Bahan Kuliah Matematika Diskrit_Fakultas Sistem Informas

1. How About Some…How About Some…
RelationsRelations

2. RelationsRelations
When (a, b) belongs to R, a is said to beWhen (a, b) belongs to R, a is said to be relatedrelated to bto b
by R.by R.
Example:Example: Let P be a set of people, C be a set of cars,Let P be a set of people, C be a set of cars,
and D be the relation describing which person drivesand D be the relation describing which person drives
which car(s).which car(s).
P = {Carl, Suzanne, Peter, Carla},P = {Carl, Suzanne, Peter, Carla},
C = {Mercedes, BMW, tricycle}C = {Mercedes, BMW, tricycle}
D = {(Carl, Mercedes), (Suzanne, Mercedes),D = {(Carl, Mercedes), (Suzanne, Mercedes),
(Suzanne, BMW), (Peter, tricycle)}(Suzanne, BMW), (Peter, tricycle)}
This means that Carl drives a Mercedes, SuzanneThis means that Carl drives a Mercedes, Suzanne
drives a Mercedes and a BMW, Peter drives a tricycle,drives a Mercedes and a BMW, Peter drives a tricycle,
and Carla does not drive any of these vehicles.and Carla does not drive any of these vehicles.

3. Functions as RelationsFunctions as Relations
You might remember that aYou might remember that a functionfunction f from a set A tof from a set A to
a set B assigns a unique element of B to eacha set B assigns a unique element of B to each
element of A.element of A.
TheThe graphgraph of f is the set of ordered pairs (a, b) suchof f is the set of ordered pairs (a, b) such
that b = f(a).that b = f(a).
Since the graph of f is a subset of ASince the graph of f is a subset of A××B, it is aB, it is a relationrelation
from A to B.from A to B.
Moreover, for each elementMoreover, for each element aa of A, there is exactlyof A, there is exactly
one ordered pair in the graph that hasone ordered pair in the graph that has aa as its firstas its first
element.element.

4. Functions as RelationsFunctions as Relations
Conversely, if R is a relation from A to B such thatConversely, if R is a relation from A to B such that
every element in A is the first element of exactly oneevery element in A is the first element of exactly one
ordered pair of R, then a function can be defined withordered pair of R, then a function can be defined with
R as its graph.R as its graph.
This is done by assigning to an element aThis is done by assigning to an element a∈∈A theA the
unique element bunique element b∈∈B such that (a, b)B such that (a, b)∈∈R.R.

5. Relations on a SetRelations on a Set
Definition:Definition: A relation on the set A is a relation from AA relation on the set A is a relation from A
to A.to A.
In other words, a relation on the set A is a subset ofIn other words, a relation on the set A is a subset of
AA××A.A.
Example:Example: Let A = {1, 2, 3, 4}. Which ordered pairs areLet A = {1, 2, 3, 4}. Which ordered pairs are
in the relation R = {(a, b) | a < b} ?in the relation R = {(a, b) | a < b} ?

6. Relations on a SetRelations on a Set
Solution:Solution: R = {R = {(1, 2),(1, 2), (1, 3),(1, 3), (1, 4),(1, 4), (2, 3),(2, 3),(2, 4),(2, 4),(3, 4)}(3, 4)}
RR 11 22 33 44
11
22
33
44
11 11
22
33
44
22
33
44
XX XX XX
XX XX
XX

7. Relations on a SetRelations on a Set
How many different relations can we define on aHow many different relations can we define on a
set A with n elements?set A with n elements?
A relation on a set A is a subset of AA relation on a set A is a subset of A××A.A.
How many elements are in AHow many elements are in A××A ?A ?
There are nThere are n22
elements in Aelements in A××A, so how many subsetsA, so how many subsets
(= relations on A) does A(= relations on A) does A××A have?A have?
The number of subsets that we can form out of a setThe number of subsets that we can form out of a set
with m elements is 2with m elements is 2mm
. Therefore, 2. Therefore, 2nn22
subsets can besubsets can be
formed out of Aformed out of A××A.A.
Answer:Answer: We can define 2We can define 2nn22
different relationsdifferent relations
on A.on A.

8. Properties of RelationsProperties of Relations
We will now look at some useful ways to classifyWe will now look at some useful ways to classify
relations.relations.
Definition:Definition: A relation R on a set A is calledA relation R on a set A is called reflexivereflexive
if (a, a)if (a, a)∈∈R for every element aR for every element a∈∈A.A.
Are the following relations on {1, 2, 3, 4} reflexive?Are the following relations on {1, 2, 3, 4} reflexive?
R = {(1, 1), (1, 2), (2, 3), (3, 3), (4, 4)}R = {(1, 1), (1, 2), (2, 3), (3, 3), (4, 4)} No.No.
R = {(1, 1), (2, 2), (2, 3), (3, 3), (4, 4)}R = {(1, 1), (2, 2), (2, 3), (3, 3), (4, 4)} Yes.Yes.
R = {(1, 1), (2, 2), (3, 3)}R = {(1, 1), (2, 2), (3, 3)} No.No.
Definition:Definition: A relation on a set A is calledA relation on a set A is called irreflexiveirreflexive ifif
(a, a)(a, a)∉∉R for every element aR for every element a∈∈A.A.

9. Properties of RelationsProperties of Relations
Definitions:Definitions:
A relation R on a set A is calledA relation R on a set A is called symmetricsymmetric if (b, a)if (b, a)∈∈RR
whenever (a, b)whenever (a, b)∈∈R for all a, bR for all a, b∈∈A.A.
A relation R on a set A is calledA relation R on a set A is called antisymmetricantisymmetric ifif
a = b whenever (a, b)a = b whenever (a, b)∈∈R and (b, a)R and (b, a)∈∈R.R.
A relation R on a set A is calledA relation R on a set A is called asymmetricasymmetric ifif
(a, b)(a, b)∈∈R implies that (b, a)R implies that (b, a)∉∉R for all a, bR for all a, b∈∈A.A.

10. Properties of RelationsProperties of Relations
Are the following relations on {1, 2, 3, 4}Are the following relations on {1, 2, 3, 4}
symmetric, antisymmetric, or asymmetric?symmetric, antisymmetric, or asymmetric?
R = {(1, 1), (1, 2), (2, 1), (3, 3), (4, 4)}R = {(1, 1), (1, 2), (2, 1), (3, 3), (4, 4)} symmetricsymmetric
R = {(1, 1)}R = {(1, 1)} sym. andsym. and
antisym.antisym.
R = {(1, 3), (3, 2), (2, 1)}R = {(1, 3), (3, 2), (2, 1)} antisym.antisym.
and asym.and asym.
R = {(4, 4), (3, 3), (1, 4)}R = {(4, 4), (3, 3), (1, 4)} antisym.antisym.

11. Properties of RelationsProperties of Relations
Definition:Definition: A relation R on a set A is calledA relation R on a set A is called transitivetransitive
if whenever (a, b)if whenever (a, b)∈∈R and (b, c)R and (b, c)∈∈R, then (a, c)R, then (a, c)∈∈R forR for
a, b, ca, b, c∈∈A.A.
Are the following relations on {1, 2, 3, 4}Are the following relations on {1, 2, 3, 4}
transitive?transitive?
R = {(1, 1), (1, 2), (2, 2), (2, 1), (3, 3)}R = {(1, 1), (1, 2), (2, 2), (2, 1), (3, 3)} Yes.Yes.
R = {(1, 3), (3, 2), (2, 1)}R = {(1, 3), (3, 2), (2, 1)} No.No.
R = {(2, 4), (4, 3), (2, 3), (4, 1)}R = {(2, 4), (4, 3), (2, 3), (4, 1)} No.No.

12. Counting RelationsCounting Relations
Example:Example: How many different reflexive relations canHow many different reflexive relations can
be defined on a set A containing n elements?be defined on a set A containing n elements?
Solution:Solution: Relations on R are subsets of ARelations on R are subsets of A××A, whichA, which
contains ncontains n22
elements.elements.
Therefore, different relations on A can be generatedTherefore, different relations on A can be generated
by choosing different subsets out of these nby choosing different subsets out of these n22
elements, so there are 2elements, so there are 2nn22
relations.relations.
AA reflexivereflexive relation, however,relation, however, mustmust contain the ncontain the n
elements (a, a) for every aelements (a, a) for every a∈∈A.A.
Consequently, we can only choose among nConsequently, we can only choose among n22
– n =– n =
n(n – 1) elements to generate reflexive relations, son(n – 1) elements to generate reflexive relations, so
there are 2there are 2n(n – 1)n(n – 1)
of them.of them.

13. Combining RelationsCombining Relations
Relations are sets, and therefore, we can apply theRelations are sets, and therefore, we can apply the
usualusual set operationsset operations to them.to them.
If we have two relations RIf we have two relations R11 and Rand R22, and both of them, and both of them
are from a set A to a set B, then we can combineare from a set A to a set B, then we can combine
them to Rthem to R11 ∪∪ RR22, R, R11 ∩∩ RR22, or R, or R11 – R– R22..
In each case, the result will beIn each case, the result will be another relation fromanother relation from
A to BA to B..

14. Combining RelationsCombining Relations
…… and there is another important way to combineand there is another important way to combine
relations.relations.
Definition:Definition: Let R be a relation from a set A to a set BLet R be a relation from a set A to a set B
and S a relation from B to a set C. Theand S a relation from B to a set C. The compositecomposite ofof
R and S is the relation consisting of ordered pairs (a,R and S is the relation consisting of ordered pairs (a,
c), where ac), where a∈∈A, cA, c∈∈C, and for which there exists anC, and for which there exists an
element belement b∈∈B such that (a, b)B such that (a, b)∈∈R andR and
(b, c)(b, c)∈∈S. We denote the composite of R and S byS. We denote the composite of R and S by
SS°° RR..
In other words, if relation R contains a pair (a, b) andIn other words, if relation R contains a pair (a, b) and
relation S contains a pair (b, c), then Srelation S contains a pair (b, c), then S°° R contains aR contains a
pair (a, c).pair (a, c).

15. Combining RelationsCombining Relations
Example:Example: Let D and S be relations on A = {1, 2, 3, 4}.Let D and S be relations on A = {1, 2, 3, 4}.
D = {(a, b) | b = 5 - a} “b equals (5 – a)”D = {(a, b) | b = 5 - a} “b equals (5 – a)”
S = {(a, b) | a < b} “a is smaller than b”S = {(a, b) | a < b} “a is smaller than b”
D = {(1, 4), (2, 3), (3, 2), (4, 1)}D = {(1, 4), (2, 3), (3, 2), (4, 1)}
S = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}S = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}
SS°° D = {D = { (2, 4),(2, 4), (3, 3),(3, 3), (3, 4),(3, 4), (4, 2),(4, 2), (4, 3),(4, 3),
D maps an element a to the element (5 – a), andD maps an element a to the element (5 – a), and
afterwards S maps (5 – a) to all elements larger than (5afterwards S maps (5 – a) to all elements larger than (5
– a), resulting in– a), resulting in SS°°D = {(a,b) | b > 5 – a}D = {(a,b) | b > 5 – a} oror SS°°D = {(a,b)D = {(a,b)
| a + b > 5}.| a + b > 5}.
(4, 4)}(4, 4)}

16. Combining RelationsCombining Relations
We already know thatWe already know that functionsfunctions are justare just
special casesspecial cases ofof relationsrelations (namely those that(namely those that
map each element in the domain onto exactlymap each element in the domain onto exactly
one element in the codomain).one element in the codomain).
If we formally convert two functions intoIf we formally convert two functions into
relations, that is, write them down as sets ofrelations, that is, write them down as sets of
ordered pairs, the composite of these relationsordered pairs, the composite of these relations
will be exactly the same as the composite ofwill be exactly the same as the composite of
the functions (as defined earlier).the functions (as defined earlier).

17. Combining RelationsCombining Relations
Definition:Definition: Let R be a relation on the set A. TheLet R be a relation on the set A. The
powers Rpowers Rnn
, n = 1, 2, 3, …, are defined inductively by, n = 1, 2, 3, …, are defined inductively by
RR11
= R= R
RRn+1n+1
= R= Rnn
°° RR
In other words:In other words:
RRnn
= R= R°° RR°° …… °° R (n times the letter R)R (n times the letter R)

18. Combining RelationsCombining Relations
Theorem:Theorem: The relation R on a set A is transitive if andThe relation R on a set A is transitive if and
only if Ronly if Rnn
⊆⊆ R for all positive integers n.R for all positive integers n.
Remember the definition of transitivity:Remember the definition of transitivity:
Definition:Definition: A relation R on a set A is called transitiveA relation R on a set A is called transitive
if whenever (a, b)if whenever (a, b)∈∈R and (b, c)R and (b, c)∈∈R, then (a, c)R, then (a, c)∈∈R forR for
a, b, ca, b, c∈∈A.A.
The composite of R with itself contains exactly theseThe composite of R with itself contains exactly these
pairs (a, c).pairs (a, c).
Therefore, for a transitive relation R, RTherefore, for a transitive relation R, R°° R does notR does not
contain any pairs that are not in R, so Rcontain any pairs that are not in R, so R°° RR ⊆⊆ R.R.
Since RSince R°° R does not introduce any pairs that are notR does not introduce any pairs that are not
already in R, it must also be true that (Ralready in R, it must also be true that (R°° R)R)°° RR ⊆⊆ R, andR, and
so on, so that Rso on, so that Rnn
⊆⊆ R.R.