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@ -3192,8 +3192,111 @@ $$ m D n \Leftrightarrow 3 | (m^2 - n^2) $$
(1) Prove that the relation is an equivalence relation.
**Proof:**
Suppose $D$ is a relation on $\mathbb{Z}$ defined as follows:
$$ \forall m, n \in \mathbb{Z}, m D n \Leftrightarrow 3 | (m^2 - n^2) $$
To prove that $D$ is an equivalence relation, it must be shown that $D$ is
reflexive, symmetric, and transitive.
_Proof ($D$ is reflexive):_
Let $x \in \mathbb{Z}$.
To prove that $D$ is reflexive, it must be shown that $(x, x) \in D$. By the
definition for $D$, this means it must be shown that:
$$ 3 | (x^2 - x^2) $$
Now, $x^2 - x^2 = 0$, and it is true that $3 | 0$, since $0 = 3 \cdot 0$. Thus
$(x, x) \in D$, and it can be concluded that $D$ is reflexive.
_Proof ($D$ is symmetric):_
Let $x, y \in \mathbb{Z}$.
To prove that $D$ is symmetric, it must be shown that
$(x, y) \in D \to (y, x) \in D$. By definition of $D$, this means it must be
shown that:
$$ [3 | (x^2 - y^2)] \to [3 | (y^2 - x^2)] $$
Suppose $3 | (x^2 - y^2)$. By the definition of divisibility, this means that:
$$ x^2 - y^2 = 3k $$
for some integer $k$.
Now, consider that:
$$ y^2 - x^2 = -1(x^2 - y^2) $$
Then, by substitution:
$$ = -1(3k) $$
$$ = 3(-k) $$
Now, $-k$ is an integer (by the product of integers), thus $3 | (y^2 - x^2)$,
and hence $(y, x) \in D$, and therefore $D$ is symmetric.
_Proof ($D$ is transitive):_
Let $x, y, z \in \mathbb{Z}$.
To prove that $D$ is transitive, it must be shown that
$[(x, y) \in D \wedge (y, z) \in D] \to [(x, z) \in D]$.
Suppose $(x, y) \in D$ and $(y, z) \in D$. Then, by the definition for $D$, this
means:
$$ 3 | (x^2 - y^2) $$
and also:
$$ 3 | (y^2 - z^2) $$
(It must be shown that $3 | (x^2 - z^2)$.)
By the definition of divisibility, this means that:
$$ x^2 - y^2 = 3k $$
and also that:
$$ y^2 - z^2 = 3p $$
for some integers $k$ and $p$.
Now, if one adds $x^2 - y^2$ and $y^2 - z^2$, this yields:
$$ x^2 - y^2 + y^2 - z^2 = x^2 - z^2 $$
Then, by substitution:
$$ x^2 - z^2 = (3k) + (3p) $$
$$ = 3(k + p) $$
Now, $k + p$ is an integer (by the sum of integers). Thus $3 | (x^2 - z^2)$ (by
the definition of divisibility). It follows that $(x, z) \in D$, and therefore
$D$ is transitive.
_Conclusion:_
Since $D$ has been shown to be reflexive, symmetric, and transitive, it follows
that $D$ is an equivalence relation. This is what was to be shown.
Q.E.D.
(2) Describe the distinct equivalence classes of each relation.
There are two distinct equivalence classes:
$$ [0] = \{\dots, -6, -3, 0, 3, 6, \dots\}, [1] = \{\dots, -5, -4, -2, -1, 1, 2, 4, 5, \dots\} $$
27. $R$ is the relation defined on $\mathbb{Z}$ as follows: For every
$(m, n) \in \mathbb{Z}$,
@ -3201,16 +3304,195 @@ $$ m R n \Leftrightarrow 4 | (m^2 - n^2) $$
(1) Prove that the relation is an equivalence relation.
**Proof:**
Suppose $R$ is a relation defined on $\mathbb{Z}$ as follows:
$$ \forall m, n \in \mathbb{Z}, m R n \Leftrightarrow 4 | (m^2 - n^2) $$
To prove that $R$ is an equivalence relation, it must be shown that $R$ is
reflexive, symmetric, and transitive.
_Proof ($R$ is reflexive):_
Let $x \in \mathbb{Z}$.
To prove that $R$ is reflexive, it must be shown that $(x, x) \in R$. By the
definition for $R$, this means it must be shown that:
$$ 4 | (x^2 - x^2) $$
Since $x^2 - x^2 = 0$, this means it must be shown that $4 | 0$. Now, $4 | 0$
because $0 = 4 \cdot 0$. Therefore $(x, x) \in R$, and it can be concluded that
$R$ is reflexive.
_Proof ($R$ is symmetric):_
Let $x, y \in \mathbb{Z}$.
To prove that $R$ is symmetric, it must be shown that
$(x, y) \in R \to (y, x) \in R$.
Suppose $(x, y) \in R$, then, by definition for $R$, this means:
$$ 4 | (x^2 - y^2) $$
By the definition of divisibility, this means that:
$$ x^2 - y^2 = 4k $$
for some integer $k$.
Now, consider that:
$$ y^2 - x^2 = -1(x^2 - y^2) $$
Then, by substitution:
$$ y^2 - x^2 = -1(4k) $$
$$ = 4(-k) $$
Now, $-k$ is an integer (by the product of integers). Hence $4 | (y^2 - x^2)$,
and it follows that $(y, x) \in R$, and therefore $R$ is symmetric.
_Proof ($R$ is transitive):_
Let $x, y, z \in \mathbb{Z}$.
To prove that $R$ is transitive, it must be shown that
$[(x, y) \in R \wedge (y, z) \in R] \to (x, z) \in R$.
Suppose $(x, y) \in R$ and $(y, z) \in R$. By the definition for $R$, this means
that:
$$ 4 | (x^2 - y^2) $$
and also that:
$$ 4 | (y^2 - z^2) $$
Now, by the definition for divisibility, this means that:
$$ x^2 - y^2 = 4k $$
and also that:
$$ y^2 - z^2 = 4p $$
for some integers $k$ and $p$.
Now, consider that:
$$ x^2 - z^2 = x^2 - y^2 + y^2 - z^2 $$
Then, by substitution:
$$ x^2 - z^2 = 4k + 4p $$
$$ x^2 - z^2 = 4(k + p) $$
Now, $k + p$ is an integer (by the sum of integers), and so it follows that
$4 | (x^2 - z^2)$. This means that $(x, z) \in R$, and therefore $R$ is
transitive.
_Conclusion:_
Since it has been shown that $R$ is reflexive, symmetric, and transitive, it can
be concluded that $R$ is an equivalence relation. This is what was to be shown.
Q.E.D.
(2) Describe the distinct equivalence classes of each relation.
There are two distinct equivalence classes:
$$ [0] = \{\dots, -8, -4, -2, 0, 2, 4, 8, \dots\} = \text{ the set of all even integers } $$
$$ [1] = \{\dots, -9, -5, -1, 1, 5, 9\dots\} = \text{ the set of all odd integers } $$
28. $I$ is the relation defined on $\mathbb{R}$ as follows:
$$ \text{For every } x, y \in \mathbb{R}, m I n \Leftrightarrow x - y \text{ is an integer} $$
(1) Prove that the relation is an equivalence relation.
**Proof:**
Suppose $I$ is a relation defined on $\mathbb{R}$ as follows:
$$ \forall x, y \in \mathbb{R}, m I n \Leftrightarrow (x - y) \in \mathbb{Z} $$
To prove that $I$ is an equivalence relation, it must be shown that $I$ is
reflexive, symmetric, and transitive.
_Proof ($I$ is reflexive):_
Let $x \in \mathbb{R}$.
To prove that $I$ is reflexive, it must be shown that $(x, x) \in I$. By the
definition for $I$, this means it must be shown that:
$$ (x - x) \in \mathbb{Z} $$
Now, $x - x = 0$, and $0 \in \mathbb{Z}$. Thus $(x, x) \in I$, and therefore $I$
is reflexive.
_Proof ($I$ is symmetric):_
Let $x, y \in \mathbb{R}$.
To prove that $I$ is symmetric, it must be shown that
$(x, y) \in I \to (y, x) \in I$.
Suppose $(x, y) \in I$, by the definition for $I$, this means that:
$$ (x - y) \in \mathbb{Z} $$
Now, consider:
$$ y - x = -1(x - y) $$
Now, $-1(x - y)$ is an integer (by the product of integers), and thus
$(y - x) \in \mathbb{Z}$. Thus $(y, x) \in I$, and therefore $I$ is symmetric.
_Proof ($I$ is transitive):_
Let $x, y, z \in \mathbb{R}$.
To prove that $I$ is transitive, it must be shown that
$[(x, y) \in I \wedge (y, z) \in I] \to (x, z) \in I$.
Suppose $(x, y) \in I$ and $(y, z) \in I$. By the definition for $I$, this means
that:
$$ (x - y) \in \mathbb{Z} $$
and also that:
$$ (y - z) \in \mathbb{Z} $$
Now, consider that:
$$ x - z = (x - y) + (y - z) $$
Thus, $(x - z) \in \mathbb{Z}$ (by the sum of integers). It follows that
$(x, z) \in I$, and therefore $I$ is transitive.
_Conclusion:_
Since it has been shown that $I$ is reflexive, symmetric, and transitive, it can
be concluded that $I$ is an equivalence relation. This is what was to be shown.
Q.E.D.
(2) Describe the distinct equivalence classes of each relation.
There is one class for each real number $x$ with $0 \leq x < 1$. The distinct
classes are all sets of the form
$[x] = y \in \mathbb{R}, | y = n + x \text{ for some integer } n$, where $x$ is
a real number such that $0 \leq x < 1$.
29. Define $P$ on the set $\mathbb{R} \times \mathbb{R}$ of ordered pairs of
real numbers as follows: For every
$(w, x), (y, z) \in \mathbb{R} \times \mathbb{R}$,
@ -3219,8 +3501,73 @@ $$ (w, x) P (y, z) \Leftrightarrow w = y $$
(1) Prove that the relation is an equivalence relation.
**Proof:**
Suppose $P$ is a relation on $\mathbb{R} \times \mathbb{R}$, defined as:
$$ \forall (w, x), (y, z) \in \mathbb{R} \times \mathbb{R}, (w, x) P (y, z) \Leftrightarrow w = y $$
To prove that $P$ is an equivalence relation, it must be shown that $P$ is
reflexive, symmetric, and transitive.
_Proof ($P$ is reflexive):_
Let $(w, x) \in \mathbb{R} \times \mathbb{R}$.
To prove that $P$ is reflexive, it must be shown that $[(w, x), (w, x)] \in P$.
By the definition for $P$, this means it must be shown that:
$$ w = w $$
This is trivially true. Thus $[(w, x), (w, x)] \in P$, and therefore $P$ is
reflexive.
_Proof ($P$ is symmetric):_
Let $(w, x), (y, z) \in \mathbb{R} \times \mathbb{R}$.
TO prove that $P$ is symmetric, it must be shown that
$[(w, x), (y, z)] \in P \to [(y, z), (w, x)] \in P$.
Suppose $[(w, x), (y, z)] \in P$. By definition for $P$, this means that:
$$ w = y $$
This means that $y = w$, by the symmetric property of equality. This means that
$[(y, z), (w, x)] \in P]$, and therefore $P$ is symmetric.
_Proof ($P$ is transitive):_
Let $(w, x), (y, z), (a, b) \in \mathbb{R} \times \mathbb{R}$.
To prove that $P$ is transitive, it must be shown that
$[[(w, x), (y, z)] \in P \wedge [(y, z), (a, b)] \in P \to [(w, x), (a, b)] \in P$.
Suppose $[(w, x), (y, z)] \in P$ and $[(y, z), (a, b)] \in P$. By the definition
for $P$, this means that:
$$ w = y $$
And also that:
$$ y = a $$
By the transitive property of equality, this means that $w = a$. It follows that
$[(w, x), (a, b)] \in P$, and therefore $P$ is transitive.
_Conclusion:_
Since it has been shown that $P$ is reflexive, symmetric, and transitive, it
follows that $P$ is an equivalence relation. This is what was to be shown.
Q.E.D.
(2) Describe the distinct equivalence classes of each relation.
There is one equivalence class for each real number. The distinct equivalence
classes are all sets of ordered pairs
$(x, y) \in \mathbb{R} \times \mathbb{R}, | x = a$ for each real number $a$.
30. Define $Q$ on the set $\mathbb{R} \times \mathbb{R}$ as follows: For every
$(w, x), (y, z) \in \mathbb{R} \times \mathbb{R}$,
@ -3228,8 +3575,12 @@ $$ (w, x) Q (y, z) \Leftrightarrow x = z $$
(1) Prove that the relation is an equivalence relation.
Omitted.
(2) Describe the distinct equivalence classes of each relation.
Omitted.
31. Let $P$ be the set of all points in the Cartesian plane except the origin.
$R$ is the relation defined on $P$ as follows: For every $p_1$ and $p_2$ in
$P$,
@ -3238,16 +3589,23 @@ $$ p_1 R p_2 \Leftrightarrow p_1 \text{ and } p_2 \text{ lie on the same half-li
(1) Prove that the relation is an equivalence relation.
Omitted.
(2) Describe the distinct equivalence classes of each relation.
Omitted.
32. Let $A$ be the set of all straight lines in the Cartesian plane. Define a
relation $\mid \mid$ on $A$ as follows: For every $l_1$ and $l_2$ in $A$,
$$ l_1 \mid \mid l_2 \Leftrightarrow l_1 \text{ is parallel to } l_2 $$
$$ l_1 \parallel l_2 \Leftrightarrow l_1 \text{ is parallel to } l_2 $$
Then $\mid \mid$ is an equivalence relation on $A$. Describe the equivalence
Then $\parallel$ is an equivalence relation on $A$. Describe the equivalence
classes of this relation.
Every possible slope is an equivalence class, including vertical lines
(undefined).
33. Let $A$ be the set of points in the rectangle with $x$ and $y$ coordinates
between $0$ and $1$. That is,
@ -3282,6 +3640,10 @@ than themselves. Then $R$ is an equivalence relation on $A$. Imagine gluing
together all the points that are in the same equivalence class. Describe the
resulting figure.
Gluing the top and the bottom edges of the rectangle together forms a cylinder,
and then gluing the left and right edges of the rectangle together forms a
doughnut shape (a torus).
34. The documentation for the computer language Java recommends that when an
"equals method" is defined for an object, it be an equivalence relation.
That is, if $R$ is defined as follows:
@ -3300,37 +3662,217 @@ where $c$ is a small positive number that depends on the resolution of the
computer display. Is the programmer's equals method an equivalence relation?
Justify your answer.
No. If points $p$, $q$, and $r$ all lie on a straight line with $q$ in the
middle, and if $p$ is $c$ units from $q$ and $q$ is $c$ units from $r$, then $p$
is more than $c$ units from $r$. In other words, the programmer's equals method
is not an equivalence relation because it is not transitive.
35. Find an additional representative circuit for the input/output table of
Example 8.3.9.
Omitted.
Let $R$ be an equivalence relation on a set $A$. Prove each of the statements in
36-41 directly from the definitions of equivalence relation and equivalence
class without using the results of Lemma 8.3.2, Lemma 8.3.3, or Theorem 8.3.4.
36. For every $a$ in $a$, $a \in [a]$.
**Proof:**
Suppose $R$ is an equivalence relation on a set $A$, and let $a \in A$.
Since $R$ is an equivalence relation, this means that $R$ is reflexive, or, in
other words, every element in $A$ is related to itself by $R$. In particular,
$a R a$, and hence, by definition of an equivalence class, $a \in [a]$. This is
what was to be shown.
Q.E.D.
37. For every $a$ and $b$ in $A$, if $b \in [a]$ then $a R b$.
**Proof:**
Suppose $R$ is an equivalence relation on a set $A$, and let $a, b \in A$.
Let $b \in [a]$.
By definition of class, this means that:
$$ b \in [a] \Leftrightarrow b R a $$
Since $R$ is an equivalence relation, $R$ is symmetric. By the definition of
symmetry, this means that:
$$ a R b $$
This is what was to be shown.
Q.E.D.
38. For every $a$, $b$, and $c$ in $A$, if $b R c$ and $c \in [a]$ then
$b \in [a]$.
**Proof:**
Suppose $R$ is an equivalence relation on a set $A$, and let $a, b, c \in A$.
Let $b R c$ and let $c \in [a]$.
We must show that $b \in [a]$.
By the definition of class, since $c \in [a]$, this means that $c R a$. Since
$R$ is an equivalence relation, and therefore transitive, and also since
$b R c$, it follows, by the definition of transitive, that $b R c$ and $c R a$.
In other words $b R a$, and by definition of class, this means that $b \in [a]$.
This is what was to be shown.
Q.E.D.
39. For every $a$ and $b$ in $A$, if $[a] = [b]$ then $a R b$.
**Proof:**
Suppose $R$ is an equivalence relation on a set $A$, and let $a, b \in A$.
Let $[a] = [b]$.
Since $R$ is reflexive (by the definition of equivalence relation), it follows
that $a \in [a]$ and $b \in [b]$.
By the supposition, $[a] = [b]$, and so it follows that $a \in [b]$. By the
definition of class, this means that $a R b$. This is what was to be shown.
Q.E.D.
40. For every $a$, $b$, and $x$ in $A$, if $a R b$ and $x \in [a]$ then
$x \in [b]$.
**Proof:**
Suppose $R$ is an equivalence relation on a set $A$, and let $a, b, x \in A$.
Let $a R b$ and $x \in [a]$.
It must be shown that $x \in [b]$.
Since $x \in [a]$, by the definition of equivalence class, this means that
$x R a$. Since $a R b$, by the definition of transitivity (since $R$ is an
equivalence relation and therefore transitive), it follows that $x R b$. By the
definition of equivalence class, this means that $x \in [b]$. This is what was
to be shown.
Q.E.D.
41. For every $a$ and $b$ in $A$, if $a \in [b]$ then $[a] = [b]$.
**Proof:**
Suppose $R$ is an equivalence relation on a set $A$, and let $a, b \in A$.
Let $a \in [b]$.
To prove that $[a] = [b]$, it must be shown that $[a] \subseteq [b]$, and that
$[b] \subseteq [a]$.
_Proof ($[a] \subseteq [b]$):_
Let $x \in [a]$.
By the definition of equivalence class, this means that $x R a$. Since
$a \in [b]$, this means that $a R b$. Since $R$ is transitive (because $R$ is an
equivalence relation), this means that $x R b$. By the definition of class, this
means that $x \in [b]$. It follows that $[a] \subseteq [b]$. This is what was to
be shown.
_Proof ($[b] \subseteq [a]$):_
Let $x \in [b]$.
By the definition of equivalence, class this means that $x R b$. Since
$a \in [b]$, this means that $a R b$. Since $R$ is symmetric (because $R$ is an
equivalence relation), this means that $b R a$. Then, since $R$ is transitive
(again, because $R$ is an equivalence relation), it follows that $x R a$. By the
definition of equivalence class, this means that $x \in [a]$. It follows that
$[b] \subseteq [a]$. This is what was to be shown.
_Conclusion:_
Since it has been shown that $[a] \subseteq [b]$ and also that
$[b] \subseteq [a]$, it follows (by the definition for subset), that
$[a] = [b]$. This is what was to be shown.
Q.E.D.
42. Let $R$ be the relation defined in Example 8.3.12.
a. Prove that $R$ is reflexive.
**Proof:**
Suppose $A$ is the set of all ordered pairs of integers for which the second
element of the pair is nonzero:
$$ A = \mathbb{Z} \times (\mathbb{Z} - \{0\}) $$
Then, define a relation $R$ on $A$ as follows:
$$ \forall (a, b), (c, d) \in A, (a, b) R (c, d) \Leftrightarrow ad = bc $$
Let $(x, y) \in A$.
To prove that $R$ is reflexive, it must be shown that $[(x, y), (x, y)] \in R$.
By the definition of $R$, this means it must be shown that:
$$ xy = yx $$
By the commutative law of product, this is true. Therefore
$[(x, y), (x, y)] \in R$, and $R$ is reflexive.
Q.E.D.
b. Prove that $R$ is symmetric.
**Proof:**
Suppose $A$ is the set of all ordered pairs of integers for which the second
element of the pair is nonzero:
$$ A = \mathbb{Z} \times (\mathbb{Z} - \{0\}) $$
Then, define a relation $R$ on $A$ as follows:
$$ \forall (a, b), (c, d) \in A, (a, b) R (c, d) \Leftrightarrow ad = bc $$
Let $(x, y), (z, a) \in A$.
To prove that $R$ is symmetric, it must be shown that
$[(x, y), (z, a)] \in R \to [(z, a), (x, y)] \in R$.
Suppose $[(x, y), (z, a)] \in R$. By the definition for $R$, this means that:
$$ xa = yz $$
(It must be shown that $zy = ax$.)
By the commutative property for product, and the symmetric property of equality,
$xa = yz$ can be rewritten as:
$$ zy = ax $$
It follows that $[(z, a), (x, y)] \in R$, and therefore $R$ is symmetric. This
is what was to be shown.
Q.E.D.
c. List four distinct elements in $[(1, 3)]$.
$$ (2, 6), (-2, -6), (3, 9), (-3, -9) $$
d. List four distinct elements in $[(2, 5)]$.
$$ (4, 10), (6, 15), (8, 20), (10, 25) $$
43. In Example 8.3.12, define operations of addition $(+)$ and multiplication
$(\cdot)$ as follows: For every $(a, b), (c, d) \in A$,
@ -3342,28 +3884,40 @@ a. Prove that this addition is well defined. That is, show that if
$[(a, b)] = [(a', b')]$ and $[(c, d)] = [(c', d')]$, then
$[(ad + bc), bd] = [(a'd' + b'c', b'd')]$.
Omitted.
b. Prove that this multiplication is well defined. That is, show that if
$[(a, b)] = [(a', b')]$ and $[(c, d)] = [(c', d')]$, then
$[(ac, bd)] = [(a'c', b'd')]$.
Omitted.
c. Show that $[(0, 1)]$ is an identity element for addition. That is, show that
for any $(a, b) \in A$,
$$ [(a, b)] + [(0, 1)] = [(0, 1)] + [(a, b)] = [(a, b)] $$
Omitted.
d. Find an identity element for multiplication. That is, find $(i, j)$ in $A$ so
that for every $(a, b)$ in $A$,
$[(a, b)] \cdot [(i, j)] = [(i, j)] \cdot [(a, b)] = [(a, b)]$.
Omitted.
e. For any $(a, b) \in A$, show that $[(-a, b)]$ is an inverse for $[(a, b)]$
for addition. That is, show that
$[(-a, b)] + [(a, b)] = [(a, b)] + [(-a, b)] = [(0, 1)]$.
Omitted.
f. Given any $(a, b) \in A$ with $a \neq 0$, find an inverse for $[(a, b)]$ for
multiplication. That is, find $(c, d)$ in $A$ so that
$[(a, b)] \cdot [(c, d)] = [(c, d)] \cdot [(a, b)] = [(i, j)]$, where $[(i, j)]$
is the identity element you found in part (d).
Omitted.
44. Let $A = \mathbb{Z}^+ \times \mathbb{Z}^+$. Define a relation $R$ on $A$ as
follows: For every $(a, b)$ and $(c, d)$ in $A$,
@ -3371,16 +3925,28 @@ $$ (a, b) R (c, d) \Leftrightarrow a + d = c + b $$
a. Prove that $R$ is reflexive.
Omitted.
b. Prove that $R$ is symmetric.
Omitted.
c. Prove that $R$ is transitive.
Omitted.
d. List five elements in $[(1, 1)]$.
Omitted.
e. List five elements in $[(3, 1)]$.
Omitted.
f. List five elements in $[(1, 2)]$.
Omitted.
g. Describe the distinct equivalence classes of $R$.
45. The following argument claims to prove that the requirement that an
@ -3393,23 +3959,53 @@ transitive. For any two elements $x$ and $y$ in $A$, if $x R y$ then $y R x$
since $R$ is symmetric. Thus it follows by transitivity that $x R x$, and hence
$R$ is reflexive."
The mistake in the argument is that just because $R$ is symmetric and transitive
does not necessarily mean it is reflexive. Recall that for $R$ to be reflexive,
$\forall x \in A, x R x$. Consider, however, a set where the relation is both
symmetric and transitive, but not reflexive:
$$ A = \{1, 2\} $$
$$ R = \{(1, 1)\} $$
Now, $R$ is symmetric, since $(1, 1) \to (1, 1)$, and $R$ is transitive, since
$(1, 1) \wedge (1, 1) \to (1, 1)$, and while $(1, 1)$ is reflexive, there is no
ordered pair in the set where $2 R 2$, so therefore $R$ is not reflexive, even
though $R$ is symmetric and transitive.
46. Let $R$ be a relation on a set $A$ and suppose $R$ is symmetric and
transitive. Prove the following: If for every $x$ in $A$ there is a $y$ in
$A$ such that $x R y$, then $R$ is an equivalence relation.
Omitted.
47. Refer to the quote at the beginning of this section to answer the following
questions.
a. What is the name of the Knight's song called?
Omitted.
b. What is the name of the Knight's song?
Omitted.
c. What is the Knight's song called?
Omitted.
d. What _is_ the Knight's song?
Omitted.
e. What is your (full, legal) name?
Omitted.
f. What are you called?
Omitted.
g. What _are_ you? (Do not answer this on paper; just think about it.)
Omitted.

20
chapter_8/notes.md
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@ -439,3 +439,23 @@ $$ d | (m - n) $$
Symbolically:
$$ m \equiv n(\mod d) \Leftrightarrow d | (m - n) $$
---
Page 542
**Example 8.3.12**
_Rational Numbers are Really Equivalence Classes
Let $A$ be the set of all ordered pairs of integers for which the second element
of the pair is nonzero. Symbolically:
$$ A = \mathbb{Z} \times (\mathbb{Z} - \{0\}) $$
Define a relation $R$ on $A$ as follows: For all pairs $(a, b)$ and $(c, d)$ in
$A$,
$$ (a, b) R (c, d) \Leftrightarrow ad = bc $$
The fact is that $R$ is an equivalence relation.