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**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.

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

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

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

b) : a = b – 2. Let R be the relation in the set N given by R = {(a. if the images of distinct elements of X under f are distinct. namely b. RELATIONS AND FUNCTIONS 7 15.2 (i) is not onto as elements e.4).2 (i) and (iv) are one-one and the function f2 and f3 in Fig 1.. i. for every x1. Let R be the relation in the set {1. The function f3 and f4 in Fig 1. polynomial function. subtraction. 2. (3. (2.. there are some elements like e and f in X2 which are not images of any element of X1 under f1. f is called many-one. 4} given by R = {(1. Consider the functions f1. i. rational function. 2). 1). . 3. 4) ∈ R (B) (3. Further. constant function. (3. f in X2 are not the image of any element in X1 under f1. there exists an element x in X such that f (x) = y. 16. 3). 8) ∈ R (C) (6. we observe that the images of distinct elements of X1 under the function f1 are distinct. 3).2 (iii). Definition 6 A function f : X → Y is said to be onto (or surjective). (A) R is reflexive and symmetric but not transitive. modulus function. (1. 2). (D) R is an equivalence relation.2 (ii) and (iii) are many-one.3 Types of Functions The notion of a function along with some special functions like identity function. (B) R is reflexive and transitive but not symmetric. for every y ∈ Y.e. (A) (2. The function f1 and f4 in Fig 1. while all elements of X3 are images of some elements of X1 under f3. (iv) are onto and the function f1 in Fig 1. but the image of two distinct elements 1 and 2 of X1 under f2 is same. we would like to extend our study about function from where we finished earlier. The above observations lead to the following definitions: Definition 5 A function f : X → Y is defined to be one-one (or injective). Choose the correct answer. Addition. Otherwise. f2. 7) ∈ R 1. (1. multiplication and division of two functions have also been studied.2. (4. 8) ∈ R (D) (8. Choose the correct answer. f (x1) = f (x2) implies x1 = x2. signum function etc. In this section. we would like to study different types of functions. x2 ∈ X. if every element of Y is the image of some element of X under f. In Fig 1. f3 and f4 given by the following diagrams. along with their graphs have been given in Class XI. b > 6}. 2)}.e. (C) R is symmetric and transitive but not reflexive. As the concept of function is of paramount importance in mathematics and among other disciplines as well.

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

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

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

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

∀ x ∈ A. 4. we will study composition of functions and the inverse of a bijective function. 15} be functions defined as f (2) = 3. 3. gof ≠ fog. by the combination of these two functions. Solution We have gof (2) = g (f (2)) = g (3) = 7. Thus.4 Composition of Functions and Invertible Function In this section. 9} and g : {3. f (4) = f (5) = 5 and g (3) = g (4) = 7 and g (5) = g (9) = 11. 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. 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. gof (3) = g (f (3)) = g (4) = 7. 5. 9} → {7. This leads to the following definition: Definition 8 Let f : A → B and g : B → C be two functions. gof (4) = g (f (4)) = g (5) = 11 and gof (5) = g (5) = 11. 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. In order to have confidentiality. 5. Let B ⊂ N be the set of all roll numbers and C ⊂ N be the set of all code numbers. Fig 1. fog (x) = f (g (x)) = f (3x2) = cos (3x2). Similarly. Note that 3cos2 x ≠ cos 3x2. Find gof. 4. Show that gof ≠ fog. each student is eventually attached a code number. f (3) = 4. the Board arranges to deface the roll numbers of students in the answer scripts and assigns a fake code number to each roll number. Then the composition of f and g. . for x = 0. who appeared in Class X of a Board Examination in 2006. Hence. 12 MATHEMATICS 1. if f : R → R and g : R → R are given by f (x) = cos x and g (x) = 3x2. 11.5 Example 15 Let f : {2. denoted by gof. is defined as the function gof : A → C given by gof (x) = g(f (x)). Solution We have gof (x) = g (f (x)) = g (cos x) = 3 (cos x)2 = 3 cos2 x. 4. Example 16 Find gof and fog. Consider the set A of all students. 5} → {3.

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

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

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

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

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

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

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

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

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

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

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

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

F. 4. Let A = N × N and ∗ be the binary operation on A defined by (a. 5} by a ∗ b = L. Show that none of the operations given above has identity. of a and b.M. b + d) . b) ∗ (c.C. 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. Is ∗ defined on the set {1.M.2). 3. Consider a binary operation ∗ on the set {1.C. Let ∗ be the binary operation on N given by a ∗ b = L. 10. 5} given by the following multiplication table (Table 1. 3. 6. (Hint: use the following table) Table 1. of a and b a binary operation? Justify your answer. 2. 4. 11. 3. Let ∗′ be the binary operation on the set {1. of a and b. 4. of a and b. RELATIONS AND FUNCTIONS 25 4. 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. Find (i) 5 ∗ 7. 8. 2.C. d) = (a + c. Is the operation ∗′ same as the operation ∗ defined in Exercise 4 above? Justify your answer. 5} defined by a ∗′ b = H. Let ∗ be the binary operation on N defined by a ∗ b = H.2 5.C. Is ∗ commutative? Is ∗ associative? Does there exist identity for this binary operation on N? 9. (i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4) (ii) Is ∗ commutative? (iii) Compute (2 ∗ 3) ∗ (4 ∗ 5). 2.F.

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

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

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

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

11. 3). (b. ∀ 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 ∗. 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. Is R an equivalence relation on P(X)? Justify your answer. (b. b. b. consider the binary operation ∗ : P(X) × P(X) → P(X) given by A ∗ B = A ∩ B ∀ A. 1). B ∈ P(X).. 3. let ∗ : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A). B in P(X). Let A = {1. 2. ∀ a. 4.30 MATHEMATICS Define the relation R in P(X) as follows: For subsets A. Define a binary operation ∗ on the set {0. Given a non-empty set X. B = {– 4. 1)} 12. x ∈ A and g ( x) = 2 x − − 1. 3}. 2. Show that ∗ is commutative but not associative. 3. (Hint: One may note that two functions f : A → B and g : A → B such that f (a) = g (a) ∀ a ∈ A. – 2. Are f and g equal? 2 Justify your answer. 1. (c. b ∈ R. 2} and f. . are called equal functions). Further. 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. 13. c ∈ R. a ∗ (b o c) = (a ∗ b) o (a ∗ b). . (Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ). 16. 1. c} and T = {1. Consider the binary operations ∗ : R × R → R and o : R × R → R defined as a ∗b = |a – b| and a o b = a.. show that ∀ a. 5} as ⎧ a + b. 2. 3) which are reflexive and symmetric but not transitive is (A) 1 (B) 2 (C) 3 (D) 4 17. 2). Find the number of all onto functions from the set {1. g : A → B be functions defined 1 by f (x) = x2 – x. we say that the operation ∗ distributes over the operation o]. Does o distribute over ∗? Justify your answer. 2. Find F–1 of the following functions F from S to T. Let A = {1. x ∈ A. 10. n} to itself. 2. 2}. ARB if and only if A ⊂ B. (c. 2) is (A) 1 (B) 2 (C) 3 (D) 4 . Let A = {– 1. Then number of relations containing (1. if it exists. o is associative but not commutative. B in P(X). Let S = {a. Given a non-empty set X. 0. 1)} (ii) F = {(a. 14. 9. 2). 2) and (1. Then number of equivalence relations containing (1. 3}. (i) F = {(a. [If it is so. where P(X) is the power set of X. 3}. 0. 15.

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

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

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