chap6 relations def 1: let a and b be sets. a binary relation from a to b is a subset of a x b a r...

Chap6 Relations Def 1: Let A and B be sets. A binary relation from A to B is a subset of A x B a R b: (a,b) R a R b: (a,b) R Figure1Def 2: A relation on the set A is a relation from A to A

Example 4 Example 6 ( How many “possible“ relations…)

Chap6 Relations

Def 3: A relation R on a set A is called reflexive if (a,a) R for every element a A .

Example 7

Chap6 Relations

Def 4:A relation R on a set A is called symmetric if (b,a) R whenever (a,b) R , for a,b A. A relation R on a set A such that (a,b) R and (b ,a) R only if a=b , for a,b A , is called antisymmetric .

Example 10 a relation can be both or neither ex both : {(1,1)} neither : { (1,2), (2,3),(3,2) }

Chap6 Relations

Def 5: A relation R on a set A is called transitive if whenever (a,b) R and (b,c) R ,then (a,c) R , for a,b,c A. Example 13 Example 16

Chap6 Relations

Def 6:Let R be a relation from a set A to a set B and S be a relation from B to a set C . The composite of R and S is the relation consisting of ordered pairs (a,c) , where a A ,cC , and for which there exists an element b B such that (a,b) R and (b,c) S.

We denote the composite of R and S by S R Example 19

Chap6 Relations

Def 7 : Let R be a relation on a set A . The power Rn , n=1,2,3,…, are defined inductively by R1=R and R n+1=Rn R Example 20

1 2 3 4

Theorem1: The relation R on a set A is transitive if and only if Rn R , for n=1,2,3,….. “ if “ Rn R , R2 R if ( a,b) R and ( b,c ) R ,( a,c ) R2

R2 R => (a,c) R => R is transitive

Chap6 Relations

“ only if “ by mathematical induction n=1,trivial assume Rk R to show Rk+1 R assume (a,b) Rk+1 since Rk+1 = Rk R

there exist x A such that (a,x) R and ( x,b) Rk since Rk R , ( x,b) R ( a,b) R Rk+1 R

Chap6 Representing Relations Using Digraphs

Def 1: -Example 7 -determine whether a relation is reflexive , symmetric ,antisymmetric and transitive

-Example 10

Chap6 Closure of Relation

- R= { (1,1), (1,2) , (2,1), (3,2) } on A = {1,2,3} is not reflexive

R’ = { (1,1), (1,2) , (2,1), (3,2) , (2,2), (3,3)} R R’ , R’ is reflexive , R’ , where is any

reflexive relation and R

R’ is the reflexive closure of R R’ = R , ={(a,a)| a A} is the diagonal relation on A

Example 1


- R= { (1,1), (1,2) , (2,2), (2,3) , (3,1), (3,2)} on {1,2,3}

R’= { (1,1), (1,2) ,(2,1), (2,2), (2,3) ,(3,1), (3,2),(1,3)}

R R’ , R’ is symmetric , R’ , where is any symmetric relation and R

R’ is the symmetric closure of R R’=R R-1, R-1={(b,a) | ( a,b) R, a=b}

Example 2


-R= { (1,3), (1,4) , (2,1) , (3,2)} on {1,2,3,4}

1 2 1 2 4 3 4 3

add (2,4), (3,1) ,(2,3) ,(1,2) to become transitive?

Chap6 Relations

Def 1: path ,cycle Example 3Theorem 1 Let R be a relation on a set A .There is a path of

length n from a to b if and only if ( a,b) Rn

Chap6 Relations

Def 2: Let R be a relation on a set A . The connectivity relation R* consists of the pairs (a,b) such that there is a path between a and b in R R* = Rn

Example 4 a Rb if a has met b R2 contains ( a,b) if c,( a,c) R and ( c,b) R

Rn contains ( a,b) if x1,x2,..,x n-1 ,( a,x1) R , (x1,x2) R ,…(x n-1,b) R

R*, dose R* contain ( you, Mongolis presidat ) ?

n=1

Chap6 Relations

Theorem 2 The transitive closure of a relation R equal the connectivity relation R*Proof: R R* have to show : 1. R* is transitive 2. R* where is transitive and R

Chap6 Relations

1. If ( a,b) R* and (b,c) R* ,then there are paths from a to b and from b to c in R

( a,c) R* R* is transitive

aab

c

Chap6 Relations

2. is transitive and R to prove R*

n is also transitive and n

since * = k and k , *

since R , R* * ( any path in R is also a path in ) R* *

R* = R R2 R3 ….. Rn

k=1

Chap6 Relations

a=x1 x2

xi+1

xi-1xj+1xi =xj

xj-2

xj-1

b=xm

m>n

Chap6 Equivalence Relations Def 1 :A relation on a set A is an equivalence relation if it is reflexive, symmetric, and transitive

– Two elements related by an equivalence relation are called equivalent.

Example 1

R: relation on the set of strings of English letters such that aRb iff l (a)=l (b), where l(x) is the length of the string X. Is R an equivalence relation ?

Chap6 Equivalence Relations

l(a) =l(a) : reflexive

Suppose a R b , l(a)= l(b) , then since

l(b)=l(a), b R a : symmetric

suppose a R b and b R c , l(a)= l(b) and l(b)=l(c) ,

l(a)=l(c), a R c : transitive


Example 2 R: relation on the set of real numbers,

aRb iff a-b is an integer .

transitive : aRb and bRc, a - b and b-c are

integers a-c=(a-b)+(b-c) is an integer

Example 3 Congruence Modulo m

show R={(a,b)|ab(mod m)} is an

equivalence relation on the set of integers

, m 1﹥


a b ( mod m ) iff m divides a – b

– a – a = 0 is divisible by m

– a – b is divisible by m, a – b = km

b – a = ( – k )m, b a ( mod m )

– a b ( mod m ) and b c ( mod m )

m divides a – b and b – c

a – b = km and b – c = lm

a – c = ( a – b ) + ( b – c ) = ( k+l )m


Equivalence classes

A: set of students at NDHU, who graduated

from high school

R on A:{(x,y)|x and y graduated from same

high school }

given a student x, we can form a set of all

students equivalent to x with respect to R ;

This set is an equivalence class of R, denoted [x]R

[x]R = { s | ( x , s ) R }


Example 6 What are the equivalence classes of 0 and 1 for congruence modulo 4 ? 0 : a 0 (mod 4), integers divisible by 4 [0]= { …,-8 ,-4, 0, 4, 8,…} 1: a 1 (mod 4) a-1 divisible by 4 integers having a remained 1 when divided by 4


[1]= { …,-7 ,-3, 1, 5, 9,…} The equivalence classes of the relation congruence modulo m are congruence classes modulo m

[a]m={…, a-2m, a-m, a, a+m, a+2m,…} x a (mod m)

x – a = km

x = a + km


Partitions

A: Students majoring in exactly one subject

R on A:{(x ,y)|x and y have same major }

R splits all students in A into a collections

of disjoint subsets, where each subset

contains students with a specified major


Theorem 1. Let R be an equivalence relation on

a set A. The following statements

are equivalent:

( i ) a R b

(ii ) [ a ] = [ b ]

(iii) [ a ] [ b ]


Proof: (i) implies (ii)

a R b, prove [a]=[b], to show [a] [b] and

[b] [a]

suppose c [a], to show c [b]

a R c, since a R b, b R a b R c c [b]

(ii) implies (iii) [a] is nonempty,

[a][b]


(iii)implies (i)

[a][b], there is an element c ,

c [a] and c [b]

a R c and b R c a R b

We have shown (i) implies (ii), (ii) implies

(iii) , and (iii) implies (i) . (i) , (ii) , (iii)

are equivalent

Chap6 Equivalence Relations a partition of a set S is a collection of disjoint

nonempty subsets of S that have S as their union

the collection of subsets Ai , iI forms a

partition of S iff

Ai for iI

Ai Aj , when ij,

and

Ai= S iI Figure 1


an equivalent relation partitions a set

[ a ]R= A aA

[a]R [b]R= when [a]R [b]R


Theorem 2 Let R be an equivalence on a set S.

Then the equivalence classes of R

form a partition of S. Conversely,

given a partition {Ai | iI }of the

set S, there is an equivalence

relation R that has the set Ai, iI,

as its equivalence classes


There are m different congruence classes

modulo m , corresponding to the m different

remainders possible when an integer is

divided by m

[ 0 ]m, [ 1 ]m,…, [ m-1]m


Example 8 What are the sets in the partition of

the integers arising from congruence

modulo 4 ?

[ 0 ]4={… , -8 , -4 , 0 , 4 , 8 , … }

[ 1 ]4={… , -7 , -3 , 1 , 5 , 9 , … }

[ 2 ]4={… , -6 , -2 , 2 , 6 , 10 , … }

[ 3 ]4={… , -5 , -1 , 3 , 7 , 11 , … }

chap6 relations def 1: let a and b be sets. a binary relation from a to b is a subset of a x b a r...

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