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(d) The relation R4 is symmetric but is neither reflexive nor transitive. For example, (1)
|{1, 2}∩{1, 2}| = 2 and (2) |{1, 2}∩{1}| = 1 and |{1}∩{1, 2}| = 1 but |{1, 2}∩{1, 2}| = 2.
Exercises for Section 5.2. Equivalence Relations
1. (a) Proof. Let (a, b) ∈ N × N. Since a + b = b + a, it follows that (a, b) R (a, b). Thus R is
reflexive. Next, assume that (a, b) R (c, d), where (a, b), (c, d) ∈ N × N. Then a + d = b + c
and so c + b = d + a. Thus (c, d) R (a, b) and R is symmetric. Finally, assume that
(a, b) R (c, d) and (c, d) R (e, f ), where (a, b), (c, d), (e, f ) ∈ N × N. Thus a + d = b + c
and c + f = d + e. Adding these two equations, we obtain
(a + d) + (c + f ) = (b + c) + (d + e).
Thus a + f = b + e and so (a, b) R (e, f ). Therefore, R is transitive.
(b) By definition,
[(3, 1)] =
=
=
=
{(x, y) ∈ N × N : (x, y) R (3, 1)}
{(x, y) ∈ N × N : x + 1 = y + 3}
{(x, y) ∈ N × N : y = x − 2}
{(3, 1), (4, 2), (5, 3), . . .}.
[(5, 5)] = {(x, y) ∈ N × N : y = x} = {(1, 1), (2, 2), (3, 3), . . .}.
[(4, 7)] = {(x, y) ∈ N × N : y = x + 3} = {(1, 4), (2, 5), (3, 6), . . .}.
2. There are four distinct equivalence classes: [1] = {1, 3, 4}, [2] = {2}, [5] = {5, 7} and [6] = {6}.
3. R = {(a, a), (a, d), (b, b), (b, f ), (c, c), (c, e), (d, a), (d, d), (e, c), (e, e), (f, b), (f, f )}.
4. Since u ∈ [x] ∩ [y], it follows that [x] ∩ [y] ̸= ∅ and so [x] = [y]. Because u ∈ [x], it also follows
that [u] = [x] = [y]. The fact that v ∈
/ [x] ∩ [y] implies that [v] ̸= [x]. Furthermore, z ∈ [v]
implies that [z] = [v]. Since w ̸∈ [u] ∪ [z], it follows that [w] ̸= [u] and [w] ̸= [z]. Therefore,
the distinct equivalence classes are
[u] = {u, x, y}, [v] = {v, z} and [w] = {w}.
Thus the relation R is
R
= {(u, u), (u, x), (u, y), (x, u), (x, x), (x, y), (y, u), (y, x),
(y, y), (v, v), (v, z), (z, v), (z, z), (w, w)}.
5. The relation R is
R =
{(1, 1), (1, 3), (1, 4), (1, 5), (2, 2), (3, 1), (3, 3), (3, 4), (3, 5),
(4, 1), (4, 3), (4, 4), (4, 5), (5, 1), (5, 3), (5, 4), (5, 5), (6, 6)}.
6. (a) The relation R is an equivalence relation. Proof. Let a ∈ R∗ . Then aa = a2 > 0 (since
a ̸= 0) and so a R a. Thus R is reflexive. Assume that a R b, where a, b ∈ R∗ . Then
ab > 0. Since ba = ab, it follows that ba > 0. Therefore, b R a and R is symmetric.
Finally, assume that a R b and b R c, where a, b, c ∈ R∗ . Hence ab > 0 and bc > 0. Thus
(ab)(bc) = ab2 c > 0. Dividing by the positive number b2 , we have ac > 0 and so R is
transitive.
(b) There are two distinct equivalence classes: [−1] = {x ∈ R∗ : (−1)x > 0} = {x ∈ R∗ : x <
0} and [1] = {x ∈ R∗ : x > 0}.
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7. (a) Proof. Let a ∈ Z. Then a − a = 0 and so a R a. Thus R is reflexive. Assume that a R b,
where a, b ∈ Z. Then a + b = 0 or a − b = 0. If a + b = 0, then b + a = a + b = 0; while if
a − b = 0, then b − a = −(a − b) = 0. Thus b + a = 0 or b − a = 0. Therefore, b R a and
R is symmetric.
We now assume that a R b and b R c, where a, b, c ∈ Z. Hence
(1) a + b = 0 or a − b = 0 and (2) b + c = 0 or b − c = 0.
We show that a + c = 0 or a − c = 0. We consider two cases.
Case 1. a + b = 0. Thus a = −b. Suppose first that b + c = 0. Then a − c = −b − c =
−(b + c) = 0; while if b − c = 0, then a + c = −b + c = −(b − c) = 0.
Case 2. a − b = 0. Thus a = b. If b + c = 0, then a + c = b + c = 0; while if b − c = 0,
then a − c = b − c = 0.
In each case, either a + c = 0 or a − c = 0 and so a R c. Therefore, R is transitive.
(b) For each a ∈ Z, [a] = {a, −a}. Thus [0] = {0}. [Also notice that a R b if a = b or a = −b.]
8. (a) Proof. Let a ∈ Z. Then a + 3a = 4a = 2(2a) is an even integer and so a R a. Thus R is
reflexive.
Assume that a R b, where a, b ∈ Z. Then a + 3b is even. Thus a + 3b = 2k for some
integer k and so a = 2k − 3b. Observe that
b + 3a = b + 3(2k − 3b) = b + 6k − 9b = 6k − 8b = 2(3k − 4b).
Since 3k − 4b ∈ Z, it follows that b + 3a is even. Therefore, b R a and R is symmetric.
Finally, assume that a R b and b R c, where a, b, c ∈ Z. Hence a + 3b and b + 3c are
both even and so a + 3b = 2x and b + 3c = 2y for some integers x and y. Adding these
two equations, we obtain
(a + 3b) + (b + 3c) = 2x + 2y,
which implies that
a + 3c = 2x + 2y − 4b = 2(x + y − 2b).
Since x + y − 2b is an integer, a + 3c is even. Therefore, a R c and R is transitive.
(b) There are two distinct equivalence classes:
[3] =
=
{x ∈ S : x R 3} = {x ∈ S : x + 9 is even}
{x ∈ S : x is odd} = {−7, −5, −1, 3, 9}.
Similarly, [4] = {x ∈ S : x is even} = {−4, 4}.
9. (a) Proof. Let a ∈ Z. Then 11a − 5a = 6a = 2(3a) is an even integer and so a R a. Thus R
is reflexive.
Assume that a R b, where a, b ∈ Z. Then 11a − 5b is even. Thus 11a − 5b = 2k for
some integer k. Observe that
11b − 5a = (11a − 5b) − 16a + 16b = 2k − 16a + 16b = 2(k − 8a + 8b).
Since k −8a+8b ∈ Z, it follows that 11b−5a is even. Therefore, b R a and R is symmetric.
Finally, assume that a R b and b R c, where a, b, c ∈ Z. Hence 11a − 5b and 11b − 5a
are both even and so 11a − 5b = 2x and 11b − 5c = 2y for some integers x and y. Adding
these two equations, we obtain
(11a − 5b) + (11b − 5c) = 2x + 2y,
which implies that
11a − 5c = 2x + 2y − 6b = 2(x + y − 3b).
Since x + y − 3b is an integer, 11a − 5c is even. Therefore, a R c and R is transitive.
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(b) There are two distinct equivalence classes:
[0] = {x ∈ Z : x is even} and [1] = {x ∈ Z : x is odd}.
10. (a) Proof. Let (a, b) ∈ Z × Z. Then a + b + a + b = 2a + 2b = 2(a + b). Since a + b is an
integer, a + b + a + b is even and so (a, b) R (a, b). Thus R is reflexive.
Next, assume that (a, b) R (c, d), where (a, b), (c, d) ∈ Z × Z. Then a + b + c + d is even.
Since a + b + c + d = c + d + a + b, it follows that c + d + a + b is even and so (c, d) R (a, b).
Hence R is symmetric.
Assume that (a, b) R (c, d) and (c, d) R (e, f ), where (a, b), (c, d), (e, f ) ∈ Z × Z. Thus
a + b + c + d is even and c + d + e + f is even. Hence
a + b + c + d = 2x and c + d + e + f = 2y
for some integers x and y. Adding these two equations, we obtain
(a + b + c + d) + (c + d + e + f ) = 2x + 2y
and so
a + b + e + f = 2x + 2y − 2(c + d) = 2(x + y − c − d).
Since x + y − c − d is an integer, a + b + e + f is even and so (a, b) R (e, f ). Therefore, R
is transitive.
(b) There are two distinct equivalence classes:
[(0, 0)] = {(x, y) ∈ Z × Z : (x, y) R (0, 0)}
= {(x, y) ∈ Z × Z : x + y is even}
= {(x, y) ∈ Z × Z : x and y are of the same parity}
Similarly, [(0, 1)] = {(x, y) ∈ Z × Z : x and y are of opposite parity }.
11. Observe that R is not reflexive since (1, 1) ̸ R (1, 1). Thus R is not an equivalence relation.
12. (b). For example, R = S × S (an equivalence relation) satisfies these conditions. On the other
hand, if z is related to y but not to z, then R is not an equivalence relation.
13. Yes. Define the relation R on S such that the resulting distinct equivalence classes are [a] =
{a, c, e}, [b] = {b, d} and [f ] = {f }. Then
R = {(a, a), (a, c), (a, e), (b, b), (b, d), (c, a), (c, c), (c, e), (d, b), (d, d), (f, f )}.
√
14. (a) First notice that if a = b, then a+a
a and aa = a. Suppose that a and b are two
2 =
√
√
√
positive real numbers such that a+b
ab. Hence a+b = 2 ab and so (a+b)2 = (2 ab)2 .
2 =
Therefore, a2 + 2ab + b2 = 4ab, which implies that a2 − 2ab + b2 = (a − b)2 = 0. Thus
a = b. Therefore, a R b if a = b. We now show that R is an equivalence relation.
Proof. Let x ∈ R+ . Since x = x, it follows that x R x and R is reflexive. Assume that
x R y, where x, y ∈ R+ . Then x = y and so y = x. Therefore, y R x and R is symmetric.
Assume that x R y and y R z, where x, y, z ∈ R+ . Then x = y and y = z. Thus x = z
and so x R z. Therefore, R is transitive.
(b) For each x ∈ R+ , [x] = {x}.
15. (a) Proof. Let x ∈ S. Since P is a partition of S, it follows that x ∈ Si for some i with
1 ≤ i ≤ k. Thus, x R x and R is reflexive.
Assume that x R y, where x, y ∈ S. Then x, y ∈ Si for some i with 1 ≤ i ≤ k. Thus
y, x ∈ Si for some i with 1 ≤ i ≤ k. Therefore, y R x and R is symmetric.
Finally, we show that R is transitive. Assume that x R y and y R z, where x, y, z ∈ S.
Thus there exist i, j with 1 ≤ i, j ≤ k such that x, y ∈ Si and y, z ∈ Sj . Since y ∈ Si and
y ∈ Sj , it follows that i = j because P is a partition of S. Thus x, z ∈ Si for some i with
1 ≤ i ≤ k. Hence x R z and R is transitive.
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(b) The distinct equivalence classes resulting from R are S1 , S2 , · · · , Sk .
16. Observe that
(1) tells us that there are three equivalence classes.
(2) tells us that no equivalence class has exactly three elements.
(3) tells us that 3 and 4 are in different equivalence classes.
(4) tells us that 1 and 7 are in the same equivalence classes.
Let P = {A1 , A2 , A3 } be a partition of A, where A1 = {1, 4, 6, 7}, A2 = {3, 5} and A3 = {2}.
Define a relation R on A by a R b if a, b ∈ Ai for some i with 1 ≤ i ≤ 3. Then R is an
equivalence relation and P is the set of equivalence classes that satisfies (1) – (4).
17. The statement is true. Proof. Assume, to the contrary, that there is an equivalence relation
on A with |R| = n + 1. Since (a, a) ∈ R for every vertex a ∈ A and |R| = n + 1, there must
exist x, y ∈ A with x ̸= y such that (x, y) ∈ R. Since R is symmetric, (y, x) ∈ R and so
|R| ≥ n + 2, which is a contradiction.
Exercises for Section 5.3. Functions
1. (a) f is a function from A to B.
(b) f is not a function from A to B, as |S| may be 1.
(c) f is not a function, as |S| may be 2 or 3.
(d) f is not a function, as the sum of elements in S may exceed 3.
(e) f is a function from A to B.
(f) f is not a function as S may consist of a single element.
2. f1 and f4 are functions from A to B, f2 is not a function from A to B as some element of A
has no image in B and f3 is not a function from A to B as some element of A has two images
in B.
3. (a) The domain of f is A, the codomain of f is B and the range of f is {x, z}.
(b) The image of d is x
(c) The element y is not an image.
(d) The set f (X) = {x, z}.
(e) One example: g = {(x, d), (y, c), (z, b)}.
4. f (x) = f (y).
5. Proof. We first show that f (A1 ∪ A2 ) ⊆ f (A1 ) ∪ f (A2 ). Let b ∈ f (A1 ∪ A2 ). Then there
exists a ∈ A1 ∪ A2 such that f (a) = b. Since a ∈ A1 ∪ A2 , it follows that a ∈ A1 or a ∈ A2 ,
say the former. Thus b = f (a) ∈ f (A1 ). Since f (A1 ) is a subset of f (A1 ) ∪ f (A2 ), we have
b ∈ f (A1 ) ∪ f (A2 ) and so f (A1 ∪ A2 ) ⊆ f (A1 ) ∪ f (A2 ).
Next, we show that f (A1 ) ∪ f (A2 ) ⊆ f (A1 ∪ A2 ). Let b ∈ f (A1 ) ∪ f (A2 ). Then b ∈ f (A1 )
or b ∈ f (A2 ), say the former. Thus there exists a ∈ A1 such that f (a) = b. Since a ∈ A1 , it
follows that a ∈ A1 ∪ A2 and so b = f (a) ∈ f (A1 ∪ A2 ). Hence f (A1 ) ∪ f (A2 ) ⊆ f (A1 ∪ A2 ).
Therefore, f (A1 ) ∪ f (A2 ) = f (A1 ∪ A2 ).
6. Proof. Let b ∈ f (A1 ∩ A2 ). Then there exists a ∈ A1 ∩ A2 such that f (a) = b. Since
a ∈ A1 ∩ A2 , it follows that a ∈ A1 and a ∈ A2 . Thus b = f (a) ∈ f (A1 ) and b = f (a) ∈ f (A2 ),
implying that b ∈ f (A1 ) ∩ f (A2 ). Therefore, f (A1 ∩ A2 ) ⊆ f (A1 ) ∩ f (A2 ).