Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide01.html Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide02.html Equivalence Relations prev | slides | next prev | slides | next Consider the following relations on the set of people in this room {( a , b ) | a and b were born in the same month}, {( a , b ) | a and b are the same sex}, Equivalence Relations {( a , b ) | a and b are from the the same state}. Observe that these relations are all reflexive, symmetric and transitive. Because of this they are all equivalent in some way. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 of 1 09/11/2003 03:48 PM 1 of 1 09/11/2003 03:48 PM Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide03.html Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide04.html Equivalence Relations Equivalence Relations prev | slides | next prev | slides | next A relation on a set A is an equivalence relation if it is reflexive, Let R be an equivalence relation on a set A . The set of all elements symmetric and transitive. that are related to an element a of A is called the equivalence class of a . This is denoted [ a ] R or just [ a ] if it is clear what R is. Suppose that R is a relation on the positive integers such that ( a , b ) R if and only if a <5 and b <5. Is R and equivalence relation? Suppose R is {( a , b ) | a and b were born in the same month} and is defined on the set of people in this room. Then Since a = a it follows that if a <5 then ( a , a ) R so we know that R is reflexive. [ a ] = { b | b was born in the same month as a }. Suppose A = {1, 2, 3, 4} and Suppose ( a , b ) R so both a <5 and b <5. In this case certainly ( b , a ) R so that R is symmetric. R = {(1,1), (1,2), (2,1), (2,2), (3,3), (3,4), (4,3), (4,4)} Finally, if ( a , b ) R and ( b , c ) R then both a and c are less than 5 so We can list the equvalence class for each element of A : ( a , c ) R showing that R is transitive. [1] = {1, 2}, [2] = {1, 2}, [3] = {3, 4}, [4] = {3,4} Thus R is an equivalence relation. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 of 1 09/11/2003 03:48 PM 1 of 1 09/11/2003 03:48 PM
Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide05.html Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide06.html Equivalence Relations Equivalence Relations prev | slides | next prev | slides | next Theorem A partition of a set S is a collection of disjoint, nonempty subsets of S that have S as their union. Let R be an equivalence relation on A . The following statements are equivalent. If S = {1, 2, 3, 4, 5, 6, 7, 8} then one partition of S is 1. a R b { {1,2}, {3}, {4, 5, 6}, {7, 8} } 2. [ a ] = [ b ] Notice that every element of S is in exactly one of the subsets. The equivalence classes of a relation on a set A form a partition of 3. [ a ] [ b ] A . We’ll prove this using a standard approach. First we’ll show that The union of all the [ a ] is equal to A . statement 1 statement 2. Next we’ll show that statement 2 statement 3. Finally we’ll show that statement 3 statement 1. [ a ] [ b ] = when [ a ] [ b ]. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 of 1 09/11/2003 03:48 PM 1 of 1 09/11/2003 03:48 PM Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide07.html Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide08.html Equivalence Relations Equivalence Relations prev | slides | next prev | slides | next Proof: Statement 1 Statement 2 Proof: Statement 2 Statement 3 Assume c [ a ] so that aRc . Because equivalence relations are Because [ a ]=[ b ] we know that a [ a ] and a [ b ]. symmetric we know that cRa . Since aRb by the transitive property Since we know at least one element common to both sets [ a ] and we can conclude that cRb so that c [ b ]. [ b ] we can conclude that [ a ] [ b ] . This argument shows that [ a ] [ b ]. This shows that [ a ]=[ b ] [ a ] [ b ] . We can reverse the argument above to show that [ b ] [ a ]. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Taken together this shows that aRb [ a ]=[ b ]. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 of 1 09/11/2003 03:48 PM 1 of 1 09/11/2003 03:48 PM
Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide09.html Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide10.html Equivalence Relations Equivalence Relations prev | slides | next prev | slides | next Extended Example: Congruence Classes Proof: Statement 3 Statement 1 The integer divsion algorithm is p = mq + r . Here r is the Suppose c [ a ] [ b ] so that c is in both sets [ a ] and [ b ]. This means remainder that results when p is divided by m . that cRa and cRb . The modulo function is a function m :( Z x Z + ) Z 0+ that returns the By the symmetric and transitive properties we can conclude that aRb remainder when one integer is divided by another. The set Z + is the set of positive integers and the set Z 0+ is the set of nonnegative This shows that [ a ] [ b ] aRb . integers. The proof is now completed. We say that a is congruent to b modulo m if 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 m | ( a - b ) which means m divides ( a - b ). Another way to express this is that a mod m = b mod m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 of 1 09/11/2003 03:48 PM 1 of 1 09/11/2003 03:48 PM Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide11.html Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide12.html Equivalence Relations Equivalence Relations prev | slides | next prev | slides | next Theorem We use the notation Let m be a positive integer greater than 1. The relation b (mod m ) a R = {( a , b ) | a b (mod m )} to indicate that a is congruent to b modulo m. is an equivalence relation on the set of integers. Example: Think of a clock: in some sense 15 minutes and 75 minutes are the same, since in both cases the minute hand is a the Proof three-o’clock position. In fact 15 75 (mod 60). This is easy to see because 75-15 = 60. We need to show that R is reflexive, symmetric and transitive. To see that R is reflexive we need to show that ( a , a ) R for all integers Example: 53 89 (mod 12). To see this observe that 89-53=36 and a . 12 | 36. We know that a a (mod m ) if m | ( a - a ). Since a - a =0 and we 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 know that m | 0 we can conclude that R is reflexive. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 of 1 09/11/2003 03:48 PM 1 of 1 09/11/2003 03:48 PM
Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide13.html Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide14.html Equivalence Relations Equivalence Relations prev | slides | next prev | slides | next Proof (continued) Proof (continued) To see that R is symmetric we assume that ( a , b ) R so that Finally, we need to show that R is transitive. Assume that ( a , b ) R and ( b , c ) R . This means that a - b = km b (mod m ) which means that a - b = km for some integer k a for some integer k . In this case c (mod m ) which means that b - c = jm for some integer j b b - a = - km Adding the equations on the right gives which is also divisible by m , but this means that ( a - b ) + ( b - c ) = km + jm a - c = ( k + j ) m b a (mod m ) In the second form we see that ( a - c ) is a multiple of m so that a is so ( b , a ) R . Thus R is symmetric. congruent to c modulo m . This means that ( a , c ) R , and so R is transitive. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 of 1 09/11/2003 03:48 PM 1 of 1 09/11/2003 03:50 PM Equivalence Relations http://localhost/~senning/courses/ma229/slides/equivalence-relations/slide15.html Equivalence Relations prev | slides | next Since the congruence modulo m relation is an equivalence relation it must partition the set of integers. Consider all the numbers that satisfy a 1 (mod 3) Any integer which has a remainder of 1 when divided by 3 will satisfy this expression; examples are numbers like 4 and 7. We denote the congruence class of a modulo m with [ a ] m . Example: [0] 3 = {..., -9, -6, -3, 0, 3, 6, 9, ...} [1] 3 = {..., -8, -5, -2, 1, 4, 7, 10, ...} [2] 3 = {..., -7, -4, -1, 2, 5, 8, 11, ...} 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 of 1 09/11/2003 03:49 PM
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