diff --git a/chapter_7/exercises.md b/chapter_7/exercises.md index 7a3dbfb..c01cef8 100644 --- a/chapter_7/exercises.md +++ b/chapter_7/exercises.md @@ -2389,7 +2389,7 @@ $$ -x_1 = -x_2 $$ $$ x_1 = x_2 $$ -Thus it has been shown that $(x_1, y_1) = (x_2) = y_2$. +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. @@ -2425,7 +2425,7 @@ $$ x = -w $$ 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 $G\left(-w, \dfrac{t}{2}\right)$: +Now, evaluating for $G(x, y)$, which is $G\left(-w, \dfrac{t}{2}\right)$: $$ G\left(-w, \frac{t}{2}\right) = \left(2\left(\frac{t}{2}, -(-w)\right)\right) $$ @@ -2447,7 +2447,7 @@ a. Is $H$ one-to-one? Prove or give a counterexample. $H$ is one-to-one. -**Proof: +**Proof:** Suppose $(x_1, y_1), (x_2, y_2) \in \mathbb{R} \times \mathbb{R}$ such that $H(x_1, y_1) = H(x_2, y_2)$. @@ -2558,7 +2558,73 @@ Omitted. a. Is $\log_{8}27 = \log_{2}3$? Why or why not? -a. Is $\log_{16}9 = \log_{4}3$? 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. @@ -2567,15 +2633,112 @@ The properties of logarithm established in 33-35 are used in Sections 11.4 and $$ \log_{b}\left(\frac{x}{y}\right) = \log_{b}x - \log_{b}y $$ +**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 @@ -2584,9 +2747,61 @@ $(f + g)(x) = f(x) + g(x)$ 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 @@ -2596,16 +2811,284 @@ $(c \cdot f)(x) = c \cdot (f(x))$ for every real number $x$. 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 $F^{-1}$.