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In simple words, a relation in set theory shows how elements from one set are connected to elements in another set. It helps us understand the link or pairing between items in different sets. Functions are a special kind of relation where each input has only one output. There are many types of relations, such as:
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In mathematics, sets and relations are closely linked. A relation helps us understand how two sets are connected.
If you have two sets, a relation tells you how the elements from one set are related to the elements of the other set.
For example, if none of the elements from one set are connected to the elements of the second set, it is called an empty relation.
A relation in math shows how values from two sets are connected.
Let’s say we have set x and set y and we write them as ordered pairs like (x, y).
Given ordered pairs:
{ (1, 2), (-3, 4), (5, 6), (-7, 8), (9, 2) }
So, a relation helps us pair elements from one set with elements from another set.
There are 9 types of relations:
Example of empty relation: if there are 300 apples in the fruit bucket. There’s no possibility of finding a relation R of getting any banana in the basket. So, R is Void as it has 300 apples and no bananas.
Condition: Let A be a non-empty set then the relation \(I_{A}\)= {(a, a): a ∈ A} on A is called the identity relation on A.
Example of Identity relation: If A = {1, 2, 3}, then the relation given by the set {(1, 1), (2, 2), (3, 3)} is called the identity relation on the given set A denoted by \(I_{A}\).
Example of Universal relation: Suppose we have a set 1 that comprises all the natural numbers and another set 2 that consists of all whole numbers. Then we can say that the relation between 1 and 2 is universal as every element of set 1 is within set 2.
An empty plus universal relation is also called a trivial relation.
Condition: Consider R as a relation in a set, and A denotes a universal relation because, in this whole relation, each component of A is related to each component of A. i.e R = A × A.
In other words, a relation R in a set A is termed a universal relation if each element of A is related to every element of A, i.e., R = A × A.
Condition: Let R signify a relation on a non-empty set A, then the relation R is said to be symmetric relation ⇔ (a, b) ∈ R ⇒ (b, a) ∈ R ∀ a, b ∈ A.
Example of Symmetric relation: If A = {3, 4}, then relation R = {(3, 3), (4, 4), (3, 4), (4, 3)} is symmetric ∵ if (a, b) ∈ R then (b, a) ∈ R ∀ a, b ∈ A.
Example of anti-symmetric relation: If A = {3, 4, 5}, then relation R = {(3, 3), (4, 4), (5, 5)} is anti-symmetric ∵ if (a, b) ∈ R and (b, a) ∈ R then a = b ∀ a, b ∈ A.
Example of Transitive relation: If A = {3, 4, 5}, then relation R = {(3, 4), (4, 5), (3, 5)} is transitive ∵ if (a, b) ∈ R and (b, c) ∈ R then (a, c) ∈ R, ∀ a, b, c ∈ A.
Conditions: If the relation R is reflexive, then all the elements of the given set are mapped with itself, such that for all a∈Q, then (a, a)∈R.
The relation R is said to be symmetric on set Q, if (a,b)∈R, then (b, a)∈R, such that a, b ∈ Q.
The relation R on set Q, if (a,b)∈R and (b,c)∈R, then (a,c)∈R for all a,b,c ∈R is said to be a transitive relation.
Example of Equivalence relation: If A = {3, 4, 5}, then relation R = {(3, 3), (4, 4), (5, 5), (3, 5), (5, 3), (3, 4), (4, 5)} is an equivalence relation ∵ the relation R is reflexive, symmetric and transitive as shown above respectively.
Condition: Let A and B be any two non-empty sets and R be the relation from A to B. Then, the inverse of R, denoted by \(R^{-1}\) is a relation from B to A and is given by; \(R^{-1}\) = {(b, a): (a, b) ∈ R}
Example of Inverse relation: If A = {a, c}, B = {d, e} and the relation R from A to B is given by: R = {(a, d), (c, e)}. Find \(R^{-1}\)?
If R is a relation from A to B where A and B are non-empty sets then inverse relation \(R^{-1}\) is given y: \(R^{-1}\) = {(b, a): (a, b) ∈ R}
⇒ \(R^{-1}\) = {(d, a), (e, c)}
Example of reflexive relation: If the set B={3, 4, 5} then the relation {(3,3),(4,4),(5,5)} is a reflexive relation.
Here, each element of set B is mapped with itself, such that 3∈A, (3,3)∈R.
The number of reflexive relations on a set with n elements is presented by \(N=2^{n\left(n-1\right)}\), where capital N is the number of reflexive relations and small n is the number of elements in the set.
Here is a brief summary of the types of relations in discrete mathematics along with their representation:
Types of Relations |
Representation |
Empty Relation |
R = ∅ ⊂ A × A |
Universal Relation |
R = A × A |
Identity Relation |
I or \(I_{A}\)= {(a, a), a ∈ A} |
Inverse Relation |
\(R^{-1}\) = {(b, a): (a, b) ∈ R} |
Reflexive Relation |
(a, a) ∈ R |
Symmetric Relation |
aRb ⇒ bRa, ∀ a, b ∈ A |
Transitive Relation |
aRb and bRc ⇒ aRc ∀ a, b, c ∈ A |
Some other types of relations depending on the mapping of two sets are as follows:
One-to-One Relations: A relation is said to be One to One relation if all the different elements of one set are outlined to different elements of another set.
One-to-Many Relations: A relation is supposed to be a One to Many relations if the identical element of one set is mapped to more than one element of a different set.
Many-to-One Relations: A relation is stated to be a Many-to-One relation if all the different elements of one set are mapped to the corresponding element of another set.
Many to Many Relations: A relation is supposed to be Many to Many relations if one or more than one component of one set is mapped with identical elements of another set.
We often address relations between any two or more objects. Below mentioned are some important terms regarding types of relations.
Domain of a Relation: Let R express a relation from set A to set B. Then, the set of all the first components of the ordered pair relating to relation R makes the domain of the relation R.
i.e. Domain (R) = {a: (a, b) ∈ R}
Example of relation: Let A = {2, 4} and B = {p, q} be two sets such that the relation R from A to B is given by:
R = {(2, p), (2, q), (4, p), (4, q)}.
Find the domain of given relation R.
The domain of a relation R is given by; Domain (R) = {a: (a, b) ∈ R}
⇒ Domain (R) = {2, 4}
Range of a Relation: Let R express a relation from set A to set B then the set of all second components of the ordered pair belonging to relation R forms the range of the relation R.
i.e. Range (R) = {b: (a, b) ∈ R}
Example of relation: If A = {a, c}, B = {d, e} and the relation R from A to B is given by: R = {(a, e), (c, e)}.
Find the range of given relation R.
The range of a relation R is given by Range (R) = {b: (a, b) ∈ R}
⇒ Range (R) = {e}
Codomain of a Relation: Let R signify a relation from set A to set B, then the set B is called the codomain of the relation R. i.e. Codomain (R) = B.
Relation defines the connection between two sets. There are 3 different methods of representing the relations. Each of them is discussed below with examples:
Representation of Relations in Set-Builder Form
A shorthand approach is applied to write sets and is often practiced for sets with an infinite number of components. It is practiced with various types of numbers, such as integers, natural numbers, whole numbers, real numbers and so on. The set-builder form is also applied to display sets with an interval or an equation.
Example: The relation of two sets P={3, 4, 5} and Q={9, 16, 25} in which elements of set P are the positive square root of components of set Q. This relation can be addressed in set-builder form as follows.
R={(a,b):a is the positive square root of b,a∈P,b∈Q,}
Representation of Relations in Roaster Form
Roster form is a description of a set that records down all of the elements existing in the set and is distributed by commas and confined within braces.
The relation can be expressed in roaster form by addressing all the possible ordered pairs of the two sets.
Example: The relation of two sets P={3, 4, 5} and Q={9, 16, 25} in which elements of set P are the square root of components of set Q. The relation can be formulated in roaster form as follows.
R = {(3,9),(4,16), (5,25)}
Representation of Relations in Arrow Diagram
In the arrow diagram method as the name suggests, the relation between sets is indicated by drawing arrows from the first elements to the second elements of all the combinations which relate to the relation.
That is the relationship between two sets is determined by using the arrow formed from one given set to another set.
Example: The relation of two sets P={3, 4, 5} and Q={9, 16, 25}, in which elements of set P are the square root of components of set the Q. The relation can be addressed by using an arrow diagram as shown below:
The important points on types of relations are given below:
Different types of relations like reflexive, symmetric, transitive, and equivalence are widely used in mathematics, computer science, and real-life modeling. They help in organizing data, building networks, defining hierarchies, and designing algorithms for sorting, scheduling, and classification.
With the knowledge of composition of relations and their various types let us check out some of the solved examples for the same.
Example 1: Given a set X = {p, q, r}. Which of the following is an identity relation on X?
1) R = {(p, q), (p, r)}
2) R = {(p, p), (q, q), (r, r)}
3) R = {(p, p), (q, q), (p, r)}
4) none of these
Solution: Concept:
Identity Relation: Let A be a non-empty set then the relation \(I_{A}\) = {(a, a): ∀ a ∈ A} on A is called the identity relation on A.
Example:
If A = {1, 2, 3}, then relation given by the set {(1, 1), (2, 2), (3, 3)} is called the identity relation on the given set A denoted by \(I_{A}\).
Calculation: Given: X = {p, q, r}.
If A is a non-empty set then the relation \(I_{A}\) = {(a, a): ∀ a ∈ A} on A is called the identity relation on A.
So, the identity relation on a given set X is given by: \(I_{x}\)= {(p, p), (q, q), (r, r)}.
Hence option 2 is the correct one.
Example 2: Which of the following statements is/are TRUE for an empty relation R on a non-empty set S?
1) R is not reflexive
2) R is not symmetric
3) R is not transitive
4) All of these
Solution: Since R is empty and S is non-empty, R is not reflexive. Since R is empty, R is both symmetric and transitive. Hence option 1 is the correct one.
Example 3: How many reflexive relations are there on a set with 4 elements?
Solution: Data:
Number of elements in a set = n = 4
Formula: Total number of reflexive relations in a set = \(2^{n^{2}-n}\).
Calculation: Total number of reflexive relations in a set = \(2^{4^{2}-4}=2^{12}\).
Example 4: Consider the following statements:
(I) If the binary relation itself is transitive, then the transitive closure is that same binary relation.
(II) An asymmetric relation is always anti-symmetric.
(III) An empty relation on a nonempty set is always symmetric.
Which of the above statement/s is/are TRUE?
1) (II) and (III) only
2) (II) only
3) (I) and (III) only
4) (I), (II) and (III)
Solution: Answer: Option 4
Explanation:
(I) If the binary relation itself is transitive, then the transitive closure is that same binary relation, otherwise it will be different.
(II) An asymmetric relation is always anti-symmetric, as anti-symmetric relation allows self-loops in digraph but asymmetric does not. So the asymmetric condition is more strict.
(III) An empty relation on a nonempty set is always symmetric, but not reflexive.
Example 5: If the number of elements in a set A is 11, then how many anti-symmetric relations are possible on A?
1) \(2^{11}\)
2)\(2^{11}\times 3^{55}\)
3) \(11!\)
4) \(2^{11}\times 3^{11}\)
Solution: Number of anti-symmetric relations on a set with n elements =\(2^{n}\times3^{\frac{n^{2}−n}{2}}=2^{11}.3^{55}\).
Hence option 2 is correct.
Example 6: If A = {2, 4, 5}, B = {5, 9, 11} and R is the relation from A to B such that a R b ⇔ b = 2a + 1 then the inverse relation of R is?
Solution: Concept:
Inverse Relation: Let R be a relation from A to B. Then the inverse relation, denoted by \(R^{-1}\), from the set B to A, is defined as, \(R^{-1}\) = {(b, a) : (a, b) ∈ R}.
Calculation: Given: A = {2, 4, 5}, B = {5, 9, 11} and R is the relation from A to B such that a R b ⇔ b = 2a + 1.
So, when a = 2 ⇒ b = 2a + 1 = 5 ∈ B ⇒ (2, 5) ∈ R
When a = 4 then b = 2a + 1 = 9 ∈ B ⇒ (4, 9) ∈ R
When a = 5 then b = 2a + 1 = 11 ∈ B ⇒ (5, 11) ∈ R
⇒ R = {(2, 5), (4, 9), (5, 11)}
If R is a relation from A to B. Then the inverse relation, denoted by\(R^{-1}\), from the set B to A, is defined as, \(R^{-1}\) = {(b, a) : (a, b) ∈ R}.
⇒ \(R^{-1}\) = {(5, 2), (9, 4), (11, 5)}
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