# Chapter 1: Sets – Exercise 1.6

## Sets – Exercise 1.6 – Q.1

The smallest set A such that

A ∪ {1,2} = {1, 2, 3, 5, 9} is {3, 5, 9}

∵ {3, 5, 9} ∪ {1, 2} = {1, 2, 3, 5, 9}

Any other set 8 such that 8 ∪ {1, 2} = {1, 2, 3, 5, 9} will contain A. For example we contake B to be {1, 3, 5, 9} or {1, 2, 3, 5, 9}. Clearly B contains A = {3, 5, 9}.

## Sets– Exercise 1.6 – Q.2(i)

i. A = (1, 2, 4, 5), B = (2, 3, 5, 6), C = (4, 5, 6, 7)

B ∩ C = {5, 6}

A ∪ (B ∩ C) = {1, 2, 4, 5, 6} ........ (1)

(A ∪ B) = {1, 2, 3, 4, 5, 6}

(A ∪ C) = {1, 2, 4, 5, 6, 7}

(A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6}

From eqn (1) and eqn (2),

we get A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

## Sets– Exercise 1.6 – Q.2(ii)

A = {1, 2, 4, 5}, B = {2, 3, 5, 6}, C = (4, 5, 6, 7)

B ∪ C = {2, 3, 4, 5, 6, 7}

A ∩(B ∪ C) = (2, 4, 5)........(1)

(A ∩ B) = (2, 5)

(A ∩ C) = (4, 5)

(A ∩ B ) ∪ (B ∩ C) = (2, 4, 5)........(2)

From eqn (1) and eqn (2), we get

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

## Sets– Exercise 1.6 – Q.2(iii)

A = {1, 2, 4, 5}, B = {2, 3, 5, 6}, C = {4, 5, 6, 7}

B - C = {2, 3}

A ∩ (B – C) = (2)                    …...(1)

(A ∩ B) = (2, 5)

(A ∩ C) = (4, 5)

(A ∩ B) - (A ∩ C) = {2}            …..(2)

From eqn (1) and eqn (2), we get

A ∩ (B-C) = (A ∩ B) - (A ∩ C)

## Sets– Exercise 1.6 – Q.2(iv)

A = {1, 2, 4, 5}, B = {2, 3, 5, 6}, C = { 4, 5, 6, 7}

B ∪ C = {2, 3, 4, 5, 6, 7)

A -(B ∪ C) = {1}                        .......(1)

(A- B) = {1, 4}

(A- C) = {1, 2}

(A - B) ∩ (A- C) = {1}           ........(2)

From eqn (1) and eqn (2), we get

A – (B ∪ C) = (A - B) ∩ (A - C)

## Sets– Exercise 1.6 – Q.2(v)

A = (1, 2, 4, 5), B =  (2, 3, 5, 6), C = (4, 5, 6, 7)

B ∩ C = {5, 6}

A - (B ∩ C) = {1, 2, 4}....(1)

(A - B) = {1, 4}

(A - C) = {1, 2}

(A – B) ∪ (A - C) = {1, 2, 4} .............(2)

From eqn (1) and eqn (2), we get

A – (B ∩ C) = (A – B) ∪ (A – C)

## Sets– Exercise 1.6 – Q.2(vi)

A = (1, 2, 4, 5), B = (2, 3, 5, 6), C = (4, 5, 6, 7)

B∆C= (B - C) ∪ (C- B) = (2, 3} ∪ {4, 7} = {2, 3, 4, 7}

A ∩ (B∆C) = {2. 4}........(i)

(A ∩ B) = {2, 5}

(A ∩ C)= {4, 5}

(A ∩ B) ∆ (A ∩ C) = [(A ∩ B) – (A ∩ C)] ∪ [(A ∩ C) – (A ∩ B)]

(A ∩ B) ∆ (A ∪ C) = {2} ∪ {4} = {24}      .............(2)

From eqn (1) and eqn(2), we get

A ∩ (B ∆ C) = (A ∩ B) ∆ (A ∪ C)

## Sets– Exercise 1.6 – Q.3(i)

U = {2, 3, 5, 7, 9} is the universal set

A = {3, 7}, B = {2, 5, 7, 9}

A ∪ B = {x : x ϵ A or x ϵ B}

= {2, 3, 5, 7, 9}

LHS = (A ∪ B)'

= {2, 3, 5, 7, 9}

RHS = A' ∩ B'

A' = {x ϵ U : x ∉ A}

B' = {x ϵ U : x ∉ B}

= {3}

∴A' ∩ B' = {2 ,5, 9} ∩ {B}

= ∮  [∵ the two sets are disjoint]

∴ LHS = RHS Proved

## Sets– Exercise 1.6 – Q.3(ii)

LHS = (A ∩ B)'

Now,

A ∩ B = {x|x ϵ A and x ϵ B}

= {7}

∴ (A ∩ B)' = {7}'

= {x ϵ U : x ∉ 7}

= {1, 2, 3, 5, 9}

RHS = A' ∪ B'

Now, A' = {2,5,9} [form (i)]

and B' = {3} [form (i)]

∴ A' ∪ B'= {2, 3, 5, 9}

Hence, LHS = RHS Proved

Let x ϵ B. Then

⟹ x ϵ B ∪ A

⟹ x ϵ A ∪ B

∴ B ⊂ (A ∩ B)

## Sets– Exercise 1.6 – Q.4(ii)

Let x ϵ A ∩ B. Then

⟹ x ϵ A and x ϵ B

⟹ x ϵ B

∴ (A ∩ B) ⊂ B.

## Sets– Exercise 1.6 – Q.4(iii)

Let x ϵ A ⊂ B. Then

⟹ x ϵ B

Let and x ϵ A ∩ B

⟺ x ϵ A and x ϵ B

⟺ x ϵ A and x ϵ A (∵A ⊂ B)

∴ (A ∩ B) = A

## Sets– Exercise 1.6 – Q.5

(i) In order to show that the following four statements are equivalent, we need to show that (1) ⇒ (2), (2) ⇒ (3), (3) ⇒ (4) and (4) ⇒ (1)

We first show that (1) ⇒ (2)

We assume that A ⊂ B, and use this to show that A -B = ∮

Now A - B = {x ϵ A : x ∉ B}, As A ⊂ B,

∴ Each element of A is an element of B,

∴ A - B = ∮

Hence, we have proved that (1) ⇒ (2).

(ii) We new show that (2) ⇒ (3)

So assume that A - B = ∮

To show: A ∪ B = B

∴  A - B = ∮

∴ Every element of A is an element of B

[∵ A - B = ∮ only when ther is some element in A which is not in B]

So A ⊂ B and therefore A ∪ B = B

So (2) ⇒ (3) is true.

(iii) We new show that (3) ⇒ (4)

Assume that A ∪ B = B

To show: A ∩ B = A

∵   A ∪ B = B

∴   A ⊂ B and so A ∩ B = A

So (3) ⇒ (4) is true.

(iv) Finally we show that (4) ⇒ (1), which will prove the equivalence of the four statements.

SO, assume that A ∩ B = A

To show: A ⊂ B

∵ A ∩ B = A, therefore A ⊂ B, and so (4) ⇒ (1) is true.

Hence, (1) ⟺ (2) ⟺ (3) ⟺ (4).

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