Relation
A connection between the elements of two or more sets is Relation. The sets must be non-empty. A subset of the Cartesian product also forms a relation R. A relation may be represented either by Roster method or by Set-builder method.
Let A and B be two sets such that A = {2, 5, 7, 8, 10, 13} and B = {1, 2, 3, 4, 5}. Then,
R = {(x, y): x = 4y – 3, x ∈ A and y ∈ B} (Set-builder form)
R = {(5, 2), (10, 3), (13, 4)} (Roster form)
Explanation
Given any two non-empty sets A and B, A relation R from A to B is a subset of the Cartesian product A x B and is derived by describing a relationship between the first element (say x) and the other element (say y) of the ordered pairs in A & B.
Consider an example of two sets, A = {2, 5, 7, 8, 9, 10, 13} and B = {1, 2, 3, 4, 5}. The Cartesian product A × B has 30 ordered pairs such as A × B = {(2, 3), (2, 5)…(10, 12)}. From this, we can obtain a subset of A × B, by introducing a relation R between the first element and the second element of the ordered pair (x, y) as
R = {(x, y): x = 4y – 3, x ∈ A and y ∈ B}
Then, R = {(5, 2), (9, 3), (13, 4)}.
(Arrow representation of the Relation R)
Representation of Relation
A relation is represented either by Roster method or by Set-builder method. Consider an example of two sets A = {9, 16, 25} and B = {5, 4, 3, -3, -4, -5}. The relation is that the elements of A are the square of the elements of B.
- In set-builder form, R = {(x, y): x is the square of y, x ∈ A and y ∈ B}.
- In roster form, R = {(9, 3), (9, -3), (16, 4), (16, -4), (25, 5), (25, -5)}.
Terminologies
- Before getting into details, let us get familiar with a few terms:
- Image: Suppose we are looking in a mirror. What do we see? An image or reflection. Similarly, for any ordered pairs, in any Cartesian product (say A × B), the second element is called the image of the first element.
- Domain: The set of all first elements of the ordered pairs in a relation R from a set A to a set B.
- Range: The set of all second elements in a relation R from a set A to a set B.
- Codomain: The whole set B. Range ⊆ Codomain.
Total Number of Relations
For two non-empty set, A and B. If the number of elements in A is h i.e., n(A) = h & that of B is k i.e., n(B) = k, then the number of ordered pair in the Cartesian product will be n(A × B) = hk. The total number of relations is 2hk.
Solved Examples for You
Problem: Let A = {5, 6, 7, 8, 9, 10} and B = {7, 8, 9, 10, 11, 13}. Define a relation R from A to B by
R = {(x, y): y = x + 2}. Write down the domain, codomain and range of R.
Solution: Here, R = {(5, 7), (6, 8), (7, 9), (8, 10), (9, 11)}.
Domain = {5, 6, 7, 8, 9}
Range = {7, 8, 9, 10, 11}
Co-domain = {7, 8, 9, 10, 11, 13}.
Types of Relations or Relationship
Let us study about the various types of relations.
Empty Relation
If no element of set X is related or mapped to any element of X, then the relation R in A is an empty relation, i.e, R = Φ. Think of an example of set A consisting of only 100 hens in a poultry farm. Is there any possibility of finding a relation R of getting any elephant in the farm? No! R is a void or empty relation since there are only 100 hens and no elephant.
Universal Relation
A relation R in a set, say A is a universal relation if each element of A is related to every element of A, i.e., R = A × A. Also called Full relation. Suppose A is a set of all natural numbers and B is a set of all whole numbers. The relation between A and B is universal as every element of A is in set B. Empty relation and Universal relation are sometimes called trivial relation.
Identity Relation
In Identity relation, every element of set A is related to itself only. I = {(a, a), ∈ A}. For example, If we throw two dice, we get 36 possible outcomes, (1, 1), (1, 2), … , (6, 6). If we define a relation as R: {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}, it is an identity relation.
Inverse Relation
Let R be a relation from set A to set B i.e., R ∈ A × B. The relation R-1 is said to be an Inverse relation if R-1 from set B to A is denoted by R-1 = {(b, a): (a, b) ∈ R}. Considering the case of throwing of two dice if R = {(1, 2), (2, 3)}, R-1 = {(2, 1), (3, 2)}. Here, the domain of R is the range of R-1 and vice-versa.
Reflexive Relation
If every element of set A maps to itself, the relation is Reflexive Relation. For every a ∈ A, (a, a) ∈ R.
Symmetric Relation
A relation R on a set A is said to be symmetric if (a, b) ∈ R then (b, a) ∈ R, for all a & b ∈ A.
Transitive Relation
A relation in a set A is transitive if, (a, b) ∈ R, (b, c) ∈ R, then (a, c) ∈ R, for all a, b, c ∈ A
Equivalence Relation
A relation is said to be equivalence if and only if it is Reflexive, Symmetric, and Transitive. For example, if we throw two dices A & B and note down all the possible outcome.
Define a relation R= {(a, b): a ∈ A, b ∈ B}, we find that {(1, 1), (2, 2), …, (6, 6) ∈ R} (reflexive). If {(a, b) = (1, 2) ∈ R} then, {(b, a) = (2, 1) ∈ R} (symmetry). ). If {(a, b) = (1, 2) ∈ R} and {(b, c) = (2, 3) ∈ R} then {(a, c) = (1, 3) ∈ R} (transitive)
Solved Example for You
Problem: Three friends A, B, and C live near each other at a distance of 5 km from one another. We define a relation R between the distances of their houses. Is R an equivalence relation?
Solution: For an equivalence Relation, R must be reflexive, symmetric and transitive.
- R is not reflexive as A cannot be 5 km away to itself.
- The relation, R is symmetric as the distance between A & B is 5 km which is the same as the distance between B & A.
- R is transitive as the distance between A & B is 5 km, the distance between B & C is 5 km and the distance between A & C is also 5 km.
Therefore, this relation is not equivalent.
