Summary Math 1st year Chapter No.1

Chapter 1
RELATIONS AND FUNCTIONS




d
he
™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




is
one when we read it. — G. H. HARDY ™

1.1 Introduction




bl
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
pu
and their graphs. The concept of the term ‘relation’ in
mathematics has been drawn from the meaning of relation
be T

in English language, according to which two objects or
quantities are related if there is a recognisable connection
re
or link between the two objects or quantities. Let A be
o R


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
tt E



(i) {(a, b) ∈ A × B: a is brother of b}, Lejeune Dirichlet
(ii) {(a, b) ∈ A × B: a is sister of b}, (1805-1859)
C




(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},
no N




(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




d
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




he
(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.




is
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.




bl
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
pu
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.
be T

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
re
o R


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.
tt E



Remark In Class XI, we have seen two ways of representing a relation, namely raster
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
C




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.
no N




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.




d
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,




he
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.




is
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.




bl
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
⇒
pu L1 is perpendicular to L2
⇒ L2 is perpendicular to L1
⇒ (L2, L1) ∈ R.
be T

R is not transitive. Indeed, if L1 is perpendicular to L2 and
re
Fig 1.1
L2 is perpendicular to L3, then L1 can never be perpendicular to
o R


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),
tt E



(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,
C




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.
no N




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.

, 4 MATHEMATICS


In Example 5, note that all even integers are related to zero, as (0, ± 2), (0, ± 4)
etc., lie in R and no odd integer is related to 0, as (0, ± 1), (0, ± 3) etc., do not lie in R.
Similarly, all odd integers are related to one and no even integer is related to one.
Therefore, the set E of all even integers and the set O of all odd integers are subsets of




d
Z satisfying following conditions:
(i) All elements of E are related to each other and all elements of O are related to




he
each other.
(ii) No element of E is related to any element of O and vice-versa.
(iii) E and O are disjoint and Z = E ∪ O.
The subset E is called the equivalence class containing zero and is denoted by




is
[0]. Similarly, O is the equivalence class containing 1 and is denoted by [1]. Note that
[0] ≠ [1], [0] = [2r] and [1] = [2r + 1], r ∈ Z. Infact, what we have seen above is true




bl
for an arbitrary equivalence relation R in a set X. Given an arbitrary equivalence
relation R in an arbitrary set X, R divides X into mutually disjoint subsets Ai called
partitions or subdivisions of X satisfying:
pu (i) all elements of Ai are related to each other, for all i.
(ii) no element of Ai is related to any element of Aj , i ≠ j.
be T

(iii) ∪ Aj = X and Ai ∩ Aj = φ, i ≠ j.
The subsets Ai are called equivalence classes. The interesting part of the situation
re
o R


is that we can go reverse also. For example, consider a subdivision of the set Z given
by three mutually disjoint subsets A1, A2 and A3 whose union is Z with
A1 = {x ∈ Z : x is a multiple of 3} = {..., – 6, – 3, 0, 3, 6, ...}
tt E



A2 = {x ∈ Z : x – 1 is a multiple of 3} = {..., – 5, – 2, 1, 4, 7, ...}
A3 = {x ∈ Z : x – 2 is a multiple of 3} = {..., – 4, – 1, 2, 5, 8, ...}
C




Define a relation R in Z given by R = {(a, b) : 3 divides a – b}. Following the
arguments similar to those used in Example 5, we can show that R is an equivalence
no N




relation. Also, A1 coincides with the set of all integers in Z which are related to zero, 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. Thus, A1 = [0], A2 = [1] and A3 = [2].
In fact, A1 = [3r], A2 = [3r + 1] and A3 = [3r + 2], for all r ∈ Z.
©




Example 6 Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7} by
R = {(a, b) : both a and b are either odd or even}. Show that R is an equivalence
relation. Further, show that all the elements of the subset {1, 3, 5, 7} are related to each
other and all the elements of the subset {2, 4, 6} are related to each other, but no
element of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6}.