presentation on inverse of relation

1. IRREFLEXIVE
Partial Order Relation
Hafiz Muhammad Shahzad

2. Fourth Type of relation: IRREFLEXIVE
RELATION:
• Let R be a binary relation on a set A. R is irreflexive iff for all a∈A,(a,a)
∉R.
• That is, R is irreflexive if no element in A is related to itself by R.

3. EXAMPLE:
Let A = {1,2,3,4} and define the following relations on A:
R1 = {(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)}
R2 = {(1,1), (1,2), (2,1), (2,2), (3,3), (4,4)} it is actually reflexive not irrflexive.
R3 = {(1,2), (2,3), (3,3), (3,4)}
Then
R1 is irreflexive since no element of A is related to itself in R1. i.e. (1,1)∉ R1 ,
(2,2) ∉ R1 , (3,3) ∉ R1 ,(4,4) ∉ R1
R2 is not irreflexive, since all elements of A are related to themselves in R2
R3 is not irreflexive since (3,3) ∈R3. Note that R3 is not reflexive.

4. DIRECTED GRAPH OF AN IRREFLEXIVE
RELATION:
• Let R be an irreflexive relation on a set A. Then by definition, no
element of A is related to itself by R. Accordingly, there is no loop at
each point of A in the directed graph of R.
EXAMPLE: Let A = {1,2,3} and R = {(1,3), (2,1), (2,3), (3,2)}
be represented by the directed graph.

5. MATRIX REPRESENTATION OF AN IRREFLEXIVE
RELATION
• its matrix representation the diagonal elements are all zero, if one of
them is not zero the we will say that the relation is not irreflexive.
• EXAMPLE: Let A = {1,2,3} and R = {(1,3), (2,1), (2,3), (3,2)} be
represented by the matrix.
• Then R is irreflexive, since all elements in the main diagonal are 0’s.

6. Fifth Type of relation ANTISYMMETRIC
RELATION:
• If (a,b) then no (b,a) but each element can be related.
Points
• If set (2,1) exist in R then (1,2) should not be in R then it is called
Antisymmetric.
• (1,1), (2,2) sets in the relation can also be called antisymmetirc.
• Symmetric and antisymmetirc are not opposite. Important point.

7. Example
Let A = {1,2,3,4} and define the following relations on A.
R1 = {(1,1),(2,2),(3,3)}
R2 = {(1,2),(2,2), (2,3), (3,4), (4,1)}
R3={(1,3),(2,2), (2,4), (3,1), (4,2)}
R4={(1,3),(2,4), (3,1), (4,3)}
Solution
R1 is anti-symmetric and symmetric .
R2 is anti-symmetric but not symmetric because (1,2) ∈ R2but (2,1) ∉ R2.
R3 is not anti-symmetric since (1,3) & (3,1) ∈ R3 but 1 ≠ 3. Note that R3 is
symmetric.
R4 is neither anti-symmetric because (1,3) & (3,1) ∈ R4 but 1 ≠ 3 nor symmetric
because (2,4) ∈ R4 but (4,2) ∉R

8. DIRECTED GRAPH OF AN ANTISYMMETRIC
RELATION:
• Let R be an anti-symmetric relation on a set A. Then by definition, no
two distinct elements of A are related to each other. Accordingly,
there is no pair of arrows between two distinct elements of A in the
directed graph of R.

9. EXAMPLE:
• Let A = {1,2,3} And R be the relation defined on A is R ={(1,1), (1,2), (2,3),
(3,1)}.Thus R is represented by the directed graph as
• R is anti-symmetric, since there is no pair of arrows between two distinct
points in A.

10. MATRIX REPRESENTATION OF AN
ANTISYMMETRIC RELATION:
• Point: Each first row and each first column must be different.
• EXAMPLE: Let A = {1,2,3} and a relation R = {(1,1), (1,2), (2,3), (3,1)}on
A be represented by the matrix.
• Then R is anti-symmetric as clear by the form of matrix M

11. PARTIAL ORDER RELATION:
• Let R be a binary relation defined on a set A. R is a partial order
relation,if and only if, R is reflexive, antisymmetric, and transitive.
• Please do not mix with EQUIVALENCE RELATION.
• Let A be a non-empty set and R a binary relation on A. R is an
equivalence relation if, and only if, R is reflexive, symmetric, and
transitive.

12. Example
• Let R be the set of real numbers and define the“less than or equal to” ,
on R as follows:
for all real numbers x and y in R. x ≤y ⇔ x.
Show that ≤ is a Let R be the set of real numbers and define the“less than or
equal to” , on R as follows: for all real numbers x and y in R.x ≤y ⇔ x < y or x = y
Show that ≤ is a partial order relation relation.
Reflexive
Real Numbers= {-3,-2,-1,0,1,2,3, 3.5, 7.1, 8,9,10,11...}
(<=) R= { (0,0), (1,1), (2,2), (3.5,3.5) } This relation is Reflexive.

13. Example ...con
• for all real numbers x and y in R. x ≤y ⇔ x. This is the condition.
Real Numbers= {-3,-2,-1,0,1,2,3, 3.5, 7.1, 8,9,10,11...}
R<= { (0,0), (1,1), (2,2), (3.5,3.5) } This relation is Reflexive.
Anti Symmetric
If I take (1,2) then (2,1) is not exist due to the condition. It means this
relation is anti symmetric.
Transitive
take (1,2) , (2,4) (1,4) It means all condition are successfully fullfilled. It
is Transitive as well.

14. Example
• Let “R” be the relation defined on the set of integers Z as follows:
for all a, b ∈Z, aRb iff b=ar for some positive integer r. Show that R is a
partial order on Z.
Solution:
Z= {-2,-1,0, 1, 2 ,3 4}
Reflexive:
Take a=0, r=1. => b=ar = (0)1 =0 ,b=0 hence , first order pair comes (0,0)
Take a=1, r=1 => b=ar = (1)1 =1 ,b= 1 hence second order pair comes (1,1)
Take a=2, r=1 => b=2 b=ar = (2)1 =2 ,b= 2 hence third order pair comes (2,2)
Hence (0,0), (1,1) and (2,2) it is reflexive.

15. Example ...con
Anti Symmetric:
Take a=1, r=2 => b=ar = (1)2 =1 ,b=1 hence , order pair comes (1,1)
Take a=2, r=2 => b=ar = (2)2 =4 ,b= 4 hence second order pair comes
(2,4) , means (4,2) should not be exit is the subset
Take a=4, r=2 b=ar = (4)2 =16 ,b= 16 hence third order pair comes
(4,16), hence (4,2) is not exist.
Or
a=4, r=1 b=ar = (4)1 =4, b= 4, hence third order pair comes (4,4), hence
(4,2) is also not exist.
it is Anti Symmetric.

16. Example ...con
Transitive:
Take a=2, r=2 => b=ar = (2)2 =4,b=4 hence , order pair comes (2,4)
Take a=4, r=2 b=ar = (4)2 =16 ,b= 16 hence second order pair comes (4,16).
Take a=4, r=2 => b=ar = (2)2 =4 ,b= 4 hence second order pair comes (2,4)
Take a=2, r=4 b=ar = (2)4 =16 ,b= 16 hence third order pair comes (2,16).
R= {(2,4),(4,16),(2,16)}
This relation is transitive as well.
Hence, it is proved that this is partial order of the given conditon. b=ar

17. Inverse of Relations
Composite Relation
Complementry Relation

18. INVERSE OF A RELATION (R-1):
• the inverse relation R-1 of R is obtained by interchanging the elements
of all the ordered pairs in R.
• E.g
• R={ (2,2), (5,6), (2,7), (3,6),(4,8)}
• R-1 ={ (2,2), (6,5), (7,2), (6,3),(8,4)}

19. 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.
• Then inverse of the above relation can be obtained simply changing
the directions of the arrows and hence the diagram is

20. 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.
• The matrix representation of inverse relation R-1 is obtained by
simply taking its transpose. (i.e., changing rows by columns and
columns by rows). Hence R-1 is represented by Mt as shown.

21. 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:
R-1 = A×B - R

22. Example:
• Find Complementry Relation;
• Let A = {1,2,3} and B={1,2}
• R = {(1,1), (1,3), (2,2), (2,3), (3,1)}
Solution
A*B= {(1,1), (1,2), (2,1),(2,2),(3,1),(3,2)}
R-1= A×B – R = { (1,2), (2,1),(3,2) }

23. Example
• Let R be a binary relation on a set A. Prove that
(i) If R is reflexive, then R-1 is reflexive.
(ii) If R is symmetrice, then R-1 is Symmetrice.
(iii) If R is transirive , then R-1 is transitive
(iv) if R is antisymmetrice, then R-1 is anti symmetric.
Let A={(1,2,3}
Reflexive R={ (1,1), (2,2),(3,3)} => R-1 is reflexive as well.
Symmetrice R={ (1,2), (2,1)} => R-1= {(2,1), (1,2)} is Symmetric as well.
Transtive R={ (1,2),(2,3), (1,3)}=> R-1 ={(2,1),(3,2),(3,1) } is transitive as well.
AntiSymmetric R= { (1,2), (3,1)} => R-1 ={ (2,1),(1,3)} is antisymmetric

24. Example
• Let R be a relation on a set A. Prove that R is reflexive iff R-1 is
irreflexive.
Solution:
Let A={1,2}
A*A={(1,1), (1,2),(2,1),(2,2)} => Reflexive =(1,1), (2,2)
R-1 =A×B – R= { (1,2), (2,1} it is irreflexive.

25. Example
• Suppose that R is a symmetric relation on a set A. Is R(Bar) also
symmetric.
Solve your own

26. Solution
• Let A={1,2}
• R= {(1,2), (2,1)}
• A*A ={ (1,1),(1,2),(2,1),(2,2)}
• R(bar)= A*A-R
• R(BAR)= {(1,1),(2,2)}
• Yes it is symmetric.

27. Composite Relation

28. Example
• R= {(a,1),(a,4),(b,3) ,(c,1),(c,4)} as a relation from A to B
• S={(1,x), (2,x), (3,y), (3,z) } Be a relation from B to C. Write Composite
relation of R and S.
SoR={ } find
E.g Second element of R and first element of S
• Composite relation of R and S is
SoR={(a,x), (b,y), (b,z),(c,x)}

29. Composite relation from 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
above diagram.

30. Solution
You have to see set B , each element of set B that
have arrow and head is pointing will be taken.
Composite relation of R and S => ROS

31. Example
• Let X = {1,2,3},Y′={a, b, c, d}, Y={a, b, c, d, e} and Z ={x, y, z}. Define
functions f:X→Y′ and g: X →Z by the arrow diagrams.

32. Solution

33. Matrix represenation in Composite relation

34. Matrix
• Find the matrix reresenting thereatliosn SoR and RoS where the
matrices representing R and S are

35. Solution

36. Example
• Let R and S be reflexive relations on a set A. Prove SoR is reflexive.

37. Solultion
• Let A= {1,2}
• R={(1,1), (2,2)}
• S=((1,1), (2,2),(2,1))
• SoR={ (1,1), (2,2), (2,1) }
• SoR is relexive.