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TIF 21101 APPLIED MATH 1 (MATEMATIKA TERAPAN 1) Matematika Terapan 1 2014/2015 M. Ilyas Hadikusuma, M.Eng Week 4 Relation and Function I
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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

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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”.

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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

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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”.

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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?????

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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.

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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

/

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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}

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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)}

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Relation and FunctionRelation and Function

Pictoral Repesentation of Relation

Arrow Diagram

Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng

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Relation and FunctionRelation and Function

Table Representation

Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng

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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:

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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

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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)}

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Relation and FunctionRelation and Function

Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng

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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

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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

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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!

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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)}

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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

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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) }

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Relation and FunctionRelation and Function

Exercises :

1

Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng

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Relation and FunctionRelation and Function

2.

Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng