Chapter 1

RELATIONS AND FUNCTIONS
™There is no permanent place in the world for ugly mathematics ... . It may
be very hard to define mathematical beauty but that is just as true of
beauty of any kind, we may not know quite what we mean by a
beautiful poem, but that does not prevent us from recognising
one when we read it. — G. H. HARDY ™

1.1 Introduction
Recall that the notion of relations and functions, domain,
co-domain and range have been introduced in Class XI
along with different types of specific real valued functions
and their graphs. The concept of the term ‘relation’ in
mathematics has been drawn from the meaning of relation
in English language, according to which two objects or
quantities are related if there is a recognisable connection
or link between the two objects or quantities. Let A be
the set of students of Class XII of a school and B be the
set of students of Class XI of the same school. Then some
of the examples of relations from A to B are
(i) {(a, b) ∈ A × B: a is brother of b}, Lejeune Dirichlet
(ii) {(a, b) ∈ A × B: a is sister of b}, (1805-1859)
(iii) {(a, b) ∈ A × B: age of a is greater than age of b},
(iv) {(a, b) ∈ A × B: total marks obtained by a in the final examination is less than
the total marks obtained by b in the final examination},
(v) {(a, b) ∈ A × B: a lives in the same locality as b}. However, abstracting from
this, we define mathematically a relation R from A to B as an arbitrary subset
of A × B.
If (a, b) ∈ R, we say that a is related to b under the relation R and we write as
a R b. In general, (a, b) ∈ R, we do not bother whether there is a recognisable
connection or link between a and b. As seen in Class XI, functions are special kind of
relations.
In this chapter, we will study different types of relations and functions, composition
of functions, invertible functions and binary operations.

2 MATHEMATICS

1.2 Types of Relations
In this section, we would like to study different types of relations. We know that a
relation in a set A is a subset of A × A. Thus, the empty set φ and A × A are two
extreme relations. For illustration, consider a relation R in the set A = {1, 2, 3, 4} given by
R = {(a, b): a – b = 10}. This is the empty set, as no pair (a, b) satisfies the condition
a – b = 10. Similarly, R′ = {(a, b) : | a – b | ≥ 0} is the whole set A × A, as all pairs
(a, b) in A × A satisfy | a – b | ≥ 0. These two extreme examples lead us to the
following definitions.
Definition 1 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.
Definition 2 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.
Both the empty relation and the universal relation are some times called trivial
relations.
Example 1 Let A be the set of all students of a boys school. Show that the relation R
in A given by R = {(a, b) : a is sister of b} is the empty relation and R′ = {(a, b) : the
difference between heights of a and b is less than 3 meters} is the universal relation.
Solution Since the school is boys school, no student of the school can be sister of any
student of the school. Hence, R = φ, showing that R is the empty relation. It is also
obvious that the difference between heights of any two students of the school has to be
less than 3 meters. This shows that R′ = A × A is the universal relation.
Remark In Class XI, we have seen two ways of representing a relation, namely
roaster method and set builder method. However, a relation R in the set {1, 2, 3, 4}
defined by R = {(a, b) : b = a + 1} is also expressed as a R b if and only if
b = a + 1 by many authors. We may also use this notation, as and when convenient.
If (a, b) ∈ R, we say that a is related to b and we denote it as a R b.
One of the most important relation, which plays a significant role in Mathematics,
is an equivalence relation. To study equivalence relation, we first consider three
types of relations, namely reflexive, symmetric and transitive.
Definition 3 A relation R in a set A is called
(i) reflexive, if (a, a) ∈ R, for every a ∈ A,
(ii) symmetric, if (a1, a2) ∈ R implies that (a2, a1) ∈ R, for all a1, a2 ∈ A.
(iii) transitive, if (a1, a2) ∈ R and (a2, a3) ∈ R implies that (a1, a3) ∈ R, for all a1, a2,
a3 ∈ A.

RELATIONS AND FUNCTIONS 3

Definition 4 A relation R in a set A is said to be an equivalence relation if R is
reflexive, symmetric and transitive.
Example 2 Let T be the set of all triangles in a plane with R a relation in T given by
R = {(T1, T2) : T1 is congruent to T2}. Show that R is an equivalence relation.
Solution R is reflexive, since every triangle is congruent to itself. Further,
(T1, T2) ∈ R ⇒ T1 is congruent to T2 ⇒ T2 is congruent to T1 ⇒ (T2, T1) ∈ R. Hence,
R is symmetric. Moreover, (T1, T2), (T2, T3) ∈ R ⇒ T1 is congruent to T2 and T2 is
congruent to T3 ⇒ T1 is congruent to T3 ⇒ (T1, T3) ∈ R. Therefore, R is an equivalence
relation.
Example 3 Let L be the set of all lines in a plane and R be the relation in L defined as
R = {(L1, L2) : L1 is perpendicular to L2}. Show that R is symmetric but neither
reflexive nor transitive.
Solution R is not reflexive, as a line L1 can not be perpendicular to itself, i.e., (L1, L1)
∉ R. R is symmetric as (L1, L2) ∈ R
⇒ L1 is perpendicular to L2
⇒ L2 is perpendicular to L1
⇒ (L2, L1) ∈ R.
R is not transitive. Indeed, if L1 is perpendicular to L2 and Fig 1.1
L2 is perpendicular to L3, then L1 can never be perpendicular to
L3. In fact, L1 is parallel to L3, i.e., (L1, L2) ∈ R, (L2, L3) ∈ R but (L1, L3) ∉ R.
Example 4 Show that the relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2),
(3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive.
Solution R is reflexive, since (1, 1), (2, 2) and (3, 3) lie in R. Also, R is not symmetric,
as (1, 2) ∈ R but (2, 1) ∉ R. Similarly, R is not transitive, as (1, 2) ∈ R and (2, 3) ∈ R
but (1, 3) ∉ R.
Example 5 Show that the relation R in the set Z of integers given by
R = {(a, b) : 2 divides a – b}
is an equivalence relation.
Solution R is reflexive, as 2 divides (a – a) for all a ∈ Z. Further, if (a, b) ∈ R, then
2 divides a – b. Therefore, 2 divides b – a. Hence, (b, a) ∈ R, which shows that R is
symmetric. Similarly, if (a, b) ∈ R and (b, c) ∈ R, then a – b and b – c are divisible by
2. Now, a – c = (a – b) + (b – c) is even (Why?). So, (a – c) is divisible by 2. This
shows that R is transitive. Thus, R is an equivalence relation in Z.

b) : 3 divides a – b}. A1 = [3r]. 3. 6.. O is the equivalence class containing 1 and is denoted by [1]. A1 = [0]. .. 3..... ± 4) etc. we can show that R is an equivalence relation. (0. 4. The subsets Ai are called equivalence classes. 8. Given an arbitrary equivalence relation R in an arbitrary set X. [0] = [2r] and [1] = [2r + 1]. note that all even integers are related to zero. Example 6 Let R be the relation defined in the set A = {1. R divides X into mutually disjoint subsets Ai called partitions or subdivisions of X satisfying: (i) all elements of Ai are related to each other. Infact. for all i. show that all the elements of the subset {1. A2 = [1] and A3 = [2]. 6}. 7} are related to each other and all the elements of the subset {2. 5. 5. ± 3) etc. 4. – 3. For example. – 5. Also. (ii) no element of Ai is related to any element of Aj . A2 = [3r + 1] and A3 = [3r + 2].. The subset E is called the equivalence class containing zero and is denoted by [0].. 4 MATHEMATICS In Example 5. 2... 1.. 5. In fact. . Note that [0] ≠ [1]. all odd integers are related to one and no even integer is related to one. 0. as (0. as (0. Similarly. – 6. what we have seen above is true for an arbitrary equivalence relation R in a set X. 3. i ≠ j.. 7} by R = {(a..} A2 = {x ∈ Z : x – 1 is a multiple of 3} = {. 6} are related to each other.} A3 = {x ∈ Z : x – 2 is a multiple of 3} = {.} Define a relation R in Z given by R = {(a. . A2 coincides with the set of all integers which are related to 1 and A3 coincides with the set of all integers in Z which are related to 2. for all r ∈ Z. Show that R is an equivalence relation. lie in R and no odd integer is related to 0. Further. but no element of the subset {1. b) : both a and b are either odd or even}.. – 4. 7} is related to any element of the subset {2. ± 2). 4. Thus. (iii) E and O are disjoint and Z = E ∪ O. – 1. Following the arguments similar to those used in Example 5. A1 coincides with the set of all integers in Z which are related to zero. (iii) ∪ Aj = X and Ai ∩ Aj = φ. . (0. r ∈ Z. A2 and A3 whose union is Z with A1 = {x ∈ Z : x is a multiple of 3} = {. 3. Similarly. – 2. 4. 7. (ii) No element of E is related to any element of O and vice-versa.. do not lie in R. ± 1). 5. The interesting part of the situation is that we can go reverse also. consider a subdivision of the set Z given by three mutually disjoint subsets A1. 6. i ≠ j. 2. Therefore... the set E of all even integers and the set O of all odd integers are subsets of Z satisfying following conditions: (i) All elements of E are related to each other and all elements of O are related to each other.

while elements of {2. 3. 5. Hence. y) : x is exactly 7 cm taller than y} (d) R = {(x. (a. Further. 5. symmetric and transitive: (i) Relation R in the set A = {1. both a and a must be either odd or even. Also. Show that the relation R in the set R of real numbers. 4. b) : a ≤ b}. as all of them are even. 3. b) : b = a + 1} is reflexive. R is an equivalence relation. RELATIONS AND FUNCTIONS 5 Solution Given any element a in A. 4. Similarly.1 1. Determine whether each of the following relations are reflexive. y) : y is divisible by x} (iv) Relation R in the set Z of all integers defined as R = {(x. 4. 6} are related to each other. 4. 6} as R = {(a. y) : 3x – y = 0} (ii) Relation R in the set N of natural numbers defined as R = {(x. 2. 2. symmetric or transitive. c) ∈ R ⇒ all elements a. a) ∈ R. 3. y) : y = x + 5 and x < 4} (iii) Relation R in the set A = {1. all the elements of the subset {2. EXERCISE 1. 5. y) : x and y live in the same locality} (c) R = {(x. no element of the subset {1. 7} can be related to any element of {2. 14} defined as R = {(x.. b) ∈ R and (b. Further. symmetric or transitive. c) ∈ R.. y) : x and y work at the same place} (b) R = {(x. 5. Check whether the relation R defined in the set {1. c. 6}.. y) : x is father of y} 2. 3. 5. 4. 3. Check whether the relation R in R defined by R = {(a. 5. Show that the relation R in R defined as R = {(a. is reflexive and transitive but not symmetric. . 4. 13. 6} are even. b) ∈ R ⇒ both a and b must be either odd or even ⇒ (b. a) ∈ R. b) : a ≤ b2} is neither reflexive nor symmetric nor transitive. Similarly. defined as R = {(a. (a. must be either even or odd simultaneously ⇒ (a. y) : x – y is an integer} (v) Relation R in the set A of human beings in a town at a particular time given by (a) R = {(x. 3. . 2. all the elements of {1. 3. as elements of {1. 7} are related to each other. b. 6} as R = {(x. b) : a ≤ b3} is reflexive. y) : x is wife of y} (e) R = {(x. 7} are odd. so that (a. as all the elements of this subset are odd.

3. 5} is related to any element of {2. 12. Show that R is an equivalence relation. 13 and T3 with sides 6. show that the set of all points related to a point P ≠ (0. L2) : L1 is parallel to L2}. 5} are related to each other and all the elements of {2. Which triangles among T1. 2. Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}. Show that the relation R in the set {1. is an equivalence relation. is an equivalence relation. is equivalence relation. (ii) Transitive but neither reflexive nor symmetric. 3. T2 with sides 5. Show that the relation R in the set A = {1. What is the set of all elements in A related to the right angle triangle T with sides 3. 4. 9. 10. b) : |a – b| is a multiple of 4} (ii) R = {(a. (v) Symmetric and transitive but not reflexive. Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1. 5. T2 and T3 are related? 13. given by R = {(x. Further. 5} given by R = {(a. (iv) Reflexive and transitive but not symmetric. y) : x and y have same number of pages} is an equivalence relation. b) : |a – b| is even}. 7. 8. is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.6 MATHEMATICS 6. 12. . b) : a = b} is an equivalence relation. 3} given by R = {(1. Show that the relation R defined in the set A of all polygons as R = {(P1. 11. (iii) Reflexive and symmetric but not transitive. 1)} is symmetric but neither reflexive nor transitive. Give an example of a relation. given by (i) R = {(a. Which is (i) Symmetric but neither reflexive nor transitive. P2) : P1 and P2 have same number of sides}. 4. Show that the relation R in the set A of points in a plane given by R = {(P. Show that the relation R in the set A of all the books in a library of a college. Show that all the elements of {1. T2) : T1 is similar to T2}. Find the set of all elements related to 1 in each case. Show that the relation R defined in the set A of all triangles as R = {(T1. 0) is the circle passing through P with origin as centre. 4}. 2. 8. (2. Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}. But no element of {1. 3. 2). 4} are related to each other. 4 and 5? 14. Consider three right angle triangles T1 with sides 3. 10.

we observe that the images of distinct elements of X1 under the function f1 are distinct. i. rational function. Addition. constant function. x2 ∈ X. Consider the functions f1. Choose the correct answer. f (x1) = f (x2) implies x1 = x2.3 Types of Functions The notion of a function along with some special functions like identity function.2 (i) is not onto as elements e. signum function etc. multiplication and division of two functions have also been studied. modulus function. Let R be the relation in the set N given by R = {(a. b > 6}. if every element of Y is the image of some element of X under f. f3 and f4 given by the following diagrams. we would like to study different types of functions. The function f3 and f4 in Fig 1. (3. polynomial function. 3). The function f1 and f4 in Fig 1. f is called many-one. Let R be the relation in the set {1.2. 3). f in X2 are not the image of any element in X1 under f1. for every x1.4). 4) ∈ R (B) (3... (C) R is symmetric and transitive but not reflexive. 16. (B) R is reflexive and transitive but not symmetric. (4. while all elements of X3 are images of some elements of X1 under f3. 8) ∈ R (C) (6. Definition 6 A function f : X → Y is said to be onto (or surjective). 2. 2). we would like to extend our study about function from where we finished earlier. In Fig 1. (1.2 (i) and (iv) are one-one and the function f2 and f3 in Fig 1. As the concept of function is of paramount importance in mathematics and among other disciplines as well. (3. 1). 2). (2. f2. for every y ∈ Y. The above observations lead to the following definitions: Definition 5 A function f : X → Y is defined to be one-one (or injective). subtraction. if the images of distinct elements of X under f are distinct.2 (iii). i. . In this section. Otherwise. RELATIONS AND FUNCTIONS 7 15. (iv) are onto and the function f1 in Fig 1. 4} given by R = {(1. 2)}. (D) R is an equivalence relation. 8) ∈ R (D) (8. Further. along with their graphs have been given in Class XI. 3. there exists an element x in X such that f (x) = y.2 (ii) and (iii) are many-one. (1. namely b. Choose the correct answer.e. 7) ∈ R 1. but the image of two distinct elements 1 and 2 of X1 under f2 is same. (A) (2. there are some elements like e and f in X2 which are not images of any element of X1 under f1. (A) R is reflexive and symmetric but not transitive. b) : a = b – 2.e.

Solution The function f is one-one. Further. . Let f : A → N be function defined by f (x) = roll number of the student x. given by f (x) = 2x. Solution No two different students of the class can have same roll number. as for 1 ∈ N. Example 7 Let A be the set of all 50 students of Class X in a school. for f (x1) = f (x2) ⇒ 2x1 = 2x2 ⇒ x1 = x2. This implies that 51 in N is not roll number of any student of the class. if f is both one-one and onto. The function f4 in Fig 1. 8 MATHEMATICS Fig 1. Therefore.2 (iv) is one-one and onto. f must be one-one. Show that f is one-one but not onto. Example 8 Show that the function f : N → N.2 (i) to (iv) Remark f : X → Y is onto if and only if Range of f = Y. is one-one but not onto. so that 51 can not be image of any element of X under f. f is not onto. We can assume without any loss of generality that roll numbers of students are from 1 to 50. there does not exist any x in N such that f (x) = 2x = 1. f is not onto. Definition 7 A function f : X → Y is said to be one-one and onto (or bijective). Hence.

y ≠ 1. defined as f (x) = x2. as given any y ∈ N. the element – 2 in the co-domain R is not image of any element x in the domain R (Why?). f ( x) = ⎨ ⎩ x − 1.if x is even is both one-one and onto. But f is onto. Hence. f is not one- one.3 Example 10 Show that the function f : N → N. Example 12 Show that f : N → N. we have f (1) = 1. as f (x1) = f (x2) ⇒ 2x1 = 2x2 ⇒ x1 = x2. Also.4 . Solution Since f (– 1) = 1 = f (1). RELATIONS AND FUNCTIONS 9 Example 9 Prove that the function f : R → R.if x is odd. given by ⎧ x + 1. is one-one and onto. f is onto. ( ) = y. Therefore f is not onto. as f (1) = f (2) = 1. given by f (x) = 2x. is onto but not one-one. 2 2 2 Fig 1. there exists in R such that f ( ) = 2 . Also for 1 ∈ N. is neither one-one nor onto. for every x > 2. Fig 1. given any real y y y number y in R. given by f (1) = f (2) = 1 and f (x) = x – 1. Also. we can choose x as y + 1 such that f (y + 1) = y + 1 – 1 = y. Solution f is one-one. Solution f is not one-one. Example 11 Show that the function f : R → R.

i.2 1 1. this may not be true. x where R∗ is the set of all non-zero real numbers. i. Note that if x1 is odd and x2 is even. 3}. is neither one-one nor onto. 3} under f. 2. 2. showing that f is not onto. Prove that the Greatest Integer Function f : R → R. Also. Thus. Example 13 Show that an onto function f : {1. both x1 and x2 must be either odd or even. Show that the function f : R∗ → R∗ defined by f (x) = is one-one and onto. then also f (x1) = f (x2) ⇒ x1 – 1 = x2 – 1 ⇒ x1 = x2.e. f has to be onto. Then f (x1) = f (x2) ⇒ x1 + 1 = x2 + 1 ⇒ x1 = x2. f is one-one. Solution Suppose f is not one-one. Then there exists two elements. 2. any odd number 2r + 1 in the co-domain N is the image of 2r + 2 in the domain N and any even number 2r in the co-domain N is the image of 2r – 1 in the domain N. three elements of {1. Is the result true. Hence. 3} must be onto. where [x] denotes the greatest integer less than or equal to x. Also. then we will have x1 + 1 = x2 – 1. Example 14 Show that a one-one function f : {1. Similarly. for every finite set X. EXERCISE 1. if the domain R∗ is replaced by N with co-domain being same as R∗? 2. x2 – x1 = 2 which is impossible. a contradiction. Therefore. Therefore.. 2. Check the injectivity and surjectivity of the following functions: (i) f : N → N given by f (x) = x2 (ii) f : Z → Z given by f (x) = x2 (iii) f : R → R given by f (x) = x2 (iv) f : N → N given by f (x) = x3 (v) f : Z → Z given by f (x) = x3 3. f is onto. 3} → {1. 10 MATHEMATICS Solution Suppose f (x1) = f (x2). In fact. 2. 3} → {1.. using the similar argument. given by f (x) = [x]. Hence. if both x1 and x2 are even. 3} must be taken to 3 different elements of the co-domain {1. the range set can have at the most two elements of the co-domain {1. the possibility of x1 being even and x2 being odd can also be ruled out. Similarly. Solution Since f is one-one. say 1 and 2 in the domain whose image in the co-domain is same. In contrast to this. the image of 3 under f can be only one element. 2. Remark The results mentioned in Examples 13 and 14 are also true for an arbitrary finite set X. Thus. 3} is always one-one. this is a characteristic difference between a finite and an infinite set. a one-one function f : X → X is necessarily onto and an onto map f : X → X is necessarily one-one. Suppose both x1 and x2 are odd.e. 2. Examples 8 and 10 show that for an infinite set. f must be one-one. .

4). 3}. 7. Let A = R – {3} and B = R – {1}. 5. 5. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto. if x is negative. 5). RELATIONS AND FUNCTIONS 11 4. Justify your answer. Justify your answer. (i) f : R → R defined by f (x) = 3 – 4x (ii) f : R → R defined by f (x) = 1 + x2 8. if n is odd 9. (3. state whether the function is one-one. Let f : R → R be defined as f (x) = 3x. Let A and B be sets. In each of the following cases. 6. ⎪ n . 7} and let f = {(1. 6)} be a function from A to B. if x > 0 ⎪ f ( x) = ⎨0. Show that f : A × B → B × A such that f (a. onto or bijective. . ⎝ x−3⎠ 11. 10. Is f one-one and onto? Justify your answer. if x is positive or 0 and | x | is – x. Choose the correct answer. B = {4. Let A = {1. (2. Show that the Signum Function f : R → R. where | x | is x. given by f (x) = | x |. Consider the function f : A → B defined by ⎛ x−2⎞ f (x) = ⎜ ⎟ . if x < 0 ⎩ is neither one-one nor onto. ⎧n +1 ⎪⎪ 2 . Let f : R → R be defined as f(x) = x4. given by ⎧1. Show that the Modulus Function f : R → R. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto. Choose the correct answer. Show that f is one-one. 2. 12. is neither one- one nor onto. 6. Let f : N → N be defined by f (n) = ⎨ for all n ∈ N. if n is even ⎪⎩ 2 State whether the function f is bijective. b) = (b. a) is bijective function. if x = 0 ⎪ –1.

Each student appearing in the Board Examination is assigned a roll number by the Board which is written by the students in the answer script at the time of examination. who appeared in Class X of a Board Examination in 2006. Thus.4 Composition of Functions and Invertible Function In this section.5 Example 15 Let f : {2. Hence. 11. the Board arranges to deface the roll numbers of students in the answer scripts and assigns a fake code number to each roll number. Let B ⊂ N be the set of all roll numbers and C ⊂ N be the set of all code numbers. This leads to the following definition: Definition 8 Let f : A → B and g : B → C be two functions. ∀ x ∈ A. Fig 1. 3. Consider the set A of all students. gof ≠ fog. for x = 0. 5. each student is eventually attached a code number. 12 MATHEMATICS 1. 15} be functions defined as f (2) = 3. 4. 9} and g : {3. gof (4) = g (f (4)) = g (5) = 11 and gof (5) = g (5) = 11. if f : R → R and g : R → R are given by f (x) = cos x and g (x) = 3x2. . In order to have confidentiality. 5. gof (3) = g (f (3)) = g (4) = 7. denoted by gof. Show that gof ≠ fog. Solution We have gof (2) = g (f (2)) = g (3) = 7. 5} → {3. by the combination of these two functions. is defined as the function gof : A → C given by gof (x) = g(f (x)). fog (x) = f (g (x)) = f (3x2) = cos (3x2). Solution We have gof (x) = g (f (x)) = g (cos x) = 3 (cos x)2 = 3 cos2 x. Note that 3cos2 x ≠ cos 3x2. In this process each student is assigned a roll number through the function f and each roll number is assigned a code number through the function g. we will study composition of functions and the inverse of a bijective function. 9} → {7. Find gof. f (3) = 4. 4. Similarly. f (4) = f (5) = 5 and g (3) = g (4) = 7 and g (5) = g (9) = 11. Then the composition of f and g. 4. Example 16 Find gof and fog. This gives rise to two functions f : A → B and g : B → C given by f (a) = the roll number assigned to the student a and g (b) = the code number assigned to the roll number b.

as g is one-one ⇒ x1 = x2. Example 19 Show that if f : A → B and g : B → C are onto. Further. Solution Given an arbitrary element z ∈ C. ∀ x ∈ B are called identity ⎩ ⎭ 5 ⎩5 ⎭ functions on sets A and B. there exists an element x in A . gof (x) = x. ∀ x ∈ A. gof is one-one. RELATIONS AND FUNCTIONS 13 ⎧7 ⎫ ⎧3⎫ 3x + 4 Example 17 Show that if f : R − ⎨ ⎬ → R − ⎨ ⎬ is defined by f ( x ) = and ⎩5 ⎭ ⎩5⎭ 5x − 7 ⎧3 ⎫ ⎧7 ⎫ 7x + 4 g : R − ⎨ ⎬ → R − ⎨ ⎬ is defined by g ( x) = . IA (x) = x. there exists a pre-image y of z under g such that g (y) = z. then gof : A → C is also one-one. ⎩5 ⎭ ⎩5⎭ 5x − 3 ⎧3⎫ ⎧7 ⎫ A = R – ⎨ ⎬ . Example 18 Show that if f : A → B and g : B → C are one-one. which implies that gof = IB and fog = IA. ∀ x ∈ B and fog (x) = x. since g is onto. ∀ x ∈ A. then fog = IA and gof = IB. IB (x) = x. then gof : A → C is also onto. where. Solution We have ⎛ (3x + 4) ⎞ 7⎜ +4 ⎛ 3x + 4 ⎞ ⎝ (5 x − 7) ⎟⎠ 21x + 28 + 20 x − 28 41x gof ( x) = g ⎜ ⎟= = = =x ⎝ 5x − 7 ⎠ ⎛ (3x + 4) ⎞ 15 x + 20 − 15 x + 21 41 5⎜ ⎟ − 3 ⎝ (5 x − 7) ⎠ ⎛ (7 x + 4) ⎞ 3⎜ +4 ⎛ 7x + 4 ⎞ ⎝ (5 x − 3) ⎟⎠ 21x + 12 + 20 x − 12 41x Similarly. as f is one-one Hence. respectively. B = R – ⎨ ⎬ . Solution Suppose gof (x1) = gof (x2) ⇒ g (f (x1)) = g(f (x 2)) ⇒ f (x1) = f (x2). for y ∈ B. fog ( x) = f ⎜ ⎟= = = =x ⎝ 5x − 3 ⎠ ⎛ (7 x + 4) ⎞ 35 x + 20 − 35 x + 21 41 5⎜ ⎟−7 ⎝ (5 x − 3) ⎠ Thus.

4. 2. 5. 2. where. i. if gof is onto? Solution Consider f : {1. Remark It can be verified in general that gof is one-one implies that f is one-one. Not only this. 2. X = {1. g (2) = 2 and g (3) = g (4) = 3. c} be one-one and onto function given by f (1) = a. 2. Remark The interesting fact is that the result mentioned in the above example is true for an arbitrary one-one and onto function f : X → Y. b. 3} such that gof = IX and fog = IY. 3.e. Show that there exists a function g : {a. 4. while in the reverse process of the composite gof. 3} as g (a) = 1. 6} as g (x) = x. b. 5. 3. 4} → {1. Solution Consider f : {1. f (2) = b and f (3) = c. ∀ x and g : {1. if f : X → Y is a function such that there exists a function g : Y → X such that gof = IX and fog = IY. 3. 2. showing that gof is onto. f (2) = 2. We observe that while composing f and g. 3. Are f and g both necessarily one-one. f (3) = f (4) = 3. then f must be one-one and onto. gof is onto implies that g is onto. 4} → {1. 2. Example 20 Consider functions f and g such that composite gof is defined and is one- one. 14 MATHEMATICS with f (x) = y. b. 2. 3. g (1) = 1. first the reverse process of g is applied and then the reverse process of f. gof (x) = x ∀ x. Then. Similarly. Solution Consider g : {a. Further. 3. c}. since f is onto. Each student appearing in Class X Examination of the Board is assigned a roll number under the function f and each roll number is assigned a code number under g. This helps in assigning mark to the student scoring that mark. examiner enters the mark against each code number in a mark book and submits to the office of the Board.. 2. 2. After the answer scripts are examined. gof (x) = g (f (x)) = g (y) = z. 3} defined as f (1) = 1. 4. 2. first f and then g was applied. 3} and Y = {a. g (b) = 2 and g (c) = 3. 2. to get gof. Example 21 Are f and g both necessarily onto. 4} → {1. Example 22 and Remark lead to the following definition: . c} → {1. for x = 1. The Board officials decode by assigning roll number back to each code number through a process reverse to g and thus mark gets attached to roll number rather than code number. even the converse is also true . 2. But g is clearly not one-one. we would like to have close look at the functions f and g described in the beginning of this section in reference to a Board Examination. 6} → {1. Now. b. which shows that gof is one-one. Therefore. 5. It can be seen that gof is onto but f is not onto. 4} and g : {1. 6} defined as f (x) = x. c} → {1. the process reverse to f assigns a roll number to the student having that roll number. 4 and g (5) = g (6) = 5. 2. 3. Example 22 Let f : {1. It is easy to verify that the composite gof = IX is the identity function on X and the composite fog = IY is the identity function on Y. The above discussion. 3. 3} → {a.

Find the inverse. This shows that gof = IN ⎝ 4 ⎠ 4 and fog = IY. f is invertible with f –1 = g. This gives a function g : Y → N . This shows that x = . then f must be one-one and onto and conversely. Hence. By the definition of Y. Find the inverse of f. is invertible. specially when the actual inverse of f is not to be determined. which implies that f is invertible and g is the inverse of f. if f is one-one and onto. Find the inverse of f. gof (x) = g (f (x)) = g (4x + 3) = = x and 4 4 ⎛ ( y − 3) ⎞ 4 ( y − 3) fog (y) = f (g (y)) = f ⎜ ⎟= + 3 = y – 3 + 3 = y. Then y = 4x2 + 12x + 15. The function g is called the inverse of f and is denoted by f –1. y −6 −3 2 . Now. S is the range of f. if f is invertible. for some x in N. defined by g (y) = y . Define g : Y → N by 4 ( y − 3) (4 x + 3 − 3) g ( y) = . where. where. Example 25 Let f : N → R be a function defined as f (x) = 4x2 + 12x + 15. Solution Consider an arbitrary element y of Y. Example 24 Let Y = {n2 : n ∈ N } ⊂ N . Show that f : N → S. Example 23 Let f : N → Y be a function defined as f (x) = 4x + 3. ( y − 3) for some x in the domain N . Solution Let y be an arbitrary element of range f. Thus. Now. RELATIONS AND FUNCTIONS 15 Definition 9 A function f : X → Y is defined to be invertible. which shows that gof = IN and fog = IY. for some n ∈ N . Y = {y ∈ N : y = 4x + 3 for some x ∈ N }. This fact significantly helps for proving a function f to be invertible by showing that f is one-one and onto. which implies that y = (2x + 3)2 + 6. if there exists a function g : Y → X such that gof = IX and fog = IY. Solution An arbitrary element y in Y is of the form n2. ( y) =( y) 2 gof (n) = g (n2) = n 2 = n and fog (y) = f = y . Consider f : N → Y as f (n) = n2. as y ≥ 6. This implies that n = y . then f must be invertible. Show that f is invertible. Show that f is invertible. This gives x = (( ) ) . y = 4x + 3.

Example 26 Consider f : N → N. 16 MATHEMATICS Let us define g : S → N by g (y) = (( ) ) y−6 −3 . Find out f –1. Example 27 Consider f : {1. 2 Now gof (x) = g (f (x)) = g (4x2 + 12x + 15) = g ((2x + 3)2 + 6) = (( (2 x + 3) 2 + 6 − 6 − 3 ) ) ( 2 x + 3 − 3) = =x 2 2 (( ) y − 6) − 3 ⎞ ⎛ 2 (( y − 6) − 3 ) + 3 ⎞⎟ 2 ⎛ and fog (y) = f ⎜⎜ ⎟⎟ = ⎜⎜ ⎟ +6 ⎝ 2 ⎠ ⎝ 2 ⎠ (( y − 6) − 3+ 3 )) + 6 = ( y − 6 ) + 6 = y – 6 + 6 = y. g(b) = ball and g(c) = cat. ball. gof = IN and fog =IS. Theorem 1 If f : X → Y. Show that f. ho(gof) = (hog) o f. Also. This shows that ho(gof) = (hog) o f. . g and gof are invertible. b. c} and g : {a. 2 2 = Hence. f (3) = c. ∀ x. Show that ho(gof ) = (hog) of. This implies that f is invertible with f –1 = g. g : N → N and h : N → R defined as f (x) = 2x. Hence. ∀ x ∈ N. cat} defined as f (1) = a. g–1 and (gof)–1 and show that (gof) –1 = f –1o g–1. f (2) = b. g : Y → Z and h : Z → S are functions. Proof We have ho(gof ) (x) = h(gof (x)) = h(g (f (x))). g (y) = 3y + 4 and h (z) = sin z. g(a) = apple. b. This result is true in general situation as well. y and z in N. ((hog) o f ) (x) = (hog) ( f (x)) = (hog) (2x) = h ( g (2x)) = h(3(2x) + 4) = h(6x + 4) = sin (6x + 4). c} → {apple. ∀ x in X. 2. then ho(gof ) = (hog) o f. ∀ x in X and (hog) of (x) = hog (f (x)) = h(g (f (x))). 3} → {a. Solution We have ho(gof) (x) = h(gof (x)) = h(g (f (x))) = h (g (2x)) = h(3(2x) + 4) = h(6x + 4) = sin (6x + 4) ∀ x ∈N.

Proof To show that gof is invertible with (gof)–1 = f –1og–1. f and g are bijective functions. 2. g and gof are invertible. 2)}. Find f –1. Determine whether the functions f : S → S defined as below have inverses. It is easy to see that (g o f)–1 o (g o f) = I{1. if it exists. we have seen that f. g –1{ball} = b and g –1{cat} = c. ball. b. Hence (gof)–1 = f –1 og–1 . b. It is easy to verify that f –1 o f = I{1. gof : {1. by Theorem 1 = (f –1o(g–1og)) of. it is enough to show that ( f –1og–1)o(gof) = IX and (gof)o( f –1og–1) = IZ. (a) f = {(1. ball. Then gof is also invertible with (gof)–1 = f –1og–1. (3. 3). RELATIONS AND FUNCTIONS 17 Solution Note that by definition. (2. . so that f is invertible with f –1 = {(3. 2. cat} → {a. by definition of g–1 = IX. c} → (1. 3} → {apple. f –1{c} = 3. g –1{apple} = a. by Theorem 1 = (f –1 o IY) of. 2). 1). (3. f –1og–1 (apple)= f –1(g–1(apple)) = f –1(a) = 1 = (gof)–1 (apple) f –1og–1 (ball) = f –1(g–1(ball)) = f –1(b) = 2 = (gof)–1 (ball) and f –1og–1 (cat) = f –1(g–1(cat)) = f –1(c) = 3 = (gof)–1 (cat). We can define (gof)–1 : {apple. 2. ball. gof (3) = cat. 3}. 1)} (c) f = {(1. Now. g –1og = I{a. f o f –1 = I{a. where. cat}. c}. c} be defined as f –1{a} = 1. 1)} Solution (a) It is easy to see that f is one-one and onto. so that f is invertible with the inverse f –1 of f given by f –1 = {(1. Example 28 Let S = {1. c} and g o g–1 = ID. cat} → {1. 2. Thus. 3} and (gof) o (gof)–1 = ID. so that f is not invertible. Similarly. (gof)–1 (ball) = 2 and (g o f)–1 (cat) = 3. 1). f is not one-one. (2. 3} by (gof)–1 (apple) = 1. cat} is given by gof (1) = apple. (f –1og –1) o (gof) = ((f –1og–1) og) of. (3. 2. f –1{b} = 2. 3}. ball. 1). Let f –1: {a. (1. 2. (2. (2. 3} and g–1 : {apple. (b) Since f (2) = f (3) = 1. gof (2) = ball. D = {apple. 2). 3)} (b) f = {(1. 3)} = f. 3). (c) It is easy to see that f is one-one and onto. 2). Now. it can be shown that (gof ) o (f –1 o g –1) = IZ. 2). b. (2. Now. (3. 1). b. The above result is true in general situation also. Theorem 2 Let f : X → Y and g : Y → Z be two invertible functions.

given by f (x) = is one-one. 3} be given by f = {(1. 2. (4. 4. State with reason whether following functions have inverse (i) f : {1. 9. (3. 3. 3 (4 x + 3) 2 2 4. 1] → Range f. g) o h = (foh) . Find the inverse of f. (goh) 3. . 8. 2. 8} → {1. 3. 3). where R+ is the set of all non-negative real numbers. Show that f is invertible. 10)} (ii) g : {5. 4). What is the (6 x − 4) 3 3 inverse of f ? 5. (4. Show that (f + g) o h = foh + goh (f . Write down gof. Let f. 7. Show that f : [–1. g and h be functions from R to R. 10). x 2y (Hint: For y ∈ Range f. x ≠ . 11). (5. i. Consider f : R → R given by f (x) = 4x + 3. 4} → {10} with f = {(1. (5. 13)} x 6. 2)} (iii) h : {2. 9). (8. 10). Find the inverse ( x + 2) of the function f : [–1. 10).18 MATHEMATICS EXERCISE 1. (3. x = ) x+2 (1 − y ) 7. 3. If f (x) = . 6.. 3). 5} → {1. 4). 1] → R. for all x ≠ . (3. for some x in [–1. 11. 1)} and g = {(1.e. 2. ∞) given by f (x) = x2 + 4. (4. show that fof (x) = x. Let f : {1. 4} → {1. Show that f is invertible with the inverse f –1 of f given by f –1(y) = y − 4 . (6. Find gof and fog. 2. 2). 3). 2. 1]. (7. y = f (x) = . 5} and g : {1. (2. 13} with h = {(2. if (i) f (x) = | x | and g(x) = | 5x – 2 | 1 (ii) f (x) = 8x and g(x) = x 3 . (2.3 1. Consider f : R+ → [4. 5). 4} with g = {(5. 3. 5} → {7. 7). 1)}.

multiplication. Find f –1 and show that (f –1)–1 = f. then fof (x) is 1 (A) x 3 (B) x 3 (C) x (D) (3 – x3). ⎧ 4⎫ 4x 14. 3} → {a.5 Binary Operations Right from the school days. Thus.. addition. 12. The main feature of these operations is that given any two numbers a and b. subtraction. Show that f has unique inverse. ⎝ 3 ⎠ 10. you must have come across four fundamental operations namely addition. ∞) given by f (x) = 9x2 + 6x – 5. we first add two numbers and the result is then added to the third number. b ≠ 0. 1 13. Consider f : {1. (f –1)–1 = f. Use one-one ness of f). c} given by f (1) = a. i. (Hint: suppose g1 and g2 are two inverses of f. The inverse of ⎩ 3⎭ 3x + 4 ⎧ 4⎫ f is the map g : Range f → R – ⎨− ⎬ given by ⎩ 3⎭ 3y 4y (A) g ( y) = (B) g ( y) = 3 − 4y 4 − 3y 4y 3y (C) g ( y) = (D) g ( y) = 3 − 4y 4 − 3y 1. Let f : X → Y be an invertible function.e. Show that f is invertible ⎛ ( y + 6 ) −1 ⎞ with f –1(y) = ⎜ ⎟. 11. f (2) = b and f (3) = c. When we need to add three numbers. Show that the inverse of f –1 is f. If f : R → R be given by f (x) = (3 − x3 ) 3 . Let f : R – ⎨− ⎬ → R be a function defined as f (x) = . It is to be noted that only two numbers can be added or b multiplied at a time. we associate another number a + b a or a – b or ab or . subtraction . b. RELATIONS AND FUNCTIONS 19 9. Consider f : R+ → [– 5. 2. Then for all y ∈ Y. multiplication and division. fog1(y) = 1Y(y) = fog2(y). Let f : X → Y be an invertible function.

b) → a ÷ b 3 ∉ N. 5) under ‘–’ is 3 – 5 = – 2 ∉ N. given by (a. is not a function and hence not a binary b a operation. as the image of (3. ÷ : R∗ × R∗ → R∗. 20 MATHEMATICS and division are examples of binary operation. they are binary operations on R. subtraction and multiplication are binary operations on R. ‘–’ and ‘×’ are functions. b) → a – b. . b) → is a function and hence a b binary operation on R∗. Solution + : R × R → R is given by (a. b) → . given by (a. Example 30 Show that subtraction and division are not binary operations on N. b) to a unique element a + 4b2 in R. given by (a. This gives rise to a general definition as follows: Definition 10 A binary operation ∗ on a set A is a function ∗ : A × A → A. is not defined. b) by a ∗ b. is not binary operation. Similarly. Example 29 Show that addition. but division is not a binary operation on R. is not a binary operation. given by (a. If we want to have a general definition which can cover all these four operations. b) → a + 4b2 is a binary operation. b a However. 5) under ÷ is 3 ÷ 5 = 5 Example 31 Show that ∗ : R × R → R given by (a. ÷ : N × N → N. a But ÷: R × R → R. Solution – : N × N → N. as the image of (3. b) → a – b × : R × R → R is given by (a. as ‘binary’ means two. as for b = 0. We denote ∗ (a. ∗ is a binary operation on R. b) → a + b – : R × R → R is given by (a. Further. b) → ab Since ‘+’. then the set of numbers is to be replaced by an arbitrary set X and then general binary operation is nothing but association of any pair of elements a. show that division is a binary operation on the set R∗ of nonzero real numbers. Solution Since ∗ carries each pair (a. b from X to another element of X.

B) → A ∪ B and ∩ : P × P → P given by (A. This can be generalised for general operation ∗ : A × A → A. B) → A ∩ B are binary operations on the set P. ‘subtract 4 from 3’. j)th entry being ai ∗ aj. 2. Solution Since union operation ∪ carries each pair (A.. 7) = 7. B) in P × P to a unique element A ∩ B in P. – 7) = – 7.e.. 3) = 3. a2. For example consider A = {1. ∨ (1. If A = {a1. an}. . b} are binary operations. For subtraction and division we have to write ‘subtract 3 from 4’. the intersection operation ∩ carries each pair (A. b) → max {a. b) → min {a. . an}. but subtraction of 3 and 4 in different order give different results. we can define a binary operation ∗ : A × A → A given by ai ∗ aj = the entry in the ith row and jth column of the operation table. Using the similar argument. addition and multiplication of 3 and 4 are meaningful. ∨ (2. Similarly. ∪ is binary operation on P. Similarly. i. ∧ (4. Then. a2.. in case of multiplication of 3 and 4. Table 1. ∨ (1. Remark ∨ (4.1 Here. ∨ (4.. Solution Since ∨ carries each pair (a. order is immaterial. . Show that ∪ : P × P → P given by (A. 3) = 3. but division of 3 and 4 in different order give different results. RELATIONS AND FUNCTIONS 21 Example 32 Let P be the set of all subsets of a given set X.. j) the entry of the table being maximum of ith and jth elements of the set A. ∩ is a binary operation on P. b) in R × R to a unique element namely maximum of a and b lying in R. Conversely. but subtraction and division of 3 and 4 are meaningless. One may note that 3 and 4 can be added in any order and the result is same. i.e. ∨ is a binary operation. – 7) = 4. ‘divide 3 by 4’ or ‘divide 4 by 3’. 3}. Thus. 3 + 4 = 4 + 3. the operation ∨ on A defined in Example 33 can be expressed by the following operation table (Table 1. b} and the ∧ : R × R → R given by (a. Example 33 Show that the ∨ : R × R → R given by (a.. Here. When number of elements in a set A is small. 2) = 2. Then the operation table will be having n rows and n columns with (i.. 3 – 4 ≠ 4 – 3.1) . one can say that ∧ is also a binary operation. B) in P × P to a unique element A ∪ B in P. we are having 3 rows and 3 columns in the operation table with (i.. given any operation table having n rows and n columns with each entry being an element of A = {a1. we can express a binary operation ∗ on the set A through a table called the operation table for the operation ∗. 7) = 4 and ∧ (4.

the expression a1 ∗ a2 ∗ . Thus. This leads to the following: Definition 12 A binary operation ∗ : A × A → A is said to be associative if (a ∗ b) ∗ c = a ∗ (b ∗ c). ∀ a. subtraction and division are not associative. But in case of addition. since 3 – 4 ≠ 4 – 3. Example 34 Show that + : R × R → R and × : R × R → R are commutative binary operations. since (8 ∗ 5) ∗ 3 = (8 + 10) ∗ 3 = (8 + 10) + 6 = 24. c ∈ R. . Therefore. association of three numbers 8. 22 MATHEMATICS This leads to the following definition: Definition 11 A binary operation ∗ on the set X is called commutative. we can write a1 ∗ a2 ∗ . 3 ÷ 4 ≠ 4 ÷ 3 shows that ‘÷’ is not commutative.. Solution Since a + b = b + a and a × b = b × a. c. b. we encounter a natural problem. 8 + 5 + 2 has the same value whether we look at it as ( 8 + 5) + 2 or as 8 + (5 + 2). Solution Addition and multiplication are associative. Recall that in the earlier classes brackets were used whenever subtraction or division operations or more than one operation occurred. unless bracket is used. ∈ A. However. as (8 – 5) – 3 ≠ 8 – (5 – 3) and (8 ÷ 5) ÷ 3 ≠ 8 ÷ (5 ÷ 3). (8 – 5) – 2 ≠ 8 – (5 – 2). Solution The operation ∗ is not associative. The expression a ∗ b ∗ c may be interpreted as (a ∗ b) ∗ c or a ∗ (b ∗ c) and these two expressions need not be same. ‘+’ and ‘×’ are commutative binary operation. Example 37 Show that ∗ : R × R → R given by a ∗ b → a + 2b is not associative. 5 and 3 through the binary operation ‘subtraction’ is meaningless. But subtraction is not associative on R. b. But in absence of this property. Division is not associative on R∗. b ∈ R.. showing that the operation ∗ is not commutative. For example. Similarly. for every a. while 8 ∗ (5 ∗ 3) = 8 ∗ (5 + 6) = 8 ∗ 11 = 8 + 22 = 30. Remark Associative property of a binary operation is very important in the sense that with this property of a binary operation. ∗ an which is not ambiguous. since (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c) ∀ a. Solution Since 3 ∗ 4 = 3 + 8 = 11 and 4 ∗ 3 = 4 + 6 = 10. association of 3 or even more than 3 numbers through addition is meaningful without using bracket. b ∈ X. if a ∗ b = b ∗ a. Example 35 Show that ∗ : R × R → R defined by a ∗ b = a + 2b is not commutative. ∗ an is ambiguous unless brackets are used. However.. but – : R × R → R and ÷ : R∗ × R∗ → R∗ are not commutative. ∀ a. Example 36 Show that addition and multiplication are associative binary operation on R. If we want to associate three elements of a set X through a binary operation on X. ‘–’ is not commutative..

as 0 ∉ N. given any a ≠ 0 in R. Remark Zero is identity for the addition operation on R but it is not identity for the addition operation on N. for a ≠ 0. there exists – a in R such that a + (– a) = 0 (identity for ‘+’) = (– a) + a. a a a . Solution a + 0 = 0 + a = a and a × 1 = a = 1 × a. if a ∗ e = a = e ∗ a. the number 1 plays this role. an element a ∈ A is said to be invertible with respect to the operation ∗. for the multiplication operation on R. ∀ a in R∗. But there is no identity element for the operations – : R × R → R and ÷ : R∗ × R∗ → R∗. we can not find any element e in R∗ such that a ÷ e = e ÷ a. This leads to the following definition: a a Definition 14 Given a binary operation ∗ : A × A → A with the identity element e in A. Further. 1 Similarly. any number remains unaltered by adding zero. as a × 1 = a = 1 × a. Example 39 Show that – a is the inverse of a for the addition operation ‘+’ on R and 1 is the inverse of a ≠ 0 for the multiplication operation ‘×’ on R. there is no element e in R with a – e = e – a. This leads to the following definition: Definition 13 Given a binary operation ∗ : A × A → A. One further notices that for the addition operation + : R × R → R. 1 1 1 Similarly. if it exists. ∀ a ∈ A. a × = 1 = × a implies that is the inverse of a for multiplication. Similarly. is called identity for the operation ∗. – a is the inverse of a for addition. In fact the addition operation on N does not have any identity. ∀ a ∈ R implies that 0 and 1 are identity elements for the operations ‘+’ and ‘×’ respectively. Example 38 Show that zero is the identity for addition on R and 1 is the identity for multiplication on R. But in case of multiplication. Hence. the interesting feature of the number zero is that a + 0 = a = 0 + a.e. we can choose a 1 1 in R such that a × = 1(identity for ‘×’) = × a. i. 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.. ∀ a. an element e ∈ A. ‘–’ and ‘÷’ do not have identity element. RELATIONS AND FUNCTIONS 23 For the binary operation ‘+’ on R. ∀ a in R. a Solution As a + (– a) = a – a = 0 and (– a) + a = 0. given any a ∈ R.

define a ∗ b = 2ab (v) On Z+. b}. (i) On Z. – a can not be inverse of a for addition operation on N. . although – a satisfies a + (– a) = 0 = (– a) + a. for a ≠ 1. define a ∗ b = ab + 1 ab (iii) On Q.4 1. 38 and 39 show that addition on R is a commutative and associative binary operation with 0 as the identity element and – a as the inverse of a in R ∀ a. define ∗ by a ∗ b = a – b (ii) On Z+. define a ∗ b = b +1 3. Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation. which implies that other than 1 no element of N a has inverse for multiplication operation on N. 3. define a ∗ b = a – b (ii) On Q. for a ≠ 1 in N. define a ∗ b = 2 (iv) On Z+. For each binary operation ∗ defined below. 1 Similarly. define ∗ by a ∗ b = a 2. define ∗ by a ∗ b = ab2 (iv) On Z+. 4. (i) On Z+. Consider the binary operation ∧ on the set {1. EXERCISE 1. define ∗ by a ∗ b = ab (iii) On R. 24 MATHEMATICS Example 40 Show that – a is not the inverse of a ∈ N for the addition operation + on 1 N and is not the inverse of a ∈ N for multiplication operation × on N. Write the operation table of the operation ∧ . a Solution Since – a ∉ N. define ∗ by a ∗ b = | a – b | (v) On Z+. 2. Examples 34. give justification for this. ∉ N. determine whether ∗ is commutative or associative. define a ∗ b = ab a (vi) On R – {– 1}. 5} defined by a ∧ b = min {a. 36.

(Hint: use the following table) Table 1. 6.F. Consider a binary operation ∗ on the set {1. Is ∗ defined on the set {1. Let ∗ be a binary operation on the set Q of rational numbers as follows: (i) a ∗ b = a – b (ii) a ∗ b = a2 + b2 (iii) a ∗ b = a + ab (iv) a ∗ b = (a – b)2 ab (v) a ∗ b = (vi) a ∗ b = ab2 4 Find which of the binary operations are commutative and which are associative.C. Find (i) 5 ∗ 7. 10. 3.2 5. Let ∗ be the binary operation on N defined by a ∗ b = H. 5} defined by a ∗′ b = H. 4.M. of a and b. 20 ∗ 16 (ii) Is ∗ commutative? (iii) Is ∗ associative? (iv) Find the identity of ∗ in N (v) Which elements of N are invertible for the operation ∗? 7. Let A = N × N and ∗ be the binary operation on A defined by (a. 5} by a ∗ b = L. 3. Is ∗ commutative? Is ∗ associative? Does there exist identity for this binary operation on N? 9.2). 2.C. 4. 2. Let ∗ be the binary operation on N given by a ∗ b = L.F. RELATIONS AND FUNCTIONS 25 4. 11. of a and b. Let ∗′ be the binary operation on the set {1.C.M. of a and b. 2.C. of a and b a binary operation? Justify your answer. (i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4) (ii) Is ∗ commutative? (iii) Compute (2 ∗ 3) ∗ (4 ∗ 5). 5} given by the following multiplication table (Table 1. 8. 4. Is the operation ∗′ same as the operation ∗ defined in Exercise 4 above? Justify your answer. 3. b + d) . d) = (a + c. Show that none of the operations given above has identity. b) ∗ (c.

7}} or {x. a) ∈ R2 ⇒ (b. Similarly. hence. ∀ (x. y} ⊂ {2. 8. 5. This shows that R is symmetric. 4. and (a. (x. v) and (u. a ∗ a = a ∀ a ∈ N. Further. b) ∈ R1 ∩ R2 and (b. v) R (a. a) ∈ R1. Solution Since R1 and R2 are equivalence relations. (A) Is ∗ both associative and commutative? (B) Is ∗ commutative but not associative? (C) Is ∗ associative but not commutative? (D) Is ∗ neither commutative nor associative? Miscellaneous Examples Example 41 If R1 and R2 are equivalence relations in a set A. y) R (a. Justify. 9}. Thus. Example 42 Let R be a relation on the set A of ordered pairs of positive integers defined by (x. y). c) ∈ R1 ∩ R2. Similarly. R is an equivalence relation. 7. b) ∈ R1 ∩ R2 ⇒ (a. 6. Let R1 be a relation in X given by R1 = {(x. y) R (u. y): {x. a) ∈ R1 ∩ R2. R1 ∩ R2 is symmetric. y} ⊂ {1. b) ⇒ xv = yu and a a b a ub = va ⇒ xv = yu ⇒ xv = yu ⇒ xb = ya and hence (x. Show that R is an equivalence relation. y) R (u. This shows that R1 ∩ R2 is transitive. R1 ∩ R2 is an equivalence relation. 4. R u u v u is transitive. . y} ⊂ {3. Consider a binary operation ∗ on N defined as a ∗ b = a3 + b3. 8} or {x. b). b) ∈ R2 ⇒ (b. a) ∈ R1 and (b. a) ∈ R2 ∀ a ∈ A. Choose the correct answer. v) R (x. then a ∗ (b ∗ c) = (c ∗ b) ∗ a 13. (a. Thus. (x. 12. c) ∈ R1 ∩ R2 ⇒ (a. since xy = yx. State whether the following statements are true or false. Thus. 3. This shows that R is reflexive. showing R1 ∩ R2 is reflexive. a) ∈ R1 ∩ R2. 5. y). v) if and only if xv = yu. y) ∈ A. c) ∈ R2 ⇒ (a. (x. Find the identity element for ∗ on A. y) R (x. ∀ a. 9}}. (ii) If ∗ is a commutative binary operation on N. Show that R1 = R2. y) R (u. 6. Solution Clearly. 2. (i) For an arbitrary binary operation ∗ on a set N. Example 43 Let X = {1. if any. 26 MATHEMATICS Show that ∗ is commutative and associative. This implies that (a. b) ∈ R1 and (a. Further. (a. y) : x – y is divisible by 3} and R2 be another relation on X given by R2 = {(x. show that R1 ∩ R2 is also an equivalence relation. c) ∈ R1 and (a. (a. v) ⇒ xv = yu ⇒ uy = vx and hence (u.

b) ∈ R ⇒ f (a) = f (b) ⇒ f (b) = f (a) ⇒ (b. 2 4 4 Hence. shows that ∗ is commutative. y} ⊂ {2. (a. Hence. R1 = R2. y} ⊂ {3. 5. Solution For every a ∈ X. b ∈ N 2 Solution (a) Clearly. Further. y) ∈ R2. a+b b+a (b) a ∗ b = = = b ∗ a. y} ⊂ {1. R1 ⊂ R2. (a. Similarly. 7} or {x. Define a relation R in X given by R = {(a. a) ∈ R. 2 2 ⎛a+b⎞ (a ∗ b) ∗ c = ⎜ ⎟ ∗ c. 7}. y} ⊂ {3. by definition a ∗ b = b ∗ a = 1. 4. y} ⊂ {1. Examine if R is an equivalence relation. RELATIONS AND FUNCTIONS 27 Solution Note that the characteristic of sets {1. c ∈ N. Also (a ∗ b) ∗ c = (1 ∗ c) =1 and a ∗ (b ∗ c) = a ∗ (1) = 1. Hence R is both associative and commutative. b ∈ N. b ∈ N (b) a ∗ b = ∀ a. 9} ⇒ x – y is divisible by 3 ⇒ {x. Therefore. 8} and {3. 6. since f (a) = f (a). ⎝ 2 ⎠ ⎛a+b⎞ ⎜ ⎟ + c a + b + 2c ⎝ 2 ⎠ = = . 7} or {x. 5. ∀ a. ∗ is not associative. 4. 6. Hence. 8} or {x. (x. a) ∈ R. which implies that R is transitive. b. c) ∈ R. Hence. R is symmetric. y} ∈ R2 ⇒ {x. Further. y) ∈ R1 ⇒ x – y is a multiple of 3 ⇒ {x. (a + b ) (a) a ∗ b = 1 ∀ a. b): f(a) = f(b)}. Example 45 Determine which of the following binary operations on the set N are associative and which are commutative. 8} or {x. 9} ⇒ (x. (a. . 4. This shows that R2 ⊂ R1. Therefore. 2 4 ⎛b+c⎞ But a ∗ (b ∗ c) = a ∗ ⎜ ⎟ ⎝ 2 ⎠ b+c a+ = 2 = 2a + b + c ≠ a + b + 2c in general. 5. 6. ∀ a. Similarly. b) ∈ R and (b. Example 44 Let f : X → Y be a function. showing that R is reflexive. {x. 9} is that difference between any two elements of these sets is a multiple of 3. y} ⊂ {2. R is an equivalence relation. {2. y} ∈ R1. c) ∈ R ⇒ f (a) = f (b) and f (b) = f (c) ⇒ f (a) = f (c) ⇒ (a.

Solution Clearly IN is onto. 1). Thus. (2. 1) to R1 at a time. we can not add any two pairs out of (2. we will be forced to add the remaining third pair in order to maintain transitivity and in the process. 1) to R1 to get R2. 1) = 1. 2). 1). (2. 2). 2). 3) which are reflexive and transitive but not symmetric is four. But IN + IN is not onto. then for symmetry we must add (3. 2) and (3. 3) and (3. 2). 1) = 2 and the only choice left is for the pair (2. 28 MATHEMATICS Example 46 Find the number of all one-one functions from set A = {1. (2. total number of one-one maps from {1. Similarly. (1. 2} is a function from {1. 2) and (3. Now. (3. to R1 to get the desired relations. 3} to itself. 3) and (3. 3). 3} containing (1. This shows that the total number of equivalence relations containing (1.e. 2. 3} to itself is simply a permutation on three symbols 1. 1) respectively. i. (2. (2. the relation will become symmetric also which is not required. 2) also and now for transitivity we are forced to add (1. 1). i. Then show that the number of relations containing (1. Solution One-one function from {1. (1. ∗ (1. say (2. as by doing so. (2. 3}. 2). 1) is two. 2. 3). 2) and (2. we can obtain R3 and R4 by adding (3. 2) = 2. 2} to {1. 3)}. (1. 1) is two. 3) to R1. 2}. Example 50 Consider the identity function IN : N → N defined as IN (x) = x ∀ x ∈ N. 2. 2)} → {1. 1). 3. 2. Thus. 2) and (2. 1) is {(1. Therefore. ∗ (1.. 3) which is reflexive and transitive but not symmetric is {(1. 1). a function from {(1. . 2). 2. Solution The smallest equivalence relation R1 containing (1. Solution The smallest relation R1 containing (1.e. transitive but not symmetric. Thus. However. Since 2 is the inverse of 2. 3 which is 3! = 6. (1. ∗ (2. Example 49 Show that the number of binary operations on {1. (3. 3). 1). If we add any one. the total number of desired relations is four. then the relation R2 will be reflexive. Show that although IN is onto but IN + IN : N → N defined as (IN + IN) (x) = IN (x) + IN (x) = x + x = 2x is not onto. 2) must be equal to 1. Example 47 Let A = {1. 2}. 3} to itself is same as total number of permutations on three symbols 1. Solution A binary operation ∗ on {1. 2. (1. ∗ (2. (3. the only equivalence relation bigger than R1 is the universal relation. 3). 2} × {1. 2) and (2. 2} having 1 as identity and having 2 as the inverse of 2 is exactly one. (3.. the number of desired binary operation is only one. Since 1 is the identity for the desired binary operation ∗. Now we are left with only 4 pairs namely (2. Example 48 Show that the number of equivalence relation in the set {1. 1). 2). 2) and (2. as we can find an element 3 in the co-domain N such that there does not exist any x in the domain N with (IN + IN) (x) = 2x = 3. 2) and (2. 1)}. 2. if we add the pair (2.

⎥ . if n is odd and f (n) = n + 1. 1+ | x | x ∈ R is one one and onto function. 3. Let f : W → W be defined as f (n) = n – 1. f + g is not one-one. Therefore. if n is even. 2. If f : R → R is defined by f(x) = x2 – 3x + 2. Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by f ( x ) = . Give examples of two functions f : N → N and g : N → N such that g o f is onto but f is not onto. 7. . Show that the function f : R → R given by f (x) = x3 is injective. but f + g is not ⎢⎣ 2 ⎥⎦ one-one. ⎧ x − 1 if x > 1 (Hint : Consider f (x) = x + 1 and g ( x ) = ⎨ ⎩ 1 if x = 1 8. Give examples of two functions f : N → Z and g : Z → Z such that g o f is injective but g is not injective. Here. ⎡ π⎤ Solution Since for any two distinct elements x1 and x2 in ⎢ 0. Find the inverse of f. Show that f and g are one-one. 6. ⎝2⎠ 2 2 Miscellaneous Exercise on Chapter 1 1. RELATIONS AND FUNCTIONS 29 ⎡ π⎤ Example 51 Consider a function f : ⎢ 0. Let f : R → R be defined as f (x) = 10x + 7. consider P(X) which is the set of all subsets of X. Find the function g : R → R such that g o f = f o g = 1R. both f and g must be one-one. find f (f (x)). ⎥ → R given by f (x) = sin x and ⎣ 2⎦ π g : ⎡ 0. sin x1 ≠ sin x2 and ⎣ 2⎦ cos x1 ≠ cos x2. Show that f is invertible. 5. W is the set of all whole numbers. ⎤ → R given by g(x) = cos x. (Hint : Consider f (x) = x and g (x) = | x |). But (f + g) (0) = sin 0 + cos 0 = 1 and ⎛ π⎞ π π (f + g) ⎜ ⎟ = sin + cos = 1 . Given a non empty set X. x 4.

B in P(X). x ∈ A. Given a non-empty set X. x ∈ A and g ( x) = 2 x − − 1. 2. B ∈ P(X). Given a non-empty set X. Are f and g equal? 2 Justify your answer. n} to itself. (c. if it exists. .30 MATHEMATICS Define the relation R in P(X) as follows: For subsets A. 0. 3}. Then number of equivalence relations containing (1. 2}. Let A = {1. 1)} (ii) F = {(a. if a + b < 6 a ∗b = ⎨ ⎩ a + b − 6 if a + b ≥ 6 Show that zero is the identity for this operation and each element a of the set is invertible with 6 – a being the inverse of a. Further. 4. 2) and (1. 2} and f. Is R an equivalence relation on P(X)? Justify your answer. 1. (i) F = {(a. B in P(X). let ∗ : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A). 0. – 2. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation ∗. Show that ∗ is commutative but not associative. c ∈ R. 2.. 3}. (Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ). 2) is (A) 1 (B) 2 (C) 3 (D) 4 . (b. (c. 1)} 12. Then number of relations containing (1. ∀ a. 2. 10. c} and T = {1. Let A = {1. B = {– 4. Does o distribute over ∗? Justify your answer. 1. Find F–1 of the following functions F from S to T. 2). we say that the operation ∗ distributes over the operation o]. consider the binary operation ∗ : P(X) × P(X) → P(X) given by A ∗ B = A ∩ B ∀ A. 3) which are reflexive and symmetric but not transitive is (A) 1 (B) 2 (C) 3 (D) 4 17. 9. 3). (b. 1). . (Hint: One may note that two functions f : A → B and g : A → B such that f (a) = g (a) ∀ a ∈ A. ∀ A. [If it is so. where P(X) is the power set of X. ARB if and only if A ⊂ B. 3}. 16. b ∈ R. 11. 3. b. Consider the binary operations ∗ : R × R → R and o : R × R → R defined as a ∗b = |a – b| and a o b = a. 13. 15. 2. b. 2). 2. Define a binary operation ∗ on the set {0. g : A → B be functions defined 1 by f (x) = x2 – x. Let A = {– 1.. 14. Let S = {a. Find the number of all onto functions from the set {1. show that ∀ a. are called equal functions). o is associative but not commutative. 3. 5} as ⎧ a + b. a ∗ (b o c) = (a ∗ b) o (a ∗ b). Show that the empty set φ is the identity for the operation ∗ and all the elements A of P(X) are invertible with A–1 = A.

we studied different types of relations and equivalence relation. a) ∈ R ∀ a ∈ X. ∃ x ∈ X such that f (x) = y. . symmetric and transitive. Let f : R → R be the Signum Function defined as ⎧ 1.  Symmetric relation R in X is a relation satisfying (a.  A function f : X → Y is invertible if ∃ g : Y → X such that gof = IX and fog = IY. b) ∈ R implies (b. if f is both one-one and onto.  Universal relation is the relation R in X given by R = X × X. x > 0 ⎪ f ( x ) = ⎨ 0.  Equivalence class [a] containing a ∈ X for an equivalence relation R in X is the subset of X containing all elements b related to a. does fog and gof coincide in (0. composition of functions. b} are (A) 10 (B) 16 (C) 20 (D ) 8 Summary In this chapter. x = 0 ⎪−1. a) ∈ R.  Reflexive relation R in X is a relation with (a. Number of binary operations on the set {a. invertible functions and binary operations.  The composition of functions f : A → B and g : B → C is the function gof : A → C given by gof (x) = g(f (x)) ∀ x ∈ A.  A function f : X → Y is one-one and onto (or bijective). The main features of this chapter are as follows:  Empty relation is the relation R in X given by R = φ ⊂ X × X.  A function f : X → Y is invertible if and only if f is one-one and onto.  Equivalence relation R in X is a relation which is reflexive. where [x] is greatest integer less than or equal to x. RELATIONS AND FUNCTIONS 31 18.  A function f : X → Y is onto (or surjective) if given any y ∈ Y.  Transitive relation R in X is a relation satisfying (a. b) ∈ R and (b. x2 ∈ X.  A function f : X → Y is one-one (or injective) if f (x1) = f (x2) ⇒ x1 = x2 ∀ x1. c) ∈ R. Then. c) ∈ R implies that (a. 1]? 19. x < 0 ⎩ and g : R → R be the Greatest Integer Function given by g (x) = [x].

But the general adoption of symbols like f.. who used the word ‘function’ in his manuscript “Geometrie” in 1637 to mean some positive integral power xn of a variable x while studying geometrical curves like hyperbola. c in X.  An element a ∈ X is invertible for binary operation ∗ : X × X → X. The element b is called inverse of a and is denoted by a–1. John Bernoulli (1667-1748) used the notation φx for the first time in 1718 to indicate a function of x. James Gregory (1636-1675) in his work “ Vera Circuli et Hyperbolae Quadratura” (1667) considered function as a quantity obtained from other quantities by successive use of algebraic operations or by any other operations. F(x). b in X. Descartes (1596-1650). was given after set theory was developed by Georg Cantor (1845-1918). This is not true for infinite set  A binary operation ∗ on a set A is a function ∗ from A × A to A.  An operation ∗ on X is commutative if a ∗ b = b ∗ a ∀ a. to represent functions was made by Leonhard Euler (1707-1783) in 1734 in the first part of his manuscript “Analysis Infinitorium”. φ (x) etc. The set theoretic definition of function known to us presently is simply an abstraction of the definition given by Dirichlet in a rigorous manner. Lejeunne Dirichlet (1805-1859) gave the definition of function which was being used till the set theoretic definition of function presently used. However. where he discussed about analytic function and used the notion f (x). Historical Note The concept of function has evolved over a long period of time starting from R. for different function of x. e is the identity for the binary operation ∗. the slope of the curve. Later G. He was the first to use the phrase ‘function of x’. in his manuscript “Historia” (1714). Leibnitz used the word ‘function’ to mean quantities that depend on a variable.  An operation ∗ on X is associative if (a ∗ b) ∗ c = a ∗ (b ∗ c) ∀ a. Later on. seu de functionibus” written in 1673 used the word ‘function’ to mean a quantity varying from point to point on a curve such as the coordinates of a point on the curve. Joeph Louis Lagrange (1736-1813) published his manuscripts “Theorie des functions analytiques” in 1793. This is the characteristic property of a finite set. parabola and ellipse. ψ . if there exists b ∈ X such that a ∗ b = e = b ∗ a where. Leibnitz (1646-1716) in his manuscript “Methodus tangentium inversa. a function f : X → X is one-one (respectively onto) if and only if f is onto (respectively one-one). Subsequently. the tangent and the normal to the curve at a point. W. b.. φ. —™ — . if a ∗ e = a = e ∗ a ∀ a ∈ X.32 MATHEMATICS  Given a finite set X.  An element e ∈ X is the identity element for binary operation ∗ : X × X → X. F.