Relations: Definition, Types, Domain and Range, Examples

Relations: In our daily life, we have used the term relations in such a way as the relation between brothers and sisters, father and son, husband and wife, or teacher and student. In Maths, the term relation relates numbers, symbols, variables, sets, and groups of sets. It is often the relationship between the \(x\) and \(y\) values in an ordered pair. Here, the \(x-\)values and the \(y-\)values are from two different sets.

In this section, let us learn about the Cartesian product of sets, the definition of relation, how the relations are formed, and the types of relations with some solved examples.

Cartesian Products of Two Sets

Let \(A\) and \(B\) be two non-empty sets. Then, the set of all ordered pairs \((a,b)\) such that \(a∈A\) and \(b∈B\) is called the Cartesian product of the sets \(A\) and \(B.\) It is denoted by \(A×B.\) Thus, 

\(A \times B = \{ (a,b):a \in A\,{\rm{and}}\,b \in B\} \)

This is also called the cross-product of sets.

Note: If \(A=ϕ\) or \(B=ϕ,\) then we define \(A×B=ϕ,\) where \(ϕ\) is an empty set.

Example: Let \(A = \{ 2,4\} \) and \(B = \{ 1,2\} ,\) then \(A \times B = \{ 2,4\} \times \{ 1,2\} \)

\( = \{ (2,1),(2,2),(4,1),(4,2\} \)

Similarly, \(B \times A = \{ 1,2\} \times \{ 2,4\} \)

\( = \{ (1,2),(1,4),(2,2),(2,4)\} \)

Thus, it is evident from the example, to write \(A×B,\) we take an element from set \(A\) as the first element and form ordered pairs with elements from set \(B\) as the second element. Next, we choose the next element from the set \(A,\) and with elements in set \(B,\) we form ordered pairs. So this process is continued till all the elements of set \(A\) are exhausted.

Definition: Relation

Relation defines the relationship between two sets. A relation is the subset of the Cartesian product of the sets.

Let \(A\) and \(B\) be two sets. Then a relation from set \(A\) to set \(B\) is a subset of \(A×B.\)

Thus, a relation from \(A\) to \(B⇔R⊆A×B\)

For a relation \(R\) from a set \(A\) to set \(B,\) if \((a,b)∈R,\) we write \(aRb\) which is read as \(‘a\) is related to \(b\) by the relation \(R.\) Here, \(A\) and \(B\) are non-empty sets.

Domain and Range of a Relation

Let \(R\) be a relation from a set \(A\) to a set \(B.\) Then, the set of all first components or coordinates of the ordered pairs is called the domain of \(R.\) The set of all second components or coordinates of the ordered pairs is called the range of \(R.\) Thus, 

\({\rm{Dom}}{\mkern 1mu} (R) = \{ a:(a,b) \in R\} \)

\({\rm{Range }}(R) = \{ b:(a,b) \in R\} \)

The set \(B\) is called the co-domain of relation \(R.\)

How are Relations Formed?

Let \(A\) and \(B\) denote the sets of male and female members in the royal family of Dasrath’s kingdom. 

\(A = \left\{ {{\rm{Dasrath, Ram, Laxman, Shatrughna, Bharat}}} \right\}\)

\(B = {\rm{\{ Kaushalya, Kaikeyi, Sumitra, Sita, Urmila, Shrutakirti, Mandavi\} }}\)

If we define the relation as “was the husband of”, then the fact that Dasrath was the husband of Kaushalya, Kaikei and Sumitra; Ram was the husband of Sita; Laxman was the husband of Urmila; Bharat was the husband of Mandvi and Shatrughan was the husband of Shrutkirti can be represented as follows.

• Dasrath \(R\) Kaushalya
• Dasrath \(R\) Kaikeyi
• Dasrath \(R\) Sumitra
• Ram \(R\) Sita
• Laxman \(R\) Urmila
• Bharat \(R\) Mandavi
• Shatrughna \(R\) Shrutakirti

A visual representation of this relation \(R\) in the form of an arrow diagram is as follows.

If we write them as ordered pairs, the above fact can also be written as a set \(R\) of ordered pairs as given below.

\(R = {\rm{\{ (Dasrath,}}\,{\rm{Kaushalya),(Dasrath,Kaikei),(Dasrath,Sumitra),(Ram,Sita)}}\)
\({\rm{(Laxman,}}\,{\rm{Urmila), (Bharat,}}\,{\rm{Mandavi), (Shatrughan,}}\,{\rm{Shrutakirti)\} }}{\rm{.}}\)

Thus, the relation can be written as

\(R⊆A×B\)

Thus, the relation “was the husband of” from set \(A\) to set \(B\) gives rise to a subset \(R\) of \(A×B\) such that \((x,y) \in R \Leftrightarrow xRy.\)

Total Number of Relations

Let \(A\) and \(B\) be two non-empty sets consisting of \(p\) and \(q\) elements, respectively. Then, \(A×B\) consists of \(pq\) ordered pairs. So, the total number of subsets of \(A×B\) is \({2^{pq}}.\) Since each subset of \(A×B\) defines a relation from \(A\) to \(B,\) so the total number of relations from \(A\) to \(B\) is \({2^{pq}}.\) Among these \({2^{pq}}\) Relations, the void relation \(ϕ,\) and universal relation \(A×B\) are the trivial relation from \(A\) to \(B.\)

Representation of Relations

A relation from set \(A\) to set \(B\) can be represented in the following ways.

Roster Form

A relation is represented by the set of all ordered pairs belonging to \(R.\)

Example: If \(R\) is a relation from a set \(A = \left\{ { – 2,\, – 1,\,0,\,1,\,2} \right\}\) to set \(B = \left\{ {0,\,1,\,4,\,9,\,10} \right\}\) by rule method, \(aRb \Leftrightarrow {a^2} = b.\) Then \(0R0, – 2R4, – 1R1,1R1\) and \(2R4\)

So, the relation \(R\) can be represented in roster form as 

\(R = \{ (0,0),( – 1,1),( – 2,4),(1,1),(2,4)\} \)

Set-Builder Form

In this form, the relation \(R\) from \(A\) to set \(B\) is represented as 

\(R = \{ (a,b):a \in A,b \in B,{\rm{and}}\,{\rm{the}}\,{\rm{rule}}\} .\)

Example: If \(A = \{ 1,2,3,4,5\} ,B = \left\{ {1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}, \ldots } \right\}\) and then relation \(R\) from set \(A\) to set \(B\) is given by \(R = \left\{ {(1,1),\left( {2,\frac{1}{2}} \right),\left( {3,\frac{1}{3}} \right),\left( {4,\frac{1}{4}} \right),\left( {5,\frac{1}{5}} \right)} \right\}\)

Then, the relation R in set-builder form can be repressed as 

\(R = \left\{ {(a,b):a \in A,b \in B} \right.,{\rm{and}}\,\left. {b = \frac{1}{a}} \right\}\)

Note that it is not possible to express every relation from a set \(A\) to set \(B\) in set-builder form. 

Example: Relation \(R = \{ (1,a),(1,c),(3,b)\} \) from set \(A = \left\{ {1,\,2,\,3,\,4} \right\}\) to set \(B = \left\{ {a,\,b,\,c} \right\}\) cannot be described in set-builder form.

Types of Relations

There are several types of relations.

Empty Relation

A relation is said to be an empty relation if the element of any set is not related to another set or itself. Sometimes, empty relation is also known as void relation.

Condition: No element of set \(A\) is mapped with another set \(B\) or set \(A\) itself. The empty relation is shown by \(R=∅.\)

Universal Relation

A relation is a universal relation if all the elements of a set are related to another set or the set itself. 

Condition: The relation \(R\) is universal if all the elements of set \(A\) are related to all the elements of set \(B\) or the set \(A\) itself.

Therefore, \(R = A \times A\) or \(R = A \times B\)

Identity Relation

If all the elements are related to itself, a relation is said to be an identity relation. It is generally denoted by \(‘I’.\)

Condition: If the relation is identity \((I),\) then all the elements are related to itself, such that \(x∈P,\) it is \((x,x)∈I.\)

Inverse of a Relation

Let \(A,B\) be two sets, and let \(R\) be a relation from a set \(A\) to a set \(B.\) Then the inverse of \(R,\) denoted by \({R^{ – 1}},\) is a relation from \(B\) to \(A\) and is defined by \({R^{ – 1}} = \{ (b,a):(a,b) \in R\} \)

Here,

\((a,b) \in R \Leftrightarrow (b,a) \in {R^{ – 1}}\)

Also,

\({\rm{Domain}}(R) = {\rm{Range}}\left( {{R^{ – 1}}} \right)\)

\({\rm{Range}}(R) = {\rm{Domain}}\left( {{R^{ – 1}}} \right)\)

Other Types of Relations

Reflexive relation

A relation \(R\) is defined on the set \(A\) is reflexive if, \((a,a)∈R,\)
\(\forall a \in A.\)

Symmetric relation

A relation \(R\) is defined on the set \(A\) is symmetric if \((a,b)∈R\) then \((b,a)∈R, ∀a,b∈A\)

Transitive relation

A relation \(R\) is defined on the set \(A\) is transitive, if \((a,b)∈R,(b,c)∈R,\) then \((a,c)∈R, ∀a,b,c∈A\)

Equivalence relation

A relation is said to be equivalence if it is reflexive, symmetric, and transitive.

Solved Examples

Q.1. Let \(R\) be the relation on the set \(N\) of natural numbers is defined by \(R = \left\{ {(a,b):a + 3b = 12,a \in N,b \in N} \right\}\) then find the domain of \(R\) and range of \(R.\)
Ans: Given: \(R = \left\{ {(a,b):a + 3b = 12,a \in N,b \in N} \right\}\)
Now,
\(a+3b=12\)
\( \Rightarrow a = 12 – 3b\)
Putting \(b=1,2,3\) in the above relation, we get
\(a=9, 6, 3\)
For \(b=4,a=12-3b\) gives \(a=0,\) but it does not belong to \(N.\) Also, values of a given by \(a=12-3b\) do not belong to \(N\) for all \(b>4\)
\(\therefore R = \left\{ {(9,1),(6,2),(3,3)} \right\}\)
Therefore, the domain of \(R\) is \(\left\{ {9,6,3} \right\},\) and the range of \(R\) is \(\{ 1,2,3\} .\)

Q.2. Let \(A = \{ 1,2,3,4,5,6\} .\) Define a relation \(R\) on set \(A\) by \( R = \{ (x,y):y = x + 1\} .\) Depict this relation using an arrow diagram.
Ans: Given: \( R = \{ (x,y):y = x + 1\} \)
Now,
\(y=x+1\)
Putting \(x=1, 2, 3, 4, 5, 6\) respectively in \(y=x+1,\) then we get
\(y=2, 3, 4, 5, 6, 7\) respectively.
\(\therefore (1,2) \in R,(2,3) \in R,(3,4) \in R,(4,5) \in R,(5,6) \in R\) and \((6,7) \notin R\)
For \(x=6,\) then we get \(y=7\) but which is not belongs to set \(A.\)
Hence, \(R = \left\{ {(1,2),(2,3),(3,4),(4,5),(5,6)} \right\}\)
The arrow diagram represents \(R\) as follows:

Q.3. Let \(A = \{ 1,2,3,4,5,6\} \). Define a relation \(R\) on set \(A\) by \(R = \left\{ {(x,y):y = x + 1} \right\}.\) Write down the domain and range of \(R.\)
Ans: Given: \(R = \left\{ {(x,y):y = x + 1} \right\}\)
Now, \(y=x+1\)
Putting \(x=1, 2, 3, 4, 5, 6\) respectively in \(y=x+1,\) then we get
\(y=2, 3, 4, 5, 6, 7\) respectively.
\(\therefore (1,2) \in R,(2,3) \in R,(3,4) \in R,(4,5) \in R,(5,6) \in R\) and \((6,7) \notin R\)
For \(x=6,\) then we get \(y=7\) but which is not belongs to set \(A.\)
Hence, \(R = \left\{ {(1,2),(2,3),(3,4),(4,5),(5,6)} \right\}\)
Therefore,
Domain\( = \{ 1,2,3,4,5\} \)
Range\( = \{ 2,3,4,5,6\} \)

Q.4. If \(A = \{ 1,2,3\} ,B = \{ 4,5,6\} ,\) then check whether the following relation \({R_1} = \{ (1,4),(1,5),(1,6)\} \) is a relation from set \(A\) to set \(B.\)
Ans: Given: \(A = \{ 1,2,3\} ,B = \{ 4,5,6\} ,\)
And also, the relation is defined as \({R_1} = \{ (1,4),(1,5),(1,6)\} \)
Now, \(A \times B = \{ (1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)\} \)
Thus, clearly, \({R_1} \subseteq A \times B.\) So it is a relation from set \(A\) to set \(B.\)

Q.5. If \(A = \{ 1,2,3\} ,B = \{ 4,5,6\} ,\) then check whether the following relation \({R_4} = \{ (4,2),(2,6),(5,1),(2,4)\} \) is a relation from set \(A\) to set \(B.\)
Ans: Given: \(A = \{ 1,2,3\} ,B = \{ 4,5,6\} ,\)
And also, the relation is defined as \({R_4} = \{ (4,2),(2,6),(5,1),(2,4)\} \)
Now, \(A \times B = \{ (1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)\} \)
Since, \((4,2) \in {R_4}\) but \((4,2) \notin A \times B.\) So \({R_4}\) is not a relation from \(A\) to \(B.\)

Summary

A relation gives the relationship between two sets. A relation \(R\) from set \(A\) to set \(B\) is a subset of \(A×B.\) The number of relations from set \(A\) to set \(B\) is equal to the number of subsets of \(A×B.\) A relation from a set \(A\) to a set \(B\) is \(R,\) then the set of all first components of the ordered pairs is called the domain of \(R,\) while the set of all second components of the ordered pairs is called the range of \(R.\) There are different ways of representing a relation such as the roster form, and the set-builder form. There are different types of relations: universal, empty, identity, inverse, reflexive, symmetric, transitive, and equivalence relations.

Frequently Asked Questions (FAQs)

Q.1. What are relations in Math?
Ans: In Mathematics, a relation between two sets is a collection of ordered pairs containing one element from each set.

Q.2. What are the different types of relations?
Ans: The different types of relations are universal, empty, identity, inverse, reflexive, transitive, symmetric, and equivalence relations.

Q.3. What shows a many-to-one relationship?
Ans: In a one-to-many relationship, one element from a set can be associated with one or more elements in another set.

Q.4. What are the examples of relations?
Ans: Consider \(A = \{ 9,16\} \) and \(B = \{ 4,3, – 3, – 4\} .\) The relation is that the elements of set \(A\) are the square the elements of set \(B.\) So in set-builder form
\(R = \{ (x,y):x\) is the square of \(y,\) where \(x \in A\) and \(y \in B\} \)
Similarly, If \(A = \{ – 1,1,2\} \) and \({\rm{B = \{ 1,4,9,10\} }}\) and \(aRb\) means \({a^2} = b,\) therefore, the relation \(R\) in roster form is \(\{ ( – 1,1),(1,1),(2,4)\} .\)

Q.5. What are the four types of relations?
Ans: The four types of relations are:
(i) Reflexive relation
(ii) Symmetric relation
(iii) Transitive relation
(iv) Equivalence relation