1392 lines
33 KiB
Markdown
1392 lines
33 KiB
Markdown
Page 411
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**Exercise Set 6.1**
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1. In each of (a) -(f), answer the following questions: Is $A \subseteq B$? Is
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$B \subseteq A$? Is either $A$ or $B$ a proper subset of the other?
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a. $A = \{2, \{2\}, (\sqrt{2})^2\}$, $B = \{2, \{2\}, \{\{2\}\}\}$
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$A \subseteq B$ ?:
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$$ A = \{2, \{2\}, (\sqrt{2})^2\} = \{2, \{2\}, 2\} = \{2, \{2\}\} $$
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Yes, every element in $A$ is in $B$.
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$B \subseteq A$ ?:
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No, because $\{\{2\}\}$ is an element of $B$, but is not an element of $A$, so
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$\B \nsubseteq A$.
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Is either $A$ or $B$ a proper subset of the other?
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Yes, $A$ is a proper subset of $B$, because every element in $A$ is in $B$, but
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not every element in $B$ is in $A$.
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b. $A = \{3, \sqrt{5^2 - 4^2}, 24 \mod 7\}$, $B = \{8 \mod 5\}$
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$A \subseteq B$ ?:
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$$ A = \{3, \sqrt{5^2 - 4^2}, 24 \mod 7\} = \{3, 3, 3\} = \{3\} $$
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$$ B = \{8 \mod 5\} = \{3\} $$
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Yes, $A$ is a subset of $B$ since every element of $A$ is in $B$.
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$B \subseteq A$ ?:
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Yes, $B$ is a subset of $A$ since every element of $B$ is in $A$.
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Is either $A$ or $B$ a proper subset of the other?
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Yes, both $A$ and $B$ are proper subsets of the other since $A = B$.
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c. $A = \{\{1, 2\}, \{2, 3\}\}$, $B = \{1, 2, 3\}$
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$A \subseteq B$ ?:
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No, because there are no elements in $A$ that are in $B$, $A \nsubseteq B$
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$B \subseteq A$ ?:
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No, because there are no elements in $B$ that are in $A$, $B \nsubseteq A$
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Is either $A$ or $B$ a proper subset of the other?
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No, since neither set share any elements, neither is a proper subset of the
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other.
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d. $A = \{a, b, c\}$, $B = \{\{a\}, \{b\}, \{c\}\}$
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$A \subseteq B$ ?:
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No, because there are no elements in $A$ that are in $B$, $A \nsubseteq B$
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$B \subseteq A$ ?:
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No, because there are no elements in $B$ that are in $A$, $B \nsubseteq A$
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Is either $A$ or $B$ a proper subset of the other?
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No, since neither set share any elements, neither is a proper subset of the
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other.
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e. $A = \{\sqrt{16}, \{4\}\}$, $B = \{4\}$
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$A \subseteq B$ ?:
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$$ A = \{\sqrt{16}, \{4\}\} = \{4, \{4\}\} $$
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No, because every element of $A$ is not an element in $B$ ($4$ is not in $B$),
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$A \nsubseteq B$.
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$B \subseteq A$ ?:
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Yes, because every element in $B$ is an element in $A$, $B \subseteq A$.
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Is either $A$ or $B$ a proper subset of the other?
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Yes, $B$ is a proper subset of $A$ since $B \subseteq A$ and $A \nsubseteq B$.
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f. $A = \{x \in \mathbb{R} | \cos x \in \mathbb{Z}\}$,
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$B = \{x \in \mathbb{R} | \sin x \in \mathbb{Z}\}$
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From trigonometry, we know that $\cos x = -1 \text{ or } 0 \text{ or } 1$ and
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$\sin x = -1 \text{ or } 0 \text{ or } 1 $. When we evaluate for $x$ in these
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cases we find:
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$$ A = \{\dots, -\frac{5\pi}{2}, -\frac{3\pi}{2}, \pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots\} $$
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$$ B = \{\dots, -\frac{5\pi}{2}, -\frac{3\pi}{2}, \pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots\} $$
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$A \subseteq B$ ?: Yes.
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$B \subseteq A$ ?: Yes.
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Yes, $B$ is a proper subset of $A$ since $B \subseteq A$ and $A \nsubseteq B$.
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Yes, since $A = B$.
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2. Complete the proof from Example 6.1.3: Prove that $B \subseteq A$ where
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$$ A = \{m \in \mathbb{Z} | m, = 2a \text{ for some integer } a\} $$
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and
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$$ B = \{n \in \mathbb{Z} | n = 2b - 2 \text{ for some integer } b\} $$
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_Part 2, Proof that $B \subseteq A$:_
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Suppose $x$ is a particular but arbitrarily chosen element of $B$.
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By definition of $B$, there is an integer, say $b$, such that $x = 2b - 2$.
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To prove that $B \subseteq A$, we must show that there is some $x$ that can
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equal both $2a$, for some integer $a$, and that same $x$ can also equal
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$2b - 2$.
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$$ 2b - 2 = 2a $$
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$$ a = b - 1 $$
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By the difference integers, $a$ is an integer. Then, by substitution:
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$$ 2a = 2(b - 1) $$
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$$ = 2b - 2 $$
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$$ = x $$
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Thus, by definition of $A$, $x$ is an element of $A$.
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Q.E.D.
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3. Let sets $R$, $S$, and $T$ be defined as follows:
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$$ R = \{x \in \mathbb{Z} | x \text{ is divisible by } 2\} $$
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$$ S = \{y \in \mathbb{Z} | y \text{ is divisible by } 3\} $$
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$$ T = \{z \in \mathbb{Z} | z \text{ is divisible by } 6\} $$
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Prove or disprove each of the following statements.
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a. $R \subseteq T$
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$R \nsubseteq T$ since $2 \in R$ since $2 \mid 2$, but $2 \notin T$ since
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$6 \cancel{\mid} 2$.
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b. $T \subseteq R$
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**Proof:**
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Suppose $n$ is any integer such that $6 \mid n$, therefore $n \in T$.
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By the definition of divisibility:
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$$ n = 6m $$
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for some integer $m$.
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$$ n = 2(3m) $$
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By the product of integers, $3m$ is an integer. It follows that
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$n = 2 \cdot (\text{some integer})$. Thus $2 \mid n$, so $n \in R$. This is what
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was to be shown.
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Q.E.D.
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c. $T \subseteq S$
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**Proof:**
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Suppose $n$ is any integer such that $6 \mid n$, therefore $n \in T$.
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By the definition of divisibility:
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$$ n = 6m $$
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for some integer $m$.
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$$ n = 3(2m) $$
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By the product of integers, $2m$ is an integer. It follows that
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$n = 3 \cdot (\text{some integer})$. Thus $3 \mid n$, so $n \in S$. This is what
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was to be shown.
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Q.E.D.
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4. Let $A = \{n \in \mathbb{Z} | n = 5r \text{ for some integer } r\}$ and
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$B = \{m \in \mathbb{Z} | m = 20s \text{ for some integer } s\}$. Prove or
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disprove each of the following statements.
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a. $A \subseteq B$
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$A \nsubseteq B$ since $5 \in A$ since $5 \mid 5$, but $5 \notin B$ since
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$5 \cancel{\mid} 20$.
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b. $B \subseteq A$
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**Proof:**
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Suppose $n$ is any integer such that $20 \mid n$, therefore $n \in B$.
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By the definition of divisibility:
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$$ n = 20m $$
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for some integer $m$.
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$$ n = 5(4m) $$
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By the product of integers, $4m$ is an integer. It follows that
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$n = 5 \cdot (\text{some integer})$. Thus $5 \mid n$, so $n \in A$. This is what
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was to be shown.
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Q.E.D.
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5. Let $C = \{n \in \mathbb{Z} | n = 6r - 5 \text{ for some integer } r\}$ and
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$D = \{m \in \mathbb{Z} | m = 3s + 1 \text{ for some integer } s\}$. Prove or
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disprove each of the following statements.
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a. $C \subseteq D$
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**Proof:**
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Suppose $n$ is any integer such that $n = 6r - 5$ for some integer $r$, which
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means that $n \in C$.
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Also suppose that $m$ is any integer such that $m = 3s + 1$ for some integer
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$s$, which means that $m \in S$.
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We must show that there exists some $r$ that when substituted for $s$ will
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satisfy the definition of $n$.
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Let $s = 2r - 2$. Then, by substitution:
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$$ m = 3(2r - 2) + 1 $$
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$$ = 6r - 6 + 1 $$
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$$ = 6r - 5 $$
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$$ = n $$
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By the product and difference of integers, $6r - 5$ is an integer, therefore
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$n \in D$. This is what was to be shown.
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Q.E.D.
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b. $D \subseteq C$
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**Disproof:**
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$D \nsubseteq C$ because there are elements in $D$ that are not in $C$. For
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example $4$ is in $D$ because $4 = 3(1) + 1$, but $4$ is not in $C$. If $4$ were
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in $C$, this would mean:
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$$ 4 = 6r - 5 $$
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for some integer $r$.
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$$ 4 + 5 = 6r $$
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$$ 9 = 6r $$
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$$ \frac{9}{6} = r $$
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$$ \frac{3}{2} = r $$
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But $\dfrac{3}{2}$ is not an integer. This is a contradiction, therefore
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$D \nsubseteq C$.
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Q.E.D.
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6. Let $A = \{x \in \mathbb{Z} | x = 5a + 2 \text{ for some integer } a\}$,
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$B = \{y \in \mathbb{Z} | y = 10b - 3 \text{ for some integer } b\}$, and
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$C = \{z \in \mathbb{Z} | z = 10c + 7 \text{ for some integer } c\}$.
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Prove or disprove each of the following statements.
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a. $A \subseteq B$
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**Disproof (by counterexample):**
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Suppose $n$ is any integer such that $n = 5a + 2$ for some integer $a$. This
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means that $n \in A$.
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Suppose also that there is some integer $m$ such that $m = 10b - 3$ for some
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integer $b$. This means that $m \in B$.
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To show that there is some integer $b$ that will satisfy $n$, we must relate it
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to $a$:
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$$ 5a + 2 = 10b - 3 $$
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$$ 5a + 5 = 10b $$
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$$ 5a + 5 = 10b $$
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$$ \frac{1}{2}a + \frac{1}{2} = b $$
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$$ \frac{a + 1}{2} = b $$
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In order for $A \subseteq B$, every element of $A$ must be in $B$. If $a = 0$,
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then $n = 2$, so $n \in A$. If $a = 0$, then $b = \dfrac{1}{2}$, which is not an
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integer, thus $2 \notin B$. Therefore $A \nsubseteq B$.
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Q.E.D.
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b. $B \subseteq A$
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**Proof:**
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Suppose $y$ is any integer such that $y = 10b - 3$ for some integer $b$. This
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means that $y \in B$.
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Let's first find $a$ as it relates to $y$.
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$$ y = 5a + 2 $$
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$$ 10b - 3 = 5a + 2 $$
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$$ 10b - 5 = 5a $$
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$$ 2b - 1 = a $$
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So let $a = 2b - 1$.
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Then substitute in for the condition for $A$:
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$$ x = 5a + 2 $$
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$$ = 5(2b - 1) + 2 $$
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$$ = 10b - 5 + 2 $$
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$$ = 10b - 3 $$
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$$ = y $$
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Therefore $y \in A$.
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Q.E.D.
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c. $B = C$
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To prove $B = C$, we must prove both that $B \subseteq C$ and $C \subseteq B$.
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_Prove $B \subseteq C$:
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Suppose $y$ is any integer such that $y = 10b - 3$ for some integer $b$. This
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means that $y \in B$.
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Let's first find some integer $c$ as it relates to $y$:
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$$ y = 10c + 7 $$
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$$ 10b - 3 = 10c + 7 $$
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$$ 10b - 10 = 10c $$
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$$ b - 1 = c $$
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So, let $c = b - 1$.
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Then substitute in for the condition for $C$:
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$$ z = 10c + 7 $$
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$$ = 10(b - 1) + 7 $$
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$$ = 10b - 10 + 7 $$
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$$ = 10b - 3 $$
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$$ = y $$
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Thus $y \in C$, and therefore $B \subseteq C$.
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_Prove $C \subseteq B$:
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Suppose $z$ is any integer such that $z = 10c + 7$ for some integer $c$. This
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means that $z \in C$.
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Let's first find some integer $b$ as it relates to $z$:
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$$ z = 10b - 3 $$
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$$ 10c + 7 = 10b - 3 $$
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$$ 10c + 10 = 10b $$
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$$ c + 1 = b $$
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So, let $b = c + 1$.
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Then substitute in for the condition for $B$:
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$$ y = 10b - 3 $$
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$$ = 10(c + 1) - 3 $$
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$$ = 10c + 10 - 3 $$
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$$ = 10c + 7 $$
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$$ = z $$
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Therefore $z \in B$.
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Thus $z \in B$, and therefore $C \subseteq B$.
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Since $B \subseteq C$ and $C \subseteq B$, it follows that $B = C$. This is what
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was to be shown.
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Q.E.D.
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7. Let $A = \{x \in \mathbb{Z} | x = 6a + 4 \text{ for some integer } a\}$,
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$B = \{y \in \mathbb{Z} | y = 18b - 2 \text{ for some integer } b\}$, and
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$C = \{z \in \mathbb{Z} | z = 18c + 16 \text{ for some integer } c\}$.
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Prove or disprove each of the following statements.
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a. $A \subseteq B$
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**Disproof (by counterexample):**
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Suppose $x$ is any integer such that $x = 6a + 4$ for some integer $a$. This
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means that $x \in A$.
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Let's first find some integer $b$ as it relates to $a$.
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$$ x = 18b - 2 $$
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$$ 6a + 4 = 18b - 2 $$
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$$ 6a + 6 = 18b $$
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$$ \frac{6}{18}a + \frac{6}{18} = b $$
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$$ \frac{1}{3}a + \frac{1}{3} = b $$
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$$ \frac{a + 1}{3} = b $$
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By definition of $b$, $b$ must always be an integer for all $a$.
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Suppose $a = 0$, then:
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$$ x = 6(0) + 4 = 4 $$
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so $4 \in A$, but:
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$$ b = \frac{0 + 1}{3} = \frac{1}{3} $$
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so $4 \notin B$. We can see this as $4 = 18b - 2$ results in $b = \dfrac{1}{3}$,
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but $b$ must be an integer.
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b. $B \subseteq A$
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**Proof:**
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Suppose $y$ is any integer such that $y = 18b - 2$ for some integer $b$. This
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means that $y \in B$.
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Let's first find some integer $a$ as it relates to $b$.
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$$ y = 6a + 4 $$
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$$ 18b - 2 = 6a + 4 $$
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$$ 18b - 6 = 6a $$
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$$ 3b - 1 = a $$
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Now, substitute $a$ in for the condition for $A$:
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$$ x = 6a + 4 $$
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$$ = 6(3b - 1) + 4 $$
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$$ = 18b - 2 $$
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$$ = y $$
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Therefore $B \subseteq A$.
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c. $B = C$
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To prove $B = C$, we must prove both that $B \subseteq C$ and $C \subseteq B$.
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_Prove $B \subseteq C$:
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Suppose $y$ is any integer such that $y = 18b - 2$ for some integer $b$. This
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means that $y \in B$.
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Let's first find some integer $c$ as it relates to $b$.
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$$ y = 18c + 16 $$
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$$ 18b - 2 = 18c + 16 $$
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$$ 18b - 18 = 18c $$
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$$ b - 1 = c $$
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So, let $c = b - 1$. Now substitute $c$ in for the condition of $C$:
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$$ z = 18c + 16 $$
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$$ = 18(b - 1) + 16 $$
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$$ = 18b - 18 + 16 $$
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$$ = 18b - 2 $$
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$$ = y $$
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Therefore $B \subseteq C$.
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_Prove $C \subseteq B$:
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Suppose $z$ is any integer such that $z = 18c + 16$ for some integer $c$.
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Let's first find some $b$ as it relates to $c$.
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$$ z = 18b - 2 $$
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$$ 18c + 16 = 18b - 2 $$
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$$ 18c + 18 = 18b $$
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$$ c + 1 = b $$
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So, let $b = c + 1$. Now, let's substitute $b$ in for the condition for $B$.
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$$ y = 18b - 2 $$
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$$ = 18(c + 1) - 2 $$
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$$ = 18c + 18 - 2 $$
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$$ = 18c + 16 $$
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$$ = z $$
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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
|
|
$C = \{x \in \mathbb{R} | 3 \leq x < 9\}$. Find each of the following:
|
|
|
|
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
|
|
$C = \{x \in \mathbb{R} | 6 < x \leq 8\}$. Find each of the following:
|
|
|
|
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
|
|
each of the following:
|
|
|
|
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. $A_1 \cup A_2 \cup A_3 \cup A_4 = \text{ ?}$
|
|
|
|
$$
|
|
A_1 = \{1, 1^2\} = \{1, 1\} = \{1\} \\
|
|
A_2 = \{2, 2^2\} = \{2, 4\} \\
|
|
A_3 = \{3, 3^2\} = \{3, 9\} \\
|
|
A_4 = \{4, 4^2\} = \{4, 16\} \\
|
|
$$
|
|
|
|
$$ A_1 \cup A_2 \cup A_3 \cup A_4 = \{1, 2, 3, 4, 9, 16\} $$
|
|
|
|
b. $A_1 \cap A_2 \cap A_3 \cap A_4 = \text{ ?}$
|
|
|
|
$$ A_1 \cap A_2 \cap A_3 \cap A_4 = \emptyset $$
|
|
|
|
c. Are $A_1, A_2, A_3$, and $A_4$ mutually disjoint? Explain.
|
|
|
|
No, since $A_2$ and $A_4$ both contain the element $4$, they are not mutually
|
|
disjoint.
|
|
|
|
20. Let $B_i = \{x \in \mathbb{R} | 0 \leq x\leq i\}$ for each integer
|
|
$i = 1, 2, 3, 4$.
|
|
|
|
a. $B_1 \cup B_2 \cup B_3 \cup B_4 = \text{ ?}$
|
|
|
|
$$ B_1 \cup B_2 \cup B_3 \cup B_4 = \{x \in \mathbb{R} | 0 \leq x \leq 4\} $$
|
|
|
|
b. $B_1 \cap B_2 \cap B_3 \cap B_4 = \text{ ?}$
|
|
|
|
$$ B_1 \cap B_2 \cap B_3 \cap B_4 = \{x \in \mathbb{R} | 0 \leq x \leq 1\} $$
|
|
|
|
c. Are $B_1, B_2, B_3$, and $B_4$ mutually disjoint? Explain.
|
|
|
|
No, since all sets include all real numbers within the range $0 \leq x \leq 1$,
|
|
they are not mutually disjoint.
|
|
|
|
21. Let $C_i = \{i, -i\}$ for each nonnegative integer $i$.
|
|
|
|
$$
|
|
C_0 = \{0, -0\} = \{0\} \\
|
|
C_1 = \{1, -1\} \\
|
|
C_2 = \{2, -2\} \\
|
|
C_3 = \{3, -3\} \\
|
|
C_4 = \{4, -4\} \\
|
|
$$
|
|
|
|
a. $\bigcup_{i = 0}^{4}C_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 0}^{4}C_i = C_0 \cup C_1 \cup C_2 \cup C_3 \cup C_4 $$
|
|
|
|
$$ \bigcup_{i = 0}^{4}C_i = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\} $$
|
|
|
|
b. $\bigcap_{i = 0}^{4}C_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 0}^{4}C_i = \emptyset $$
|
|
|
|
c. Are $C_0, C_1, C_2, \dots$ mutually disjoint? Explain.
|
|
|
|
Yes, since none of the sets have any elements in common, they are mutually
|
|
disjoint.
|
|
|
|
d. $\bigcup_{i = 0}^{n}C_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 0}^{n}C_i = \{-n, -(n - 1), \dots -2, -1, 0, 1, 2, \dots (n - 1), n\} $$
|
|
|
|
e. $\bigcap_{i = 0}^{n}C_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 0}^{n}C_i = \emptyset $$
|
|
|
|
f. $\bigcup_{i = 0}^{\infty}C_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 0}^{\infty}C_i = \{-\infty, \dots, -2, -1, 0, 1, 2, \dots, \infty\} = \mathbb{Z} $$
|
|
|
|
g. $\bigcap_{i = 0}^{\infty}C_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 0}^{\infty}C_i = \emptyset $$
|
|
|
|
22. Let $D_i = \{x \in \mathbb{R} | -i \leq x \leq i\} = [-i, i]$ for each
|
|
nonnegative integer $i$.
|
|
|
|
$$
|
|
D_0 = [-0, 0] = \{0\} \\
|
|
D_1 = [-1, 1] \\
|
|
D_2 = [-2, 2] \\
|
|
D_3 = [-3, 3] \\
|
|
D_4 = [-4, 4] \\
|
|
$$
|
|
|
|
a. $\bigcup_{i = 0}^{4}D_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 0}^{4}D_i = \{x \in \mathbb{R} | -4 \leq x \leq 4\} = [-4, 4] $$
|
|
|
|
b. $\bigcap_{i = 0}^{4}D_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 0}^{4}D_i = \{0\} $$
|
|
|
|
c. Are $D_0, D_1, D_2, \dots$ mutually disjoint? Explain.
|
|
|
|
No, in fact all sets have at least $\{0}$ in common , as $i$ increases, so does
|
|
the amount of elements all sets have in common, or $D_k \subseteq D_{k + 1}$.
|
|
|
|
d. $\bigcup_{i = 0}^{n}D_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 0}^{n}D_i = \{x \in \mathbb{R} | -n \leq x \leq n\} = [-n, n] $$
|
|
|
|
e. $\bigcap_{i = 0}^{n}D_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 0}^{n}D_i = \{0\} $$
|
|
|
|
f. $\bigcup_{i = 0}^{\infty}D_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 0}^{\infty}D_i = (-\infty, \infty) = \mathbb{R} $$
|
|
|
|
g. $\bigcap_{i = 0}^{\infty}D_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 0}^{\infty}D_i = \{0\} $$
|
|
|
|
23. Let
|
|
$V_i = \{x \in \mathbb{R} | -\dfrac{1}{i} \leq x \leq \dfrac{1}{i}\} = \left[-\dfrac{1}{i}, \dfrac{1}{i}\right]$
|
|
for each positive integer $i$.
|
|
|
|
$$
|
|
V_1 = \left[-\frac{1}{1}, \frac{1}{1}\right] = [-1, 1] \\
|
|
V_2 = \left[-\frac{1}{2}, \frac{1}{2}\right] \\
|
|
V_3 = \left[-\frac{1}{3}, \frac{1}{3}\right] \\
|
|
V_4 = \left[-\frac{1}{4}, \frac{1}{4}\right] \\
|
|
$$
|
|
|
|
a. $\bigcup_{i = 1}^{4}V_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 1}^{4}V_i = [-1, 1] $$
|
|
|
|
b. $\bigcap_{i = 1}^{4}V_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 1}^{4}V_i = \left[-\frac{1}{4}, \frac{1}{4}\right] $$
|
|
|
|
c. Are $V_1, V_2, V_3, \dots$ mutually disjoint? Explain.
|
|
|
|
No, every set includes $0$.
|
|
|
|
d. $\bigcup_{i = 1}^{n}V_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 1}^{n}V_i = [-1, 1] $$
|
|
|
|
e. $\bigcap_{i = 1}^{n}V_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 1}^{n}V_i = \left[-\frac{1}{n}, \frac{1}{n}\right] $$
|
|
|
|
f. $\bigcup_{i = 1}^{\infty} = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 1}^{\infty} = [-1, 1] $$
|
|
|
|
g. $\bigcap_{i = 1}^{\infty} = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 1}^{\infty} = \{0\} \text{ because as } i \to \infty \text{ then } \frac{1}{i} \to 0 $$
|
|
|
|
24. Let $W_i = \{x \in \mathbb{R} | x > i\} = (i, \infty)$ for each nonnegative
|
|
integer $i$.
|
|
|
|
$$
|
|
W_0 = (0, \infty) \\
|
|
W_1 = (1, \infty) \\
|
|
W_2 = (2, \infty) \\
|
|
W_3 = (3, \infty) \\
|
|
W_4 = (4, \infty) \\
|
|
$$
|
|
|
|
a. $\bigcup_{i = 0}^{4}W_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 0}^{4}W_i = (0, \infty) $$
|
|
|
|
b. $\bigcap_{i = 0}^{4}W_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 0}^{4}W_i = (4, \infty) $$
|
|
|
|
c. Are $W_0, W_1, W_2, \dots$ mutually disjoint? Explain.
|
|
|
|
No, because they all have $(i, \infty)$ in common, or $W_{i + 1} \subseteq W_i$.
|
|
|
|
d. $\bigcup_{i = 0}^{n}W_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 0}^{n}W_i = (0, \infty) $$
|
|
|
|
e. $\bigcap_{i = 0}^{n}W_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 0}^{n}W_i = (n, \infty) $$
|
|
|
|
f. $\bigcup_{i = 0}^{\infty}W_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 0}^{\infty}W_i = (0, \infty) $$
|
|
|
|
g. $\bigcap_{i = 0}^{\infty}W_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 0}^{\infty}W_i = \emptyset $$
|
|
|
|
There is no real number greater than every positive integer, so no element
|
|
belongs to all $W_i$.
|
|
|
|
25. Let
|
|
$R_i = \{x \in \mathbb{R} | 1 \leq x \leq 1 + \dfrac{1}{i}\} = \left[1, 1 + \dfrac{1}{i}\right]$
|
|
for each positive integer $i$.
|
|
|
|
$$
|
|
R_1 = \left[1, 1 + \frac{1}{1}\right] = [1, 2] \\
|
|
R_2 = \left[1, 1 + \frac{1}{2}\right] = \left[1, \frac{3}{2}\right] \\
|
|
R_3 = \left[1, 1 + \frac{1}{3}\right] = \left[1, \frac{4}{3}\right] \\
|
|
R_4 = \left[1, 1 + \frac{1}{4}\right] = \left[1, \frac{5}{4}\right] \\
|
|
$$
|
|
|
|
a. $\bigcup_{i = 1}^{4}R_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 1}^{4}R_i = [1, 2] $$
|
|
|
|
b. $\bigcap_{i = 1}^{4}R_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 1}^{4}R_i = \left[1, \frac{5}{4}\right] $$
|
|
|
|
c. Are $R_1, R_2, R_3, \dots$ mutually disjoint? Explain.
|
|
|
|
No, they all include the element $1$.
|
|
|
|
d. $\bigcup_{i = 1}^{n}R_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 1}^{n}R_i = [1, 2] $$
|
|
|
|
e. $\bigcap_{i = 1}^{n}R_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 1}^{n}R_i = \left[1, 1 + \frac{1}{n}\right]$$
|
|
|
|
f. $\bigcup_{i = 1}^{\infty}R_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 1}^{\infty}R_i = [1, 2] $$
|
|
|
|
g. $\bigcap_{i = 1}^{\infty}R_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 1}^{\infty}R_i = \{1\} $$
|
|
|
|
Because $\dfrac{1}{\infty} \to 0$ and
|
|
$\left(1 + \dfrac{1}{\infty}\right) \to 1$.
|
|
|
|
26. Let
|
|
$S_i = \{x \in \mathbb{R} | 1 < x < 1 + \dfrac{1}{i}\} = \left(1, 1 + \dfrac{1}{i}\right)$
|
|
for each positive integer $i$.
|
|
|
|
$$
|
|
S_1 = \left(1, 1 + \frac{1}{1}\right) = (1, 2) \\
|
|
S_2 = \left(1, 1 + \frac{1}{2}\right) = \left(1, \frac{3}{2}\right) \\
|
|
S_3 = \left(1, 1 + \frac{1}{3}\right) = \left(1, \frac{4}{3}\right) \\
|
|
S_4 = \left(1, 1 + \frac{1}{4}\right) = \left(1, \frac{5}{4}\right) \\
|
|
$$
|
|
|
|
a. $\bigcup_{i = 1}^{4}S_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 1}^{4}S_i = (1, 2) $$
|
|
|
|
b. $\bigcap_{i = 1}^{4}S_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 1}^{4}S_i = \left(1, \frac{5}{4}\right) $$
|
|
|
|
c. Are $S_1, S_2, S_3, \dots$ mutually disjoint? Explain.
|
|
|
|
No, any element sufficiently close to $1$ are in all the sets.
|
|
|
|
d. $\bigcup_{i = 1}^{n}S_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 1}^{n}S_i = (1, 2) $$
|
|
|
|
e. $\bigcap_{i = 1}^{n}S_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 1}^{n}S_i = \left(1, 1 + \frac{1}{n}\right) $$
|
|
|
|
f. $\bigcup_{i = 1}^{\infty}S_i = \text{ ?}$
|
|
|
|
$$ \bigcup_{i = 1}^{\infty}S_i = (1, 2) $$
|
|
|
|
g. $\bigcap_{i = 1}^{\infty}S_i = \text{ ?}$
|
|
|
|
$$ \bigcap_{i = 1}^{\infty}S_i = \emptyset $$
|
|
|
|
Because the range converges on $1$, but cannot include $1$, the set is empty.
|
|
|
|
27.
|
|
|
|
a. Is $\{\{a, d, e\}, \{b, c\}, \{d, f\}\}$ a partition of
|
|
$\{a, b, c, d, e, f\}$?
|
|
|
|
No, since $d$ is an element in two sets, the sets are not mutually disjoint, and
|
|
so therefore is not a partition.
|
|
|
|
b. Is $\{\{w, x, v\}, \{u, y, q\}, \{p, z\}\}$ a partition of
|
|
$\{p, q, u, v, w, x, y, z\}$?
|
|
|
|
$$ \{w, x, v\} \cup \{u, y, q\} \cup \{p, z\} = \{p, q, u, v, w, x, y, z\} $$
|
|
|
|
and:
|
|
|
|
$$ \{w, x, v\} \cap \{u, y, q\} \cap \{p, z\} = \emptyset $$
|
|
|
|
So yes, the given sets are a partition of the overall set.
|
|
|
|
c. Is $\{\{5, 4\}, \{7, 2\}, \{1, 3, 4\}, \{6, 8\}\}$ a partition of
|
|
$\{1, 2, 3, 4, 5, 6, 7, 8\}$?
|
|
|
|
No, as $4$ is an element in two of the given sets, and so the given sets are not
|
|
a partition of the overall set.
|
|
|
|
d. Is $\{\{3, 7, 8\}, \{2, 9\}, \{1, 4, 5\}\}$ a partition of
|
|
$\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$?
|
|
|
|
No, since none of the sets contain $6$.
|
|
|
|
e. Is $\{\{1, 5\}, \{4, 7\}, \{2, 8, 6, 3\}\}$ a partition of
|
|
$\{1, 2, 3, 4, 5, 6, 7, 8\}$?
|
|
|
|
Yes, since none of the elements in each of the given sets are in any other of
|
|
the given sets and all of the elements make up the overall set.
|
|
|
|
28. Let $E$ be the set of all even integers and $O$ the set of all odd integers.
|
|
Is $\{E, O\}$ a partition of $\mathbb{Z}$, the set of all integers? Explain
|
|
your answer.
|
|
|
|
Yes, since no integer is both even and odd, and all integers are either even or
|
|
odd, $\{E, O\}$ is a partition of $\mathbb{Z}$.
|
|
|
|
29. Let $\mathbb{R}$ be the set of all real numbers. Is
|
|
$\{\mathbb{R}^+, \mathbb{R}^-, \{0\}\}$ a partition of $\mathbb{R}$? Explain
|
|
your answer.
|
|
|
|
Yes, since all real numbers are either positive, negative, or $0$, and
|
|
$\mathbb{R}^+$, $\mathbb{R}^-$ and $\{0\}$ do not have any elements in common,
|
|
these subsets all form a partition of $\mathbb{R}$.
|
|
|
|
30. Let $\mathbb{Z}$ be the set of all integers and let
|
|
|
|
$$ A_0 = \{n \in \mathbb{Z} | n = 4k, \text{ for some integer } k\} $$
|
|
|
|
$$ A_1 = \{n \in \mathbb{Z} | n = 4k + 1, \text{ for some integer } k\} $$
|
|
|
|
$$ A_2 = \{n \in \mathbb{Z} | n = 4k + 2, \text{ for some integer } k\} $$
|
|
|
|
and
|
|
|
|
$$ A_3 = \{n \in \mathbb{Z} | n = 4k + 3, \text{ for some integer } k\} $$
|
|
|
|
Is $\{A_0, A_1, A_2, A_3\}$ a partition of $\mathbb{Z}$? Explain your answer.
|
|
|
|
Yes. These sets are mutually disjoint, and by the quotient-remainder theorem,
|
|
every integer has exactly one of the forms $n = 4k$, $n = 4k + 1$, $n = 4k + 2$,
|
|
$n = 4k + 3$.
|
|
|
|
31. Suppose $A = \{1, 2\}$ and $B = \{2, 3\}$. Find each of the following:
|
|
|
|
a. $\mathscr{P}(A \cap B)$
|
|
|
|
$$ A \cap B = \{2\} $$
|
|
|
|
$$ \mathscr{P}(A \cap B) = \{\emptyset, \{2\}\} $$
|
|
|
|
b. $\mathscr{P}(A)$
|
|
|
|
$$ \mathscr{P}(A) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}} $$
|
|
|
|
c. $\mathscr{P}(A \cup B)$
|
|
|
|
$$ A \cup B = \{1, 2, 3\} $$
|
|
|
|
$$ \mathscr{P}(A \cup B) = \{\emptyset, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\} $$
|
|
|
|
d. $\mathscr{P}(A \times B)$
|
|
|
|
$$ A \times B = \{(1, 2), (1, 3), (2, 2), (2, 3)\} $$
|
|
|
|
$$ \mathscr{P}(A \times B) = \{\emptyset, \{(1, 2)\}, \{(1, 3)\}, \{(2, 2)\}, \{(2, 3)\}, \{(1, 2), (1, 3)\}, \{(1, 2), (2, 2)\}, \{(1, 2,), (2, 3)\}, \{(1, 3), (2, 2)\}, \{(1, 3), (2, 3)\}, \{(2, 2), (2, 3)\}, \{(1, 2), (1, 3), (2, 2)\}, \{(1, 2), (1, 3), (2, 3)\}, \{(1, 2), (2, 2), (2, 3)\}, \{(1, 3), (2, 2), (2, 3)\}, \{(1, 2), (1, 3), (2, 2), (2, 3)\}\} $$
|
|
|
|
32.
|
|
|
|
a. Suppose $A = \{1\}$ and $B = \{u, v\}$. Find $\mathscr{P}(A \times B)$.
|
|
|
|
$$ A \times B = \{(1, u), (1, v)\} $$
|
|
|
|
$$ \mathscr{P}(A \times B) = \{\emptyset, \{(1, u)\}, \{(1, v)\}, \{(1, u), (1, v)\}\} $$
|
|
|
|
b. Suppose $X = \{a, b\}$ and $Y = \{x, y\}$. Find $\mathscr{P}(X \times Y)$.
|
|
|
|
$$ X \times Y = \{(a, x), (a, y), (b, x), (b, y)\} $$
|
|
|
|
$$ \mathscr{P}(X \times Y) = \{\emptyset, \{(a, x)\}, \{(a, y)\}, \{(b, x)\}, \{(b, y)\}, \{(a, x), (a, y)\}, \{(a, x), (b, x)\}, \{(a, x), (b, y)\}, \{(a, y), (b, x)\}, \{(a, y), (b, y)\}, \{(b, x), (b, y)\}, \{(a, x), (a, y), (b, x)\}, \{(a, x), (a, y), (b, y)\}, \{(a, x), (b, x), (b, y)\}, \{(a, y), (b, x), (b, y)\} \{(a, x), (a, y), (b, x), (b, y)\}\} $$
|
|
|
|
33.
|
|
|
|
a. Find $\mathscr{P}(\emptyset)$.
|
|
|
|
$$ \mathscr{P}(\emptyset) = \{\emptyset\} $$
|
|
|
|
b. Find $\mathscr{P}(\mathscr{P}(\emptyset))$.
|
|
|
|
$$ \mathscr{P}(\mathscr{P}(\emptyset)) = \{\emptyset, \{\emptyset\}\} $$
|
|
|
|
b. Find $\mathscr{P}(\mathscr{P}(\mathscr{P}(\emptyset)))$.
|
|
|
|
$$ \mathscr{P}(\mathscr{P}(\mathscr{P}(\emptyset))) = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}, \{\{\emptyset\}\}\} $$
|
|
|
|
34. let $A_1 = \{1\}$, $A_2 = \{u, v\}$, and $A_3 = \{m, n\}$. Find each of the
|
|
following sets:
|
|
|
|
a. $A_1 \cup (A_2 \times A_3)$
|
|
|
|
$$ A_2 \times A_3 = \{(u, m), (u, n), (v, m), (v, n)\} $$
|
|
|
|
$$ A_1 \cup (A_2 \times A_3) = \{1, (u, m), (u, n), (v, m), (v, n)\} $$
|
|
|
|
b. $(A_1 \cup A_2) \times A_3$
|
|
|
|
$$ A_1 \cup A_2 = \{1, u, v\} $$
|
|
|
|
$$ (A_1 \cup A_2) \times A_3 = \{(1, m), (1, n), (u, m), (u, n), (v, m), (v, n)\} $$
|
|
|
|
35. let $A = \{a, b\}$, $B = \{1, 2\}$, and $C = \{2, 3\}$. Find each of the
|
|
following sets:
|
|
|
|
a. $A \times (B \cup C)$
|
|
|
|
$$ B \cup C = \{1, 2, 3\} $$
|
|
|
|
$$ A \times (B \cup C) = \{(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)\} $$
|
|
|
|
b. $(A \times B) \cup (A \times C)$
|
|
|
|
$$ A \times B = \{(a, 1), (a, 2), (b, 1), (b, 2)\} $$
|
|
|
|
$$ A \times C = \{(a, 2), (a, 3), (b, 2), (b, 3)\} $$
|
|
|
|
$$ (A \times B) \cup (A \times C) = \{(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)\} $$
|
|
|
|
c. $A \times (B \cap C)$
|
|
|
|
$$ B \cap C = \{2\} $$
|
|
|
|
$$ A \times (B \cap C) = \{(a, 2), (b, 2)\} $$
|
|
|
|
d. $(A \times B) \cap (A \times C)$
|
|
|
|
$$ A \times B = \{(a, 1), (a, 2), (b, 1), (b, 2)\} $$
|
|
|
|
$$ A \times C = \{(a, 2), (a, 3), (b, 2), (b, 3)\} $$
|
|
|
|
$$ (A \times B) \cap (A \times C) = \{(a, 2), (b, 2)\} $$
|
|
|
|
36. Trace the action of Algorithm 6.1.1 on the variables $i$, $j$,
|
|
$\text{found}$, and $\text{answer}$ for $m = 3$, $n = 3$, and sets $A$ and
|
|
$B$ represented as the arrays
|
|
$a[1] = u, a[2] = v, a[3] = w, b[1] = w, b[2] = u,$ and $b[3] = v$.
|
|
|
|
Omitted.
|
|
|
|
37. Trace the action of Algorithm 6.1.1 on the variables $i$, $j$,
|
|
$\text{found}$, and $\text{answer}$ for $m = 4$, $n = 4$ and sets $A$ and
|
|
$B$ represented as the arrays
|
|
$a[1] = u, a[2] = v, a[3] = w, a[4] = x, b[1] = r, b[2] = u, b[3] = y, b[4] = z$.
|
|
|
|
Omitted.
|
|
|
|
38. Write an algorithm to determine whether a given element $x$ belongs to a
|
|
given set that is represented as the array $a[1], a[2], \dots, a[n]$.
|
|
|
|
Omitted.
|