1. Lecture # 16 Inverse of Relations
Discrete Mathematics
Lecture # 16
Inverse of Relations

2. R-1 = {(b,a) BA | (a,b) R}
Inverse of Relation
Let R be a relation from A to B. The inverse relation R-1 from B to A is defined as:
R-1 = {(b,a) BA | (a,b) R}
More simply, the inverse relation R-1 of R is obtained by interchanging the elements of all the ordered pairs in R.

3. Example Let A = {2, 3, 4} and B = {2,6,8}
and let R be the “divides” relation from A to B i.e.
for all (a,b)  A  B, a R b  a | b (a divides b)
Then R = {(2,2), (2,6), (2,8), (3,6), (4,8)}
and
R-1= {(2,2), (6,2), (8,2), (6,3), (8,4)}
In words, R-1may be defined as:
for all (b,a) B  A, b R a  b is a multiple of a.

4. Arrow Diagram of an Inverse Relation
The relation R = {(2,2), (2,6), (2,8), (3,6), (4,8)} is represented by the arrow diagram.

5. Arrow Diagram of an Inverse Relation
Then inverse of the above relation can be obtained simply changing the directions of the arrows and hence the diagram is

6. Matrix Representation of Inverse Relation
The relation
R = {(2, 2), (2, 6), (2, 8), (3, 6), (4, 8)}
from A = {2, 3,4} to B = {2, 6, 8}
is defined by the matrix M below:

7. Exercise Let R be a binary relation on a set A. Prove that:
If R is reflexive, then R-1 is reflexive.
If R is symmetric, then R-1 is symmetric.
If R is transitive, then R-1 is transitive.
If R is antisymmetric, then R-1 is antisymmetric.

8. Solution If R is reflexive, then R-1 is reflexive.
Suppose that the relation R on A is reflexive.
 a  A, (a, a) R.
Since R-1 consists of exactly those ordered pairs which are obtained by interchanging the first and second element of ordered pairs in R, therefore,
if (a, a)  R then (a, a)  R-1.
Accordingly,  a  A, (a, a)  R-1.
Hence R-1is reflexive as well.

9. Solution If R is symmetric, then R-1 is symmetric.
We will take (a, b)  R-1 and we will show that (b, a)  R-1
Let (a, b)  R-1for a,b A.
, (b, a) R. (By definition of R-1)
Since R is symmetric, therefore
(a, b) R.
b, a) R-1. (By definition of R-1)
Accordingly R-1is symmetric.

10. Solution If R is transitive, then R-1 is transitive.
Let (a, b)  R-1and (b, c)  R-1.
(b, a) R and (c, b) R. (by definition of R-1)
Now R is transitive, therefore
(c, b) R and (b, a) R then (c, a) R.
(a, c)  R (by definition of R-1)
for all a, b, c  A, if (a, b)  R-1and (b, c)  R-1then (a, c)  R-1.
R-1 is taransitive.

11. Solution if R is anti-symmetric. Then R-1 is anti-symmetric.
Let (a,b) R-1 and (b,a) R-1
(b,a) R and (a,b) R (by definition of R-1 )
we have to show that a=b.
R is anti-symmetric.
(a,b)R and (b,a)R then a = b. Thus
(a,b)  R-1 and (b,a)  R-1 then a=b.
Accordingly R-1 is antisymmetric.

12. Exercise
Show that the relation R on a set A is symmetric if, and only if,
R= R-1.

13. Solution Suppose the relation R on A is symmetric. To prove R = R-1
Let (a,b)R.
Since R is symmetric, so
(b,a)  R.
if (b,a)  R then (a,b)  R-1. (by definition of R-1)
Since (a,b) is an arbitrary element of R, so
R  R-1 …………(1)
Next, let (c,d)  R-1. By definition of R-1 (d,c) R. Since R is symmetric,
so (c,d) R. Thus we have shown that if (c,d)  R-1 then (c,d) R.Hence
R-1  R…………..(2)

14. Solution By (1) and (2) it follows that R= R-1. Conversely
suppose R = R-1.
We have to show that R is symmetric. Let (a,b)R.
Now by definition of R-1 (b,a)  R-1 .Since R = R-1, so (b,a)  R-1= R
Thus we have shown that if (a,b)R then (b,a)R
Accordingly R is symmetric.

15. Complementry Relation
Let R be a relation from a set A to a set B. The complementry relation R of R is the set of all those ordered pairs in AB that do not belong to R.
Symbolically:
= AB - R = {(a,b)  AB| (a,b)R}

16. Example Let A = {1,2,3} and R = {(1,1), (1,3), (2,2), (2,3), (3,1)}
be a relation on A
Then
= {(1,2), (2,1), (3,2), (3,3)}

17. Exercise
Let R be the relation R = {(a,b)| a<b} on the set of integers. Find
Comp(R)
R-1

18. Solution a) = ZZ - R = {(a,b)| a < b} = {(a,b)| a  b}
b) R-1= {(a,b) | a > b}

19. Exercise
Let R be a relation on a set A. Prove that R is reflexive iff is irreflexive
SOLUTION:
Suppose R is reflexive. Then by definition, for all aA, (a,a) R
But then by definition of the complementry relation (a,a)  ,  aA.
Accordingly is irreflexive.
Conversely
If is irreflexive, then (a,a)  ,  aA.
Hence by definition of , it follows that (a,a) R,  aA
Accordingly R is reflexive.

20. Exercise
Suppose that R is a symmetric relation on a set A. Is also symmetric.
SOLUTION:
Let (a,b) . Then by definition of , (a,b) R. Since R is symmetric, so
if (a,b) R then (b,a)  R.
{for (b,a)  R and (a,b)  R will contradict the symmetry property of R}
ow (b,a) R (b,a)  .Hence if (a,b)  then (b,a) 
Thus is also symmetric.

21. Composite Relation
Let R be a relation from a set A to a set B and S a relation from B to a set C. The composite of R and S denoted SoR is the relation from A to C, consisting of ordered pairs (a,c) where a A, c C, and for which there exists an element b B such that (a,b) R and (b,c) S.
Symbolically:
SoR = {(a,c)|a A, c C, b B, (a,b) R and (b,c) S}

22. Example
Define R = {(a,1), (a,4), (b,3),(c,1), (c,4)} as a relation from A to B
and S = {(1,x),(2,x), (3,y), (3,z)} be a relation from B to C.
Hence
SoR = {(a,x), (b,y), (b,z), (c,x)}

23. Composite Relation of Arrow Diagram

24. Composite Relation of Arrow Diagram
Let A = {a,b,c},B = {1,2,3,4}and C = {x,y,z}. Define relation R from A to B and S from B to C by the following arrow diagram.

25. Matrix representation of composite relation:
The matrix representation of the composite relation can be found using the Boolean product of the matrices for the relations. Thus if MR and MS are the matrices for relations R (from A to B) and S (from B to C), then
MSoR = MR OMS
is the matrix for the composite relation SoR from A to C.

26. Rules for Boolean Operations
BOOLEAN ADDITION
BOOLEAN MULTIPLICATION
1 + 1 = 1
1 . 1 = 1
1 + 0 = 1
1 . 0 = 0
0 + 0 = 0
0 . 0 = 0

27. Exercise
Find the matrix representing the relations SoR and RoS where the matrices representing R and S are

28. Solution
The matrix representation for SoR is

29. Solution
The matrix representation for RoS is

30. Exercise
Let R and S be reflexive relations on a set A. Prove SoR is reflexive.
SOLUTION:
Since R and S are reflexive relations on A, so
 a A, (a,a) R and (a,a) S
and by definition of the composite relation SoR, it is clear that
(a,a) SoR  a A.
Accordingly SoR is also reflexive flexive.