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Chapter 5: Relations and Functions
Exercises for Section 5.1. Relations
1. R = {(1, 5), (2, 5), (2, 6)}
2. (a) A × B = {(w, x), (w, y), (w, z), (x, x), (x, y), (x, z), (y, x), (y, y), (y, z)}
(b) R = {(w, x), (w, y), (w, z), (x, x)}
(c) R = {(w, x), (w, y), (x, x), (x, y), (y, x)}
3. See Figure 27.
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Figure 27: The relation R in Exercise 3
4. R = {(1, 3), (1, 5), (2, 3), (2, 4), (2, 5), (3, 2), (3, 5)}
5. R = {(1, 2), (2, 3), (2, 5), (4, 3), (5, 2)}.
6. Since 7, 11, 13, 17, 31 and 61 are primes, (7, 7), (1, 1), (1, 3), (1, 7), (3, 1) and (6, 1) are elements
of R. Because there exists no prime whose first digit is 2 and whose last digit is 4, it follows
that (2, 4) ∈
/ R. Similarly, (2, 6), (3, 2), (4, 4) and (1, 0) are not elements of R.
7. (a) symmetric,
(d) transitive
(b) transitive, (c) reflexive,
(e) reflexive, transitive.
8. The relation R possesses all three properties.
9. (a) 3 R 1 but 1 ̸ R 3,
(b) reflexive, transitive.
10. (a) The relation R1 is reflexive and symmetric, but not transitive. For example, |0 − 1| =
1 ≤ 1, |1 − 2| = 1 ≤ 1, but |0 − 2| > 1. Thus 0 R1 1 and 1 R1 2, but 0 ̸ R1 2.
(b) The relation R2 possess none of the properties reflexive, symmetric, transitive. For example, (1) −2 ̸ R2 − 2; (2) 2 R2 0, but 0 ̸ R2 2; and (3) 0 R2 1 and 1 R2 2 but 0 ̸ R2 2.
(c) The relation R3 possess none of the properties reflexive, symmetric,
transitive. For ex√
ample,
(1)
−1
̸
R
−
1;
(2)
−1R
1,
but
1
̸
R
−
1;
and
(3)
2
R
3
3
3
3 2 and 2 R3 4 but
√
2 ̸ R3 4.
(d) The relation R4 is symmetric; but it is not reflexive (0 ̸ R4 0, for example) and it is not
transitive (0 R4 3 and 3 R4 0, but 0 ̸ R4 0, for example).
11. (a) The relation R1 is reflexive and transitive but not symmetric. For example, {1} ⊆ {1, 2}
but {1, 2} ̸⊆ {1}.
(b) The relation R2 is symmetric, while it is neither reflexive ({1, 2} ∩{1, 2} ̸= ∅, for example)
nor transitive ({1} ∩ {2} = ∅ and {2} ∩ {1} = ∅ but {1} ∩ {1} =
̸ ∅, for example).
(c) The relation R3 is reflexive and transitive, but it is not symmetric. For example, 1 =
|{1}| ≤ |{1, 2}| = 2 but |{1, 2}| ̸≤ |{1}|.
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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}.