🚧 Almost done with chapter 1
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@ -753,3 +753,431 @@ c. $R \times S \times T$
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**Solution**
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xxxxy, xxxyx, xxyxx, xyxxx, yxxxx
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---
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**Exercise Set 1.3**
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Page 45
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1. Let $A = \{2, 3, 4\}$ and $B = \{6, 8, 10\}$ and define a relation $R$ from
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$A$ to $B$ as follows: For every $(x, y) \in A \times B$,
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$$ (x, y) \in R \quad \text{ means that } \frac{y}{x} \text{ is an integer.} $$
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**Solution**
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a.
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Is 4 _R_ 6?
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No, $\dfrac{6}{4} = \dfrac{3}{2}$, which is not an integer.
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Is 4 _R_ 8?
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Yes, $\dfrac{8}{4} = 2$, which is an integer.
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Is $(3, 8) \in R$?
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No, $\dfrac{8}{3}$ is not an integer.
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Is $(2, 10) \in R$?
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Yes, $\dfrac{10}{2} = 5$ which is an integer.
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b. Write _R_ as a set of ordered pairs.
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$$ R = \{(2, 6), (2, 8), (2, 10), (3, 6), (4, 8)\} $$
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c. Write the domain and co-domain of _R_.
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The domain of _R_ is $\{2, 3, 4\}$.
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The co-domain of _R_ is $\{6, 8, 10\}$.
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d. Draw an arrow diagram for _R_.
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2. Let $C = D = \{-3, -2, -1, 1, 2, 3\}$ and define a relation $S$ from $C$ to
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$D$ as follows: For every $(x, y) \in C \times D$,
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$$ (x, y) \in S \quad \text{ means that } \frac{1}{x} - \frac{1}{y} \text{ is an integer.} $$
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**Solution**
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a.
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Is 2 _S_ 2?
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Yes, $\dfrac{1}{2} - \dfrac{1}{2} = 0 \in \mathbb{Z}$.
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Is -1 _S_ -1?
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Yes $\dfrac{1}{-1} - \dfrac{1}{-1} = 0 \in \mathbb{Z} $
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Is $(3, 3) \in S$?
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Yes $\dfrac{1}{3} - \dfrac{1}{3} = 0 \in \mathbb{Z} $
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Is $(3, -3) \in S$?
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No, $\dfrac{1}{3} - \dfrac{1}{-3} = \dfrac{2}{3} \notin \mathbb{Z} $
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b. Write _S_ as a set of ordered pairs.
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$$ S = \{(-3, -3), (-2, -2), (-1, -1), (1, 1), (2, 2), (3, 3)\} $$
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c. Write the domain and co-domain of _S_.
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The domain and co-domain of _S_ is $\{-3, -2, -1, 1, 2, 3\}$.
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d. Draw an arrow diagram for _S_.
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3. Let $E = \{1, 2, 3\}$ and $F = \{-2, -1, 0\}$ and define a relation $T$ from
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$E$ to $F$ as follows: For every $(x, y) \in E \times F$,
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$$ (x, y) \in T \quad \text{ means that } \frac{x - y}{3} \text{ is an integer.} $$
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**Solution**
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a.
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Is 3 _T_ 0?
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Yes, $\dfrac{3 - 0}{3} = 1 \in \mathbb{Z}$.
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Is 1 _T_ (-1)?
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No, $\dfrac{(1) - (-1)}{3} = \dfrac{2}{3} \notin \mathbb{Z}$.
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Is $(2, -1) \in T$?
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Yes, $\dfrac{(2) - (-1)}{3} = 1 \in \mathbb{Z}$.
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Is $(3, -2) \in T$?
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No, $\dfrac{(3) - (-2)}{3} = \dfrac{5}{3} \notin \mathbb{Z}$.
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b. Write $T$ as a set of ordered pairs.
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$$ T = \{(1, -2), (2, -1), (3, 0)\} $$
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c. Write the domain and co-domain of $T$.
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The domain of $T$ is $\{1, 2, 3\}$, and the co-domain of $T$ is $\{-2, -1, 0\}$.
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d. Draw an arrow diagram for $T$.
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4. Let $G = \{-2, 0, 2\}$ and $H = \{4, 6< 8\}$ and define a relation $V$ from
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$G$ to $H$ as follows: For every $(x, y) \in G \times H$,
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$$ (x, y) \in V \quad \text{ means that } \frac{x - y}{4} \text{ is an integer.} $$
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**Solution**
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a.
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Is 2 _V_ 6?
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Yes, $\dfrac{(2) - (6)}{4} = -1 \in \mathbb{Z}$.
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Is (-2) _V_ (8)?
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No, $\dfrac{(-2) - (8)}{4} = -\dfrac{10}{4} = -\dfrac{5}{2} \notin \mathbb{Z}$.
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Is $(0, 6) \in V$?
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No, $\dfrac{(0) - (6)}{4} = -\dfrac{6}{4} = -\dfrac{3}{2} \notin \mathbb{Z}$.
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Is $(2, 4) \in V$?
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No, $\dfrac{(2) - (4)}{4} = -\dfrac{1}{2} \notin \mathbb{Z}$.
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b. Write $V$ as a set of ordered pairs.
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$$ V = \{(-2, 6), (0, 4), (0, 8), (2, 6)\} $$
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c. Write the domain and co-domain of _V_.
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The domain of _V_ is $\{-2, 0, 2\}$ and the co-domain of _V_ is $\{4, 6, 8\}$
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d. Draw an arrow diagram for _V_.
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5. Define a relation $S$ from $\mathbb{R}$ to $\mathbb{R}$ as follows: For every
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$(x, y) \in \mathbb{R} \times \mathbb{R}$,
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$$ (x, y) \in S \quad \text{ means that } x \geq y $$
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**Solution**
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a.
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Is $(2, 1) \in S$?
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Yes, $(2) \geq (1)$.
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Is $(2, 2) \in S$?
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Yes, $(2) \geq (2)$.
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Is 2 _S_ 3?
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No, $(2) \cancel{\geq} (3)$.
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Is (-1) _S_ (-2)?
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Yes, $(-1) \geq (-2)$.
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b. Draw the graph of _S_ in the Cartesian plane.
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6. Define a relation $R$ from $\mathbb{R}$ to $\mathbb{R}$ as follows: For every
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$(x, y) \in \mathbb{R} \times \mathbb{R}$,
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$$ (x, y) \in R \quad \text{ means that } y = x^2 $$
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**Solution**
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a.
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Is $(2, 4) \in R$?
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Yes, $(4) = (2)^2$.
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Is $(4, 2) \in R$?
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No, $(2) \neq (4)^2$.
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Is (-3) _R_ 9?
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Yes, $(9) = (-3)^2$.
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Is 9 _R_ (-3)?
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No, $(-3) \neq (9)^2$.
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b. Draw the graph of _R_ in the Cartesian plane.
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7. Let $A = \{4, 5, 6\}$ and $B = \{5, 6, 7\}$ and define relations $R$, $S$,
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and $T$ from $A$ to $B$ as follows: For every $(x, y) \in A \times B$:
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$$ (x, y) \in R \quad \text{ means that } x \geq y $$
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$$ (x, y) \in S \quad \text{ means that } \frac{x - y}{2} \text{ is an integer.} $$
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$$ T = \{(4, 7), (6, 5), (6, 7)\} $$
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**Solution**
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a. Draw arrow diagrams for $R$, $S$, and $T$.
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b. Indicate whether any of the relations $R$, $S$, and $T$ are functions.
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$R$ is not a function because it satisfies neither property (1) nor property (2)
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of the definition. It fails property (1) because $(4, y) \not in R$, for any $y$
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in $B$. It fails property (2) because $(6, 5) \in R$ and $(6, 6) \in R$ and
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$5 \neq 6$.
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$S$ is not a function because $(5, 5) \in S$ and $(5, 7) \in S$ and $5 \neq 7$.
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So $S$ does not satisfy property (2) of the definition of a function.
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$T$ is not a function both because $(5, x) \notin T$ for any $x$ in $B$ and
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because $(6, 5) \in T$ and $(6, 7) \in T$ and $5 \neq 7$. So $T$ does not
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satisfy either property (1) or property (2) of the definition of a function.
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8. Let $A = \{2, 4\}$ and $B = \{1, 3, 5\}$ and define relations $U$, $V$, and
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$W$ from $A$ to $B$ as follows: For every $(x, y) \in A \times B$:
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$$ (x, y) \in U \quad \text{ means that } y - x > 2 $$
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$$ (x, y) \in V \quad \text{ means that } y - 1 = \frac{x}{2} $$
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W = \{(2, 5), (4, 1), (2, 3)\}
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**Solution**
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a. Draw arrow diagrams for $U$, $V$, and $W$.
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b. Indicate whether any of the relations $U$, $V$, and $W$ are functions.
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$U$ is not a function by property (1), as $(4, y) \notin B$.
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$V$ is not a function by property (1) as $(2, y) \notin B$.
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$T$ is not a function by property (2) as $(2, 3) \in B$ and $(2, 5) \in B$ and
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$3 \neq 5$.
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9.
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**Solution**
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a. Find all functions from $\{0, 1\}$ to $\{1\}$.
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$$ \{(0, 1), (1, 1)\} $$
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b. Find two relations form $\{0, 1\}$ to $\{1\}$ that are not functions.
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$$ \{(0, 1)\}, \{(1, 1)\} $$
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10. Find four relations from $\{a, b\}$ to $\{x, y\}$ that are not functions
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from $\{a, b\}$ to $\{x, y\}$.
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**Solution**
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$$ \{(a, x)\}, \{(a, y)\}, \{(b, x)\}, \{(b, y)\} $$
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11. Let $A = \{0, 1, 2\}$ and let $S$ be the set of all strings over $A$. Define
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a relation $L$ from $S$ to $\mathbb{Z}^{\text{nonneg}}$ as follows: For
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every string $s$ in $S$ and every nonnegative integer $n$,
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$$ (s, n) \in L \quad \text{ means that the length of } s \text{ is } n $$
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Then $L$ is a function because every string in $S$ has one and only one length.
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Find $L(0201)$ and $L(12)$.
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**Solution**
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$$ L(0201) = 4 $$
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$$ L(12) = 2 $$
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12. Let $A = \{x, y\}$ and let $S$ be the set of all strings over $A$. Define a
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relation $C$ from $S$ to $S$ as follows: For all strings $s$ and $t$ in $S$,
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$$ (s, t) \in C \quad \text{ means that } t = ys $$
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Then $C$ is a function because every string in $S$ consists entirely of $x$'s
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and $y$'s and adding an additional $y$ on the left creates a single new string
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that consists of $x$'s and $y$'s and is, therefore, also in $S$. Find $C(x)$ and
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$C(yyxyx)$.
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**Solution**
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$$ C(x) = yx $$
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$$ C(yyxyx) = yyyxyx $$
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13. Let $A = \{-1, 0, 1\}$ and $B = \{t, u, v, w\}$. Define a function
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$F: A \to B$ by the following arrow diagram:
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**Solution**
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a. Write the domain and co-domain of $F$.
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The domain of $F$ is $\{-1, 0, 1\}$, and the co-domain of $F$ is
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$\{t, u, v, w\}$.
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b. Find $F(-1)$, $F(0)$, and $F(1)$.
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$$ F(-1) = u $$
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$$ F(0) = w $$
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$$ F(1) = u $$
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14. Let $C = \{1, 2, 3, 4\}$ and $D = \{a, b, c, d\}$. Define a function
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$G: C \to D$ by the following diagram:
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**Solution**
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a. Write the domain and co-domain of $G$.
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The domain of $G$ is $\{1, 2, 3, 4\}$, and the co-domain of $G$ is
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$\{a, b, c, d\}$.
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b. Find $G(1)$, $G(2)$, $G(3)$, and $G(4)$.
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$$ G(1) = c $$
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$$ G(2) = c $$
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$$ G(3) = c $$
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$$ G(4) = c $$
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15. Let $X = \{2, 4, 5\}$ and $Y = \{1, 2, 4, 6\}$. Which of the following arrow
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diagrams determine functions from $X$ to $Y$?
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**Solution**
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Only (d) is a function.
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(a) is not a function by property (2), as $(2, 1) \in X \to Y$ and
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$(2, 6) \in X \to Y$, and $1 \neq 6$.
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(b) is not a function by property (1), as $(5, y) \notin X \to Y$.
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(c) is not a function by property (2), as $(4, 1) \in X \to Y$ and
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$(4, 2) \in X \to Y$ and $(1 \neq 2)$.
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(e) is not a function by property (1), as $(2, y) \notin X \to Y$.
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16. Let $f$ be the squaring function defined in Example 1.3.6. Find $f(-1)$,
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$f(0)$, and $f\left(\dfrac{1}{2}\right)$.
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**Solution**
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$$ f(-1) = (-1)^2 = 1 $$
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$$ f(0) = (0)^2 = 0 $$
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$$ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} $$
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17. Let $g$ be the successor function defined in Example 1.3.6. Find $g(-1000)$,
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$g(0)$, and $g(999)$.
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**Solution**
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$$ g(-1000) = (-1000) + 1 = -999 $$
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$$ g(0) = (0) + 1 = 1 $$
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$$ g(999) = (999) + 1 = 1000 $$
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18. Let $h$ be the constant function defined in Example 1.3.6. Find
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$h\left(-\dfrac{12}{5}\right)$, $h\left(\dfrac{0}{1}\right)$, and
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$h\left(\dfrac{9}{17}\right)$.
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**Solution**
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$$ h\left(-\frac{12}{5}\right) = 2 $$
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$$ h\left(\frac{0}{1}\right) = 2 $$
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$$ h\left(\frac{9}{17}\right) = 2 $$
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19. Define functions $f$ and $g$ from $\mathbb{R}$ to $\mathbb{R}$ by the
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following formulas: For every $x \in \mathbb{R}$,
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$$ f(x) = 2x \quad \text{ and } \quad g(x) = \frac{2x^3 + 2x}{x^2 + 1} $$
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Does $f = g$? Explain.
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**Solution**
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Yes, by factoring out $2x$ from the numerator of $g(x)$ we find they are the
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same function:
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$$ g(x) = \frac{2x^3 + 2x}{x^2 + 1} = \frac{2x(x^2 + 1)}{(x^2 + 1)} = 2x = f(x) $$
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This means that for every input $x$ to both $g$ and $f$, $f(x) = g(x)$, and so
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$f = g$ by definition of equality of functions.
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20. Define functions $H$ and $K$ from $\mathbb{R}$ to $\mathbb{R}$ by the
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following formulas: For every $x \in \mathbb{R}$,
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$$ H(x) = (x - 2)^2 \quad \text{ and } \quad K(x) = (x - 1)(x - 3) + 1 $$
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Does $H = K$? Explain.
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**Solution**
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$$ H(x) = (x - 2)^2 = (x - 2)(x - 2) = x^2 - 4x + 4 $$
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$$ K(x) = (x - 1)(x - 3) + 1 = x^2 - 4x + 3 + 1 = x^2 - 4x + 4 $$
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Therefore $H(x) = K(x)$ by the definition of equality of functions.
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