问题1：给定A = {1,2,3}，B = {3，4}，C = {4，5，6}，求出(A×B)∩(B×C)。
解决方案：
Given:
A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}
Let us find: (A × B) ∩ (B × C)
(A × B) = {1, 2, 3} × {3, 4}
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
(B × C) = {3, 4} × {4, 5, 6}
= {(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
Therefore,
(A × B) ∩ (B × C) = {(3, 4)}
问题2：如果A = {2，3}，B = {4，5}，C = {5，6}求A×(B∪C)，(A×B)∪(A×C)。
解决方案：
Given:
A = {2, 3}, B = {4, 5} and C = {5, 6}
Let us find: A x (B ∪ C) and (A x B) ∪ (A x C)
(B ∪ C) = {4, 5, 6}
A × (B ∪ C) = {2, 3} × {4, 5, 6}
= {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A × B) = {2, 3} × {4, 5}
= {(2, 4), (2, 5), (3, 4), (3, 5)}
(A × C) = {2, 3} × {5, 6}
= {(2, 5), (2, 6), (3, 5), (3, 6)}
Therefore,
(A × B) ∪ (A × C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
A × (B ∪ C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
问题3：如果A = {1,2,3}，B = {4}，C = {5}，请验证：
(i)A×(B∪C)=(A×B)∪(A×C)
(ii)A×(B∩C)=(A×B)∩(A×C)
(iii)A×(B – C)=(A×B)–(A×C)
解决方案：
Given:
A = {1, 2, 3}, B = {4} and C = {5}
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
Let’s assume LHS: (B ∪ C)
(B ∪ C) = {4, 5}
A × (B ∪ C) = {1, 2, 3} × {4, 5}
= {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
Now, RHS
(A × B) = {1, 2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}
(A × C) = {1, 2, 3} × {5}
= {(1, 5), (2, 5), (3, 5)}
(A × B) ∪ (A × C) = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)}
Therefore,
LHS = RHS
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
Let’s assume LHS: (B ∩ C)
(B ∩ C) = ∅ (No common element)
A × (B ∩ C) = {1, 2, 3} × ∅
= ∅
Now, RHS
(A × B) = {1, 2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}
(A × C) = {1, 2, 3} × {5}
= {(1, 5), (2, 5), (3, 5)}
(A × B) ∩ (A × C) = ∅
Therefore,
LHS = RHS
(iii) A × (B − C) = (A × B) − (A × C)
Let’s assume LHS: (B − C)
(B − C) = ∅
A × (B − C) = {1, 2, 3} × ∅
= ∅
Now, RHS
(A × B) = {1, 2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}
(A × C) = {1, 2, 3} × {5}
= {(1, 5), (2, 5), (3, 5)}
(A × B) − (A × C) = ∅
Therefore,
LHS = RHS
问题4：设A = {1,2,3，B = {1,2,3,4}，C = {5，6}，D = {5，6，7，8}。验证：
(i)A×C⊂B×D
(ii)A×(B×C)=(A×B)×(A×C)
解决方案：
Given:
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
(i) A x C ⊂ B x D
Let us consider LHS A x C
A × C = {1, 2} × {5, 6}
= {(1, 5), (1, 6), (2, 5), (2, 6)}
Now, RHS
B × D = {1, 2, 3, 4} × {5, 6, 7, 8}
= {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
Since, all elements of A × C is in B × D.
Therefore,
We can say A × C ⊂ B × D
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
Let’s assume LHS A × (B ∩ C)
(B ∩ C) = ∅
A × (B ∩ C) = {1, 2} × ∅
= ∅
Now, RHS
(A × B) = {1, 2} × {1, 2, 3, 4}
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
(A × C) = {1, 2} × {5, 6}
= {(1, 5), (1, 6), (2, 5), (2, 6)}
Hence, there is no common element between A × B and A × C
(A × B) ∩ (A × C) = ∅
Therefore,
A × (B ∩ C) = (A × B) ∩ (A × C)
问题5：如果A = {1,2,3}，B = {3，4}，C = {4，5，6}，则找到
(i)A×(B∩C)
(ii)(A×B)∩(A×C)
(iii)A×(B∪C)
(iv)(A×B)∪(A×C)
解决方案：
Given:
A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}
(i) A × (B ∩ C)
(B ∩ C) = {4}
A × (B ∩ C) = {1, 2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}
(ii) (A × B) ∩ (A × C)
(A × B) = {1, 2, 3} × {3, 4}
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
(A × C) = {1, 2, 3} × {4, 5, 6}
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}
(iii) A × (B ∪ C)
(B ∪ C) = {3, 4, 5, 6}
A × (B ∪ C) = {1, 2, 3} × {3, 4, 5, 6}
= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}
(iv) (A × B) ∪ (A × C)
(A × B) = {1, 2, 3} × {3, 4}
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
(A × C) = {1, 2, 3} × {4, 5, 6}
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A × B) ∪ (A × C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}
问题6：证明：
(i)(A∪B)×C =(A×C)=(A×C)∪(B×C)
(ii)(A∩B)×C =(A×C)∩(B×C)
解决方案：
(i) (A ∪ B) × C = (A × C) = (A × C) ∪ (B × C)
Let (x, y) be an arbitrary element of (A ∪ B) × C
(x, y) ∈ (A ∪ B) C
Since, (x, y) are elements of Cartesian product of (A ∪ B) × C
x ∈ (A ∪ B) and y ∈ C
(x ∈ A or x ∈ B) and y ∈ C
(x ∈ A and y ∈ C) or (x ∈ Band y ∈ C)
(x, y) ∈ A × C or (x, y) ∈ B × C
(x, y) ∈ (A × C) ∪ (B × C) … (1)
Let (x, y) be an arbitrary element of (A × C) ∪ (B × C).
(x, y) ∈ (A × C) ∪ (B × C)
(x, y) ∈ (A × C) or (x, y) ∈ (B × C)
(x ∈ A and y ∈ C) or (x ∈ B and y ∈ C)
(x ∈ A or x ∈ B) and y ∈ C
x ∈ (A ∪ B) and y ∈ C
(x, y) ∈ (A ∪ B) × C … (2)
From 1 and 2, we get: (A ∪ B) × C = (A × C) ∪ (B × C)
(ii) (A ∩ B) × C = (A × C) ∩ (B × C)
Let (x, y) be an arbitrary element of (A ∩ B) × C.
(x, y) ∈ (A ∩ B) × C
Since, (x, y) are elements of Cartesian product of (A ∩ B) × C
x ∈ (A ∩ B) and y ∈ C
(x ∈ A and x ∈ B) and y ∈ C
(x ∈ A and y ∈ C) and (x ∈ Band y ∈ C)
(x, y) ∈ A × C and (x, y) ∈ B × C
(x, y) ∈ (A × C) ∩ (B × C) … (1)
Let (x, y) be an arbitrary element of (A × C) ∩ (B × C).
(x, y) ∈ (A × C) ∩ (B × C)
(x, y) ∈ (A × C) and (x, y) ∈ (B × C)
(x ∈A and y ∈ C) and (x ∈ Band y ∈ C)
(x ∈A and x ∈ B) and y ∈ C
x ∈ (A ∩ B) and y ∈ C
(x, y) ∈ (A ∩ B) × C … (2)
From 1 and 2, we get: (A ∩ B) × C = (A × C) ∩ (B × C)
问题7：如果A×B⊆C×D和A∩B∈∅，证明A⊆C和B⊆D.
解决方案：
Given:
A × B ⊆ C x D and A ∩ B ∈ ∅
A × B ⊆ C x D denotes A × B is subset of C × D that is every element A × B is in C × D.
And A ∩ B ∈ ∅ denotes A and B does not have any common element between them.
A × B = {(a, b): a ∈ A and b ∈ B}
Therefore,
We can say (a, b) ⊆ C × D [Since, A × B ⊆ C x D is given]
a ∈ C and b ∈ D
a ∈ A = a ∈ C
A ⊆ C
And
b ∈ B = b ∈ D
B ⊆ D
Hence proved.