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Section 7.5 - Equivalence Relations - PowerPoint PPT Presentation

Equivalence Relations

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**1. **1 Section 7.5 Equivalence Relations

**2. **2 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

**3. **3 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

**4. **4 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

**5. **5 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

**6. **6 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?

**7. **7 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

**8. **8 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

**9. **9 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, …}

**10. **10 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, …}

**11. **11 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

**12. **12 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 ? ?

**13. **13 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]

**14. **14 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

**15. **15 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

**16. **16 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

**17. **17 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

**18. **18 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

**19. **19 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

**20. **20 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)

**21. **21 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

**22. **22 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

**23. **23 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