diff --git a/chapter_7/exercises.md b/chapter_7/exercises.md index 2c3f30f..7a8be90 100644 --- a/chapter_7/exercises.md +++ b/chapter_7/exercises.md @@ -9,16 +9,30 @@ Page 458 a. Write the domain of $f$ and the co-domain of $f$. +Domain: $\{1, 3, 5\}$ + +Co-domain: $\{s, t, u, v\}$ + b. Find $f(1)$, $f(3)$, and $f(5)$. +$f(1) = v, f(3) = s, f(5) = v$ + c. What is the range of $f$? +$\{s, v\}$ + d. Is $3$ an inverse image of $s$? Is $1$ an inverse image of $u$? +yes; no + e. What is the inverse image of $s$? of $u$? of $v$? +$\{3\}$;$\emptyset$;$\{1, 5\}$ + f. Represent $f$ as a set of ordered pairs. +$\{(1, v), (3, s), (5, v)\}$ + 2. Let $X = \{1, 3, 5\}$ and $Y = \{a, b, c, d\}$. Define $g: X \to Y$ by the following arrow diagram. @@ -26,56 +40,105 @@ f. Represent $f$ as a set of ordered pairs. a. Write the domain of $g$ and the co-domain of $g$. +Domain: $\{1, 3, 5\}$ + +Co-domain: $\{a, b, c, d\}$ + b. Find $g(1)$, $g(3)$, and $g(5)$. +$g(1) = b, g(3) = b, g(5) = b$ + c. What is the range of $g$? +$\{b\}$ + d. Is $3$ an inverse image of $a$? Is $1$ an inverse image of $b$? +no;yes + e. What is the inverse image of $b$? of $c$? +$\{1, 3, 5\}, \emptyset$ + f. Represent $g$ as a set of ordered pairs. +$$ \{(1, b), (3, b), (5, b)\} $$ + 3. Indicate whether the statements in parts (a)-(d) are true or false for all functions. Justify your answers. a. If two elements in the domain of a function are equal, then their images in the co-domain are equal. +True. The definition of a function states that every input element in the domain +must have an output element in the co-domain. Since two elements in the domain +of the function are equal, then their outputs in the co-domain must be equal by +this definition. + b. If two elements in the co-domain of a function are equal, then their preimages in the domain are also equal. +This is false. A function can have the same output for two different inputs. + c. A function can have the same output for more than one input. +True, the definition of a function only states that every input to the function +must have an output, not necessarily unique outputs. + d. A function can have the same input for more than one output. +This is false. A single input can only map to a single output, not multiple +outputs. + 4. a. Find all functions from $X = \{a, b\}$ to $Y = \{u, v\}$. +$$ f(a) = u, f(a) = v, f(b) = u, f(b) = v $$ + b. Find all functions from $X = \{a, b, c\}$ to $Y = \{u\}$. +$$ f(a) = u, f(b) = u, f(c) = u $$ + c. Find all functions from $X = \{a, b, c\}$ to $Y = \{u, v\}$. +$$ f(a) = u, f(a) = v, f(b) = u, f(b) = v, f(c) = u, f(c) k v $$ + 5. Let $I_{\mathbb{z}}$ bee the identity function defined on the set of all integers, and suppose that $e$, $b_i^{jk}$, $K(t)$, and $u_{kj}$ all represent integers. Find the following: a. $I_{\mathbb{Z}}(e)$ +$$ I_{\mathbb{Z}}(e) = e $$ + b. $I_{\mathbb{Z}}\left(b_i^{jk}\right)$ +$$ I_{\mathbb{Z}}\left(b_i^{jk}\right) = b_i^{jk}\right $$ + c. $I_{\mathbb{Z}}(K(t))$ +$$ I_{\mathbb{Z}}(K(t)) = K(t) $$ + d. $I_{\mathbb{Z}}(u_{kj})$ +$$ I_{\mathbb{Z}}(u_{kj}) = u_{kj} $$ + 6. Find functions defined on the set of nonnegative integers that can be used to define the sequences whose first six terms are given below. a. $1, -\dfrac{1}{3}, \dfrac{1}{5}, -\dfrac{1}{7}, \dfrac{1}{9}, -\dfrac{1}{11}$ +$$ f: \mathbb{Z}^{\text{nonneg}} \to \mathbb{R} $$ + +$$ f(n) = \frac{(-1)^n}{2n + 1} $$ + b. $0, -2, 4, -6, 8, -10$ +$$ f: \mathbb{Z}^{\text{nonneg}} \to \mathbb{R} $$ + +$$ f(n) = (-1)^n \cdot 2n $$ + 7. Let $A = \{1, 2, 3, 4, 5\}$, and define a function $F: \mathscr{P}(A) \to \mathbb{Z}$ as follows: For each set $X$ in $\mathscr{P}(A)$, @@ -92,12 +155,28 @@ Find the following: a. $F(\{1, 3, 4\})$ +$$ F(\{1, 3, 4\}) = 1 $$ + +because $\{1, 3, 4\}$ has an odd number of elements. + b. $F(\emptyset)$ +$$ F(\emptyset) = 0 $$ + +because $\emptyset$ has an even number of elements. + c. $F(\{2, 3\})$ +$$ F(\{2, 3\}) = 0 $$ + +because $\{2, 3\}$ has an even number of elements. + d. $F(\{2, 3, 4, 5\})$ +$$ F(\{2, 3, 4, 5\}) = 0 $$ + +because $\{2, 3, 4, 5\}$ has an even number of elements. + 8. Let $J_5 = \{0, 1, 2, 3, 4\}$, and define a function $F: J_5 \to J_5$ as follows: For each $x \in J_5$, $F(x) = (x^3 + 2x + 4) \mod 5$. @@ -105,14 +184,54 @@ Find the following: a. $F(0)$ +$$ F(0) = ((0)^3 + 2(0) + 4) \mod 5 $$ + +$$ = (0 + 0 + 4) \mod 5 $$ + +$$ = 4 \mod 5 $$ + +$$ = 4 $$ + b. $F(1)$ +$$ F(1) = ((1)^3 + 2(1) + 4) \mod 5 $$ + +$$ = (1 + 2 + 4) \mod 5 $$ + +$$ = 7 \mod 5 $$ + +$$ = 2 $$ + c. $F(2)$ +$$ F(2) = ((2)^3 + 2(2) + 4) \mod 5 $$ + +$$ = (8 + 4 + 4) \mod 5 $$ + +$$ = 16 \mod 5 $$ + +$$ = 1 $$ + d. $F(3)$ +$$ F(3) = ((3)^3 + 2(3) + 4) \mod 5 $$ + +$$ = (27 + 6 + 4) \mod 5 $$ + +$$ = 37 \mod 5 $$ + +$$ = 2 $$ + e. $F(4)$ +$$ F(4) = ((4)^3 + 2(4) + 4) \mod 5 $$ + +$$ = (64 + 8 + 4) \mod 5 $$ + +$$ = 76 \mod 5 $$ + +$$ = 1 $$ + 9. Define a function $S: \mathbb{Z}^+ \to \mathbb{Z}^+$ as follows: For each positive integer $n$, @@ -122,16 +241,28 @@ Find the following: a. $S(1)$ +$$ S(1) = 1 $$ + b. $S(15)$ +$$ S(15) = 1 + 3 + 5 + 15 = 24 $$ + c. $S(17)$ +$$ S(17) = 1 + 17 = 18 $$ + d. $S(5)$ +$$ S(5) = 1 + 5 = 6 $$ + e. $S(18)$ +$$ S(18) = 1 + 2 + 3 + 6 + 9 + 18 = 39 $$ + f. $S(21)$ +$$ S(21) = 1 + 3 + 7 + 21 = 32 $$ + 10. Let $D$ be the set of all finite subsets of positive integers. Define a function $T: \mathbb{Z}^+ \to D$ as follows: For each positive integer @@ -141,16 +272,28 @@ Find the following: a. $T(1)$ +$$ T(1) = \{1\} $$ + b. $T(15)$ +$$ T(15) = \{1, 3, 5, 15\} $$ + c. $T(17)$ +$$ T(17) = \{1, 17\} $$ + d. $T(5)$ +$$ T(5) = \{1, 5\} $$ + e. $T(18)$ +$$ T(18) = \{1, 2, 3, 6, 9, 18\} $$ + f. $T(21)$ +$$ T(21) = \{1, 3, 7, 21\} $$ + 11. Define $F: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$ as follows: For every ordered pair $(a, b)$ of integers, $F(a, b) = (2a + 1, 3b - 2)$. @@ -159,12 +302,36 @@ Find the following: a. $F(4, 4)$ +$$ F(4, 4) = (2(4) + 1, 3(4) - 2) $$ + +$$ = (8 + 1, 12 - 2) $$ + +$$ = (9, 10) $$ + b. $F(2, 1)$ +$$ F(2, 1) = (2(2) + 1, 3(1) - 2) $$ + +$$ = (4 + 1, 3 - 2) $$ + +$$ = (5, 1) $$ + c. $F(3, 2)$ +$$ F(3, 2) = (2(3) + 1, 3(2) - 2) $$ + +$$ = (6 + 1, 6 - 2) $$ + +$$ = (7, 4) $$ + d. $F(1, 5)$ +$$ F(1, 5) = (2(1) + 1, 3(5) - 2) $$ + +$$ = (2 + 1, 15 - 2) $$ + +$$ = (3, 13) $$ + 12. Let $J_5 = \{0, 1, 2, 3, 4\}$, and define $G: J_5 \times J_5 \to J_5 \times J_5$ as follows: For each $(a, b) \in J_5 \times J_5$, @@ -175,12 +342,44 @@ Find the following: a. $G(4, 4)$ +$$ G(4, 4) = ((2(4) + 1) \mod 5, (3(4) - 2) \mod 5) $$ + +$$ = ((8 + 1) \mod 5, (12 - 2) \mod 5) $$ + +$$ = (9 \mod 5, 10 \mod 5) $$ + +$$ = (4, 0) $$ + b. $G(2, 1)$ +$$ G(2, 1) = ((2(2) + 1) \mod 5, (3(1) - 2) \mod 5) $$ + +$$ = ((4 + 1) \mod 5, (3 - 2) \mod 5) $$ + +$$ = (5 \mod 5, 1 \mod 5) $$ + +$$ = (0, 1) $$ + c. $G(3, 2)$ +$$ G(3, 2) = ((2(3) + 1) \mod 5, (3(2) - 2) \mod 5) $$ + +$$ = ((6 + 1) \mod 5, (6 - 2) \mod 5) $$ + +$$ = (7 \mod 5, 4 \mod 5) $$ + +$$ = (2, 4) $$ + d. $G(1, 5)$ +$$ G(1, 5) = ((2(1) + 1) \mod 5, (3(5) - 2) \mod 5) $$ + +$$ = ((2 + 1) \mod 5, (15 - 2) \mod 5) $$ + +$$ = (3 \mod 5, 13 \mod 5) $$ + +$$ = (3, 3) $$ + 13. Let $J_5 = \{0, 1, 2, 3, 4\}$, and define functions $f: J_5 \to J_5$ and $g: J_5 \to J_5$ as follows: For each $x \in J_5$, @@ -188,6 +387,17 @@ $$ f(x) = (x + 4)^2 \mod 5 \quad \text{ and } \quad g(x) = (x^2 + 3x + 1) \mod 5 Is $f = g$? Explain. +| $x$ | $f(x)$ | $g(x)$ | +| --- | ------ | ------ | +| $0$ | $1$ | $1$ | +| $1$ | $0$ | $0$ | +| $2$ | $1$ | $1$ | +| $3$ | $4$ | $4$ | +| $4$ | $4$ | $4$ | + +The table shows that $f(x) = g(x)$ for every $x \in J_5$. Therefore $f = g$ by +definition of equality of functions. + 14. Define functions $H$ and $K$ from $\mathbb{R}$ to $\mathbb{R}$ by the following formulas: @@ -197,6 +407,10 @@ $$ H(x) = \lfloor x \rfloor + 1 \quad \text{ and } \quad K(x) = \lceil x \rceil Does $H = K$? Explain. +No. For example say $x = 0$, then $H(0) = \lfloor 0 \rfloor + 1 = 0 + 1 = 1$ and +$K(0) = \lceil 0 \rceil = 0$. Therefore it cannot be said that for every +$x \in \mathbb{R}$ that $H(x) = K(x)$, and thus $H \neq K$. + 15. Let $F$ and $G$ be functions from the set of all real numbers to itself. Define the product functions $F \cdot G: \mathbb{R} \to \mathbb{R}$ and $G \cdot F: \mathbb{R} \to \mathbb{R}$ as follows: For every @@ -208,6 +422,14 @@ $$ (G \cdot F)(x) = G(x) \cdot F(x) $$ Does $F \cdot G = G \cdot F$? Explain. +Yes, by the commutative law of multiplication of Real numbers: + +$$ (F \cdot G)(x) = F(x) \cdot G(x) = G(x) \cdot F(x) = (G \cdot F)(x) $$ + +Therefore, since $(F \cdot G)(x) = (G \cdot F)(x)$ for all $x \in \mathbb{R}$, +it can be concluded that $F \cdot G = G \cdot F$ by the definition of equality +of functions. + 16. Let $F$ and $G$ be function sfrom the set of all real numbers to itself. Define new functions $F - G: \mathbb{R} \to \mathbb{R}$ and $G - F: \mathbb{R} \to \mathbb{R}$ as follows: For every $x \in \mathbb{R}$, @@ -218,41 +440,94 @@ $$ (G - F)(x) = G(x) - F(x) $$ Does $F - G = G - F$? Explain. +No. Consider the definition of the difference of sets: + +$$ (F - G)(x) = F(x) - G(x) = F(x) $$ + +and: + +$$ (G - F)(x) = G(x) - F(x) = G(x) $$ + +Since $F(x) \neq G(x)$ for all $x \in \mathbb{R}$, it can be concluded that +$F - G \neq G - F$ by the definition of the equality of functions. + 17. Use the definition of logarithm to fill in the blanks below. a. $\log_28 = 3$ because _____. +$$ 2^3 = 8 $$ + b. $\log_5\left(\dfrac{1}{25}\right) = -2$ because _____. +$$ 5^{-2} = \frac{1}{5^2} = \frac{1}{25} $$ + c. $\log_44 = 1$ because _____. +$$ 4^1 = 4 $$ + d. $\log_3(3^n) = n$ because _____. +$$ 3^n = 3^n $$ + e. $\log_41 = 0$ because _____. +$$ 4^0 = 1 $$ + 18. Find exact values for each of the following quantities without using a calculator. a. $\log_{3}81$ +$$ 3^{\text{?}} = 81 $$ + +$$ \log_{3}81 = 4 $$ + b. $\log_{2}1024$ +$$ 2^{\text{?}} = 1024 $$ + +$$ \log_{2}1024 = 10 $$ + c. $\log_{3}\left(\dfrac{1}{27}\right)$ +$$ \log_{3}\left(\frac{1}{27}\right) = -3 $$ + d. $\log_{2}1$ +$$ \log_{2}1 = 0 $$ + e. $\log_{10}\left(\dfrac{1}{10}\right)$ +$$ \log_{10}\left(\dfrac{1}{10}\right) = -1 $$ + f. $\log_{3}3$ +$$ \log_{3}3 = 1 $$ + g. $\log_{2}(2^k)$ +$$\log_{2}(2^k) = k $$ + 19. Use the definition of logarithm to prove that for any positive real number $b$ with $b \neq 1$, $\log_{b}b = 1$. +**Proof:** + +Let $b$ be any positive real number with $b \neq 1$. Since $b^1 = b$, then +$\log_{b}b = 1$ by definition of logarithm. + +Q.E.D. + 20. Use the definition of logarithm to prove that for any positive real number $b$ with $b \neq 1$, $\log_{b}1 = 0$. +**Proof:** + +Let $b$ be any positive real number with $b \neq 1$. Since $b^0 = 1$, then +$\log_{b}1 = 0$ by definition of logarithm. + +Q.E.D. + 21. If $b$ is any positive real number with $b \neq 1$ and $x$ is any real number, $b^{-x}$ is defined as follows: @@ -260,15 +535,144 @@ $b^{-x} = \dfrac{1}{b^x}$. 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 @@ -276,8 +680,24 @@ real numbers $u$ and $b$, with $b \neq 1$. 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 @@ -293,13 +713,25 @@ 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 $f: S \to Z$ as follows: For each string $s$ in $S$ +a. Define $f: S \to \mathbb{Z}$ as follows: For each string $s$ in $S$ $$ f(s) = @@ -311,25 +743,51 @@ $$ 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 $H(10101, 00011$. +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. @@ -344,6 +802,8 @@ a. | 0 | 1 | 0 | | 0 | 0 | 1 | +Omitted. + b. | Input | Intput | Input | Output | @@ -359,15 +819,17 @@ b. | 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 | | | | | | | | | | | | | | | | | -| 1 | 0 | | | | | | | | | | | | | | | | | -| 0 | 1 | | | | | | | | | | | | | | | | | -| 0 | 0 | | | | | | | | | | | | | | | | | +| 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, @@ -376,8 +838,35 @@ $$ f(x_1, x_2, x_3) = (4x_1 + 3x_2 + 2x_3) \mod 2 $$ 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 @@ -386,6 +875,21 @@ $n \neq 0$. 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 @@ -394,14 +898,33 @@ $n \neq 0$. Student D claims that $h$ is not well defined. Justify student D's claim. -35. Let $U = \{1, 2, 3, 4\}$. Student A tries to define a function $R: U \to Z$ - as follows: For each $x \in U$, +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 $x \in V$, @@ -410,28 +933,63 @@ $S(x)$ 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 $F(A) \cup F(B) \subseteq F(A \cup B)$. @@ -457,6 +1015,18 @@ 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 @@ -464,29 +1034,236 @@ 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 $u \in S$, @@ -504,9 +1281,13 @@ 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 $n$ @@ -518,18 +1299,34 @@ $\pm 1$ - namely, $1$, $3$, $7$, and $9$. 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. diff --git a/chapter_7/test_yourself.md b/chapter_7/test_yourself.md index 681359d..640a492 100644 --- a/chapter_7/test_yourself.md +++ b/chapter_7/test_yourself.md @@ -4,25 +4,44 @@ Page 458 1. Given a function $f$ from a set $X$ to a set $Y$, $f(x)$ is _____. +the unique output element in $Y$ that is related to $x$ by $f$. + 2. Given a function $f$ from a set $X$ to a set $Y$, if $f(x) = y$ then $y$ is called _____ or _____ or _____. +the value of $f$ at $x$; the image of $x$ under $f$; the output of $f$ for the +input $x$ + 3. Given a function $f$ from a set $X$ to a set $Y$, the range of $f$ (or the image of $X$ under $f$) is _____. +the set of all $y$ in $Y$ such that $f(x) = y$ + 4. Given a function $f$ from a set $X$ to $Y$, if $f(x) = y$ then $x$ is called _____ or _____. +an inverse image of $y$ under $f$; a preimage of $y$ + 5. Given a function $f$ from a set $X$ to a set $Y$, if $y \in Y$ then $f^{-1}(y) =$ _____ and is called _____. +$\{x \in X | f(x) = y\}$; the inverse image of $y$ + 6. Given functions $f$ and $g$ from a set $X$ to a set $Y$, $f = g$ if, and only if, _____. +$f(x) = g(x)$ for every $x \in X$ + 7. Given positive real numbers $x$ and $b$ with $b \neq 1$, $\log_b(x) =$ _____. +the exponent to which $b$ must be raised to obtain $x$. + 8. Given a function $f$ from a set $X$ to a set $Y$ and a subset $A$ of $X$, $f(A) =$ _____. +$\{y \in Y | y = f(x) \text{ for some } x \in A\}$ + 9. Given a function $f$ from a set $X$ to a set $Y$ and a subset $C$ of $Y$, $f^{-1}(C) =$ _____. + +$\{x \in X | f(x) \in C\}$