discrete structures chapter 5 relations nurul amelina nasharuddin multimedia department
Post on 19-Dec-2015
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Objectives
On completion of this topic, student should be able to:
a. Determine the properties of relations – reflexive, symmetric, transitive, and antisymmetric
b. Determine equivalence and partial order relations
c. Represent relations using matrix and graph
Outline
• Properties of relations• Matrix and graph representation of relations• Equivalence relations• Partial order relations
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• The most basic relation is “=” (e.g. x = y)
• Generally x R y TRUE or FALSE‒ R(x,y) is a more generic representation‒ R is a binary relation between elements of some
set A to some set B, where xA and yB
• Binary relations: x R yOn sets xX, yY R X Y
Recall: Relations
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Example
• “less than” relation from A={0,1,2} to B={1,2,3}
• Use traditional notation0 < 1, 0 < 2, 0 < 3, 1 < 2, 1 < 3, 2 < 3
• Or use set notationAB={(0,1),(0,2),(0,3),(1,1),(1,2),(1,3),(2,1),(2,2), (2,3)}R={(0,1),(0,2),(0,3),(1,2),(1,3),(2,3)} AB
• Or use Arrow Diagrams
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Reflexive: Let R be a binary relation on a set A
•Eg: Let A = {1,2,3,4}
R1 ={(1,1),(1,2),(2,2),(2,3),(3,3),(4,4)}
R1 is reflexive
R2 = {(1,1),(2,2),(3,3)}
R2 is not reflexive (why?)
Properties of Relations
A, xRxx R Reflexiveis
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Symmetric: Let R be a binary relation on a set A
•Eg: Let A = {1,2,3}
R1 ={(1,2),(2,1),(1,3),(3,1)}
R1 is symmetric
R2 = {(1,1),(2,2),(3,3),(2,3)}.
R2 is not symmetric because (3,2) R2
Properties of Relations
yRxA, xRyyx R , Symmetricis
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Transitive: Let R be a binary relation on a set A
•Let A = {1,2,3,4}R = {(2,1),(3,1),(3,2),(4,1),(4,2),(4,3)}R is transitive because (3,2) & (2,1) → (3,1) (4,2) & (2,1) → (4,1) (4,3) & (3,1) → (4,1) (4,3) & (3,2) → (4,2)
Properties of Relations
xRzyRzA, xRyzyx R ,, Transitiveis
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Properties of Relations
xRy or (x,y) R for all x and y in A,
• R is reflexive for all x in A, (x,x) R
• R is symmetric for all x and y in A, if (x,y) R then (y,x) R
• R is transitive for all x, y and z in A, if (x,y) R and (y,z) R then (x,z) R
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Example (1)• Define a relation of A called R
A = {2,3,4,5,6,7,8,9}R = {(4,4),(4,7),(7,4),(7,7),(2,2),(3,3),(3,6),(3,9), (6,6),(6,3),(6,9),(9,9),(9,3),(9,6)}
• Is RReflexive? NoSymmetric? YesTransitive? Yes
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Example (2)• A = {0,1,2,3}• R over A = {(0,0),(0,1),(0,3),(1,0),
(1,1),(2,2),(3,0),(3,3)}
• Is RReflexive? Yes.Symmetric? Yes.Transitive? No. (1,0),(0,3) R but (1,3) R
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Proving Properties on Infinite Sets -“equal” relation (1)
• Define a relation R on R (the set of all real numbers):
For all x, y R, x R y ↔ x = y
• Is R reflexive? symmetric? transitive?• R is reflexive iff x R, x R x. By definition of R,
this means x = x, for all x R. This is true, since every real number is equal to itself. Hence, R is reflexive
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Proving Properties on Infinite Sets -“equal” relation (1)
• R is symmetric iff x,y R, if x R y then y R x. By definition of R, this means that x,y R, if x = y then y = x. So this is true; if one number is equal to a second, then the second is equal to the first. Hence, R is symmetric
• R is transitive iff x,y R, if x = y and y = z, then x = z. So this is true; if one real number equals to a second, and the second equals a third, then the first equals the third
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Proving Properties on Infinite Sets -“less than” relation (1)
• Define a relation R on R (the set of all real numbers):
For all x, y R, x R y ↔ x < y
• Is R reflexive? symmetric? transitive?
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Proving Properties on Infinite Sets -“less than” relation (2)
• R is reflexive iff x R, x R x. By definition of R, this means x < x, for all x R. But this is false. Hence, R is not reflexive
• R is symmetric iff x,y R, if x R y then y R x. By definition of R. This means that x,y R, if x < y then y > x. But this is false. Hence, R is not symmetric
• R is transitive iff x,y,z R, if x < y and y < z, then x < z. Hence, R is transitive
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Properties of Congruence Modulo 3 (1)
• Define a relation R on Z: For all m,n Z, m R n 3|(m – n)
R is called congruence modulo 3
• Is R reflexive?• Is R symmetric?• Is R transitive?
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Properties of Congruence Modulo 3 (2)
• R is reflexive iff for all m in Z, m R m. By definition of R, this means 3|m – m or 3|0. This is true since 0 = 0 . 3, so R is reflexive
• R is symmetric iff for all m,n in Z, m R n then n R m. By definition of R, this means if 3|(m – n) then 3|(n – m). This is true
m – n = 3k, for some integer kn – m = - (m – n) = 3(-k)Hence 3|(n – m). Therefore R is symmetric
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Properties of Congruence Modulo 3 (2)
• R transitive iff for all m,n,p Z, if m R n and n R p then m R p. By definition of R, if 3|(m – n) and 3|(n – p) then 3|(m – p). So by definition of divide
m – n = 3r for some rn – p = 3s for some s
It is crucial to observe that (m – n) + (n – p) = m – p(m – n) + (n – p) = m – p = 3r + 3s
m – p = 3(r + s)Hence 3|(m – p). Therefore, R is transitive
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Matrix Representation of a Relation
• MR = [mij] (where i=row, j=col) mij={1 iff (i,j) R and 0 iff (i,j) R}
• Eg: R : {1,2,3}{1,2} where x > y R = {(2,1),(3,1),(3,2)}
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110
321
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RM
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Example• Example:
A = {1,2,3,4}, B = {w,x,y,z} R = {(1,x),(2,x),(3,y),(3,z)}
00004
11003
00102
00101
zyxw
M R = zero-one matrix
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Graph Representation of a Relation
• Let A = {1,2,3,4} and R is relation on A where R = {(1,1),(1,2),(2,3),(3,2),(3,3),(3,4),(4,2)}
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Union, Intersection, Difference and Composition of Relations
• R: AB and S: AB
• R: AB and S: BC
}),(),(|),{( SyxRyxBAyxSR }),(),(|),{( SyxRyxBAyxSR }),(),(|),{( SyxRyxBAyxSR }),(),(|),{( RyxSyxBAyxRS
}),(),(,|),{( ScbRbaBbCAcaRS
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Composition of Relations
• Example: A = {1,2,3,4}, B = {w,x,y,z}, C = {5,6,7},R1: AB = {(1,x),(2,x),(3,y),(3,z)} andR2: BC = {(w,5),(x,6)}
Therefore, R2 o R1 = {(1,6),(2,6)}
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Equivalence Relations
• Any binary relation that is:ReflexiveSymmetricTransitive
• Eg: A = {1,2,3}, R = {(1,1),(2,2),(2,3),(3,2),(3,3)}R is reflexive, symmetric and transitiveTherefore, R is an equivalence relation
• Recall example earlier: Congruence modulo 3 is an equivalence
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Antisymmetry Relations
• Let R be a relation on a set A. R is antisymmetric iff for all a and b in A, if a R b and b R a then a = b
• In other words, a relation is antisymmetric iff there are no pairs of distinct elements a and b with a related to b and b related to a
• Eg: Let A = {0,1,2}R1= {(0,2),(1,2),(2,0)}. R1 is not antisymmetricR2 = {(0,0),(0,1),(0,2),(1,1),(1,2)}. R2 is antisymmetric
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Example (1)
• Let A = {1,2,3}, R = {(1,2),(2,1),(2,3)}
R is not symmetric, (3,2) RR is also not antisymmetric because (1,2),(2,1) R
• Let A = {1,2,3}, S = {(1,1),(2,2)}
S is both symmetric and antisymmetric
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Example (2)
• Let R1 be the divides relation (a|b) on Z+
• Is R1 antisymmetric? Prove or give counterexample
If a R1 b and b R1 a, then a = ba R1 b means a|b → b = k1ab R1 a means b|a → a = k2bIt follows that, b = k1a = k1(k2b) =
(k1k2)bThus, k1 = k2 = 1. Hence a = b
R1 antisymmetric
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Example (2)
• Let R2 be the divides relation on Z
• Is R2 antisymmetric? Prove or give counterexample.
Let a = 2, b = -2Hence, a|b (-2 = -1(2)) and b|a (2 = -1(-2)) but a b
R2 is not antisymmetric
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Partial Order Relations
• Any binary relation that is:ReflexiveAntisymmetricTransitive
• Partial Order Set (POSET)(S,R) = R is a partial order relation on set S
• Examples: (Z, ) (Z+,|) {note: | symbolizes divides}