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.

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

€

(a,a)∈ R for all a in S.

€

(a,b)∈ R implies (b,a) ∈ R

€

(a,b)∈ R and (b,c)∈ R imply (a,c)∈ R

Reflexive

Symmetric

Transitive

Properties of Equivalence Relations

a

Reflexive

a b

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 €

(a,b)∈ R

€

(x,1)∈ =

€

(p,q)∈ ⇔

€

p⇔ q

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 thata–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) = 0So 0 = adf – bcf + bcf – bde = d(af – be)Since d ≠ 0, af–be = 0Hence (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!