Miscellaneous Problems based on the Algebra of Sets

Introduction

Algebra of sets include:

Problems

Q1. In each of the following, determine whether the statement is true or false. Justify with an example?

(i) If x ∈ A and A ∈ B, then x ∈ B

(ii) If A ⊂ B and B ∈ C, then A ∈ C

(iii) If A ⊂ B and x ∉ B, then x ∉ A

(iv) If A ⊄ B and B ⊄ C, then A ⊄ C

(v) If x ∈ A and A ⊄ B, then x ∈ B

(vi)If A ⊂ B and B ⊂ C, then A ⊂ C

Q2. Show that for any sets A and B,

A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)

Q3. According to a survey, 21 respondents preferred product A, 26 preferred product B, and 29 preferred product C. If 14 individuals liked A and B, 12 people liked C and A, 14 people liked B and C, and 8 people enjoyed all three goods, now count how many people liked product C?

Q4. Let, N = {1, 2, 3, …, 100}, solve the following?

(i) Write the subset A of N, when elements are odd numbers.

(ii) Write the subset B of N, when elements are represented as x + 2, where x ∈ N.

Q5. If A and B are subsets of the universal set U, then prove

(i) B ⊂ A ∪ B

(ii) A ∩ B ⊂ A

(iii) A ⊂ B ⇒ A ∩ B = A

Solutions

Ans 1:

(i) False

Let,

A = {1, 2} and B = {1, {1, 2}, {3}}

Now, we have,

2 ∈ {1, 2} and {1, 2} ∈ {1, {1, 2}, {3}}

Hence, we get, A ∈ B. We also know, {2} ∉ {1, {1, 2}, {3}}

(ii) False

Let,A {2}, B = {0, 2}, C = {1, {0, 2}, 3}

From the question,

A ⊂ B

Hence, B ∈ C, But, we know, A ∉ C

(iii)True

Let, A ⊂ B. Also, x ∉ B

Let,x ∈ A,

We have, x ∈ B and,we have, x ∉ B

∴ x ∉ A

(iv) False

A ⊄ B . Also, B ⊄ C

Let, A = {1, 2}, B = {0, 6, 8}, C = {0, 1, 2, 6, 9}

∴ A ⊂ C

(v) False

Let, x ∈ A. Also, A ⊄ B

Now,

A = {3, 5, 7} and B = {3, 4, 6}

We know that, A ⊄ B

∴ 5 ∉ B

(vi)  True

Now, A ⊂ B and B ⊂ C

Let, x ∈ A.Then, we have, x ∈ B. And, x ∈ C

Therefore, A ⊂ C

Ans 2:

To Prove,

A = (A ∩ B) ∪ (A – B)

Proof:

 Let x ∈ A

Case I,

X ∈ (A ∩ B)

⇒ X ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)

Case II,

X ∉A ∩ B

⇒ X ∉ B or X ∉ A

⇒ X ∉ B (X ∉ A)

⇒ X ∉ A – B ⊂ (A ∪ B) ∪ (A – B)

∴A ⊂ (A ∩ B) ∪ (A – B) (i)

It can be concluded that, A ∩ B ⊂ A and (A – B) ⊂ A

Thus, (A ∩ B) ∪ (A – B) ⊂ A (ii)

Equating (i) and (ii),

A = (A ∩ B) ∪ (A – B)

We also have to show,

A ∪ (B – A) ⊂ A ∪ B

Let us assume,

X ∈ A ∪ (B – A)

X ∈ A or X ∈ (B – A)

⇒ X ∈ A or (X ∈ B and X ∉A)

⇒ (X ∈ A or X ∈ B) and (X ∈ A and X ∉A)

⇒ X ∈ (B ∪A)

∴ A ∪ (B – A) ⊂ (A ∪ B) (iii)

According to the question,

To prove:

(A ∪ B) ⊂ A ∪ (B – A)

Let y ∈ A∪B

Y ∈ A or y ∈ B

(y ∈ A or y ∈ B) and (X ∈ A and X ∉A)

⇒ y ∈ A or (y ∈ B and y ∉A)

⇒ y ∈ A ∪ (B – A)

Thus, A ∪ B ⊂ A ∪ (B – A) (iv)

∴ From equations (iii) and (iv), we get:

A ∪ (B – A) = A ∪ B

Ans 3:

Let set A, set B, and set C = people who like product A, product B, and product C respectively.

Now,

People who like product A, n (A) = 21

People who like product B, n (B) = 26

People who like product C, n (C) = 29

People who like both products A and B, n (A ∩ B) = 14

People who like both products A and C, n(C ∩ A) = 12

People who like both products C and B, n (B ∩ C) = 14

People who like all three products, n (A ∩ B ∩ C) = 8

From the Venn diagram,

People who only like product C = {29 – (4 + 8 + 6)}

= {29 – 18}

= 11 students

Ans 4:

(i) A = {x | x ∈ N and x is odd}

= {1, 3, 5, 7, …, 99}

(ii) B = {y | y = x + 2, x ∈ N}

So, for 1 ∈ N, y = 1 + 2 = 3

            2 ∈ N, y = 2 + 2 = 4,……..

Therefore, B = {3, 4, 5, 6, … , 100}

Ans 5:

(i) A ∪ B = {x | x ∈ A or x ∈ B}

Thus x ∈ B ⇒ x ∈ A ∪ B

Hence, B ⊂ A ∪ B

(ii) A ∩ B = {x | x ∈ A and x ∈ B}

Thus x ∈ A ∩ B ⇒ x ∈ A

Hence A ∩ B ⊂ A

(iii) Let, x ∈ A ∩ B ⇒ x ∈ A

Thus A ∩ B ⊂ A

Also, since A ⊂ B, x ∈ A ⇒ x ∈ B ⇒ x ∈ A ∩ B

so that A ⊂ A ∩ B

Hence, A = A ∩ B.