Math 1120b –
March 26, 2015
Homework 7 – solutions
18:1,2(i) Solution: S = R and a ∼ b ⇔ a = b or − b is an equivalence relation.
• Clearly a ∼ a for all a ∈ R, so ∼ is reflexive.
• If a ∼ b then either a = b or a = −b. In the first case, b = a, so b ∼ a. In
the second case, b = −a, so b ∼ a again, which proves ∼ is symmetric.
• If a ∼ b and b ∼ c, then a = b and b = µc, where , µ ∈ {1, −1}. Substituting, a = µ · c. Since the product µ = ±1 as well, a ∼ c, so ∼ is
transitive.
Now we determine the equivalence classes. Since a ∼ b ⇔ b = a or b = −a, we
see cl(a) = seta, −a, for each a ∈ R.
18:1,2(ii) Solution: This is not an equivalence relation, since it fails to be reflexive. For
example, 1 6∼ 1, because 1 · 1 6= 0.
18:1,2(iii) Solution: This is an equivalence relation.
• Since a2 + a = a2 + a, we see a ∼ a for all a ∈ R, so ∼ is reflexive.
• If a ∼ b then
a2 + a = b2 + b
⇐⇒ b2 + b = a2 + a
⇐⇒ b ∼ a,
so ∼ is symmetric.
• If a ∼ b and b ∼ c, then
a2 + a = b2 + b
= c2 + c
so a ∼ c as well, which shows ∼ is transitive.
To determine the equivalence classes, we notice that a ∼ b if and only if
a2 + a = b2 + b
⇐⇒
⇐⇒
⇐⇒
⇐⇒
a2 − b2 = b − a
(a − b)(a + b) = −(a − b)
a + b = −1
provided a 6= b;
b = −a − 1.
That is, a ∼ b if and only if b = a or b = −a − 1, so
cl(a) = {a, −a − 1}
for all a ∈ R.
18:1,2(iv) Solution: We can see this is not an equivalence relation by finding three people
a, b, and c for which b lives close to the middle of a line joining a and c, and the
distance from a to c is slightly greater than 100 miles, so that “∼” is seen not to
be transitive.
1
18:1,2(v, viii) Solution: First, these are two different descriptions of the same relation. Both
equivalence relations. To prove this and compute the equivalence classes, we
should reproduce the very similar example from class, where we showed that
z ∼ w ⇔ |z| = |w|
was an equivalence relation on C whose equivalence classes are circles centered
at the origin.
18:1,2(vi) Solution: This is an equivalence relation. Reflexive and symmetric are easy. To
see it’s transitive, suppose a ∼ b and b ∼ c. By definition, ab = k 2 and bc = l2
for some positive integers k, l.
Then
ab2 c = k 2 l2 , so
2
kl
, since b > 0.
ac =
b
If kl/b is not an integer, then neither is its square. But ac is an integer, so kl/b
must be an integer, which implies a ∼ c.
To describe the equivalence classes, we should look at prime factorizations. If
a, b are positive integers, for some k ≥ 0, there exist prime numbers p1 , . . . , pk for
mk
1
and b = pn1 1 · · · pnk k , were the exponents m1 , . . . , mk and
which a = pm
1 · · · pk
n1 , . . . , nk are nonnegative.
1 +n1
Now ab = pm
· · · pkmk +nk is a perfect square if and only the exponents
1
mi + ni are even, for all 1 ≤ i ≤ k.
So a ∼ b if and only if mi ≡ ni mod 2, for 1 ≤ i ≤ k. In order to describe the
classes nicely, if a is a natural number, if pi is the highest power of p that divides
a, let νp (a) = i. (For all but finitely many primes p, of course νp (a) = 0.)
Then, for all natural numbers a, we can say
cl(a) = {b : νp (a) ≡ νp (b)
mod 2 for all primes p}.
18:4 Solution: Suppose S is a set and S1 , S2 , . . . , Sk is a partition of S. We need to
show two things: that there is an equivalence relation ∼ for which these sets
are the equivalence classes, and that ∼ is the only equivalence relation with this
property.
First, we construct an equivalence relation. For s, t ∈ S, let s ∼ t if and only if
there exists some integer i for which s, t ∈ Si . It is easy to see that ∼ is reflexive,
symmetric and transitive, so it is indeed an equivalence relation.
We compute its equivalence classes. If s ∈ S, then s ∈ Si for a unique integer
i satisfying 1 ≤ i ≤ k, because we assumed S1 , §2 , . . . , Sk were a partition of S.
Then
cl(s) = {t ∈ S : s ∼ t}
= Si
by definition of “∼”. So each equivalence class is one of the sets Si .
Because the sets form a partition, each set Si is nonempty, so there exists some
element si ∈ Si for 1 ≤ i ≤ k. Since cl(si ) = Si , each set Si is an equivalence
class of ∼.
This shows that the equivalence classes of ∼ are the sets S1 , S2 , . . . , Sk , which
is what we wanted to prove.
Now we prove the second claim, that ∼ is the only equivalence relation with this
property. For this, suppose “'” is also an equivalence relation with equivalence
classes S1 , S2 , . . . , Sk .
For all s, t ∈ S, by definition, s ∼ t if and only if s, t ∈ Si for some integer
1 ≤ i ≤ k. By assumption, Si is an equivalence class of ', so s, t ∈ Si if and only
if s ' t. So s ∼ t ⇔ s ' t, which is to say ∼ and ' are the same relation.
18:6 Solution: Suppose S = {1, 2, 3, 4} and ∼ is an equivalence relation on S with
1 ∼ 2 and 2 ∼ 3.
By transitivity, cl(1) ⊇ {1, 2, 3}. This means that either cl(1) = {1, 2, 3}, or
cl(1) = S = {1, 2, 3, 4}. In the second case, ∼ has only one equivalence class: i.e.,
all elements of S are related.
In the first case, we see 4 6∼ i for 1 ≤ i ≤ 3, so cl(4) = {4}, and the equivalence
relation has two equivalence classes, S1 = {1, 2, 3} and S2 = {4}.
By problem 4, knowing the equivalence classes uniquely determines the equivalence relation, so we have found the only two possibilities for ∼.