Section 3.3 Equivalence Relations

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1 Section 3.3 Purpose of Section To introduce the concept of an equivalence relation and show how it subdivides or partitions a set into distinct categories. Introduction Classifying objects and placing similar objects into groups provides a way to organize information and focus attention on the similarities of like objects and not on the dissimilarities of dislike objects. Mathematicians have been classifying objects for millennia. Lines in the plane can be subdivided into groups of parallel lines. Lines in a given group many not be the same line, but they are considered equivalent insofar as our classification is concerned. Triangles can also be classified as being congruent 1 or noncongruent. In modular arithmetic we say two integers are equivalent if they have the same remainder when divided by a given number, say 7 for example. Here, the numbers -12, -5, 2, 9, 16, 23, 30, would be considered equivalent, whereas and 3 and 5 are not. This classification based on division by 7 subdivides the entire set of natural numbers into seven distinct equivalence classes depending on whether a number s remainder is 0,1,2,3,4,5, or 6 when divided by 7. So what are the properties a relation must have in order that it subdivide set into distinct categories? Keep reading. Equivalence Relation The reader is already familiar with one equivalence relation, the relation of two things being equal. That s right " = ". However, since ambiguity restricts us from using the equal sign to denote the equivalence relation, we have adopted the nearby symbol 2 " ". So what properties does an equivalence relation possess? Definition A relation " " on a set A is an equivalence relation if for all x, y, z in A, the following RST properties hold: Reflexive: x x Symmetric: if x y, then y x Transitive: if x y and y z, then x z. 1 Recall that two triangles are congruent their corresponding sides and angles have the same measurements. 2 Another common notation for the equivalence relation is " ".

2 Note: Remember, a relation on a set A is a subset of all ordered pairs of A, so x y means x, y. However, the keep in mind is a set. For example, ( ) set notation looks awkward so we prefer to simply write x Example 1 y. The following relations are equivalence relations on the given sets. x y means equality ( = ) between numbers or sets. x y means x is congruent to y where x, y are triangles. x y means x y for logical sentences x, y. x y means x has the same birthday as y where x and y are people. x y means x differs from y by an integer multiple of 5 ) on the set Z of integers. Example 2 Non The following relations are not equivalence relations on the given sets since at least one of the three RST conditions fails. x y means x is in love with y on the set of all people. (Not likely symmetric; at least for two persons, one loves the other but not Vice-versa.) x y means " x y" on the real numbers. (Not symmetric; 2 3 does not imply 3 2.) x y means x and y have a common factor greater than 1 on the set of integers. (Not transitive; 2 and 6 have a common factor, 6 and 3 have a common factor, but 2 and 3 do not.) x y means x y on a family of sets. (Not symmetric; A B B A.)

3 Equivalence Broadens the Concept of Equality Partitioning of Sets into Equivalence Classes The equivalence relation provides a way for mathematicians to subdivide collections of objects into distinct subsets called partitions. Definition: Let A be a set. A partition of A is a (finite or infinite) collection A, A,... of nonempty subsets of A such that { } 1 2 the union of all the A i is A the sets in the collection are pairwise disjoint; that is Ai Aj distinct pair A and A. i j = for every Example 3 The following are partitions of a set. The even and odd integers are a partition the integers into two distinct sets.

4 A collection of people partitioned into groups according to the first letter in their surname. A set of students in a classroom partitioned into males and females. (If there are no members of one sex, then the partition is the set itself.) The following theorem lies at the heart of the importance of the equivalence relation. Theorem 1 Equivalence Classes Let A be any set and " " an equivalence relation on A. For every x A, x, by define a set, denoted by [ ] [ x] = { y A : y x} This set is called the equivalence ence class of x and consists of all elements of A equivalen to x. Under these conditions each element of the set A belongs to one and only one equivalence class, which means the equivalence relation partitions the set A into distinct subsets. Proof: First observe every element x A belongs to at least one equivalence class since x x x x. Also note the union of the equivalence, which means [ ] classes is A since every element x A belongs to some equivalence class. Now show the equivalence classes are disjoint. We do this by showing if two equivalence classes intersect, then they are the same equivalence class. Let s, t A and assume that the equivalence classes [ ] define, intersect. That is, there exists [ s] [ t] s and [ ] t which they y (as drawn in the diagram).

5 We now show[ s] = [ t]. We first prove [ s] [ t] by letting [ s] that x [ t]. Letting [ s] of the equivalence relation, we have i) x s [ ] [ ] x and showing x and using the symmetric and transitive properties ii) but y s and so y s, hence s y iii) since x s and s y we have x y. iv) but y t and so y t v) since x y and y t we have x t This shows x [ t] and hence [ s] [ t]. The proof [ s] [ t] [ s] = [ t]. is similar and so Significance of Theorem 1 (Partitioning Sets) An equivalence relation defined on a set A, partitions the set into disjoint subsets, called equivalence classes. The elements of each equivalent class are equivalent to each other, but not to any element in a different equivalence class. One can think of the equivalence relation as a kind of generalized equality where objects in a given class are equal. The set of equivalences classes of A partitioned by is called the quotient set of A modulo, and denoted by A/. See Figure 1. Example 4 (Modular Arithmetic) Partitioning of A into Equivalence Classes Figure 1 Let A =Z be the integers. We define two numbers, x y as equivalent if they have the same remainder when divided by a given number, say 5, and we

6 write this as x y ( mod 5) find the equivalence classes of this relation. Solution reflexive: x x( mod 5) reflexive: symmetric: If x y ( mod 5) symmetric:. Show this relation is an equivalence relation and since 5 divides x x = 0. then 5 divides x y. This means there exists an integer k such that x y = 5k. But this implies y x = 5k = 5( k ) which means 5 divides y x Hence is a symmetric relation.. Hence y x ( mod 5). transitive: If x y ( mod 5) and y z ( mod 5) transitive:, then 5 divides x y and 5 divides y z. Hence, there exist integers k 1, k 2 that satisfy x y = 5k1 and y z = 5k2. Adding these equations yields or ( x y) + ( y z) = 5k1 + 5k2 ( ) x z = 5 k + k = 5k 1 2 3 which shows that 5 divides x z or x z ( mod 5). Hence is a transitive relation. The above equivalence relation divides the integers Z into the equivalence classes, where each equivalence class consists of integers with remainders of 0,1,2,3,4 when divided by 5. We give the numbers 0,1,2,3,4 special status and assign them as representatives of each equivalence class 3 listed below. ble 1 Equivalence Classes for mod ( 5) Table 1 Equivalence Classes for [ 0] = { 5 n : n Z} = { 10, 5, 0, 5, 10 } [ 1] = { 5n + 1: n Z} = { 9, 4, 1, 6, 11 } [ 2] = { 5n + 2 : n Z} = { 8, 3, 2, 7, 12 } [ 3] = { 5n + 3: n Z} = { 7, 2, 3, 8, 13 } [ 4] = { 5n + 4 : n Z} = { 6, 1, 4, 9, 14 } 3 In number theory these equivalence classes are also called residue classes.

7 In this example, the quotient set Z / (set of equivalence classes) would be {[ ] [ ] [ ] [ ] [ ]} Z / = 0, 1, 2, 3, 4 Note: Some people do not understand how the remainder of 3 / 5 can be 2. Remainders are defined as non negative integers, so we write 3 / 5 = 5 + 2 / 5 = 1+ 2 / 5. ( ) Modular Arithmetic: Modular arithmetic occurs in many areas of mathematics as well as application in our daily lives. Instead of working with an infinite number of integers, it is only necessary to work with a few. The smallest number is mod ( 2 ), which subdivides the integers into two types, even and odd integers. Public key cryptography allows computers to do exact arithmetic mod ( n ), where n is the product of two prime numbers each 100 digits long. Example 5 (Equivalence Classes in the Plane) The Cartesian product A = N N defines points in the first quadrant of the Cartesian plane with integer coordinates. Define a relationship between, c, d as ( a b ) and ( ) (, ) (, ) a b c d b a = d c. Given that this relation defines an equivalence relation (See Problem 7), find its equivalence classes. Solution This relation classifies points (, ) x y as equivalent if their y coordinate minus their x coordinate are equal. This means two points (with positive integer coefficients), are equivalent if they both lie on 45 degree lines y = x + n, where n is an integer. Each value of n defines a different equivalence class. Some typical equivalence classes are

8 Table 3 Equivalence Classes...... ( 2,0 ) = {(2,0),(3,1),(4, 2),... } ( n = 2) ( 1,0 ) = {(1,0),(2,1),(3,2),... } ( n = 1) ( 0,0 ) = {(0,0),(1,1),(2, 2),... } ( n = 0) ( 0,1 ) = {(0,1),(1, 2),(2,3),... } ( n = 1) ( 0, 2 ) = {(0,2),(1,3),(2, 4),... } ( n = 2) These equivalence classes are illustrated in Figure 2 as points with integer coordinates lying on 45 degree lines in the first quadrant. Equivalence Classes as Grid Points on Lines y = x + n Figure 2 Margin Note: a is equivalent to b means a is equal to b, not in every respect, but as far as a particular property is concerned. Graph of an Equivalence Relation R The graph of the relation R = {( 1,1 ),( 2, 2 ),( 3,3)( 4, 4 ),( 1,3 ),( 3,1 ),( 3,4 ),( 4,3) }

9 on the set A = { 1,2,3, 4} is shown in Figure 3. It is an easy matter to verify that R is an equivalence relation. What are the equivalence classes of this relation? Solution The relation R says 1 1, 2 2, 3 3, 4 4, 1 3, 3 1, 3 4, 4 3 Using the transitivity property, we can show 1 3 4, and that the number 2 is only equivalent to itself. Hence the set A has been subdivided into two 2. Note that the two equivalence classes equivalence classes { 1,3, 4} and { } are disjoint and their union is A. Graph of an Equivalence Relation Figure 3

10 Problems 1. Let A be the set of students at a university and let x and y be students; i.e. elements of A. Tell if the following relations on A are equivalent relations. a) x is related to y iff x and y have the same major. b) x is related to y iff x and y have the GPA. c) x is related to y iff x and y are from the same country. d) x is related to y iff x and y e) x is related to y iff x and y have the same major. 2. (?) Tell if the following relations R are equivalence relations on a set A. If the relation is an equivalence relation, find the equivalence classes. a) xry iff y 2 = x ( A =R ) b) xry iff x is a factor of y ( A =N ) c) xry iff x and y are multiples of 5 ( A =N ) (Modular Arithmetic) Given the set { 1,2,3,4,5,6,7,8 } have x y ( mod 4), i.e. 4 divides x y 3. (Modular Arithmetic) relation and find the equivalence classes of the relation. 3. (True or False) a) 8 3( mod 5) b) 12 1( mod 6) c) 21 0( mod 3) d) 10 10( mod12) e) 99 9( mod 9) A = and for x, y A we. Prove that is an equivalence 4. (Defining New ) Let " " and " " be two equivalence relations on a set. Define two more relations by ( ) ( ) ( ) ( ) a b a b a b a b a b a b Determine if " " and " " are equivalence relations.

11 5. (An Old Favorite) The equals relation " = " is the most familiar equivalence relation. What are the equivalence classes of the equals relation on the set A = 1, 2,3,4,5? { } 6.. (Equivalence Relation) The Cartesian product A = N N defines the points in the first quadrant with integer coordinates. Show the relationship between, c, d defined by ( a b ) and ( ) is an equivalence relation. (, ) (, ) a b c d b a = d c 7. (Graph of an Equivalence Relation) Figure 4 shows an equivalence relation A = 1,2,3, 4. What are the equivalence classes? on the set { } 8. (Projective Plane) Graph of an Equivalence Relation Figure 4 { } 2 (Projective Plane) Define A = ( 0,0) minus the origin. Define a relation between two points (, ) R the set of all points in the plane by x y and ( x, y ) saying they are related if and only if they lie on the same line passing through the origin. Show that this relation is an equivalence relation and draw a graphical representation of the equivalence classes by picking a representative from each class. The space of all equivalence classes under this relation is called the projective plane.

12 9. (Equivalence Classes in Logic) Define an equivalence relation on logical sentences by saying two sentences are equivalent if they have the same truth value. Find the equivalence classes in the following collection of sentences. 1+ 2 = 3 3 < 5 2 7 x 2 < 0 for some real number. 2 2 sin x + cos x = 1 Georg Cantor was born in 1845. Leopold Kronecker was a big fan of Cantor. Cantor's theorem guarantees larger and larger infinite sets. 10. (Similar Matrices) Two matrices A, B if there is an invertible matrix M such that MAM B 1 =. a) Show that similarity of matrices is an equivalence relation. b) Let 1 1 1 0 A =, M 4 1 = 1 1 find B. What common properties do A, B share? 11. (Equivalence Relation on R ) Let f : R R. Define a relation on R by a b f ( a) = f ( b) a) Show is an equivalence relation. f x = x? b) What are the equivalence classes for ( ) 2 c) What are the equivalence classes for f ( x) = sin x? 12. (Arithmetic in Modular Arithmetic) Suppose Show a c b d ( mod 5) ( mod 5)

13 ( ) ( ) ( ) a) a + b c + d mod 5 b) a b c d mod 5 c) ab cd mod 5 13. (Equivalent Angles) We say two angles x, y are related if and only if 2 2 sin cos 1 xry x + y = Show that R is an equivalence relation and identify the equivalence classes. (Moduli in Modular Arithmetic) Note that 2 2 0( mod 4) 14. (Moduli in Modular Arithmetic) = but neither factor is zero. What is 2 4 modulo 8? Is there any product a b with neither factor a b 0 mod3? Can you form a conjecture about the moduli related to 0, but ( ) this phenomenon?