 Mathematics RELATIONS - Introduction and Types of Relations For CBSE-NCERT

### Topics Covered star Introduction
star Empty relation and Universal relation
star Reflexive Relation
star Symmetric Relation
star Transitive Relation
star Equivalence Relation

### Introduction

\color{green} ✍️ The concept of the term color{green} "‘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.

\color{green} ✍️ 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" },
(ii) {(a, b) ∈ A × B: "a is sister of b"},
(iii) {(a, b) ∈ A × B: "age of a is greater than age of 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.

### Empty relation and Universal relation \color{green} ✍️ we would like to study different types of relations. We know that a relation in a set A is a subset of A × A.

\color{purple} "★ DEFINITION ALERT"

\color{fuchsia} \mathbf\ul"EMPTY RELATION"

\color{purple} ✍️ 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 = phi ⊂ A × A.

\color{purple} ✍️ 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.

\color{purple} "★ DEFINITION ALERT"

\color{fuchsia} \mathbf\ul"UNIVERSAL RELATION"

\color{purple} ✍️ 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.

\color{green} ★ \color{green} \mathbf(KEY \ CONCEPT)
\color{navy} "If (a, b) ∈ R, we say that a is related to b and we denote it as" a \ \R \ \b.

Q 3184756657 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 xx A  is the universal relation.

### Reflexive Relation \color{purple} "★ DEFINITION ALERT"

\color{fuchsia} \mathbf\ul"REFLEXIVE RELATION"

\color{purple} ✍️ A relation R in a set A is called color{green}ul"reflexive" if (a, a) ∈ R, for every a∈ A,

\color{purple} ✍️ It means R is color{green}ul"reflexive" if every element of a∈ A is related to itself.

\color{green} ✍️ A relation R in a set A is color{green}ul"not reflexive" if there be at least one element a ∈ A such that (a, a) ∉ R.

color{green}"Example -" a set A = {p, q, r, s}.

\color{green} ✍️ The relation R_1 = {(p, p), (p, r), (q, q), (r, r), (r, s), (s, s)} in A is reflexive,

since every element in A is R_1-related to itself.

\color{green} ✍️ But the relation R_2 = {(p, p), (p, r), (q, r), (q, s), (r, s)} is not reflexive in A .

since q, r, s ∈ A but (q, q) ∉ R_2, (r, r) ∉ R_2 and (s, s) ∉ R_2.

### Symmetric Relation \color{purple} "★ DEFINITION ALERT"

\color{fuchsia} \mathbf\ul"SYMMETRIC RELATION"

\color{purple} ✍️ A relation R in a set A is called color{green}ul"symmetric", if (a_1, a_2) ∈ R implies that (a_2, a_1)∈ R, for all a_1, a_2 ∈ A.

\color{purple} ✍️ It means if a is related to b then b also related to a.

color{green}"Example -" a set A = {1, 2, 3, 4}.

\color{green} ✍️ The relation R_1 = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 4), (4, 1), (4, 4) }

R_1 is not symmetric because (3, 4) in R_1  and (4, 3) ∉ R_1

\color{green} ✍️ But the relation R_2 ={(1, 1), (1, 2), (2, 1)}

R_1 is symmetric because (1, 2) in R_1  and (2, 1) ∉ R_1

### Transitive Relation \color{purple} "★ DEFINITION ALERT"

\color{fuchsia} \mathbf\ul"TRANSITIVE RELATION"

\color{purple} ✍️ A relation R in a set A is called color{green}ul"transitive", if (a_1, a_2) ∈ R and (a_2, a_3)∈ R implies that (a_1, a_3)∈ R, for all a_1, a_2, a_3 ∈ A.

\color{purple} ✍️ Let A be a set in which the relation R defined.

R is said to be transitive, if

(a, b) ∈ R and (b, a) ∈ R ⇒ (a, c) ∈ R,

That is a \ \R\ \b and b\ \ R\ \ c ⇒ a \ \ R\ \ c where a, b, c ∈ A.

\color{purple} ✍️ The relation is said to be color{green}ul"non-transitive", if

(a, b) ∈ R and (b, c) ∈ R do not imply (a, c ) ∈ R.

color{green}"Example -" A={ 0,1,2,3} and a relation R on A be given by

R={ (0,0), (0,1), (0,3), (1,0), (1,1), (2,2), (3,0), (3,3) }

R is not transitive because (1,0) ∈ R, \ \ (0,3) ∈ R and (1,3) ∉R imply R is not transitive.
Q 3154856754 Let L be the set of all lines in a plane and R be the relation in L defined as
R = {(L_1, L_2) : L_1 is perpendicular to L_2}. 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., (L_1, L_1)
∉ R. R is symmetric as (L_1, L_2) ∈ R

⇒ L_1 is perpendicular to L_2

⇒ L_2 is perpendicular to L_1

⇒ (L_2, L_1) ∈ R.

R is not transitive. Indeed, if L_1 is perpendicular to L_2 and

L_2 is perpendicular to L_3, then L_1 can never be perpendicular to
L_3. In fact, L_1 is parallel to L_3, i.e., (L_1, L_2) ∈ R, (L_2, L_3) ∈ R but (L_1, L_3) ∉ R.
Q 3184856757 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.

### Equivalence Relation

\color{purple} "★ DEFINITION ALERT"

\color{fuchsia} \mathbf\ul"EQUIVALENCE RELATION"

\color{purple} ✍️ A relation R in a set A is said to be an color{green}ul"equivalence relation" if R is reflexive, symmetric and transitive.

\color{purple} ✍️ A relation R in a set A is called
"(i) reflexive," if (a, a) ∈ R, for every a∈ A,

"(ii) symmetric," if (a_1, a_2) ∈ R implies that (a_2, a_1)∈ R, for all a_1, a_2 ∈ A.

"(iii) transitive," if (a_1, a_2) ∈ R and (a_2, a_3)∈ R implies that (a_1, a_3)∈ R, for all a_1, a_2, a_3 ∈ A.

color{green}"Example -" A={1,2,3,4} and a relation R on A be given by

R={ (1,1), (1,2), (2,1), (2,2), (3,4), (4,3), (3,3), (4,4) }

then R is a equivalence relation.
Q 3114756659 Let T be the set of all triangles in a plane with R a relation in T given by
R = {(T_1, T_2) : T_1 is congruent to T_2}. Show that R is an equivalence relation. Solution:

R is reflexive, since every triangle is congruent to itself. Further,

(T_1, T_2) ∈ R ⇒ T_1 is congruent to T_2 ⇒ T_2 is congruent to T_1⇒ (T_2, T_1) ∈ R. Hence,
R is symmetric. Moreover, (T_1, T_2), (T_2, T_3) ∈ R ⇒ T_1 is congruent to T_2 and T_2 is
congruent to T_3⇒ T_1 is congruent to T_3⇒ (T_1, T_3) ∈ R. Therefore, R is an equivalence
relation.
Q 3114856759 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. 