🚧 Midway through 6.1

chapter_6/exercises.md

@@ -7,17 +7,106 @@ Page 411
 
 a. $A = \{2, \{2\}, (\sqrt{2})^2\}$, $B = \{2, \{2\}, \{\{2\}\}\}$
 
+$A \subseteq B$ ?:
+
+$$ A = \{2, \{2\}, (\sqrt{2})^2\} = \{2, \{2\}, 2\} = \{2, \{2\}\} $$
+
+Yes, every element in $A$ is in $B$.
+
+$B \subseteq A$ ?:
+
+No, because $\{\{2\}\}$ is an element of $B$, but is not an element of $A$, so
+$\B \nsubseteq A$.
+
+Is either $A$ or $B$ a proper subset of the other?
+
+Yes, $A$ is a proper subset of $B$, because every element in $A$ is in $B$, but
+not every element in $B$ is in $A$.
+
 b. $A = \{3, \sqrt{5^2 - 4^2}, 24 \mod 7\}$, $B = \{8 \mod 5\}$
 
+$A \subseteq B$ ?:
+
+$$ A = \{3, \sqrt{5^2 - 4^2}, 24 \mod 7\} = \{3, 3, 3\} = \{3\} $$
+
+$$ B = \{8 \mod 5\} = \{3\} $$
+
+Yes, $A$ is a subset of $B$ since every element of $A$ is in $B$.
+
+$B \subseteq A$ ?:
+
+Yes, $B$ is a subset of $A$ since every element of $B$ is in $A$.
+
+Is either $A$ or $B$ a proper subset of the other?
+
+Yes, both $A$ and $B$ are proper subsets of the other since $A = B$.
+
 c. $A = \{\{1, 2\}, \{2, 3\}\}$, $B = \{1, 2, 3\}$
 
+$A \subseteq B$ ?:
+
+No, because there are no elements in $A$ that are in $B$, $A \nsubseteq B$
+
+$B \subseteq A$ ?:
+
+No, because there are no elements in $B$ that are in $A$, $B \nsubseteq A$
+
+Is either $A$ or $B$ a proper subset of the other?
+
+No, since neither set share any elements, neither is a proper subset of the
+other.
+
 d. $A = \{a, b, c\}$, $B = \{\{a\}, \{b\}, \{c\}\}$
 
+$A \subseteq B$ ?:
+
+No, because there are no elements in $A$ that are in $B$, $A \nsubseteq B$
+
+$B \subseteq A$ ?:
+
+No, because there are no elements in $B$ that are in $A$, $B \nsubseteq A$
+
+Is either $A$ or $B$ a proper subset of the other?
+
+No, since neither set share any elements, neither is a proper subset of the
+other.
+
 e. $A = \{\sqrt{16}, \{4\}\}$, $B = \{4\}$
 
+$A \subseteq B$ ?:
+
+$$ A = \{\sqrt{16}, \{4\}\} = \{4, \{4\}\} $$
+
+No, because every element of $A$ is not an element in $B$ ($4$ is not in $B$),
+$A \nsubseteq B$.
+
+$B \subseteq A$ ?:
+
+Yes, because every element in $B$ is an element in $A$, $B \subseteq A$.
+
+Is either $A$ or $B$ a proper subset of the other?
+
+Yes, $B$ is a proper subset of $A$ since $B \subseteq A$ and $A \nsubseteq B$.
+
 f. $A = \{x \in \mathbb{R} | \cos x \in \mathbb{Z}\}$,
 $B = \{x \in \mathbb{R} | \sin x \in \mathbb{Z}\}$
 
+From trigonometry, we know that $\cos x = -1 \text{ or } 0 \text{ or } 1$ and
+$\sin x = -1 \text{ or } 0 \text{ or } 1 $. When we evaluate for $x$ in these
+cases we find:
+
+$$ A = \{\dots, -\frac{5\pi}{2}, -\frac{3\pi}{2}, \pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots\} $$
+
+$$ B = \{\dots, -\frac{5\pi}{2}, -\frac{3\pi}{2}, \pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots\} $$
+
+$A \subseteq B$ ?: Yes.
+
+$B \subseteq A$ ?: Yes.
+
+Yes, $B$ is a proper subset of $A$ since $B \subseteq A$ and $A \nsubseteq B$.
+
+Yes, since $A = B$.
+
 2. Complete the proof from Example 6.1.3: Prove that $B \subseteq A$ where
 
 $$ A = \{m \in \mathbb{Z} | m, = 2a \text{ for some integer } a\} $$
@@ -26,6 +115,32 @@ and
 
 $$ B = \{n \in \mathbb{Z} | n = 2b - 2 \text{ for some integer } b\} $$
 
+_Part 2, Proof that $B \subseteq A$:_
+
+Suppose $x$ is a particular but arbitrarily chosen element of $B$.
+
+By definition of $B$, there is an integer, say $b$, such that $x = 2b - 2$.
+
+To prove that $B \subseteq A$, we must show that there is some $x$ that can
+equal both $2a$, for some integer $a$, and that same $x$ can also equal
+$2b - 2$.
+
+$$ 2b - 2 = 2a $$
+
+$$ a = b - 1 $$
+
+By the difference integers, $a$ is an integer. Then, by substitution:
+
+$$ 2a = 2(b - 1) $$
+
+$$ = 2b - 2 $$
+
+$$ = x $$
+
+Thus, by definition of $A$, $x$ is an element of $A$.
+
+Q.E.D.
+
 3. Let sets $R$, $S$, and $T$ be defined as follows:
 
 $$ R = \{x \in \mathbb{Z} | x \text{ is divisible by } 2\}  $$
@@ -38,26 +153,135 @@ Prove or disprove each of the following statements.
 
 a. $R \subseteq T$
 
+$R \nsubseteq T$ since $2 \in R$ since $2 \mid 2$, but $2 \notin T$ since
+$6 \cancel{\mid} 2$.
+
 b. $T \subseteq R$
 
+**Proof:**
+
+Suppose $n$ is any integer such that $6 \mid n$, therefore $n \in T$.
+
+By the definition of divisibility:
+
+$$ n = 6m $$
+
+for some integer $m$.
+
+$$ n = 2(3m) $$
+
+By the product of integers, $3m$ is an integer. It follows that
+$n = 2 \cdot (\text{some integer})$. Thus $2 \mid n$, so $n \in R$. This is what
+was to be shown.
+
+Q.E.D.
+
 c. $T \subseteq S$
 
+**Proof:**
+
+Suppose $n$ is any integer such that $6 \mid n$, therefore $n \in T$.
+
+By the definition of divisibility:
+
+$$ n = 6m $$
+
+for some integer $m$.
+
+$$ n = 3(2m) $$
+
+By the product of integers, $2m$ is an integer. It follows that
+$n = 3 \cdot (\text{some integer})$. Thus $3 \mid n$, so $n \in S$. This is what
+was to be shown.
+
+Q.E.D.
+
 4. Let $A = \{n \in \mathbb{Z} | n = 5r \text{ for some integer } r\}$ and
    $B = \{m \in \mathbb{Z} | m = 20s \text{ for some integer } s\}$. Prove or
    disprove each of the following statements.
 
 a. $A \subseteq B$
 
+$A \nsubseteq B$ since $5 \in A$ since $5 \mid 5$, but $5 \notin B$ since
+$5 \cancel{\mid} 20$.
+
 b. $B \subseteq A$
 
+**Proof:**
+
+Suppose $n$ is any integer such that $20 \mid n$, therefore $n \in B$.
+
+By the definition of divisibility:
+
+$$ n = 20m $$
+
+for some integer $m$.
+
+$$ n = 5(4m) $$
+
+By the product of integers, $4m$ is an integer. It follows that
+$n = 5 \cdot (\text{some integer})$. Thus $5 \mid n$, so $n \in A$. This is what
+was to be shown.
+
+Q.E.D.
+
 5. Let $C = \{n \in \mathbb{Z} | n = 6r - 5 \text{ for some integer } r\}$ and
    $D = \{m \in \mathbb{Z} | m = 3s + 1 \text{ for some integer } s\}$. Prove or
    disprove each of the following statements.
 
 a. $C \subseteq D$
 
+**Proof:**
+
+Suppose $n$ is any integer such that $n = 6r - 5$ for some integer $r$, which
+means that $n \in C$.
+
+Also suppose that $m$ is any integer such that $m = 3s + 1$ for some integer
+$s$, which means that $m \in S$.
+
+We must show that there exists some $r$ that when substituted for $s$ will
+satisfy the definition of $n$.
+
+Let $s = 2r - 2$. Then, by substitution:
+
+$$ m = 3(2r - 2) + 1 $$
+
+$$ = 6r - 6 + 1 $$
+
+$$ = 6r - 5 $$
+
+$$ = n $$
+
+By the product and difference of integers, $6r - 5$ is an integer, therefore
+$n \in D$. This is what was to be shown.
+
+Q.E.D.
+
 b. $D \subseteq C$
 
+**Disproof:**
+
+$D \nsubseteq C$ because there are elements in $D$ that are not in $C$. For
+example $4$ is in $D$ because $4 = 3(1) + 1$, but $4$ is not in $C$. If $4$ were
+in $C$, this would mean:
+
+$$ 4 = 6r - 5 $$
+
+for some integer $r$.
+
+$$ 4 + 5 = 6r $$
+
+$$ 9 = 6r $$
+
+$$ \frac{9}{6} = r $$
+
+$$ \frac{3}{2} = r $$
+
+But $\dfrac{3}{2}$ is not an integer. This is a contradiction, therefore
+$D \nsubseteq C$.
+
+Q.E.D.
+
 6. Let $A = \{x \in \mathbb{Z} | x = 5a + 2 \text{ for some integer } a\}$,
    $B = \{y \in \mathbb{Z} | y = 10b - 3 \text{ for some integer } b\}$, and
    $C = \{z \in \mathbb{Z} | z = 10c + 7 \text{ for some integer } c\}$.
@@ -66,10 +290,141 @@ Prove or disprove each of the following statements.
 
 a. $A \subseteq B$
 
+**Disproof (by counterexample):**
+
+Suppose $n$ is any integer such that $n = 5a + 2$ for some integer $a$. This
+means that $n \in A$.
+
+Suppose also that there is some integer $m$ such that $m = 10b - 3$ for some
+integer $b$. This means that $m \in B$.
+
+To show that there is some integer $b$ that will satisfy $n$, we must relate it
+to $a$:
+
+$$ 5a + 2 = 10b - 3 $$
+
+$$ 5a + 5 = 10b $$
+
+$$ 5a + 5 = 10b $$
+
+$$ \frac{1}{2}a + \frac{1}{2} = b $$
+
+$$ \frac{a + 1}{2} = b $$
+
+In order for $A \subseteq B$, every element of $A$ must be in $B$. If $a = 0$,
+then $n = 2$, so $n \in A$. If $a = 0$, then $b = \dfrac{1}{2}$, which is not an
+integer, thus $2 \notin B$. Therefore $A \nsubseteq B$.
+
+Q.E.D.
+
 b. $B \subseteq A$
 
+**Proof:**
+
+Suppose $y$ is any integer such that $y = 10b - 3$ for some integer $b$. This
+means that $y \in B$.
+
+Let's first find $a$ as it relates to $y$.
+
+$$ y = 5a + 2 $$
+
+$$ 10b - 3 = 5a + 2 $$
+
+$$ 10b - 5 = 5a $$
+
+$$ 2b - 1 = a $$
+
+So let $a = 2b - 1$.
+
+Then substitute in for the condition for $A$:
+
+$$ x = 5a + 2 $$
+
+$$ = 5(2b - 1) + 2 $$
+
+$$ = 10b - 5 + 2 $$
+
+$$ = 10b - 3 $$
+
+$$ = y $$
+
+Therefore $y \in A$.
+
+Q.E.D.
+
 c. $B = C$
 
+To prove $B = C$, we must prove both that $B \subseteq C$ and $C \subseteq B$.
+
+_Prove $B \subseteq C$:
+
+Suppose $y$ is any integer such that $y = 10b - 3$ for some integer $b$. This
+means that $y \in B$.
+
+Let's first find some integer $c$ as it relates to $y$:
+
+$$ y = 10c + 7 $$
+
+$$ 10b - 3 = 10c + 7 $$
+
+$$ 10b - 10 = 10c $$
+
+$$ b - 1 = c $$
+
+So, let $c = b - 1$.
+
+Then substitute in for the condition for $C$:
+
+$$ z = 10c + 7 $$
+
+$$ = 10(b - 1) + 7 $$
+
+$$ = 10b - 10 + 7 $$
+
+$$ = 10b - 3 $$
+
+$$ = y $$
+
+Thus $y \in C$, and therefore $B \subseteq C$.
+
+_Prove $C \subseteq B$:
+
+Suppose $z$ is any integer such that $z = 10c + 7$ for some integer $c$. This
+means that $z \in C$.
+
+Let's first find some integer $b$ as it relates to $z$:
+
+$$ z = 10b - 3 $$
+
+$$ 10c + 7 = 10b - 3 $$
+
+$$ 10c + 10 = 10b $$
+
+$$ c + 1 = b $$
+
+So, let $b = c + 1$.
+
+Then substitute in for the condition for $B$:
+
+$$ y = 10b - 3 $$
+
+$$ = 10(c + 1) - 3 $$
+
+$$ = 10c + 10 - 3 $$
+
+$$ = 10c + 7 $$
+
+$$ = z $$
+
+Therefore $z \in B$.
+
+Thus $z \in B$, and therefore $C \subseteq B$.
+
+Since $B \subseteq C$ and $C \subseteq B$, it follows that $B = C$. This is what
+was to be shown.
+
+Q.E.D.
+
 7. Let $A = \{x \in \mathbb{Z} | x = 6a + 4 \text{ for some integer } a\}$,
    $B = \{y \in \mathbb{Z} | y = 18b - 2 \text{ for some integer } b\}$, and
    $C = \{z \in \mathbb{Z} | z = 18c + 16 \text{ for some integer } c\}$.
@@ -78,49 +433,234 @@ Prove or disprove each of the following statements.
 
 a. $A \subseteq B$
 
+**Disproof (by counterexample):**
+
+Suppose $x$ is any integer such that $x = 6a + 4$ for some integer $a$. This
+means that $x \in A$.
+
+Let's first find some integer $b$ as it relates to $a$.
+
+$$ x = 18b - 2 $$
+
+$$ 6a + 4 = 18b - 2 $$
+
+$$ 6a + 6 = 18b $$
+
+$$ \frac{6}{18}a + \frac{6}{18} = b $$
+
+$$ \frac{1}{3}a + \frac{1}{3} = b $$
+
+$$ \frac{a + 1}{3} = b $$
+
+By definition of $b$, $b$ must always be an integer for all $a$.
+
+Suppose $a = 0$, then:
+
+$$ x = 6(0) + 4 = 4 $$
+
+so $4 \in A$, but:
+
+$$ b = \frac{0 + 1}{3} = \frac{1}{3} $$
+
+so $4 \notin B$. We can see this as $4 = 18b - 2$ results in $b = \dfrac{1}{3}$,
+but $b$ must be an integer.
+
 b. $B \subseteq A$
 
+**Proof:**
+
+Suppose $y$ is any integer such that $y = 18b - 2$ for some integer $b$. This
+means that $y \in B$.
+
+Let's first find some integer $a$ as it relates to $b$.
+
+$$ y = 6a + 4 $$
+
+$$ 18b - 2 = 6a + 4 $$
+
+$$ 18b - 6 = 6a $$
+
+$$ 3b - 1 = a $$
+
+Now, substitute $a$ in for the condition for $A$:
+
+$$ x = 6a + 4 $$
+
+$$ = 6(3b - 1) + 4 $$
+
+$$ = 18b - 2 $$
+
+$$ = y $$
+
+Therefore $B \subseteq A$.
+
 c. $B = C$
 
+To prove $B = C$, we must prove both that $B \subseteq C$ and $C \subseteq B$.
+
+_Prove $B \subseteq C$:
+
+Suppose $y$ is any integer such that $y = 18b - 2$ for some integer $b$. This
+means that $y \in B$.
+
+Let's first find some integer $c$ as it relates to $b$.
+
+$$ y = 18c + 16 $$
+
+$$ 18b - 2 = 18c + 16 $$
+
+$$ 18b - 18 = 18c $$
+
+$$ b - 1 = c $$
+
+So, let $c = b - 1$. Now substitute $c$ in for the condition of $C$:
+
+$$ z = 18c + 16 $$
+
+$$ = 18(b - 1) + 16 $$
+
+$$ = 18b - 18 + 16 $$
+
+$$ = 18b - 2 $$
+
+$$ = y $$
+
+Therefore $B \subseteq C$.
+
+_Prove $C \subseteq B$:
+
+Suppose $z$ is any integer such that $z = 18c + 16$ for some integer $c$.
+
+Let's first find some $b$ as it relates to $c$.
+
+$$ z = 18b - 2 $$
+
+$$ 18c + 16 = 18b - 2 $$
+
+$$ 18c + 18 = 18b $$
+
+$$ c + 1 = b $$
+
+So, let $b = c + 1$. Now, let's substitute $b$ in for the condition for $B$.
+
+$$ y = 18b - 2 $$
+
+$$ = 18(c + 1) - 2 $$
+
+$$ = 18c + 18 - 2 $$
+
+$$ = 18c + 16 $$
+
+$$ = z $$
+
+Therefore $C \subseteq B$.
+
+Since $B \subseteq C$ and $C \subseteq B$, we conclude that $B = C$. This is
+what was to be shown.
+
+Q.E.D.
+
 8. Write in words to read each of the following out loud. Then write each set
    using the symbols for union, intersection, set difference, or set complement.
 
 a. $\{x \in U | x \in A \text{ and } x \in B\}$
 
+_In words:_
+
+The set of all $x$ in $U$ such that $x$ is in $A$ and $x$ is in $B$.
+
+_In symbolic notation:_
+
+$$ A \cap B $$
+
 b. $\{x \in U | x \in A \text{ or } x \in B\}$
 
+_In words:_
+
+The set of all $x$ in $U$ such that $x$ is in $A$ or $x$ is in $B$.
+
+_In symbolic notation:_
+
+$$ A \cup B $$
+
 c. $\{x \in U | x \in A \text{ and } x \notin B\}$
 
+_In words:_
+
+The set of all $x$ in $U$ such that $x$ is in $A$ and $x$ is not in $B$.
+
+_In symbolic notation:_
+
+$$ A - B $$
+
 d. $\{x \in U | x \notin A\}$
 
+_In words:_
+
+The set of all $x$ in $U$ such that $x$ is not in $A$.
+
+_In symbolic notation:_
+
+$$ A^c $$
+
 9. Complete the following sentences without using the symbols $\cup$, $\cap$, or
    $-$.
 
 a. $x \notin A \cup B$ if, and only if, _____.
 
+$x$ is not in $A$ and $x$ is not in $B$.
+
 b. $x \notin A \cap B$ if, and only if, _____.
 
+$x$ is not in $A$ or $x$ is not in $B$.
+
 c. $x \notin A - B$ if, and only if, _____.
 
+$x$ is not in $A$, or $x$ is in $B$, or both.
+
+Note: recall that the negation of an "and", which is $A - B$, is an "or", thus:
+
+$$ x \in (A - B) \to x \in A \wedge x \notin B $$
+
+so:
+
+$$ \neg(x \in (A - B)) \to \neg(x \in A \wedge x \notin B) \to x \notin A \vee x \in B $$
+
 10. Let $A = \{1, 3, 5, 7, 9\}$, $b = \{3, 6, 9\}$, and $C = \{2, 4, 6, 8\}$.
     Find each of the following:
 
 a. $A \cup B$
 
+$$ A \cup B = \{1, 3, 5, 6, 7, 9\} $$
+
 b. $A \cap B$
 
+$$ A \cap B = \{3, 9\} $$
+
 c. $A \cup C$
 
+$$ A \cup C = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} $$
+
 d. $A \cap C$
 
+$$ A \cap C = \emptyset $$
+
 e. $A - B$
 
+$$ A - B = \{1, 5, 7\} $$
+
 f. $B - A$
 
+$$ B - A = \{6\} $$
+
 g. $B \cup C$
 
+$$ B \cup C = \{2, 3, 4, 6, 8, 9\} $$
+
 h. $B \cap C$
 
+$$ B \cap C = \{6\} $$
+
 11. Let the universal set $\mathbb{R}$, the set of all real numbers, and let
     $A = \{x \in \mathbb{R} | 0 < x \leq 2\}$,
     $B = \{x \in \mathbb{R} | 1 \leq x < 4\}$, and
@@ -128,24 +668,56 @@ h. $B \cap C$
 
 a. $A \cup B$
 
+$$ A \cup B = \{x \in \mathbb{R} | 0 < x < 4\} $$
+
 b. $A \cap B$
 
+$$ A \cap B = \{x \in \mathbb{R} | 1 \leq x \leq 2\} $$
+
 c. $A^c$
 
+$$ A^c = \{x \in \mathbb{R} | x \leq 0 \text{ or } x > 2\} $$
+
 d. $A \cup C$
 
+$$ A \cup C = \{x \in \mathbb{R} | 0 < x \leq 2 \text{ or } 3 \leq x < 9 \} $$
+
 e. $A \cap C$
 
+$$ A \cap C = \emptyset $$
+
 f. $B^c$
 
+$$ B^c = \{x \in \mathbb{R} | x < 1 \text{ or } x \geq 4\} $$
+
 g. $A^c \cap B^c$
 
+$$ A^c = \{x \in \mathbb{R} | x \leq 0 \text{ or } x > 2\} $$
+
+$$ B^c = \{x \in \mathbb{R} | x < 1 \text{ or } x \geq 4\} $$
+
+$$ A^c \cap B^c = \{x \in \mathbb{R} | x \leq 0 \text{ or } x \geq 4\}  $$
+
 h. $A^c \cup B^c$
 
+$$ A^c = \{x \in \mathbb{R} | x \leq 0 \text{ or } x > 2\} $$
+
+$$ B^c = \{x \in \mathbb{R} | x < 1 \text{ or } x \geq 4\} $$
+
+$$ A^c \cup B^c = \{x \in \mathbb{R} | x < 1 \text{ or } x > 2\} $$
+
 i. $(A \cap B)^c$
 
+$$ A \cap B = \{x \in \mathbb{R} | 1 \leq x \leq 2\} $$
+
+$$ (A \cap B)^c = \{x \in \mathbb{R} | x < 1 \text{ or } x > 2 \} $$
+
 j. $(A \cup B)^c$
 
+$$ A \cup B = \{x \in \mathbb{R} | 0 < x < 4\} $$
+
+$$ (A \cup B)^c = \{x \in \mathbb{R} | x \leq 0 \text{ or } x \geq 4\} $$
+
 12. Let the universal set be $\mathbb{R}$, the set of all real numbers, and let
     $A = \{x \in \mathbb{R} | -3 \leq x \leq 0\}$,
     $B = \{x \in \mathbb{R} | -1 < x < 2\}$, and
@@ -153,24 +725,56 @@ j. $(A \cup B)^c$
 
 a. $A \cup B$
 
+$$ A \cup B = \{x \in \mathbb{R} | -3 \leq x < 2\} $$
+
 b. $A \cap B$
 
+$$ A \cap B = \{x \in \mathbb{R} | -1 < x \leq 0\} $$
+
 c. $A^c$
 
+$$ A^c = \{x \in \mathbb{R} | x < -3 \text{ or } x > 0 \} $$
+
 d. $A \cup C$
 
+$$ A \cup C = \{x \in \mathbb{R} | -3 \leq x \leq 0 \text{ or } 6 < x \leq 8\} $$
+
 e. $A \cap C$
 
+$$ A \cap C = \emptyset $$
+
 f. $B^c$
 
+$$ B^c = \{x \in \mathbb{R} | x \leq -1 \text{ or } x \geq 2 \} $$
+
 g. $A^c \cap B^c$
 
+$$ A^c = \{x \in \mathbb{R} | x < -3 \text{ or } x > 0 \} $$
+
+$$ B^c = \{x \in \mathbb{R} | x \leq -1 \text{ or } x \geq 2 \} $$
+
+$$ A^c \cap B^c = \{x \in \mathbb{R} | x < -3 \text{ or } x \geq 2 \} $$
+
 h. $A^c \cup B^c$
 
+$$ A^c = \{x \in \mathbb{R} | x < -3 \text{ or } x > 0 \} $$
+
+$$ B^c = \{x \in \mathbb{R} | x \leq -1 \text{ or } x \geq 2 \} $$
+
+$$ A^c \cup B^c = \{x \in \mathbb{R} | x \leq -1 \text{ or } x > 0 \} $$
+
 i. $(A \cap B)^c$
 
+$$ A \cap B = \{x \in \mathbb{R} | -1 < x \leq 0\} $$
+
+$$ (A \cap B)^c = \{x \in \mathbb{R} | x \leq -1 \text{ or } x > 0\} $$
+
 j. $(A \cup B)^c$
 
+$$ A \cup B = \{x \in \mathbb{R} | -3 \leq x < 2\} $$
+
+$$ (A \cup B)^c = \{x \in \mathbb{R} | x < -3 \text{ or } x \geq 2 \}$$
+
 13. Let $S$ be the set of all strings of $0$'s and $1$'s of length $4$, and let
     $A$ and $B$ be the following subsets of $S$:
     $A = \{1110, 1111, 1000, 1001\}$ and $B = \{1100, 0100, 1111, 0111\}$. Find
@@ -178,66 +782,146 @@ j. $(A \cup B)^c$
 
 a. $A \cap B$
 
+$$ A \cap B = \{1111\} $$
+
 b. $A \cup B$
 
+$$ A \cup B = \{1100, 0100, 1110, 1111, 0111, 1000, 1001\} $$
+
 c. $A - B$
 
+$$ A - B = \{1110, 1000, 1001\} $$
+
 d. $B - A$
 
+$$ B - A = \{1100, 0100, 0111\} $$
+
 14. In each of the following, draw a Venn diagram for sets $A$, $B$, and $C$
     that satisfy the given conditions.
 
 a. $A \subseteq B$, $C \subseteq B$, $A \cap C = \emptyset$
 
+Done physically.
+
 b. $C \subseteq A$, $B \cap C = \emptyset$
 
+Done physically.
+
 15. In each of the following, draw a Venn diagram for sets $A$, $B$, and $C$
     that satisfy the given conditions.
 
 a. $A \cap B = \emptyset$, $A \subseteq C$, $C \cap B \neq \emptyset$
 
+Done physically.
+
 b. $A \subseteq B$, $C \subseteq B$, $A \cap C \neq \emptyset$
 
+Done physically.
+
 c. $A \cap B \neq \emptyset$, $B \cap C \neq \emptyset$, $A \cap C = \emptyset$,
 $A \nsubseteq B$, $C \nsubseteq B$
 
+Done physically.
+
 16. Let $A = \{a, b, c\}$, $B = \{b, c, d\}$, and $C = \{b, c, e\}$.
 
 a. Find $A \cup (B \cap C)$, $(A \cup B) \cap C$, and
 $(A \cup B) \cap (A \cup C)$. Which of these sets are equal?
 
+$$ B \cap C = \{b, c\} $$
+
+$$ A \cup (B \cap C) = \{a, b, c\} $$
+
+$$ A \cup B = \{a, b, c, d\} $$
+
+$$ (A \cup B) \cap C = \{b, c\} $$
+
+$$ A \cup C = \{a, b, c, e\} $$
+
+$$ (A \cup B) \cap (A \cup C) = \{a, b, c\} $$
+
+$$ A \cup (B \cap C) = (A \cup B) \cap (A \cup C) $$
+
 b. Find $A \cap (B \cup C)$, $(A \cap B) \cup C$, and
 $(A \cap B) \cup (A \cap C)$. Which of these sets are equal?
 
+$$ B \cup C = \{b, c, d, e\} $$
+
+$$ A \cap (B \cup C) = \{b, c\} $$
+
+$$ A \cap B = \{b, c\} $$
+
+$$ (A \cap B) \cup C = \{b, c, e\} $$
+
+$$ A \cap C = \{b, c\} $$
+
+$$ (A \cap B) \cup (A \cap C) = \{b, c\} $$
+
+$$ A \cap (B \cup C) = A \cap C = (A \cap B) \cup (A \cap C) $$
+
 c. Find $(A - B) - C$ and $A - (B - C)$. Are these sets equal?
 
+$$ A - B = \{a\} $$
+
+$$ (A - B) - C = \{a\} $$
+
+$$ B - C = \{d\} $$
+
+$$ A - (B - C) = \{a, b, c\} $$
+
+$$ (A - B) - C \neq A - (B - C) $$
+
 17. Consider the following Venn diagram. For each of (a)-(f), copy the diagram
     and shade the region corresponding to the indicated set.
 
 a. $A \cap B$
 
+Omitted.
+
 b. $B \cup C$
 
+Omitted.
+
 c. $A^c$
 
+Omitted.
+
 d. $A - (B \cup C)$
 
+Omitted.
+
 e. $(A \cup B)^c$
 
+Omitted.
+
 f. $A^c \cap B^c$
 
+Omitted.
+
 (See page 412 for image)
 
 18.
 
 a. Is the number $0$ in $\emptyset$? Why?
 
+No, by the definition of $\emptyset$, there are no elements in $\emptyset$. In
+other words $\emptyset \neq \{0\}$.
+
 b. Is $\emptyset = \{\emptyset\}$? Why?
 
+No, by the definition of $\emptyset$, there are no elements in $\emptyset$. In
+other words $\emptyset \neq \{\emptyset\}$.
+
 c. Is $\emptyset \in \{\emptyset\}$ Why?
 
+Yes, because $\emptyset$ itself can be an element in a set, it is true that
+$\emptyset \in \{\emptyset\}$.
+
 d. Is $\emptyset \in \emptyset$? Why?
 
+No, by the definition of $\emptyset$, it is empty, it has no elements, therefore
+$\emptyset$ cannot contain itself. $\emptyset \notin \emptyset$.
+
 19. Let $A_i = \{i, i^2\}$ for each integer $i = 1, 2, 3, 4$.
 
 a. $A_1 \cup A_2 \cup A_3 \cup A_4 = \text{ ?}$

chapter_6/test_yourself.md

@@ -4,25 +4,51 @@ Page 411$a
 
 1. The notation $A \subseteq B$ is read "_____" and means that _____.
 
+The set $A$ is a subset of the set $B$; if $x \in A$ then $x \in B$
+
 2. To use an element argument for proving that a set $X$ is a subset of a set
    $Y$, you suppose that _____ and show that _____.
 
+$x$ is a particular but arbitrarily chosen element of $X$; $x$ is an element of
+$Y$.
+
 3. To disprove that a set $X$ is a subset of a set $Y$, you show that there is
    _____.
 
+an element in $X$ that is not in $Y$.
+
 4. An element $x$ is in $A \cup B$ if, and only if, _____.
 
+$x$ is in either $A$ or $B$.
+
 5. An element $x$ is in $A \cap B$ if, and only if, _____.
 
+$x$ is in both $A$ and $B$.
+
 6. An element $x$ is in $B - A$ if, and only if, _____.
 
+$x$ is in $B$ but not in $A$.
+
 7. An element $x$ is in $A^c$ if, and only if, _____.
 
+$x$ is in the universal set and is not in $A$.
+
 8. The empty set is a set with _____.
 
+no elements.
+
 9. The power set of a set $A$ is _____.
 
+the set of all subsets of $A$.
+
 10. Sets $A$ and $B$ are disjoint if, and only if, _____.
 
+they have no elements in common, or $A \cap B = \emptyset$.
+
 11. A collection of nonempty sets $A_1, A_2, A_3, \dots$ is a partition of a set
     $A$ if, and only if, _____.
+
+all $A_i$ are a subset of $A$, but are also disjoint.
+
+$A$ is the union of all the sets $A_1, A_2, A_3, \dots$ and
+$A_i \cap A_j = \emptyset$ whenever $i \neq j$.