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.

 
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