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TIF 21101
APPLIED MATH 1
(MATEMATIKA TERAPAN 1)
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Week 4
Relation and Function I
Relation and FunctionRelation and Function
OverviewObviously, we do not realize that there many connectionsare happened in our circumtances. For examples, day andnight happens because of earth rotation, all students inmath are also connected to other subjects and so on.Strictly speaking, something happens because of other
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Strictly speaking, something happens because of othersubject called “reason”.
Relations can be used to solve problems such asdetermining which pairs of cities are linked by airline flightsin a network, finding a viable order for the different phasesof a complicated project, or producing a useful way to storeinformation in computer databases.
For couple weeks later, you all will be introduced this“connection” in mathematic’s view. And we shall learn to“map” or “transform” the “connection”.
Relation and FunctionRelation and Function
Objectives
� Cartesian Product
� Relation
� Invers Relation
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
� Invers Relation
� Pictoral Repesentation of Relation
� Composition of Relation
� Relation Properties
Relation and FunctionRelation and Function
Cartesian Product
Consider two sets A and B. The set of all ordered
pairs (a, b) where a∈A and b∈B is called the
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
pairs (a, b) where a∈A and b∈B is called the
product, or Cartesian product, of A and B.
The short designation of this product is A x B,
which is read “A cross B”.
Relation and FunctionRelation and FunctionEx.Let A={1, 2} and B={a, b, c}.
Then AxB {(1,a},(1,b),(1,c),(2,a),(2,b),(2,c)} BxA {(a, 1), (a,2), (b, 1), (b,2), (c,1),(c,2)} AxA {(1, 1), (1,2), (2,1), (2,2)}
From the example above we can conclude, that, First,
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
First,
A x B ≠≠≠≠ B x AThe Cartesian product deals with ordered pairs, so naturally the order in which the sets are considered is important.
Second, using n(s) for the number of elements in a set S, we have
n(A x B) = n(A) . n(B) = 2 x 3 = 6Therefore, there will be 26 = 64 relation from A to B
So…..what is relation?????
Relation and FunctionRelation and Function
RelationRelation is just a subset of the Cartesian product of the sets.
Definition.
Let A and B be sets. A binary relation or, simply,
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Let A and B be sets. A binary relation or, simply, relation from A to B is a subset of A x B.
In other words, a binary relation from A to B is a set R of ordered pairs where the first element (domain) of each ordered pair comes from A and the second element (codomain or range) comes from B.
Relation and FunctionRelation and Function
We use the notation a R b to denote that (a, b) ∈ R and a R b to denote that (a, b) ∉ R.
Moreover, when (a, b) belongs to R, a is said to be related to b by R.
/
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Assume C= {1,2,3} and D ={x,y,z} and let R {(1,y), (1,z), (3,y)}. Put the R or R for the followings:
1…X 1…Y 1…Z
2…X 2…Y 2…Z
3…X 3…Y 3…Z
/
Relation and FunctionRelation and Function
Invers Relation The invers relation of set is defined as the opposite mapping of relation itself.
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Let R be any relation from a set A to a set B. The inverse of R, denoted by R-1, is the relation from B to A which consists of those ordered pairs which, when reversed, belong to R; that is,
R-1= {(b,a): (a,b) ∈ R}
Relation and FunctionRelation and Function
Ex.
Let R = {(1,y), (1,z), (3,y)} from A = {1,2,3} to
B = {x,y,z}, then
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
R-1 = {(y, 1), (z, 1), (y,3)}
Relation and FunctionRelation and Function
Pictoral Repesentation of Relation
Arrow Diagram
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Relation and FunctionRelation and Function
Table Representation
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Relation and FunctionRelation and Function
Matrice Representation
Suppose R is the relation from A to B, where
A={ a1,a2,a3,…,am} and B={ b1,b2,b3,…,bn}.
The relation can be describe in matrice M=[mij] as
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
The relation can be describe in matrice M=[mij] as
folow:
Relation and FunctionRelation and Function
Ex.
a1 = 2
a2 = 3
a3 = 4
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
a3 = 4
b1 = 2
b2 = 4
b3 = 8
b4 = 9
b5 = 15
Relation and FunctionRelation and Function
Directed Graph
First we write down the elements of the set, and
then we drawn an arrow from each element x to
each element y whenever x is related to y.
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
each element y whenever x is related to y.
The point is, directed graph does not show the
relation between one set to the other. It just shows
the relation among the element inside the set.
Ex. R is relation on the set A = {1,2,3,4}
R = {(1,2), (2,2), (2,4), (3,2), (3,4), (4,1), (4,3)}
Relation and FunctionRelation and Function
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Relation and FunctionRelation and Function
Prac.
Show the relation from
the directed graph
Bandung
Jakarta Surabaya
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Medan
Makassar
Kupang
Relation and FunctionRelation and Function
Composition of Relation
Suppose A, B, and C be sets, and let R be a relation from A to B and let S be a relation
from B to C. R ⊆ A x B and S ⊆ B x C.
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
from B to C. R ⊆ A x B and S ⊆ B x C.
Then R and S give rise to a relation from A to C, which is denoted by RoS and defined as
Relation and FunctionRelation and Function
Ex.
Assume A= {1,2,3,4}, B ={a,b,c,d}, C ={x,y,z} and let R= {(1,a), (2,d), (3,a) (3,b), (3,d)} and S ={(b,x), (b,z), (c,y), (d,z)} . Show the relation a(RoS)c!
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
relation a(RoS)c!
Relation and FunctionRelation and Function
From the picture we can observe that there is an arrow from 2 to d which is followed by an arrow from d to z. We can view these two arrows as a “path” which “connects” the element 2 ∈ A to the element z ∈ C. Thus,
2(R o S)z since 2Rd and dSz
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Similarly there is a path from 3 to x and a path from 3 to z. Hence,
3(R o S)x and 3(R o S)z
No other element of A is connected to an element of C. Therefore, the composition of relations R o S gives
RoS= {(2,z), (3,x), (3,z)}
Relation and FunctionRelation and Function
Soal :
R = {(1, 2), (1, 6), (2, 4), (3, 4), (3, 6), (3, 8)}
S = {(2, u), (4, s), (4, t), (6, t), (8, u)}
Gambarkan grafiknya dan tentukan R o S
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Gambarkan grafiknya dan tentukan R o S
Relation and FunctionRelation and Function
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
R o S = {(1, u), (1, t), (2, s), (2, t), (3, s), (3, t), (3, u) }
Relation and FunctionRelation and Function
Exercises :
1
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Relation and FunctionRelation and Function
2.
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng