4895 lines
125 KiB
Markdown
4895 lines
125 KiB
Markdown
Page 458
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**Exercise Set 7.1**
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1. Let $X = \{1, 3, 5\}$ and $Y = \{s, t, u, v\}$. Define $f: X \to Y$ by the
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following arrow diagram.
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(See page 458 for image)
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a. Write the domain of $f$ and the co-domain of $f$.
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Domain: $\{1, 3, 5\}$
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Co-domain: $\{s, t, u, v\}$
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b. Find $f(1)$, $f(3)$, and $f(5)$.
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$f(1) = v, f(3) = s, f(5) = v$
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c. What is the range of $f$?
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$\{s, v\}$
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d. Is $3$ an inverse image of $s$? Is $1$ an inverse image of $u$?
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yes; no
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e. What is the inverse image of $s$? of $u$? of $v$?
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$\{3\}$;$\emptyset$;$\{1, 5\}$
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f. Represent $f$ as a set of ordered pairs.
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$\{(1, v), (3, s), (5, v)\}$
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2. Let $X = \{1, 3, 5\}$ and $Y = \{a, b, c, d\}$. Define $g: X \to Y$ by the
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following arrow diagram.
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(See page 459 for image)
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a. Write the domain of $g$ and the co-domain of $g$.
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Domain: $\{1, 3, 5\}$
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Co-domain: $\{a, b, c, d\}$
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b. Find $g(1)$, $g(3)$, and $g(5)$.
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$g(1) = b, g(3) = b, g(5) = b$
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c. What is the range of $g$?
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$\{b\}$
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d. Is $3$ an inverse image of $a$? Is $1$ an inverse image of $b$?
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no;yes
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e. What is the inverse image of $b$? of $c$?
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$\{1, 3, 5\}, \emptyset$
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f. Represent $g$ as a set of ordered pairs.
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$$ \{(1, b), (3, b), (5, b)\} $$
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3. Indicate whether the statements in parts (a)-(d) are true or false for all
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functions. Justify your answers.
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a. If two elements in the domain of a function are equal, then their images in
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the co-domain are equal.
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True. The definition of a function states that every input element in the domain
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must have an output element in the co-domain. Since two elements in the domain
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of the function are equal, then their outputs in the co-domain must be equal by
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this definition.
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b. If two elements in the co-domain of a function are equal, then their
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preimages in the domain are also equal.
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This is false. A function can have the same output for two different inputs.
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c. A function can have the same output for more than one input.
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True, the definition of a function only states that every input to the function
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must have an output, not necessarily unique outputs.
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d. A function can have the same input for more than one output.
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This is false. A single input can only map to a single output, not multiple
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outputs.
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4.
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a. Find all functions from $X = \{a, b\}$ to $Y = \{u, v\}$.
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$$ f(a) = u, f(a) = v, f(b) = u, f(b) = v $$
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b. Find all functions from $X = \{a, b, c\}$ to $Y = \{u\}$.
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$$ f(a) = u, f(b) = u, f(c) = u $$
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c. Find all functions from $X = \{a, b, c\}$ to $Y = \{u, v\}$.
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$$ f(a) = u, f(a) = v, f(b) = u, f(b) = v, f(c) = u, f(c) k v $$
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5. Let $I_{\mathbb{z}}$ bee the identity function defined on the set of all
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integers, and suppose that $e$, $b_i^{jk}$, $K(t)$, and $u_{kj}$ all
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represent integers. Find the following:
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a. $I_{\mathbb{Z}}(e)$
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$$ I_{\mathbb{Z}}(e) = e $$
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b. $I_{\mathbb{Z}}\left(b_i^{jk}\right)$
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$$ I_{\mathbb{Z}}\left(b_i^{jk}\right) = b_i^{jk}\right $$
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c. $I_{\mathbb{Z}}(K(t))$
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$$ I_{\mathbb{Z}}(K(t)) = K(t) $$
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d. $I_{\mathbb{Z}}(u_{kj})$
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$$ I_{\mathbb{Z}}(u_{kj}) = u_{kj} $$
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6. Find functions defined on the set of nonnegative integers that can be used to
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define the sequences whose first six terms are given below.
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a. $1, -\dfrac{1}{3}, \dfrac{1}{5}, -\dfrac{1}{7}, \dfrac{1}{9}, -\dfrac{1}{11}$
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$$ f: \mathbb{Z}^{\text{nonneg}} \to \mathbb{R} $$
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$$ f(n) = \frac{(-1)^n}{2n + 1} $$
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b. $0, -2, 4, -6, 8, -10$
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$$ f: \mathbb{Z}^{\text{nonneg}} \to \mathbb{R} $$
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$$ f(n) = (-1)^n \cdot 2n $$
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7. Let $A = \{1, 2, 3, 4, 5\}$, and define a function
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$F: \mathscr{P}(A) \to \mathbb{Z}$ as follows: For each set $X$ in
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$\mathscr{P}(A)$,
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$$
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F(x) =
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\begin{cases}
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0& \text{if } X \text{ has an even number of elements} \\
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1 & \text{if } X \text{ has an odd number of elements}
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\end{cases}
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$$
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Find the following:
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a. $F(\{1, 3, 4\})$
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$$ F(\{1, 3, 4\}) = 1 $$
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because $\{1, 3, 4\}$ has an odd number of elements.
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b. $F(\emptyset)$
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$$ F(\emptyset) = 0 $$
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because $\emptyset$ has an even number of elements.
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c. $F(\{2, 3\})$
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$$ F(\{2, 3\}) = 0 $$
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because $\{2, 3\}$ has an even number of elements.
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d. $F(\{2, 3, 4, 5\})$
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$$ F(\{2, 3, 4, 5\}) = 0 $$
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because $\{2, 3, 4, 5\}$ has an even number of elements.
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8. Let $J_5 = \{0, 1, 2, 3, 4\}$, and define a function $F: J_5 \to J_5$ as
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follows: For each $x \in J_5$, $F(x) = (x^3 + 2x + 4) \mod 5$.
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Find the following:
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a. $F(0)$
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$$ F(0) = ((0)^3 + 2(0) + 4) \mod 5 $$
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$$ = (0 + 0 + 4) \mod 5 $$
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$$ = 4 \mod 5 $$
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$$ = 4 $$
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b. $F(1)$
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$$ F(1) = ((1)^3 + 2(1) + 4) \mod 5 $$
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$$ = (1 + 2 + 4) \mod 5 $$
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$$ = 7 \mod 5 $$
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$$ = 2 $$
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c. $F(2)$
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$$ F(2) = ((2)^3 + 2(2) + 4) \mod 5 $$
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$$ = (8 + 4 + 4) \mod 5 $$
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$$ = 16 \mod 5 $$
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$$ = 1 $$
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d. $F(3)$
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$$ F(3) = ((3)^3 + 2(3) + 4) \mod 5 $$
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$$ = (27 + 6 + 4) \mod 5 $$
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$$ = 37 \mod 5 $$
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$$ = 2 $$
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e. $F(4)$
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$$ F(4) = ((4)^3 + 2(4) + 4) \mod 5 $$
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$$ = (64 + 8 + 4) \mod 5 $$
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$$ = 76 \mod 5 $$
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$$ = 1 $$
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9. Define a function $S: \mathbb{Z}^+ \to \mathbb{Z}^+$ as follows: For each
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positive integer $n$,
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$$ S(n) = \text{ the sum of the positive divisors of } n $$
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Find the following:
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a. $S(1)$
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$$ S(1) = 1 $$
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b. $S(15)$
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$$ S(15) = 1 + 3 + 5 + 15 = 24 $$
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c. $S(17)$
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$$ S(17) = 1 + 17 = 18 $$
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d. $S(5)$
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$$ S(5) = 1 + 5 = 6 $$
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e. $S(18)$
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$$ S(18) = 1 + 2 + 3 + 6 + 9 + 18 = 39 $$
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f. $S(21)$
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$$ S(21) = 1 + 3 + 7 + 21 = 32 $$
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10. Let $D$ be the set of all finite subsets of positive integers.
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Define a function $T: \mathbb{Z}^+ \to D$ as follows: For each positive integer
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$n$, $T(n) =$ the set of positive divisors of $n$.
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Find the following:
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a. $T(1)$
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$$ T(1) = \{1\} $$
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b. $T(15)$
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$$ T(15) = \{1, 3, 5, 15\} $$
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c. $T(17)$
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$$ T(17) = \{1, 17\} $$
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d. $T(5)$
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$$ T(5) = \{1, 5\} $$
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e. $T(18)$
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$$ T(18) = \{1, 2, 3, 6, 9, 18\} $$
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f. $T(21)$
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$$ T(21) = \{1, 3, 7, 21\} $$
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11. Define $F: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$ as
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follows: For every ordered pair $(a, b)$ of integers,
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$F(a, b) = (2a + 1, 3b - 2)$.
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Find the following:
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a. $F(4, 4)$
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$$ F(4, 4) = (2(4) + 1, 3(4) - 2) $$
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$$ = (8 + 1, 12 - 2) $$
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$$ = (9, 10) $$
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b. $F(2, 1)$
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$$ F(2, 1) = (2(2) + 1, 3(1) - 2) $$
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$$ = (4 + 1, 3 - 2) $$
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$$ = (5, 1) $$
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c. $F(3, 2)$
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$$ F(3, 2) = (2(3) + 1, 3(2) - 2) $$
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$$ = (6 + 1, 6 - 2) $$
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$$ = (7, 4) $$
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d. $F(1, 5)$
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$$ F(1, 5) = (2(1) + 1, 3(5) - 2) $$
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$$ = (2 + 1, 15 - 2) $$
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$$ = (3, 13) $$
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12. Let $J_5 = \{0, 1, 2, 3, 4\}$, and define
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$G: J_5 \times J_5 \to J_5 \times J_5$ as follows: For each
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$(a, b) \in J_5 \times J_5$,
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$$ G(a, b) = ((2a + 1) \mod 5, (3b - 2) \mod 5) $$
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Find the following:
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a. $G(4, 4)$
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$$ G(4, 4) = ((2(4) + 1) \mod 5, (3(4) - 2) \mod 5) $$
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$$ = ((8 + 1) \mod 5, (12 - 2) \mod 5) $$
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$$ = (9 \mod 5, 10 \mod 5) $$
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$$ = (4, 0) $$
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b. $G(2, 1)$
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$$ G(2, 1) = ((2(2) + 1) \mod 5, (3(1) - 2) \mod 5) $$
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$$ = ((4 + 1) \mod 5, (3 - 2) \mod 5) $$
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$$ = (5 \mod 5, 1 \mod 5) $$
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$$ = (0, 1) $$
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c. $G(3, 2)$
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$$ G(3, 2) = ((2(3) + 1) \mod 5, (3(2) - 2) \mod 5) $$
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$$ = ((6 + 1) \mod 5, (6 - 2) \mod 5) $$
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$$ = (7 \mod 5, 4 \mod 5) $$
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$$ = (2, 4) $$
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d. $G(1, 5)$
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$$ G(1, 5) = ((2(1) + 1) \mod 5, (3(5) - 2) \mod 5) $$
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$$ = ((2 + 1) \mod 5, (15 - 2) \mod 5) $$
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$$ = (3 \mod 5, 13 \mod 5) $$
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$$ = (3, 3) $$
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13. Let $J_5 = \{0, 1, 2, 3, 4\}$, and define functions $f: J_5 \to J_5$ and
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$g: J_5 \to J_5$ as follows: For each $x \in J_5$,
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$$ f(x) = (x + 4)^2 \mod 5 \quad \text{ and } \quad g(x) = (x^2 + 3x + 1) \mod 5 $$
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Is $f = g$? Explain.
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| $x$ | $f(x)$ | $g(x)$ |
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| --- | ------ | ------ |
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| $0$ | $1$ | $1$ |
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| $1$ | $0$ | $0$ |
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| $2$ | $1$ | $1$ |
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| $3$ | $4$ | $4$ |
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| $4$ | $4$ | $4$ |
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The table shows that $f(x) = g(x)$ for every $x \in J_5$. Therefore $f = g$ by
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definition of equality of functions.
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14. Define functions $H$ and $K$ from $\mathbb{R}$ to $\mathbb{R}$ by the
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following formulas:
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For every $x \in \mathbb{R}$,
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$$ H(x) = \lfloor x \rfloor + 1 \quad \text{ and } \quad K(x) = \lceil x \rceil $$
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Does $H = K$? Explain.
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No. For example say $x = 0$, then $H(0) = \lfloor 0 \rfloor + 1 = 0 + 1 = 1$ and
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$K(0) = \lceil 0 \rceil = 0$. Therefore it cannot be said that for every
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$x \in \mathbb{R}$ that $H(x) = K(x)$, and thus $H \neq K$.
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15. Let $F$ and $G$ be functions from the set of all real numbers to itself.
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Define the product functions $F \cdot G: \mathbb{R} \to \mathbb{R}$ and
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$G \cdot F: \mathbb{R} \to \mathbb{R}$ as follows: For every
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$x \in \mathbb{R}$,
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$$ (F \cdot G)(x) = F(x) \cdot G(x) $$
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$$ (G \cdot F)(x) = G(x) \cdot F(x) $$
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Does $F \cdot G = G \cdot F$? Explain.
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Yes, by the commutative law of multiplication of Real numbers:
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$$ (F \cdot G)(x) = F(x) \cdot G(x) = G(x) \cdot F(x) = (G \cdot F)(x) $$
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Therefore, since $(F \cdot G)(x) = (G \cdot F)(x)$ for all $x \in \mathbb{R}$,
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it can be concluded that $F \cdot G = G \cdot F$ by the definition of equality
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of functions.
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16. Let $F$ and $G$ be function sfrom the set of all real numbers to itself.
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Define new functions $F - G: \mathbb{R} \to \mathbb{R}$ and
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$G - F: \mathbb{R} \to \mathbb{R}$ as follows: For every $x \in \mathbb{R}$,
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$$ (F - G)(x) = F(x) - G(x) $$
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$$ (G - F)(x) = G(x) - F(x) $$
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Does $F - G = G - F$? Explain.
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No. Consider the definition of the difference of sets:
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$$ (F - G)(x) = F(x) - G(x) = F(x) $$
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and:
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$$ (G - F)(x) = G(x) - F(x) = G(x) $$
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Since $F(x) \neq G(x)$ for all $x \in \mathbb{R}$, it can be concluded that
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$F - G \neq G - F$ by the definition of the equality of functions.
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17. Use the definition of logarithm to fill in the blanks below.
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a. $\log_28 = 3$ because _____.
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$$ 2^3 = 8 $$
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b. $\log_5\left(\dfrac{1}{25}\right) = -2$ because _____.
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$$ 5^{-2} = \frac{1}{5^2} = \frac{1}{25} $$
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c. $\log_44 = 1$ because _____.
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$$ 4^1 = 4 $$
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d. $\log_3(3^n) = n$ because _____.
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$$ 3^n = 3^n $$
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e. $\log_41 = 0$ because _____.
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$$ 4^0 = 1 $$
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18. Find exact values for each of the following quantities without using a
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calculator.
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a. $\log_{3}81$
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$$ 3^{\text{?}} = 81 $$
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$$ \log_{3}81 = 4 $$
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b. $\log_{2}1024$
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$$ 2^{\text{?}} = 1024 $$
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$$ \log_{2}1024 = 10 $$
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c. $\log_{3}\left(\dfrac{1}{27}\right)$
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$$ \log_{3}\left(\frac{1}{27}\right) = -3 $$
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d. $\log_{2}1$
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$$ \log_{2}1 = 0 $$
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e. $\log_{10}\left(\dfrac{1}{10}\right)$
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$$ \log_{10}\left(\dfrac{1}{10}\right) = -1 $$
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f. $\log_{3}3$
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$$ \log_{3}3 = 1 $$
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g. $\log_{2}(2^k)$
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$$\log_{2}(2^k) = k $$
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19. Use the definition of logarithm to prove that for any positive real number
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$b$ with $b \neq 1$, $\log_{b}b = 1$.
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**Proof:**
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Let $b$ be any positive real number with $b \neq 1$. Since $b^1 = b$, then
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$\log_{b}b = 1$ by definition of logarithm.
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Q.E.D.
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20. Use the definition of logarithm to prove that for any positive real number
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$b$ with $b \neq 1$, $\log_{b}1 = 0$.
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**Proof:**
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Let $b$ be any positive real number with $b \neq 1$. Since $b^0 = 1$, then
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$\log_{b}1 = 0$ by definition of logarithm.
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Q.E.D.
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21. If $b$ is any positive real number with $b \neq 1$ and $x$ is any real
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number, $b^{-x}$ is defined as follows:
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$b^{-x} = \dfrac{1}{b^x}$. Use this definition and the definition of logarithm
|
|
to prove that $\log_{b}\left(\dfrac{1}{u}\right) = -\log_{b}u$ for all positive
|
|
real numbers $u$ and $b$, with $b \neq 1$.
|
|
|
|
**Proof:**
|
|
|
|
Let $b$ be any positive real number with $b \neq 1$. Let $u$ be any positive
|
|
real number.
|
|
|
|
Let $v = \log_{b}\left(\dfrac{1}{u}\right)$. By the definition of logarithm,
|
|
this means that $b^v = \dfrac{1}{u}$. It follows by algebra that:
|
|
|
|
$$ b^v = \frac{1}{u} $$
|
|
|
|
$$ u \cdot b^v = 1 $$
|
|
|
|
$$ u = \frac{1}{b^v} $$
|
|
|
|
$$ u = b^{-v} $$
|
|
|
|
Hence, by the definition of logarithm:
|
|
|
|
$$ -v = \log_{b}(u) $$
|
|
|
|
and by algebra:
|
|
|
|
$$ v = -\log_{b}(u) $$
|
|
|
|
Since $v = \log_{b}\left(\dfrac{1}{u}\right)$ and $v = -\log_{b}(u)$, it follows
|
|
by the definition of equality that:
|
|
|
|
$$ \log_{b}\left(\frac{1}{u}\right) = -\log{b}(u) $$
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
22. Use the unique factorization for the integers theorem (Section 4.4) and the
|
|
definition of logarithm to prove that $\log_{3}(7)$ is irrational.
|
|
|
|
_Hint:_ Use a proof by contradiction. Suppose $\log_{3}7$ is rational. Then
|
|
$\log_{3}7 = \dfrac{a}{b}$ for some integers $a$ and $b$ with $b \neq 0$.
|
|
|
|
Apply the definition of logarithm and rewrite $\log_{3}7 = \dfrac{a}{b}$ in
|
|
exponential form.
|
|
|
|
**Proof (by contradiction):**
|
|
|
|
Suppose $\log_{3}(7)$ is rational, that is $\log_{3}(7) = \dfrac{a}{b}$ for some
|
|
integers $a$ and $b$ where $b \neq 0$.
|
|
|
|
By the definition of logarithm, this would mean that:
|
|
|
|
$$ 3^{\frac{a}{b}} = 7 $$
|
|
|
|
Then by algebra:
|
|
|
|
$$ 3^a = 7^b $$
|
|
|
|
Since $b \neq 0$, we know that $7^b \neq 1$, and by equality it follows that
|
|
$3^a \neq 1$. Additionally, by the definition of exponentiation, it is known
|
|
that $7^b > 0$ and $3^a > 0$ (they are both positive numbers).
|
|
|
|
But, by the unique factorization for integers theorem, this means that $7^b$ and
|
|
$3^a$ are two different prime factorizations of the same positive integer. This
|
|
is only possible if the positive integer is equal to $1$.
|
|
|
|
Hence $3^a = 7^b = 1$, but it has already been established that
|
|
$3^a = 7^b \neq 1$. This is a contradiction.
|
|
|
|
Therefore the supposition is false, and $\log_{3}(7)$ is irrational.
|
|
|
|
Q.E.D.
|
|
|
|
23. If $b$ and $y$ are positive real numbers such that $\log_{b}y = 3$, what is
|
|
$\log_{\frac{1}{b}}y$? Explain.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $b$ and $y$ are positive real numbers such that $\log_{b}y = 3$.
|
|
|
|
By the definition of logarithm, this means that:
|
|
|
|
$$ b^3 = y $$
|
|
|
|
To find $\log_{\frac{1}{b}}y$, first, replace $y$ by substitution:
|
|
|
|
$$ \log_{\frac{1}{b}}y $$
|
|
|
|
$$ = \log_{\frac{1}{b}}(b^3) $$
|
|
|
|
Then notice that $\dfrac{1}{b} = b^{-1}$, and then substitute:
|
|
|
|
$$ = \log_{b^{-1}}(b^3) $$
|
|
|
|
By the definition of logarithm, this means that:
|
|
|
|
$$ (b^{-1})^x = b^3 $$
|
|
|
|
Where $x$ is $\log_{\frac{1}{b}}y$, or our answer. By the multiplication of
|
|
exponents, this means that:
|
|
|
|
$$ b^{-1 \cdot x} = b^3 $$
|
|
|
|
And by multiplication of negative numbers:
|
|
|
|
$$ b^{-1 \cdot -3} = b^3 $$
|
|
|
|
Therefore $x = -3$, or:
|
|
|
|
$$ \log_{\frac{1}{b}}y = -3 $$
|
|
|
|
This is what was to be found.
|
|
|
|
Q.E.D.
|
|
|
|
24. If $b$ and $y$ are positive real numbers such that $\log_{b}y = 2$, what is
|
|
$\log_{b^2}(y)$? Explain.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $b$ and $y$ are positive real numbers such that $\log_{b}y = 2$. By the
|
|
definition of logarithm, this means that:
|
|
|
|
$$ \log_{b}y = 2 $$
|
|
|
|
$$ b^2 = y $$
|
|
|
|
To find $\log_{b^2}(y)$, first substitute in for $y$:
|
|
|
|
$$ \log_{b^2}(b^2) $$
|
|
|
|
By the definition of logarithm, this means that:
|
|
|
|
$$ \log_{b^2}(b^2) = 1 $$
|
|
|
|
because $(b^2)^1 = b^2$.
|
|
|
|
This is what was to be found.
|
|
|
|
Q.E.D.
|
|
|
|
25. Let $A = \{2, 3, 5\}$ and $B = \{x, y\}$. Let $p_1$ and $p_2$ be the
|
|
**projections of $A \times B$ onto the first and second coordinates.** That
|
|
is, for each pair $(a, b) \in A \times B$, $p_1(a, b) = a$ and
|
|
$p_2(a, b) = b$.
|
|
|
|
a. Find $p_1(2, y)$ and $p_1(5, x)$. What is the range of $p_1$?
|
|
|
|
$$ p_1(2, y) = 2 $$
|
|
|
|
$$ p_1(5, x) = 5 $$
|
|
|
|
Range of $p_1$:
|
|
|
|
$$ \{2, 3, 5\} $$
|
|
|
|
b. Find $p_2(2, y)$ and $p_2(5, x)$. What is the range of $p_2$?
|
|
|
|
$$ p_2(2, y) = y $$
|
|
|
|
$$ p_2(5, x) = x $$
|
|
|
|
Range of $p_2$:
|
|
|
|
$$ \{x, y\} $$
|
|
|
|
26. Observe that $\mod$ and $\text{div}$ can be defined as functions from
|
|
$\mathbb{Z}^{\text{nonneg}}$ \times \mathbb{Z}^+$ to $\mathbb{Z}$. For each
|
|
ordered pair $(n, d)$ consisting of a nonnegative integer $n$ and a positive
|
|
integer $d$, let
|
|
|
|
$\mod(n, d) = n \mod d$ (the nonnegative remainder obtained when $n$ is divided
|
|
by $d$).
|
|
|
|
$\text{div}(n, d) = n \text{ div } d$ (the integer quotient obtained when $n$ is
|
|
divided by $d$).
|
|
|
|
Find each of the following:
|
|
|
|
a. $\mod(67, 10)$ and $\text{div}(67, 10)$
|
|
|
|
$$ \mod(67, 10) = 7 $$
|
|
|
|
$$ \text{div}(67, 10) = 6 $$
|
|
|
|
b. $\mod(59, 8)$ and $\text{div}(59, 8)$
|
|
|
|
$$ \mod(59, 8) = 3 $$
|
|
|
|
$$ \text{div}(59, 8) = 7 $$
|
|
|
|
c. $\mod(30, 5)$ and $\text{div}(30, 5)$
|
|
|
|
$$ \mod(30, 5) = 0 $$
|
|
|
|
$$ \text{div}(30, 5) = 6 $$
|
|
|
|
27. Let $S$ be the set of all strings of $a$'s and $b$'s.
|
|
|
|
a. Define $f: S \to \mathbb{Z}$ as follows: For each string $s$ in $S$
|
|
|
|
$$
|
|
f(s) =
|
|
\begin{cases}
|
|
& \text{ the number of b's to the left-most a in s} \\
|
|
0 & \text{if s contains no a's}
|
|
\end{cases}
|
|
$$
|
|
|
|
Find $f(aba)$, $f(bbab)$, and $f(b)$. What is the range of $f$?
|
|
|
|
$$ f(aba) = 0 $$
|
|
|
|
$$ f(bbab) = 2 $$
|
|
|
|
$$ f(b) = 0 $$
|
|
|
|
The range of $f$: $\mathbb{Z}^{\text{nonneg}}$
|
|
|
|
b. Define $g: S \to S$ as follows: For each string $s$ in $S$,
|
|
|
|
$$ g(s) = \text{ the string obtained by writing the characters of s in reverse order} $$
|
|
|
|
Find $g(aba)$, $g(bbab)$, and $g(b)$. What is the range of $g$?
|
|
|
|
$$ g(aba) = aba $$
|
|
|
|
$$ g(bbab) = babb $$
|
|
|
|
The range of $g$ is $S$.
|
|
|
|
28. Consider the coding and decoding functions $E$ and $D$ defined in Example
|
|
7.1.9.
|
|
|
|
a. Find $E(0110)$ and $D(111111000111)$.
|
|
|
|
$$ E(0110) = 000111111000 $$
|
|
|
|
$$ D(111111000111) = 1101 $$
|
|
|
|
b. Find $E(1010)$ and $D(000000111111)$.
|
|
|
|
$$ E(1010) = 111000111000 $$
|
|
|
|
$$ D(000000111111) = 0011 $$
|
|
|
|
29. Consider the Hamming distance function defined in Example 7.1.10.
|
|
|
|
a. Find $H(10101, 00011)$.
|
|
|
|
$$ H(10101, 00011) = 3 $$
|
|
|
|
b. Find $H(00110, 10111)$.
|
|
|
|
$$ H(00110, 10111) = 2 $$
|
|
|
|
30. Draw arrow diagrams for the Boolean functions defined by the following
|
|
input/output tables.
|
|
|
|
a.
|
|
|
|
| Input | Intput | Output |
|
|
| ------- | ------ | ------ |
|
|
| $P$ | $Q$ | $R$ |
|
|
| ------- | - | |
|
|
| 1 | 1 | 0 |
|
|
| 1 | 0 | 1 |
|
|
| 0 | 1 | 0 |
|
|
| 0 | 0 | 1 |
|
|
|
|
Omitted.
|
|
|
|
b.
|
|
|
|
| Input | Intput | Input | Output |
|
|
| ----- | ------ | ----- | ------ |
|
|
| $P$ | $Q$ | $R$ | $S$ |
|
|
| - | - | - | - |
|
|
| 1 | 1 | 1 | 1 |
|
|
| 1 | 1 | 0 | 0 |
|
|
| 1 | 0 | 1 | 1 |
|
|
| 1 | 0 | 0 | 1 |
|
|
| 0 | 1 | 1 | 0 |
|
|
| 0 | 1 | 0 | 0 |
|
|
| 0 | 0 | 1 | 0 |
|
|
| 0 | 0 | 0 | 1 |
|
|
|
|
Omitted.
|
|
|
|
31. Fill in the following table to show the values of all possible two-place
|
|
Boolean functions.
|
|
|
|
| Input | Input | $f_1$ | $f_2$ | $f_3$ | $f_4$ | $f_5$ | $f_6$ | $f_7$ | $f_8$ | $f_9$ | $f_{10}$ | $f_{11}$ | $f_{12}$ | $f_{13}$ | $f_{14}$ | $f_{15}$ | $f_{16}$ |
|
|
| ----- | ----- | ----- | ----- | ----- | ----- | ----- | ----- | ----- | ----- | ----- | -------- | -------- | -------- | -------- | -------- | -------- | -------- |
|
|
| 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
|
|
| 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
|
|
| 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |
|
|
| 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 |
|
|
|
|
32. Consider the three-place Boolean function $f$ defined by the following rule:
|
|
For each triple $(x_1, x_2, x_3)$ of $0$'s and $1$'s,
|
|
|
|
$$ f(x_1, x_2, x_3) = (4x_1 + 3x_2 + 2x_3) \mod 2 $$
|
|
|
|
a. Find $f(1, 1, 1)$ and $f(0, 0, 1)$.
|
|
|
|
$$ f(1, 1, 1) = (4(1) + 3(1) + 2(1)) \mod 2 $$
|
|
|
|
$$ f(1, 1, 1) = (4 + 3 + 2) \mod 2 $$
|
|
|
|
$$ f(1, 1, 1) = 9 \mod 2 $$
|
|
|
|
$$ f(1, 1, 1) = 1 $$
|
|
|
|
$$ f(0, 0, 1) = (4(0) + 3(0) + 2(1)) \mod 2 $$
|
|
|
|
$$ f(0, 0, 1) = (0 + 0 + 2) \mod 2 $$
|
|
|
|
$$ f(0, 0, 1) = 2 \mod 2 $$
|
|
|
|
$$ f(0, 0, 1) = 0 $$
|
|
|
|
b. Describe $f$ using an input/output table.
|
|
|
|
| $x_1$ | $x_2$ | $x_3$ | $f(x_1, x_2, x_3)$ |
|
|
| ----- | ----- | ----- | ------------------ |
|
|
| $0$ | $0$ | $0$ | $0$ |
|
|
| $0$ | $0$ | $1$ | $0$ |
|
|
| $0$ | $1$ | $0$ | $1$ |
|
|
| $0$ | $1$ | $1$ | $1$ |
|
|
| $1$ | $0$ | $0$ | $0$ |
|
|
| $1$ | $0$ | $1$ | $0$ |
|
|
| $1$ | $1$ | $0$ | $1$ |
|
|
| $1$ | $1$ | $1$ | $1$ |
|
|
|
|
33. Student A tries to define a function $g: \mathbb{Q} \to \mathbb{Z}$ by the
|
|
rule
|
|
|
|
$g\left(\dfrac{m}{n}\right) = m - n$, for all integers $m$ and $n$ with
|
|
$n \neq 0$.
|
|
|
|
Student B claims that $g$ is not well defined. Justify student B's claim.
|
|
|
|
Suppose $\dfrac{m}{n} = \dfrac{1}{2}$, this would mean that
|
|
$g\left(\dfrac{m}{n}\right) = 1 - 2 = -1$.
|
|
|
|
Since $\dfrac{m}{n} = \dfrac{1}{2}$, this means that
|
|
$\dfrac{m}{n} = \dfrac{1}{2} = \dfrac{2}{4}$. Since they are equivalent, this
|
|
means that
|
|
$g\left(\dfrac{1}{2}\right) = g\left(\dfrac{2}{4}\right) = 2 - 4 = -2$.
|
|
|
|
But notice that:
|
|
|
|
$$ g\left(\frac{1}{2}\right) = -1 \neq -2 = g\left(\frac{2}{4}\right) $$
|
|
|
|
Since the function $g$ gives two different outputs for the same input, the
|
|
function $g$ is not well defined.
|
|
|
|
34. Student C tries to define a function $h: \mathbb{Q} \to \mathbb{Q}$ by the
|
|
rule
|
|
|
|
$h\left(\dfrac{m}{n}\right) = \dfrac{m^2}{n}$, for all integers $m$ and $n$ with
|
|
$n \neq 0$.
|
|
|
|
Student D claims that $h$ is not well defined. Justify student D's claim.
|
|
|
|
Suppose $\dfrac{m}{n} = \dfrac{2}{3}$, then
|
|
$h\left(\dfrac{2}{3}\right) = \dfrac{(2)^2}{3} = \dfrac{4}{3}$.
|
|
|
|
Notice that $\dfrac{2}{3} = \dfrac{4}{6}$, so
|
|
$h\left(\dfrac{4}{6}\right) = \dfrac{(4)^2}{6} = \dfrac{16}{6} = \dfrac{8}{3}$.
|
|
|
|
Notice that:
|
|
|
|
$$ h\left(\frac{2}{3}\right) = \frac{4}{3} \neq \frac{8}{3} = h\left(\frac{4}{6}\right) $$
|
|
|
|
Since the function $h$ does not produce the same output given the same input,
|
|
the function is not well defined.
|
|
|
|
35. Let $U = \{1, 2, 3, 4\}$. Student A tries to define a function
|
|
$R: U \to \mathbb{Z}$ as follows: For each $x \in U$,
|
|
|
|
$R(x)$ is the integer $y$ so that $(xy) \mod 5 = 1$.
|
|
|
|
Student B claims that $R$ is not well defined. Who is correct: student A or
|
|
student B? Justify your answer.
|
|
|
|
Consider $R(3) = 2$ since $(3 \cdot 2) \mod 5 = 1$. On the other hand,
|
|
$R(3) = 7$ since $(3 \cdot 7) \mod 5 = 1$.
|
|
|
|
Since $R$ returns multiple outputs for the same input, it is not well defined,
|
|
and Student B is correct.
|
|
|
|
36. Let $V = \{1, 2, 3\}$. Student C tries to define a function $S: V \to V$ as
|
|
follows: For each $x \in V$,
|
|
|
|
$S(x)$ is the integer $y$ in $V$ so that $(xy) \mod 4 = 1$.
|
|
|
|
Student D claims that $S$ is not well defined. Who is right: student C or
|
|
student D? Justify your answer.
|
|
|
|
Consider $S(1) = 17$ since $(1 \cdot 17) \mod 4 = 1$. On the other hand
|
|
$S(1) = 13$ since $(1 \cdot 13) \mod 4 = 1$.
|
|
|
|
Since $S$ returns multiple outputs for the same input, it is not well defined,
|
|
and Student D is correct.
|
|
|
|
37. On certain computers the integer data type goes from $-2,147,483,648$
|
|
through $2,147,483,647$. Let $S$ be the set of all integers from
|
|
$-2,147,483,648$ through $2,147,483,647$. Try to define a function
|
|
$f: S \to S$ by the rule $f(n) = n^2$ for each $n$ in $S$. Is $f$ well
|
|
defined? Explain.
|
|
|
|
No, $2,147,483,247 = 2^{31} - 1$, so for values of $n$ greater than $2^{16}$,
|
|
$f(n) = n^2$ will be greater than $2^{32}$, which falls outside of $S$.
|
|
|
|
38. Let $X = \{a, b, c\}$ and $Y = \{r, s, t, u, v, w\}$. Define $f: X \to Y$ as
|
|
follows: $f(a) = v$, $f(b) = v$, and $f(c) = t$.
|
|
|
|
a. Draw an arrow diagram for $f$.
|
|
|
|
Omitted.
|
|
|
|
b. Let $A = \{a, b\}$, $C = \{t\}$, $D = \{u, v\}$, and $E = \{r, s\}$. Find
|
|
$f(A)$, $f(X)$, $f^{-1}(C)$, $f^{-1}(D)$, $f^{-1}(E)$, and $f^{-1}(Y)$.
|
|
|
|
$$ f(A) = \{v\} $$
|
|
|
|
$$ f(X) = $\{t, v\} $$
|
|
|
|
$$ f^{-1}(C) = \{c\} $$
|
|
|
|
$$ f^{-1}(D) = \{a, b\} $$
|
|
|
|
$$ f^{-1}(E) = \emptyset $$
|
|
|
|
$$ f^{-1}(Y) = \{a, b, c\} $$
|
|
|
|
39. Let $X = \{1, 2, 3, 4\}$ and $Y = \{a, b, c, d, e\}$. Define $g: X \to Y$ as
|
|
follows: $g(1) = a$, $g(2) = a$, $g(3) = a$, and $g(4) = d$.
|
|
|
|
a. Draw an arrow diagram for $g$.
|
|
|
|
Omitted.
|
|
|
|
b. Let $A = \{2, 3\}$, $C = \{a\}$, and $D = \{b, c\}$. Find $g(A)$, $g(X)$,
|
|
$g^{-1}(C)$, $g^{-1}(D)$, and $g^{-1}(Y)$.
|
|
|
|
$$ g(A) = \{a\} $$
|
|
|
|
$$ g(X) = \{a, d\} $$
|
|
|
|
$$ g^{-1}(C) = \{1, 2, 3\} $$
|
|
|
|
$$ g^{-1}(D) = \emptyset $$
|
|
|
|
$$ g^{-1}(Y) = \{1, 2, 3, 4\} $$
|
|
|
|
40. Let $X$ and $Y$ be sets, let $A$ and $B$ be any subsets of $X$, and let $F$
|
|
be a function from $X$ to $Y$. Fill in the blanks in the following proof
|
|
that $F(A) \cup F(B) \subseteq F(A \cup B)$.
|
|
|
|
**Proof:**
|
|
|
|
Let $y$ be any element in $F(A) \cup F(B)$. _[We must show that $y$ is in
|
|
$F(A \cup B)$.]_ By definition of union, __ (i) __.
|
|
|
|
_Case 1 $y \in F(A)$:_
|
|
|
|
In this case, by definition of $F(A)$, $y = F(x)$ for __ (ii) __ $x \in A$.
|
|
Since $A \subseteq A \cup B$, it follows from the definition of union that
|
|
$x \in$ __ (iii) __. Hence, $y = F(x)$ for some $x \in A \cup B$, and thus, by
|
|
definition of $F(A \cup B)$, $y \in$ __ (iv) __.
|
|
|
|
_Case 2, $y \in F(B)$:_
|
|
|
|
In this case, by definition of $F(B)$, __ (v) __ for some $x \in B$. Since
|
|
$B \subseteq A \cup B$ it follows from the definition of union that __ (vi) __.
|
|
Thus $y \in F(A \cup B)$.
|
|
|
|
Therefore, regardless of whether $y \in F(A)$ or $y \in F(B)$, we have that
|
|
$y \in F(A \cup B)$ _[as was to be shown]_.
|
|
|
|
i. $y \in F(A) \cup F(B)$
|
|
|
|
ii. some
|
|
|
|
iii. $A \cup B$
|
|
|
|
iv. $F(A \cup B)$
|
|
|
|
v. $y = F(x)$
|
|
|
|
vi. $x \in A \cup B$
|
|
|
|
In 41-49 let $X$ and $Y$ be sets, let $A$ and $B$ be any subsets of $X$, and let
|
|
$C$ and $D$ be any subsets of $Y$. Determine which of the properties are true
|
|
for every function $F$ from $X$ to $Y$ and which are false for at least one
|
|
function $F$ from $X$ to $Y$. Justify your answers.
|
|
|
|
41. If $A \subseteq B$ then $F(A) \subseteq F(B)$
|
|
|
|
**Proof:**
|
|
|
|
Let $F$ be a function from $X$ to $Y$ and suppose $A \subseteq X$,
|
|
$B \subseteq X$, and $A \subseteq B$.
|
|
|
|
Then, let $y$ be some element such that $y \in F(A)$.
|
|
|
|
By definition of image of a set, $y = F(x)$ for some $x \in A$. Thus since
|
|
$A \subseteq B$, $x \in B$, and so $y = F(x)$ for some $x \in B$. Hence
|
|
$y \in F(B)$, and therefore $F(A) \subseteq F(B)$.
|
|
|
|
Q.E.D.
|
|
|
|
42. $F(A \cap B) \subseteq F(A) \cap F(B)$
|
|
|
|
**Proof:**
|
|
|
|
Suppose $y$ is some element such that $y \in F(A \cap B)$.
|
|
|
|
By the supposition and the definition of $A \cap B$, this means that $y = F(x)$
|
|
for some $x \in A \cap B$.
|
|
|
|
By the definition of intersection, it follows that $x \in A$ and $x \in B$.
|
|
|
|
By the definition of $F(A)$ and $F(B)$, $y = F(x)$ is in $F(A)$ and in $F(B)$.
|
|
|
|
Hence, by the definition of intersection, $y \in F(A) \cap F(B)$.
|
|
|
|
Since $y \in F(A) \cap F(B)$, it can be concluded that
|
|
$F(A \cap B) \subseteq F(A) \cap F(B)$.
|
|
|
|
Q.E.D.
|
|
|
|
43. $F(A) \cap F(B) \subseteq F(A \cap B)$
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Let $X = \{1, 2, 3\}$ and $Y = \{a, b\}$. Then, define a function $F: X \to Y$
|
|
such that $F(1) = a, F(2) = b, F(3) = b$.
|
|
|
|
Let $A = \{1, 2\}$ and $B = \{1, 3\}$. Then $F(A) = \{a, b\}$ and
|
|
$F(B) = \{a, b\}$.
|
|
|
|
So $F(A) \cap F(B) = \{a, b\}$, and $F(A \cap B) = F(\{1\}) = \{a\}$.
|
|
|
|
Since $\{a\} \neq \{a, b\}$, the given statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
44. For all subsets $A$ and $B$ of $X$, $F(A - B) = F(A) - F(B)$.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Let $X = \{1, 2\}$ and $Y = \{a\}$. Then, define a function $F: X \to Y$ such
|
|
that $F(1) = a$ and $F(2) = a$.
|
|
|
|
Let $A = \{1\}$ and $B = \{2\}$. Then $F(A - B) = F(\{1\}) = \{a\}$.
|
|
|
|
Then $F(A) - F(B) = F(\{1\}) - F(\{2\}) = \{a\} - \{a\} = \emptyset$.
|
|
|
|
Since $\{a\} \neq \emptyset$, the given statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
45. For all subsets $C$ and $D$ of $Y$, if $C \subseteq D$, then
|
|
$F^{-1}(C) \subseteq F^{-1}(D)$.
|
|
|
|
**Proof:**
|
|
|
|
Let $F$ be a function from a set $X$ to a set $Y$, and suppose $C \subseteq Y$,
|
|
$D \subseteq Y$, and $C \subseteq D$.
|
|
|
|
Suppose $x \in F^{-1}(C)$. Then $F(x) \in C$. Since $C \subseteq D$,
|
|
$F(x) \in D$ also. Hence, by definition of inverse image, $x \in F^{-1}(D)$.
|
|
Therefore $F^{-1}(C) \subseteq F^{-1}(D)$.
|
|
|
|
Q.E.D.
|
|
|
|
46. For all subsets $C$ and $D$ of $Y$,
|
|
|
|
$$ F^{-1}(C \cup D) = F^{-1}(C) \cup F^{-1}(D) $$
|
|
|
|
**Proof:**
|
|
|
|
In order to prove:
|
|
|
|
$$ F^{-1}(C \cup D) = F^{-1}(C) \cup F^{-1}(D) $$
|
|
|
|
We must prove:
|
|
|
|
$$ F^{-1}(C \cup D) \subseteq F^{-1}(C) \cup F^{-1}(D) $$
|
|
|
|
and:
|
|
|
|
$$ F^{-1}(C) \cup F^{-1}(D) \subseteq F^{-1}(C \cup D) $$
|
|
|
|
_Proof $F^{-1}(C \cup D) \subseteq F^{-1}(C) \cup F^{-1}(D)$:_
|
|
|
|
Suppose $x \in F^{-1}(C \cup D)$. Then $F(x) \in C \cup D$. By the definition of
|
|
union, this means that $F(x) \in C$ or $F(x) \in D$.
|
|
|
|
_Case $F(x) \in C$:_
|
|
|
|
Since $F(x) \in C$, this means that $x \in F^{-1}(C)$. By the definition of
|
|
union, this means that $x \in F^{-1}(C) \cup F^{-1}(D)$.
|
|
|
|
_Case $F(x) \in D$:_
|
|
|
|
Since $F(x) \in D$, this means that $x \in F^{-1}(D)$. By the definition of
|
|
union, this means that $x \in F^{-1}(C) \cup F^{-1}(D)$.
|
|
|
|
In both cases, $x \in F^{-1}(C) \cup F^{-1}(D)$. Therefore, any element in
|
|
$F^{-1}(C \cup D)$ is also in $F^{-1}(C)$, and
|
|
$F^{-1}(C \cup D) \subseteq F^{-1}(C) \cup F^{-1}(D)$, as was to be shown.
|
|
|
|
_Proof $F^{-1}(C) \cup F^{-1}(D) \subseteq F^{-1}(C \cup D)$:_
|
|
|
|
Suppose $x \in F^{-1}(C) \cup F^{-1}(D)$. By definition of union, this means
|
|
that $x \in F^{-1}(C)$ or $x \in F^{-1}(D)$.
|
|
|
|
_Case $x \in F^{-1}(C)$:_
|
|
|
|
Since $x \in F^{-1}(C)$, this means that $F(x) \in C$. It follows by definition
|
|
of union that $F(x) \in C \cup D$, or $x \in F^{-1}(C \cup D)$.
|
|
|
|
_Case $x \in F^{-1}(D)$:_
|
|
|
|
Since $x \in F^{-1}(D)$, this means that $F(x) \in D$. It follows by definition
|
|
of union that $F(x) \in C \cup D$, or $x \in F^{-1}(C \cup D)$.
|
|
|
|
In both cases, $x \in F^{-1}(C \cup D)$. Therefore any element in
|
|
$F^{-1}(C) \cup F^{-1}(D)$ is in $F^{-1}(C \cup D)$, and so
|
|
$F^{-1}(C) \cup F^{-1}(D) \subseteq F^{-1}(C \cup D)$. This is what was to be
|
|
shown.
|
|
|
|
_Conclusion:_
|
|
|
|
Since both subset relations have been proved, it can be concluded that
|
|
$F^{-1}(C \cup D) = F^{-1}(C) \cup F^{-1}(D)$. This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
47. For all subsets $C$ and $D$ of $Y$,
|
|
|
|
$$ F^{-1}(C \cap D) = F^{-1}(C) \cap F^{-1}(D) $$
|
|
|
|
**Proof:**
|
|
|
|
In order to prove:
|
|
|
|
$$ F^{-1}(C \cap D) = F^{-1}(C) \cap F^{-1}(D) $$
|
|
|
|
it must be shown that:
|
|
|
|
$$ F^{-1}(C \cap D) \subseteq F^{-1}(C) \cap F^{-1}(D) $$
|
|
|
|
and also that:
|
|
|
|
$$ F^{-1}(C) \cap F^{-1}(D) \subseteq F^{-1}(C \cap D) $$
|
|
|
|
_Proof $F^{-1}(C \cap D) \subseteq F^{-1}(C) \cap F^{-1}(D)$:_
|
|
|
|
Suppose $x \in F^{-1}(C \cap D)$, or $F(x) \in C \cap D$. By definition of
|
|
intersection, this means that $F(x) \in C$ and $F(x) \in D$, or
|
|
$x \in F^{-1}(C) \cap F^{-1}(D)$. This is what was to be shown.
|
|
|
|
_Proof $F^{-1}(C) \cap F^{-1}(D) \subseteq F^{-1}(C \cap D)$:_
|
|
|
|
Suppose $x \in F^{-1}(C) \cap F^{-1}(D)$, or $F(x) \in C$ and $F(x) \in D$. By
|
|
definition of intersection, $F(x) \in C \cap D$, or $x \in F^{-1}(C \cap D)$.
|
|
This is what was to be shown.
|
|
|
|
_Conclusion:_
|
|
|
|
Since both subset relations have been proved, it can be concluded that
|
|
$F^{-1}(C \cap D) = F^{-1}(C) \cap F^{-1}(D)$. This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
48. For all subsets $C$ and $D$ of $Y$,
|
|
|
|
$$ F^{-1}(C - D) = F^{-1}(C) - F^{-1}(D) $$
|
|
|
|
**Proof:**
|
|
|
|
In order to prove:
|
|
|
|
$$ F^{-1}(C - D) = F^{-1}(C) - F^{-1}(D) $$
|
|
|
|
it must be shown that:
|
|
|
|
$$ F^{-1}(C - D) \subseteq F^{-1}(C) - F^{-1}(D) $$
|
|
|
|
and also that:
|
|
|
|
$$ F^{-1}(C) - F^{-1}(D) \subseteq F^{-1}(C - D) $$
|
|
|
|
_Proof $F^{-1}(C - D) \subseteq F^{-1}(C) - F^{-1}(D)$:_
|
|
|
|
Suppose $x \in F^{-1}(C - D)$, or $F(x) \in C - D$. By definition of difference
|
|
of sets, this means that $F(x) \in C$ and $F(x) \notin D$. By the definition of
|
|
inverse image, this means $x \in F^{-1}(C)$ and $x \notin F^{-1}(D)$. By the
|
|
definition of difference, this is $x \in F^{-1}(C) - F^{-1}(D)$. Thus
|
|
$F^{-1}(C - D) \subseteq F^{-1}(C) - F^{-1}(D)$, which is what was to be shown.
|
|
|
|
_Proof $F^{-1}(C) - F^{-1}(D) \subseteq F^{-1}(C - D)$:_
|
|
|
|
Suppose $x \in F^{-1}(C) - F^{-1}(D)$, or $F(x) \in C$ and $F(x) \notin D$. By
|
|
the definition of inverse image, this means that $F(x) \in C - D$, or
|
|
$x \in F^{-1}(C - D)$. Thus $F^{-1}(C) - F^{-1}(D) \subseteq F^{-1}(C - D)$,
|
|
which is what was to be shown.
|
|
|
|
_Conclusion:_
|
|
|
|
Since both subset relations have been proved, it can be concluded that
|
|
$F^{-1}(C - D) = F^{-1}(C) - F^{-1}(D)$, which is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
49. $F(F^{-1}(C)) \subseteq C$
|
|
|
|
**Proof:**
|
|
|
|
Suppose $x \in F(F^{-1}(C))$. By definition of image, there exists some
|
|
$a \in F^{-1}(C)$ such that $F(a) = x$. By definition of inverse image,
|
|
$a \in F^{-1}(C)$ means $F(a) \in C$. Since $F(a) = x$, we have $x \in C$.
|
|
Therefore $F(F^{-1}(C)) \subseteq C$.
|
|
|
|
Q.E.D.
|
|
|
|
50. Given a set $S$ and a subset $A$, the **characteristic function of $A$**,
|
|
denoted $\chi_A$, is the function defined from $S$ to $\mathbb{Z}$ with the
|
|
property that for each $u \in S$,
|
|
|
|
$$
|
|
\chi_{A}(u) =
|
|
\begin{cases}
|
|
1 & \text{if } u \in A \\
|
|
0 & \text{if } u \notin A
|
|
\end{cases}
|
|
$$
|
|
|
|
Show that each of the following holds for all subsets $A$ and $B$ of $S$ and
|
|
every $u \in S$.
|
|
|
|
a. \chi_{A \cap B}(u) = \chi_{A}(u) \cdot \chi_{B}(u)
|
|
|
|
Omitted.
|
|
|
|
b.
|
|
$\chi_{A \cup B}(u) = \chi_{A}(u) + \chi_{B}(u) - \chi_{A}(u) \cdot \chi_{B}(u)$
|
|
|
|
Omitted.
|
|
|
|
Each of exercises 51-53 refers to the Euler phi function, denoted $\phi$, which
|
|
is defined as follows: For each integer $n \geq 1$, $\phi(n)$ is the number of
|
|
positive integers less than or equal to $n$ that have no common factors with $n$
|
|
except $\pm 1$. For example $\phi(10) = 4$ because there are four positive
|
|
integers less than or equal to $10$ that have no common factors with $10$ except
|
|
$\pm 1$ - namely, $1$, $3$, $7$, and $9$.
|
|
|
|
51. Find each of the following:
|
|
|
|
a. $\phi(15)$
|
|
|
|
Omitted.
|
|
|
|
b. $\phi(2)$
|
|
|
|
Omitted.
|
|
|
|
c. $\phi(5)$
|
|
|
|
Omitted.
|
|
|
|
d. $\phi(12)$
|
|
|
|
Omitted.
|
|
|
|
e. $\phi(11)$
|
|
|
|
Omitted.
|
|
|
|
f. $\phi(1)$
|
|
|
|
Omitted.
|
|
|
|
52. Prove that if $p$ is a prime number and $n$ is an integer with $n \geq 1$,
|
|
then $\phi(p^n) = p^n - p^{n - 1}$.
|
|
|
|
Omitted.
|
|
|
|
53. Prove that there are infinitely many integers $n$ for which $\phi(n)$ is a
|
|
perfect square.
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
Page 480
|
|
|
|
**Exercise Set 7.2**
|
|
|
|
1. The definition of one-to-one is stated in two ways:
|
|
|
|
$$ \forall x_1, x_2 \in X, \text{ if } F(x_1) = F(x_2) \text{ then } x_1 = x_2 $$
|
|
|
|
and
|
|
|
|
$$ \forall x_1, x_2 \in X, \text{ if } x_1 \neq x_2 \text{ then } F(x_1) \neq F(x_2) $$
|
|
|
|
Why are these two statements logically equivalent?
|
|
|
|
Because the second statement is the contrapositive of the first.
|
|
|
|
2. Fill in each blank with the word _most_ or _least_.
|
|
|
|
a. A function $F$ is one-to-one if, and only if, each element in the co-domain
|
|
of $F$ is the image of at _____ one element in the domain of $F$.
|
|
|
|
most
|
|
|
|
b. A function $F$ is onto if, and only if, each element in the co-domain of $F$
|
|
is the image of at _____ one element in the domain of $F$.
|
|
|
|
least
|
|
|
|
3. When asked to state the definition of one-to-one, a student replies, "A
|
|
function $f$ is one-to-one if, and only if, every element of $X$ is sent by
|
|
$f$ to exactly one element of $Y$." Give a counterexample to show that the
|
|
student's reply is incorrect.
|
|
|
|
Suppose $X = \{a, b\}$ and $Y = \{1, 2\}$, and that $f: X \to Y$ such that
|
|
$f(a) = 1$ and $f(b) = 1$. This fulfills the students definition as every
|
|
element in $X$ is sent by $f$ to exactly one element of $Y$. Note that $f$ is
|
|
not one-to-one though, as $f(a) = f(b)$, but $a \neq b$.
|
|
|
|
4. Let $f: X \to Y$ be a function. True or false? A sufficient condition for $f$
|
|
to be one-to-one is that for every element $y$ in $Y$, there is at most one
|
|
$x$ in $X$ with $f(x) = y$. Explain your answer.
|
|
|
|
This is true. This is the definition for one-to-one, since every element $y$ in
|
|
$Y$ has at most one element $x$ in $X$, this means that, given any $x_1$ and
|
|
$x_2$ in $X$, if $x_1 \neq x_2$, then $F(x_1) \neq F(x_2)$. The key wording that
|
|
makes this true is "at most one."
|
|
|
|
5. All but two of the following statements are correct ways to express the fact
|
|
that a function $f$ is onto. Find the two that are incorrect.
|
|
|
|
a. $f$ is onto $\Leftrightarrow$ every element in its co-domain is the image of
|
|
some element in its domain.
|
|
|
|
true.
|
|
|
|
b. $f$ is onto $\Leftrightarrow$ every element in its domain has a corresponding
|
|
image in its co-domain.
|
|
|
|
false.
|
|
|
|
c. $f$ is onto $\Leftrightarrow \forall y \in Y, \exists x \in X$ such that
|
|
$f(x) = y$.
|
|
|
|
true.
|
|
|
|
d. $f$ is onto $\Leftrightarrow \forall x \in X, \exists y \in Y$ such that
|
|
$f(x) = y$.
|
|
|
|
false.
|
|
|
|
e. $f$ is onto $\Leftrightarrow$ the range of $f$ is the same as the co-domain
|
|
of $f$.
|
|
|
|
true.
|
|
|
|
6. Let $X = \{1, 5, 9\}$ and $Y = \{3, 4, 7\}$.
|
|
|
|
a. Define $f: X \to Y$ by specifying that
|
|
|
|
$$ f(1) = 4, f(5) = 7, f(9) = 4 $$
|
|
|
|
Is $f$ one-to-one? Is $f$ onto? Explain your answers.
|
|
|
|
$f$ is not one-to-one, as $f(1) = 4$ and $f(9) = 4$, but $1 \neq 9$.
|
|
|
|
$f$ is not onto, as there is no $x \in X$ such that $f(x) = 3$
|
|
|
|
b. Define $g: X \to Y$ by specifying that
|
|
|
|
$$ g(1) = 7, g(5) = 3, g(9) = 4 $$
|
|
|
|
Is $g$ one-to-one? Is $g$ onto? Explain your answers.
|
|
|
|
$g$ is one-to-one, as $g(1) \neq g(5) \neq g(9)$.
|
|
|
|
$g$ is onto, as every $y$ in $Y$ is an image of at least one $x$ in $X$.
|
|
|
|
7. Let $X = \{a, b, c, d\}$ and $Y = \{e, f, g\}$. Define functions $F$ and $G$
|
|
by the arrow diagrams below.
|
|
|
|
(See page 481) for images.
|
|
|
|
a. Is $F$ one-to-one? Why or why not? Is it onto? Why or why not?
|
|
|
|
$F$ is not one-to-one, as $F(c) = e$ and $F(d) = e$, but $c \neq d$.
|
|
|
|
$F$ is onto, as every $y$ in $Y$ is an image of at least one $x$ in $X$.
|
|
|
|
b. Is $G$ one-to-one? Why or why not? Is it onto? Why or why not?
|
|
|
|
$G$ is not one-to-one, as $G(a) = f$, $G(b) = f$, and $G(d) = f$, but
|
|
$a \neq b \neq d$.
|
|
|
|
$G$ is not onto, as $g \in Y$, but there is no $x$ in $X$ such that $G(x) = g$.
|
|
|
|
8. Let $X = \{a, b, c\}$ and $Y = \{d, e, f, g\}$. Define functions $H$ and $K$
|
|
by the arrow diagrams below.
|
|
|
|
(See page 481) for images.
|
|
|
|
a. Is $H$ one-to-one? Why or why not? Is it onto? Why or why not?
|
|
|
|
$H$ is not one-to-one, as $H(b) = f$ and $H(c) = f$, but $b \neq a$.
|
|
|
|
$H$ is not onto, as both $e$ and $g$ are in $Y$, but there is no $x$ in $X$ such
|
|
that $H(x) = e$ nor $H(x) = g$.
|
|
|
|
b. Is $K$ one-to-one? Why or why not? Is it onto? Why or why not?
|
|
|
|
$K$ is one-to-one, as $K(a) \neq K(b) \neq K(c)$.
|
|
|
|
$K$ is not onto, as $g \in Y$, but $\nexists x \in X$ such that $K(x) = g$.
|
|
|
|
9. Let $X = \{1, 2, 3\}$, $Y = \{1, 2, 3, 4\}$, and $Z = \{1, 2\}$.
|
|
|
|
a. Define a function $f: X \to Y$ that is one-to-one but not onto.
|
|
|
|
Let $f: X \to Y$ such that $f(1) = 1$, $f(2) = 2$, and $f(3) = 3$.
|
|
|
|
b. Define a function $g: X \to Z$ that is onto but not one-to-one.
|
|
|
|
Let $g: X \to Z$ such that $g(1) = 1$, $g(2) = 2$, and $g(3) = 2$.
|
|
|
|
c. Define a function $h: X \to X$ that is neither one-to-one nor onto.
|
|
|
|
Let $h: X \to X$ such that $h(1) = 1$, $h(2) = 1$, and $h(3) = 1$.
|
|
|
|
d. Define a function $k: X \to X$ that is one-to-one and onto but is not the
|
|
identity function on $X$.
|
|
|
|
Let $k: X \to X$, such that $k(1) = 3$, $k(2) =1$, $k(3) = 2$.
|
|
|
|
10.
|
|
|
|
a. Define $f: \mathbb{Z} \to \mathbb{Z}$ by the rule $f(n) = 2n$, for every
|
|
integer $n$.
|
|
|
|
i. Is $f$ one-to-one? Prove or give a counterexample.
|
|
|
|
$f$ is one-to-one.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $f(n_1) = f(n_2)$.
|
|
|
|
To prove $f$ is one-to-one, it must be shown that $n_1 = n_2$.
|
|
|
|
By definition of $f$, $f(n_1) = f(n_2)$ can be substituted with:
|
|
|
|
$$ 2n_1 = 2n_2 $$
|
|
|
|
Then, by algebra:
|
|
|
|
$$ n_1 = n_2 $$
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
ii. Is $f$ onto? prove or give a counterexample.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Consider $1 \in \mathbb{Z}$. It is claimed that $1 \neq f(n)$ for any integer
|
|
$n$.
|
|
|
|
For if there were an integer $n$ such that $1 = f(n)$, then, by definition of
|
|
$f$, $1 = 2n$.
|
|
|
|
Then, by division:
|
|
|
|
$$ n = \frac{1}{2} $$
|
|
|
|
.
|
|
|
|
Note then that $n$ is not an integer. Hence $1 \neq f(n)$ for any integer $n$.
|
|
|
|
Therefore, it can be concluded that $f$ is not onto.
|
|
|
|
Q.E.D.
|
|
|
|
b. Let $2\mathbb{Z}$ denote the set of all even integers. That is,
|
|
$2\mathbb{Z} = \{n \in \mathbb{Z} | n = 2k \text{, for some integer } k\}$.
|
|
Define $h: \mathbb{Z} \to 2\mathbb{Z}$ by the rule $h(n) = 2n$, for each integer
|
|
$n$. Is $h$ onto? Prove or give a counterexample.
|
|
|
|
$h$ is onto.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $m$ is an integer such that $m \in 2\mathbb{Z}$.
|
|
|
|
To prove that $h$ is onto, it must be shown that there is some integer which
|
|
when passed through $h$ equals $m$.
|
|
|
|
By definition of $2\mathbb{Z}$, this means that:
|
|
|
|
$$ m = 2k $$
|
|
|
|
for some integer $k$.
|
|
|
|
Then:
|
|
|
|
$$ h(k) = 2k = m $$
|
|
|
|
Hence there is an integer, namely $k$, such that $h(k) = m$.
|
|
|
|
Q.E.D.
|
|
|
|
11.
|
|
|
|
a. Define $g: \mathbb{Z} \to \mathbb{Z}$ by the rule $g(n) = 4n - 5$, for each
|
|
integer $n$.
|
|
|
|
i. Is $g$ one-to-one? Prove or give a counterexample.
|
|
|
|
$g$ is one-to-one.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n_1, n_2 \in \mathbb{Z}$ such that $g(n_1) = g(n_2)$.
|
|
|
|
To prove $g$ is one-to-one, it must be shown that $n_1 = n_2$.
|
|
|
|
By definition of $g$, $g(n_1) = g(n_2)$ can be expressed by substitution as:
|
|
|
|
$$ 4n_1 - 5 = 4n_2 - 5 $$
|
|
|
|
Then, by algebra:
|
|
|
|
$$ 4n_1 = 4n_2 $$
|
|
|
|
$$ n_1 = n_2 $$
|
|
|
|
This is what was to be shown, and it can therefore be concluded that $g$ is
|
|
one-to-one.
|
|
|
|
Q.E.D.
|
|
|
|
ii. Is $g$ onto? Prove or give a counterexample.
|
|
|
|
$g$ is not onto.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Suppose $m \in \mathbb{Z}$.
|
|
|
|
To prove that $g$ is onto, it must be shown that there exists some integer $n$
|
|
such that $g(n) = m$.
|
|
|
|
By the definition of $g$, $g(n) = m$ can be expressed by substitution as:
|
|
|
|
$$ 4n - 5 = m $$
|
|
|
|
Then, by algebra:
|
|
|
|
$$ 4n = m + 5 $$
|
|
|
|
$$ n = \frac{m + 5}{4} $$
|
|
|
|
But then $n$ is not necessarily an integer, say in the case of $m = 0$. Note
|
|
that $0 \in \mathbb{Z}$, but if $m = 0$, then $n = \dfrac{5}{4}$, and
|
|
$\dfrac{5}{4}$ is not an integer.
|
|
|
|
Hence there is no $n$, such that $g(n) = 0$.
|
|
|
|
Therefore it can be concluded that $g$ is not onto.
|
|
|
|
Q.E.D.
|
|
|
|
b. Define $G: \mathbb{R} \to \mathbb{R}$ by the rule $G(x) = 4x - 5$ for every
|
|
real number $x$. Is $G$ onto? Prove or give a counterexample.
|
|
|
|
$G$ is onto.
|
|
|
|
**Proof:**
|
|
|
|
Suppose there exists some $y \in \mathbb{R}$.
|
|
|
|
To prove $G$ is onto, it must be shown that there exists some $x \in \mathbb{R}$
|
|
such that $G(x) = y$.
|
|
|
|
By the given definition for $G$, $G(x) = y$ can be expressed by substitution as:
|
|
|
|
$$ 4x - 5 = y $$
|
|
|
|
$$ 4x = y + 5 $$
|
|
|
|
$$ x = \frac{y + 5}{4} $$
|
|
|
|
Now, $\dfrac{y + 5}{4}$ is a real number by the addition and division of real
|
|
numbers. Hence $x = \dfrac{y + 5}{4} \in \mathbb{R}$.
|
|
|
|
Then, evaluate $G\left(\dfrac{y + 5}{4}\right)$:
|
|
|
|
$$ G\left(\frac{y + 5}{4}\right) = 4\left(\frac{y + 5}{4}\right) - 5 $$
|
|
|
|
$$ = y + 5 - 5 $$
|
|
|
|
$$ = y $$
|
|
|
|
Hence, it has been shown that $G(x) = y$ for some $x$.
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $G$ is onto.
|
|
|
|
Q.E.D.
|
|
|
|
12.
|
|
|
|
a. Define $F: \mathbb{Z} \to \mathbb{Z}$ by the rule $F(n) = 2 - 3n$, for each
|
|
integer $n$.
|
|
|
|
i. Is $F$ one-to-one? Prove or give a counterexample.
|
|
|
|
$F$ is one-to-one.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n_1, n_2 \in \mathbb{Z}$ such that $F(n_1) = F(n_2)$.
|
|
|
|
To prove $F$ is one-to-one, it must be shown that $n_1 = n_2$.
|
|
|
|
By the given definition of $F$, $F(n_1) = F(n_2)$ can be expressed by
|
|
substitution as:
|
|
|
|
$$ 2 - 3n_1 = 2 - 3n_2 $$
|
|
|
|
Then, by algebra:
|
|
|
|
$$ -3n_1 = -3n_2 $$
|
|
|
|
$$ n_1 = n_2 $$
|
|
|
|
Hence it has been shown that $n_1 = n_2$ when $F(n_1) = F(n_2)$.
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $F$ is
|
|
one-to-one.
|
|
|
|
Q.E.D.
|
|
|
|
ii. Is $F$ onto? Prove or give a counterexample.
|
|
|
|
$F$ is not onto.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
To prove that $F$ is onto, it must be shown that there exists some
|
|
$m \in \mathbb{Z}$ such that $m = 2 - 3n$.
|
|
|
|
Evaluating for $n$ shows:
|
|
|
|
$$ m = 2 - 3n $$
|
|
|
|
$$ 3n = 2 - m $$
|
|
|
|
$$ n = \dfrac{2 - m}{3} $$
|
|
|
|
But since $n$ must be an integer by the definition for $F$, this evaluation
|
|
shows that there exists at least one $m \in \mathbb{Z}$ that is not in the
|
|
co-domain of $F$.
|
|
|
|
Take $m = 1$, for example, note that $1 \in \mathbb{Z}$. But, when $m = 1$, then
|
|
$n = \dfrac{1}{3}$, which is not an integer.
|
|
|
|
Therefore, it can be concluded that $F$ is not onto.
|
|
|
|
Q.E.D.
|
|
|
|
b. Define $G: \mathbb{R} \to \mathbb{R}$ by the rule $G(x) = 2 - 3x$ for each
|
|
real number $x$. Is $G$ onto? Prove or give a counterexample.
|
|
|
|
$G$ is onto.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $y \in \mathbb{R}$.
|
|
|
|
To prove that $G$ is onto, it must be shown that $G(x) = y$ for some
|
|
$x \in \mathbb{R}$.
|
|
|
|
By the given definition for $G$, $G(x) = y$ can be expressed by substitution as:
|
|
|
|
$$ 2 - 3x = y $$
|
|
|
|
Then, by algebra:
|
|
|
|
$$ -3x = y - 2 $$
|
|
|
|
$$ x = -\left(\frac{y - 2}{3}\right) $$
|
|
|
|
$$ x = \frac{2 - y}{3} $$
|
|
|
|
Now, $\dfrac{2 - y}{3}$ by the product, division, and addition of real numbers.
|
|
It follows that $x \in \mathbb{R}$ since $x = \dfrac{2 - y}{3}$.
|
|
|
|
Now, evaluating $G\left(\dfrac{2 - y}{3}\right)$:
|
|
|
|
$$ G\left(\frac{2 - y}{3}\right) = 2 - 3\left(\frac{2 - y}{3}\right) $$
|
|
|
|
$$ = 2 - (2 - y) $$
|
|
|
|
$$ = 2 - 2 + y $$
|
|
|
|
$$ = y $$
|
|
|
|
Hence it has been shown that $G(x) = y$ for some $x \in \mathbb{R}$.
|
|
|
|
This is what was to be shown, and therefore it can be concluded that $G$ is
|
|
onto.
|
|
|
|
Q.E.D.
|
|
|
|
13.
|
|
|
|
a. Define $H: \mathbb{R} \to \mathbb{R}$ by the rule $H(x) = x^2$, for each real
|
|
number $x$.
|
|
|
|
i. Is $H$ one-to-one? Prove or give a counterexample.
|
|
|
|
$H$ is not one-to-one.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Suppose $x_1, x_2 \in \mathbb{R}$ such that $H(x_1) = H(x_2)$.
|
|
|
|
To prove that $H$ is one-to-one, it must be shown that $x_1 = x_2$.
|
|
|
|
Substituting $H(x_1) = H(x_2)$ by the given definition for $H$:
|
|
|
|
$$ (x_1)^2 = (x_2)^2 $$
|
|
|
|
$$ \sqrt{(x_1)^2} = \sqrt{(x_2)^2} $$
|
|
|
|
$$ \pm x_1 = \pm x_2 $$
|
|
|
|
But $\pm x_1 = x_1$ or $\pm x_1 = -x_1$. Similarly, $\pm x_2 = x_2$ or
|
|
$\pm x_2 = -x_2$. It follows then that there exists some $-x_1 = x_2$ or
|
|
$x_1 = -x_2$, but this cannot be the case when $H(x_1) = H(x_2)$.
|
|
|
|
Consider $x_1 = -2$,and $x_2 = 2$. Note that $x_1, x_2 \in \mathbb{R}$.
|
|
|
|
Then:
|
|
|
|
$$ H(x_1) = (-2)^2 = 4 = (2)^2 = H(x_2) $$
|
|
|
|
So, $H(-2) = H(2)$, but $-2 \neq 2$. Therefore, by the definition of one-to-one,
|
|
it can be concluded that $H$ is not one-to-one.
|
|
|
|
Q.E.D.
|
|
|
|
ii. Is $H$ onto? Prove or give a counterexample.
|
|
|
|
$H$ is not onto.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Suppose there is some $y$ such that $y \in \mathbb{R}$.
|
|
|
|
To prove that $H$ is onto, it must be shown that $H(x) = y$ for some
|
|
$x \in \mathbb{R}$.
|
|
|
|
By substitution of the given definition for $H$:
|
|
|
|
$$ x^2 = y $$
|
|
|
|
$$ x = \sqrt{y} $$
|
|
|
|
Now, $\sqrt{y} \in \mathbb{R}$, but only if $y \geq 0$. If $y < 0$, then
|
|
$\sqrt{y}$ is a complex or imaginary number.
|
|
|
|
Consider $y = -1$. Note that $-1 \in \mathbb{R}$.
|
|
|
|
Then, by substitution into $H(x)$:
|
|
|
|
$$ x^2 = -1 $$
|
|
|
|
$$ x = \sqrt{-1} $$
|
|
|
|
$$ x = i \notin \mathbb{R} $$
|
|
|
|
Thus it has been shown that there is no such $x \in \mathbb{R}$ such that
|
|
$H(x) = -1$.
|
|
|
|
By the definition of onto, it can be concluded that $H$ is not onto.
|
|
|
|
Q.E.D.
|
|
|
|
b. Define $K: \mathbb{R}^{\text{nonneg}} \to \mathbb{R}^{\text{nonneg}}$ by the
|
|
rule $K(x) = x^2$, for each nonnegative real number $x$. Is $K$ onto? Prove or
|
|
give a counterexample.
|
|
|
|
$K$ is onto.
|
|
|
|
**Proof:**
|
|
|
|
Suppose there exists some $y$ such that $y \in \mathbb{R}^{\text{nonneg}}$.
|
|
|
|
To prove that $K$ is onto, it must be shown that $K(x) = y$ for some
|
|
$x \in \mathbb{R}^{\text{nonneg}}$.
|
|
|
|
By substitution of the given definition for $K$:
|
|
|
|
$$ x^2 = y $$
|
|
|
|
$$ x = \sqrt{y} $$
|
|
|
|
Now, $\sqrt{y} \in \mathbb{R}^{\text{nonneg}}$ by the square root of positive
|
|
real numbers.
|
|
|
|
Evaluating for $K(\sqrt{y})$:
|
|
|
|
$$ K(\sqrt{y}) = (\sqrt{y})^2 $$
|
|
|
|
$$ = y $$
|
|
|
|
Thus it has been shown that $K(x) = y$ for some
|
|
$x \in \mathbb{R}^{\text{nonneg}}$.
|
|
|
|
This is what was to be shown, and therefore it can be concluded that $K$ is
|
|
onto.
|
|
|
|
Q.E.D.
|
|
|
|
14. Explain the mistake in the following "proof."
|
|
|
|
**Theorem:** The function $f: \mathbb{Z} \to \mathbb{Z}$ defined by the formula
|
|
$f(n) = 4n + 3$, for each integer $n$, is one-to-one.
|
|
|
|
"**Proof:** Suppose any integer $n$ is given. Then by definition of $f$, there
|
|
is only one possible value for $f(n)$ - namely, $4n + 3$. Hence $f$ is
|
|
one-to-one."
|
|
|
|
This "proof" makes the mistake of assuming the conclusion. In order to prove
|
|
that a function is one-to-one, it must be shown that given any two inputs, say
|
|
$n_1, n_2 \in \mathbb{Z}$ such that $f(n_1) = f(n_2)$, then $n_1 = n_2$.
|
|
|
|
Alternatively, one could show that given any two outputs, say
|
|
$f(n_1), f(n_2) \in \mathbb{Z}$, that if $f(n_1) \neq f(n_2)$, then
|
|
$n_1 \neq n_2$.
|
|
|
|
In each of 15-18 a function $f$ is defined on a set of real numbers. Determine
|
|
whether or not $f$ is one-to-one and justify your answer.
|
|
|
|
15. $f(x) = \dfrac{x + 1}{x}$, for each number $x \neq 0$
|
|
|
|
Scratch Proof:
|
|
|
|
$$ \frac{x_1 + 1}{x_1} = \frac{x_2 + 1}{x_2} $$
|
|
|
|
$$ (x_2)(x_1 + 1) = (x_1)(x_2 + 1) $$
|
|
|
|
$$ x_2x_1 + x_2 = x_2x_1 + x_1 $$
|
|
|
|
$$ x_2 = x_1 $$
|
|
|
|
$f$ is one-to-one.
|
|
|
|
16. $f(x) = \dfrac{x}{x^2 + 1}$, for each real number $x$
|
|
|
|
$$ \frac{x_1}{x_1^2 + 1} = \frac{x_2}{x_2^2 + 1} $$
|
|
|
|
$$ (x_2^2 + 1)x_1 = (x_1^2 + 1)x_2 $$
|
|
|
|
$$ x_2^2x_1 + x_1 = x_1^2x_2 + x_2 $$
|
|
|
|
$f$ is not one-to-one since $x_1 \neq x_2$. Take $x_1 = 2$ and
|
|
$x_2 = \dfrac{1}{2}$:
|
|
|
|
$$ \frac{2}{2^2 + 1} = \frac{\dfrac{1}{2}}{\left(\dfrac{1}{2}\right)^2 + 1} $$
|
|
|
|
$$ \frac{2}{4 + 1} = \frac{\dfrac{1}{2}}{\dfrac{1}{4} + 1} $$
|
|
|
|
$$ \frac{2}{5} = \frac{\dfrac{1}{2}}{\dfrac{5}{4}} $$
|
|
|
|
$$ \frac{2}{5} = \frac{1}{2} \cdot \frac{4}{5} $$
|
|
|
|
$$ \frac{2}{5} = \frac{4}{10} $$
|
|
|
|
$$ \frac{2}{5} = \frac{2}{5} $$
|
|
|
|
Since $f(2) = f\left(\dfrac{1}{2}\right)$, but $2 \neq \dfrac{1}{2}$, it can be
|
|
concluded that $f$ is not one-to-one.
|
|
|
|
17. $f(x) = \dfrac{3x - 1}{x}$, for each real number $x \neq 0$
|
|
|
|
$$ \frac{3x_1 - 1}{x_1} = \frac{3x_2 - 1}{x_2} $$
|
|
|
|
$$ x_2(3x_1 - 1) = x_1(3x_2 - 1) $$
|
|
|
|
$$ 3x_1x_2 - x_2 = 3x_1x_2 - x_1 $$
|
|
|
|
$$ -x_2 = -x_1 $$
|
|
|
|
$$ x_2 = x_1 $$
|
|
|
|
Since $x_1 = x_2$, $f$ is one-to-one.
|
|
|
|
18. $f(x) = \dfrac{x + 1}{x - 1}$, for each real number $x \neq 1$
|
|
|
|
$$ \frac{x_1 + 1}{x_1 - 1} = \frac{x_2 + 1}{x_2 - 1} $$
|
|
|
|
$$ (x_1 + 1)(x_2 - 1) = (x_2 + 1)(x_1 - 1) $$
|
|
|
|
$$ x_1x_2 + x_2 - x_1 - 1 = x_1x_2 + x_1 - x_2 - 1 $$
|
|
|
|
$$ x_1x_2 + x_2 - x_1 - 1 = x_1x_2 + x_1 - x_2 - 1 $$
|
|
|
|
$$ x_2 - x_1 = x_1 - x_2 $$
|
|
|
|
$$ 2x_2 = 2x_1 $$
|
|
|
|
$$ x_2 = x_1 $$
|
|
|
|
$f$ is one-to-one.
|
|
|
|
19. Referring to Example 7.2.3, assume that records with the following ID
|
|
numbers are to be placed in sequence into Table 7.2.1. Find the position
|
|
into which each record is placed.
|
|
|
|
a. $417302072$
|
|
|
|
$$ 417302072 \mod 11 = 0 $$
|
|
|
|
Since position $0$ is empty, $417302072$ is placed in position $0$.
|
|
|
|
b. $364981703$
|
|
|
|
$$ 364981703 \mod 11 = 9 $$
|
|
|
|
Since position $9$ is empty, $364981703$ is placed in position $9$.
|
|
|
|
c. $283090787$
|
|
|
|
$$ 283090787 \mod 11 = 1 $$
|
|
|
|
Since position $1$ is not empty, position $2$ is checked. Since position $2$ is
|
|
not empty, position $3$ is checked. Since position $3$ is empty, $283090787$ is
|
|
placed in position $3$.
|
|
|
|
20. Define $\text{Floor}: \mathbb{R} \to \mathbb{Z}$ by the formula
|
|
$\text{Floor}(x) = \lfloor x \rfloor$, for every real number $x$.
|
|
|
|
a. Is $\text{Floor}$ one-to-one? Prove or give a counterexample.
|
|
|
|
$\text{Floor}$ is not one-to-one.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Consider $x_1, x_2 \in \mathbb{R}$ such that $x_1 = 1.1$ and $x_2 = 1.2$.
|
|
|
|
By the definition of $\text{Floor}$:
|
|
|
|
$$ \text{Floor}(1.1) = \lfloor 1.1 \rfloor = 1 $$
|
|
|
|
and
|
|
|
|
$$ \text{Floor}(1.2) = \lfloor 1.2 \rfloor = 1 $$
|
|
|
|
Thus $\text{Floor}(1.1) = \text{Floor}(1.2)$, but $1.1 \neq 1.2$.
|
|
|
|
By the definition of one-to-one, it can be concluded that $\text{Floor}$ is not
|
|
one-to-one.
|
|
|
|
Q.E.D.
|
|
|
|
b. Is $\text{Floor}$ onto? Prove or give a counterexample.
|
|
|
|
$\text{Floor}$ is onto.
|
|
|
|
**Proof:**
|
|
|
|
Suppose there exists some $y$ such that $y \in \mathbb{Z}$.
|
|
|
|
To prove that $\text{Floor}$ is onto, it must be shown that
|
|
$\text{Floor}(x) = y$ for some $x \in \mathbb{R}$.
|
|
|
|
Now, let $x = y$.
|
|
|
|
By substitution of the given definition for $\text{Floor}$, and the supposition
|
|
that $x = y$:
|
|
|
|
$$ \lfloor x \rfloor = y $$
|
|
|
|
By substitution for $x$:
|
|
|
|
$$ \lfloor y \rfloor = y $$
|
|
|
|
$$ y = y $$
|
|
|
|
Thus it has been shown that $\text{Floor}(x) = y$ for some $x \in \mathbb{R}$.
|
|
|
|
This is what was to be shown, and therefore, by the definition of onto, it can
|
|
be concluded that $\text{Floor}$ is onto.
|
|
|
|
Q.E.D.
|
|
|
|
21. Let $S$ be the set of all strings of $0$'s and $1$'s, and define
|
|
$L: S \to \mathbb{Z}^{\text{nonneg}}$ by
|
|
|
|
$$ L(s) = \text{ the length of } s \text{, for every string } s \text{ in } S $$
|
|
|
|
a. Is $L$ one-to-one? Prove or give a counterexample.
|
|
|
|
$L$ is not one-to-one.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Suppose $s_1, s_2 \in S$ such that $s_1 = 10$ and $s_2 = 01$.
|
|
|
|
Then, by definition of $L$:
|
|
|
|
$$ L(s_1) = 2 = L(s_2) $$
|
|
|
|
Hence $L(s_1) = L(s_2)$ and $s_1 \neq s_2$.
|
|
|
|
Therefore it can be concluded, by the definition of one-to-one, that $L$ is not
|
|
one-to-one.
|
|
|
|
b. Is $L$ onto? Prove or give a counterexample.
|
|
|
|
$L$ is onto.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is some integer such that $n \in \mathbb{Z}^{\text{nonneg}}$.
|
|
|
|
To prove that $L$ is onto, it must be shown that $L(s) = n$ for some string
|
|
$s \in S$.
|
|
|
|
Let $s$ be some string such that $s \in S$.
|
|
|
|
Since $s \in S$, this means that the $s$ is either $\lambda$ (where $\lambda$ is
|
|
the null string), or some combination of all strings of $0$'s and $1$'s.
|
|
|
|
This means that the length of $s$ is at least $0$ (when $s = \lambda$), and
|
|
otherwise is an ever increasing integer. Therefore for every $s$ passed through
|
|
$L$, there will always be a corresponding nonnegative integer $n$.
|
|
|
|
By the definition of onto, it can therefore be concluded that $L$ is onto.
|
|
|
|
Q.E.D.
|
|
|
|
22. Let $S$ be the set of all strings of $0$'s and $1$'s, and define
|
|
$D: S \to \mathbb{Z}$ as follows: For every $s \in S$,
|
|
|
|
$$ D(s) = \text{ the number of 1's in } s \text{ minus the number of 0's in } s $$
|
|
|
|
a. Is $D$ one-to-one? Prove or give a counterexample.
|
|
|
|
$D$ is not one-to-one.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Suppose $s_1, s_2 \in S$ such that $s_1 = 01$ and $s_2 = 10$.
|
|
|
|
By definition of $D$:
|
|
|
|
$$ D(s_1) = 0 = D(s_2) $$
|
|
|
|
So $D(s_1) = D(s_2)$, but $s_1 \neq s_2$.
|
|
|
|
Therefore, by the definition of one-to-one, $D$ is not one-to-one.
|
|
|
|
Q.E.D.
|
|
|
|
b. Is $D$ onto? Prove or give a counterexample.
|
|
|
|
$D$ is onto.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n \in \mathbb{Z}$.
|
|
|
|
To prove $D$ is onto, it must be shown that $D(s) = n$ for some string
|
|
$s \in S$.
|
|
|
|
Consider three cases:
|
|
|
|
_Case $n = 0$:_
|
|
|
|
Let $s = \lambda$. Then $D(s) = 0 = n$.
|
|
|
|
_Case $n > 0$:_
|
|
|
|
Let $s$ be a string of $n$ ones. Then $D(s) = n - 0 = n$.
|
|
|
|
_Case $n < 0$:_
|
|
|
|
Let $s$ be a string of $|n|$ ones. Then $D(s) = 0 - |n| = n$.
|
|
|
|
In all cases, there exists some $s \in S$ such that $D(s) = n$.
|
|
|
|
Therefore, by definition of onto, $D$ is onto.
|
|
|
|
Q.E.D.
|
|
|
|
23. Define $F: \mathscr{P}(\{a, b, c\}) \to \mathbb{Z}$ as follows: For every
|
|
$A$ in $\mathscr{P}(\{a, b, c\})$,
|
|
|
|
$$ F(A) = \text{ the number of elements in } A $$
|
|
|
|
a. Is $F$ one-to-one? Prove or give a counterexample.
|
|
|
|
$F$ is not one-to-one.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Suppose $A_1 = \{a\}$, and $A_2 = \{b\}$.
|
|
|
|
Then, by the definition of $F$:
|
|
|
|
$$ F(A_1) = 1 = F(A_2) $$
|
|
|
|
So $F(A_1) = F(A_2)$, but $A_1 \neq A_2$.
|
|
|
|
By the definition of one-to-one, it can be concluded that $F$ is not one-to-one.
|
|
|
|
Q.E.D.
|
|
|
|
b. Is $F$ onto? Prove or give a counterexample.
|
|
|
|
$F$ is not onto.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Consider $-1 \in \mathbb{Z}$.
|
|
|
|
To prove that $F$ is onto, it would have to be shown that $F(A) = -1$ for some
|
|
$A \in \mathscr{P}(\{a, b, c\})$, but:
|
|
|
|
$$ \mathscr{P}(\{a, b, c\}) = \{\emptyset, \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a, b, c\}\} $$
|
|
|
|
This shows that there is no element in $\mathscr{P}(\{a, b, c\})$ such that
|
|
$F(A) = -1$ even though $-1 \in \mathbb{Z}$.
|
|
|
|
Therefore, $F$ is not onto.
|
|
|
|
Q.E.D.
|
|
|
|
24. Let $S$ be the set of all strings of $a$'s and $b$'s, and define
|
|
$N: S \to \mathbb{Z}$ by
|
|
|
|
$$ N(s) = \text{ the number of a's in } s \text{, for each } s \in S $$
|
|
|
|
a. Is $N$ one-to-one? Prove or give a counterexample.
|
|
|
|
$N$ is not one-to-one.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Consider $s_1, s_2 \in S$ such that $s_1 = ab$ and $s_2 = ba$.
|
|
|
|
By the given definition for $N$:
|
|
|
|
$$ N(s_1) = 1 = N(s_2) $$
|
|
|
|
Thus $N(s_1) = N(s_2)$, but $s_1 \neq s_2$.
|
|
|
|
By the definition of one-to-one, $N$ is not one-to-one.
|
|
|
|
Q.E.D.
|
|
|
|
b. Is $N$ onto? Prove or give a counterexample.
|
|
|
|
$N$ is not onto.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Consider $-1 \in \mathbb{Z}$.
|
|
|
|
To prove that $N$ is onto, it would have to be shown that $N(s) = -1$ for some
|
|
$s \in S$, but by definition of string, and by the definition of $s \in S$, $s$
|
|
can have at a minimum $0$ $a$'s in it.
|
|
|
|
Therefore, $N$ is not onto.
|
|
|
|
Q.E.D.
|
|
|
|
25. Let $S$ be the set of all strings in $a$'s and $b$'s, and define
|
|
$C: S \to S$ by
|
|
|
|
$$ C(s) = as \text{, for each } s \in S $$
|
|
|
|
($C$ is called **concatenation** by $a$ on the left.)
|
|
|
|
a. Is $C$ one-to-one? Prove or give a counterexample.
|
|
|
|
$C$ is one-to-one.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $s_1, s_2 \in S$ such that $C(s_1) = C(s_2)$.
|
|
|
|
To prove $C$ is one to one, it must be shown that $s_1 = s_2$.
|
|
|
|
By the given definition of $C$:
|
|
|
|
$$ as_1 = as_2 $$
|
|
|
|
Since the strings $as_1$ and $as_2$ are equal and share the same first character
|
|
$a$, the remaining portions $s_1$ and $s_2$ must also be equal.
|
|
|
|
$$ s_1 = s_2 $$
|
|
|
|
Since $C(s_1) = C(s_2)$ and $s_1 = s_2$, by the definition of one-to-one, it can
|
|
be concluded that $C$ is one-to-one.
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
b. Is $C$ onto? Prove or give a counterexample.
|
|
|
|
$C$ is not onto.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Consider some string $t \in S$ such that $t = b$.
|
|
|
|
To prove that $C$ is onto, it must be shown that $C(s) = b$ for some $s \in S$.
|
|
|
|
But, by definition of $C$, $C(s) = as$ for each $s \in S$, but $b$ does not have
|
|
a concatenated $a$ on the left.
|
|
|
|
Therefore, by definition of onto, it can be concluded that $C$ is not onto.
|
|
|
|
Q.E.D.
|
|
|
|
26. Define $S: \mathbb{Z}^+ \to \mathbb{Z}^+$ by the rule: For each integer $n$,
|
|
|
|
$$ S(n) = \text{ the sum of the positive divisors of } n $$
|
|
|
|
a. Is $S$ one-to-one? Prove or give a counterexample.
|
|
|
|
$S$ is not one-to-one.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Consider $n_1, n_2 \in \mathbb{Z}^+$ where $n_1 = 6$ and $n_2 = 11$.
|
|
|
|
By definition of $S$:
|
|
|
|
$$ S(n_1) = 6 + 3 + 2 + 1 = 12 = 11 + 1 = S(n_2) $$
|
|
|
|
So $S(n_1) = S(n_2)$, but $n_1 \neq n_2$.
|
|
|
|
By the definition of one-to-one, $S$ is not one-to-one.
|
|
|
|
Q.E.D.
|
|
|
|
b. Is $S$ onto? Prove or give a counterexample.
|
|
|
|
$S$ is not onto.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Consider $5 \in \mathbb{Z}^+$.
|
|
|
|
To prove $S$ is onto, it would have to be shown that $S(n) = 5$ for some
|
|
$n \in \mathbb{Z}^+$.
|
|
|
|
In order for $S(n) = 5$, note that it must be the case that $n < 5$.
|
|
|
|
But $S(1) = 1$, $S(2) = 3$, $S(3) = 4$, and $S(4) = 7$.
|
|
|
|
Hence there is no positive integer $n$ such that $S(n) = 5$.
|
|
|
|
Q.E.D.
|
|
|
|
27. Let $D$ be the set of all finite subsets of positive integers, and define
|
|
$T: \mathbb{Z}^+ \to D$ by the following rule:
|
|
|
|
For every integer $n$,
|
|
$T(n) = \text{ the set of all of the positive divisors of } n$.
|
|
|
|
a. Is $T$ one-to-one? Prove or give a counterexample.
|
|
|
|
$T$ is one-to-one.
|
|
|
|
**Proof (by contradiction):**
|
|
|
|
Suppose $n_1, n_2 \in \mathbb{Z}^+$ such that $n_1 \neq n_2$ and
|
|
$T(n_1) = T(n_2)$.
|
|
|
|
Since $n_1 \neq n_2$, it follows that $n_1 < n_2$ or $n_1 > n_2$.
|
|
|
|
_Case $n_1 < n_2$:_
|
|
|
|
By the definition of $T$, $n_2$ is a positive divisor of $n_2$, so
|
|
$n_2 \in T(n_2)$.
|
|
|
|
But, since $T(n_1) = T(n_2)$, this means that $n_2 \in T(n_1)$.
|
|
|
|
This means that $n_2$ is a positive divisor of $n_1$, or $n_1 = n_2$. This is a
|
|
contradiction.
|
|
|
|
_Case $n_1 > n_2$:_
|
|
|
|
By the definition of $T$, $n_1$ is a positive divisor of $n_1$, so
|
|
$n_1 \in T(n_1)$.
|
|
|
|
But, since $T(n_1) = T(n_2)$, this means that $n_1 \in T(n_2)$.
|
|
|
|
This means that $n_1$ is a positive divisor of $n_2$, or $n_1 = n_2$. This is a
|
|
contradiction.
|
|
|
|
In both cases, it has been shown that $n_1 = n_2$, which contradicts the
|
|
supposition.
|
|
|
|
Therefore it can be concluded that $T$ is one-to-one.
|
|
|
|
b. Is $T$ onto? Prove or give a counterexample.
|
|
|
|
$T$ is not onto.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Consider the set $\{1, 2, 3\}$. Note that $\{1, 2, 3\} \in D$.
|
|
|
|
To prove that $T$ is onto, it must be shown that $T(n) = \{1, 2, 3\}$, but the
|
|
set $\{1, 2, 3\}$ would also include $6$ since any such $n$ would also be
|
|
divisible by $6$ (by the given definition of $T$).
|
|
|
|
Since $6 \notin \{1, 2, 3\}$, it can be concluded that $T$ is not onto.
|
|
|
|
Q.E.D.
|
|
|
|
28. Define $G: \mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}$ as
|
|
follows:
|
|
|
|
$$ G(x, y) = (2y, -x) \text{ for every } (x, y) \in \mathbb{R} \times \mathbb{R} $$
|
|
|
|
a. Is $G$ one-to-one? Prove or give a counterexample.
|
|
|
|
$G$ is one-to-one.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $(x_1, y_1), (x_2, y_2) \in \mathbb{R} \times \mathbb{R}$ such that
|
|
$G(x_1, y_1) = G(x_2, y_2)$.
|
|
|
|
To prove that $G$ is one-to-one, it must be shown that
|
|
$(x_1, y_1) = (x_2, y_2)$.
|
|
|
|
By the definition for $G$:
|
|
|
|
$$ (2(y_1), -(x_1)) = (2(y_2), -(x_2)) $$
|
|
|
|
$$ (2y_1, -x_1) = (2y_2, -x_2) $$
|
|
|
|
By the definition of ordered pair (and algebra), this means that:
|
|
|
|
$$ 2y_1 = 2y_2 $$
|
|
|
|
$$ y_1 = y_2 $$
|
|
|
|
and:
|
|
|
|
$$ -x_1 = -x_2 $$
|
|
|
|
$$ x_1 = x_2 $$
|
|
|
|
Thus it has been shown that $(x_1, y_1) = (x_2, y_2)$.
|
|
|
|
By the definition of one-to-one, it can be concluded that $G$ is one-to-one.
|
|
|
|
Q.E.D.
|
|
|
|
b. Is $G$ onto? Prove or give a counterexample.
|
|
|
|
$G$ is onto.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $(t, w) \in \mathbb{R} \times \mathbb{R}$.
|
|
|
|
To prove that $G$ is onto, it must be shown that $G(x, y) = (t, w)$ for some
|
|
$(x, y) \in \mathbb{R} \times \mathbb{R}$.
|
|
|
|
By the definition for $G$:
|
|
|
|
$$ (2y, -x) = (t, w) $$
|
|
|
|
By the definition of ordered pairs (and algebra), this means that:
|
|
|
|
$$ 2y = t $$
|
|
|
|
$$ y = \frac{t}{2} $$
|
|
|
|
and:
|
|
|
|
$$ -x = w $$
|
|
|
|
$$ x = -w $$
|
|
|
|
Now, note that $\dfrac{t}{2} \in \mathbb{R}$, and $-w \in \mathbb{R}$. It
|
|
follows that $\left(\dfrac{t}{2}, -w\right) \in \mathbb{R} \times \mathbb{R}$.
|
|
|
|
Now, evaluating for $G(x, y)$, which is $G\left(-w, \dfrac{t}{2}\right)$:
|
|
|
|
$$ G\left(-w, \frac{t}{2}\right) = \left(2\left(\frac{t}{2}, -(-w)\right)\right) $$
|
|
|
|
$$ = (t, w) $$
|
|
|
|
Hence it has been shown that $G(x, y) = (t, w)$ for some
|
|
$(x, y) \in \mathbb{R} \times \mathbb{R}$.
|
|
|
|
Therefore, by the definition of onto, it can be concluded that $G$ is onto.
|
|
|
|
Q.E.D.
|
|
|
|
29. Define $H: \mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}$ as
|
|
follows:
|
|
|
|
$$ H(x, y) = (x + 1, 2 - y) \text{ for every } (x, y) \in \mathbb{R} \times \mathbb{R} $$
|
|
|
|
a. Is $H$ one-to-one? Prove or give a counterexample.
|
|
|
|
$H$ is one-to-one.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $(x_1, y_1), (x_2, y_2) \in \mathbb{R} \times \mathbb{R}$ such that
|
|
$H(x_1, y_1) = H(x_2, y_2)$.
|
|
|
|
To prove $H$ is one-to-one. It must be shown that $(x_1, y_1) = (x_2, y_2)$.
|
|
|
|
By the given definition of $H$:
|
|
|
|
$$ (x_1 + 1, 2 - y_1) = (x_2 + 1, 2 - y_2) $$
|
|
|
|
By the definition of ordered pair (and algebra):
|
|
|
|
$$ x_1 + 1 = x_2 + 1 $$
|
|
|
|
$$ x_1 = x_2 $$
|
|
|
|
and:
|
|
|
|
$$ 2 - y_1 = 2 - y_2 $$
|
|
|
|
$$ -y_1 = -y_2 $$
|
|
|
|
$$ y_1 = y_2 $$
|
|
|
|
It follows then that $(x_1, y_1) = (x_2, y_2)$.
|
|
|
|
Therefore, by the definition of one-to-one, it can be concluded that $H$ is
|
|
one-to-one.
|
|
|
|
Q.E.D.
|
|
|
|
b. Is $H$ onto? Prove or give a counterexample.
|
|
|
|
$H$ is onto.
|
|
|
|
**Proof:
|
|
|
|
Suppose $(u, v) \in \mathbb{R} \times \mathbb{R}$.
|
|
|
|
To prove $H$ is onto, it must be shown that $H(x, y) = (u, v)$ for some
|
|
$(x, y) \in \mathbb{R} \times \mathbb{R}$.
|
|
|
|
By the given definition of $H$:
|
|
|
|
$$ (x + 1, 2 - y) = (u, v) $$
|
|
|
|
By the definition of ordered pair (and algebra):
|
|
|
|
$$ x + 1 = u $$
|
|
|
|
$$ x = u - 1 $$
|
|
|
|
and:
|
|
|
|
$$ 2 - y = v $$
|
|
|
|
$$ -y = v - 2 $$
|
|
|
|
$$ y = 2 - v $$
|
|
|
|
Now, $u - 1 \in \mathbb{R}$ by the difference of real numbers, and
|
|
$2 - v \in \mathbb{R}$ by the difference of real numbers. It follows that
|
|
$(u - 1, 2 - v) \in \mathbb{R} \times \mathbb{R}$.
|
|
|
|
Evaluating for $H(u - 1, 2 - v)$:
|
|
|
|
$$ H(u - 1, 2 - v) = ((u - 1) + 1, 2 - (2 - v)) $$
|
|
|
|
$$ = (u - 1 + 1, 2 - 2 + v) $$
|
|
|
|
$$ = (u, v) $$
|
|
|
|
Thus it has been shown that $H(x, y) = (u, v)$ for some
|
|
$(x, y) \in \mathbb{R} \times \mathbb{R}$.
|
|
|
|
Therefore, by the definition of onto, it can be concluded that $H$ is onto.
|
|
|
|
Q.E.D.
|
|
|
|
30. Define $J: \mathbb{Q} \times \mathbb{Q} \to \mathbb{R}$ by the rule
|
|
|
|
$$ J(r, s) = r + \sqrt{2}s \text{ for each } (r, s) \in \mathbb{Q} \times \mathbb{Q} $$
|
|
|
|
a. Is $J$ one-to-one? Prove or give a counterexample.
|
|
|
|
Omitted.
|
|
|
|
b. Is $J$ onto? Prove or give a counterexample.
|
|
|
|
Omitted.
|
|
|
|
31. Define $F: \mathbb{Z}^+ \times \mathbb{Z}^+ \to \mathbb{Z}^+$ and
|
|
$G: \mathbb{Z}^+ \times \mathbb{Z}^+ \to \mathbb{Z}^+$ as follows:
|
|
|
|
For each $(n, m) \in \mathbb{Z}^+ \times \mathbb{Z}^+$,
|
|
|
|
$$ F(n, m) = 3^n5^m \text{ and } G(n, m) = 3^n6^m $$
|
|
|
|
a. Is $F$ one-to-one? Prove or give a counterexample.
|
|
|
|
Omitted.
|
|
|
|
b. Is $G$ one-to-one? Prove or give a counterexample.
|
|
|
|
Omitted.
|
|
|
|
32.
|
|
|
|
a. Is $\log_{8}27 = \log_{2}3$? Why or why not?
|
|
|
|
Let $x = \log_{8}27$, and let $y = \log_{2}3$. By definition of logarithms:
|
|
|
|
$$ 8^x = 27 \text{ and } 2^y = 3 $$
|
|
|
|
Now, $8 = 2^3$, so:
|
|
|
|
$$ 8^x = (2^3)^x = 2^{3x} $$
|
|
|
|
Also, $27 = 3^3, so:$
|
|
|
|
$$ 27 = 3^3 = (2^y)^3 = 2^{3y}$$
|
|
|
|
Hence, since $8^x = 27$:
|
|
|
|
$$ 8^x = 2^{3x} = 27 = 2^{3y} $$
|
|
|
|
Since:
|
|
|
|
$$ 2^{3x} = 2^{3y} $$
|
|
|
|
By the laws of exponents:
|
|
|
|
$$ 3x = 3y $$
|
|
|
|
Then, by algebra:
|
|
|
|
$$ x = y $$
|
|
|
|
Now, we back-substitute our original definitions of $x$ and $y$, and find that:
|
|
|
|
$$ \log_{8}27 = \log_{2}3 $$
|
|
|
|
It can therefore be concluded that the answer to the query is yes.
|
|
|
|
b. Is $\log_{16}9 = \log_{4}3$? Why or why not?
|
|
|
|
Let $x = \log_{16}9$ and $y = \log_{4}3$. Then by definition of log:
|
|
|
|
$$ 16^x = 9 \text{ and } 4^y = 3 $$
|
|
|
|
Note that $16 = 4^2$, so:
|
|
|
|
$$ 9 = (4^2)^x = 4^{2x} $$
|
|
|
|
Note that $9 = 3^2$, so:
|
|
|
|
$$ 9 = 3^2 = (4^y)^2 = 4^{2y} $$
|
|
|
|
So, by the laws of equivalency:
|
|
|
|
$$ 4^{2x} = 9 = 4^{2y} $$
|
|
|
|
$$ 4^{2x} = 4^{2y} $$
|
|
|
|
By the laws of exponents then:
|
|
|
|
$$ 2x = 2y $$
|
|
|
|
Then, by algebra:
|
|
|
|
$$ x = y $$
|
|
|
|
Back-substituting in the definitions for $x$ and $y$:
|
|
|
|
$$ \log_{16}9 = \log_{4}3 $$
|
|
|
|
Therefore the answer to the given question is yes.
|
|
|
|
The properties of logarithm established in 33-35 are used in Sections 11.4 and
|
|
11.5.
|
|
|
|
33. Prove that for all positive real numbers $b$, $x$, and $y$ with $b \neq 1$,
|
|
|
|
$$ \log_{b}\left(\frac{x}{y}\right) = \log_{b}x - \log_{b}y $$
|
|
|
|
**Proof:**
|
|
|
|
Suppose that $b$, $x$, and $y$ are any positive real numbers with $b \neq 1$.
|
|
|
|
Let $u = \log_{b}x$ and $v = \log_{b}y$. By definition of logarithm then:
|
|
|
|
$$ b^u = x \text{ and } b^v = y $$
|
|
|
|
By substitution:
|
|
|
|
$$ \frac{x}{y} = \frac{b^u}{b^v} $$
|
|
|
|
By the laws of exponents:
|
|
|
|
$$ = b^{u - v} $$
|
|
|
|
Taking the logarithm base $b$ of both sides now gives:
|
|
|
|
$$ \log_{b}\left(\frac{x}{y}\right) = \log_{b}(b^{u - v}) $$
|
|
|
|
$$ = u - v $$
|
|
|
|
Back-substituting the definitions of $u$ and $v$ yields:
|
|
|
|
$$ = \log_{b}x - \log_{b}y $$
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
34. Prove that for all positive real numbers $b$, $x$, and $y$ with $b \neq 1$,
|
|
|
|
$$ \log_{b}(xy) = \log_{b}x + \log_{b}y $$
|
|
|
|
**Proof:**
|
|
|
|
Suppose $b$, $x$, and $y$ are any positive real numbers with $b \neq 1$.
|
|
|
|
Let $u = \log_{b}x$, and $v = \log_{b}y$.
|
|
|
|
By definition of logarithms, this means that:
|
|
|
|
$$ b^u = x \text{ and } b^v = y $$
|
|
|
|
By substitution, this means that:
|
|
|
|
$$ xy = b^u \cdot b^v $$
|
|
|
|
$$ = b^{u + v} $$
|
|
|
|
Taking the logarithm of base $b$ of both sides yields:
|
|
|
|
$$ \log_{b}(xy) = \log_{b}(b^{u + v}) $$
|
|
|
|
$$ = u + v $$
|
|
|
|
Back-substituting in the values for $u$ and $v$ shows:
|
|
|
|
$$ \log_{b}(xy) = \log_{b}x + \log_{b}y $$
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
35. Prove that for all real numbers $a$, $b$, and $x$ with $b$ and $x$ positive
|
|
and $b \neq 1$,
|
|
|
|
$$ \log_{b}(x^a) = a\log_{b}x $$
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$, $b$, and $x$ are any real numbers with $x$ and $b$ being positive
|
|
and $b \neq 1$.
|
|
|
|
Let $r = \log_{b}(x^a)$ and $s = \log_{b}x$.
|
|
|
|
By definition of logarithms, this means that:
|
|
|
|
$$ b^r = x^a \text{ and } b^s = x $$
|
|
|
|
Since $b^s = x$, by substitution:
|
|
|
|
$$ b^r = x^a = (b^s)^a = b^{sa} $$
|
|
|
|
So:
|
|
|
|
$$ x^a = b^{sa} $$
|
|
|
|
Now, applying $\log_{b}$ to both sides:
|
|
|
|
$$ \log_{b}(x^a) = \log_{b}(b^{sa}) $$
|
|
|
|
$$ = sa $$
|
|
|
|
Back-substituting in the definition for $s$, this yields:
|
|
|
|
$$ \log_{b}(x^a) = \log_{b}x \cdot a $$
|
|
|
|
Or:
|
|
|
|
$$ \log_{b}(x^a) = a\log_{b}x $$
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
Exercises 36 and 37 use the following definition: If
|
|
$f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ are functions,
|
|
then the function $(f + g): \mathbb{R} \to \mathbb{R}$ is defined by the formula
|
|
$(f + g)(x) = f(x) + g(x)$ for every real number $x$.
|
|
|
|
36. If $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ are
|
|
both one-to-one, is $f + g$ also one-to-one? Justify your answer.
|
|
|
|
No.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Suppose $f$ and $g$ are functions such that $f: \mathbb{R} \to \mathbb{R}$ and
|
|
$g: \mathbb{R} \to \mathbb{R}$ and both $f$ and $g$ are one-to-one functions.
|
|
Furthermore, suppose $(f + g)$ is a function where
|
|
$(f + g): \mathbb{R} \to \mathbb{R}$ such that $(f + g)(x) = f(x) + g(x)$.
|
|
|
|
Consider $f(x) = x$ and $g(x) = -x$. Note that $f$ and $g$ are one-to-one
|
|
functions still follow the definitions of $f$ and $g$ in the supposition.
|
|
|
|
Then, by definition of $(f + g)$, $(f + g)(x) = f(x) + g(x) = x + (-x) = 0$.
|
|
|
|
Then consider $x_1 = 1$, and $x_2 = 2$, then:
|
|
|
|
$$ f(x_1) = 1 \text{ and } g(x_1) = -1 \text{ and } (f + g)(x_1) = 1 + (-1) = 0 $$
|
|
|
|
$$ f(x_2) = 2 \text{ and } g(x_2) = -2 \text{ and } (f + g)(x_2) = 2 + (-2) = 0 $$
|
|
|
|
So $(f + g)(x_1) = (f + g)(x_2)$, but $x_1 \neq x_2$.
|
|
|
|
By the definition of one-to-one, it can therefore be concluded that $(f + g)$ is
|
|
not one-to-one.
|
|
|
|
Q.E.D.
|
|
|
|
37. If $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ are
|
|
both onto, is $f + g$ also onto? Justify your answer.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Suppose $f$ and $g$ are both functions where $f: \mathbb{R} \to \mathbb{R}$, and
|
|
$g: \mathbb{R} \to \mathbb{R}$. Furthermore, suppose
|
|
$(f + g): \mathbb{R} \to \mathbb{R}$ where $(f + g)(x) = f(x) + g(x)$ for some
|
|
$x \in \mathbb{R}$.
|
|
|
|
Consider $f(x) = x$ and $g(x) = -x$. Note that both $f$ and $g$ are still onto
|
|
based off the definition of onto as required by the supposition.
|
|
|
|
Then by definition of $(f + g)$:
|
|
|
|
$$ (f + g)(x) = x + (-x) = 0 $$
|
|
|
|
Since no matter what the value for $x$ will always output $0$, while
|
|
$0 \in \mathbb{R}$, by the definition of onto, every element in the co-domain of
|
|
$\mathbb{R}$ must have a corresponding input image.
|
|
|
|
Consider that $1 \in \mathbb{R}$, but there is no input image $x$ such that
|
|
$(f + g)(x) = 1$.
|
|
|
|
Therefore, by the definition of onto, $(f + g)$ is not onto.
|
|
|
|
Q.E.D.
|
|
|
|
Exercises 38 and 39 use the following definition: If
|
|
$f: \mathbb{R} \to \mathbb{R}$ and $c$ is a nonzero real number, the function
|
|
$(c \cdot f): \mathbb{R} \to \mathbb{R}$ is defined by the formula
|
|
$(c \cdot f)(x) = c \cdot (f(x))$ for every real number $x$.
|
|
|
|
38. Let $f: \mathbb{R} \to \mathbb{R}$ be a function and $c$ a nonzero real
|
|
number. If $f$ is one-to-one, is $c \cdot f$ also one-to-one? Justify your
|
|
answer.
|
|
|
|
Yes, $(c \cdot f)$ is one-to-one.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $f: \mathbb{R} \to \mathbb{R}$ is a one-to-one function, and that $c$ is
|
|
a nonzero real number such that $(c \cdot f): \mathbb{R} \to \mathbb{R}$ is
|
|
defined as $(c \cdot f)(x) = c \cdot (f(x))$ for any real number $x$.
|
|
|
|
To prove $(c \cdot f)$ is one-to-one, it must be shown that there are some
|
|
$x_1, x_2 \in \mathbb{R}$ such that if $(c \cdot f)(x_1) = (c \cdot f)(x_2)$,
|
|
then $x_1 = x_2$.
|
|
|
|
By definition of $(c \cdot f)$:
|
|
|
|
$$ (c \cdot f)(x_1) = c \cdot (f(x_1)) = c \cdot (f(x_2)) = (c \cdot f)(x_2) $$
|
|
|
|
$$ c \cdot (f(x_1)) = c \cdot (f(x_2)) $$
|
|
|
|
By arithmetic:
|
|
|
|
$$ f(x_1) = f(x_2) $$
|
|
|
|
By the supposition, $f$ is a one-to-one function, so therefore, by definition of
|
|
one-to-one:
|
|
|
|
$$ x_1 = x_2 $$
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
39. Let $f: \mathbb{R} \to \mathbb{R}$ be a function and $c$ a nonzero real
|
|
number. If $f$ is onto, is $c \cdot f$ also onto? Justify your answer.
|
|
|
|
$c \cdot f$ is onto.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $f: \mathbb{R} \to \mathbb{R}$ such that $f$ is onto. Furthermore,
|
|
suppose $c$ is a nonzero real number, where
|
|
$(c \cdot f): \mathbb{R} \to \mathbb{R}$ is defined as
|
|
$(c \cdot f)(x) = c \cdot (f(x))$ for any real number $x$.
|
|
|
|
To prove that $(c \cdot f)(x)$ is onto, it must be shown that there exists some
|
|
$y \in \mathbb{R}$, such that $(c \cdot f)(x) = y$.
|
|
|
|
By definition for $c \cdot f$:
|
|
|
|
$$ (c \cdot f)(x) = c \cdot (f(x)) = y $$
|
|
|
|
$$ c \cdot (f(x)) = y $$
|
|
|
|
By algebra:
|
|
|
|
$$ f(x) = \frac{y}{c} $$
|
|
|
|
Since $f$ is onto (by the supposition), this means that there exists some
|
|
$z \in \mathbb{R}$ such that $f(z) = \dfrac{y}{c}$.
|
|
|
|
Let $x = z$, then:
|
|
|
|
$$ (c \cdot f)(x) = c \cdot (f(x)) $$
|
|
|
|
$$ = c \cdot (f(z)) $$
|
|
|
|
$$ = c \cdot \left(\frac{y}{c}\right) $$
|
|
|
|
$$ = y $$
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $(c \cdot f)$
|
|
is onto.
|
|
|
|
Q.E.D.
|
|
|
|
40. Suppose $F: X \to Y$ is one-to-one.
|
|
|
|
a. Prove that for every subset $A \subseteq X$, $F^{-1}(F(A)) = A$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $A \subseteq X$.
|
|
|
|
To prove that $F^{-1}(F(A)) = A$, it must be shown that:
|
|
|
|
$$ F^{-1}(F(A)) \subseteq A $$
|
|
|
|
and also that:
|
|
|
|
$$ A \subseteq F^{-1}(F(A)) $$
|
|
|
|
_Proof ($F^{-1}(F(A)) \subseteq A$):_
|
|
|
|
Let $x \in F^{-1}(F(A))$.
|
|
|
|
By the definition of inverse image:
|
|
|
|
$$ F^{-1}(F(A)) = \{x \in X | F(x) \in F(A)\} $$
|
|
|
|
By the definition for $F(A)$, there exists $r \in A$ such that $F(r) = F(x)$.
|
|
|
|
Since $F(r) = F(x)$, and since $F$ is one-to-one, it follows that $x \in A$
|
|
|
|
Since $x \in F^{-1}(F(A))$ and $x \in A$, it can be concluded that
|
|
$F^{-1}(F(A)) \subseteq A$.
|
|
|
|
This is what was to be shown.
|
|
|
|
_Proof ($A \subseteq F^{-1}(F(A))$):_
|
|
|
|
Let $x \in A$.
|
|
|
|
Since $x \in A$, then $F(x) \in F(A)$, by the definition of $F(A)$.
|
|
|
|
By the definition of inverse image:
|
|
|
|
$$ x \in F^{-1}(F(A)) $$
|
|
|
|
Since $x \in A$ and $x \in F^{-1}(F(A))$, it can be concluded that
|
|
$A \subseteq F^{-1}(F(A))$.
|
|
|
|
This is what was to be shown.
|
|
|
|
_Conclusion:_
|
|
|
|
Since both subset definitions have been shown, it can be concluded that
|
|
$F^{-1}(F(A)) = A$.
|
|
|
|
Q.E.D.
|
|
|
|
b. Prove that for all subsets $A_1$ and $A_2$ in $X$,
|
|
$F(A_1 \cap A_2) = F(A_1) \cap F(A_2)$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $A_1, A_2 \in X$.
|
|
|
|
To prove $F(A_1 \cap A_2) = F(A_1) \cap F(A_2)$, it must be shown that:
|
|
|
|
$$ F(A_1 \cap A_2) \subseteq F(A_1) \cap F(A_2) $$
|
|
|
|
and that:
|
|
|
|
$$ F(A_1) \cap F(A_2) \subseteq F(A_1 \cap A_2) $$
|
|
|
|
_Proof ($F(A_1 \cap A_2) \subseteq F(A_1) \cap F(A_2)$):_
|
|
|
|
Suppose $y \in F(A_1 \cap A_2)$.
|
|
|
|
It must be shown that $y \in F(A_1) \cap F(A_2)$.
|
|
|
|
By the definition of $F(A_1 \cap A_2)$, there exists some $x \in A_1 \cap A_2$
|
|
such that $F(x) = y$.
|
|
|
|
By the definition of intersection:
|
|
|
|
$$ x \in A_1 \text{ and } x \in A_2 $$
|
|
|
|
Since $x \in A_1$ and $x \in A_2$, it follows that:
|
|
|
|
$$ y \in F(A_1) \text{ and } y \in F(A_2) $$
|
|
|
|
By the definition of intersection, this means that:
|
|
|
|
$$ y \in F(A_1) \cap F(A_2) $$
|
|
|
|
Since $y \in F(A_1 \cap A_2)$ and $y \in F(A_1) \cap F(A_2)$, it can be
|
|
concluded that $F(A_1 \cap A_2) \subseteq F(A_1) \cap F(A_2)$.
|
|
|
|
This is what was to be shown.
|
|
|
|
_Proof ($F(A_1) \cap F(A_2) \subseteq F(A_1 \cap A_2)$):_
|
|
|
|
Suppose $y \in F(A_1) \cap F(A_2)$.
|
|
|
|
It must be shown that $y \in F(A_1 \cap A_2)$.
|
|
|
|
By the definition of intersection:
|
|
|
|
$$ y \in F(A_1) \text{ and } y \in F(A_2) $$
|
|
|
|
By the definition of $F(A_1)$, there exists some $x_1 \in A_1$ such that:
|
|
|
|
$$ F(x_1) = y $$
|
|
|
|
Similarly, by definition of $F(A_2)$, there exists some $x_2 \in A_2$ such that:
|
|
|
|
$$ F(x_2) = y $$
|
|
|
|
Since $F$ is one-to-one (by the supposition), and since $F(x_1) = y = F(x_2)$,
|
|
or $F(x_1) = F(x_2)$, this means that:
|
|
|
|
$$ x_1 = x_2 $$
|
|
|
|
By the definition of intersection:
|
|
|
|
$$ x_1 \in A_1 \cap A_2 $$
|
|
|
|
It follows then that since $y = F(x_1)$, that:
|
|
|
|
$$ y \in F(A_1) \cap F(A_2) $$
|
|
|
|
Since $y \in F(A_1) \cap F(A_2)$ and $y \in F(A_1) \cap F(A_2)$, it can be
|
|
concluded that $F(A_1) \cap F(A_2) \subseteq F(A_1 \cap A_2)$.
|
|
|
|
This is what was to be shown.
|
|
|
|
_Conclusion:_
|
|
|
|
Since both subset relations have been shown, it can be concluded that
|
|
$F(A_1 \cap A_2) = F(A_1) \cap F(A_2)$.
|
|
|
|
Q.E.D.
|
|
|
|
41. Suppose $F: X \to Y$ is onto. Prove that for every subset $B \subseteq Y$,
|
|
$F(F^{-1}(B)) = B$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $F: X \to Y$ such that $F$ is onto.
|
|
|
|
Let $B \subseteq Y$.
|
|
|
|
To prove that $F(F^{-1}(B)) = B$, it must be shown that:
|
|
|
|
$$ F(F^{-1}(B)) \subseteq B $$
|
|
|
|
and that:
|
|
|
|
$$ B \subseteq F(F^{-1}(B)) $$
|
|
|
|
_Proof ($F(F^{-1}(B)) \subseteq B$):_
|
|
|
|
Suppose $y \in F(F^{-1}(B))$.
|
|
|
|
It must be shown that $y \in B$.
|
|
|
|
By definition of $F$, there exists some $x \in F^{-1}(B)$ such that $F(x) = y$.
|
|
|
|
By definition of inverse image, since $x \in F^{-1}(B)$, this means that:
|
|
|
|
$$ F(x) \in B $$
|
|
|
|
Since $F(x) = y$, it follows then that:
|
|
|
|
$$ y \in B $$
|
|
|
|
Since $y \in F(F^{-1}(B))$ and $y \in B$, it can be concluded that
|
|
$F(F^{-1}(B)) \subseteq B$.
|
|
|
|
_Proof ($B \subseteq F(F^{-1}(B))$):_
|
|
|
|
Suppose $y \in B$.
|
|
|
|
It must be shown that $y \in F(F^{-1}(B))$.
|
|
|
|
Since $y \in B$, and since $B \subseteq Y$, it follows that $y \in Y$.
|
|
|
|
By the supposition, $F$ is onto. It follows that since $y \in Y$, there exists
|
|
some $x \in X$ such that $F(x) = y$.
|
|
|
|
Since $F(x) = y$ and $y \in B$, by the definition of inverse function:
|
|
|
|
$$ x \in F^{-1}(B) $$
|
|
|
|
It follows then that:
|
|
|
|
$$ y \in F(F^{-1}(B)) $$
|
|
|
|
Since $y \in B$ and $y \in F(F^{-1}(B))$, it can be concluded that
|
|
$B \subseteq F(F^{-1}(B))$.
|
|
|
|
_Conclusion:_
|
|
|
|
Since both subset relations have been shown, it can be concluded that
|
|
$F(F^{-1}(B)) = B$.
|
|
|
|
Q.E.D.
|
|
|
|
Let $X = \{a, b, c, d, e\}$ and $Y = \{s, t, u, v, w\}$. In each of 42 and 43 a
|
|
one-to-one correspondence $F: X \to Y$ is defined by an arrow diagram. In each
|
|
case draw an arrow diagram for $F^{-1}$.
|
|
|
|
42.
|
|
|
|
(See page 483 for image.)
|
|
|
|
Omitted.
|
|
|
|
43.
|
|
|
|
(See page 483 for image.)
|
|
|
|
Omitted.
|
|
|
|
In 44-55 indicate which of the functions in the referenced exercise are
|
|
one-to-one correspondences. For each function that is a one-to-one
|
|
correspondence, find the inverse function.
|
|
|
|
44. Exercise 10a
|
|
|
|
The exercise is not a one-to-one correspondence because it is not onto.
|
|
|
|
45. Exercise 10b
|
|
|
|
Exercise 10b shows that the function $h$ is onto.
|
|
|
|
To prove that $h$ is one-to-one, it must be shown that there exists some
|
|
$n_1, n_2 \in \mathbb{Z}$ such that if $h(n_1) = h(n_2)$, then $n_1 = n_2$.
|
|
|
|
By definition of $h$, this implies that:
|
|
|
|
$$ 2n_1 = 2n_2 $$
|
|
|
|
Then, by algebra:
|
|
|
|
$$ n_1 = n_2 $$
|
|
|
|
This is what was to be shown, and therefore it can be concluded that $h$ is a
|
|
one-to-one correspondence.
|
|
|
|
Now, to find the inverse function.
|
|
|
|
Given any integer $m \in 2\mathbb{Z}$ (where $2\mathbb{Z}$ is the set of all
|
|
even integers) such that $h(n) = m$, by the definition of $h$, it follows that:
|
|
|
|
$$ h(n) = m = 2n $$
|
|
|
|
The inverse can be found by evaluating for $n$ as it relates to $m$.
|
|
|
|
$$ m = 2n $$
|
|
|
|
$$ n = \frac{m}{2} $$
|
|
|
|
Thus:
|
|
|
|
$$ h^{-1}(m) = \frac{m}{2} $$
|
|
|
|
for some $m \in 2\mathbb{Z}$.
|
|
|
|
46. Exercise 11a
|
|
|
|
The exercise is not a one-to-one correspondence because it is not onto.
|
|
|
|
47. Exercise 11b
|
|
|
|
Exercise 11b shows that $G$ is onto.
|
|
|
|
To prove that $G$ is one-to-one, it must be shown that there exists some
|
|
$x_1, x_2 \in \mathbb{R}$ such that when $G(x_1) = G(x_2)$, then $x_1 = x_2$.
|
|
|
|
By the definition of $G$:
|
|
|
|
$$ 4x_1 - 5 = 4x_2 - 5 $$
|
|
|
|
By algebra:
|
|
|
|
$$ 4x_1 = 4x_2 $$
|
|
|
|
$$ x_1 = x_2 $$
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $G$ is
|
|
one-to-one.
|
|
|
|
Now to find the inverse.
|
|
|
|
Suppose there is some $y \in \mathbb{R}$ such that $y = 4x - 5$, then evaluating
|
|
for $x$:
|
|
|
|
$$ x = \frac{y + 5}{4} $$
|
|
|
|
Replacing $x$ with $G^{-1}(y)$:
|
|
|
|
$$ G^{-1}(y) = \frac{y + 5}{4} $$
|
|
|
|
By definition of inverse, this is true if and only if
|
|
$G\left(\dfrac{y + 5}{4}\right) = y$. By the definition for $G$:
|
|
|
|
$$ G\left(\frac{y + 5}{4}\right) = 4\left(\frac{y + 5}{4}\right) - 5 $$
|
|
|
|
$$ = (y + 5) - 5 $$
|
|
|
|
$$ = y $$
|
|
|
|
Therefore, it can be concluded that $G^{-1}(y) = \dfrac{y + 5}{4}$ for every
|
|
$y \in \mathbb{R}$.
|
|
|
|
48. Exercise 12a
|
|
|
|
The function $F$ is not a one-to-one correspondence, because $F$ is not onto.
|
|
|
|
49. Exercise 12b
|
|
|
|
Exercise 12b shows that $G$ is onto. To prove that $G$ is one-to-one, it must be
|
|
shown that there exists some $x_1, x_2 \in \mathbb{R}$ such that when
|
|
$G(x_1) = G(x_2)$, then $x_1 = x_2$.
|
|
|
|
By the definition of $G$, this means that:
|
|
|
|
$$ 2 - 3x_1 = 2 - 3x_2 $$
|
|
|
|
$$ -3x_1 = -3x_2 $$
|
|
|
|
$$ x_1 = x_2 $$
|
|
|
|
This is what was to be shown. Therefore, it can be concluded that $G$ is
|
|
one-to-one.
|
|
|
|
Now, to find the inverse. Suppose there is some $y = 2 - 3x$. Solving for $x$:
|
|
|
|
$$ 3x = 2 - y $$
|
|
|
|
$$ x = \frac{2 - y}{3} $$
|
|
|
|
Then substituting for $x$ with $G^{-1}(y)$:
|
|
|
|
$$ G^{-1}(y) = \dfrac{2 - y}{3} $$
|
|
|
|
By the definition of inverse, this can only be true if
|
|
$G\left(\dfrac{2 - y}{3}\right) = y$. By the definition for $G$:
|
|
|
|
$$ G\left(\frac{2 - y}{3}\right) = 2 - 3\left(\frac{2 - y}{3}\right) $$
|
|
|
|
$$ = 2 - (2 - y) $$
|
|
|
|
$$ = 2 - 2 + y $$
|
|
|
|
$$ = y $$
|
|
|
|
Therefore, it can be concluded that:
|
|
|
|
$$ G^{-1}(y) = \frac{2 - y}{3} $$
|
|
|
|
for any $y \in \mathbb{R}$.
|
|
|
|
50. Exercise 21
|
|
|
|
The function $L$ is not a one-to-one correspondence, because $L$ is not
|
|
one-to-one.
|
|
|
|
51. Exercise 22
|
|
|
|
The function $D$ is not a one-to-one correspondence, because $D$ is not
|
|
one-to-one.
|
|
|
|
52. Exercise 15 with the co-domain taken to be the set of all real numbers not
|
|
equal to $1$.
|
|
|
|
Omitted.
|
|
|
|
53. Exercise 16 with the co-domain taken to be the set of all real numbers.
|
|
|
|
Omitted.
|
|
|
|
54. Exercise 17 with the co-domain taken to be the set of all real numbers not
|
|
equal to $3$
|
|
|
|
Omitted.
|
|
|
|
55. Exercise 18 with the co-domain taken to be the set of all real numbers not
|
|
equal to 1.
|
|
|
|
Omitted.
|
|
|
|
56. In Example 7.2.8 a one-to-one correspondence was defined from the power set
|
|
of $\{a, b\}$ to the set of all strings of $0$'s and $1$'s that have length
|
|
$2$. Thus the elements of these two sets can be matched up exactly, and so
|
|
the two sets have the same number of elements.
|
|
|
|
a. Let $X = \{x_1, x_2, \dots, x_n\}$ be a set with $n$ elements. Use Example
|
|
7.2.8 as a model to define a one-to-one correspondence from $\mathscr{P}(X)$,
|
|
the set of all subsets of $X$, to the set of all strings of $0$'s and $1$'s that
|
|
have length $n$.
|
|
|
|
Omitted.
|
|
|
|
b. In Section 9.2 we show that there are $2^n$ strings of $0's$ and $1$'s that
|
|
have length $n$. What does this allow you to conclude about the number of
|
|
subsets of $\mathscr{P}(X)$? (This provides an alternative proof of Theorem
|
|
6.3.1.)
|
|
|
|
Omitted.
|
|
|
|
57. Write a computer algorithm to check whether a function from one finite set
|
|
to another is one-to-one. Assume the existence of an independent algorithm
|
|
to compute values of the function.
|
|
|
|
Omitted.
|
|
|
|
58. Write a computer algorithm to check whether a function from one finite set
|
|
to another is onto. Assume the existence of an independent algorithm to
|
|
compute values of the function.
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
Page 494
|
|
|
|
**Exercise Set 7.3**
|
|
|
|
In each of 1 and 2, functions $f$ and $g$ are defined by arrow diagrams. Find
|
|
$g \circ f$ and $f \circ g$ and determine whether $g \circ f$ equals
|
|
$f \circ g$.
|
|
|
|
1. (See page 494 for image)
|
|
|
|
$$ f(1) = 5, f(3) = 3, f(5) = 1 $$
|
|
|
|
$$ g(1) = 3, g(3) = 5, g(5) = 1 $$
|
|
|
|
$$ g(f(1)) = g(5) = 1, g(f(3)) = g(3) = 5, g(f(5)) = g(1) = 3 $$
|
|
|
|
$$ f(g(1)) = f(3) = 3, f(g(3)) = f(5) = 1, f(g(5)) = f(1) = 5 $$
|
|
|
|
Since not all elements of $g(f(x))$ do not equal $f(g(x))$ (such as
|
|
$g(f(1)) = 1 \neq 3 = f(g(1))$), it can be concluded that:
|
|
|
|
$$ g \circ f \neq f \circ g $$
|
|
|
|
2. (See page 494 for image)
|
|
|
|
$$ f(1) = 3, f(3) = 1, f(5) = 5 $$
|
|
|
|
$$ g(1) = 1, g(3) = 1, g(5) = 1 $$
|
|
|
|
$$ g(f(1)) = g(3) = 1, g(f(3)) = g(1) = 1, g(f(5)) = g(5) = 1 $$
|
|
|
|
$$ f(g(1)) = f(1) = 3, f(g(3)) = f(1) = 3, f(g(5)) = f(1) = 3 $$
|
|
|
|
Since not all elements of $g(f(x))$ do not equal $f(g(x))$ (such as
|
|
$g(f(1)) = 1 \neq 3 = f(g(1))$), it can be concluded that:
|
|
|
|
$$ g \circ f \neq f \circ g $$
|
|
|
|
In 3 and 4, functions $F$ and $G$ are defined by formulas. Find $G \circ F$ and
|
|
$F \circ G$ and determine whether $G \circ F$ equals $F \circ G$.
|
|
|
|
3. $F(x) = x^3$ and $G(x) = x - 1$, for each real number $x$.
|
|
|
|
$$ (G \circ F)(x) = G(F(x)) = G(x^3) = x^3 - 1 $$
|
|
|
|
$$ (F \circ G)(x) = F(G(x)) = F(x - 1) = (x - 1)^3 $$
|
|
|
|
$$ = (x - 1)(x - 1)(x - 1) $$
|
|
|
|
$$ = (x^2 - 2x + 1)(x - 1) $$
|
|
|
|
$$ = x^2(x - 1) - 2x(x - 1) + 1(x - 1) $$
|
|
|
|
$$ = x^3 - x^2 - 2x^2 - 2x + x - 1 $$
|
|
|
|
$$ = x^3 - 3x^2 - x - 1 $$
|
|
|
|
As $x^3 - 1 \neq x^3 - 3x^2 - x - 1 \forall x \in \mathbb{R}$
|
|
|
|
Consider $x = 2$, then:
|
|
|
|
$$ (G \circ F)(2) = (2)^3 - 1 = 8 - 1 = 7 $$
|
|
|
|
$$ (F \circ G)(2) = (2 - 1)^3 = (2 - 1)(2 - 1)(2 - 1) = (1)(1)(1) = 1 $$
|
|
|
|
Note that:
|
|
|
|
$$ 7 \neq 1 $$
|
|
|
|
So $(G \circ F)(2) \neq (F \circ G)(2)$.
|
|
|
|
Hence it can be concluded then that, for all real numbers:
|
|
|
|
$$ G \circ F \neq F \circ G $$
|
|
|
|
4. $F(x) = x^5$ and $G(x) = x^{\frac{1}{5}}$ for each real number $x$.
|
|
|
|
$$ (G \circ F)(x) = G(F(x)) = G(x^5) = (x^5)^{\frac{1}{5}} = x^{5 \cdot \frac{1}{5}} = x $$
|
|
|
|
$$ (F \circ G)(x) = F(G(x)) = F(x^{\frac{1}{5}}) = (x^{\frac{1}{5}})^5 = x^{\frac{1}{5} \cdot 5} = x $$
|
|
|
|
Since both $(G \circ F)(x) = x = (F \circ G)(x)$, it can be concluded that for
|
|
all $x \in \mathbb{R}$:
|
|
|
|
$$ G \circ F = F \circ G $$
|
|
|
|
5. Define $f: \mathbb{R} \to \mathbb{R}$ by the rule $f(x) = -x$ for every real
|
|
number $x$. Find $(f \circ f)(x)$.
|
|
|
|
$$ (f \circ f)(x) = f(f(x)) = f(-x) = -(-x) = x $$
|
|
|
|
6. Define $F: \mathbb{Z} \to \mathbb{Z}$ and $G: \mathbb{Z} \to \mathbb{Z}$ by
|
|
the rules $F(a) = 7a$ and $G(a) = a \mod 5$ for each integer $a$. Find
|
|
$(G \circ F)(0)$, $(G \circ F)(1)$, $(G \circ F)(2)$, $(G \circ F)(3)$, and
|
|
$(G \circ F)(4)$.
|
|
|
|
$$ (G \circ F)(0) = G(F(0)) = G(7(0)) = G(0) = 0 \mod 5 = 0 $$
|
|
|
|
$$ (G \circ F)(1) = G(F(1)) = G(7(1)) = G(7) = 7 \mod 5 = 2 $$
|
|
|
|
$$ (G \circ F)(2) = G(F(2)) = G(7(2)) = G(14) = 14 \mod 5 = 4 $$
|
|
|
|
$$ (G \circ F)(3) = G(F(3)) = G(7(3)) = G(21) = 21 \mod 5 = 1 $$
|
|
|
|
$$ (G \circ F)(4) = G(F(4)) = G(7(4)) = G(28) = 28 \mod 5 = 3 $$
|
|
|
|
7. Define $L: \mathbb{Z} \to \mathbb{Z}$ and $M: \mathbb{Z} \to \mathbb{Z}$ by
|
|
the rules $L(a) = a^2$ and $M(a) = a \mod 5$ for each integer $a$.
|
|
|
|
a. Find $(L \circ M)(12)$, $(M \circ L)(12)$, $(L \circ M)(9)$, and
|
|
$(M \circ L)(9)$.
|
|
|
|
$$ (L \circ M)(12) = L(M(12)) = L(12 \mod 5) = L(2) = 2^2 = 4 $$
|
|
|
|
$$ (M \circ L)(12) = M(L(12)) = M(12^2) = M(144) = 144 \mod 5 = 4 $$
|
|
|
|
$$ (L \circ M)(9) = L(M(9)) = L(9 \mod 5) = L(4) = 4^2 = 16 $$
|
|
|
|
$$ (M \circ L)(9) = M(L(9)) = M(9^2) = M(81) = 81 \mod 5 = 1 $$
|
|
|
|
b. Is $L \circ M = M \circ L$?
|
|
|
|
No, since $(L \circ M)(9) = 16 \neq 1 = (M \circ L)(9)$, it can be concluded
|
|
that $L \circ M \neq M \circ L$ for all integers.
|
|
|
|
8. Let $S$ be the set of all strings in _a_'s and _b_'s and let
|
|
$L: S \to \mathbb{Z}$ be the length function:
|
|
|
|
For all strings $s \in S$ ,
|
|
|
|
$$ L(s) = \text{ the number of characters in } s $$
|
|
|
|
Let $T: \mathbb{Z} \to \{0, 1, 2\}$ be the $\mod 3$ function:
|
|
|
|
$$ \text{For every integer } n, \quad T(n) = n \mod 3 $$
|
|
|
|
a. $(T \circ L)(abaa) = \text{ ?}$
|
|
|
|
$$ (T \circ L)(abaa) = T(L(abaa)) = T(4) = 4 \mod 3 = 1 $$
|
|
|
|
b. $(T \circ L)(baaab) = \text{ ?}$
|
|
|
|
$$ (T \circ L)(baaab) = T(L(baaab)) = T(5) = 5 \mod 3 = 2 $$
|
|
|
|
c. $(T \circ L)(aaa) = \text{ ?}$
|
|
|
|
$$ (T \circ L)(aaa) = T(L(aaa)) = T(3) = 3 \mod 3 = 0 $$
|
|
|
|
9. Define $F: \mathbb{R} \to \mathbb{R}$ and $G: \mathbb{R} \to \mathbb{Z}$ by
|
|
the following formulas: $F(x) = \dfrac{x^2}{3}$ and
|
|
$G(x) = \lfloor x \rfloor$ for every $x \in \mathbb{R}$.
|
|
|
|
a. $(G \circ F)(2) = \text{ ?}$
|
|
|
|
$$ (G \circ F)(2) = G(F(2)) = G\left(\frac{(2)^2}{3}\right) = G\left(\frac{4}{3}\right) = \left\lfloor v\frac{4}{3} \right\rfloor = 1 $$
|
|
|
|
b. $(G \circ F)(-3) = \text{ ?}$
|
|
|
|
$$ (G \circ F)(-3) = G(F(-3)) = G\left(\frac{(-3)^2}{3}\right) = G\left(\frac{9}{3}\right) = G(3) = \lfloor 3 \rfloor = 3 $$
|
|
|
|
c. $(G \circ F)(5) = \text{ ?}$
|
|
|
|
$$ (G \circ F)(5) = G(F(5)) = G\left(\frac{(5)^2}{3}\right) = G\left(\frac{25}{3}\right) = \left\lfloor \frac{25}{3} \right\rfloor = 8 $$
|
|
|
|
10. Define $F: \mathbb{Z} \to \mathbb{Z}$ and $G: \mathbb{Z} \to \mathbb{Z}$ by
|
|
the rules $F(n) = 2n$ and $G(n) = \left\lfloor \dfrac{n}{2} \right\rfloor$
|
|
for every integer $n$.
|
|
|
|
a. Find $(G \circ F)(8)$, $(F \circ G)(8)$, $(G \circ F)(3)$, and
|
|
$(F \circ G)(3)$.
|
|
|
|
$$ (G \circ F)(8) = G(F(8)) = G(2(8)) = G(16) = \left\lfloor \frac{(16)}{2} \right\rfloor = \lfloor 8 \rfloor = 8 $$
|
|
|
|
$$ (F \circ G)(8) = F(G(8)) = F\left(\left\lfloor \frac{(8)}{2} \right\rfloor\right) = F(\lfloor 4 \rfloor) = F(4) = 2(4) = 8 $$
|
|
|
|
$$ (G \circ F)(3) = G(F(3)) = G(2(3)) = G(6) = \left \lfloor \frac{(6)}{2} \right\rfloor = \lfloor 3 \rfloor = 3 $$
|
|
|
|
$$ (F \circ G)(3) = F(G(3)) = F\left(\left\lfloor \frac{(3)}{2} \right\rfloor\right) = F(1) = 2(1) = 2 $$
|
|
|
|
b. Is $G \circ F = F \circ G$? Explain.
|
|
|
|
No, since $(G \circ F)(3) = 3 \neq 2 = (F \circ G)(3)$, it can be concluded that
|
|
$G \circ F \neq F \circ G$ for all integers.
|
|
|
|
11. Define $F: \mathbb{R} \to \mathbb{R}$ and $G : \mathbb{R} \to \mathbb{R}$ by
|
|
the rules $F(n) = 3x$ and $G(n) = \left\lceil \dfrac{x}{3} \right\rceil$ for
|
|
every real number $x$.
|
|
|
|
a. Find $(G \circ F)(6)$, $(F \circ G)(6)$, $(G \circ F)(1)$, and
|
|
$(F \circ G)(1)$.
|
|
|
|
$$ (G \circ F)(6) = G(F(6)) = G(3(6)) = G(18) = \left\lceil \frac{(18)}{3} \right\rceil = \lceil 6 \rceil = 6 $$
|
|
|
|
$$ (F \circ G)(6) = F(G(6)) = F\left(\left\lceil \frac{(6)}{3} \right\rceil \right) = F(\lceil 2 \rceil) = F(2) = 3(2) = 6 $$
|
|
|
|
$$ (G \circ F)(1) = G(F(1)) = G(3(1)) = G(3) = \left\lceil \frac{(3)}{3} \right\rceil = \lceil 1 \rceil = 1 $$
|
|
|
|
$$ (F \circ G)(1) = F(G(1)) = F\left(\left\lceil \frac{(1)}{3} \right\rceil \right) = F(1) = 3(1) = 3 $$
|
|
|
|
b. Is $G \circ F = F \circ G$? Explain.
|
|
|
|
No, since $(G \circ F)(1) = 1 \neq 3 = (F \circ G)(1)$, it can be concluded that
|
|
$G \circ F \neq F \circ G$ for all real numbers.
|
|
|
|
The functions of each pair in 12-14 are inverse to each other. For each pair,
|
|
check that both compositions give the identity function.
|
|
|
|
12. $F: \mathbb{R} \to \mathbb{R}$ and $F^{-1}: \mathbb{R} \to \mathbb{R}$ are
|
|
defined by
|
|
|
|
$$ F(x) = 3x + 2 \quad \text{ and } \quad F^{-1}(y) = \frac{y - 2}{3} $$
|
|
|
|
for every $y \in \mathbb{R}$.
|
|
|
|
$$ (F^{-1} \circ F)(x) = F^{-1}(F(x)) = F^{-1}(3x + 2) = \frac{(3x + 2) - 2}{3} = \frac{3x}{3} = x = I_{\mathbb{R}}(x) $$
|
|
|
|
Hence, for every $x \in \mathbb{R}$, $F^{-1} \circ F = I_{\mathbb{R}}$ by
|
|
definition of the equality of functions.
|
|
|
|
$$ (F \circ F^{-1})(y) = F(F^{-1}(y)) = F\left(\frac{y - 2}{3}\right) = 3\left(\frac{y - 2}{3}\right) + 2 = y - 2 + 2 = y = I_{\mathbb{R}(y)} $$
|
|
|
|
Hence, for every $y \in \mathbb{R}$, $F \circ F^{-1} = I_{\mathbb{R}}$ by
|
|
definition of the equality of functions.
|
|
|
|
13. $G: \mathbb{R}^+ \to \mathbb{R}^+$ and
|
|
$G^{-1}: \mathbb{R}^+ \to \mathbb{R}^+$ are defined by
|
|
|
|
$$ G(x) = x^2 \quad \text{ and } \quad G^{-1}(x) = \sqrt{x} $$
|
|
|
|
for every $x \in \mathbb{R}^+$.
|
|
|
|
$$ (G^{-1} \circ G)(x) = G^{-1}(G(x)) = G^{-1}(x^2) = \sqrt{(x^2)} = x = I_{\mathbb{R}^+}(x) $$
|
|
|
|
Hence, for every $x \in \mathbb{R}^+$, $G^{-1} \circ G = I_{\mathbb{R}^+}$ by
|
|
definition of the equality of functions.
|
|
|
|
$$ (G \circ G^{-1})(y) = G(G^{-1}(y)) = G\left(\sqrt{y}\right) = \left(\sqrt{y}\right)^2 = y = I_{\mathbb{R}^+}(y) $$
|
|
|
|
Hence, for every $y \in \mathbb{R}^+$, $G \circ G^{-1} = I_{\mathbb{R}^+}$ by
|
|
definition of the equality of functions.
|
|
|
|
14. $H$ and $H^{-1}$ are both defined from $\mathbb{R} - \{1\}$ to
|
|
$\mathbb{R} - \{1\}$ by the formula
|
|
|
|
$$ H(x) = H^{-1}(x) = \frac{x + 1}{x - 1}, \quad \text{ for each } x \in \mathbb{R} - \{1\} $$
|
|
|
|
$$ (H^{-1} \circ H)(x) = H^{-1}(H(x)) = H^{-1}\left(\frac{x + 1}{x - 1}\right) $$
|
|
|
|
$$ = \frac{\dfrac{x + 1}{x - 1} + 1}{\dfrac{x + 1}{x - 1} - 1} $$
|
|
|
|
$$ = \frac{\dfrac{x + 1 + (x - 1)}{x - 1}}{\dfrac{x + 1 - (x - 1)}{x - 1}} $$
|
|
|
|
$$ = \frac{x + 1 + (x - 1)}{x + 1 - (x - 1)}$$
|
|
|
|
$$ = \frac{x + 1 + x - 1}{x + 1 - x + 1}$$
|
|
|
|
$$ = \frac{2x}{2} $$
|
|
|
|
$$ = x = I_{\mathbb{R} - \{1\}}(x) $$
|
|
|
|
Hence, for every $x \in \mathbb{R} - \{1\}$,
|
|
$H^{-1} \circ H = I_{\mathbb{R} - \{1\}}$ by definition of the equality of
|
|
functions.
|
|
|
|
$$ (H \circ H^{-1})(y) = H(H^{-1}(y)) = H\left(\frac{y + 1}{y - 1}\right) $$
|
|
|
|
$$ = \frac{\dfrac{y + 1}{y - 1} + 1}{\dfrac{y + 1}{y - 1} - 1} $$
|
|
|
|
$$ = \frac{\dfrac{y + 1 + (y - 1)}{y - 1}}{\dfrac{y + 1 - (y - 1)}{y - 1}} $$
|
|
|
|
$$ = \frac{y + 1 + (y - 1)}{y + 1 - (y - 1)}$$
|
|
|
|
$$ = \frac{y + 1 + y - 1}{y + 1 - y + 1}$$
|
|
|
|
$$ = \frac{2y}{2} $$
|
|
|
|
$$ = y = I_{\mathbb{R} - \{1\}}(y) $$
|
|
|
|
Hence, for every $y \in \mathbb{R} - \{1\}$,
|
|
$H \circ H^{-1} = I_{\mathbb{R} - \{1\}}$ by definition of the equality of
|
|
functions.
|
|
|
|
15. Explain how it follows from the definition of logarithm that
|
|
|
|
a. $\log_{b}(b^x) = x$, for every real number $x$.
|
|
|
|
By definition of logarithm with base $b$, for each real number $x$,
|
|
$\log_{b}(b^x)$ is the exponent to which $b$ must be raised to obtain $b^x$. But
|
|
this exponent is just $x$. So $\log_{b}(b^x) = x$.
|
|
|
|
b. $b^{\log_{b}x} = x$, for every positive real number $x$.
|
|
|
|
By definition of logarithm with base $b$, for each real number r$x$, $\log_{b}x$
|
|
is the exponent to which $b$ must be raised to obtain $x$. So
|
|
$b^{\log_{b}x} = x$.
|
|
|
|
16. Prove Theorem 7.3.1(b): If $f$ is any function from a set $X$ to a set $Y$,
|
|
then $I_y \circ f = f$, where $I_y$ is the identity function on $Y$.
|
|
|
|
_Hint:_ Suppose $f$ is any function from a set $X$ to a set $Y$, and show that
|
|
for every $x$ in $X$, $(I_Y \circ f)(x) = f(x)$.
|
|
|
|
**Proof:**
|
|
|
|
_Part (b):_
|
|
|
|
Suppose $f$ is any function from a set $X$ to a set $Y$.
|
|
|
|
To prove that $I_Y \circ f = f$, it must be shown that for every $x \in X$,
|
|
$(I_Y \circ f)(x) = f(x)$.
|
|
|
|
By the definition of the composition of functions:
|
|
|
|
$$ (I_Y \circ f)(x) = I_Y(f(x)) $$
|
|
|
|
By the definition of the Identity function, since $f(x) = y$:
|
|
|
|
$$ I_Y(f(x)) = I_Y(y) = y = f(x) $$
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
17. Prove Theorem 7.3.2(b): If $f: X \to Y$ is a one-to-one and onto function
|
|
with inverse function $f^{-1}: Y \to X$, then $f \circ f^{-1} = I_Y$, where
|
|
$I_Y$ is the identity function on $Y$.
|
|
|
|
**Proof:**
|
|
|
|
_Part (b):_
|
|
|
|
Suppose $f: X \to Y$ is a one-to-one and onto function with inverse function
|
|
$f^{-1}: Y \to X$.
|
|
|
|
To prove that $f \circ f^{-1} = I_Y$, we must show that for each $y \in Y$,
|
|
$(f \circ f^{-1})(y) = y$.
|
|
|
|
By the definition of the composition of functions:
|
|
|
|
$$ (f \circ f^{-1})(y) = f(f^{-1}(y)) $$
|
|
|
|
Since $f$ is one-to-one and onto, this implies that there exists a unique
|
|
$x \in X$ such that $f(x) = y$. Therefore, by the definition of inverse
|
|
functions:
|
|
|
|
$$ f^{-1}(y) = x $$
|
|
|
|
Substituting this in to our composition of functions:
|
|
|
|
$$ (f \circ f^{-1})(y) = f(x) = y = I_Y $$
|
|
|
|
This is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
18. Suppose $Y$ and $Z$ are sets and $g: Y \to Z$ is a one-to-one function. This
|
|
means that if $g$ takes the same value on any two elements of $Y$, then
|
|
those elements are equal. Thus, for example, if $a$ and $b$ are elements of
|
|
$Y$ and $g(a) = g(b)$, then it can be inferred that $a = b$. What can be
|
|
inferred in the following situations?
|
|
|
|
a. $s_k$ and $s_m$ are elements of $Y$ and $g(s_k) = g(s_m)$.
|
|
|
|
It can be inferred that $s_k = s_m$.
|
|
|
|
b. $\dfrac{z}{2}$ and $\dfrac{t}{2}$ are elements of $Y$ and
|
|
$g\left(\dfrac{z}{2}\right) = g\left(\dfrac{t}{2}\right)$.
|
|
|
|
It can be inferred that $\dfrac{z}{2} = \dfrac{t}{2}$, and furthermore, by
|
|
algebra, that $z = t$.
|
|
|
|
c. $f(x_1)$ and $f(x_2)$ are elements of $Y$ and $g(f(x_1)) = g(f(x_2))$.
|
|
|
|
We can infer that $f(x_1) = f(x_2)$. Of note here is that we cannot infer that
|
|
$x_1 = x_2$ since we do not know if $f$ is one-to-one.
|
|
|
|
19. If $f: X \to Y$ and $g: Y \to Z$ are functions and $g \circ f$ is
|
|
one-to-one, must $g$ be one-to-one? Prove or give a counterexample.
|
|
|
|
No, $g$ is not necessarily one-to-one.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Suppose $X = \{a, b\}$, $Y = \{1, 2, 3\}$, and $Z = \{x, y\}$. Then suppose:
|
|
|
|
$$ f(a) = 2, f(b) = 3, g(1) = x, g(2) = x, and g(3) = y $$
|
|
|
|
So $g \circ f$ is one-to-one since $(g \circ f)(a) = g(f(a)) = g(2) = x$ and
|
|
$(g \circ f)(b) = g(f(b)) = g(3) = y$.
|
|
|
|
Thus it has been shown that for some sets $X$, $Y$, and $Z$, there are functions
|
|
$f: X \to Y$ and $g: Y \to Z$ such that $g \circ f$ is one-to-one, but $g$ is
|
|
not one-to-one.
|
|
|
|
Q.E.D.
|
|
|
|
20. If $f: X \to Y$ and $g: Y \to Z$ are functions and $g \circ f$ is onto, must
|
|
$f$ be onto? Prove or give a counterexample.
|
|
|
|
No, $f$ is not necessarily onto.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Suppose $X = \{a, b, c, d\}$, $Y = \{1, 2, 3, 4, 5\}$, and $Z = \{x, y, z\}$.
|
|
Then, define $f$ and $g$ as:
|
|
|
|
$$ f(a) = 1, f(b) = 2, f(c) = 3, f(d) = 4, g(1) = x, g(2) = y, g(3) = z, g(4) = z $$
|
|
|
|
Then, $g\circ f$ is onto, as
|
|
$(g \circ f)(a) = x, (g \circ f)(b) = y, (g \circ f)(c) = z, (g \circ f)(d) = z$.
|
|
|
|
But, notice that $f$ is not onto, as the element $5$ is in the co-domain of $f$,
|
|
but is not in the range of $f$.
|
|
|
|
Hence $f$ is not onto.
|
|
|
|
Q.E.D.
|
|
|
|
21. If $f: X \to Y$ and $g: Y \to Z$ are functions and $g \circ f$ is
|
|
one-to-one, must $f$ be one? Prove or give a counterexample.
|
|
|
|
_Hint:_
|
|
|
|
Suppose $f: X \to Y$ and $g: Y \to Z$ are functions and $g \circ f$ is
|
|
one-to-one. Given $x_1$ and $x_2$ in $X$, if $f(x_1) = f(x_2)$ then
|
|
$(g \circ f)(x_1) = (g \circ f)(x_2)$. (Why?) Then use the fact that $g \circ f$
|
|
is one-to-one.
|
|
|
|
Yes, $f$ is one-to-one.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $f: X \to Y$ and $g: Y \to Z$ are functions and that $g \circ f$ is
|
|
one-to-one.
|
|
|
|
Let $x_1, x_2 \in X$ such that $f(x_1) = f(x_2)$. To prove that $f$ is
|
|
one-to-one, it must be shown that $x_1 = x_2$.
|
|
|
|
By the definition of the composition of functions:
|
|
|
|
$$ (g \circ f)(x_1) = g(f(x_1)) $$
|
|
|
|
And also by the definition of the composition of functions:
|
|
|
|
$$ (g \circ f)(x_2) = g(f(x_2)) $$
|
|
|
|
Now, since $f(x_1) = f(x_2)$, it follows that:
|
|
|
|
$$ g(f(x_1)) = g(f(x_2)) $$
|
|
|
|
Furthermore, since $g \circ f$ is one-to-one, it also follows that:
|
|
|
|
$$ x_1 = x_2 $$
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $f$ is
|
|
one-to-one.
|
|
|
|
Q.E.D.
|
|
|
|
22. If $f: X \to Y$ and $g: Y \to Z$ are functions and $g \circ f$ is onto, must
|
|
$g$ be onto? Prove or give a counterexample.
|
|
|
|
_Hint:_
|
|
|
|
Suppose $f: X \to Y$ and $g: Y \to Z$ are functions and $g \circ f$ is onto.
|
|
Given $z \in Z$, there is an element $x$ in $X$ such that $(g \circ f)(x) = z$.
|
|
(Why?) If $y = f(x)$, what can you deduce about $g(y)$?
|
|
|
|
Yes, $g$ is onto.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $f: X \to Y$ and $g: Y \to Z$ are functions and $g \circ f$ is onto.
|
|
|
|
Let $z \in Z$.
|
|
|
|
To prove that $g$ is onto, it must be shown that there exists some $y \in Y$
|
|
such that $g(y) = z$.
|
|
|
|
Since $g \circ f$ is onto, this implies that there exists some $x \in X$ such
|
|
that $(g \circ f)(x) = z$. By the definition of the composition of functions,
|
|
this can be expressed as:
|
|
|
|
$$ (g \circ f)(x) = g(f(x)) = z $$
|
|
|
|
Now, let $f(x) = y$ where $y \in Y$. Then, it follows that:
|
|
|
|
$$ g(y) = z $$
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $g$ is onto.
|
|
|
|
Q.E.D.
|
|
|
|
23. Let $f: W \to X$, $g: X \to Y$, and $h: Y \to Z$ be functions. Must
|
|
$h \circ (g \circ f) = (h \circ g) \circ f$? Prove or give a counterexample.
|
|
|
|
The stated equality is true.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $f: W \to X$, $g: X \to Y$, and $h: Y \to Z$ are functions.
|
|
|
|
Let $w \in W$.
|
|
|
|
To prove $h \circ (g \circ f) = (h \circ g) \circ f$, it must be shown that
|
|
$(h \circ (g \circ f))(w) = ((h \circ g) \circ f)(w)$.
|
|
|
|
By the definition of the composition of functions:
|
|
|
|
$$ (h \circ (g \circ f))(w) = h((g \circ f)(w)) = h(g(f(w))) $$
|
|
|
|
Also by the definition of the composition of functions:
|
|
|
|
$$ ((h \circ g) \circ f)(w) = (h \circ g)(f(w)) = h(g(f(w))) $$
|
|
|
|
Thus it has been shown that the two sides of the given proposed equality are
|
|
indeed equal since $h(g(f(w))) = h(g(f(w)))$.
|
|
|
|
Therefore $h \circ (g \circ f) = (h \circ g) \circ f$.
|
|
|
|
Q.E.D.
|
|
|
|
24. True or False? Given any set $X$ and given any functions $f: X \to X$,
|
|
$g: X \to X$, and $h: X \to X$, if $h$ is one-to-one and
|
|
$h \circ f = h \circ g$, then $f = g$. Justify your answer.
|
|
|
|
True.
|
|
|
|
**Proof:**
|
|
|
|
Suppose given any set $X$ such that $f: X \to X$, $g: X \to X$, and $h: X \to X$
|
|
are functions. Furthermore, suppose $h$ is one-to-one and
|
|
$h \circ f = h \circ g$.
|
|
|
|
Let $x \in X$.
|
|
|
|
To prove $f = g$, it must be shown that $f(x) = g(x)$.
|
|
|
|
By the definition of the composition of functions:
|
|
|
|
$$ (h \circ f)(x) = h(f(x)) $$
|
|
|
|
And also by the definition of the composition of functions:
|
|
|
|
$$ (h \circ g)(x) = h(g(x)) $$
|
|
|
|
By the supposition, this means that:
|
|
|
|
$$ h(f(x)) = h(g(x)) $$
|
|
|
|
Now, since $h$ is one-to-one, and since $h(f(x)) = h(g(x))$, it follows that:
|
|
|
|
$$ f(x) = g(x) $$
|
|
|
|
This is what was to be shown. Therefore $f = g$.
|
|
|
|
Q.E.D.
|
|
|
|
25. True or False? Given any set $X$ and given any functions $f: X \to X$,
|
|
$g: X \to X$, and $h: X \to X$, if $h$ is one-to-one and
|
|
$f \circ h = g \circ h$, then $f = g$. Justify your answer.
|
|
|
|
Omitted.
|
|
|
|
In 26 and 27 find $(g \circ f)^{-1}$, $g^{-1}$, $f^{-1}$, and
|
|
$f^{-1} \circ g^{-1}$, and state how $(g \circ f)^{-1}$ and
|
|
$f^{-1} \circ g^{-1}$ are related.
|
|
|
|
26. Let $X = \{a, b, c\}$, $Y = \{x, y, z\}$, and $Z = \{u, v, w\}$. Define
|
|
$f: X \to Y$ and $g: Y \to Z$ by the arrow diagrams below.
|
|
|
|
(See page 495 for image.)
|
|
|
|
Omitted.
|
|
|
|
27. Define $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ by
|
|
the formulas
|
|
|
|
$$ f(x) = x + 3 \quad \text{ and } \quad g(x) = -x \quad \text{ for each } x \in \mathbb{R} $$
|
|
|
|
Omitted.
|
|
|
|
28. Prove or give a counterexample: If $f: X \to Y$ and $g: Y \to X$ are
|
|
functions such that $g \circ f = I_x$ and $f \circ g = I_y$, then $f$ and
|
|
$g$ are both one-to-one and onto and $g = f^{-1}$.
|
|
|
|
Omitted.
|
|
|
|
29. Suppose $f: X \to Y$ and $g: Y \to Z$ are both one-to-one and onto. Prove
|
|
that $(g \circ f)^{-1}$ exists and that
|
|
$(g \circ f)^{-1} = f^{-1} \circ g^{-1}$.
|
|
|
|
Omitted.
|
|
|
|
30. Let $f: X \to Y$ and $g: Y \to Z$. Is the following property true or false?
|
|
For every subset $C$ in $Z$, $(g \circ f)^{-1}(C) = f^{-1}(g^{-1}(C))$.
|
|
Justify your answer.
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
Page 507
|
|
|
|
**Exercise Set 7.4**
|
|
|
|
1. When asked what it means to say that set $A$ has the same cardinality as set
|
|
$B$, a student replies, "$A$ and $B$ are one-to-one and onto." What _should_
|
|
the student have replied? Why?
|
|
|
|
Since $A$ and $B$ are sets and not functions, the student's statement is
|
|
incorrect, since sets cannot have the properties of being one-to-one and onto,
|
|
only a function can have these properties. Instead, the student have claimed
|
|
that $A$ and $B$ have the same cardinality if, and only if, there is a function
|
|
from $A$ to $B$ that is both one-to-one and onto (a one-to-one correspondence).
|
|
|
|
2. Show that "there are as many squares as there are numbers" by exhibiting a
|
|
one-to-one correspondence from the positive integers, $\mathbb{Z}^+$, to the
|
|
set $S$ of all squares of positive integers:
|
|
|
|
$$ S = \{n \in \mathbb{Z}^+ | n = k^2, \text{ for some positive integer } k\} $$
|
|
|
|
**Proof:**
|
|
|
|
Suppose there is a set $S$ that is the set of all squares:
|
|
|
|
$$ S = \{n \in \mathbb{Z}^+ | n = k^2, \text{ for some positive integer } k\} $$
|
|
|
|
To prove that $S$ and the set of all positive integers, $\mathbb{Z}^+$, have the
|
|
same cardinality, it must be shown that there exists some function, $f$, such
|
|
that $f$ is a one-to-one correspondence from $\mathbb{Z}^+$ to $S$. In other
|
|
words, it must be shown that there exists some function $f: \mathbb{Z}^+ \to S$
|
|
such that $f$ is one-to-one and onto.
|
|
|
|
_Proof ($f$ is one-to-one):_
|
|
|
|
Suppose $k_1, k_2 \in \mathbb{Z}^+$ such that $f(k_1) = f(k_2)$.
|
|
|
|
To prove that $f$ is one-to-one, it must be shown that $k_1 = k_2$.
|
|
|
|
By definition of $f$:
|
|
|
|
$$ k_1^2 = k_2^2 $$
|
|
|
|
By algebra:
|
|
|
|
$$ \sqrt{k_1^2} = \sqrt{k_2^2} $$
|
|
|
|
$$ \pm k_1 = \pm k_2 $$
|
|
|
|
Recall that $k_1, k_2 \in \mathbb{Z}^+$, so it follows that:
|
|
|
|
$$ k_1 = k_2 $$
|
|
|
|
This is what was to be shown. Therefore, it can be concluded that $f$ is
|
|
one-to-one.
|
|
|
|
_Proof ($f$ is onto):_
|
|
|
|
Suppose there is some $m \in S$.
|
|
|
|
To prove that $f$ is onto, it must be shown that there exists some
|
|
$k \in \mathbb{Z}^+$ such that $f(k) = m$.
|
|
|
|
By definition of $S$:
|
|
|
|
$$ m = k^2 $$
|
|
|
|
for some integer $k$.
|
|
|
|
Then, by definition of $f$:
|
|
|
|
$$ f(k) = k^2 = m $$
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $f$ is onto.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $f$ is both one-to-one and onto, it follows by the
|
|
properties of cardinality that the sets $S$ and $\mathbb{Z}^+$ have the same
|
|
cardinality (the same number of elements in them).
|
|
|
|
Q.E.D.
|
|
|
|
3. Let
|
|
$3\mathbb{Z} = \{n \in \mathbb{Z} | n = 3k, \text{ for some integer } k\}$.
|
|
Prove that $\mathbb{Z}$ and $3\mathbb{Z}$ have the same cardinality.
|
|
|
|
**Proof:**
|
|
|
|
Suppose there is a set $3\mathbb{Z}$ that represents the set of all integers
|
|
divisible by $3$:
|
|
|
|
$$ 3\mathbb{Z} = \{n \in \mathbb{Z} | n = 3k, \text{ for some integer } k\} $$
|
|
|
|
Define $f: \mathbb{Z} \to 3\mathbb{Z}$ as $f(n) = 3n$ for some integer $n$.
|
|
|
|
To prove that $\mathbb{Z}$ and $3\mathbb{Z}$ have the same cardinality, it must
|
|
shown that $f$ is a one-to-one correspondence from $\mathbb{Z}$ to 3\mathbb{Z}.
|
|
|
|
_Proof ($f$ is one-to-one):_
|
|
|
|
Suppose there exists some $x_1, x_2 \in 3\mathbb{Z}$ such that
|
|
$f(x_1) = f(x_2)$.
|
|
|
|
To prove that $f$ is one-to-one, it must be shown that $x_1 = x_2$.
|
|
|
|
By definition of $f$:
|
|
|
|
$$ 3x_1 = 3x_2 $$
|
|
|
|
By algebra:
|
|
|
|
$$ x_1 = x_2 $$
|
|
|
|
This is what was to be shown. Therefore, it can be concluded that $f$ is
|
|
one-to-one.
|
|
|
|
_Proof ($f$ is onto):_
|
|
|
|
Suppose there is some $m \in 3\mathbb{Z}$.
|
|
|
|
To prove that $f$ is onto, it must be shown that $f(k) = m$ for some integer
|
|
$k$.
|
|
|
|
By definition of $3\mathbb{Z}$:
|
|
|
|
$$ f(k) = 3k = m $$
|
|
|
|
for some integer $k$.
|
|
|
|
This is what was to be shown. Therefore, it can be concluded that $f$ is onto.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $f$ is a one-to-one correspondence from
|
|
$\mathbb{Z}$ to $3\mathbb{Z}$, it can be concluded that $\mathbb{Z}$ and
|
|
$3\mathbb{Z}$ have the same cardinality.
|
|
|
|
Q.E.D.
|
|
|
|
4. Let $\mathbb{O}$ be the set of all odd integers. Prove that $\mathbb{O}$ has
|
|
the same cardinality as $2\mathbb{Z}$, the set of all even integers.
|
|
|
|
**Proof:**
|
|
|
|
Suppose there is a set $\mathbb{O}$ that represents the set of all odd integers:
|
|
|
|
$$ \mathbb{O} = \{n \in \mathbb{Z} | n = 2k + 1 \text{ for some integer } k\} $$
|
|
|
|
Let $f: \mathbb{O} \to 2\mathbb{Z}$ (where $2\mathbb{Z}$ is the set of all even
|
|
integers) such that $f(n) = n - 1$ for some integer $n$.
|
|
|
|
To prove that $\mathbb{O}$ has the same cardinality as $2\mathbb{Z}$, it must be
|
|
shown that $f$ is a one-to-one correspondence from $\mathbb{O}$ to
|
|
$2\mathbb{Z}$.
|
|
|
|
_Proof ($f$ is one-to-one):_
|
|
|
|
Suppose $x_1, x_2 \in \mathbb{O}$ such that $f(x_1) = f(x_2)$.
|
|
|
|
To prove that $f$ is one-to-one, it must be shown that $x_1 = x_2$.
|
|
|
|
By the definition of $f$:
|
|
|
|
$$ x_1 - 1 = x_2 - 1 $$
|
|
|
|
By algebra:
|
|
|
|
$$ x_1 = x_2 $$
|
|
|
|
This is what was to be shown. Therefore, it can be concluded that $f$ is
|
|
one-to-one.
|
|
|
|
_Proof ($f$ is onto):_
|
|
|
|
Suppose $m \in 2\mathbb{Z}$.
|
|
|
|
To prove that $f$ is onto, it must be shown that $f(k) = m$ for some integer
|
|
$k \in \mathbb{O}$.
|
|
|
|
Let $k = m + 1$. Since $m \in 2\mathbb{Z}$, $m + 1$ is odd (by definition of
|
|
odd). This means that $k$ is odd, so $k \in \mathbb{O}$. Then:
|
|
|
|
$$ f(k) = k - 1 = (m + 1) - 1 = m $$
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $f$ is onto.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $f$ is a one-to-one correspondence from
|
|
$\mathbb{O}$ to $2\mathbb{Z}$, it can be concluded that $\mathbb{O}$ and
|
|
$2\mathbb{Z}$ have the same cardinality.
|
|
|
|
5. Let $25\mathbb{Z}$ be the set of all integers that are multiples of $25$.
|
|
Prove that $25\mathbb{Z}$ has the same cardinality as $2\mathbb{Z}$, the set
|
|
of all even integers.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $25\mathbb{Z}$ represents the set of all integers that are multiples of
|
|
$25$.
|
|
|
|
$$ 25\mathbb{Z} = \{n \in \mathbb{Z} | n = 25k \text{ for some integer } k\} $$
|
|
|
|
Let $f: 25\mathbb{Z} \to 2\mathbb{Z}$ (where $2\mathbb{Z}$ represents the set of
|
|
all even integers), and define $f$ as $f(n) = \dfrac{2}{25}n$ for some integer
|
|
$n$.
|
|
|
|
To prove that $25\mathbb{Z}$ has the same cardinality as $2\mathbb{Z}$, it must
|
|
be shown that $f$ is a one-to-one correspondence from
|
|
$25\mathbb{Z} \to 2\mathbb{Z}$.
|
|
|
|
_Proof ($f$ is one-to-one):_
|
|
|
|
Suppose $x_1, x_2 \in 25\mathbb{Z}$ such that $f(x_1) = f(x_2)$.
|
|
|
|
To prove that $f$ is one-to-one, it must be shown that $x_1 = x_2$.
|
|
|
|
By the definition for $f$:
|
|
|
|
$$ \frac{2}{25}x_1 = \frac{2}{25}x_2 $$
|
|
|
|
By algebra:
|
|
|
|
$$ x_1 = x_2 $$
|
|
|
|
This is what was to be shown. Therefore, it can be concluded that $f$ is
|
|
one-to-one.
|
|
|
|
_Proof ($f$ is onto):_
|
|
|
|
Let $m \in 2\mathbb{Z}.
|
|
|
|
To show that $f$ is onto, it must be shown that $f(k) = m$ for some
|
|
$k \in 25\mathbb{Z}$.
|
|
|
|
Let $k = \dfrac{25m}{2}$.
|
|
|
|
Since $m$ is even, $\dfrac{m}{2}$ is an integer, so $k = 25 \cdot \dfrac{m}{2}$,
|
|
which is a multiple of $25$. It follows that $k \in 25\mathbb{Z}$.
|
|
|
|
Then:
|
|
|
|
$$ f(k) = \frac{2}{25}\left(\frac{25m}{2}\right) = m $$
|
|
|
|
This is what was to be shown. Therefore, it can be concluded that $f$ is onto.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $f$ is a one-to-one correspondence from
|
|
$25\mathbb{Z} \to 2\mathbb{Z}$, it can be concluded that $25\mathbb{Z}$ and
|
|
$2\mathbb{Z}$ have the same cardinality.
|
|
|
|
6. Use the functions $I$ and $J$ defined in the paragraph following Example
|
|
7.4.1 to show that even though there is a one-to-one correspondence, $H$,
|
|
from $2\mathbb{Z}$ to $\mathbb{Z}$, there is also a function from
|
|
$2\mathbb{Z}$ to $\mathbb{Z}$ that is one-to-one but not onto and a function
|
|
from $\mathbb{Z}$ to $2\mathbb{Z}$ that is onto but not one-to-one. In other
|
|
words, show that $I$ is one-to-one but not onto, and show that $J$ is onto
|
|
but not one-to-one.
|
|
|
|
_Hint:_ If $m \in 2\mathbb{Z}$, show that $J(m) = J(m + 1) = m$.
|
|
|
|
Suppose $I: 2\mathbb{Z} \to \mathbb{Z}$, and define $I$ as $I(n) = n$ for some
|
|
even integer $n$.
|
|
|
|
Furthermore, suppose $J: \mathbb{Z} \to 2\mathbb{Z}$, and define $J$ as $J(m) =
|
|
2\left\lfloor \dfrac{m}{2} \right\rfloor$ for some integer $m$.
|
|
|
|
**Proof ($I$ is one-to-one, but not onto):**
|
|
|
|
_Proof ($I$ is one-to-one):_
|
|
|
|
Suppose $x_1, x_2 \in 2\mathbb{Z}$ such that $I(x_1) = I(x_2)$. To prove that
|
|
$I$ is one-to-one, it must be shown that $x_1 = x_2$.
|
|
|
|
By definition for $I$:
|
|
|
|
$$ x_1 = x_2 $$
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $I$ is
|
|
one-to-one.
|
|
|
|
_Proof ($I$ is not onto):_
|
|
|
|
Suppose $m \in \mathbb{Z}$.
|
|
|
|
To prove that $I$ is onto, it would need to be shown that $I(k) = m$, for some
|
|
$k \in \mathbb{Z}$.
|
|
|
|
But since $m \in \mathbb{Z}$, $m$ could be an odd integer. If $m$ is odd, then
|
|
$m = 2k + 1$ for some integer $k$, and so $I$ is not onto.
|
|
|
|
Consider $m = 3$, then there would exist no $k$ for which $I(k) = 3$, since $I$.
|
|
|
|
Therefore $I$ is not onto.
|
|
|
|
**Proof ($J$ is onto, but not one-to-one):**
|
|
|
|
_Proof ($J$ is onto):_
|
|
|
|
Suppose $m \in 2\mathbb{Z}$.
|
|
|
|
To prove that $J$ is onto, it must be shown that $J(k) = m$ for some
|
|
$k \in \mathbb{Z}$.
|
|
|
|
Let $k = m$.
|
|
|
|
By the definition for $J$:
|
|
|
|
$$ J(k) = 2\lfloor \frac{k}{2} \rfloor $$
|
|
|
|
Since $k$ is even (since $k = m$ and $m \in 2\mathbb{Z}$), $k = 2p$ for some
|
|
integer $p$. By substitution:
|
|
|
|
$$ J(k) = 2\lfloor \frac{2p}{2} \rfloor $$
|
|
|
|
$$ = 2\lfloor p \rfloor $$
|
|
|
|
Then, by definition of floor:
|
|
|
|
$$ = 2p $$
|
|
|
|
Since $p$ is an integer, $2p \in 2\mathbb{Z}$, by the definition of even. It
|
|
follows that:
|
|
|
|
$$ J(k) = 2p = k = m $$
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $J$ is onto.
|
|
|
|
_Proof ($J$ is not one-to-one):_
|
|
|
|
Were it to be proven that $J$ is one-to-one, it would have to assumed that for
|
|
some $x_1, x_2 \in \mathbb{Z}$, such that $J(x_1) = J(x_2)$, and then shown that
|
|
$x_1 = x_2$.
|
|
|
|
By the definition for $J$:
|
|
|
|
$$ 2\lfloor \frac{x_1}{2} \rfloor = 2\lfloor \frac{x_2}{2} \rfloor $$
|
|
|
|
But the floor function does not necessarily have to take the same image to
|
|
generate the same elements of its co-domain.
|
|
|
|
Consider $x_1 = 2$ and $x_2 = 3$, then by definition for $J$:
|
|
|
|
$$ J(x_1) = 2\lfloor \frac{2}{2} \rfloor = 2\lfloor 1 \rfloor = 2(1) = 2 $$
|
|
|
|
$$ J(x_2) = 2\lfloor \frac{3}{2} \rfloor = 2(1) = 2 $$
|
|
|
|
So, $J(x_1) = J(x_2)$, but $x_1 \neq x_2$. Therefore $J$ is not one-to-one.
|
|
|
|
7.
|
|
|
|
a. Check that the formula for $F$ given at the end of Example 7.4.2 produces the
|
|
correct values for $n = 1, 2, 3, \text{ and } 4$.
|
|
|
|
The formula for $F$ is as follows:
|
|
|
|
$$
|
|
F(n) =
|
|
\begin{cases}
|
|
\dfrac{n}{2} & \text{if } n \text{ is an even positive integer} \\
|
|
-\dfrac{n - 1}{2} & \text{if } n \text{ is an odd positive integer}
|
|
\end{cases}
|
|
$$
|
|
|
|
_Case $n = 1$:_
|
|
|
|
$$ F(1) = -\frac{(1) - 1}{2} = -\frac{0}{2} = (-1)0 = 0 $$
|
|
|
|
_Case $n = 2$:_
|
|
|
|
$$ F(2) = \frac{(2)}{2} = 1 $$
|
|
|
|
_Case $n = 3$:_
|
|
|
|
$$ F(3) = -\frac{(3) - 1}{2} = (-1)\left(\frac{2}{2}\right) = (-1)(1) = -1 $$
|
|
|
|
_Case $n = 4$:_
|
|
|
|
$$ F(4) = \frac{(4)}{2} = 2 $$
|
|
|
|
b. Use the floor function to write a formula for $F$ as a single algebraic
|
|
expression for each positive integer $n$.
|
|
|
|
$$ F(n) = (-1)^n\lfloor \frac{n}{2} \rfloor $$
|
|
|
|
8. Use the result of exercise 3 to prove that $3\mathbb{Z}$ is countable.
|
|
|
|
Recall that a set is countable if, and only if, it is finite or countably
|
|
infinite.
|
|
|
|
Additionally, recall that a set is countably infinite if, and only if, it has
|
|
the same cardinality as the set of positive integers $\mathbb{Z}^+$ (_i.e._
|
|
there is a function that is a one-to-one correspondence from the given set to
|
|
$\mathbb{Z}^+$).
|
|
|
|
**Proof:**
|
|
|
|
Suppose $3\mathbb{Z}$ represents the set of all integers that are multiples of
|
|
$3$:
|
|
|
|
$$ 3\mathbb{Z} = \{n \in \mathbb{Z} | n = 3k, \text{ for some integer } k\} $$
|
|
|
|
To prove that $3\mathbb{Z}$ is countable, it must be shown that $3\mathbb{Z}$ is
|
|
finite or countably infinite.
|
|
|
|
$3\mathbb{Z}$ is not finite, thus it must be shown that $3\mathbb{Z}$ is
|
|
countably infinite.
|
|
|
|
As was shown in Example 7.4.3, $\mathbb{Z}$ is countably infinite (_i.e._
|
|
$\mathbb{Z}$ has the same cardinality as $\mathbb{Z}^+$). Additionally, by
|
|
problem 3, it has been shown that $3\mathbb{Z}$ has the same cardinality has
|
|
$\mathbb{Z}$. By the transitive property of cardinality, it follows that
|
|
$3\mathbb{Z}$ has the same cardinality as $\mathbb{Z}^+$, and therefore
|
|
$3\mathbb{Z}$ is countably infinite.
|
|
|
|
9. Show that the set of all nonnegative integers is countable by exhibiting a
|
|
one-to-one correspondence between $\mathbb{Z}^+$ and
|
|
$\mathbb{Z}^{\text{nonneg}}$.
|
|
|
|
Recall that a set is countable if, and only if, it is finite or countably
|
|
infinite.
|
|
|
|
Additionally, recall that a set is countably infinite if, and only if, it has
|
|
the same cardinality as the set of positive integers $\mathbb{Z}^+$ (_i.e._
|
|
there is a function that is a one-to-one correspondence from the given set to
|
|
$\mathbb{Z}^+$).
|
|
|
|
**Proof:**
|
|
|
|
Suppose $\mathbb{Z}^+$ represents the set of all positive integers and that
|
|
$\mathbb{Z}^{\text{nonneg}}$ represents the set of all nonnegative integers.
|
|
|
|
To prove that $\mathbb{Z}^{\text{nonneg}}$ is countable, it must be shown that
|
|
$\mathbb{Z}^{\text{nonneg}}$ is finite or countably infinite. Since
|
|
$\mathbb{Z}^{\text{nonneg}}$ is not finite, it follows that it must be shown
|
|
that $\mathbb{Z}^{\text{nonneg}}$ is countably infinite (_i.e._ has the same
|
|
cardinality as $\mathbb{Z}^+$).
|
|
|
|
Let $f: \mathbb{Z}^+ \to \mathbb{Z}^{\text{nonneg}}$ be defined as
|
|
$f(n) = n - 1$ for some positive integer $n$.
|
|
|
|
_Proof ($f$ is one-to-one):_
|
|
|
|
Suppose $x_1, x_2 \in \mathbb{Z}^+$ such that $f(x_1) = f(x_2)$.
|
|
|
|
To prove that $f$ is one-to-one, it must be shown that $x_1 = x_2$.
|
|
|
|
By definition of $f$:
|
|
|
|
$$ x_1 - 1 = x_2 - 1 $$
|
|
|
|
By algebra:
|
|
|
|
$$ x_1 = x_2 $$
|
|
|
|
This is what was to be shown. Therefore, it can be concluded that $f$ is
|
|
one-to-one.
|
|
|
|
_Proof ($f$ is onto):_
|
|
|
|
Suppose $m \in \mathbb{Z}^{\text{nonneg}}$.
|
|
|
|
To prove that $f$ is onto, it must be shown that $f(n) = m$ for some
|
|
$n \in \mathbb{Z}^+$.
|
|
|
|
By definition for $f$:
|
|
|
|
$$ f(n) = n - 1 $$
|
|
|
|
Let $n = m + 1$. Since $m \in \mathbb{Z}^{\text{nonneg}}$, it follows that
|
|
$n \in \mathbb{Z}^+$ (since adding $1$ to any nonnegative integer is positive).
|
|
|
|
Then, by substitution:
|
|
|
|
$$ = (m + 1) - 1 $$
|
|
|
|
By algebra:
|
|
|
|
$$ = m $$
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $f$ is onto.
|
|
|
|
_Conclusion:_
|
|
|
|
Since $f$ has been shown to be a one-to-one correspondence, it can be concluded
|
|
that $\mathbb{Z}^{\text{nonneg}}$ is countable.
|
|
|
|
In 10-14 $S$ denotes the set of real numbers strictly between $0$ and $1$. That
|
|
is, $S = \{x \in \mathbb{R} | 0 < x < 1\}$.
|
|
|
|
10. Let $U = \{x \in \mathbb{R} | 0 < x < 2\}$. Prove that $S$ and $U$ have the
|
|
same cardinality.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $S$ represents the set of real numbers strictly between $0$ and $1$:
|
|
|
|
$$ S = \{x \in \mathbb{R} | 0 < x < 1\} $$
|
|
|
|
Additionally, suppose $U$ is the set of real numbers strictly between $0$ and
|
|
$2$:
|
|
|
|
$$ U = \{x \in \mathbb{R} | 0 < x < 2\} $$
|
|
|
|
Let $f: S \to U$, and define $f$ as $f(n) = 2n$ for some $n \in S$.
|
|
|
|
To prove that $S$ and $U$ have the same cardinality, it must be shown that $f$
|
|
is a one-to-one correspondence from $S \to U$.
|
|
|
|
_Proof ($f$ is one-to-one):_
|
|
|
|
Suppose $n_1, n_2 \in S$ such that $f(n_1) = f(n_2)$.
|
|
|
|
To prove that $f$ is one-to-one, it must be shown that $n_1 = n_2$.
|
|
|
|
By definition of $f$:
|
|
|
|
$$ 2n_1 = 2n_2 $$
|
|
|
|
By algebra:
|
|
|
|
$$ n_1 = n_2 $$
|
|
|
|
This is what was to be shown. Therefore, it can be concluded that $f$ is
|
|
one-to-one.
|
|
|
|
_Proof ($f$ is onto):_
|
|
|
|
Let $m \in U$.
|
|
|
|
To prove that $f$ is onto, it must be shown that $f(n) = m$ for some $n \in S$.
|
|
|
|
Let $n = \frac{m}{2}$.
|
|
|
|
Since $m \in U$, $0 < m < 2$. It follows that $0 < \dfrac{m}{2} < 1$. By
|
|
substitution, this means that $0 < n < 1$, which means that $n \in S$.
|
|
|
|
Then, by definition for $f$:
|
|
|
|
$$ f(n) = 2n $$
|
|
|
|
By substitution:
|
|
|
|
$$ = 2\left(\frac{m}{2}\right) $$
|
|
|
|
By algebra:
|
|
|
|
$$ = m $$
|
|
|
|
This is what was to be shown. Therefore, it can be concluded that $f$ is onto.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $f$ is a one-to-one correspondence, it can be
|
|
concluded that $S$ and $U$ have the same cardinality.
|
|
|
|
Q.E.D.
|
|
|
|
11. Let $V = \{x \in \mathbb{R} | 2 < x < 5\}$. Prove that $S$ and $V$ have the
|
|
same cardinality.
|
|
|
|
_Hint:_ Define $h: S \to V$ as follows: $h(x) = 3x + 2$, for every $x \in S$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $S$ represents the set of real numbers strictly between $0$ and $1$:
|
|
|
|
$$ S = \{x \in \mathbb{R} | 0 < x < 1\} $$
|
|
|
|
Additionally, suppose $V$ is the set of real numbers strictly between $2$ and
|
|
$5$:
|
|
|
|
$$ V = \{x \in \mathbb{R} | 2 < x < 5\} $$
|
|
|
|
Let $h: S \to V$, and define $h$ as $h(x) = 3x + 2$ for some $x \in S$.
|
|
|
|
To prove that $S$ and $V$ have the same cardinality, it must be shown that $h$
|
|
is a one-to-one correspondence for $S \to V$.
|
|
|
|
_Proof ($h$ is one-to-one):_
|
|
|
|
Suppose $x_1, x_2 \in S$ such that $h(x_1) = h(x_2)$.
|
|
|
|
To prove $h$ is one-to-one, it must be shown that $x_1 = x_2$.
|
|
|
|
By the definition of $h$:
|
|
|
|
$$ 3x_1 + 2 = 3x_2 + 2 $$
|
|
|
|
By algebra:
|
|
|
|
$$ 3x_1 = 3x_2 $$
|
|
|
|
$$ x_1 = x_2 $$
|
|
|
|
This is what was to be shown. Therefore, it can be concluded that $h$ is
|
|
one-to-one.
|
|
|
|
_Proof ($h$ is onto):_
|
|
|
|
Let $v \in V$.
|
|
|
|
To prove that $h$ is onto, it must be shown that $f(x) = v$ for some $x \in S$.
|
|
|
|
Let $x = \dfrac{v - 2}{3}$.
|
|
|
|
Since $v \in V$, $2 < v < 5$. It follows that:
|
|
|
|
$$ 2 - 2 < v - 2 < 5 - 2 $$
|
|
|
|
$$ 0 < v - 2 < 3 $$
|
|
|
|
$$ \frac{0}{3} < \frac{v - 2}{3} < \frac{3}{3} $$
|
|
|
|
$$ 0 < \frac{v - 2}{3} < 1 $$
|
|
|
|
By substitution, this means that:
|
|
|
|
$$ 0 < x < 1 $$
|
|
|
|
This means that $x \in S$.
|
|
|
|
Then, by definition for $h$:
|
|
|
|
$$ h(x) = 3x + 2 $$
|
|
|
|
By substitution:
|
|
|
|
$$ h(x) = 3\left(\frac{v - 2}{3}\right) + 2 $$
|
|
|
|
By algebra:
|
|
|
|
$$ = (v - 2) + 2 $$
|
|
|
|
$$ = v $$
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $h$ is onto.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $h$ is a one-to-one correspondence for $S \to V$,
|
|
it can be concluded that $S$ and $V$ have the same cardinality.
|
|
|
|
Q.E.D.
|
|
|
|
12. Let $a$ and $b$ be real numbers with $a < b$, and suppose that
|
|
$W = \{x \in \mathbb{R} | a < x < b\}$. Prove that $S$ and $W$ have the same
|
|
cardinality.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $S$ represents the set of real numbers strictly between $0$ and $1$:
|
|
|
|
$$ S = \{x \in \mathbb{R} | 0 < x < 1\} $$
|
|
|
|
Additionally, suppose $a, b \in \mathbb{R}$ such that $a < b$. Then, suppose $W$
|
|
is a set defined as:
|
|
|
|
$$ W = \{x \in \mathbb{R} | a < x < b\} $$
|
|
|
|
Let $f: S \to W$ be defined as $f(x) = (b - a)x + a$ for some $x \in S$.
|
|
|
|
To prove that $S$ and $W$ have the same cardinality, it must be shown that $f$
|
|
is a one-to-one correspondence for $S \to W$.
|
|
|
|
_Proof ($f$ is one-to-one):_
|
|
|
|
Let $x_1, x_2 \in S$ such that $f(x_1) = f(x_2)$.
|
|
|
|
To prove that $f$ is one-to-one, it must be shown that $x_1 = x_2$.
|
|
|
|
By the definition of $f$:
|
|
|
|
$$ (b - a)x_1 + a = (b - a)x_2 + a $$
|
|
|
|
By algebra:
|
|
|
|
$$ (b - a)x_1 = (b - a)x_2 $$
|
|
|
|
$$ x_1 = x_2 $$
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $f$ is
|
|
one-to-one.
|
|
|
|
_Proof ($f$ is onto):_
|
|
|
|
Let $w \in W$.
|
|
|
|
To prove that $f$ is onto, it must be shown that $f(x) = w$ for some $x \in S$.
|
|
|
|
Let $x = \dfrac{w - a}{b - a}$.
|
|
|
|
Since $w \in W$, this means that $a < w < b$. It follows that:
|
|
|
|
$$ a - a < w - a < b - a $$
|
|
|
|
$$ 0 < w - a < b - a $$
|
|
|
|
$$ \frac{0}{b - a} < \frac{w - a}{b - a} < \frac{b - a}{b - a} $$
|
|
|
|
$$ 0 < \frac{w - a}{b - a} < 1 $$
|
|
|
|
By substitution:
|
|
|
|
$$ 0 < x < 1 $$
|
|
|
|
It follows that $x \in S$.
|
|
|
|
By definition for $f$:
|
|
|
|
$$ f(x) = (b - a)x + a $$
|
|
|
|
By substitution:
|
|
|
|
$$ f(x) = (b - a)\left(\frac{w - a}{b - a}\right) + a $$
|
|
|
|
$$ = (w - a) + a $$
|
|
|
|
$$ = w $$
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $f$ is onto.
|
|
|
|
_Conclusion:_
|
|
|
|
Since it has been shown that $f$ is a one-to-one correspondence for $S \to W$,
|
|
it can be concluded that $S$ and $W$ have the same cardinality.
|
|
|
|
Q.E.D.
|
|
|
|
13. Draw the graph of the function $f$ defined by the following formula:
|
|
|
|
For each real number $x$ with $0 < x < 1$,
|
|
|
|
$$ f(x) = \tan\left(\pi x - \frac{\pi}{2}\right) $$
|
|
|
|
Use the graph to explain why $S$ and $\mathbb{R}$ have the same cardinality.
|
|
|
|
Omitted.
|
|
|
|
14. Define a function $g$ from the set of real numbers to $S$ by the following
|
|
formula:
|
|
|
|
For each real number $x$,
|
|
|
|
$$ g(x) = \frac{1}{2} \cdot \left(\frac{x}{1 + |x|}\right) + \frac{1}{2} $$
|
|
|
|
Prove that $g$ is a one-to-one correspondence. (It is possible to prove this
|
|
statement either with calculus or without it.) What conclusion can you draw from
|
|
this fact?
|
|
|
|
Omitted.
|
|
|
|
15. Show that the set of all bit strings (strings of $0$'s and $1$'s) is
|
|
countable.
|
|
|
|
Recall that a set is countable if, and only if, it is finite or countably
|
|
infinite.
|
|
|
|
Additionally, recall that a set is countably infinite if, and only if, it has
|
|
the same cardinality as the set of positive integers $\mathbb{Z}^+$ (_i.e._
|
|
there is a function that is a one-to-one correspondence from the given set to
|
|
$\mathbb{Z}^+$).
|
|
|
|
**Proof:**
|
|
|
|
Suppose $B$ is the set of all bit strings (strings of $0$'s and $1$'s).
|
|
|
|
To prove that $B$ is countable, it must be shown that $B$ is finite or countably
|
|
infinite. Since $B$ is not finite, it must be shown that $B$ is countably
|
|
infinite.
|
|
|
|
To show that $B$ is countably infinite, it must be shown that $B$ has the same
|
|
cardinality as the set of all positive integers, $\mathbb{Z}^+$.
|
|
|
|
To show that $B$ and $\mathbb{Z}^+$ have the same cardinality, it must be shown
|
|
that there exists some one-to-one correspondence for $B \to \mathbb{Z}^+$.
|
|
|
|
Consider a function, $f$ that maps the bit strings by their length to some
|
|
positive integer. For example, say $f(\lambda) = 1$ (where $\lambda$ represents
|
|
the null string). Additional examples would include $f(0) = 2, f(1) = 3$ for bit
|
|
strings of length $2$. Further examples for length $3$ include
|
|
$f(00) = 4, f(01) = 5, f(10) = 6, f(11) = 7$, and so on.
|
|
|
|
Generally, for each integer $n \geq 0$, there are $2^n$ bit strings of length
|
|
$n$, and $f$ maps them to the positive integers between $2^n$ (inclusive) and
|
|
$2^{n + 1} - 1$ (inclusive).
|
|
|
|
$f$ is one-to-one since two bit strings never map to the same positive integer.
|
|
|
|
$f$ is onto as every positive integer is in the range for $f$ (namely
|
|
$\mathbb{Z}^+$).
|
|
|
|
Therefore, since $f$ is a one-to-one correspondence for $B \to \mathbb{Z}^+$, it
|
|
can be concluded that $B$ and $\mathbb{Z}^+$ have the same cardinality, and
|
|
therefore $B$ is countable.
|
|
|
|
Q.E.D.
|
|
|
|
16. Show that $\mathbb{Q}$, the set of all rational numbers, is countable.
|
|
|
|
Omitted.
|
|
|
|
17. Show that $\mathbb{Q}$, the set of all rational numbers, is dense along the
|
|
number line by showing that given any two rational numbers $r_1$ and $r_2$
|
|
with $r_2 < r_2$, there exists a rational number $x$ such that
|
|
$r_1 < x < r_2$.
|
|
|
|
_Hint:_ See the hints for exercises 18 and 19 in Section 4.3.
|
|
|
|
18. _Hint:_
|
|
|
|
$$ \frac{\dfrac{a}{b} + \dfrac{c}{d}}{2} = \frac{\dfrac{(ad + bc)}{(bd)}{2} =
|
|
\frac{ad + bc}{2bd} $$
|
|
|
|
19. _Hint:_ If $a < b$ then $a + a < a + b$ (by T19 of Appendix A), or
|
|
equivalently, $2a < a + b$. Thus $a < \dfrac{a + b}{2}$ (by T20 of Appendix
|
|
A).
|
|
|
|
**Proof:**
|
|
|
|
Let $r_1, r_2 \in \mathbb{Q}$ such that $r_1 < r_2$.
|
|
|
|
To prove $\mathbb{Q}$ is dense along the number line, it must be shown there
|
|
exists some $x \in \mathbb{Q}$ such that $r_1 < x < r_2$.
|
|
|
|
Let $x = \dfrac{r_1 + r_2}{2}$.
|
|
|
|
Since $r_1, r_2 \in \mathbb{Q}$, it follows that
|
|
$\dfrac{r_1 + r_2}{2} \in \mathbb{Q}$, hence $x \in \mathbb{Q}$.
|
|
|
|
Since $r_1 < r_2$, it follows that $r_1$ is less than their average:
|
|
|
|
$$ r_1 < \frac{r_1 + r_2}{2} $$
|
|
|
|
Similarly, since $r_2 > r_1$, it follows that $r_2$ is greater than their
|
|
average:
|
|
|
|
$$ \frac{r_1 + r_2}{2} < r_2 $$
|
|
|
|
This means that:
|
|
|
|
$$ r_1 < \frac{r_1 + r_2}{2} < r_2 $$
|
|
|
|
Now, by substitution:
|
|
|
|
$$ r_1 < x < r_2 $$
|
|
|
|
This is what was to be shown. Therefore it can be concluded that $\mathbb{Q}$ is
|
|
dense along the number line.
|
|
|
|
Q.E.D.
|
|
|
|
20. Must the average of two irrational numbers always be irrational? Prove or
|
|
give a counterexample.
|
|
|
|
**Disproof (by counterexample):**
|
|
|
|
Consider $r_1, r_2 \notin \mathbb{Q}$ where $r_1 = \sqrt{2}$ and
|
|
$r_2 = -\sqrt{2}$.
|
|
|
|
Then, their average would be:
|
|
|
|
$$ \frac{r_1 + r_2}{2} = \frac{\sqrt{2} + (-\sqrt{2})}{2} = \frac{0}{2} = 0 $$
|
|
|
|
Now, $0 \in \mathbb{Q}$.
|
|
|
|
This shows that the average of two irrational numbers is not always irrational.
|
|
|
|
Q.E.D.
|
|
|
|
21. Show that the set of all irrational numbers is dense along the number line
|
|
by showing that given any two real numbers, there is an irrational number in
|
|
between.
|
|
|
|
22. Give two examples of functions from $\mathbb{Z}$ to $\mathbb{Z}$ that are
|
|
one-to-one but not onto.
|
|
|
|
23. Give two examples of functions from $\mathbb{Z}$ to $\mathbb{Z}$ that are
|
|
onto but not one-to-one.
|
|
|
|
24. Define a function: $g: \mathbb{Z}^+ \times \mathbb{Z}y+ \to \mathbb{Z}^+$ by
|
|
the formula $g(m, n) = 2^m3^n$ for all
|
|
$(m, n) \in \mathbb{Z}^+ \times \mathbb{Z}^+$. Show that $g$ is one-to-one
|
|
and use this result to prove that $\mathbb{Z}^+ \times \mathbb{Z}^+$ is
|
|
countable.
|
|
|
|
25.
|
|
|
|
a. Explain how to use the following diagram to show that
|
|
$\mathbb{Z}^{\text{nonneg}} \times \mathbb{Z}^{\text{nonneg}}$ and
|
|
$\mathbb{Z}^{\text{nonneg}}$ have the same cardinality.
|
|
|
|
(See Page 508 for image.)
|
|
|
|
b. Define a function
|
|
$H: \mathbb{Z}^{\text{nonneg}} \times \mathbb{Z}^{\text{nonneg}} \to \mathbb{Z}^{\text{nonneg}}$
|
|
by the formula
|
|
|
|
$$ H(m, n) = n + \frac{(m + n)(m + n + 1)}{2} $$
|
|
|
|
for all nonnegative integers $m$ and $n$. Interpret the action of $H$
|
|
geometrically using the diagram of part (a).
|
|
|
|
24. Prove that the function $H$ defined analytically in exercise 23b is a
|
|
one-to-one correspondence.
|
|
|
|
25. Prove that $0.1999 \dots = 0.2$.
|
|
|
|
26. Prove that any infinite set contains a countably infinite subset.
|
|
|
|
27. Prove that if $A$ is any countably infinite set, $B$ is any set, and
|
|
$g: A \to B$ is onto, then $B$ is countable.
|
|
|
|
28. Prove that a disjoint union of any finite set and any countably infinite set
|
|
is countably infinite.
|
|
|
|
29. Prove that a union of any two countably infinite sets is countably infinite.
|
|
|
|
30. Use the result of exercise 29 to prove that the set of all irrational
|
|
numbers is uncountable.
|
|
|
|
31. Use the results of exercises 28 and 29 to prove that a union of any two
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countable sets is countable.
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32. Prove that $\mathbb{Z} \times \mathbb{Z}$, the Cartesian product of the set
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of integers with itself, is countably infinite.
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33. Use the results of exercises 27, 31, and 32 to prove the following: If $R$
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is the set of all solutions to all equations of the form $x^2 + bx + c = 0$,
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where $b$ and $c$ are integers, then $R$ is countable.
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34. Let $\mathscr{P}(S)$ be the set of all subsets of set $S$, and let $T$ be
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the set of all functions from $S$ to $\{0, 1\}$. Show that $\mathscr{P}(S)$
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and $T$ have the same cardinality.
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35. Let $S$ be a set and let $\mathscr{P}(S)$ be the set of all subsets of $S$.
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Show that $S$ is "smaller than" $\mathscr{P}(S)$ in the sense that there is
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a one-to-one function from $S$ to $\mathscr{P}(S)$ but there is no onto
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function from $S$ to $\mathscr{P}(S)$.
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36. The Schroeder-Bernstein theorem states the following: if $A$ and $B$ are any
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sets with the property that there is a one-to-one function from $A$ to $B$
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and a one-to-one function from $B$ to $A$, then $A$ and $B$ have the same
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cardinality. Use this theorem to prove that there are as many functions from
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$\mathbb{Z}^+$ to $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ as there functions from
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$\mathbb{Z}^+$ to $\{0, 1\}$.
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37. Prove that if $A$ and $B$ are any countably infinite sets, then $A \times B$
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is countably infinite.
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38. Suppose $A_1, A_2, A_3, \dots$ is an infinite sequence of countable sets.
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Recall that
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$$ \bigcup_{i = 1}^{\infty}A_i = \{x | x \in A_i \text{ for some positive integer } i\} $$
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Prove that $\bigcup_{i = 1}^{\infty}A_i$ is countable. (In other words, prove
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that a countably infinite union of countable sets is countable.)
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