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Summary Relation
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RelationRELATIONS
INTRODUCTION :
Let A and B be two sets. Then a relation R from A to B is a subset of A × B.
thus, R is a relation from A to B R A × B.
Ex. If A = {1, 2, 3} and B = {a, b, c}, then R = {(1, b), (2, c), (1, a), (3, a)} being a subset of A × B, is a relation
from A to B. Here (1, b), (2, c), (1, a) and (3, a) R, so we write 1 Rb, 2Rc, 1Ra and 3Ra. But (2, b) R, so
we write 2 R b
Total Number of Realtions : Let A and B be two non-empty finite sets consisting of m and n elements
respectively. Then A × B consists of mn ordered pairs. So, total number of subsets of A × B is 2mn.
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 belonging to R is called to domain of R, while the set of all
second components or coordinates of the ordered pairs in R is called the range of R.
Thus, Dom (R) = {a : (a, b) R}
and, Range (R) = {b : (a, b) R}
It is evident from the definition that the domain of a relation from A to B is a subset of A and its range is a
subset of B.
Ex. Let A = {1, 3, 5, 7} and B = {2, 4, 6, 8} be two sets and let R be a relation from A to B defined by the
phrase "(x, y) R x > y". Under this relation R, we have
3R2, 5R2, 5R4, 7R2, 7R4 and 7R6
i.e. R = {(3, 2), (5, 2), (5, 4), (7, 2), (7, 4), (7, 6)}
Dom (R) = {3, 5, 7} and Range (R) = {2, 4, 6}
Inverse 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) R}
Clearly, (a, b) R (b, a) R–1
Also, Dom(R) = Range(R ) and Range (R) = Dom (R–1)
–1
Ex.1 Let A be the set of first ten natural numbers and let R be a relation on A defined by (x, y) R x + 2y = 10,
i.e. R = {(x, y) : x A, y A and x + 2y = 10}. Express R and R –1 as sets of ordered pairs. Determine also (i)
domain of R and R–1 (ii) range of R and R–1 A- { 112,3 4,5 6,7 8,9 10 }
= , , , ,
✗ +24=10
10 x
Sol. We have (x, y) R x + 2y = 10 y= , x, y A ✗ 7- Y
2 2
' '
2
3
12={4,444,3%(6,2%40,1)}
3
where A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 12-1={144.419,4/63,483}
4
4
= 5
6 6
7- 7
10 1 9 g
8 8
Now, x=1 y= = A.
y
10 10
2 2
This shows that 1 is not related to any element in A. Similarly we can observe. that 3, 5, 7, 9 and 10 are not
related to any element of A under the defined relation
node06\B0AH-AI\Kota\JEE(Advanced)\Enthusiast\Advance Batch\Maths\Sheet\Det.\Eng.p65
Further we find that :
10 2
For x = 2, y = =4 A (2, 4) R
2
10 4
For x = 4, y = =3 A (4, 3) R
2
10 6
For x = 6, y = =2 A (6, 2) R
2
10 8
For x = 8, y = =1 A (8, 1) R
2
E 25
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