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Equivalence Relations. Fractions vs. Rationals. Question: Are 1/2, 2/4, 3/6, 4/8, 5/10, … the same or different? Answer: They are different symbols that stand for the same rational number.
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Fractions vs. Rationals Question: Are 1/2, 2/4, 3/6, 4/8, 5/10, … the same or different? Answer: They are different symbolsthat stand for the same rational number. When algebraists have a set of objects and wish to think of more than one of them as the same object, they define an equivalence relation.
Familiar Equivalence Relations • From arithmetic: Equals (=) • From logic: If and only if (<=>)
In this session we will: • Carefully define the notion of an equivalence relation • Show how an equivalence relation gives rise to equivalence classes • Give an important example of an equivalence relation and its classes.
Definition • An equivalence relation on a set S is a set R of ordered pairs of elements of S such that Reflexive Symmetric Transitive
Properties of Equivalence Relations a b a Reflexive Symmetric a b c Transitive
Notation • Given a relation R, we usually write a R b instead of • For example: x = 1 instead of instead of
Properties Revisited • ~ is an equivalence relation on S if ~ is: • Reflexive: a~a for all a in S • Symmetric: a~b implies b~a for all a, b in S • Transitive: a~b and b~c implies a~c for all a,b,c in S
Is equality an equivalence relation on the integers? • a = a for all a in Z • a = b implies b = a for all a,b in Z • a = b and b = c implies a = c for all a,b,c, in Z. • = is reflexive, symmetric, and transitive So = is an equivalence relation on Z!
Is ≤ an equivalence relation on the integers? • 1 ≤ 2, but 2 ≤ 1, so ≤ is not symmetric Hence, ≤ is not an equivalence relation on Z. • (Note that ≤ is reflexive and transitive.)
Say a ~ b if 2 | a – b • Choose any integer a. 2 | 0 = a – a, so a~ a for all a. (~ is reflexive) • Choose any integers a, b with a ~ b. 2 | a–b so a–b = 2n for some integer n. Then b–a = 2(–n), and 2 | b–a. Hence b ~ a. (~ is symmetric)
a ~ b if 2 | a – b (Con't) • Choose any integers a, b, c with a~b and b~c. • Now 2 | a–b and 2 | b–c means that there exist integers m and n such that a–b = 2m and b–c = 2n. a–c = a–b + b–c = 2m + 2n = 2(m + n) So 2 | a–c. Hence a~c. ~ is transitive. • Since ~ is reflexive, symmetric, and transitive ~ is an equivalence relation on the integers.
Equivalence Classes • Let ~ be given by a ~ b if 2 | a–b. • Let [n] be the set of all integers related to n • [0] = { …-4, -2, 0, 2, 4 …} • [1] = { …-3, -1, 1, 3, 5 …} • There are many different names for these equivalence classes, but only two distinct equivalence classes. Even Odd
Theorem 0.6 (paraphrased) • Every equivalence relation R on a set S partitions S into disjoint equivalence classes. • Conversely, every partition of S defines an equivalence relation on S whose equivalence classes are precisely the sets of the partition.
Example 14 (my version) • Let S = {(a,b) | a,b are integers, b≠0} • Define (a,b) ~ (c,d) if ad–bc = 0 • Show ~ is an equivalence relation. • For (a,b) in S, ab–ba = 0, so (a,b)~(a,b). Hence ~ is reflexive. • (a,b)~(c,d) implies ad–bc = 0 so cb–da = 0 which implies (c,d)~(a,b) Hence ~ is symmetric.
Example 14 (con't) Suppose (a,b)~(c,d) and (c,d)~(e,f), where b,d, and f are not zero. Then ad–bc = 0 and cf–de = 0. It follows that (ad–bc)f + b(cf–de) = 0 So 0 = adf – bcf + bcf – bde = d(af – be) Since d ≠ 0, af–be = 0 Hence (a,e) ~ (f,b), and ~ is transitive. Since ~ is reflexive, symmetric, and transitive, ~ is an equivalence relation.
The equivalence classes of ~ • [(1,2)] = [(2,4)] = [(3,6)] = [(4,8)] = … • [(3,4)] = [(6,8)] = [(9,12)] = … • Replace commas by slashes and drop the parentheses to get: • 1/2 = 2/4 = 3/6 = 4/8 = … • 3/4 = 6/8 = 9/12 = … • Each rational number is an equivalence class of ~ on the set of fractions!
