1. Equivalence Relations
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Section 7.5
Equivalence Relations
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Definition
A relation on a set is called an equivalence relation if it is:
Reflexive,
Symmetric and
Transitive
Two elements related by an equivalence relation are said to be equivalent
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3. Example 1 Congruence modulo m is a widely used equivalence relation:
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Example 1
Congruence modulo m is a widely used equivalence relation:
Let m be a positive integer (m > 1); show that R = {(a,b) | a  b (mod m)} is an equivalence relation on the set of integers
a  b (mod m) means a is congruent to b mod m; this is true if and only if m divides (is a factor of) a-b
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Example 1
To show that R is an equivalence relation, we must prove it is:
Reflexive: a-a is divisible by m, since 0=0 * m;  a  a (mod m), and the relation is reflexive
Symmetric: suppose a  b (mod m); then a-b is divisible by m, so a-b = km; it follows that b-a = (-k)m, so b  a (mod m), and the relation is symmetric
Transitive: see next slide
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5. Example 1  a  b (mod m) is an equivalence relation
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Example 1
Transitive: suppose a  b (mod m) and b  c (mod m); so m divides both a-b and b-c; this means there exist integers k and n with a-b = km and b-c = nm; adding these equations, we get a-c = (a-b)+(b-c) = km+nm = (k+n)m; thus, a  c (mod m) and the relation is transitive
 a  b (mod m) is an equivalence relation
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Example 2
Suppose A is a nonempty set, and f is a function that has A as its domain; let R be a relation on A consisting of all ordered pairs (x,y) where f(x) = f(y)
Is R an equivalence relation on A?
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7. Example 2 R is reflexive: f(x) = f(x) for all x  A
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Example 2
R is reflexive: f(x) = f(x) for all x  A
R is symmetric: if f(x) = f(y), then f(y) = f(x) (by the fundamental property of equality)
R is transitive: if f(x) = f(y) and f(y) = f(z), then f(x) = f(z) (also by the fundamental property of equality)
 R is an equivalence relation on A
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Equivalence classes
Let R be an equivalence relation on set A; the set of all elements that are related to an element e  A is called the equivalence class of e
The equivalence class of e with respect to R is denoted [e]R
If R is an equivalence relation on A, the equivalence class of element e  A is [e]R = {s | (e,s)  R}
If b  [e]R b is a representative of this equivalence class
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Example 3
What are equivalence classes of 0 and 1 for congruence modulo 4?
For 0, the equivalence class contains all integers e such that e  0 (mod 4); in other words, a-0 is divisible by 4, so [0] = {…, -8, -4, 0, 4, 8, …}
For 1, the equivalence class is all integers e such that e  1 (mod 4); so [1] = {…, -7, -3, 1, 5, 9, …}
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Generalizing example 3
We can replace 4 in the previous example with any integer m
The equivalence classes of the relation congruence modulo m are called congruence classes modulo m; such a congruence class of integer a is denoted [a]m and [a]m = {…, a-2m, a-m, a, a+m, a+2m, …}
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Example 4
Suppose R is the equivalence relation on non-empty set A such that R={(x,y)|f(x)=f(y)}; what are the equivalence classes on R?
The equivalence class of x is the set of all yA such that f(y)=f(x)
By definition, this is the inverse image of f(x);  the equivalence classes are those sets f-1(b) for every b in the range of f
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Equivalence classes of 2 elements of a set are either identical or disjoint
Let R be an equivalence relation on set A. The following statements are equivalent:
(i) aRb
(ii) [a]R = [b]R
(iii) [a]R  [b]R ≠ 
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13. Proof Assuming aRb, prove [a]=[b] by showing [a]≤[b] and [b]≤[a]:
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Proof
Assuming aRb, prove [a]=[b] by showing [a]≤[b] and [b]≤[a]:
Suppose c[a] – so aRc;
Since aRb and R is symmetric, we know that bRa;
Since R is transitive, bRa, and aRc, so bRc;
Hence c[b];
This shows that [a]≤[b]
We can use the same logic to show b]≤[a]
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Proof
Assume [a]=[b]
It follows that [a]  [b] ≠  because [a] is nonempty (since a[a] because R is reflexive)
Assume [a]R  [b]R ≠ ; then there is some element c with c[a] and c[b]:
 aRc and bRc
Since R is symmetric, cRb
Since R is transitive, aRc and cRb, so aRb
So aRb  [a]R = [b]R , [a]R = [b]R  [a]R[b]R≠, and [a]R  [b]R ≠   aRb; thus the statements are equivalent
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15. Partitions Let R be an equivalence relation on set A
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Partitions
Let R be an equivalence relation on set A
the union of the equivalence classes of R is all of A, since each aA is its own equivalence class, [a]R
in other words, [a]R = A
aA
From the theorem previously proved, it follows that equivalence classes are either equal or disjoint, so [a]R[b]R =  only when [a]R  [b]R
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Partitions
Equivalence classes split set A into disjoint subsets, called partitions: collections of disjoint non-empty subsets of a set which have the set as their union
In other words, the collection of subsets Ai forms a partition of S if and only if:
Ai  
Ai  Aj =  when i  j and
Ai = S
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17. Example 5 Suppose S = {a,b,c,d,e,f}
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Example 5
Suppose S = {a,b,c,d,e,f}
The collection of sets A1 = {a,b,c}, A2 = {d} and A3 = {e,f} form partitions of S, since:
they are disjoint and
their union is S
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18. Equivalence classes & partitions
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Equivalence classes & partitions
Equivalence classes of an equivalence relation on a set form a partition of the set; subsets in this partition are the equivalence classes
Conversely, every partition of a set can be used to form an equivalence relation:
2 elements are equivalent with respect to this relation if and only if they are in the same subset of the partition
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19. Example 6 Assume {Ai} is a partition on S
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Example 6
Assume {Ai} is a partition on S
Let R be a relation on S consisting of the pair (x,y) where x,y  Ai
To prove R is an equivalence relation, we must show that it is reflexive, symmetric, and transitive
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Example 6
Reflexive: (a,a)  R for every a  S, since a is in the same subset as itself;
Symmetric: if (a,b)  R then b and a are in the same subset, so (b,a)  R
Transitive: if (a,b)  R and (b,c)  R then a and b are both in the same subset (x) and b and c are both in the same subset (y) (continued next slide)
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Example 6
Since by definition the subsets are disjoint, and b is in both x and y, x and y must be equal
So a and c are in the same subset and R is transitive
Therefore R is an equivalence relation
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22. One more theorem Let R be an equivalence relation on a set S
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One more theorem
Let R be an equivalence relation on a set S
Then the equivalence classes of R form a partition of S
Conversely, given a partition {Ai} or set S, there is an equivalence relation R that has the sets Ai, where Ai = S, as its equivalence classes
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Example 7
What are the ordered pairs in the equivalence relation of {0,1,2,3,4,5} in the following partitions?
A1= {0}, A2 = {1,2}, A3 = {3,4,5}:
(0,0) belongs to R since A1 is an equivalence class;
(1,1), (1,2), (2,1) and (2,2) belong to R because A2 is an equivalence class;
(3,3), (3,4), (3,5), (4,4), (4,3), (4,5), (5,5), (5,3,), (5,4) belong to R because A3 is an equivalence class
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