diff --git a/chapter_6/exercises.md b/chapter_6/exercises.md index 7f03b95..6bb0810 100644 --- a/chapter_6/exercises.md +++ b/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. $A_1 \cup A_2 \cup A_3 \cup A_4 = \text{ ?}$ diff --git a/chapter_6/test_yourself.md b/chapter_6/test_yourself.md index d8461f1..0c32179 100644 --- a/chapter_6/test_yourself.md +++ b/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 $i \neq j$.