Reflexive, Symmetric,Transitive & Equivalence Relation, Number of Relations

According to english language, two objects/quantities/persons are related if there is some sort of recognisable connection between them. Also, these relations can be of multiple/various types so is the case in mathematics also.

Let us say that we are given that A & B are brothers , B & C are also brothers.From this information we can conclude that A & C are also brothers as depicted in the diagram below. In mathematics also we establish such kinds of relations from the information given to us.
So, let’s try to understand this in detail in this article.

Table of Contents:

• Reflexive Relation
• Symmetric Relation
• Transitive relation
• Equivalence Relation
• Number of Relations
• Equivalence class
• Practice Problems
• FAQs

Reflexive Relation

A relation R on set A is said to be a reflexive relation, if every element of A is related to itself, i.e., a relation R is reflexive if $(a, a)\in R \forall a\in A$

A relation R on set A is not reflexive if there exist an element a ∈ A such that $(a, a)\notin R$

Note: 

(i) The identity relation on a non-empty set is always reflexive relation.

For example, Let A={a,b,c}, the relation R₁={ (a,a),(b,b),(c,c) } on set A is identity as well as reflexive relation.

(ii) However, a reflexive relation on is not necessarily the identity relation. 

For example, Let A={a,b,c}, the relation R₂ ={(a,a),(b,b),(c,c),(a,b)} is a reflexive relation on set A but not the identity relation on A.

Example:Consider a set A={1, 2, 3}.Check if the following relations are reflexive.

1. R₁={(1, 1), (1, 2), (2, 2), (3, 1)}
2. R₂={(1, 1), (2, 2), (3, 3)}

Solution:

(a)The arrow diagram for R₁ can be represented as:

We have to check if every element of A is related to itself or not, i.e., if (1, 1), (2, 2), (3, 3) are present in the relation R₁ or not. As we can see that (3, 3) is not there in R₁. Therefore, it is not a reflexive relation.

(b)The arrow diagram for R₂ can be represented as:

We have to check if every element of A is related to itself or not, i.e.,if (1, 1), (2, 2), (3, 3) are there in the relation R₂ or not. As we can see, every element of set A is related to itself only, therefore, it is an identity relation and we know that every identity relation is a reflexive relation, therefore, R₂ is a reflexive relation.

Symmetric Relation

A relation R on a set A is said to be symmetric if $(a, b)\in R\Rightarrow (b, a)\in R \forall a,b \in A$ i.e.,

If a$a R b\Rightarrow b R a \forall a,b \in A$. It is read as if ‘a is related to b implies that b is related to a for all a,b A then the relation is symmetric.

Example: Consider a set, A={1, 2, 3}. Check whether R is symmetric or not.

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

Solution

For a relation to be symmetric, it should have (a, b) & (b, a)R

Here, we have, (1, 2) and (2, 1) & (1, 3) and (3, 1) in R.

For (1, 1), we know that on interchanging the position of the first and the second element in the ordered pair (1, 1), we will get (1, 1) only, and we can not have repeated entries in a set, therefore, the ordered pair (1, 1) is there only once.Thus, R is a symmetric relation.

Note: 

• The relation R is said to be antisymmetric on a set A, if x R y & y R x holds when x = y. or it can be defined as, relation R is antisymmetric if either (x,y) ∉ R or (y,x) ∉ R whenever x ≠ y.

• If a relation is antisymmetric that does not mean it is not symmetric. For example, every identity relation is an antisymmetric relation as well as symmetric relation.

Transitive Relation

A relation R on set A is said to be transitive if 

$(a, b) \in \mathrm{R }an\mathrm{d }(b, c) \in R \Rightarrow (a, c) \in R, \forall a, b, \mathrm{c }\in A$ or we can say that

 If a  $a R \mathrm{b }an\mathrm{d }b R c\Rightarrow a R c, \forall a, b, \mathrm{c }\in A$

Example: Show that relation R defined on the set of real numbers such that R={(a, b):a<b} is transitive.

Solution:

Let (a, b)R and (b, c)R.

a<b and b<c

a<b<c

Hence, a<c (a, c)R.

Therefore, R is a transitive relation.

Note: If a R b but b is not further related to any element then also, the relation is a transitive.

relation.Example:The relation R={(1,2)(3,5)} on set {1,2,3,5} is a transitive relation.

Equivalence Relation

A relation R is said to be equivalence if it is reflexive, symmetric as well as transitive, i.e., a relation R on a set A is said to be an equivalence relation on A if,

• It is reflexive, i.e., (a, a) R a A
• It is symmetric, i.e., (a, b) R (b, a) R a, b A
• It is transitive, i.e., (a, b) R & (b, c) R (a, c) R a, b, cA

Note:

(i)Equivalence class determined by an element is the set of all those elements which are related that particular element whose equivalence class is to be determined.

Let R={(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(4,4),(5,5)} be an equivalence relation on set A={1,2,3,4,5}.Then,The equivalence class determined by 1 would be {1,2,3}. Similarly equivalence class determined by “ 2 “ or “ 3 ” can also be found.

Also note that the union of all the equivalence classes would yield the original set on which the relation has been defined.

(ii)An identity relation is also an equivalence relation.

Example: Let us assume that R is a relation on the set of real numbers defined by x R y iff x-y is an integer. Prove that R is an equivalence relation on .

Solution:

To prove a relation to be equivalence, we have to prove the conditions of all three i.e. reflexive, symmetric and transitive relation.

1. Reflexive: Let x ,then x-x=0 is an integer. Therefore, x R x ∀  x ∈ R  
2. Symmetric: Let x,y such that x R y. Then x-y is an integer. Thus, y – x = – ( x – y), is also an integer. Therefore y R x. Hence, x R y =>y R x ∀  x, y ∈ R 
3. Transitive: Let x,y , such that x R y and y R z. Therefore x-y and y-z are integers. As we know that the sum of two integers is also an integer, so ( x – y ) + ( y – z ) = x – z is also an integer. Therefore, x R z.

Since, R is reflexive, symmetric as well as transitive.Thus, R is an equivalence relation on .

Number of Relations:

1. Number of relations from A to A is $2^{n^2}$.
2. Number of reflexive relations from A to A is $2^{n^2-n}.$.
3. Number of symmetric relations from A to A is $2^{\frac{n(n+1)}{2}}$.
4. Number of relations from A to A which are not symmetric is .
5. Number of relations from A to A which are both reflexive and symmetric is .
6. Number of relations from A to A which are symmetric but not reflexive is

Practice Problems

Example: R is a relation on a set A where A = {a, b, c} &R = {(a, a), (a, b), (a, c), (b, c), (c, a)}. Determine the elements which should be added to R to make it a symmetric relation.

Solution

To make R a symmetric relation, we will check for each element (x,y)R if (y,x)R or not. .

(a, a) ∈ R ⇒ (a, a) ∈ R for symmetric relation

(a, b) ∈ R ⇒ (b, a) ∈ R for symmetric relation but over here (b, a) ∉ R

(a, c) ∈ R ⇒ (c, a) ∈ R

(b, c) ∈ R ⇒ (c, b) ∈ R, for symmetric relation but over here (c, b) ∉ R

Hence, (b, a) and (c, b) should also be added to R to make it a symmetric relation.

Example

Show that the relation R is an equivalence relation on the set A = { 1, 2, 3, 4, 5 } given by the relation R = { (a, b):|a-b| is even }.

Solution

R = { (a, b):|a-b| is even }, a,b A

Reflexive

Let a A. From the given relation,|a – a| = | 0 |=0 and 0 is always even.

Therefore, (a, a)R a AHence R is Reflexive.

Symmetric

Let a,b A such that (a,b)R. We know that |a – b| = |-(b – a)|= |b – a|

As |a – b| is even, therefore |b – a| is also even (b,a)R

Hence, (a,b)R(b,a)R.therefore R is symmetric.

Transitive 

Let a,b,c A such that (a,b) & (b,c)R

|a-b| & |b-c| are even.

a-b & b-c are also even.

We know that sum of even numbers is even

a-b+b-c is even

a-c is even 

|a-c| is even.

Therefore, if (a, b) ∈ R and (b, c) ∈ R, then (a, c)R

Hence R is transitive.

As R is reflexive, symmetric and transitive, so R is an equivalence relation.

Example:Let S be the set of all squares in a plane with R as a relation given by R = {(S₁, S₂): S₁ is congruent to S2}. Show that R is an equivalence relation.

Solution:

Reflexive

Every square is congruent to itself.

Hence,(S, S) R .Therefore, R is reflexive. 

Symmetric

Let S₁,S₂S such that (S₁, S₂)R 

S₁ is congruent to S₂

S₂ is congruent to S₁

(S₁, S₂) R. Therefore, R is symmetric.

Transitive

Let S₁,S₂,S₃S such that (S₁, S₂) R and (S₂, S₃) R

S₁ is congruent to S₂ and S₂ is congruent to S₃

Hence, S₁ is congruent to S₃ (S₁, S₃) R

Therefore, R is transitive.Thus, the given relation R is an equivalence relation.

FAQs

Q 1. Can an element present in a relation have more than one output? 
Answer: Yes this is possible because a relation can be any subset of the cartesian product.

Q 2.What is trivial relation?
Answer: A relation which equals cartesian product of the sets on which it is defined is known as a trivial relation.

Q 3.What is the cardinality of a reflexive relation defined on a set A which has m elements?
Answer: The cardinality of such a reflexive relation would be equal to m as there are m elements and every element has to be related to itself

Q 4.Is universal relation on a set A transitive ?
Answer: Yes a universal relation on a given set A will contain all the possible ordered pairs.Hence, it will be reflexive, symmetric as well as transitive and thereby will be will be an equivalence relation.