Chapter 1: Sets – Exercise 1.7

Sets – Exercise 1.7 – Q.1

To show A' - B' = B - A

We show that A' – B' = ⊆ B - A and vice versa

Let, x ϵ A' – B'

⟹ x ϵ A' and x ∉ B'

⟹ x ∉ A and x ϵ B           [∵ A ∩ A' = ∮ and B ∩ B' = ∮]

⟹ x ϵ B and x ∉ A

⟹ x ϵ B - A

This is true for all x ϵ A' - B'

Hence A' - B' ⊆ B - A

Conversely,

Let, x ϵ B - A

⟹ x ϵ B and x ∉ A

⟹ x ∉ B' and x ϵ A'

⟹ x ϵ A' and x ∉ B'  [∵ B ∩ B' = ∮ and A ∩ A' = ∮]

⟹ x ϵ A' - B'

This is true for all x ϵ B - A

Hence B- A ⊆ A' - B'

∴ A' - B' = B - A  Proved.

Sets – Exercise 1.7 – Q.2(i)

LHS = A ∩ (A' ∪ B)

= (A ∩ A') ∪ (A ∩ B)     [∴ ∩ distributes over (i)]

= ∮ ∪ (A ∩ B)             [∵ A ∩ A' = ∮]       

= A ∩ B                      [∵ ∮ ∪ x = x for any set x]  

= RHS

∴ LHS = RHS Proved.

Sets – Exercise 1.7 – Q.2(ii)

For any sets A and B we have by De-morgan's laws

(A ∪ B)' =  A' ∩ B', (A ∩ B)' = A' ∪ B'

Also

LHS = A - (A - B)

= A ∩ (A - B)'

= A ∩ (A ∩ B')'

= A ∩ (A' ∩ (B')')                  [By De-morgan's law]

= A ∩ (A' ∪ B)                     [∵ (B')' = B]

= (A ∩ A') ∪(A ∩ B)

= ∮ ∪ (A ∩ B)                       [∵ A ∩ A' = ∮]

= A ∩ B                                 [∵ ∮ ∪ x = x, for and set x]       

= RHS

∴ LHS = RHS Proved.

Sets – Exercise 1.7 – Q.2(iii)

LHS = A ∩ (A ∪ B')

= A ∩ (A' ∩ B')        [By De-morgan's law]

= (A ∩ A') ∩ B'        [By associative law]   

= ∮ ∩ B'                    [∵ A ∩ A' = ∮]             

= ∮

= RHS

∴ LHS = RHS Proved.

Sets – Exercise 1.7 – Q.2(iv)

RHS = A ∆ (A ∩ 8)

= (A - (A ∩ B)) ∪ (A ∩ B - A)                         [∵ E∆F = (E - F) ∪ (F - E)]

= (A ∩ (A ∩ B)') ∪ (A ∩ B ∩ A')                   [∵ E - F = E ∩ F']

= (A ∩ (A' ∪ B')) ∪ (A ∩ A' ∩ B)                  [By de-morgan's law & associative law] 

= (A ∩ A') ∪ (A ∩ B') ∪ (∮ ∩ B)                   [∵ ∩ distributes over ∪ and A ∩ A' = ∮]

= ∮ ∪ (A ∩ B') ∪ ∮                                       [∵ ∮ ∩ B = ∮]

= A ∩ B'                                                                [∵ ∮ ∪ x = x for any set x]

= A - B                                                                   [∵ A ∩ B' = A - B]

= LHS

∴ LHS = RHS Proved.

Sets – Exercise 1.7 – Q.3

We have, ACB

To show: C - B ⊂ C - A

Let, x ϵ C - B

⟹ x ϵ C and x ∉ B

⟹ x ϵ C and x ∉ A           [∵ A ⊂ B]

 ⟹ x ϵ C - A

Thus, x ϵ C - B ⟹ x ϵ C- A

This is true for all x ϵ C-B

∴ C - B ⊂ C- A

Sets – Exercise 1.7 – Q.4(i)

(A ∪ B) – B = (A - B) ∪ (A - B)

= ∮ ∩ (A - B)

= A - B

Sets – Exercise 1.7 – Q.4(ii)

A – (A ∩ B) = (A - A) ∩ (A - B)

= φ ∩ (A – B)

= A – B

Sets – Exercise 1.7 – Q.4(iii)

Let x ϵ A - (A - B) ⟺ x ϵ A and x ∉ (A – B)

⟺ x ϵ A and X ϵ (A ∩ B)

⟺ x ϵ A ∩ (A ∩ B)

⟺ x ϵ (A ∩ B)

∴ A - (A – B) = (A ∩ B)

Sets – Exercise 1.7 – Q.4(iv)

Let x ϵ A ∪ (B - A) ⟹ x ϵ A or x ϵ (B - A)

⟹ x ϵ A or x ϵ B and x ∉ A

⟹ x ϵ B

⟹ x ϵ (A ∪ B)    [∵ B ⊂ (A ∪ B)]

This is true for all x ϵ A ∪ (B - A)

∴ A ∪ (B - A) ⊂ (A ∪ B)                .........(1)

Conversely,

Let, x ϵ (A ∪ B)

⟹ x ϵ A or x ϵ B

⟹ x ϵ A or x ϵ (B – A)       [∵ B ⊂ (B – A)]

∴  (A ∪ B) ⊂ A ∪ (B - A)                 ..... .(2)

From (1) and (2), we get

A ∪ (B - A) = (A ∪ B)

Sets – Exercise 1.7 – Q.4(v)

Let x ϵ A.

Then either x ϵ (A- B) or x ϵ (A ∩ B)

⟹ x ϵ(A – B) ∪ (A ∩ B)

∴ A ⊂ (A – B) ∪ (A ∩ B)..........(1)

Conversely,

Let x ϵ (A - B) ∪ (A ∩ B)

⟹ x ϵ (A – B) or x (A ∩ B)

⟹ x ϵ A and x ∉ B or x ϵ A and x ϵ B  

⟹ x ϵ A

∴(A-B) ∪ (A ∩ B) ⊂ A     ......(2)

From (1) and (2), we get

(A - B) ∪ (A ∩ B) = A