Page 235 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Equivalence Relations 221
Y ∈ F such that x ∈ Y. But then by Lemma 4.6.8, [x] = Y. Thus X = Y ∈ F,
so A/R ⊆ F.
Now suppose X ∈ F. Then since F is a partition, X = ∅, so we can choose
some x ∈ X. Therefore by Lemma 4.6.8, X = [x] ∈ A/R,so F ⊆ A/R. Thus,
A/R = F.

Commentary. We prove that A/R = F by proving that A/R ⊆ F and
F ⊆ A/R. For the first, we take an arbitrary X ∈ A/R and prove that X ∈ F.
Because X ∈ A/R means ∃x ∈ A(X = [x]), we immediately introduce the
new variable x to stand for an element of A such that X = [x]. The proof that
X ∈ F now proceeds by the slightly roundabout route of finding a set Y ∈ F
such that X = Y. This is motivated by Lemma 4.6.8, which suggests a way of
showing that an element of F is equal to [x] = X. The proof that F ⊆ A/R
also relies on Lemma 4.6.8.

We have seen how an equivalence relation R on a set A can be used to
define a partition A/R of A and also how a partition F of A can be used to
define an equivalence relation ∪ X∈F (X × X)on A. The proof of Theorem
4.6.6 demonstrates an interesting relationship between these operations. If you
start with a partition F of A, use F to define the equivalence relation R =
∪ X∈F (X × X) and then use R to define a partition A/R, then you end up back
where you started. In other words, the final partition A/R is the same as the
original partition F. You might wonder if the same idea would work in the other
order. In other words, suppose you start with an equivalence relation R on A,
use R to define a partition F = A/R, and then use F to define an equivalence
relation S =∪ X∈F (X × X). Would the final equivalence relation S be the same
as the original equivalence relation R? You are asked in exercise 9 to show that
the answer is yes.
We end this section by considering a few more examples of equivalence
relations. A very useful family of equivalence relations is given by the next
definition.

Definition 4.6.9. Suppose m is a positive integer. For any integers x and y,we
will say that x is congruent to y modulo m if ∃k ∈ Z(x − y = km). In other
words, x is congruent to y modulo m iff m | (x − y). We will use the notation
x ≡ y (mod m) to mean that x is congruent to y modulo m.

For example, 12 ≡ 27 (mod 5), since 12 − 27 =−15 = (−3) · 5. Now
it turns out that for every positive integer m, the relation C m ={(x, y) ∈
Z × Z | x ≡ y (mod m)} is an equivalence relation on Z. We will check tran-
sitivity for C m and let you check reflexivity and symmetry in exercise 10.