RELATIONS AND FUNCTIONS : MATHEMATICS XII CLASS

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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 ⍷ 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 = ⍷ 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 = ⍷, 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✱
a 3 ⍷  A.
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 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, then2 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.