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@ -603,7 +603,7 @@ d. Determine whether the relation is transitive.
No, $1 R_1 0$ and $0 R_1 3$, but $1 \cancel{R_1} 3$
2. $R_2$ = \{(0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 3)\}
2. $R_2 = \{(0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 3)\}$
a. Draw the directed graph.
@ -1493,25 +1493,173 @@ statement.
34. If $R$ is reflexive, then $R^{-1}$ is reflexive.
**Proof:**
Suppose $R$ is any relation on a set $A$, such that $R$ is reflexive.
By the definition of reflexive, this means that $\forall x \in A, (x, x) \in R$,
or $\forall x \in A, x R x$. Then, by definition of an inverse relation, it
follows that $(x, x) \in R^{-1}$, or $x R^{-1} x$.
Therefore, $R^{-1}$ is reflexive.
Q.E.D.
35. If $R$ is symmetric, then $R^{-1}$ is symmetric.
**Proof:**
Suppose $R$ is any relation on a set $A$, such that $R$ is symmetric.
By the definition of symmetric, this means that
$\forall (x, y) \in A, (x, y) \in R \to (y, x) \in R$. Since $(y, x) \in R$, it
follows, by definition of inverse relation, that $(x, y) \in R^{-1}$.
Furthermore, since $(x, y) \in R$, it follows that $(y, x) \in R^{-1}$.
Therefore $R^{-1}$ is symmetric.
Q.E.D.
36. If $R$ is transitive, then $R^{-1}$ is transitive.
**Proof:**
Suppose $R$ is any relation on a set $A$ such that $R$ is transitive.
By the definition of transitive, this means that
$\forall x, y, z \in A, [(x, y) \in R \wedge (y, z) \in R] \to (x, z) \in R$.
Since $(x, y), (y, z), (x, z) \in R$, it follows by the definition of inverse
that $(y, x), (z, y), (z, x) \in R^{-1}$. This means that
$\forall x, y, z \in A, [(z, y) \in R^{-1} \wedge (y, x) \in R^{-1}] \to (z, x) \in R^{-1}$.
Therefore $R^{-1}$ is transitive.
Q.E.D.
In 37-42, assume that $R$ and $S$ are relations on a set $A$. Prove or disprove
each statement.
37. If $R$ and $S$ are reflexive, is $R \cap S$ reflexive? Why?
$R \cap S$ is reflexive.
**Proof:**
Suppose $R$ and $S$ are any relations on some set $A$ such that $R$ and $S$ are
reflexive.
By the definition of reflexive, this means that $\forall x \in A, (x, x) \in R$,
and $\forall x \in A, (x, x) \in S$.
Since $(x, x) \in R$ and $(x, x) \in S$, it follows (by the definition of
intersection), that $(x, x) \in R \cap S$.
Therefore $R \cap S$ is reflexive.
Q.E.D.
38. If $R$ and $S$ are symmetric, is $R \cap S$ symmetric? Why?
$R \cap S$ is symmetric.
**Proof:**
Suppose $R$ and $S$ are any relations on a set $A$ such that $R$ and $S$ are
symmetric.
By the definition of symmetric, this means that
$\forall x, y \in A, (x, y) \in R \to (y, x) \in R$. Similarly,
$\forall x, y \in A, (x, y) \in S \to (y, x) \in S$.
Since $(x, y) \in R$, $(y, x) \in R$, $(x, y) \in S$, $(y, x) \in S$, it follows
by the definition of intersection that $(x, y) \in R \cap S$ and
$(y, x) \in R \cap S$.
Therefore $R \cap S$ is symmetric.
Q.E.D.
39. If $R$ and $S$ are transitive, is $R \cap S$ transitive? Why?
$R \cap S$ is transitive.
**Proof:**
Suppose $R$ and $S$ are any relations on a set $A$ such that $R$ and $S$ are
transitive.
By the definition of transitive, this means that
$\forall x, y, z \in A, [(x, y) \in R \wedge (y, z) \in R] \to (x, z) \in R$.
Similarly,
$\forall x, y, z \in A, [(x, y) \in S \wedge (y, z) \in S] \to (x, z) \in S$.
Since $(x, y), (y, z), (x, z) \in R$ and $(x, y), (y, z), (x, z) \in S$, it
follows by the definition of intersection that
$(x, y), (y, z), (x, z) \in (R \cap S)$. Furthermore, this means that
$\forall x, y, z \in A, [(x, y) \in (R \cap S) \wedge (y, z) \in (R \cap S)] \to (x, z) \in (R \cap S)$.
Therefore, by the definition of transitive, $R \cap S$ is transitive.
Q.E.D.
40. If $R$ and $S$ are reflexive, is $R \cup S$ reflexive? Why?
$R \cup S$ is reflexive.
**Proof:**
Suppose $R$ and $S$ are any relations on some set $A$ such that $R$ and $S$ are
reflexive.
By the definition of reflexive, this means that $\forall x \in A, (x, x) \in R$,
and $\forall x \in A, (x, x) \in S$.
Since $(x, x) \in R$ and $(x, x) \in S$, it follows (by the definition of
union), that $(x, x) \in R \cup S$ (since in order to satisfy the definition of
union, $(x, x) \in R$ _or_ $(x, x) \in S$).
Therefore $R \cup S$ is reflexive.
Q.E.D.
41. If $R$ and $S$ are symmetric, is $R \cup S$ symmetric? Why?
$R \cup S$ is symmetric.
**Proof:**
Suppose $R$ and $S$ are any relations on a set $A$ such that $R$ and $S$ are
symmetric.
By the definition of symmetric, this means that
$\forall x, y \in A, (x, y) \in R \to (y, x) \in R$. Similarly,
$\forall x, y \in A, (x, y) \in S \to (y, x) \in S$.
Since $(x, y) \in R$, $(y, x) \in R$, $(x, y) \in S$, $(y, x) \in S$, it follows
by the definition of union that $(x, y) \in R \cup S$ and $(y, x) \in R \cup S$
(since in order to satisfy the definition of union, $(x, y) \in R$ and
$(y, x) \in R$ _or_ $(x, y ) \in S$ and $(y, x) \in S$).
Therefore $R \cup S$ is symmetric.
Q.E.D.
42. If $R$ and $S$ are transitive, is $R \cup S$ transitive? Why?
**Disproof (by counterexample):**
Let $A = \{a, b, c, d\}$, $R = {(a, b), (b, c), (a, c)}$, and
$S = \{(b, c), (c, d), (b, d)\}$. Note that $R$ and $S$ are transitive. However,
when we take the union, $R \cup S$:
$$ (R \cup S) = \{(a, b), (b, c), (a, c), (c, d), (b, d)\} $$
Note that $(a, b), (b, d) \in (R \cup S)$, but $(a, d) \notin (R \cup S)$. By
the definition of transitive, it follows that $R \cup S$ is not transitive.
Q.E.D.
In 43-50, the following definitions are used: A relation on a set $A$ is defined
to be
@ -1528,45 +1676,284 @@ relation is irreflexive, asymmetric, intransitive, or none of these.
43. Exercise 1
$R_1 = \{(0, 0), (0, 1), (0, 3), (1, 1), (1, 0), (2, 3), (3, 3)\}$
a. Irreflexive?:
No, since $0 R_1 0$, $R_1$ is not irreflexive.
b. Asymmetric?:
No, since $(0, 1) \in R_1$ and $(1, 0) \in R_1$, $R_1$ is not asymmetric.
c. Intransitive?:
No, since $(0, 1), (1, 0), (0, 0) \in R_1$, $R_1$ is not intransitive.
44. Exercise 2
$R_2 = \{(0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 3)\}$
a. Irreflexive?:
No, since $0 R_2 0$, $R_2$ is not irreflexive.
b. Asymmetric?:
No, since $(0, 0) \in R_2$ and $(0, 0) \in R_2$, $R_2$ is not asymmetric.
c. Intransitive?:
No, since $(1, 1), (1, 2), (2, 2) \in R_2$, $R_2$ is not intransitive.
45. Exercise 3
$R_3 = \{(2, 3), (3, 2)\}$
a. Irreflexive?:
Yes, $R_3$ is irreflexive.
b. Asymmetric?:
No, since $(2, 3), (3, 2) \in R_3$, $R_3$ is not asymmetric.
c. Intransitive?:
Yes, since $(2, 3), (3, 2) \in R_3$, but $(2, 2) \notin R_3$, $R_3$ is
intransitive.
46. Exercise 4
$R_4 = \{(1, 2), (2, 1), (1, 3), (3, 1)\}$
a. Irreflexive?:
Yes, $R_4$ is irreflexive.
b. Asymmetric?:
No, since $(1, 2), (2, 1) \in R_4$, $R_4$ is not asymmetric.
c. Intransitive?:
Yes, $R_4$ is intransitive.
$$ (1, 2), (2, 1) \in R_4, \text{ but } (1, 1) \notin R_4 $$
$$ (2, 1), (1, 3) \in R_4, \text{ but } (2, 3) \notin R_4 $$
$$ (1, 3), (3, 1) \in R_4 , \text{ but } (1, 1) \notin R_4 $$
etc. (note that a more rigorous proof would check all examples.)
47. Exercise 5
$R_5 = \{(0, 0), (0, 1), (0, 2), (1, 2)\}$
a. Irreflexive?:
No, since $(0, 0) \in R_5$
b. Asymmetric?:
No, since $(0, 0) \in R_5$ and $(0, 0) \in R_5$.
c. Intransitive?:
No, since $(0, 1), (1, 2), (0, 2) \in R_5$.
48. Exercise 6
$R_6 = \{(0, 1), (0, 2)\}$
a. Irreflexive?:
Yes.
b. Asymmetric?:
Yes.
c. Intransitive?:
Yes, since there is no $(1, x)$ for some element $x$, nor is there $(2, y)$ for
some element $y$, the supposition is always false, and is therefore the if/then
proposition is vacuously true.
49. Exercise 7
$R_7 = \{(0, 3), (2, 3)\}$
a. Irreflexive?:
Yes.
b. Asymmetric?:
Yes.
c. Intransitive?:
Yes (see Exercise 48 for vacuous truth explanation, which applies here as well.)
50. Exercise 8
$R_8 = \{(0, 0), (1, 1)\}$
a. Irreflexive?:
No, since $(0, 0) \in R_8$.
b. Asymmetric?:
No, since $(0, 0) \in R_8$.
c. Intransitive?:
No, since $(0, 0), (0, 0), (0, 0) \in R_8$.
In 51-53, $R$, $S$, and $T$ are relations defined on $A = \{0, 1, 2, 3\}$.
51. Let $R = \{(0, 1), (0, 2), (1, 1), (1, 3), (2, 2), (3, 0)\}$.
Find $R^t$, the transitive closure of $R$.
First, by definition of the transitive closure, $R \subseteq R^t$, so (building
$R^t$, _i.e._ not finished):
$$ R^t = \{(0, 1), (0, 2) (1, 1), (1, 3), (2, 2), (3, 0)\} $$
Since $R^t$ must be transitive, every ordered pair triple must have a transitive
"third":
$$ (0, 1), (1, 1) \to (0, 1) $$
$$ (0, 1), (1, 3) \to (0, 3) $$
$$ (0, 2), (2, 2) \to (0, 2) $$
$$ (1, 1), (1, 3) \to (1, 3) $$
$$ (1, 3), (3, 0) \to (1, 0) $$
$$ (3, 0), (0, 1) \to (3, 1) $$
$$ (3, 0), (0, 2) \to (3, 2) $$
Now, add all missing ordered pairs to $R^t$:
$$ R^t = \{(0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 3), (2, 2), (3, 0), (3, 1), (3, 2)\} $$
Now, check to be sure all ordered triples yields:
$$ (3, 1), (1, 3) \to (3, 3) $$
$$ \boxed{R^t = \{(0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 3), (2, 2), (3, 0), (3, 1), (3, 2), (3, 3)\}} $$
52. Let $S = \{(0, 0), (0, 3), (1, 0), (1, 2), (2, 0), (3, 2)\}$.
Find $S^t$, the transitive closure of $S$.
$$ S^t = \{(0, 0), (0, 3), (1, 0), (1, 2), (2, 0), (3, 2)\} $$
Then:
$$ (0, 0), (0, 3) \to (0, 3) $$
$$ (0, 3), (3, 2) \to (0, 2) $$
$$ (1, 0), (0, 0) \to (1, 0) $$
$$ (1, 0), (0, 3) \to (1, 3) $$
$$ (1, 2), (2, 0) \to (1, 0) $$
$$ (2, 0), (0, 0) \to (2, 0) $$
$$ (2, 0), (0, 3) \to (2, 3) $$
$$ (3, 2), (2, 0) \to (3, 0) $$
New:
$$ S^t = \{(0, 0), (0, 2), (0, 3), (1, 0), (1, 2), (1, 3), (2, 0), (2, 3), (3, 0), (3, 2)\} $$
Check again:
$$ (2, 0), (0, 2) \to (2, 2) $$
$$ (3, 0), (0, 3) \to (3, 3) $$
Finally:
$$ S^t = \{(0, 0), (0, 2), (0, 3), (1, 0), (1, 2), (1, 3), (2, 0), (2, 2), (2, 3), (3, 0), (3, 2), (3, 3)\} $$
53. Let $T = \{(0, 2), (1, 0), (2, 3), (3, 1)\}$.
Find $T^t$, the transitive closure of $T$.
$$ $T^t = \{(0, 2), (1, 0), (2, 3), (3, 1)\}. $$
$$ (0, 2), (2, 3) \to (0, 3) $$
$$ (1, 0), (0, 2) \to (1, 2) $$
$$ (2, 3), (3, 1) \to (2, 1) $$
$$ (3, 1), (1, 0) \to (3, 0) $$
Now:
$$ $T^t = \{(0, 2), (0, 3), (1, 0), (1, 2), (2, 1), (2, 3), (3, 0), (3, 1)\}. $$
Furthermore:
$$ (0, 2), (2, 1) \to (0, 1) $$
$$ (0, 3), (3, 0) \to (0, 0) $$
$$ (1, 0), (0, 3) \to (1, 3) $$
$$ (1, 2), (2, 1) \to (1, 1) $$
$$ (2, 3), (3, 0) \to (2, 0) $$
Now:
$$ $T^t = \{(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3), (2,
0), (2, 1), (2, 3), (3, 0), (3, 1)\}. $$
And:
$$ (2, 1), (1, 2) \to (2, 2) $$
$$ (3, 1), (1, 2) \to (3, 2) $$
$$ (3, 1), (1, 3) \to (3, 3) $$
So:
$$ $T^t = \{(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3), (2,
0), (2, 1), (2, 2), (2, 3), (3, 0), (3, 1), (3, 2), (3, 3)\}. $$
54. Write a computer algorithm to test whether a relation $R$ defined on a
finite set $A$ is reflexive, where
$$ A = \{a[1], a[2], \dots, a[n]\} $$
Omitted.
55. Write a computer algorithm to test whether a relation $R$ defined on a
finite set $A$ is symmetric, where
$$ A = \{a[1], a[2], \dots, a[n]\} $$
Omitted.
56. Write a computer algorithm to test whether a relation $R$ defined on a
finite set $A$ is transitive, where
$$ A = \{a[1], a[2], \dots, a[n]\} $$
Omitted.