Given below are some of the concepts in relation.
Total Number of Relations:Let the number of elements in sets 'A' and 'B' be 'm' and 'n' respectively, then, the total number of relations is 2
^{mn}.
Representation of a Relation:A relation can be represented in four ways as follows:
Roster Form: We will demonstrate this with the help of an example.
Let A = {-2, -1, 0, 1, 2} and B = {0, 1, 4, 9, 10}.
Let 'R' be defined by:
a R b => a
^{2} = b
Then, R = {(0, 0), (-1, 1), (-2, 4), (1, 1), (2, 4)}
Set-Builder Form:
R = {(a, b) : a $\epsilon $ A, b $\epsilon $ B and a, b satisfies the rule which associates a and b}
Let A = {1, 2, 3, 4, 5} and B = {1, $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$, $\frac{1}{5}$, $\frac{1}{6}$, ..}
Then, R = {(1, 1), (2, $\frac{1}{2}$), (3, $\frac{1}{3}$), (4, $\frac{1}{4}$), (5, $\frac{1}{5}$)} or we can say
R = {(a, b) : a $\epsilon $ A, b $\epsilon $ B and b = 1/a}
By Arrow Form: We draw arrows from first component's top the second components of the ordered pairs belonging to 'R'.
By Lattice: In this form, a representation of relation 'R' is done by darkening the dots in the lattice for A x B which is the ordered pairs in R.
Domain and Range of a Relation:
Let 'R' be a relation from the set 'A' to 'B'. The set of all first components of the ordered pairs which are there in 'R' is termed as the Domain of 'R'. The set of all the second components of the ordered pairs belonging to 'R' is called the Range of 'R'.
Dom (R) = {a : (a, b) $\epsilon $ R} and Range (R) = {b : (a, b) $\epsilon $ R}
Relation on a Set:
If 'A' is any non-void set, then a relation from 'A' to itself (i.e. a subset of A x A) is called a relation on a set.
Inverse of a Relation:
Let 'A' and 'B' be two sets. Let 'R' be a relation from 'A' to 'B'. Then, the inverse of 'R', denoted by 'R^{-1}' is a relation from 'B' to 'A'. R^{-1} is defined by:
R^{-1} = {(b, a) : (a, b) $\epsilon $ R}
Dom(R) = Range (R^{-1}) and Range (R) = Dom (R^{-1}) Solved Example
Question: Let R be the relation on the set N of natural numbers defined by R = {(a, b) : a + 3b = 12, a, b $\epsilon $ N}. Find
Solution:
a + 3b = 12
a = 12 - 3b
Substituting b = 1, 2, 3, we get a = 9, 6, 3 respectively.
For b = 4, we get a = 0 $\notin$ N. And for b > 4, a $\notin$ N.
So, R = {(9, 1), (6, 2), (3, 3)}
Domain of R = {9, 6, 3} and Range of R = {1, 2, 3}.