diff --git a/chapter_7/exercises.md b/chapter_7/exercises.md index bc48247..5f317d1 100644 --- a/chapter_7/exercises.md +++ b/chapter_7/exercises.md @@ -3330,3 +3330,182 @@ Omitted. 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) + +2. (See page 494 for image) + +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 $x$. + +4. $F(x) = x^5$ and $G(x) = x^{\frac{1}{5}}$ for each real number $x$. + +5. Define $f: \mathbb{R} \to \mathbb{R}$ by the rule $f(x) = -x$ for every real + number $x$. Find $(f \circ f)(x)$. + +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 + $(G \circ F)(4)$. + +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 +$(M \circ L)(9)$. + +b. Is $L \circ M = M \circ L$? + +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. $(T \circ L)(abaa) = \text{ ?}$ + +b. $(T \circ L)(baaab) = \text{ ?}$ + +c. $(T \circ L)(aaa) = \text{ ?}$ + +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. $(G \circ F)(2) = \text{ ?}$ + +b. $(G \circ F)(-3) = \text{ ?}$ + +c. $(G \circ F)(5) = \text{ ?}$ + +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 +$(F \circ G)(3)$. + +b. Is $G \circ F = F \circ G$? Explain. + +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 +$(F \circ G)(1)$. + +b. Is $G \circ F = F \circ G$? Explain. + +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 $y \in \mathbb{R}$. + +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 $x \in \mathbb{R}^+$. + +14. $H$ and $H^{-1}$ are both defined from $\mathbb{R} - \{1\}$ to + $\mathbb{R} - \{1\}$ by the formula + +$$ H(x) = H^{-1}(x) = \frac{x + 1}{x - 1}, \quad \text{ for each } x \in \mathbb{R} - \{1\} $$ + +15. Explain how it follows from the definition of logarithm that + +a. $\log_{b}(b^x) = x$, for every real number $x$. + +b. $b^{\log_{b}x} = x$, for every positive real number $x$. + +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 $Y$. + +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 $Y$. + +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 $g(s_k) = g(s_m)$. + +b. $\dfrac{z}{2}$ and $\dfrac{t}{2}$ are elements of $Y$ and +$g\left(\dfrac{z}{2}\right) = g\left(\dfrac{t}{2}\right)$. + +c. $f(x_1)$ and $f(x_2)$ are elements of $Y$ and $g(f(x_1)) = g(f(x_2))$. + +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. + +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. + +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. + +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. + +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. + +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. + +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. + +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.) + +27. Define $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ by + the formulas + +$$ f(x) = x + 3 \quad \text{ and } \quad g(x) = -x \quad \text{ for each } x \in \mathbb{R} $$ + +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 $g = f^{-1}$. + +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 + $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$. + +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. diff --git a/chapter_7/notes.md b/chapter_7/notes.md index f6b2ebf..912f301 100644 --- a/chapter_7/notes.md +++ b/chapter_7/notes.md @@ -258,3 +258,158 @@ be shown.]_ Suppose $x \in X$. _[We must show that there exists an element $y$ in $Y$ such that $F^{-1}(y) = x$.]_ Let $y = F(x)$. Then $y \in Y$, and by definition of $F^{-1}$, $F^{-1}(y) = x$ _[as was to be shown.]_ + +--- + +Page 485 + +**Definition** + +Let $f: X \to Y$ and $g: Y' \to Z$ be functions with the property that the range +of $f$ is a subset of the domain of $g$. Define a new function +$g \circ f: X \to Z$ as follows: + +$$ (g \circ f)(x) = g(f(x)) \quad \text{ for each } x \in X $$ + +where $g \circ f$ is read "$g$ circle $f$" and $g(f(x))$ is read "$g$ of $f$ of +$x$." The function $g \circ f$ is called the **composition of $f$ and $g$**. + +--- + +Page 487 + +**Theorem 7.3.1 Composition with an Identity Function** + +If $f$ is a function from a set $X$ to a set $Y$, and $I_x$ is the identity +function on $X$, and $I_y$ is the identity function on $Y$, then + +$$ \text{(a) } f \circ I_x = f \quad \text{ and } \quad \text{(b) } I_y \circ f = f $$ + +**Proof:** + +_Part (a):_ + +Suppose $f$ is a function from a set $X$ to a set $Y$ and $I_x$ is the identity +function on $X$. Then, for each $x$ in $X$, + +$$ (f \circ I_x)(x) = f(I_x(x)) = f(x) $$ + +Hence, by the definition of equality of functions, $f \circ I_x = f$, as was to +be shown. + +_Part (b):_ + +This is exercise 16 at the end of this section. + +--- + +Page 488 + +**Theorem 7.3.2 Composition of a Function with Its Inverse** + +If $f: X \to Y$ is a one-to-one and onto function with inverse function +$f^{-1}: Y \to X$, then + +$$ \text{(a) } f^{-1} \circ f = I_x \quad \text{ and } \quad \text{(b) } f \circ f^{-1} = I_y $$ + +**Proof:** + +_Part (a):_ + +Suppose $f: X \to Y$ is a one-to-one and onto function with inverse function +$f^{-1}: Y \to X$. _[To show that $f^{-1} \circ f = I_x$, we must show that for +each $x \in X$, $(f^{-1} \circ f)(x) = x$.]_ Let $x$ be any element in $X$. +Then, by definition of composition of functions, + +$$ (f^{-1} \circ f)(x) = f^{-1}(f(x)) $$ + +Let + +$$ z = f^{-1}(f(x)) $$ + +By the definition of inverse function, + +$$ f(z) = f(x) $$ + +and, because $f$ is one-to-one, this implies that + +$$ z = x $$ + +Now $z = f^{-1}(f(x))$ also, and so, by substitution, + +$$ f^{-1}(f(x)) = x $$ + +Or, equivalently, + +$$ (f^{-1} \circ f)(x) = x $$ + +_[as was to be shown]._ + +Since $x$ is any element of $X$ and since $I_x(x) = x$, this proves that +$f^{-1} \circ f = I_x$. + +_Part (b):_ + +This is exercise 17 at the end of this section. + +--- + +Page 490 + +**Theorem 7.3.3** + +If $f: X \to Y$ and $g: Y \to Z$ are both one-to-one functions, then $g \circ f$ +is one-to-one. + +--- + +Page 491 + +**Proof of Theorem 7.3.3:** + +Suppose $f: X \to Y$ and $g: Y \to Z$ are both one-to-one functions. _[We must +show that $g \circ f$ is one-to-one.]_ Suppose $x_1$ and $x_2$ are elements of +$X$ such that + +$$ (g \circ f)(x_1) = (g \circ f)(x_2) $$ + +_[We must show that $x_1 = x_2$.]_ By definition of composition of functions, + +$$ g(f(x_1)) = g(f(x_2)) $$ + +Since $g$ is one-to-one, + +$$ f(x_1) = f(x_2) $$ + +And since $f$ is one-to-one, + +$$ x_1 = x_2 $$ + +_[as was to be shown]._ Hence $g \circ f$ is one-to-one. + +--- + +Page 491 + +**Theorem 7.3.4** + +If $f: X \to Y$ and $g: Y \to Z$ are both onto functions, then $g \circ f$ is +onto. + +--- + +Page 493 + +**Proof of Theorem 7.3.4** + +Suppose $f: X \to Y$ and $g: Y \to Z$ are both onto functions. _[We must show +that $g \circ f$ is onto.]_ Let $z$ be any _[particular but arbitrarily chosen]_ +element of $Z$. _[We must show the existence of an element in $X$ such that +$g \circ f$ of that element equals $z$.]_ Since $g$ is onto, there is an +element, say $y$, in $Y$ such that $g(y) = z$. And since $f$ is onto, there is +an element, say $x$, in $X$ such that $f(x) = y$. Hence there is an element $x$ +in $X$ such that + +$$ (g \circ f)(x) = g(f(x)) = g(y) = z $$ + +_[as was to be shown]._ It follows that $g \circ f$ is onto. diff --git a/chapter_7/test_yourself.md b/chapter_7/test_yourself.md index 8d6d1aa..c044625 100644 --- a/chapter_7/test_yourself.md +++ b/chapter_7/test_yourself.md @@ -116,3 +116,28 @@ function from $X$ to $Y$; both one-to-one and onto the unique element $x$ in $X$ such that $F(x) = y$ (in other words, $F^{-1}(y)$ is the unique preimage of $y$ in $X$) + +--- + +Page 494 + +**Test Yourself** + +1. If $f$ is a function from $X$ to $Y'$, $g$ is a function from $Y \to Z$, and + $Y' \subseteq Y$, then $g \circ f$ is a function from _____ to _____, and + $(g \circ f)(x) =$ _____ for every $x$ in $X$. + +2. If $f$ is a function from $X$ to $Y$ and $I_x$ and $I_y$ are the identity + functions from $X$ to $X$ and $Y$ to $Y$, respectively, then $f \circ I_x =$ + _____ and $I_y \circ f =$ _____. + +3. If $f$ is a one-to-one correspondence from $X$ to $Y$, then + $f^{-1} \circ f =$ _____ and $f \circ f^{-1} =$ _____. + +4. If $f$ is a one-to-one function from $X$ to $Y$ and $g$ is a one-to-one + function from $Y$ to $Z$, you prove that $g \circ f is one-to-one by + supposing that _____ and then showing that _____. + +5. If $f$ is an onto function from $X$ to $Y$ and $g$ is an onto function from + $Y$ to $Z$, you prove that $g \circ f$ is onto by supposing that _____ and + then showing that _____.