Relations

In everyday life, we frequently speak of relations between two or more objects. To learn the concept properly, consider the following example.
Let A={1,2,3,5} and B={2,4}, then
A⨯B={(1,2),(1,4),(2,2),(2,4),(3,2),(3,4),(5,2),(5,4)}

We can obtain a subset of A⨯B by introducing a relation between the elements of set A and B.
Let the relation R be 'is less than'
This relation R means 'first element is less than second element'. Let's find the pairs from A⨯B which satisfy this condition.

(1,2)	(1,4)	(2,2)	(2,4)	(3,2)	(3,4)	(5,2)	(5,4)
1<2 R is True	1<4 R is True	2=2 R is False	2<4 R is True	3>2 R is False	3<4 R is True	5>2 R is False	5>4 R is False

Now we write only the pairs which satisfy the relation R.
R={(1,2),(1,4),(2,4),(3,4)}
Thus the relation 'is less than' from the set A to the set B gives rise to a subset R of A⨯B such that (x,y)∈R if and only if x<y.

Definition: Let A and B be two non empty sets. Then, a relation R from A to B is a subset of A⨯B i.e. R⊂A⨯B.
Relation from A to B is represented as R:A→B
Relation from B to A is represented as R:B→A

Let R be a relation from A to B,
If R=ϕ, then R is called the empty relation.
If R=A⨯B, then R is called the universal relation.

Representation of a relation
(i) Roaster form
In this form, a relation is represented by the set of all ordered pairs which belong to the given relation.
From the above example, R={(1,2),(1,4),(2,4),(3,4)}

(ii) Set-builder form
In this form, the relation is represented as {(x,y): x∈A, y∈B, x__y}, blank space is to be replaced by the relation which associates x and y.
From the above example, R={(x,y): x∈A, y∈B, x<y}

(iii) By arrow diagram
In this form, the relation is represented by drawing arrows from first set elements to the second set elements which belong to the given relation.

Domain, Range, Co-domain
Let A,B be non-empty sets and R be a relation from A to B, then
Domain of the relation R, is the set of all first component of the ordered pairs which belong to R.
Range of the relation R is the set of all second components of the ordered pairs which belong to R.
Co-domain is the set of all elements of second set i.e. co-domain is the whole set B.

Example:
A={-1,2,5,8} and B={0,1,3,6,7}
A⨯B={(-1,0),(-1,1),(-1,3),(-1,6),(-1,7), (2,0),(2,1),(2,3),(2,6),(2,7),(5,0),(5,1),(5,3),(5,6),(5,7),(8,0),(8,1),(8,3),(8,6),(8,7)}
B⨯A={(0,-1),(0,2),(0,5),(0,8),(1,-1),(1,2),(1,5),(1,8),(3,-1),(3,2),(3,5),(3,8),(6,-1),(6,2),(6,5),(6,8),(7,-1),(7,2),(7,5),(7,8)}

$\displaystyle \small R_{1}$:A→B
$\displaystyle \small R_{1}$={(x,y): x∈A, y∈B, y=x+1}
x is element of set A.
∴y= x+1⇒ -1+1=0 (-1,0), 2+1=3 (2,3), 5+1=6 (5,6), 8+1=9 (8,9)
$\displaystyle \small R_{1}$= {(-1,0),(2,3),(5,6)} [(8,9) is not considered because 9 is not an element of set B]
Domain= {-1,2,5}
Range= {0,3,6}
Co-domain= set B={0,1,3,6,7}

$\displaystyle \small R_{2}$:A→B
$\displaystyle \small R_{2}$={(x,y): x∈A, y∈B, x<y}
$\displaystyle \small R_{2}$={(-1,0),(-1,1),(-1,3),(-1,6),(-1,7),(2,3),(2,6),(2,7),(5,6),(5,7)}
Domain= {-1,2,5}
Range= {0,1,3,6,7}
Co-domain= set B={0,1,3,6,7}

$\displaystyle \small R_{3}$:A→B
$\displaystyle \small R_{3}$={(x,y): x∈A, y∈B, y=$\displaystyle \small x^{2}$}
$\displaystyle \small R_{3}$={(-1,1)}
Domain= {-1}
Range= {1}
Co-domain= set B={0,1,3,6,7}

$\displaystyle \small R_{4}$:A→B
$\displaystyle \small R_{4}$={(x,y): x∈A, y∈B, y=x+3}
$\displaystyle \small R_{4}$= ϕ
Domain= ϕ
Range= ϕ
Co-domain= set B={0,1,3,6,7}

$\displaystyle \small R_{5}$:B→A
$\displaystyle \small R_{5}$={(x,y): x∈B, y∈A, y=x+2}
$\displaystyle \small R_{5}$={(0,2),(3,5),(6,8)}
Domain= {0,3,6}
Range= {2,5,8}
Co-domain= set A={-1,2,5,8}

Number of Relations
Let A and B be any two non-empty finite sets containing p and q elements respectively i.e. n(A)=p, n(B)=q, then
Number of ordered pairs in A⨯B = pq
Total number of subsets of A⨯B = $\displaystyle \small 2^{pq}$
Total number of relations from A to B = $\displaystyle \small 2^{pq}$
because, each relation from A to B is a subset of A⨯B.

Example: A={-1,2} and B={3,4,5}
n(A)=2, n(B)=3
Number of ordered pairs in A⨯B = 2*3=6
Total number of subsets of A⨯B = $\displaystyle \small 2^{pq}$= $\displaystyle \small 2^{6}$ =64
Total number of relations from A to B = $\displaystyle \small 2^{pq}$= $\displaystyle \small 2^{6}$ =64