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# RD Chapter 1- Sets Ex-1.7 Interview Questions Answers

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Question 1 : For any two sets A and B, prove that: A‘ – B‘ = B – A

To prove, A’ – B’ = B – A
Firstly we need to show
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’
∴ A’ – B’ = B – A
Hence Proved.

Question 2 :
For any two sets A and B, prove the following:
(i) A ∩ (A‘ ∪ B) = A ∩ B
(ii) A – (A – B) = A ∩ B
(iii) A ∩ (A ∪ B’) = ϕ
(iv) A – B = A Δ (A ∩ B)

(i) A ∩ (A’ ∪ B) = A ∩ B
Let us consider LHS A ∩ (A’ ∪ B)
Expanding
(A ∩ A’) ∪ (A ∩ B)
We know, (A ∩ A’) =ϕ
⇒ ϕ ∪ (A∩ B)
⇒ (A ∩ B)
∴ 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’
Consider 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
∴ LHS=RHS
Hence proved.
(iii) A ∩ (A ∪ B’) = ϕ
Let us consider 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
∴ LHS=RHS
Hence proved.
(iv) A – B = A Δ (A ∩ B)
Let us consider 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
∴ LHS=RHS
Hence Proved

Question 3 : If A, B, C are three sets such that A ⊂ B, then prove that C – B ⊂ C – A.

Given, ACB
To prove: C – B ⊂ C – A
Let us consider, 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
∴ C – B ⊂ C – A
Hence proved.

Question 4 :
For any two sets A and B, prove that
(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

(i) (A ∪ B) – B = A – B
Let us consider 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 us consider 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 us consider 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 us consider 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)…… (1)
Conversely,
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)…… (2)
From 1 and 2 we get,
A ∪ (B – A) = A ∪ B
Hence proved.
(v) (A – B) ∪ (A ∩ B) = A
Let us consider 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)…. (1)
Conversely,
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………. (2)
∴ From (1) and (2), We get
(A–B) ∪ (A ∩ B) = A
Hence proved.

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