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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.

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

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

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

(A) R is reflexive and symmetric but not transitive. (4. RELATIONS AND FUNCTIONS 7 15. Otherwise. for every y ∈ Y. constant function. Addition. 3). 7) ∈ R 1. 4} given by R = {(1.2 (iii). 2). i. namely b.2 (ii) and (iii) are many-one. (1. modulus function. we would like to study different types of functions. there exists an element x in X such that f (x) = y. The function f1 and f4 in Fig 1. (C) R is symmetric and transitive but not reflexive. In this section. if every element of Y is the image of some element of X under f.e. along with their graphs have been given in Class XI. 4) ∈ R (B) (3. . 2. As the concept of function is of paramount importance in mathematics and among other disciplines as well. Choose the correct answer. In Fig 1. (iv) are onto and the function f1 in Fig 1. for every x1. Let R be the relation in the set N given by R = {(a. f2. f is called many-one.3 Types of Functions The notion of a function along with some special functions like identity function. there are some elements like e and f in X2 which are not images of any element of X1 under f1. polynomial function. x2 ∈ X. i. 2)}. 3). b > 6}. while all elements of X3 are images of some elements of X1 under f3. Choose the correct answer. 8) ∈ R (D) (8. Definition 6 A function f : X → Y is said to be onto (or surjective). (B) R is reflexive and transitive but not symmetric. signum function etc. f3 and f4 given by the following diagrams.. (3. (A) (2. Consider the functions f1. subtraction. but the image of two distinct elements 1 and 2 of X1 under f2 is same. 1). multiplication and division of two functions have also been studied. (1. f (x1) = f (x2) implies x1 = x2.2 (i) and (iv) are one-one and the function f2 and f3 in Fig 1. Let R be the relation in the set {1. 8) ∈ R (C) (6.4).2 (i) is not onto as elements e. The above observations lead to the following definitions: Definition 5 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. (D) R is an equivalence relation. we observe that the images of distinct elements of X1 under the function f1 are distinct. (2. 2). b) : a = b – 2. Further. The function f3 and f4 in Fig 1. 16. we would like to extend our study about function from where we finished earlier. rational function. f in X2 are not the image of any element in X1 under f1.e.2. 3. (3.

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

if x is even is both one-one and onto. Example 12 Show that f : N → N. is neither one-one nor onto.if x is odd. Therefore f is not onto. given any real y y y number y in R. Solution f is not one-one. Solution f is one-one. Fig 1. as given any y ∈ N. f ( x) = ⎨ ⎩ x − 1. Example 11 Show that the function f : R → R. 2 2 2 Fig 1.3 Example 10 Show that the function f : N → N. the element – 2 in the co-domain R is not image of any element x in the domain R (Why?). given by f (x) = 2x. given by f (1) = f (2) = 1 and f (x) = x – 1. RELATIONS AND FUNCTIONS 9 Example 9 Prove that the function f : R → R. y ≠ 1. Solution Since f (– 1) = 1 = f (1). is onto but not one-one. there exists in R such that f ( ) = 2 . Hence. defined as f (x) = x2.4 . Also. we can choose x as y + 1 such that f (y + 1) = y + 1 – 1 = y. f is onto. as f (1) = f (2) = 1. But f is onto. given by ⎧ x + 1. is one-one and onto. we have f (1) = 1. ( ) = y. f is not one- one. for every x > 2. Also. Also for 1 ∈ N. as f (x1) = f (x2) ⇒ 2x1 = 2x2 ⇒ x1 = x2.

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

if x is positive or 0 and | x | is – x. (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. 7. Let f : R → R be defined as f (x) = 3x. where | x | is x. Let f : N → N be defined by f (n) = ⎨ for all n ∈ N. b) = (b. if x > 0 ⎪ f ( x) = ⎨0. Let A and B be sets. ⎧n +1 ⎪⎪ 2 . Justify your answer. Let f : R → R be defined as f(x) = x4. Justify your answer. 6)} be a function from A to B. Show that f is one-one. Let A = R – {3} and B = R – {1}. Show that the Modulus Function f : R → R. 7} and let f = {(1. Show that f : A × B → B × A such that f (a. (i) f : R → R defined by f (x) = 3 – 4x (ii) f : R → R defined by f (x) = 1 + x2 8. 5. Is f one-one and onto? Justify your answer. ⎪ n . onto or bijective. given by f (x) = | x |. Let A = {1. 6. . if n is even ⎪⎩ 2 State whether the function f is bijective. if x < 0 ⎩ is neither one-one nor onto. if x is negative. Show that the Signum Function f : R → R. is neither one- one nor onto. RELATIONS AND FUNCTIONS 11 4. In each of the following cases. Consider the function f : A → B defined by ⎛ x−2⎞ f (x) = ⎜ ⎟ . ⎝ x−3⎠ 11. Choose the correct answer. if n is odd 9. (2. (3. 4). 10. 3}. B = {4. 5). if x = 0 ⎪ –1. a) is bijective function. state whether the function is one-one. 2. 12. 6. Choose the correct answer. 5. given by ⎧1. (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.

3. 5. Similarly. 11. Example 16 Find gof and fog. f (4) = f (5) = 5 and g (3) = g (4) = 7 and g (5) = g (9) = 11. . 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. Show that gof ≠ fog. 15} be functions defined as f (2) = 3. each student is eventually attached a code number. Hence. 9} → {7. 5. 4. 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. denoted by gof. gof (3) = g (f (3)) = g (4) = 7. who appeared in Class X of a Board Examination in 2006. Find gof. 12 MATHEMATICS 1. In order to have confidentiality. if f : R → R and g : R → R are given by f (x) = cos x and g (x) = 3x2. Note that 3cos2 x ≠ cos 3x2. f (3) = 4. 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. Solution We have gof (x) = g (f (x)) = g (cos x) = 3 (cos x)2 = 3 cos2 x. for x = 0. ∀ x ∈ A. 4. Fig 1.4 Composition of Functions and Invertible Function In this section. Let B ⊂ N be the set of all roll numbers and C ⊂ N be the set of all code numbers. Solution We have gof (2) = g (f (2)) = g (3) = 7. fog (x) = f (g (x)) = f (3x2) = cos (3x2). 4. is defined as the function gof : A → C given by gof (x) = g(f (x)). by the combination of these two functions. Thus.5 Example 15 Let f : {2. Then the composition of f and g. Consider the set A of all students. the Board arranges to deface the roll numbers of students in the answer scripts and assigns a fake code number to each roll number. gof (4) = g (f (4)) = g (5) = 11 and gof (5) = g (5) = 11. 5} → {3. 9} and g : {3. gof ≠ fog. This leads to the following definition: Definition 8 Let f : A → B and g : B → C be two functions.

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

4} → {1. while in the reverse process of the composite gof. i. if f : X → Y is a function such that there exists a function g : Y → X such that gof = IX and fog = IY. Then. to get gof. 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. 2. 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. It can be seen that gof is onto but f is not onto. first the reverse process of g is applied and then the reverse process of f. c}. f (2) = 2. 3} → {a. Now. Are f and g both necessarily one-one. Solution Consider f : {1. b. X = {1. 4} → {1. g (1) = 1. 3. which shows that gof is one-one. 3. The above discussion. 4. 5. Similarly. Remark It can be verified in general that gof is one-one implies that f is one-one. But g is clearly not one-one. 6} as g (x) = x. 3} as g (a) = 1. 2. c} be one-one and onto function given by f (1) = a. ∀ x and g : {1. This helps in assigning mark to the student scoring that mark. 2. c} → {1. 3} and Y = {a. even the converse is also true . 4. Example 22 Let f : {1. gof (x) = x ∀ x. 4} → {1. 2. 3. 3. f (2) = b and f (3) = c. 2. 3. 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. Example 20 Consider functions f and g such that composite gof is defined and is one- one. Show that there exists a function g : {a. 2. 4} and g : {1. b. 6} defined as f (x) = x. 2. We observe that while composing f and g. first f and then g was applied. 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. 2. 2. 2. Example 22 and Remark lead to the following definition: .e. then f must be one-one and onto. 14 MATHEMATICS with f (x) = y. Not only this. b. 6} → {1. c} → {1. Further. gof is onto implies that g is onto. gof (x) = g (f (x)) = g (y) = z. 3. for x = 1. showing that gof is onto. 2. 4. Example 21 Are f and g both necessarily onto. the process reverse to f assigns a roll number to the student having that roll number. if gof is onto? Solution Consider f : {1. 3} such that gof = IX and fog = IY. 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. 2. b. 3. After the answer scripts are examined. 5. Solution Consider g : {a. 4 and g (5) = g (6) = 5. Therefore. where. since f is onto. f (3) = f (4) = 3. 3} defined as f (1) = 1. 3. examiner enters the mark against each code number in a mark book and submits to the office of the Board. g (2) = 2 and g (3) = g (4) = 3.. 5. g (b) = 2 and g (c) = 3. 2.

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

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

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

3. (4. (4. 3. . 10). x 2y (Hint: For y ∈ Range f. (5. 2).. If f (x) = . 4} → {10} with f = {(1. Find gof and fog. for some x in [–1. 4} → {1. 10). (2. 2. (8. (3. y = f (x) = . Show that f : [–1. 9). 2. g) o h = (foh) . 2)} (iii) h : {2. 2. 5} → {1. State with reason whether following functions have inverse (i) f : {1. Show that f is invertible. (3. 4). 1] → R.e. i. 10). 3. 11. 4). 13)} x 6. x ≠ . 3). Let f. 3). if (i) f (x) = | x | and g(x) = | 5x – 2 | 1 (ii) f (x) = 8x and g(x) = x 3 . where R+ is the set of all non-negative real numbers. 8. 5). (5. 4} with g = {(5. 1]. 5} → {7. x = ) x+2 (1 − y ) 7. Write down gof. 11). 1)} and g = {(1. for all x ≠ . (2. (3. 7). 3 (4 x + 3) 2 2 4. 13} with h = {(2. 3. 1)}.18 MATHEMATICS EXERCISE 1. What is the (6 x − 4) 3 3 inverse of f ? 5. show that fof (x) = x. 1] → Range f. 4. 6.3 1. Show that f is invertible with the inverse f –1 of f given by f –1(y) = y − 4 . 2. 5} and g : {1. 9. 8} → {1. Find the inverse of f. 3} be given by f = {(1. Consider f : R+ → [4. Let f : {1. (6. (goh) 3. (7. 2. 10)} (ii) g : {5. (4. ∞) given by f (x) = x2 + 4. Find the inverse ( x + 2) of the function f : [–1. given by f (x) = is one-one. Show that (f + g) o h = foh + goh (f . g and h be functions from R to R. Consider f : R → R given by f (x) = 4x + 3. 7. 3).

subtraction. 11. Let f : R – ⎨− ⎬ → R be a function defined as f (x) = . you must have come across four fundamental operations namely addition. we first add two numbers and the result is then added to the third number. addition. Find f –1 and show that (f –1)–1 = f. we associate another number a + b a or a – b or ab or . Consider f : {1. i. Thus. ⎝ 3 ⎠ 10. Show that the inverse of f –1 is f. ⎧ 4⎫ 4x 14. 1 13. 12. Show that f is invertible ⎛ ( y + 6 ) −1 ⎞ with f –1(y) = ⎜ ⎟. Consider f : R+ → [– 5. multiplication and division. 2. If f : R → R be given by f (x) = (3 − x3 ) 3 . The main feature of these operations is that given any two numbers a and b. Use one-one ness of f).e. f (2) = b and f (3) = c. ∞) given by f (x) = 9x2 + 6x – 5. b. (Hint: suppose g1 and g2 are two inverses of f. Let f : X → Y be an invertible function. Let f : X → Y be an invertible function. c} given by f (1) = a. 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.. b ≠ 0. Show that f has unique inverse. multiplication. then fof (x) is 1 (A) x 3 (B) x 3 (C) x (D) (3 – x3). RELATIONS AND FUNCTIONS 19 9.5 Binary Operations Right from the school days. fog1(y) = 1Y(y) = fog2(y). subtraction . 3} → {a. Then for all y ∈ Y. (f –1)–1 = f. When we need to add three numbers. It is to be noted that only two numbers can be added or b multiplied at a time.

is not defined. ÷ : R∗ × R∗ → R∗. 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. b) → a – b × : R × R → R is given by (a. Example 29 Show that addition. as for b = 0. 5) under ÷ is 3 ÷ 5 = 5 Example 31 Show that ∗ : R × R → R given by (a. is not a function and hence not a binary b a operation. b a However. b) → a + b – : R × R → R is given by (a. . given by (a. ∗ is a binary operation on R. is not binary operation. but division is not a binary operation on R. Further. b) → . ‘–’ and ‘×’ are functions. b from X to another element of X. Solution Since ∗ carries each pair (a. given by (a. as ‘binary’ means two. ÷ : N × N → N. subtraction and multiplication are binary operations on R. b) → is a function and hence a b binary operation on R∗. given by (a. b) by a ∗ b. 5) under ‘–’ is 3 – 5 = – 2 ∉ N. Example 30 Show that subtraction and division are not binary operations on N. b) to a unique element a + 4b2 in R. as the image of (3. show that division is a binary operation on the set R∗ of nonzero real numbers. b) → ab Since ‘+’. Similarly. a But ÷: R × R → R. This gives rise to a general definition as follows: Definition 10 A binary operation ∗ on a set A is a function ∗ : A × A → A. We denote ∗ (a. given by (a. b) → a – b. 20 MATHEMATICS and division are examples of binary operation. Solution + : R × R → R is given by (a. as the image of (3. Solution – : N × N → N. is not a binary operation. b) → a ÷ b 3 ∉ N. b) → a + 4b2 is a binary operation. they are binary operations on R. If we want to have a general definition which can cover all these four operations.

∩ is a binary operation on P. . 2. If A = {a1. order is immaterial... the intersection operation ∩ carries each pair (A. B) in P × P to a unique element A ∪ B in P. b) → min {a. Similarly. RELATIONS AND FUNCTIONS 21 Example 32 Let P be the set of all subsets of a given set X. Table 1.1) . – 7) = – 7. an}.. i. Thus. we are having 3 rows and 3 columns in the operation table with (i. B) in P × P to a unique element A ∩ B in P. 2) = 2. Then. B) → A ∪ B and ∩ : P × P → P given by (A. Here. Conversely. addition and multiplication of 3 and 4 are meaningful. a2. Remark ∨ (4. an}. ∨ (1.. the operation ∨ on A defined in Example 33 can be expressed by the following operation table (Table 1. b) → max {a. 3) = 3. i. .. Show that ∪ : P × P → P given by (A. Solution Since union operation ∪ carries each pair (A. ∨ (2.. 3) = 3. Similarly. b} are binary operations. j)th entry being ai ∗ aj. Solution Since ∨ carries each pair (a. one can say that ∧ is also a binary operation. 7) = 7. b} and the ∧ : R × R → R given by (a. 3}.. ∪ is binary operation on P. ∨ (4.e. 3 – 4 ≠ 4 – 3. ∨ is a binary operation. but division of 3 and 4 in different order give different results. b) in R × R to a unique element namely maximum of a and b lying in R. .1 Here. ‘divide 3 by 4’ or ‘divide 4 by 3’. ∨ (1. B) → A ∩ B are binary operations on the set P. given any operation table having n rows and n columns with each entry being an element of A = {a1. 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. 7) = 4 and ∧ (4. we can express a binary operation ∗ on the set A through a table called the operation table for the operation ∗. 3 + 4 = 4 + 3. Example 33 Show that the ∨ : R × R → R given by (a. When number of elements in a set A is small. ∧ (4. a2. ‘subtract 4 from 3’. One may note that 3 and 4 can be added in any order and the result is same. in case of multiplication of 3 and 4. For example consider A = {1. – 7) = 4. Then the operation table will be having n rows and n columns with (i. For subtraction and division we have to write ‘subtract 3 from 4’. Using the similar argument.e. This can be generalised for general operation ∗ : A × A → A. j) the entry of the table being maximum of ith and jth elements of the set A. but subtraction and division of 3 and 4 are meaningless..

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

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

define ∗ by a ∗ b = | a – b | (v) On Z+. Determine whether or not each of the definition of ∗ given below gives a binary operation. which implies that other than 1 no element of N a has inverse for multiplication operation on N. (i) On Z+. define a ∗ b = 2ab (v) On Z+. 5} defined by a ∧ b = min {a. although – a satisfies a + (– a) = 0 = (– a) + a. (i) On Z. EXERCISE 1. define a ∗ b = ab a (vi) On R – {– 1}. For each binary operation ∗ defined below. give justification for this. 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. Consider the binary operation ∧ on the set {1. define ∗ by a ∗ b = ab2 (iv) On Z+. 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. a Solution Since – a ∉ N. b}. define ∗ by a ∗ b = a 2. define a ∗ b = ab + 1 ab (iii) On Q. define a ∗ b = 2 (iv) On Z+. ∉ N. define ∗ by a ∗ b = a – b (ii) On Z+. define a ∗ b = b +1 3. define ∗ by a ∗ b = ab (iii) On R. – a can not be inverse of a for addition operation on N. In the event that ∗ is not a binary operation. determine whether ∗ is commutative or associative. 1 Similarly. 3.4 1. 36. Examples 34. for a ≠ 1. define a ∗ b = a – b (ii) On Q. for a ≠ 1 in N. Write the operation table of the operation ∧ . . 2. 4.

8. 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. of a and b a binary operation? Justify your answer. 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.C. of a and b. of a and b. b) ∗ (c. Let ∗ be the binary operation on N given by a ∗ b = L. 2. 2. 6.2 5.2).M. 3. Is ∗ commutative? Is ∗ associative? Does there exist identity for this binary operation on N? 9. Let ∗′ be the binary operation on the set {1. Is the operation ∗′ same as the operation ∗ defined in Exercise 4 above? Justify your answer. 5} given by the following multiplication table (Table 1. 10.C.F. d) = (a + c. (Hint: use the following table) Table 1.C. 5} defined by a ∗′ b = H. 5} by a ∗ b = L. 4.M.F. 4. of a and b. RELATIONS AND FUNCTIONS 25 4. Show that none of the operations given above has identity.C. Find (i) 5 ∗ 7. Let A = N × N and ∗ be the binary operation on A defined by (a. (i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4) (ii) Is ∗ commutative? (iii) Compute (2 ∗ 3) ∗ (4 ∗ 5). 11. 3. 4. 2. Consider a binary operation ∗ on the set {1. b + d) . Let ∗ be the binary operation on N defined by a ∗ b = H. Is ∗ defined on the set {1. 3.

v) and (u. 12. State whether the following statements are true or false. 3. Thus. . y} ⊂ {3. 26 MATHEMATICS Show that ∗ is commutative and associative. c) ∈ R1 ∩ R2 ⇒ (a. y) R (u. Solution Since R1 and R2 are equivalence relations. (a. y} ⊂ {1. 4. (a. Find the identity element for ∗ on A. and (a. Consider a binary operation ∗ on N defined as a ∗ b = a3 + b3. Let R1 be a relation in X given by R1 = {(x. hence. 2. a) ∈ R1 ∩ R2. Thus. y} ⊂ {2. (x. Solution Clearly. (ii) If ∗ is a commutative binary operation on N. 9}}. a ∗ a = a ∀ a ∈ N. Justify. b) ∈ R1 ∩ R2 ⇒ (a. b). b) ∈ R1 and (a. c) ∈ R2 ⇒ (a. Choose the correct answer. v) R (x. Further. Similarly. 6. This shows that R1 ∩ R2 is transitive. Example 42 Let R be a relation on the set A of ordered pairs of positive integers defined by (x. Show that R is an equivalence relation. R u u v u is transitive. R1 ∩ R2 is an equivalence relation. ∀ a. then a ∗ (b ∗ c) = (c ∗ b) ∗ a 13. y). y) R (u. R is an equivalence relation. Similarly. 8. ∀ (x. Further. v) if and only if xv = yu. Example 43 Let X = {1. R1 ∩ R2 is symmetric. v) ⇒ xv = yu ⇒ uy = vx and hence (u. y) R (a. if any. (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. a) ∈ R1. a) ∈ R2 ⇒ (b. b) ⇒ xv = yu and a a b a ub = va ⇒ xv = yu ⇒ xv = yu ⇒ xb = ya and hence (x. (x. This implies that (a. a) ∈ R1 and (b. a) ∈ R2 ∀ a ∈ A. c) ∈ R1 and (a. show that R1 ∩ R2 is also an equivalence relation. y): {x. a) ∈ R1 ∩ R2. c) ∈ R1 ∩ R2. showing R1 ∩ R2 is reflexive. Thus. This shows that R is reflexive. b) ∈ R1 ∩ R2 and (b. 7}} or {x. (a. b) ∈ R2 ⇒ (b. (x. 9}. y) R (x. 5. y) : x – y is divisible by 3} and R2 be another relation on X given by R2 = {(x. Show that R1 = R2. since xy = yx. This shows that R is symmetric. v) R (a. 5. y). y) ∈ A. 7. 8} or {x. y) R (u. (i) For an arbitrary binary operation ∗ on a set N. 4. 6.

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

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

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

2} and f. 2}. we say that the operation ∗ distributes over the operation o]. Find F–1 of the following functions F from S to T. 2) is (A) 1 (B) 2 (C) 3 (D) 4 . Let A = {– 1.30 MATHEMATICS Define the relation R in P(X) as follows: For subsets A. B in P(X). ∀ A. ∀ a. a ∗ (b o c) = (a ∗ b) o (a ∗ b). [If it is so. 3}. 0. 2.. consider the binary operation ∗ : P(X) × P(X) → P(X) given by A ∗ B = A ∩ B ∀ A. . 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. 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 ∗. 9. 3) which are reflexive and symmetric but not transitive is (A) 1 (B) 2 (C) 3 (D) 4 17. ARB if and only if A ⊂ B. Then number of equivalence relations containing (1. 16. (Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ). b. 1)} (ii) F = {(a. 1. Show that ∗ is commutative but not associative. B in P(X). Define a binary operation ∗ on the set {0. show that ∀ a. B = {– 4. (b. 3}. Is R an equivalence relation on P(X)? Justify your answer. where P(X) is the power set of X. x ∈ A. Let A = {1. 4. – 2. c ∈ R. 2. . if it exists. 11. 2) and (1. (i) F = {(a. Further. (c. 1)} 12. 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. let ∗ : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A). o is associative but not commutative. (b. Find the number of all onto functions from the set {1. b ∈ R. 2). (c. 2. Then number of relations containing (1. B ∈ P(X). 3. are called equal functions).. g : A → B be functions defined 1 by f (x) = x2 – x. b. 15. Let A = {1. 3. 3). 10. 5} as ⎧ a + b. Given a non-empty set X. c} and T = {1. Are f and g equal? 2 Justify your answer. 2. 14. 3}. 2. x ∈ A and g ( x) = 2 x − − 1. Consider the binary operations ∗ : R × R → R and o : R × R → R defined as a ∗b = |a – b| and a o b = a. 13. n} to itself. (Hint: One may note that two functions f : A → B and g : A → B such that f (a) = g (a) ∀ a ∈ A. Given a non-empty set X. 1. Let S = {a. 0. Does o distribute over ∗? Justify your answer. 2). 1).

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

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