | x ∈ N, x is a factor of 24}, B = {x | x ∈ N, x is a factor of 36}, C = {x | x ∈ N, x is a factor of 48}. Find C′.
Answer
ξ = {x | x ∈ N, x is a factor of 144}
144 = 1 × 144 = 2 × 72 = 3 × 48 = 4 × 36 = 6 × 24 = 8 × 18 = 9 × 16 = 12 × 12
ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}
C = {x | x ∈ N, x is a factor of 48}
48 = 1 × 48 = 2 × 24 = 3 × 16 = 4 × 12 = 6 × 8
C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
C′ - contains all the elements in universal set which are not in C.
C′ = ξ - C
C′ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144} - {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
Hence, C′ = {9, 18, 36, 72, 144}.
Let ξ = {x | x ∈ N, x is a factor of 144}, A = {x | x ∈ N, x is a factor of 24}, B = {x | x ∈ N, x is a factor of 36}, C = {x | x ∈ N, x is a factor of 48}. Find A ∪ B.
Answer
A = {x | x ∈ N, x is a factor of 24}
A = {1, 2, 3, 4, 6, 8, 12, 24}
B = {x | x ∈ N, x is a factor of 36}
B = {1, 2, 3, 4, 6, 9, 12, 18, 36}
A ∪ B - contains all the elements in set A and B.
A ∪ B = {1, 2, 3, 4, 6, 8, 12, 24} ∪ {1, 2, 3, 4, 6, 9, 12, 18, 36}
Hence, A ∪ B = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}.
Let ξ = {x | x ∈ N, x is a factor of 144}, A = {x | x ∈ N, x is a factor of 24}, B = {x | x ∈ N, x is a factor of 36}, C = {x | x ∈ N, x is a factor of 48}. Find B ∪ C.
Answer
B = {x | x ∈ N, x is a factor of 36}
B = {1, 2, 3, 4, 6, 9, 12, 18, 36}
C = {x | x ∈ N, x is a factor of 48}
C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
B ∪ C - contains all the elements in set B and C.
B ∪ C = {1, 2, 3, 4, 6, 9, 12, 18, 36} ∪ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
Hence, B ∪ C = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48}.
Let ξ = {x | x ∈ N, x is a factor of 144}, A = {x | x ∈ N, x is a factor of 24}, B = {x | x ∈ N, x is a factor of 36}, C = {x | x ∈ N, x is a factor of 48}. Find A ∪ C.
Answer
A = {x | x ∈ N, x is a factor of 24}
A = {1, 2, 3, 4, 6, 8, 12, 24}
C = {x | x ∈ N, x is a factor of 48}
C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
A ∪ C - contains all the elements in set A and C.
A ∪ C = {1, 2, 3, 4, 6, 8, 12, 24} ∪ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
Hence, A ∪ C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}.
Let ξ = {x | x ∈ N, x is a factor of 144}, A = {x | x ∈ N, x is a factor of 24}, B = {x | x ∈ N, x is a factor of 36}, C = {x | x ∈ N, x is a factor of 48}. Find A ∩ B′.
Answer
ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}
A = {x | x ∈ N, x is a factor of 24}
A = {1, 2, 3, 4, 6, 8, 12, 24}
B = {x | x ∈ N, x is a factor of 36}
B = {1, 2, 3, 4, 6, 9, 12, 18, 36}
B′ - contains all the elements in universal set which are not in B.
B′ = ξ - B
B′ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144} - {1, 2, 3, 4, 6, 9, 12, 18, 36}
B′ = {8, 16, 24, 48, 72, 144}
A ∩ B′ - contains all the common elements in set A and B′.
A ∩ B′ = {1, 2, 3, 4, 6, 8, 12, 24} ∩ {8, 16, 24, 48, 72, 144}
Hence, A ∩ B′ = {8, 24}.
Let ξ = {x | x ∈ N, x is a factor of 144}, A = {x | x ∈ N, x is a factor of 24}, B = {x | x ∈ N, x is a factor of 36}, C = {x | x ∈ N, x is a factor of 48}. Find B ∩ C′.
Answer
ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}
B = {x | x ∈ N, x is a factor of 36}
B = {1, 2, 3, 4, 6, 9, 12, 18, 36}
C = {x | x ∈ N, x is a factor of 48}
C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
C′ - contains all the elements in universal set which are not in C.
C′ = ξ - C
C′ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144} - {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
C′ = {9, 18, 36, 72, 144}
B ∩ C′ - contains all the common elements in set B and C′.
B ∩ C′ = {1, 2, 3, 4, 6, 9, 12, 18, 36} ∩ {9, 18, 36, 72, 144}
Hence, B ∩ C′ = {9, 18, 36}.
Let ξ = {x | x ∈ N, x is a factor of 144}, A = {x | x ∈ N, x is a factor of 24}, B = {x | x ∈ N, x is a factor of 36}, C = {x | x ∈ N, x is a factor of 48}. Find C - A.
Answer
A = {x | x ∈ N, x is a factor of 24}
A = {1, 2, 3, 4, 6, 8, 12, 24}
C = {x | x ∈ N, x is a factor of 48}
C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
C - A - contains all the elements in set C but not in A.
C - A = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} - {1, 2, 3, 4, 6, 8, 12, 24}
Hence, C - A = {16, 48}.
Let ξ = {x | x ∈ N, x is a factor of 144}, A = {x | x ∈ N, x is a factor of 24}, B = {x | x ∈ N, x is a factor of 36}, C = {x | x ∈ N, x is a factor of 48}. Find A - (B ∩ C).
Answer
A = {x | x ∈ N, x is a factor of 24}
A = {1, 2, 3, 4, 6, 8, 12, 24}
B = {x | x ∈ N, x is a factor of 36}
B = {1, 2, 3, 4, 6, 9, 12, 18, 36}
C = {x | x ∈ N, x is a factor of 48}
C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
B ∩ C - contains all the common elements in set B and C.
B ∩ C = {1, 2, 3, 4, 6, 9, 12, 18, 36} ∩ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
B ∩ C = {1, 2, 3, 4, 6, 12}
A - (B ∩ C) - contains all the elements in set A but not in B ∩ C.
A - (B ∩ C) = {1, 2, 3, 4, 6, 8, 12, 24} - {1, 2, 3, 4, 6, 12}
Hence, A - (B ∩ C) = {8, 24}.
Considering the sets given in Q.6, state whether the following statement is true or false:
A ∩ (B ∪ C) = A
Answer
A = {1, 2, 3, 4, 6, 8, 12, 24}
B = {1, 2, 3, 4, 6, 9, 12, 18, 36}
C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
B ∪ C - contains all the elements in set B and C.
B ∪ C = {1, 2, 3, 4, 6, 9, 12, 18, 36} ∪ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
B ∪ C = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48}
A ∩ (B ∪ C) - contains all the common elements in set A and B ∪ C.
A ∩ (B ∪ C) = {1, 2, 3, 4, 6, 8, 12, 24} ∩ {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48}
A ∩ (B ∪ C) = {1, 2, 3, 4, 6, 8, 12, 24} = A
Hence, the given statement is true.
Considering the sets given in Q.6, state whether the following statement is true or false:
A ⊂ C
Answer
A = {1, 2, 3, 4, 6, 8, 12, 24}
C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
Each of 1, 2, 3, 4, 6, 8, 12 and 24 is an element of C, so every element of A is in C.
Also, 16 ∈ C but 16 ∉ A, so A ≠ C.
∴ A ⊂ C
Hence, the given statement is true.
Considering the sets given in Q.6, state whether the following statement is true or false:
B ⊆ C
Answer
B = {1, 2, 3, 4, 6, 9, 12, 18, 36}
C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
As 9 ∈ B but 9 ∉ C, every element of B is not in C.
∴ B ⊄ C
Hence, the given statement is false.
Considering the sets given in Q.6, state whether the following statement is true or false:
A ∩ C′ = φ
Answer
ξ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}
A = {1, 2, 3, 4, 6, 8, 12, 24}
C = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
C′ - contains all the elements in universal set which are not in C.
C′ = ξ - C
C′ = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144} - {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
C′ = {9, 18, 36, 72, 144}
A ∩ C′ - contains all the common elements in set A and C′.
A ∩ C′ = {1, 2, 3, 4, 6, 8, 12, 24} ∩ {9, 18, 36, 72, 144}
As there is no element common to A and C′,
∴ A ∩ C′ = φ
Hence, the given statement is true.
Let A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g}. Then verify the following identity:
B ∪ C = C ∪ B
Answer
B = {a, c, e, g}
C = {b, e, f, g}
B ∪ C - contains all the elements in set B and C.
B ∪ C = {a, c, e, g} ∪ {b, e, f, g} = {a, b, c, e, f, g}
C ∪ B - contains all the elements in set C and B.
C ∪ B = {b, e, f, g} ∪ {a, c, e, g} = {a, b, c, e, f, g}
∴ B ∪ C = C ∪ B = {a, b, c, e, f, g}
Hence, B ∪ C = C ∪ B.
Let A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g}. Then verify the following identity:
B ∩ C = C ∩ B
Answer
B = {a, c, e, g}
C = {b, e, f, g}
B ∩ C - contains all the common elements in set B and C.
B ∩ C = {a, c, e, g} ∩ {b, e, f, g} = {e, g}
C ∩ B - contains all the common elements in set C and B.
C ∩ B = {b, e, f, g} ∩ {a, c, e, g} = {e, g}
∴ B ∩ C = C ∩ B = {e, g}
Hence, B ∩ C = C ∩ B.
Let A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g}. Then verify the following identity:
A ∪ (B ∪ C) = (A ∪ B) ∪ C
Answer
A = {a, b, c, d, e}
B = {a, c, e, g}
C = {b, e, f, g}
B ∪ C = {a, c, e, g} ∪ {b, e, f, g} = {a, b, c, e, f, g}
∴ A ∪ (B ∪ C) = {a, b, c, d, e} ∪ {a, b, c, e, f, g} = {a, b, c, d, e, f, g}
Again, A ∪ B = {a, b, c, d, e} ∪ {a, c, e, g} = {a, b, c, d, e, g}
∴ (A ∪ B) ∪ C = {a, b, c, d, e, g} ∪ {b, e, f, g} = {a, b, c, d, e, f, g}
∴ A ∪ (B ∪ C) = (A ∪ B) ∪ C = {a, b, c, d, e, f, g}
Hence, A ∪ (B ∪ C) = (A ∪ B) ∪ C.
Let A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g}. Then verify the following identity:
A ∩ (B ∩ C) = (A ∩ B) ∩ C
Answer
A = {a, b, c, d, e}
B = {a, c, e, g}
C = {b, e, f, g}
B ∩ C = {a, c, e, g} ∩ {b, e, f, g} = {e, g}
∴ A ∩ (B ∩ C) = {a, b, c, d, e} ∩ {e, g} = {e}
Again, A ∩ B = {a, b, c, d, e} ∩ {a, c, e, g} = {a, c, e}
∴ (A ∩ B) ∩ C = {a, c, e} ∩ {b, e, f, g} = {e}
∴ A ∩ (B ∩ C) = (A ∩ B) ∩ C = {e}
Hence, A ∩ (B ∩ C) = (A ∩ B) ∩ C.
Let A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g}. Then verify the following identity:
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
Answer
A = {a, b, c, d, e}
B = {a, c, e, g}
C = {b, e, f, g}
B ∩ C = {a, c, e, g} ∩ {b, e, f, g} = {e, g}
∴ A ∪ (B ∩ C) = {a, b, c, d, e} ∪ {e, g} = {a, b, c, d, e, g}
Again, A ∪ B = {a, b, c, d, e} ∪ {a, c, e, g} = {a, b, c, d, e, g}
and A ∪ C = {a, b, c, d, e} ∪ {b, e, f, g} = {a, b, c, d, e, f, g}
∴ (A ∪ B) ∩ (A ∪ C) = {a, b, c, d, e, g} ∩ {a, b, c, d, e, f, g} = {a, b, c, d, e, g}
∴ A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) = {a, b, c, d, e, g}
Hence, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
Let A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g}. Then verify the following identity:
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Answer
A = {a, b, c, d, e}
B = {a, c, e, g}
C = {b, e, f, g}
B ∪ C = {a, c, e, g} ∪ {b, e, f, g} = {a, b, c, e, f, g}
∴ A ∩ (B ∪ C) = {a, b, c, d, e} ∩ {a, b, c, e, f, g} = {a, b, c, e}
Again, A ∩ B = {a, b, c, d, e} ∩ {a, c, e, g} = {a, c, e}
and A ∩ C = {a, b, c, d, e} ∩ {b, e, f, g} = {b, e}
∴ (A ∩ B) ∪ (A ∩ C) = {a, c, e} ∪ {b, e} = {a, b, c, e}
∴ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) = {a, b, c, e}
Hence, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
Let A = {b, c, d, e} and B = {d, e, f, g} be two subsets of the universal set ξ = {b, c, d, e, f, g}. Then, verify the following:
(A ∪ B)′ = (A′ ∩ B′)
Answer
ξ = {b, c, d, e, f, g}
A = {b, c, d, e}
B = {d, e, f, g}
A ∪ B = {b, c, d, e} ∪ {d, e, f, g} = {b, c, d, e, f, g}
(A ∪ B)′ - contains all the elements in universal set which are not in A ∪ B.
(A ∪ B)′ = ξ - (A ∪ B) = {b, c, d, e, f, g} - {b, c, d, e, f, g} = φ
Again, A′ = ξ - A = {b, c, d, e, f, g} - {b, c, d, e} = {f, g}
and B′ = ξ - B = {b, c, d, e, f, g} - {d, e, f, g} = {b, c}
∴ A′ ∩ B′ = {f, g} ∩ {b, c} = φ
∴ (A ∪ B)′ = (A′ ∩ B′) = φ
Hence, (A ∪ B)′ = (A′ ∩ B′).
Let A = {b, c, d, e} and B = {d, e, f, g} be two subsets of the universal set ξ = {b, c, d, e, f, g}. Then, verify the following:
(A ∩ B)′ = (A′ ∪ B′)
Answer
ξ = {b, c, d, e, f, g}
A = {b, c, d, e}
B = {d, e, f, g}
A ∩ B = {b, c, d, e} ∩ {d, e, f, g} = {d, e}
(A ∩ B)′ - contains all the elements in universal set which are not in A ∩ B.
(A ∩ B)′ = ξ - (A ∩ B) = {b, c, d, e, f, g} - {d, e} = {b, c, f, g}
Again, A′ = ξ - A = {b, c, d, e, f, g} - {b, c, d, e} = {f, g}
and B′ = ξ - B = {b, c, d, e, f, g} - {d, e, f, g} = {b, c}
∴ A′ ∪ B′ = {f, g} ∪ {b, c} = {b, c, f, g}
∴ (A ∩ B)′ = (A′ ∪ B′) = {b, c, f, g}
Hence, (A ∩ B)′ = (A′ ∪ B′).
Fill in the blanks:
(i) A ∪ A = .......
(ii) A ∩ A = .......
(iii) A ∪ φ = .......
(iv) A ∩ φ = .......
(v) (A ∪ B)′ = .......
(vi) (A ∩ B)′ = .......
Answer
(i) A ∪ A = A.
As A ∪ A contains all the elements of A and of A, it contains exactly the elements of A.
(ii) A ∩ A = A.
As every element of A is common to A and A, A ∩ A contains exactly the elements of A.
(iii) A ∪ φ = A.
As the empty set φ has no element, A ∪ φ contains exactly the elements of A.
(iv) A ∩ φ = φ.
As the empty set φ has no element, there can be no element common to A and φ.
(v) (A ∪ B)′ = A′ ∩ B′.
This is De-Morgan's law for the complement of the union of two sets.
(vi) (A ∩ B)′ = A′ ∪ B′.
This is De-Morgan's law for the complement of the intersection of two sets.
Let ξ = {x | x ∈ N, 4 ≤ x < 18} and A, B, C be subsets of ξ given by A = {x | x is a multiple of 2}, B = {x | x is a multiple of 3} and C = {x | x ∈ N, x < 11}. Then, verify the following:
(A ∪ B)′ = (A′ ∩ B′)
Answer
ξ = {x | x ∈ N, 4 ≤ x < 18}
ξ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}
A = {x | x is a multiple of 2}
A = {4, 6, 8, 10, 12, 14, 16}
B = {x | x is a multiple of 3}
B = {6, 9, 12, 15}
A ∪ B = {4, 6, 8, 10, 12, 14, 16} ∪ {6, 9, 12, 15} = {4, 6, 8, 9, 10, 12, 14, 15, 16}
(A ∪ B)′ - contains all the elements in universal set which are not in A ∪ B.
(A ∪ B)′ = ξ - (A ∪ B)
(A ∪ B)′ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17} - {4, 6, 8, 9, 10, 12, 14, 15, 16}
(A ∪ B)′ = {5, 7, 11, 13, 17}
Again, A′ = ξ - A = {5, 7, 9, 11, 13, 15, 17}
and B′ = ξ - B = {4, 5, 7, 8, 10, 11, 13, 14, 16, 17}
∴ A′ ∩ B′ = {5, 7, 9, 11, 13, 15, 17} ∩ {4, 5, 7, 8, 10, 11, 13, 14, 16, 17} = {5, 7, 11, 13, 17}
∴ (A ∪ B)′ = (A′ ∩ B′) = {5, 7, 11, 13, 17}
Hence, (A ∪ B)′ = (A′ ∩ B′).
Let ξ = {x | x ∈ N, 4 ≤ x < 18} and A, B, C be subsets of ξ given by A = {x | x is a multiple of 2}, B = {x | x is a multiple of 3} and C = {x | x ∈ N, x < 11}. Then, verify the following:
(A ∩ B)′ = (A′ ∪ B′)
Answer
ξ = {x | x ∈ N, 4 ≤ x < 18}
ξ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}
A = {x | x is a multiple of 2}
A = {4, 6, 8, 10, 12, 14, 16}
B = {x | x is a multiple of 3}
B = {6, 9, 12, 15}
A ∩ B = {4, 6, 8, 10, 12, 14, 16} ∩ {6, 9, 12, 15} = {6, 12}
(A ∩ B)′ - contains all the elements in universal set which are not in A ∩ B.
(A ∩ B)′ = ξ - (A ∩ B)
(A ∩ B)′ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17} - {6, 12}
(A ∩ B)′ = {4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17}
Again, A′ = ξ - A = {5, 7, 9, 11, 13, 15, 17}
and B′ = ξ - B = {4, 5, 7, 8, 10, 11, 13, 14, 16, 17}
∴ A′ ∪ B′ = {5, 7, 9, 11, 13, 15, 17} ∪ {4, 5, 7, 8, 10, 11, 13, 14, 16, 17}
A′ ∪ B′ = {4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17}
∴ (A ∩ B)′ = (A′ ∪ B′) = {4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17}
Hence, (A ∩ B)′ = (A′ ∪ B′).
Let ξ = {x | x ∈ N, 4 ≤ x < 18} and A, B, C be subsets of ξ given by A = {x | x is a multiple of 2}, B = {x | x is a multiple of 3} and C = {x | x ∈ N, x < 11}. Then, verify the following:
A - B = A ∩ B′
Answer
ξ = {x | x ∈ N, 4 ≤ x < 18}
ξ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}
A = {x | x is a multiple of 2}
A = {4, 6, 8, 10, 12, 14, 16}
B = {x | x is a multiple of 3}
B = {6, 9, 12, 15}
A - B - contains all the elements in set A but not in B.
A - B = {4, 6, 8, 10, 12, 14, 16} - {6, 9, 12, 15} = {4, 8, 10, 14, 16}
Again, B′ = ξ - B = {4, 5, 7, 8, 10, 11, 13, 14, 16, 17}
∴ A ∩ B′ = {4, 6, 8, 10, 12, 14, 16} ∩ {4, 5, 7, 8, 10, 11, 13, 14, 16, 17} = {4, 8, 10, 14, 16}
∴ A - B = A ∩ B′ = {4, 8, 10, 14, 16}
Hence, A - B = A ∩ B′.
Let ξ = {x | x ∈ N, 4 ≤ x < 18} and A, B, C be subsets of ξ given by A = {x | x is a multiple of 2}, B = {x | x is a multiple of 3} and C = {x | x ∈ N, x < 11}. Then, verify the following:
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
Answer
ξ = {x | x ∈ N, 4 ≤ x < 18}
ξ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}
A = {x | x is a multiple of 2}
A = {4, 6, 8, 10, 12, 14, 16}
B = {x | x is a multiple of 3}
B = {6, 9, 12, 15}
C = {x | x ∈ N, x < 11}
C = {4, 5, 6, 7, 8, 9, 10}
B ∩ C = {6, 9, 12, 15} ∩ {4, 5, 6, 7, 8, 9, 10} = {6, 9}
∴ A ∪ (B ∩ C) = {4, 6, 8, 10, 12, 14, 16} ∪ {6, 9} = {4, 6, 8, 9, 10, 12, 14, 16}
Again, A ∪ B = {4, 6, 8, 10, 12, 14, 16} ∪ {6, 9, 12, 15} = {4, 6, 8, 9, 10, 12, 14, 15, 16}
and A ∪ C = {4, 6, 8, 10, 12, 14, 16} ∪ {4, 5, 6, 7, 8, 9, 10} = {4, 5, 6, 7, 8, 9, 10, 12, 14, 16}
∴ (A ∪ B) ∩ (A ∪ C) = {4, 6, 8, 9, 10, 12, 14, 15, 16} ∩ {4, 5, 6, 7, 8, 9, 10, 12, 14, 16}
(A ∪ B) ∩ (A ∪ C) = {4, 6, 8, 9, 10, 12, 14, 16}
∴ A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) = {4, 6, 8, 9, 10, 12, 14, 16}
Hence, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).