# 2.2 Equivalence Relations ```2.2
2.2
J.A.Beachy
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Equivalence Relations
from A Study Guide for Beginner’s by J.A.Beachy,
a supplement to Abstract Algebra by Beachy / Blair
13. For the function f : R → R defined by f (x) = x2 , for all x ∈ R, describe the
equivalence relation ∼f on R that is determined by f .
Solution: The equivalence relation determined by f is defined by setting a ∼f b if
f (a) = f (b), so a ∼ b if and only if a2 = b2 . Probably a better way to understand ∼f
is to note that a ∼f b if and only if |a| = |b|.
14. (a) Define the function f : Z12 → Z18 by setting f ([x]12 ) = [12x]18 . Find the image
f (Z12 ) of f and the factor set Z12 /f of Z12 determined by f and exhibit the one-to-one
correspondence between them.
Solution: It follows from Exercise 2.1.11 that f is a function. (That exercise shows
that f : Zn → Zk defined by f ([x]n ) = [mx]k is a function if and only if k | mn.)
We have f (Z12 ) = {18 , 18 , 18 }. Furthermore, f ([n]12 ) = f ([m]12 ) if and only
if [12n]18 = [12m]18 if and only if 18|12(n − m) if and only if 3|(n − m). Hence
[n]12 ∼ [m]12 if and only if n ≡ m (mod 3). Thus Z12 /f = {[12 ], [12 ], [12 ]},
where [12 ] = {12 , 12 , 12 , 12 }, [12 ] = {12 , 12 , 12 , 12 }, and [12 ] =
{12 , 12 , 12 , 12 }. Furthermore, f&macr; : Z12 /f → f (Z12 ) is given by f&macr;([[n]12 ]) =
f ([n]12 ) = [12n]18 . Hence f&macr;([12 ]) = 18 , f&macr;([12 ]) = 18 , f&macr;([12 ]) = 18 .
(b) Define the formula f : Z12 → Z12 by f ([x]12 ) = [x]212 , for all [x]12 ∈ Z12 . Show
that the formula f defines a function. Find the image f (Z12 ) of f and the factor
set Z12 /f of Z12 determined by f and exhibit the one-to-one correspondence between
them.
Solution: The formula for f is well-defined since if [x1 ]12 = [x2 ]12 , then x1 ≡
x2 (mod 12), and so x21 ≡ x22 (mod 12), which shows that f ([x1 ]12 ) = f ([x2 ]12 ).
To compute the image of f we have f (12 ) = 212 = 12 , f ([&plusmn;1]12 ) = [&plusmn;1]212 = 12 ,
f ([&plusmn;2]12 ) = [&plusmn;2]212 = 12 , f ([&plusmn;3]12 ) = [&plusmn;3]212 = 12 , f ([&plusmn;4]12 ) = [&plusmn;4]212 = 12 ,
f ([&plusmn;5]12 ) = [&plusmn;5]212 = 12 , and f ([&plusmn;6]12 ) = 212 = 12 . Thus the image of f is
f (Z12 ) = {12 , 12 , 12 , 12 }.
The equivalence classes in Z12 /f that correspond to these elements are, respectively,
{12 , 12 }, {[&plusmn;1]12 , [&plusmn;5]12 }, {[&plusmn;2]12 , [&plusmn;4]12 }, {[&plusmn;3]12 }.
15. For the linear transformation L : R3 → R3 defined by
L(x, y, z) = (x + y + z, x + y + z, x + y + z) ,
for all (x, y, z) ∈ R3 , give a geometric description of the partition of R3 that is
determined by L.
Solution: Since (a1 , a2 , a3 ) ∼L (b1 , b2 , b3 ) if L(a1 , a2 , a3 ) = L(b1 , b2 , b3 ), it follows from
the definition of L that (a1 , a2 , a3 ) ∼L (b1 , b2 , b3 ) if and only if a1 +a2 +a3 = b1 +b2 +b3 .
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For example, {(x, y, z) | L(x, y, z) = (0, 0, 0)} is the plane through the origin whose
equation is x+y+z = 0, with normal vector (1, 1, 1). The other subsets in the partition
of R3 defined by L are planes parallel to this one. Thus the partition consists of the
planes perpendicular to the vector (1, 1, 1).
16. For each of the following relations on the given set, determine which of the three
conditions of Definition 2.2.1 hold.
(a) For a, b ∈ Z, define a ∼ b if a + b is even.
Solution: We first check the reflexive law. Given a ∈ Z, it follows that a + a is even,
regardless of whether a is odd or even. Thus a ∼ a, and the reflexive law holds.
We next check the symmetric law. Suppose that a, b ∈ Z are given, with a ∼ b. Then
a + b is even, and so b + a is even, showing that b ∼ a. Thus the symmetric law holds.
Finally, we will check the transitive law. Let a, b, c ∈ Z, with a ∼ b and b ∼ c. Then
a + b and b + c are even, and since 2b is even, we also have that c − b = (b + c) − 2b
is even. The sum a + c = (a + b) + (c − b) is even, showing that a ∼ c, and so the
transitive law holds.
(b) For a, b ∈ Z, define a ∼ b if a + b is odd.
Solution: The reflexive law does not hold, since 0 + 0 is not odd, and thus 0 is not
equivalent to itself.
The symmetric law holds, since a + b = b + a.
The transitive law does not hold. For example: 0 ∼ 1 and 1 ∼ 2, but 0 ∼ 2 is false.
a
(c) On R&times; , define a ∼ b if ∈ Q.
b
a
a
&times;
Solution: Let a, b, c ∈ R . We have a ∼ a since = 1 ∈ Q. If a ∼ b, then ∈ Q,
a
b
b
a
b
and it follows that ∈ Q, so b ∼ a. If a ∼ b and b ∼ c, then ∈ Q and ∈ Q, so
a
b
c
a b
a
= &middot; ∈ Q, and thus a ∼ c. Therefore ∼ is an equivalence relation.
c
b c
a
(d) On R&times; , define a ∼ b if ∈ Z.
b
a
Solution: Let a, b, c ∈ R&times; . We have a ∼ a since = 1 ∈ Z. The reflexive law does
a
1
not hold, since 2 ∼ 1 but 1 ∼ 2 is false because 6∈ Z. The transitive law holds, as
2
shown in part (c), since the product of two integers is an integer.
17. On the set {(a, b)} of all ordered pairs of positive integers, define (x1 , y1 ) ∼ (x2 , y2 ) if
x1 y2 = x2 y1 . Show that this defines an equivalence relation.
Solution: We first show that the reflexive law holds. Given an ordered pair (a, b), we
have ab = ba, and so (a, b) ∼ (a, b).
We next check the symmetric law. Given (a1 , b1 ) and (a2 , b2 ) with (a1 , b1 ) ∼ (a2 , b2 ),
we have a1 b2 = a2 b1 , and so a2 b1 = a1 b2 , which shows that (a2 , b2 ) ∼ (a1 , b1 ).
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Finally, we verify the transitive law. Given (a1 , b1 ), (a2 , b2 ), and (a3 , b3 ) with (a1 , b1 ) ∼
(a2 , b2 ) and (a2 , b2 ) ∼ (a3 , b3 ), we have the equations a1 b2 = a2 b1 and a2 b3 = a3 b2 . If
we multiply the first equation by b3 and the second equation by b1 , we get a1 b2 b3 =
a2 b1 b3 = a3 b1 b2 . Since b2 6= 0 we can cancel to obtain a1 b3 = a3 b1 , showing that
(a1 , b1 ) ∼ (a3 , b3 ).
18. On the set R2 , define (x1 , y1 ) ∼ (x2 , y2 ) if x1 y1 = x2 y2 . Show that ∼ is an equivalence
relation, and give a geometric description of the equivalence classes of ∼.
Solution: For any (a, b) ∈ R2 , we have (a, b) ∼ (a, b) since ab = ab, which shows
that ∼ is reflexive. If (a, b) ∼ (c, d), then ab = cd, and so cd = ab, which implies
that (c, d) ∼ (a, b), and so ∼ is symmetric. If (a, b) ∼ (c, d) and (c, d) ∼ (e, f ), then
ab = cd and cd = ef . The transitive law for equality implies that ab = ef , and
therefore (a, b) ∼ (e, f ), so ∼ is transitive. We conclude that ∼ is an equivalence
relation.
Given any (a, b) ∈ R2 , the equivalence class of (a, b) is {(x, y) | xy = ab}. If a = 0
or b = 0, this set is the union of the x-axis and the y-axis. If ab 6= 0, then solving
ab
of a hyperbola. Thus the equivalence classes of ∼
for y yields the equation y =
x
k
consist of all hyperbolas of the form y = , where k can be any nonzero real number,
x
together with the “degenerate” hyperbola consisting of the two axes.
19. On C, define z1 ∼ z2 if |z1 | = |z2 |. Show that ∼ is an equivalence relation, and find
the equivalence classes of ∼.
Solution: The reflexive, symmetric, and transitive laws can be easily verified since ∼
is defined in terms of an equality, and equality is itself an equivalence relation. In the
complex plane, the equivalence classes are concentric circles, with center at the origin.
Alternate solution:
Define f : C → R by f (z) = |z|, for all z ∈ C. That is,
√
2
2
f (a + bi) = a + b , for all complex numbers a + bi. The given relation is ∼f , which
we know to be an equivalence relation.
20. Let u be a fixed vector in R3 , and assume that u has length 1. For vectors v and w,
define v ∼ w if v &middot; u = w &middot; u, where &middot; denotes the standard dot product. Show that ∼
is an equivalence relation, and give a geometric description of the equivalence classes
of ∼.
Solution: The reflexive, symmetric, and transitive laws for the relation ∼ really
depend on equality, and can easily be verified. Since u has length 1, the scalar v &middot; u
represents the length of the projection of v onto the line determined by u. Thus two
vectors are equivalent if and only if they lie in the same plane perpendicular to u. For
example, the plane through the origin perpendicular to u is {v | v &middot; u = 0}. It follows
that the equivalence classes of ∼ are the planes in R3 that are perpendicular to u.
21. Let f : S → T be a function. Given an equivalence relation ≃ on T , define ∼ on S
by setting x1 ∼ x2 if f (x1 ) ≃ f (x2 ), for all x1 , x2 ∈ S. Prove that ∼ is an equivalence
relation on S.
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Solution: Given a ∈ S, we have f (a) ≃ f (a), since ≃ is reflexive, and therefore a ∼ a.
If a ∼ b for a, b ∈ S, then f (a) ≃ f (b), and so f (b) ≃ f (a) since ≃ is symmetric. This
implies that b ∼ a.
If a ∼ b and b ∼ c for a, b, c ∈ S, then f (a) ≃ f (b) and f (b) ≃ f (c), which implies
that f (a) ≃ f (c), since ≃ is transitive. It follows that a ∼ c.
We have shown that ∼ is reflexive, symmetric, and transitive, so it is an equivalence
relation.
Comment: If the equivalence relation ≃ on T is ordinary equality, then the corresponding equivalence relation on S (as defined in the problem) is ∼f .
22. Let f : S → T be an onto function. Let {Pα }α∈I be a partition of T . Prove that
P = {f −1 (Pα ) | α ∈ I} is a partition of S.
Solution: Since f is an onto function, and each set Pα is nonempty, it follows that
each set f −1 (Pα ) is nonempty. For any element s ∈ S, the image f (s) belongs to
some set Pα , and then s ∈ f −1 (Pα ). Finally, for any indices α 6= β, it follows from
Problem 2.1.32 (c) that f −1 (Pα ) ∩ f −1 (Pβ ) = f −1 (Pα ∩ (Pβ ) = f −1 (∅) = ∅. This
shows that P is a partition of S.
23. Define f : Z8 → Z12 by f ([x]8 ) = [3x]12 , for all [x]8 ∈ Z8 .
(b) Find the image f (Z8 ) and the set of equivalence classes Z8 /f defined by f , and
exhibit the one-to-one correspondence between these sets.
Answer: The image of f is the subset {12 , 12 , 12 , 12 }. The corresponding
preimages, in order, are the sets {8 , 8 }, {8 , 8 }, {8 , 8 }, {8 , 8 }, and
these make up the factor set Z8 /f .
25. For [x]15 , let f ([x]15 ) = [3x]35 .
&times;
(b) Find f (Z&times;
15 ) and Z15 /f and exhibit the one-to-one correspondence between them.
&times;
15 ) = {5 , 5 , [−2]5 , [−1]5 } = Z5 , the corresponding preimages,
in order, are {15 , 15 }, {15 , [−4]15 }, {[−1]15 , 15 }, {[−2]15 , [−7]15 }, and these
make up the factor set Z&times;
15 /f .
26. For each of the following relations on R, determine which of the three conditions of
Definition 2.2.1 hold.
(a) Define a ∼ b if b = a2 .
Answer: None of the conditions hold.
(d) Define a ∼ b if b = a + qπ, for some q ∈ Q+ .
Answer: Only the transitive law holds.
27. For each of the following relations on the given set, determine which of the three
conditions of Definition 2.2.1 hold.
(a) For (x1 , y1 ), (x2 , y2 ) ∈ R2 , define (x1 , y1 ) ∼ (x2 , y2 ) if 2(y1 − x1 ) = 3(y2 − x2 ).