问题1：对于任何两个集合A和B，证明：A’– B’= B – A
解决方案：
To prove:
A’ – B’ = B – A
Firstly show that
A’ – B’ ⊆ B – A
Let, x ∈ A’ – B’
⇒ x ∈ A’ and x ∉ B’
⇒ x ∉ A and x ∈ B (since, A ∩ A’ = ϕ)
⇒ x ∈ B – A
It is true for all x ∈ A’ – B’
Therefore,
A’ – B’ = B – A
Hence, Proved.
问题2：对于任何两个A和B集，请证明以下内容：
(i)A∩(A’∪B)= A∩B
(ii)A –(A – B)= A∩B
(iii)A∩(A∪B’)= ϕ
(iv)A – B = AΔ (A∩B)
解决方案：
(i) A ∩ (A’ ∪ B) = A ∩ B
Here, LHS A ∩ (A’ ∪ B)
Upon expanding
(A ∩ A’) ∪ (A ∩ B)
We know, (A ∩ A’) =ϕ
⇒ ϕ ∪ (A∩ B)
⇒ (A ∩ B)
Therefore,
LHS = RHS
Hence, proved.
(ii) A – (A – B) = A ∩ B
For any sets A and B we have De-Morgan’s law
(A ∪ B)’ = A’ ∩ B’, (A ∩ B) ‘ = A’ ∪ B’
Take LHS
= A – (A–B)
= A ∩ (A–B)’
= A ∩ (A∩B’)’
= A ∩ (A’ ∪ B’)’) (since, (B’)’ = B)
= A ∩ (A’ ∪ B)
= (A ∩ A’) ∪ (A ∩ B)
= ϕ ∪ (A ∩ B) (since, A ∩ A’ = ϕ)
= (A ∩ B) (since, ϕ ∪ x = x, for any set)
= RHS
Therefore,
LHS = RHS
Hence, proved.
(iii) A ∩ (A ∪ B’) = ϕ
Here, LHS A ∩ (A ∪ B’)
= A ∩ (A ∪ B’)
= A ∩ (A’∩ B’) {By De–Morgan’s law}
= (A ∩ A’) ∩ B’ (since, A ∩ A’ = ϕ)
= ϕ ∩ B’
= ϕ (since, ϕ ∩ B’ = ϕ)
= RHS
Therefore,
LHS = RHS
Hence, proved.
(iv) A – B = A Δ (A ∩ B)
Here, RHS A Δ (A ∩ B)
A Δ (A ∩ B) (since, E Δ F = (E–F) ∪ (F–E))
= (A – (A ∩ B)) ∪ (A ∩ B –A) (since, E – F = E ∩ F’)
= (A ∩ (A ∩ B)’) ∪ (A ∩ B ∩ A’)
= (A ∩ (A’ ∪ B’)) ∪ (A ∩ A’ ∩ B) {by using De-Morgan’s law and associative law}
= (A ∩ A’) ∪ (A ∩ B’) ∪ (ϕ ∩ B) (by using distributive law)
= ϕ ∪ (A ∩ B’) ∪ ϕ
= A ∩ B’ (since, A ∩ B’ = A – B)
= A – B
= LHS
Therefore,
LHS = RHS
Hence, Proved
问题3：如果A，B，C是三组，使得A⊂B，则证明C – B⊂C –A。
解决方案：
Given: A ⊂ B
To prove: C – B ⊂ C – A
Let’s assume, x ∈ C – B
⇒ x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A
⇒ x ∈ C – A
Thus, x ∈ C–B ⇒ x ∈ C – A
This is true for all x ∈ C – B
Therefore,
C – B ⊂ C – A
Hence, proved.
问题4：对于任何两个集合A和B，证明
(i)(A∪B)– B = A – B
(ii)A –(A∩B)= A – B
(iii)A –(A – B)= A∩B
(iv)A∪(B – A)= A∪B
(v)(A – B)∪(A∩B)= A
解决方案：
(i) (A ∪ B) – B = A – B
Let’s assume LHS (A ∪ B) – B
= (A–B) ∪ (B–B)
= (A–B) ∪ ϕ (since, B–B = ϕ)
= A–B (since, x ∪ ϕ = x for any set)
= RHS
Hence, proved.
(ii) A – (A ∩ B) = A – B
Let’s assume LHS A – (A ∩ B)
= (A–A) ∩ (A–B)
= ϕ ∩ (A – B) (since, A-A = ϕ)
= A – B
= RHS
Hence, proved.
(iii) A – (A – B) = A ∩ B
Let’s assume LHS A – (A – B)
Let, x ∈ A – (A–B) = x ∈ A and x ∉ (A–B)
x ∈ A and x ∉ (A ∩ B)
= x ∈ A ∩ (A ∩ B)
= x ∈ (A ∩ B)
= (A ∩ B)
= RHS
Hence, proved.
(iv) A ∪ (B – A) = A ∪ B
Let’s assume LHS A ∪ (B – A)
Let, x ∈ A ∪ (B –A) ⇒ x ∈ A or x ∈ (B – A)
⇒ x ∈ A or x ∈ B and x ∉ A
⇒ x ∈ B
⇒ x ∈ (A ∪ B) (since, B ⊂ (A ∪ B))
This is true for all x ∈ A ∪ (B–A)
∴ A ∪ (B–A) ⊂ (A ∪ B) —(equation 1)
Contrarily,
Let x ∈ (A ∪ B) ⇒ x ∈ A or x ∈ B
⇒ x ∈ A or x ∈ (B–A) (since, B ⊂ (A ∪ B))
⇒ x ∈ A ∪ (B–A)
∴ (A ∪ B) ⊂ A ∪ (B–A) —(equation 2)
From equation 1 and 2 we get,
A ∪ (B – A) = A ∪ B
Hence, proved.
(v) (A – B) ∪ (A ∩ B) = A
Let’s assume LHS (A – B) ∪ (A ∩ B)
Let, x ∈ A
Then either x ∈ (A–B) or x ∈ (A ∩ B)
⇒ x ∈ (A–B) ∪ (A ∩ B)
∴ A ⊂ (A – B) ∪ (A ∩ B) —(equation 1)
Contrarily,
Let x ∈ (A–B) ∪ (A ∩ B)
⇒ x ∈ (A–B) or x ∈ (A ∩ B)
⇒ x ∈ A and x ∉ B or x ∈ B
⇒ x ∈ A
(A–B) ∪ (A ∩ B) ⊂ A —(equation 2)
∴ From equation (1) and (2), We get
(A–B) ∪ (A ∩ B) = A
Hence, proved