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**Equivalence Relations**

Rosen 7.5

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Equivalence Relation A relation on a set A is called an equivalence relation if it is Reflexive Symmetric Transitive Two elements that are related by an equivalence relation are called equivalent.

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Some preliminaries Let a be an integer and m be a positive integer. We denote by a MOD m the remainder when a is divided by m. If r = a MOD m, then a = qm + r and 0 r < m, qZ Examples Let a = 12 and m = 5, 12MOD5 = 2 Let a = -12 and m = 5, –12MOD5 = 3

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ab(MOD m) If a and b are integers and m is a positive integer, then a is congruent to b modulo m if m divides a-b. (a-b)MODm = 0 (a-b) = qm for some qZ Notation is ab(MOD m) aMODm = bMODm iff ab(MOD m) 12MOD5 = 17MOD5 = 2 (12-17)MOD 5 = -5MOD5 = 0 So, it’s the same as saying that a and b have the same remainder.

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**Prove that a MOD m = b MOD m iff ab(MOD m)**

Proof: We must show that aMODm = bMODm ab(MOD m) and that ab(MOD m) aMOD m = b MOD m First we will show that aMODm = bMODm ab(MOD m) Suppose aMODm = bMODm, then q1,q2,rZ such that a = q1m + r and b = q2m + r. a-b = q1m+r – (q2m+r) =m(q1-q2) so m divides a-b.

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**Prove that a MOD m = b MOD m iff ab(MOD m)**

Next we will show that ab(MOD m) aMOD m = bMODm. Assume that ab(MOD m) . This means that m divides a-b, so a-b = mc for cZ. Therefore a = b+mc. We know that b = qm + r for some r < m, so that bMODm = r . What is aMODm? a = b+mc = qm+r + mc = (q+c)m + r. So aMODm = r = bMODm

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Equivalence Relation A relation on a set A is called an equivalence relation if it is Reflexive Symmetric Transitive Two elements that are related by an equivalence relation are called equivalent. Example A = {2,3,4,5,6,7} and R = {(a,b) : a MOD 2 = b MOD 2} aMOD2 = aMOD2 aMOD2 = bMOD2 bMOD2=aMOD2 aMOD2=bMOD2, bMOD2=cMOD2 aMOD2=cMOD2

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Let R be the relation on the set of ordered pairs of positive integers such that ((a,b), (c,d))R iff ad=bc. Is R an equivalence relation? Proof: We must show that R is reflexive, symmetric and transitive. Reflexive: We must show that ((a,b),(a,b)) R for all pairs of positive integers. Clearly ab = ab for all positive integers. Symmetric: We must show that ((a,b),(c,d) R, then ((c,d),(a,b)) R. If ((a,b),(c,d) R, then ad = bc and cb = da since multiplication is commutative. Therefore ((c,d),(a,b)) R,

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**Proof: We must show that R is reflexive, symmetric and transitive. **

Let R be the relation on the set of ordered pairs of positive integers such that ((a,b), (c,d))R iff ad=bc. Is R an equivalence relation? Proof: We must show that R is reflexive, symmetric and transitive. Transitive: We must show that if ((a,b), (c,d))R and ((c,d), (e,f)) R, then ((a,b),(e,f) R. Assume that ((a,b), (c,d))R and ((c,d), (e,f)) R. Then ad = cb and cf = de. This implies that a/b = c/d and that c/d = e/f, so a/b = e/f which means that af = be. Therefore ((a,b),(e,f)) R. (remember we are using positive integers.) Positive integers because a/b and (-a)/(-b) would work equally well. Therefore, can’t distinguish between positive and negative pairs.

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**Prove that R = ab(MOD m) is an equivalence relation on the set of integers.**

Proof: We must show that R is reflexive, symmetric and transitive. (Remember that ab(MOD m) means that (a-b) is divisible by m. First we will show that R is reflexive. a-a = 0 and 0*m, so a-a is divisible by m.

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**Prove that R = ab(MOD m) is an equivalence relation on the set of integers.**

We will show that R is symmetric. Assume that ab(MOD m). Then (a-b) is divisible by m so (a-b) = qm for some integer q. -(a-b) = (b-a) = -qm. Therefore ba(MOD m).

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**(a-b)+(b-c) = (a-c) = jm+km = (j+k)m **

Prove that R = ab(MOD m) is an equivalence relation on the set of integers. We will show that R is transitive. Assume that ab(MOD m) and that bc(MOD m). Then integers j,k such that (a-b) = jm, and (b-c) = km. (a-b)+(b-c) = (a-c) = jm+km = (j+k)m Since j+k is an integer, then m divides (a-c) so ac(MOD m).

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Equivalence Class Let R be an equivalence relation on a set A. The set of all elements that are related to an element of A is called the equivalence class of a. The equivalence class of a with respect to R is denoted [a]R. I.e., [a]R = {s | (a,s) R} Note that an equivalence class is a subset of A created by R. If b [a]R, b is called a representative of this equivalence class.

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Example Let A be the set of all positive integers and let R = {(a,b) | a MOD 3 = b MOD 3} How many distinct equivalence classes (rank) does R create? 3 The equivalence classes for [1], [2], and [3].

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**Digraph Representation**

It is easy to recognize equivalence relations using digraphs. The subset of all elements related to a particular element forms a universal relation (contains all possible arcs) on that subset. The (sub)digraph representing the subset is called a complete (sub)digraph. All arcs are present. The number of such subsets is called the rank of the equivalence relation

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**Let A = {1,2,3,4,5,6,7,8} and let R = {(a,b)|ab(MOD 3)} be a relation on A.**

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Partition Let S1, S2, …, Sn be a collection of subsets of A. Then the collection forms a partition of A if the subsets are nonempty, disjoint and exhaust A. Si SiSj = if i j Si = A If R is an equivalence relation on a set S, then the equivalence classes of R form a partition of S.

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**How many equivalence relations can there be on a set A with n elements?**

A has two elements A has one element. One equivalence class, rank =1. rank = 2 These are the only possible equivalence classes for this number of elements. rank = 1

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**How many equivalence relations can there be on a set A with n elements?**

A has three elements Rank = 3 Rank = 2 Rank = 1

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**How many equivalence relations can there be on a set A with n elements?**

A has four elements Rank = 4 Rank = 1

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**How many equivalence relations can there be on a set A with n elements?**

A has four elements Rank = 2

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**How many equivalence relations can there be on a set A with n elements?**

A has four elements Rank = 2

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**How many equivalence relations can there be on a set A with n elements?**

A has four elements Rank = 3

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**How many equivalence relations can there be on a set A with n elements?**

1 for n = 1 2 for n = 2 5 for n = 3 15 for n = 4 ? for n = 5 Is there recurrence relation or a closed form solution?

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