# Equivalence Relations ```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
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From arithmetic: Equals (=)
From logic: If and only if (&lt;=&gt;)
In this session we will:
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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
(a,a)  R for all a in S.
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(a,b)  R implies (b, a)  R
Symmetric
(a,b)  R and (b,c)  R imply (a,c)  R
Transitive
Properties of Equivalence
Relations
a
a
Reflexive
a
b
Symmetric
b
Transitive
c
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Notation
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Given a relation R, we usually write
a R b instead of (a,b)  R
For example:
x = 1 instead of (x,1)  

of ( p,q)  
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Properties Revisited
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~ 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?
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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?
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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
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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)
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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
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Let ~ be given by a ~ b if 2 | a–b.
Let [n] be the set of all integers related
to n
Even
 = { …-4, -2, 0, 2, 4 …}
Odd
 = { …-3, -1, 1, 3, 5 …}
There are many different names for
these equivalence classes, but only two
distinct equivalence classes.
Theorem 0.6 (paraphrased)
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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)
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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.
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 ~
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[(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!
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