CBSE Class 11 Applied Mathematics Sets Notes Set 01

Sets and Their Representations

Set

A given collection of objects is said to be well defined if we can definitely say whether a given particular object belongs to the collection or not.
Set: A well-defined collection of objects is called a set. The objects in a set are its members or elements or points.
Sets are usually denoted by capital letters A, B, C, X, Y, Z etc. The elements of a set are represented by small letters a, b, c, x, y, z etc.
If a is an element of a set A, we write a ∈ A, which means that a belongs to A or that a is an element of A. If a does not belong to A, we write a ∉ A.

Illustrations:
(i) The collection of all vowels in the English alphabet contains five elements, namely a, e, i, o, u. So this collection is well defined and therefore it is a set.
(ii) The collection of all odd natural numbers less than 10 contains the numbers 1, 3, 5, 7, 9. So this collection is well defined and therefore it is a set.
(iii) The collection of all rivers of India is clearly well defined and therefore it is a set. Clearly, the river Ganga belongs to this set while the river Nile does not belong to it.

Representation of a Set

There are two methods of representing a set:
(i) Roster form or Tabular form: Under this method we list all the members of the set within braces { } and separate them by commas.
Note: In roster form, the order in which the elements are listed is immaterial.

Example 1. Write each of the following sets in the roster form:
(i) A = Set of all factors of 24.
(ii) B = Set of all integers between \( -\frac{3}{2} \) and \( \frac{11}{2} \).
(iii) C = Set of all letters in the word 'TRIGONOMETRY'.
(iv) D = Set of all months having 30 days.
Answer:
(i) All factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. ∴ A = {1, 2, 3, 4, 6, 8, 12, 24}.
(ii) All integers between \( -\frac{3}{2} \) and \( \frac{11}{2} \) are −1, 0, 1, 2, 3, 4, 5. ∴ B = {−1, 0, 1, 2, 3, 4, 5}.
(iii) It may be noted here that the repeated letters are taken only once each. ∴ C = {T, R, I, G, O, N, M, E, Y}.
(iv) We know that the months having 30 days are April, June, September and November. ∴ D = {April, June, September, November}.

Note: We denote the set of all natural numbers, all integers, all rational numbers and all real numbers by N, Z, Q and R respectively.

(ii) Set-builder form: Under this method of describing a set, we list the property or properties satisfied by all the elements of the set.
We write {x : x has property P} and read it as 'the set of all those x such that each x satisfies property P'.

Example 2. Write the set A = {1, 2, 3, 4, 5, 6, 7} in the set-builder form.
Answer:
Clearly, A = set of all natural numbers less than 8.
Thus, in the set-builder form we write it as A = {x : x ∈ N and x < 8}.

Example 3. Write the set A = {2, 4, 8, 16, 32} in the set-builder form.
Answer:
Clearly, A = {2¹, 2², 2³, 2⁴, 2⁵}.
Thus, in the set-builder form we write it as A = {x : x = 2ⁿ, where n ∈ N and 1 ≤ n ≤ 5}.

Example 4. Write the set A = {1, 4, 9, 16, 25, …} in set-builder form.
Answer:
We may write the set A as A = {x : x is the square of a natural number}.
Alternatively, we can write A = {x : x = n², where n ∈ N}.

Types of Sets

Empty Set

A set containing no element at all is called the empty set or the null set or the void set. It is denoted by ∅ or { }.
Example 1: {x : x ∈ N and 2 < x < 3} = ∅
Example 2: {x : x ∈ R and x² = −1} = ∅
Example 3: {x : x² = 4, x is odd} = ∅

Singleton Set

A set containing exactly one element is called a singleton set.
Example 1: {0} is a singleton set whose only element is 0.
Example 2: {x : x ∈ N and x² = 4} = {2}, since 2 is the only natural number whose square is 4.

Finite and Infinite Sets

A set which is empty or consists of a definite number of elements is called finite; otherwise, the set is called infinite.
The number of distinct elements contained in a finite set A is denoted by n(A).
Example: Let W be the set of the days of the week. Then W is finite.
Example: Let S be the set of solutions of the equation x² − 16 = 0. Then S is finite.
Note: All infinite sets cannot be described in roster form, because the number of elements of such a set is not finite. So we represent some infinite sets in the roster form by writing a few elements which clearly indicate the structure of the set, followed by three dots.
Example 1: {1, 2, 3, …} is the set of natural numbers.
Example 2: {…, −3, −2, −1, 0, 1, 2, 3, …} is the set of integers.

Equal Sets

Two sets A and B are said to be equal if they have exactly the same elements, and we write A = B. Otherwise, the sets are said to be unequal and we write A ≠ B. In other words, A = B ⇔ A ⊆ B and B ⊆ A.
Example 1: Let A = {1, 2, 3, 4} and B = {3, 1, 4, 2}. Then A = B.
Example 2: A set does not change if one or more elements of the set are repeated. The sets A = {1, 2, 3} and B = {2, 2, 1, 3, 3} are equal, since each element of A is in B and vice-versa.

Equivalent Sets

Two finite sets A and B are said to be equivalent if n(A) = n(B).
Equal sets are always equivalent, but equivalent sets need not be equal.
Example: Let A = {1, 3, 5} and B = {2, 4, 6}. Then n(A) = n(B) = 3. So A and B are equivalent, but clearly A ≠ B. Hence A and B are equivalent sets but not equal sets.

Subsets

Subset and Superset

Subset: A set A is said to be a subset of a set B if every element of A is also an element of B, and we write A ⊆ B.
Superset: If A ⊆ B, then B is called a superset of A and we write B ⊇ A.
Proper subset: If A ⊆ B and A ≠ B, then A is called a proper subset of B and we write A ⊂ B.
In other words, A ⊆ B if whenever a ∈ A, then a ∈ B. It is often convenient to use the symbol '⇒', which means implies. Using this symbol, we can write the definition of subset as follows: A ⊆ B if a ∈ A ⇒ a ∈ B.
If A is not a subset of B, we write A ⊄ B. We may note that for A to be a subset of B, all that is needed is that every element of A is in B. It is possible that every element of B may or may not be in A. If it so happens that every element of B is also in A, then we shall also have B ⊆ A. In this case A and B are the same sets, so that we have A ⊆ B and B ⊆ A ⇔ A = B, where '⇔' is a symbol for two-way implication and is usually read as 'if and only if' (briefly written as 'iff').
It follows from the above definition that every set A is a subset of itself, i.e. A ⊆ A. Since the empty set ∅ has no elements, we agree to say that ∅ is a subset of every set.
Note: The total number of subsets of a set containing n elements is 2ⁿ.

Subsets of the Set of Real Numbers

The set of natural numbers N = {1, 2, 3, 4, 5, …}
The set of integers Z = {…, −3, −2, −1, 0, 1, 2, 3, …}
The set of rational numbers \( Q = \left\{x : x = \frac{p}{q},\ p, q \in Z \text{ and } q \neq 0\right\} \)
The set of irrational numbers T = {x : x ∈ R and x ∉ Q}
Some of the obvious relations among these subsets are: N ⊂ Z ⊂ Q, Q ⊂ R, T ⊂ R, N ⊄ T.

Intervals as Subsets of R

Let a, b ∈ R and a < b. Then the set of real numbers {y : a < y < b} is called an open interval and is denoted by (a, b). All the points between a and b belong to the open interval (a, b), but a, b themselves do not belong to this interval.
The interval which contains the end points also is called a closed interval and is denoted by [a, b]. Thus [a, b] = {x : a ≤ x ≤ b}.
We can also have intervals closed at one end and open at the other, i.e.
[a, b) or [a, b[ = {x : a ≤ x < b} is a right half-open interval;
(a, b] or ]a, b] = {x : a < x ≤ b} is a left half-open interval.
The set [0, ∞) defines the set of non-negative real numbers, while the set (−∞, 0) defines the set of negative real numbers. The set (−∞, ∞) describes the set of real numbers in relation to a line extending from −∞ to ∞.
On the real number line, the various types of intervals described above as subsets of R are shown below:

Hence, we note that an interval contains infinitely many points.
Length of an interval: The length of each of the intervals (a, b), [a, b], [a, b) and (a, b] is (b − a).

Power Set

The collection of all subsets of a set A is called the power set of A. It is denoted by P(A). In P(A), every element is a set.
In general, if A is a set with n(A) = m, then n[P(A)] = 2^m.
Example: If A = {1, 2}, then P(A) = {∅, {1}, {2}, {1, 2}}. Also note that n[P(A)] = 2² = 4.

Universal Set

Usually, in a particular context, we have to deal with the elements and subsets of a basic set which is relevant to that particular context. This basic set is called the universal set.
The universal set is usually denoted by U and all its subsets by the letters A, B, C etc. For example, for the set of all integers, the universal set can be the set of rational numbers or, for that matter, the set R of real numbers.
Example: When we discuss sets of lines, triangles or circles in two-dimensional geometry, the plane in which these lines, triangles or circles lie is the universal set.

Example 5. Write down the power set of ∅.
Answer:
∅ has only one subset, namely ∅.
∴ P(∅) = {∅}

Example 6. Let A = {1, {2}, {3, 4}, 5}. Which of the following are incorrect statements? Rectify each:
(i) 2 ∈ A
(ii) {2} ⊂ A
(iii) {1, 2} ⊂ A
(iv) {3, 4} ⊂ A
(v) {1, 5} ⊂ A
(vi) {∅} ⊂ A
(vii) 1 ⊂ A
(viii) {1, 2, 3, 4} ⊂ A
Answer:
Clearly, A contains four elements, namely 1, {2}, {3, 4} and 5.
(i) 2 ∈ A is incorrect. The correct statement would be {2} ∈ A.
(ii) {2} ⊂ A is incorrect. The correct statement is {{2}} ⊂ A.
(iii) Clearly, 2 ∉ A and therefore {1, 2} ⊂ A is incorrect. The correct statement would be {1, {2}} ⊂ A.
(iv) Clearly, {3, 4} is an element of A. So {3, 4} ⊂ A is incorrect, and {{3, 4}} ⊂ A is correct.
(v) Since 1 and 5 are both elements of A, {1, 5} ⊂ A is correct.
(vi) Since ∅ ∉ A, {∅} ⊂ A is incorrect, while ∅ ⊂ A is correct.
(vii) Since 1 ∈ A, 1 ⊂ A is incorrect; the correct statement is {1} ⊂ A.
(viii) Since 2 ∉ A and 3 ∉ A, {1, 2, 3, 4} ⊂ A is incorrect. The correct statement would be {1, {2}, {3, 4}} ⊂ A.

Operations on Sets

Union of Sets

The union of two sets A and B, denoted by A ∪ B, is the set of all those elements which are either in A or in B or in both A and B.
Thus A ∪ B = {x : x ∈ A or x ∈ B}, ∴ x ∈ A ∪ B ⇔ x ∈ A or x ∈ B.

Example 7. Let A = {x : x is a positive integer} and B = {x : x is a negative integer}. Find A ∪ B.
Answer:
Clearly, A ∪ B = {x : x is an integer and x ≠ 0}.

Example 8. If A = {x : x = 2n + 1, n ∈ Z} and B = {x : x = 2n, n ∈ Z}, then find A ∪ B.
Answer:
We have A ∪ B = {x : x is an odd integer} ∪ {x : x is an even integer}
= {x : x is an integer} = Z

Intersection of Sets

The intersection of two sets A and B, denoted by A ∩ B, is the set of all elements which are common to both A and B.
Thus A ∩ B = {x : x ∈ A and x ∈ B}
∴ x ∈ A ∩ B ⇔ x ∈ A and x ∈ B; x ∉ A ∩ B ⇒ x ∉ A or x ∉ B.

Example 9. If A = {x : x ∈ N, x is a factor of 12} and B = {x : x ∈ N, x is a factor of 18}, find A ∩ B.
Answer:
We have A = {x : x ∈ N, x is a factor of 12} = {1, 2, 3, 4, 6, 12}
B = {x : x ∈ N, x is a factor of 18} = {1, 2, 3, 6, 9, 18}
∴ A ∩ B = {1, 2, 3, 4, 6, 12} ∩ {1, 2, 3, 6, 9, 18} = {1, 2, 3, 6}

Example 10. If A = (2, 4) and B = [3, 5), find A ∩ B.
Answer:
We have A = (2, 4) = {x : x ∈ R, 2 < x < 4}
B = [3, 5) = {x : x ∈ R, 3 ≤ x < 5}
Clearly, A ∩ B = {x : x ∈ R, 3 ≤ x < 4} = [3, 4)

Disjoint Sets

Two sets A and B are said to be disjoint if A ∩ B = ∅.

Example 11. If A = {1, 3, 5, 7, 9} and B = {2, 4, 6, 8}, find A ∩ B.
Answer:
We have A ∩ B = {1, 3, 5, 7, 9} ∩ {2, 4, 6, 8} = ∅.
So A and B are disjoint sets.

Difference of Sets

For any sets A and B, their difference (A − B) is defined as (A − B) = {x : x ∈ A and x ∉ B}.
Thus x ∈ (A − B) ⇒ x ∈ A and x ∉ B.

Example 12. If A = {x : x ∈ N, x is a factor of 6} and B = {x : x ∈ N, x is a factor of 8}, then find (i) A − B (ii) B − A.
Answer:
We have A = {x : x ∈ N, x is a factor of 6} = {1, 2, 3, 6} and B = {x : x ∈ N, x is a factor of 8} = {1, 2, 4, 8}
(i) A − B = {1, 2, 3, 6} − {1, 2, 4, 8} = {3, 6}
(ii) B − A = {1, 2, 4, 8} − {1, 2, 3, 6} = {4, 8}

Symmetric Difference of Two Sets

The symmetric difference of two sets A and B, denoted by A Δ B, is defined as A Δ B = (A − B) ∪ (B − A).

Example 13. Let A = {a, b, c, d} and B = {b, d, f, g}. Find A Δ B.
Answer:
We have (A − B) = {a, b, c, d} − {b, d, f, g} = {a, c}
(B − A) = {b, d, f, g} − {a, b, c, d} = {f, g}
∴ A Δ B = (A − B) ∪ (B − A) = {a, c} ∪ {f, g} = {a, c, f, g}

Complement of a Set

Let U be the universal set and let A ⊂ U. Then the complement of A, denoted by A′ or (U − A), is defined as
A′ = {x ∈ U : x ∉ A}
Clearly, x ∈ A′ ⇔ x ∉ A.

Example 14. If U = {1, 2, 3, 4, 5, 6, 7, 8} and A = {2, 4, 6, 8}, find (i) A′ (ii) (A′)′.
Answer:
(i) A′ = U − A = {1, 2, 3, 4, 5, 6, 7, 8} − {2, 4, 6, 8} = {1, 3, 5, 7}
(ii) (A′)′ = U − A′ = {1, 2, 3, 4, 5, 6, 7, 8} − {1, 3, 5, 7} = {2, 4, 6, 8} = A

Example 15. If A ⊂ U, prove that (i) U′ = ∅ (ii) ∅′ = U (iii) (A′)′ = A (iv) A ∪ A′ = U (v) A ∩ A′ = ∅.
Answer:
(i) U′ = U − U = {x ∈ U : x ∉ U} = ∅
(ii) ∅′ = {x ∈ U : x ∉ ∅} = U
(iii) (A′)′ = {x ∈ U : x ∉ A′} = {x ∈ U : x ∈ A} = A
(iv) A ∪ A′ = {x ∈ U : x ∈ A or x ∈ A′} = {x ∈ U : x ∈ A or x ∉ A} = U
(v) A ∩ A′ = {x ∈ U : x ∈ A and x ∈ A′} = {x ∈ U : x ∈ A and x ∉ A} = ∅

Laws of Operations on Sets

Theorem 1 (Idempotent laws): For any set A, (i) A ∪ A = A (ii) A ∩ A = A
Theorem 2 (Identity laws): For any set A, (i) A ∪ ∅ = A (ii) A ∩ U = A
Note: ∅ and U are the identity elements for union and intersection of sets respectively.
Theorem 3 (Commutative laws): For any two sets A and B, (i) A ∪ B = B ∪ A (ii) A ∩ B = B ∩ A
Theorem 4 (Associative laws): For any three sets A, B, C, (i) (A ∪ B) ∪ C = A ∪ (B ∪ C) (ii) (A ∩ B) ∩ C = A ∩ (B ∩ C)
Theorem 5 (Distributive laws): For any three sets A, B, C, (i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) (ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Theorem 6 (De Morgan's laws): For any two sets A and B, (i) (A ∪ B)′ = A′ ∩ B′ (ii) (A ∩ B)′ = A′ ∪ B′

Venn Diagrams

Venn Diagrams

Most of the relationships between sets can be represented by means of diagrams which are known as Venn diagrams. These diagrams consist of rectangles and closed curves, usually circles. The universal set is usually represented by a rectangle and its subsets by circles.
In Venn diagrams, the elements of the sets are written in their respective circles.

Below, U = {1, 2, 3, …, 10}, A = {2, 4, 6, 8, 10} and B = {4, 6}. A and B are subsets of U and also B ⊂ A.

Venn Diagrams in Different Situations

1. When two intersecting subsets of U are given: The regions A − B, A ∩ B and B − A are as shown.

2. When two disjoint subsets of a set are given:
U = {1, 2, 3, 4, 5, 6, 7}, A = {1, 3, 5}, B = {2, 4}
Clearly A ∩ B = ∅, (A − B) = {1, 3, 5} = A, (B − A) = {2, 4} = B
A′ = {2, 4, 6, 7} and B′ = {1, 3, 5, 6, 7}

Practical Problems on Union and Intersection

Results on the Number of Elements

Let A and B be finite sets.
(i) If A ∩ B = ∅, then n(A ∪ B) = n(A) + n(B).
(ii) In general, if A and B are finite sets, then n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
Note that the sets (A − B), (A ∩ B) and (B − A) are disjoint and their union is A ∪ B. Therefore
n(A ∪ B) = n(A − B) + n(A ∩ B) + n(B − A)
= n(A − B) + n(A ∩ B) + n(B − A) + n(A ∩ B) − n(A ∩ B)
= n(A) + n(B) − n(A ∩ B)
(iii) If A, B and C are finite sets, then
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C)

Example 16. In a school there are 20 teachers who teach Mathematics or Physics. Of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teach Physics?
Answer:
Let M denote the set of teachers who teach Mathematics and P denote the set of teachers who teach Physics. In the statement of the problem, the word 'or' gives us a clue of union and the word 'and' gives us a clue of intersection.
We therefore have n(M ∪ P) = 20, n(M) = 12 and n(M ∩ P) = 4. We wish to determine n(P).
Using the result n(M ∪ P) = n(M) + n(P) − n(M ∩ P), we obtain 20 = 12 + n(P) − 4
Thus n(P) = 12. Hence 12 teachers teach Physics.

Example 17. In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice.
Answer:
Let U denote the set of surveyed students, A denote the set of students taking apple juice and B denote the set of students taking orange juice. Then
n(U) = 400, n(A) = 100, n(B) = 150 and n(A ∩ B) = 75
Now n(A′ ∩ B′) = n[(A ∪ B)′] = n(U) − n(A ∪ B)
= n(U) − n(A) − n(B) + n(A ∩ B) = 400 − 100 − 150 + 75 = 225
Hence 225 students were taking neither apple juice nor orange juice.

Example 18. In a group of 850 persons, 600 can speak Hindi and 340 can speak Tamil. Find (i) how many can speak both Hindi and Tamil (ii) how many can speak Hindi only (iii) how many can speak Tamil only.
Answer:
Let A = set of persons who can speak Hindi and B = set of persons who can speak Tamil. ∴ n(A) = 600, n(B) = 340 and n(A ∪ B) = 850.
(i) Set of persons who can speak both Hindi and Tamil = A ∩ B.
n(A ∩ B) = n(A) + n(B) − n(A ∪ B) = 600 + 340 − 850 = 90. Thus 90 persons can speak both Hindi and Tamil.
(ii) Set of persons who can speak Hindi only = (A − B).
n(A − B) + n(A ∩ B) = n(A) ⇒ n(A − B) = n(A) − n(A ∩ B) = 600 − 90 = 510. Thus 510 persons can speak Hindi only.
(iii) Set of persons who can speak Tamil only = (B − A).
n(B − A) + n(A ∩ B) = n(B) ⇒ n(B − A) = n(B) − n(A ∩ B) = 340 − 90 = 250. Hence 250 persons can speak Tamil only.

Example 19. In a survey of 25 students, it was found that 12 have taken Physics, 11 have taken Chemistry and 15 have taken Mathematics; 4 have taken Physics and Chemistry, 9 have taken Physics and Mathematics, 5 have taken Chemistry and Mathematics, while 3 have taken all the three subjects. Find the number of students who have taken
(i) Physics only
(ii) Chemistry only
(iii) Mathematics only
(iv) Physics and Chemistry but not Mathematics
(v) Physics and Mathematics but not Chemistry
(vi) only one of the subjects
(vii) at least one of the three subjects
(viii) none of the three subjects.
Answer:
Let P, C and M be the sets of students who have taken Physics, Chemistry and Mathematics respectively. Let a, b, c, d, e, f and g denote the numbers of students in the respective regions.

As per the data given, we have: a + b + c + d = 12, b + c + e + f = 11, c + d + f + g = 15, b + c = 4, c + d = 9, c + f = 5, c = 3
From these equations we get: c = 3, f = 2, d = 6, b = 1.
Now c + d + f + g = 15 ⇒ 3 + 6 + 2 + g = 15 ⇒ g = 4
b + c + e + f = 11 ⇒ 1 + 3 + e + 2 = 11 ⇒ e = 5
a + b + c + d = 12 ⇒ a + 1 + 3 + 6 = 12 ⇒ a = 2
∴ a = 2, b = 1, c = 3, d = 6, e = 5, f = 2 and g = 4. So we have:
(i) Number of students who offered Physics only = a = 2
(ii) Number of students who offered Chemistry only = e = 5
(iii) Number of students who offered Mathematics only = g = 4
(iv) Number of students who offered Physics and Chemistry but not Mathematics = b = 1
(v) Number of students who offered Physics and Mathematics but not Chemistry = d = 6
(vi) Number of students who offered only one of the given subjects = (a + e + g) = (2 + 5 + 4) = 11
(vii) Number of students who offered at least one of the given subjects = (a + b + c + d + e + f + g) = (2 + 1 + 3 + 6 + 5 + 2 + 4) = 23
(viii) Number of students who offered none of the three given subjects = (25 − 23) = 2