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Properties of Relations

In document Applied Discrete Structures (Page 126-137)

6.3.1 Individual Properties

Consider the set B ={1,2,3,4,6,12,36,48} and the relations “divides” and

on B. We notice that these two relations on B have three properties in common:

• Every element inB divides itself and is less than or equal to itself. This is called the reflexive property.

• If we search for two elements from B where the first divides the second and the second divides the first, then we are forced to choose the two numbers to be the same. In other words, no two different numbers are related in both directions. The reader can verify that a similar fact is true for the relationonB. This is called the antisymmetric property. • Next if we choose three values (not necessarily distinct) from B such that the first divides the second and the second divides the third, then we always find that the first number divides the third. Again, the same is true if we replace “divides” with “is less than or equal to.” This is called the transitive property.

Relations that satisfy these properties are of special interest to us. Formal definitions of the properties follow.

Definition 6.3.1 Reflexive Relation. LetAbe a set and letrbe a relation onA. Thenrisreflexiveif and only ifarafor alla∈A.Definition 6.3.2 Antisymmetric Relation. LetA be a set and letr be a relation onA. Thenr is antisymmetric if and only if whenever arb and

=b thenbrais false. ♢

An equivalent condition for antisymmetry is that ifarbandbrathena=b. You are encouraged to convince yourself that this is true. This condition is often more convenient to prove than the definition, even though the definition is probably easier to understand.

A word of warning about antisymmetry: Students frequently find it difficult to understand this definition. Keep in mind that this term is defined through an “If...then...” statement. The question that you must ask is: Is it true that whenever there are elements aand b from A where arb and =b, it follows thatb is not related toa? If so, then the relation is antisymmetric.

Another way to determine whether a relation is antisymmetric is to examine (or imagine) its digraph. The relation is not antisymmetric if there exists a pair of vertices that are connected by edges in both directions.

Definition 6.3.3 Transitive Relation. LetAbe a set and letrbe a relation onA. ristransitive if and only if wheneverarb andbrcthenarc.


6.3.2 Partial Orderings

Not all relations have all three of the properties discussed above, but those that do are a special type of relation.

Definition 6.3.4 Partial Ordering. A relation on a setAthat is reflexive, antisymmetric, and transitive is called a partial ordering on A. A set on which there is a partial ordering relation defined is called apartially ordered


Example 6.3.5 Set Containment as a Partial Ordering. LetA be a set. ThenP(A)together with the relation(set containment) is a poset. To prove this we observe that the three properties hold, as discussed in Chapter 4.

• LetB∈ P(A). The fact thatB⊆Bfollows from the definition of subset. Hence, set containment is reflexive.

• LetB1, B2 ∈ P(A) and assume that B1 ⊆B2 and B1 ̸=B2 . Could it

be that B2 ⊆B1? No. There must be some element a A such that

a∈/B1, buta∈B2. This is exactly what we need to conclude thatB2is

not contained inB1. Hence, set containment is antisymmetric.

• LetB1, B2, B3 ∈ P(A)and assume that B1 ⊆B2 andB2 ⊆B3 . Does

it follow thatB1⊆B3 ? Yes, if a∈B1, thena∈B2 becauseB1⊆B2.

Now that we havea∈B2 and we have assumed B2 ⊆B3, we conclude

thata∈B3. Therefore,B1⊆B3 and so set containment is transitive.

Figure 6.2.6is the graph for the “set containment” relation on the power

set of{1,2}. □

Figure 6.2.6 is helpful insofar as it reminds us that each set is a subset of itself and shows us at a glance the relationship between the various subsets in

P({1,2}). However, when a relation is a partial ordering, we can streamline a graph like this one. The streamlined form of a graph is called a Hasse diagram or ordering diagram. A Hasse diagram takes into account the following facts.

• By the reflexive property, each vertex must be related to itself, so the arrows from a vertex to itself (called “self-loops”) are not drawn in a Hasse diagram. They are simply assumed.

• By the antisymmetry property, connections between two distinct ele- ments in a directed graph can only go one way, if at all. When there is a connection, we agree to always place the second element above the first (as we do above with the connection from{1}to{1,2}). For this reason, we can just draw a connection without an arrow, just a line.

• By the transitive property, if there are edges connecting one element up to a second element and the second element up to a third element, then there will be a direct connection from the first to the third. We see this in Figure 6.2.6with connected to{1}and then {1} connected to

{1,2}. Notice the edge connectingto{1,2}. Whenever we identify this situation, remove the connection from the first to the third in a Hasse diagram and simply observe that an upward path of any length implies that the lower element is related to the upper one.

CHAPTER 6. RELATIONS 114 Using these observations as a guide, we can draw a Hasse diagram for

on{1,2}as inFigure 6.3.6.

Figure 6.3.6: Hasse diagram for set containment on subsets of {1,2} Example 6.3.7 Definition of a relation using a Hasse diagram. Consider the partial ordering relationswhose Hasse diagram is Figure 6.3.8.


Figure 6.3.8: Hasse diagram for the pentagonal poset

How do we read this diagram? What is A? What is s? What does the digraph of s look like? Certainly A = {1,2,3,4,5} and 1s2, 3s4, 1s4, 1s5, etc., Notice that1s5is implied by the fact that there is a path of length three upward from 1 to 5. This follows from the edges that are shown and the transitive property that is presumed in a poset. Since1s3 and 3s4, we know that 1s4. We then combine 1s4 with 4s5 to infer 1s5. Without going into details why, here is a complete list of pairs defined bys.

s={(1,1),(2,2),(3,3),(4,4),(5,5),(1,3),(1,4),(1,5),(1,2),(3,4),(3,5),(4,5),(2,5)} A digraph for s is Figure 6.3.9. It is certainly more complicated to read and difficult to draw than the Hasse diagram.


Figure 6.3.9: Digraph for the pentagonal poset

□ A classic example of a partial ordering relation is on the real numbers, R. Indeed, when graphing partial ordering relations, it is natural to “plot” the elements from the given poset starting with the “least” element to the “greatest” and to use terms like “least,” “greatest,” etc. Because of this the reader should be forewarned that some texts use the symbol for arbitrary partial orderings. This can be quite confusing for the novice, so we continue to use generic lettersr,s, etc.

6.3.3 Equivalence Relations

Another common property of relations is symmetry.

Definition 6.3.10 Symmetric Relation. Letrbe a relation on a setA. r issymmetricif and only if wheneverarb, it follows thatbra. ♢ Consider the relation of equality defined on any set A. Certainly a = b implies thatb=aso equality is a symmetric relation onA.

Surprisingly, equality is also an antisymmetric relation on A. This is due to the fact that the condition that defines the antisymmetry property,a =b and=b, is a contradiction. Remember, a conditional proposition is always true when the condition is false. So a relation can be both symmetric and antisymmetric on a set! Again recall that these terms arenotnegatives of one other. That said, there are very few important relations other than equality that are both symmetric and antisymmetric.

Definition 6.3.11 Equivalence Relation. A relationron a setAis called an equivalence relation if and only if it is reflexive, symmetric, and transitive. ♢ The classic example of an equivalence relation is equality on a set A. In fact, the term equivalence relation is used because those relations which satisfy the definition behave quite like the equality relation. Here is another important equivalence relation.

CHAPTER 6. RELATIONS 117 Example 6.3.12 Equivalent Fractions. Let Z∗ be the set of nonzero integers. One of the most basic equivalence relations in mathematics is the relationqonZ×Zdefined by(a, b)q(c, d)if and only ifad=bc. We will leave it to the reader to, verify that q is indeed an equivalence relation. Be aware that since the elements ofZ×Zare ordered pairs, proving symmetry involves four numbers and transitivity involves six numbers. Two ordered pairs,(a, b) and(c, d), are related if the fractions a

b and c

d are numerically equal. □

Our next example involves the following fundamental relations on the set of integers.

Definition 6.3.13 Congruence Modulom. Letm be a positive integer, m≥2. We definecongruence modulo mto be the relation≡m defined on

the integers by


♢ We observe the following about congruence modulo m:

• This relation is reflexive, for if a∈Z,m|(a−a)⇒a≡ma.

• This relation is symmetric. We can prove this through the following chain of implications. a≡mb⇒m|(a−b) For some k∈Z, a−b=mk ⇒b−a=m(−k) ⇒m|(b−a) ⇒b≡ma .

• Finally, this relation is transitive. We leave it to the reader to prove that ifa≡mb andb≡mc, thena≡mc.

Frequently, you will see the equivalent notation a≡b(modm)for congru- ence modulom.

Example 6.3.14 Random Relations usually have no properties. Consider the relation s described by the digraph inFigure 6.3.15. This was created by randomly selecting whether or not two elements from{a, b, c}were related or not. Convince yourself that the following are true:

• This relation is not reflexive. • It is not antisymmetric. • Also, it is not symmetric. • It is not transitive.


Figure 6.3.15: Digraph of a random relationr

Not every random choice of a relation will be so totally negative, but as the underlying set increases, the likelihood any of the properties are true begins to

vanish. □

6.3.4 Exercises for Section 6.3


(a) Let B={a, b} andU =P(B). Draw a Hasse diagram foronU. (b) LetA={1,2,3,6}. Show that divides,|, is a partial ordering onA. (c) Draw a Hasse diagram for divides on A.

(d) Compare the graphs of parts a and c.

2. Repeat Exercise 1 withB ={a, b, c}andA={1,2,3,5,6,10,15,30}. Hint. Here is a Hasse diagram for the part (a).


Figure 6.3.16

3. Consider the relations defined by the digraphs inFigure 6.3.17.

(a) Determine whether the given relations are reflexive, symmetric, anti- symmetric, or transitive. Try to develop procedures for determining the validity of these properties from the graphs,

(b) Which of the graphs are of equivalence relations or of partial order- ings?


Figure 6.3.17: Some digraphs of relations

CHAPTER 6. RELATIONS 121 ordering relations for the given sets:

(a) A ={lines in the plane}, and r defined by xry if and only if xis parallel to y. Assume every line is parallel to itself.

(b) A=Randrdefined byxry if and only if|x−y| ≤7.

5. Consider the relation on{1,2,3,4,5,6}defined by r={(i, j) : |i−j|= 2}.

(a) Isrreflexive? (b) Isrsymmetric? (c) Isrtransitive? (d) Draw a graph ofr.

6. For the set of cities on a map, consider the relationxry if and only if city xis connected by a road to city y. A city is considered to be connected to itself, and two cities are connected even though there are cities on the road between them. Is this an equivalence relation or a partial ordering? Explain.

7. Equivalence Classes. LetA={0,1,2,3}and let

r={(0,0),(1,1),(2,2),(3,3),(1,2),(2,1),(3,2),(2,3),(3,1),(1,3)} (a) Verify that ris an equivalence relation onA.

(b) Let a A and define c(a) = {b A | arb}. c(a) is called the equivalence class ofaunderr. Findc(a)for each elementa∈A. (c) Show that{c(a)|a∈A} forms a partition ofAfor this setA. (d) Letr be an equivalence relation on an arbitrary setA. Prove that

the set of all equivalence classes under rconstitutes a partition of A.

8. Definer on the power set of{1,2,3}byArB⇔ |A|=|B|. Prove thatr is an equivalence relation. What are the equivalence classes underr? 9. Consider the following relations on Z8 ={0,1, ...,7}. Which are equiva-

lence relations? For the equivalence relations, list the equivalence classes. (a) arb iff the English spellings ofaandbbegin with the same letter. (b) asbiffa−b is a positive integer.

(c) atbiffa−bis an even integer. 10. Building onExercise

(a) Prove that congruence modulom is transitive.

(b) What are the equivalence classes under congruence modulo 2? (c) What are the equivalence classes under congruence modulo 10?

CHAPTER 6. RELATIONS 122 11. In this exercise, we prove that implication is a partial ordering. LetAbe

any set of propositions.

(a) Verify that q q is a tautology, thereby showing that is a re- flexive relation onA.

(b) Prove thatis antisymmetric onA. Note: we do not use = when speaking of propositions, but rather equivalence,.

(c) Prove that is transitive onA.

(d) Given thatqiis the propositionn < ionN, draw the Hasse diagram

for the relation on{q1, q2, q3, . . .}.

12. LetS={1,2,3,4,5,6,7}be a poset(S,)with the Hasse diagram shown below. Another relationr⊆S×S is defined as follows: (x, y)∈r if and only if there existsz∈S such thatz < xandz < yin the poset(S,).

(a) Prove that ris reflexive. (b) Prove thatris symmetric.

(c) A compatible with respect to relationris any subsetQof setSsuch that x∈Qandy ∈Q⇒(x, y)∈r. A compatible g is a maximal compatible ifQis not a proper subset of another compatible. Give all maximal compatibles with respect to relationr defined above. (d) Discuss a characterization of the set of maximal compatibles for

relation r when (S,) is a general finite poset. What conditions, if any, on a general finite poset (S,) will make r an equivalence relation?


Figure 6.3.18: Hasse diagram forrin exercise 12.

In document Applied Discrete Structures (Page 126-137)