equivalence relations
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Equivalence Relations. Rosen 7.5. 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. Some preliminaries. - PowerPoint PPT PresentationTRANSCRIPT
Equivalence Relations
Rosen 7.5
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.
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
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.
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 mFirst 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.
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
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
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,
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.)
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.
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).
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).
Equivalence ClassLet 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.
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
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
Let A = {1,2,3,4,5,6,7,8} and let R = {(a,b)|ab(MOD 3)} be a relation on A.
14
7
25
8
36
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.
How many equivalence relations can there be on a set A with n elements?
A has one element. One equivalence class, rank =1.
A has two elements
rank = 2
rank = 1
How many equivalence relations can there be on a set A with n elements?
A has three elements
Rank = 2Rank = 3
Rank = 1
How many equivalence relations can there be on a set A with n elements?
A has four elements
Rank = 4Rank = 1
How many equivalence relations can there be on a set A with n elements?
A has four elements
Rank = 2
How many equivalence relations can there be on a set A with n elements?
A has four elements
Rank = 2
How many equivalence relations can there be on a set A with n elements?
A has four elements
Rank = 3
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?