# Class 11 RD Sharma Solutions – Chapter 1 Sets – Exercise 1.7

• Last Updated : 15 Dec, 2020

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

Solution:

To prove:

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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.

### (iv) A – B = A Δ (A ∩ B)

Solution:

(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

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

Solution:

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.

### (v) (A – B) ∪ (A ∩ B) = A

Solution:

(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

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