Chapter 1: Sets

Exercise 1.1 – Solutions

Question 1

List the elements of the following sets:

(a) {x : x is an integer and x² = 9}

We need the integer values of x whose square is 9.

x² = 9
x = ±3

Therefore, the required set is:

{−3, 3}

(b) {x : x is a positive integer less than 5}

The positive integers less than 5 are 1, 2, 3 and 4.

Therefore,

{1, 2, 3, 4}

(c) {x : x is an even natural number divisible by 5}

The natural numbers divisible by 5 are 5, 10, 15, 20, ... Among these, the even numbers are 10, 20, 30, ...

Therefore,

{10, 20, 30, 40, ...}

(d) {x : x ∈ N and x < −1}

Natural numbers are non-negative/positive counting numbers and none of them is less than −1.

Therefore, the set is:

∅

Question 2

Given:

A = {−5, −√3, −1/2, 0, 2/5, π, 13.4, 1/3, 19/2}

(a) Natural numbers

There is no natural number in the given set.

Answer: ∅

(b) Whole numbers

The only whole number in the given set is 0.

Answer: {0}

(c) Integers

The integers in the given set are −5 and 0.

Answer: {−5, 0}

(d) Rational numbers

A rational number can be expressed in the form p/q, where p and q are integers and q ≠ 0.

−5, −1/2, 0, 2/5, 13.4, 1/3 and 19/2 are rational numbers. Note that 13.4 = 67/5.

Answer:

{−5, −1/2, 0, 2/5, 13.4, 1/3, 19/2}

(e) Real numbers

Every number in the given set is a real number, including −√3 and π, which are irrational real numbers.

Answer:

{−5, −√3, −1/2, 0, 2/5, π, 13.4, 1/3, 19/2}

Question 3

Write the following sets in roster form:

(a) {x : x is a two digit number and the sum of the digits is 5}

The two-digit numbers whose digits add up to 5 are:

14, 23, 32, 41 and 50.

Answer:

{14, 23, 32, 41, 50}

(b) {x : x is an integer and |x| ≤ 9}

The integers whose absolute value is less than or equal to 9 are all the integers from −9 to 9.

Answer:

{−9, −8, −7, −6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(c) {x : x is a letter of the word "SWEET"}

Repeated elements are written only once in a set.

Answer:

{S, W, E, T}

(d) {x : x = (n + 1)/n, where n is a natural number and n < 6}

Since n is a natural number and n < 6:

n = 1, 2, 3, 4, 5

For n = 1: x = (1 + 1)/1 = 2

For n = 2: x = (2 + 1)/2 = 3/2

For n = 3: x = (3 + 1)/3 = 4/3

For n = 4: x = (4 + 1)/4 = 5/4

For n = 5: x = (5 + 1)/5 = 6/5

Answer:

{2, 3/2, 4/3, 5/4, 6/5}

(e) {x : x is a composite number}

The composite numbers are positive integers greater than 1 that have factors other than 1 and themselves.

Answer:

{4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, ...}

Question 4

Write the following sets in set-builder form:

(i) {2, 4, 6, 8, ...}

These are the positive even natural numbers.

Answer:

{x : x = 2n, where n ∈ N}

(ii) {3, 6, 9, 12, 15}

These numbers are the first five positive multiples of 3.

Answer:

{x : x = 3n, where n ∈ N and n ≤ 5}

(iii) {1, 4, 9, 16, ...}

These are the squares of natural numbers.

Answer:

{x : x = n², where n ∈ N}

(iv) {8, 9, 10, 11, ...}

These are all natural numbers greater than or equal to 8.

Answer:

{x : x ∈ N and x ≥ 8}

(v) {1, 2, 3, 6}

These are the positive divisors of 6.

Answer:

{x : x ∈ N and x divides 6}

Can two different sets have the same roster form?

No. The roster form lists the elements of a set. If two sets have exactly the same elements, they represent the same set. Therefore, two different sets cannot have exactly the same roster form.

Question 5

Which of the following pairs of sets are equal?

(i) {D, E, C, E, N, T} and {C, E, N, T, D}

Repetition does not matter in a set, and the order of elements does not matter.

Both sets contain exactly the same elements: D, E, C, N and T.

Answer: The sets are equal.

(ii) {a, b, π, √2} and {a, π, √2, b}

Both sets contain exactly the same four elements.

Answer: The sets are equal.

(iii) {x : x is zero of the polynomial x²} and {x : x is the root of the equation x² = 0}

For both cases:

x² = 0
x = 0

Hence, both sets are {0}.

Answer: The sets are equal.

(iv) {x : x has numerical value less than or equal to 1} and {x : x is the root of the equation x² − 1 = 0}

The roots of x² − 1 = 0 are:

x² = 1
x = ±1

Therefore, the second set is {−1, 1}.

The first set contains numbers satisfying the stated numerical-value condition and is not the same set as {−1, 1}.

Answer: The sets are not equal.

(v) {5, 10, 15, 20} and {5, 10, 15, 20, ...}

The first set contains only four elements, whereas the second set continues with further multiples of 5.

Answer: The sets are not equal.

(vi) ∅ and {∅}

The symbol ∅ represents the empty set and has no elements. The set {∅} contains one element, namely the empty set itself.

Therefore, ∅ has 0 elements, whereas {∅} has 1 element.

Answer: The sets are not equal.

Question 6

State which of the following sets are finite or infinite.

(i) {x : x ∈ Z and (x − 1)(x + 2)(x − 3) = 0}

For the product to be zero:

x − 1 = 0  ⇒  x = 1

x + 2 = 0  ⇒  x = −2

x − 3 = 0  ⇒  x = 3

Therefore, the set is:

{−2, 1, 3}

Answer: Finite set.

(ii) {x : x and 2 are coprime}

A number is coprime to 2 when its greatest common divisor with 2 is 1. There are infinitely many such numbers, for example:

1, 3, 5, 7, 9, 11, ...

Answer: Infinite set.

(iii) {x : x is a rational number between 3 and 4}

There are infinitely many rational numbers between any two distinct numbers. For example:

3.1, 3.01, 3.001, 7/2, 15/4, ...

Answer: Infinite set.

(iv) {x : x is an integer and |x| ≥ 5}

The integers satisfying |x| ≥ 5 include:

..., −8, −7, −6, −5, 5, 6, 7, 8, ...

Since the numbers continue indefinitely in both directions, the set has infinitely many elements.

Answer: Infinite set.

Exercise 1.2 – Solutions

Question 1

Fill in the blanks with the symbol ⊂ or ⊄.

(i) {2, 3, 4} ............ {1, 2, 3, 4, 5}

Every element of {2, 3, 4} is also an element of {1, 2, 3, 4, 5}. Therefore, {2, 3, 4} is a subset of {1, 2, 3, 4, 5}.

Answer: ⊂

(ii) {x | x are triangles in a plane} ............ {x | x are polygons in a plane}

Every triangle is a polygon. Therefore, the set of triangles is a subset of the set of polygons.

Answer: ⊂

(iii) {x | x is an integer} ............ {x : x is a multiple of 4}

Not every integer is a multiple of 4. For example, 1 is an integer but 1 is not a multiple of 4.

Answer: ⊄

(iv) ∅ ............ {∅}

The empty set is a subset of every set. Also, ∅ and {∅} are different sets.

Answer: ⊂

(v) {x | x = (m − 1)/m, where m is a non-zero integer} ............ {x | x is a rational number}

We have

x = (m − 1)/m

Here, m and m − 1 are integers and m ≠ 0. Therefore, x is a rational number.

Hence, every element of the first set belongs to the set of rational numbers.

Answer: ⊂

(vi) {x | x = n²} ............ {x | x = n³}, where n is a natural number

The set of squares of natural numbers is not equal to the set of cubes of natural numbers, but the question asks whether the first set is a subset of the second set.

For example, 4 = 2², but 4 is not a cube of a natural number. Therefore, 4 belongs to the first set but not to the second set.

Answer: ⊄

(vii) {x | x ∈ R} ............ {x | x = 2ⁿ}, where R is the set of real numbers

The first set contains all real numbers, whereas the second set contains only numbers of the form 2ⁿ. For example, 3 is a real number but it cannot be written as 2ⁿ for an integer natural number n.

Answer: ⊄

Question 2

Determine whether the following statements are true or false.

(i) 1 ∈ {1}

The set {1} contains 1 as its element.

Answer: True

(ii) {2} ∈ {2}

The set {2} contains the number 2 as its element, not the set {2}.

Answer: False

(iii) {2} ∈ {{2}}

The set {{2}} has {2} as its only element.

Answer: True

(iv) ∅ ∈ {1, 2, 3}

The elements of {1, 2, 3} are 1, 2 and 3. The empty set is not an element of this set.

Answer: False

(v) ∅ ⊂ {1, 2, 3}

The empty set is a subset of every set.

Answer: True

Question 3

Write the power set of the following sets.

The power set of a set A, denoted by P(A), is the set of all subsets of A.

(i) {1}

The subsets are ∅ and {1}.

P({1}) = {∅, {1}}

(ii) {p, q}

The subsets are:

∅, {p}, {q}, {p, q}

P({p, q}) = {∅, {p}, {q}, {p, q}}

(iii) {1, 2, 5}

The subsets are:

∅, {1}, {2}, {5}, {1, 2}, {1, 5}, {2, 5}, {1, 2, 5}

P({1, 2, 5}) = {∅, {1}, {2}, {5}, {1, 2}, {1, 5}, {2, 5}, {1, 2, 5}}

(iv) {∅, {∅}}

The set has two elements:

∅ and {∅}

Therefore, its four subsets are:

∅, {∅}, {{∅}}, {∅, {∅}}

P({∅, {∅}}) = {∅, {∅}, {{∅}}, {∅, {∅}}

Question 4

Find the cardinality of the following sets.

(i) {a}

The set contains only one element, a.

Answer: n({a}) = 1

(ii) {a, {a}}

The elements are a and {a}. These are two different elements because a is an object, whereas {a} is a set containing a.

Answer: n({a, {a}}) = 2

(iii) {∅, 1, 2, {1, 2}}

The four elements are ∅, 1, 2 and {1, 2}. All four are distinct.

Answer: n({∅, 1, 2, {1, 2}}) = 4

(iv) {1, {1}, {1, {1}}}

The set contains the following three distinct elements:

• 1
• {1}
• {1, {1}}

Answer: n({1, {1}, {1, {1}}}) = 3

(v) {∅, {∅}, {∅, {∅}}}

The set contains three distinct elements:

• ∅
• {∅}
• {∅, {∅}}

Answer: n({∅, {∅}, {∅, {∅}}}) = 3

Question 5

Let A be a set and n(A) = 10. Find n(P(A)). What if A has 100 elements?

If a set A has n elements, then its power set contains 2ⁿ subsets.

Therefore,

n(P(A)) = 2^n(A)

When n(A) = 10:

n(P(A)) = 2¹⁰

n(P(A)) = 1024

Answer: 1024

When A has 100 elements:

n(P(A)) = 2¹⁰⁰

n(P(A)) = 1,267,650,600,228,229,401,496,703,205,376

Answer: 2¹⁰⁰

Exercise 1.3 – Solutions

Question 1

Find the union of sets A and B, i.e. A ∪ B, in each of the following pairs.

(i) A = {1, 2, 3, 7}, B = {2, 7, 9}

The union contains all the elements that belong to A or B, without repeating any element.

A ∪ B = {1, 2, 3, 7, 9}

Answer: {1, 2, 3, 7, 9}

(ii) A = {a, b, d, e}, B = {a, e, i, o, u}

Combining all distinct elements of A and B:

A ∪ B = {a, b, d, e, i, o, u}

Answer: {a, b, d, e, i, o, u}

(iii) A = {x : x is a natural number > 5}, B = {x : x is a natural number < 5}

A contains natural numbers greater than 5, while B contains natural numbers less than 5. Therefore, their union contains all natural numbers except 5.

A ∪ B = {x : x is a natural number and x ≠ 5}

Answer: {x : x is a natural number and x ≠ 5}

(iv) A = ∅, B = {2, √2, −1, 0}

The union of the empty set with any set is the set itself.

A ∪ B = B

Answer: {2, √2, −1, 0}

Question 2

Evaluate each of the following.

(i) {1, 2} ∩ {1, 2, 5}

The common elements of the two sets are 1 and 2.

Answer: {1, 2}

(ii) {1, 3, 5, 7, 9} ∩ {2, 4, 6, 8}

The first set contains odd numbers and the second set contains even numbers. There is no common element.

Answer: ∅

(iii) {g, o, a, t} ∩ {c, a, t}

The elements common to both sets are a and t.

Answer: {a, t}

(iv) {x : x is an integer} ∩ {x : x is a negative integer}

Every negative integer is an integer. Therefore, the intersection is the set of all negative integers.

Answer: {x : x is a negative integer}

Question 3

Which of the following pairs of sets are disjoint?

Two sets are called disjoint sets if they have no common element, that is, their intersection is the empty set.

(i) {x : x is a multiple of 2} and {x : x is a multiple of 3}

There are numbers that are multiples of both 2 and 3. For example, 6 is a multiple of both 2 and 3.

Therefore, the two sets have common elements.

Answer: Not disjoint.

(ii) {e, π, √2, 0} and {e², π/2, √3, 1}

The two sets have no common element.

Answer: Disjoint.

(iii) {x : x is a real number} and {x : x is an irrational number}

Every irrational number is a real number. Hence, the two sets have all irrational numbers in common.

Answer: Not disjoint.

Question 4

Find A − B in each of the following.

(i) A = {1, 3, 5, 8}, B = {3, 7, 8, 9}

A − B contains the elements of A which are not present in B.

The elements 3 and 8 are common to both A and B, so they are removed.

Answer: A − B = {1, 5}

(ii) A = {3, 0, 8}, B = {1, 3, 0, 8, 9}

Every element of A is also present in B.

Therefore, there is no element left in A after removing the elements belonging to B.

Answer: A − B = ∅

(iii) A = {2, 6}, B = {1, 3, 5, 9}

Neither 2 nor 6 is present in B.

Answer: A − B = {2, 6}

Question 5

From the given Venn diagram, the universal set and the three sets are:

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}

A = {1, 4, 5, 8, 11}

B = {2, 3, 4, 5}

C = {3, 5, 6, 8, 9}

(i) A′

A′ contains all elements of U that are not in A.

A′ = {2, 3, 6, 7, 9, 10, 12, 13}

Answer: {2, 3, 6, 7, 9, 10, 12, 13}

(ii) B′

B′ contains all elements of U that are not in B.

B′ = {1, 6, 7, 8, 9, 10, 11, 12, 13}

Answer: {1, 6, 7, 8, 9, 10, 11, 12, 13}

(iii) (A ∩ B)′

From the Venn diagram:

A ∩ B = {4, 5}

Therefore,

(A ∩ B)′ = U − {4, 5}

Answer: {1, 2, 3, 6, 7, 8, 9, 10, 11, 12, 13}

(iv) A′ ∪ B′

Using the results above:

A′ ∪ B′ = {2, 3, 6, 7, 9, 10, 12, 13} ∪ {1, 6, 7, 8, 9, 10, 11, 12, 13}

A′ ∪ B′ = {1, 2, 3, 6, 7, 8, 9, 10, 11, 12, 13}

Answer: {1, 2, 3, 6, 7, 8, 9, 10, 11, 12, 13}

(v) A ∩ B ∩ C

The element common to all three sets A, B and C is 5.

Answer: {5}

(vi) A ∩ (B ∪ C)

First,

B ∪ C = {2, 3, 4, 5, 6, 8, 9}

Now find the elements common to A and B ∪ C:

A ∩ (B ∪ C) = {4, 5, 8}

Answer: {4, 5, 8}

Question 6

Verify A − B = A ∩ B′ using the given Venn diagram.

From the Venn diagram:

A = {1, 5, 6, 8, 13}

B = {2, 3, 5, 7, 13}

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 13}

Step 1: Find A − B

A − B contains the elements of A that are not in B.

A − B = {1, 6, 8}

Step 2: Find B′

B′ = U − B

B′ = {1, 4, 6, 8, 9}

Step 3: Find A ∩ B′

A ∩ B′ = {1, 5, 6, 8, 13} ∩ {1, 4, 6, 8, 9}

A ∩ B′ = {1, 6, 8}

Therefore,

A − B = A ∩ B′ = {1, 6, 8}

Hence, the required result is verified.

Question 7

Let M be the set of students who opted for the Mathematics Mock Test and S be the set of students who opted for the Science Mock Test.

Given:

• n(M) = 85% of students
• n(S) = 75% of students

(a) What is the minimum possible percentage of students who opted for both tests?

For two sets:

n(M ∪ S) = n(M) + n(S) − n(M ∩ S)

The maximum possible value of n(M ∪ S) is 100%.

Therefore,

100 = 85 + 75 − n(M ∩ S)

n(M ∩ S) = 160 − 100 = 60

Answer: 60%

(b) If 10% opted for neither, how does the minimum percentage for both change?

If 10% opted for neither test, then:

n(M ∪ S) = 100% − 10% = 90%

Therefore,

90 = 85 + 75 − n(M ∩ S)

n(M ∩ S) = 160 − 90 = 70

Answer: The minimum percentage increases to 70%.

Question 8

Let E be the set of people who speak English and H be the set of people who speak Hindi.

Given:

• n(S) = 50
• n(E) = 35
• n(H) = 25

Since every person speaks at least one of the two languages:

n(E ∪ H) = 50

Using the formula:

n(E ∪ H) = n(E) + n(H) − n(E ∩ H)

50 = 35 + 25 − n(E ∩ H)

n(E ∩ H) = 10

Therefore, the number of people who speak only English is:

k = 35 − 10 = 25

Answer: k = 25

Question 9

Let the three activities be Origami, Instrumental Music and Fine Arts.

Given:

• Total students who received at least one certificate = 56
• Origami = 17
• Instrumental music = 28
• Fine arts = 25
• All three activities = 4

Let x be the number of students who received certificates for exactly two activities.

The total of the numbers in the three activity sets is:

17 + 28 + 25 = 70

A student receiving certificates for exactly two activities is counted twice in this total, while a student receiving certificates for all three activities is counted three times.

Therefore,

70 = 56 + x + 2(4)

70 = 56 + x + 8

x = 6

Answer: 6 students

Question 10

Let E, G and S represent the sets of students who speak English, German and Spanish respectively.

Given:

• n(E) = 60
• n(G) = 50
• n(S) = 35
• n(E ∩ G) = 40
• n(G ∩ S) = 30
• n(E ∩ S) = 25
• n(E ∩ G ∩ S) = 25
• Total students = 100

First find the numbers in exactly two languages.

English and German only:

40 − 25 = 15

German and Spanish only:

30 − 25 = 5

English and Spanish only:

25 − 25 = 0

Now find the numbers who speak only one language.

English only:

60 − 15 − 0 − 25 = 20

German only:

50 − 15 − 5 − 25 = 5

Spanish only:

35 − 0 − 5 − 25 = 5

(a) How many students could speak at least two languages?

Students speaking exactly two languages:

15 + 5 + 0 = 20

Students speaking all three languages = 25.

Therefore,

At least two languages = 20 + 25 = 45

Answer: 45 students

(b) How many students could speak at most one language?

At most one language means either exactly one language or no language.

Students speaking exactly one language:

20 + 5 + 5 = 30

From part (a), 45 students speak at least two languages. Therefore, the remaining students speak at most one language:

100 − 45 = 55

Answer: 55 students

(c) How many students could not speak any of the three languages?

First, find the number of students who speak at least one language:

n(E ∪ G ∪ S) = n(E) + n(G) + n(S) − n(E ∩ G) − n(G ∩ S) − n(E ∩ S) + n(E ∩ G ∩ S)

= 60 + 50 + 35 − 40 − 30 − 25 + 25

= 75

Therefore, the number of students who could not speak any of the three languages is:

100 − 75 = 25

Answer: 25 students