From 976b8b9af62f1bf665135c484a87b4b359685da6 Mon Sep 17 00:00:00 2001 From: tomit4 Date: Sat, 8 Aug 2026 18:41:16 -0700 Subject: [PATCH] :construction: Fin 7.2 --- appendix_b.txt | 2 +- chapter_7/exercises.md | 176 +++++++++++++++++++++++++++++++++++++++++ leftoff.txt | 2 +- 3 files changed, 178 insertions(+), 2 deletions(-) diff --git a/appendix_b.txt b/appendix_b.txt index 63e305a..3d78d3c 100644 --- a/appendix_b.txt +++ b/appendix_b.txt @@ -1 +1 @@ -968 +971 diff --git a/chapter_7/exercises.md b/chapter_7/exercises.md index c01cef8..bc48247 100644 --- a/chapter_7/exercises.md +++ b/chapter_7/exercises.md @@ -1695,6 +1695,32 @@ 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. @@ -3097,41 +3123,183 @@ case draw an arrow diagram for $F^{-1}$. (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 @@ -3142,15 +3310,23 @@ a. Let $X = \{x_1, x_2, \dots, x_n\}$ be a set with $n$ elements. Use Example 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. diff --git a/leftoff.txt b/leftoff.txt index bf7aeeb..f52aaac 100644 --- a/leftoff.txt +++ b/leftoff.txt @@ -1 +1 @@ -462 +484