From 418bc0bd432c2078c9673e4deda8b708b1ac4dbd Mon Sep 17 00:00:00 2001 From: tomit4 Date: Sat, 22 Aug 2026 00:18:29 -0700 Subject: [PATCH] :construction: Fin 8.3 --- chapter_8/exercises.md | 600 ++++++++++++++++++++++++++++++++++++++++- chapter_8/notes.md | 20 ++ 2 files changed, 618 insertions(+), 2 deletions(-) diff --git a/chapter_8/exercises.md b/chapter_8/exercises.md index c851a65..bc57b17 100644 --- a/chapter_8/exercises.md +++ b/chapter_8/exercises.md @@ -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)\} $$ + +Now, $R$ is symmetric, since $(1, 1) \to (1, 1)$, and $R$ is transitive, since +$(1, 1) \wedge (1, 1) \to (1, 1)$, and while $(1, 1)$ is reflexive, there is no +ordered pair in the set where $2 R 2$, so therefore $R$ is not reflexive, even +though $R$ is symmetric and transitive. + 46. Let $R$ be a relation on a set $A$ and suppose $R$ is symmetric and transitive. Prove the following: If for every $x$ in $A$ there is a $y$ in $A$ such that $x R y$, then $R$ is an equivalence relation. +Omitted. + 47. Refer to the quote at the beginning of this section to answer the following questions. a. What is the name of the Knight's song called? +Omitted. + b. What is the name of the Knight's song? +Omitted. + c. What is the Knight's song called? +Omitted. + d. What _is_ the Knight's song? +Omitted. + e. What is your (full, legal) name? +Omitted. + f. What are you called? +Omitted. + g. What _are_ you? (Do not answer this on paper; just think about it.) + +Omitted. diff --git a/chapter_8/notes.md b/chapter_8/notes.md index bc8b5bf..5c1e411 100644 --- a/chapter_8/notes.md +++ b/chapter_8/notes.md @@ -439,3 +439,23 @@ $$ 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.