Chapter 1
RELATIONS AND FUNCTIONS
1.1 Overview
1.1.1 Relation
A relation R from a non-empty set A to a non empty set B is a subset of the Cartesian
product A × B. The set of all first elements of the ordered pairs in a relation R from a
set A to a set B is called the domain of the relation R. The set of all second elements in
a relation R from a set A to a set B is called the range of the relation R. The whole set
B is called the codomain of the relation R. Note that range is always a subset of
codomain.
1.1.2 Types of Relations
A relation R in a set A is subset of A × A. Thus empty set φ and A × A are two extreme
relations.
(i) A relation R in a set A is called empty relation, if no element of A is related to any
element of A, i.e., R = φ ⊂ A × A.
(ii) A relation R in a set A is called universal relation, if each element of A is related
to every element of A, i.e., R = A × A.
(iii) A relation R in A is said to be reflexive if aRa for all a∈A, R is symmetric if
aRb ⇒ bRa, ∀ a, b ∈ A and it is said to be transitive if aRb and bRc ⇒ aRc
∀ a, b, c ∈ A. Any relation which is reflexive, symmetric and transitive is called
an equivalence relation.
Note: An important property of an equivalence relation is that it divides the set
into pairwise disjoint subsets called equivalent classes whose collection is called
a partition of the set. Note that the union of all equivalence classes gives
the whole set.
1.1.3 Types of Functions
(i) A function f : X → Y is defined to be one-one (or injective), if the images of
distinct elements of X under f are distinct, i.e.,
x1 , x2 ∈ X, f (x1) = f (x2) ⇒ x1 = x2.
(ii) A function f : X → Y is said to be onto (or surjective), if every element of Y is the
image of some element of X under f, i.e., for every y ∈ Y there exists an element
x ∈ X such that f (x) = y.
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(iii) A function f : X → Y is said to be one-one and onto (or bijective), if f is both one-
one and onto.
1.1.4 Composition of Functions
(i) Let f : A → B and g : B → C be two functions. Then, the composition of f and
g, denoted by g o f, is defined as the function g o f : A → C given by
g o f (x) = g (f (x)), ∀ x ∈ A.
(ii) If f : A → B and g : B → C are one-one, then g o f : A → C is also one-one
(iii) If f : A → B and g : B → C are onto, then g o f : A → C is also onto.
However, converse of above stated results (ii) and (iii) need not be true. Moreover,
we have the following results in this direction.
(iv) Let f : A → B and g : B → C be the given functions such that g o f is one-one.
Then f is one-one.
(v) Let f : A → B and g : B → C be the given functions such that g o f is onto. Then
g is onto.
1.1.5 Invertible Function
(i) A function f : X → Y is defined to be invertible, if there exists a function
g : Y → X such that g o f = Ix and f o g = IY. The function g is called the inverse
of f and is denoted by f –1.
(ii) A function f : X → Y is invertible if and only if f is a bijective function.
(iii) If f : X → Y, g : Y → Z and h : Z → S are functions, then
h o (g o f) = (h o g) o f.
(iv) Let f : X → Y and g : Y → Z be two invertible functions. Then g o f is also
invertible with (g o f)–1 = f –1 o g–1.
1.1.6 Binary Operations
(i) A binary operation * on a set A is a function * : A × A → A. We denote * (a, b)
by a * b.
(ii) A binary operation * on the set X is called commutative, if a * b = b * a for every
a, b ∈ X.
(iii) A binary operation * : A × A → A is said to be associative if
(a * b) * c = a * (b * c), for every a, b, c ∈ A.
(iv) Given a binary operation * : A × A → A, an element e ∈ A, if it exists, is called
identity for the operation *, if a * e = a = e * a, ∀ a ∈ A.
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(v) Given a binary operation * : A × A → A, with the identity element e in A, an
element a ∈ A, is said to be invertible with respect to the operation *, if there
exists an element b in A such that a * b = e = b * a and b is called the inverse of
a and is denoted by a–1.
1.2 Solved Examples
Short Answer (S.A.)
Example 1 Let A = {0, 1, 2, 3} and define a relation R on A as follows:
R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}.
Is R reflexive? symmetric? transitive?
Solution R is reflexive and symmetric, but not transitive since for (1, 0) ∈ R and
(0, 3) ∈ R whereas (1, 3) ∉ R.
Example 2 For the set A = {1, 2, 3}, define a relation R in the set A as follows:
R = {(1, 1), (2, 2), (3, 3), (1, 3)}.
Write the ordered pairs to be added to R to make it the smallest equivalence relation.
Solution (3, 1) is the single ordered pair which needs to be added to R to make it the
smallest equivalence relation.
Example 3 Let R be the equivalence relation in the set Z of integers given by
R = {(a, b) : 2 divides a – b}. Write the equivalence class [0].
Solution [0] = {0, ± 2, ± 4, ± 6,...}
Example 4 Let the function f : R → R be defined by f (x) = 4x – 1, ∀ x ∈ R. Then,
show that f is one-one.
Solution For any two elements x1, x2 ∈ R such that f (x1) = f (x2), we have
4x1 – 1 = 4x2 – 1
⇒ 4x1 = 4x2, i.e., x1 = x2
Hence f is one-one.
Example 5 If f = {(5, 2), (6, 3)}, g = {(2, 5), (3, 6)}, write f o g.
Solution f o g = {(2, 2), (3, 3)}
Example 6 Let f : R → R be the function defined by f (x) = 4x – 3 ∀ x ∈ R. Then
write f –1.
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Solution Given that f (x) = 4x – 3 = y (say), then
4x = y + 3
y +3
⇒ x=
4
y +3 x+3
Hence f –1
(y) = ⇒ f –1
(x) =
4 4
Example 7 Is the binary operation * defined on Z (set of integer) by
m * n = m – n + mn ∀ m, n ∈ Z commutative?
Solution No. Since for 1, 2 ∈ Z, 1 * 2 = 1 – 2 + 1.2 = 1 while 2 * 1 = 2 – 1 + 2.1 = 3
so that 1 * 2 ≠ 2 * 1.
Example 8 If f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}, write the range of f and g.
Solution The range of f = {2, 3} and the range of g = {5, 6}.
Example 9 If A = {1, 2, 3} and f, g are relations corresponding to the subset of A × A
indicated against them, which of f, g is a function? Why?
f = {(1, 3), (2, 3), (3, 2)}
g = {(1, 2), (1, 3), (3, 1)}
Solution f is a function since each element of A in the first place in the ordered pairs
is related to only one element of A in the second place while g is not a function because
1 is related to more than one element of A, namely, 2 and 3.
Example 10 If A = {a, b, c, d} and f = {a, b), (b, d), (c, a), (d, c)}, show that f is one-
one from A onto A. Find f –1.
Solution f is one-one since each element of A is assigned to distinct element of the set
A. Also, f is onto since f (A) = A. Moreover, f –1 = {(b, a), (d, b), (a, c), (c, d)}.
Example 11 In the set N of natural numbers, define the binary operation * by m * n =
g.c.d (m, n), m, n ∈ N. Is the operation * commutative and associative?
Solution The operation is clearly commutative since
m * n = g.c.d (m, n) = g.c.d (n, m) = n * m ∀ m, n ∈ N.
It is also associative because for l, m, n ∈ N, we have
l * (m * n) = g. c. d (l, g.c.d (m, n))
= g.c.d. (g. c. d (l, m), n)
= (l * m) * n.
RELATIONS AND FUNCTIONS
1.1 Overview
1.1.1 Relation
A relation R from a non-empty set A to a non empty set B is a subset of the Cartesian
product A × B. The set of all first elements of the ordered pairs in a relation R from a
set A to a set B is called the domain of the relation R. The set of all second elements in
a relation R from a set A to a set B is called the range of the relation R. The whole set
B is called the codomain of the relation R. Note that range is always a subset of
codomain.
1.1.2 Types of Relations
A relation R in a set A is subset of A × A. Thus empty set φ and A × A are two extreme
relations.
(i) A relation R in a set A is called empty relation, if no element of A is related to any
element of A, i.e., R = φ ⊂ A × A.
(ii) A relation R in a set A is called universal relation, if each element of A is related
to every element of A, i.e., R = A × A.
(iii) A relation R in A is said to be reflexive if aRa for all a∈A, R is symmetric if
aRb ⇒ bRa, ∀ a, b ∈ A and it is said to be transitive if aRb and bRc ⇒ aRc
∀ a, b, c ∈ A. Any relation which is reflexive, symmetric and transitive is called
an equivalence relation.
Note: An important property of an equivalence relation is that it divides the set
into pairwise disjoint subsets called equivalent classes whose collection is called
a partition of the set. Note that the union of all equivalence classes gives
the whole set.
1.1.3 Types of Functions
(i) A function f : X → Y is defined to be one-one (or injective), if the images of
distinct elements of X under f are distinct, i.e.,
x1 , x2 ∈ X, f (x1) = f (x2) ⇒ x1 = x2.
(ii) A function f : X → Y is said to be onto (or surjective), if every element of Y is the
image of some element of X under f, i.e., for every y ∈ Y there exists an element
x ∈ X such that f (x) = y.
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(iii) A function f : X → Y is said to be one-one and onto (or bijective), if f is both one-
one and onto.
1.1.4 Composition of Functions
(i) Let f : A → B and g : B → C be two functions. Then, the composition of f and
g, denoted by g o f, is defined as the function g o f : A → C given by
g o f (x) = g (f (x)), ∀ x ∈ A.
(ii) If f : A → B and g : B → C are one-one, then g o f : A → C is also one-one
(iii) If f : A → B and g : B → C are onto, then g o f : A → C is also onto.
However, converse of above stated results (ii) and (iii) need not be true. Moreover,
we have the following results in this direction.
(iv) Let f : A → B and g : B → C be the given functions such that g o f is one-one.
Then f is one-one.
(v) Let f : A → B and g : B → C be the given functions such that g o f is onto. Then
g is onto.
1.1.5 Invertible Function
(i) A function f : X → Y is defined to be invertible, if there exists a function
g : Y → X such that g o f = Ix and f o g = IY. The function g is called the inverse
of f and is denoted by f –1.
(ii) A function f : X → Y is invertible if and only if f is a bijective function.
(iii) If f : X → Y, g : Y → Z and h : Z → S are functions, then
h o (g o f) = (h o g) o f.
(iv) Let f : X → Y and g : Y → Z be two invertible functions. Then g o f is also
invertible with (g o f)–1 = f –1 o g–1.
1.1.6 Binary Operations
(i) A binary operation * on a set A is a function * : A × A → A. We denote * (a, b)
by a * b.
(ii) A binary operation * on the set X is called commutative, if a * b = b * a for every
a, b ∈ X.
(iii) A binary operation * : A × A → A is said to be associative if
(a * b) * c = a * (b * c), for every a, b, c ∈ A.
(iv) Given a binary operation * : A × A → A, an element e ∈ A, if it exists, is called
identity for the operation *, if a * e = a = e * a, ∀ a ∈ A.
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(v) Given a binary operation * : A × A → A, with the identity element e in A, an
element a ∈ A, is said to be invertible with respect to the operation *, if there
exists an element b in A such that a * b = e = b * a and b is called the inverse of
a and is denoted by a–1.
1.2 Solved Examples
Short Answer (S.A.)
Example 1 Let A = {0, 1, 2, 3} and define a relation R on A as follows:
R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}.
Is R reflexive? symmetric? transitive?
Solution R is reflexive and symmetric, but not transitive since for (1, 0) ∈ R and
(0, 3) ∈ R whereas (1, 3) ∉ R.
Example 2 For the set A = {1, 2, 3}, define a relation R in the set A as follows:
R = {(1, 1), (2, 2), (3, 3), (1, 3)}.
Write the ordered pairs to be added to R to make it the smallest equivalence relation.
Solution (3, 1) is the single ordered pair which needs to be added to R to make it the
smallest equivalence relation.
Example 3 Let R be the equivalence relation in the set Z of integers given by
R = {(a, b) : 2 divides a – b}. Write the equivalence class [0].
Solution [0] = {0, ± 2, ± 4, ± 6,...}
Example 4 Let the function f : R → R be defined by f (x) = 4x – 1, ∀ x ∈ R. Then,
show that f is one-one.
Solution For any two elements x1, x2 ∈ R such that f (x1) = f (x2), we have
4x1 – 1 = 4x2 – 1
⇒ 4x1 = 4x2, i.e., x1 = x2
Hence f is one-one.
Example 5 If f = {(5, 2), (6, 3)}, g = {(2, 5), (3, 6)}, write f o g.
Solution f o g = {(2, 2), (3, 3)}
Example 6 Let f : R → R be the function defined by f (x) = 4x – 3 ∀ x ∈ R. Then
write f –1.
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Solution Given that f (x) = 4x – 3 = y (say), then
4x = y + 3
y +3
⇒ x=
4
y +3 x+3
Hence f –1
(y) = ⇒ f –1
(x) =
4 4
Example 7 Is the binary operation * defined on Z (set of integer) by
m * n = m – n + mn ∀ m, n ∈ Z commutative?
Solution No. Since for 1, 2 ∈ Z, 1 * 2 = 1 – 2 + 1.2 = 1 while 2 * 1 = 2 – 1 + 2.1 = 3
so that 1 * 2 ≠ 2 * 1.
Example 8 If f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}, write the range of f and g.
Solution The range of f = {2, 3} and the range of g = {5, 6}.
Example 9 If A = {1, 2, 3} and f, g are relations corresponding to the subset of A × A
indicated against them, which of f, g is a function? Why?
f = {(1, 3), (2, 3), (3, 2)}
g = {(1, 2), (1, 3), (3, 1)}
Solution f is a function since each element of A in the first place in the ordered pairs
is related to only one element of A in the second place while g is not a function because
1 is related to more than one element of A, namely, 2 and 3.
Example 10 If A = {a, b, c, d} and f = {a, b), (b, d), (c, a), (d, c)}, show that f is one-
one from A onto A. Find f –1.
Solution f is one-one since each element of A is assigned to distinct element of the set
A. Also, f is onto since f (A) = A. Moreover, f –1 = {(b, a), (d, b), (a, c), (c, d)}.
Example 11 In the set N of natural numbers, define the binary operation * by m * n =
g.c.d (m, n), m, n ∈ N. Is the operation * commutative and associative?
Solution The operation is clearly commutative since
m * n = g.c.d (m, n) = g.c.d (n, m) = n * m ∀ m, n ∈ N.
It is also associative because for l, m, n ∈ N, we have
l * (m * n) = g. c. d (l, g.c.d (m, n))
= g.c.d. (g. c. d (l, m), n)
= (l * m) * n.
