From e7e657950b9d69c2da3d1647507a446a0ee33f4f Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Sat, 13 May 2023 06:38:55 -0600 Subject: [PATCH] Aggregate Apostol LaTeX into single file. --- Bookshelf/Apostol.tex | 1595 ++++++++++++++++++++++++++++ Bookshelf/Apostol/Chapter_1_07.tex | 589 ---------- Bookshelf/Apostol/Chapter_1_11.tex | 471 -------- Bookshelf/Apostol/Chapter_I_03.tex | 405 ------- Bookshelf/Apostol/Glossary.tex | 57 - Common/Real/Geometry/Area.tex | 93 -- DocGen4/Output/Index.lean | 6 +- preamble.tex | 8 +- 8 files changed, 1603 insertions(+), 1621 deletions(-) create mode 100644 Bookshelf/Apostol.tex delete mode 100644 Bookshelf/Apostol/Chapter_1_07.tex delete mode 100644 Bookshelf/Apostol/Chapter_1_11.tex delete mode 100644 Bookshelf/Apostol/Chapter_I_03.tex delete mode 100644 Bookshelf/Apostol/Glossary.tex delete mode 100644 Common/Real/Geometry/Area.tex diff --git a/Bookshelf/Apostol.tex b/Bookshelf/Apostol.tex new file mode 100644 index 0000000..857ddc0 --- /dev/null +++ b/Bookshelf/Apostol.tex @@ -0,0 +1,1595 @@ +\documentclass{report} + +\usepackage{graphicx} + +\input{../preamble} + +\graphicspath{{./Apostol/images/}} + +\newcommand{\lean}[2]{\leanref{../#1.html\##2}{#2}} + +\begin{document} + +\header + {One-Variable Calculus, with an Introduction to Linear Algebra} + {Tom M. Apostol} + +\tableofcontents + +\chapter{A Set of Axioms for the Real-Number System}% +\label{chap:set-axioms-real-number-system} + +\section{\verified{Lemma 1}}% +\label{sec:lemma-1} + +Nonempty set $S$ has supremum $L$ if and only if set $-S$ has infimum $-L$. + +\begin{proof} + + \lean{Chapter\_I\_03} + {Apostol.Chapter\_I\_03.is\_lub\_neg\_set\_iff\_is\_glb\_set\_neg} + + \divider + + Suppose $L = \sup{S}$ and fix $x \in S$. + By definition of the supremum, $x \leq L$ and $L$ is the smallest value + satisfying this inequality. + Negating both sides of the inequality yields $-x \geq -L$. + Furthermore, $-L$ must be the largest value satisfying this inequality. + Therefore $-L = \inf{-S}$. + +\end{proof} + +\section{\verified{Theorem I.27}}% +\label{sec:theorem-i.27} + +Every nonempty set $S$ that is bounded below has a greatest lower bound; that + is, there is a real number $L$ such that $L = \inf{S}$. + +\begin{proof} + + \lean{Chapter\_I\_03} + {Apostol.Chapter\_I\_03.exists\_isGLB} + + \divider + + Let $S$ be a nonempty set bounded below by $x$. + Then $-S$ is nonempty and bounded above by $x$. + By the completeness axiom, there exists a supremum $L$ of $-S$. + By \nameref{sec:lemma-1}, $L$ is a supremum of $-S$ if and only if $-L$ is an + infimum of $S$. + +\end{proof} + +\section{\verified{Theorem I.29}}% +\label{sec:theorem-i.29} + +For every real $x$ there exists a positive integer $n$ such that $n > x$. + +\begin{proof} + + \lean{Chapter\_I\_03} + {Apostol.Chapter\_I\_03.exists\_pnat\_geq\_self} + + \divider + + Let $n = \abs{\ceil{x}} + 1$. + It is trivial to see $n$ is a positive integer satisfying $n \geq 1$. + Thus all that remains to be shown is that $n > x$. + If $x$ is nonpositive, $n > x$ immediately follows from $n \geq 1$. + If $x$ is positive, + $$x = \abs{x} \leq \abs{\ceil{x}} < \abs{\ceil{x}} + 1 = n.$$ + +\end{proof} + +\section{\verified{Theorem I.30}}% +\label{sec:theorem-i.30} + +If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive + integer $n$ such that $nx > y$. + +\note{This is known as the "Archimedean Property of the Reals."} + +\begin{proof} + + \lean{Chapter\_I\_03} + {Apostol.Chapter\_I\_03.exists\_pnat\_mul\_self\_geq\_of\_pos} + + \divider + + Let $x > 0$ and $y$ be an arbitrary real number. + By \nameref{sec:theorem-i.29}, there exists a positive integer $n$ such that + $n > y / x$. + Multiplying both sides of the inequality yields $nx > y$ as expected. + +\end{proof} + +\section{\verified{Theorem I.31}}% +\label{sec:theorem-i.31} + +If three real numbers $a$, $x$, and $y$ satisfy the inequalities + $$a \leq x \leq a + \frac{y}{n}$$ for every integer $n \geq 1$, then $x = a$. + +\begin{proof} + + \lean{Chapter\_I\_03} + {Apostol.Chapter\_I\_03.forall\_pnat\_leq\_self\_leq\_frac\_imp\_eq} + + \divider + + By the trichotomy of the reals, there are three cases to consider: + + \paragraph{Case 1}% + + Suppose $x = a$. + Then we are immediately finished. + + \paragraph{Case 2}% + + Suppose $x < a$. + But by hypothesis, $a \leq x$. + Thus $a < a$, a contradiction. + + \paragraph{Case 3}% + + Suppose $x > a$. + Then there exists some $c > 0$ such that $a + c = x$. + By \nameref{sec:theorem-i.30}, there exists an integer $n > 0$ such that + $nc > y$. + Rearranging terms, we see $y / n < c$. + Therefore $a + y / n < a + c = x$. + But by hypothesis, $x \leq a + y / n$. + Thus $a + y / n < a + y / n$, a contradiction. + + \paragraph{Conclusion}% + + Since these cases are exhaustive and both case 2 and 3 lead to + contradictions, $x = a$ is the only possibility. + +\end{proof} + +\section{\verified{Lemma 2}}% +\label{sec:lemma-2} + +If three real numbers $a$, $x$, and $y$ satisfy the inequalities + $$a - y / n \leq x \leq a$$ for every integer $n \geq 1$, then $x = a$. + +\begin{proof} + + \lean{Chapter\_I\_03} + {Apostol.Chapter\_I\_03.forall\_pnat\_frac\_leq\_self\_leq\_imp\_eq} + + \divider + + By the trichotomy of the reals, there are three cases to consider: + + \paragraph{Case 1}% + + Suppose $x = a$. + Then we are immediately finished. + + \paragraph{Case 2}% + + Suppose $x < a$. + Then there exists some $c > 0$ such that $x = a - c$. + By \nameref{sec:theorem-i.30}, there exists an integer $n > 0$ such that + $nc > y$. + Rearranging terms, we see that $y / n < c$. + Therefore $a - y / n > a - c = x$. + But by hypothesis, $x \geq a - y / n$. + Thus $a - y / n < a - y / n$, a contradiction. + + \paragraph{Case 3}% + + Suppose $x > a$. + But by hypothesis $x \leq a$. + Thus $a < a$, a contradiction. + + \paragraph{Conclusion}% + + Since these cases are exhaustive and both case 2 and 3 lead to + contradictions, $x = a$ is the only possibility. + +\end{proof} + +\section{Theorem I.32}% +\label{sec:theorem-i.32} + +Let $h$ be a given positive number and let $S$ be a set of real numbers. + +\subsection{\verified{Theorem I.32a}}% +\label{sub:theorem-i.32a} + +If $S$ has a supremum, then for some $x$ in $S$ we have $x > \sup{S} - h$. + +\begin{proof} + + \lean{Chapter\_I\_03} + {Apostol.Chapter\_I\_03.sup\_imp\_exists\_gt\_sup\_sub\_delta} + + \divider + + By definition of a supremum, $\sup{S}$ is the least upper bound of $S$. + For the sake of contradiction, suppose for all $x \in S$, + $x \leq \sup{S} - h$. + This immediately implies $\sup{S} - h$ is an upper bound of $S$. + But $\sup{S} - h < \sup{S}$, contradicting $\sup{S}$ being the \textit{least} + upper bound. + Therefore our original hypothesis was wrong. + That is, there exists some $x \in S$ such that $x > \sup{S} - h$. + +\end{proof} + +\subsection{\verified{Theorem I.32b}}% +\label{sub:theorem-i.32b} + +If $S$ has an infimum, then for some $x$ in $S$ we have $x < \inf{S} + h$. + +\begin{proof} + + \lean{Chapter\_I\_03} + {Apostol.Chapter\_I\_03.inf\_imp\_exists\_lt\_inf\_add\_delta} + + \divider + + By definition of an infimum, $\inf{S}$ is the greatest lower bound of $S$. + For the sake of contradiction, suppose for all $x \in S$, + $x \geq \inf{S} + h$. + This immediately implies $\inf{S} + h$ is a lower bound of $S$. + But $\inf{S} + h > \inf{S}$, contradicting $\inf{S}$ being the + \textit{greatest} lower bound. + Therefore our original hypothesis was wrong. + That is, there exists some $x \in S$ such that $x < \inf{S} + h$. + +\end{proof} + +\section{Theorem I.33}% +\label{sec:theorem-i.33} + +Given nonempty subsets $A$ and $B$ of $\mathbb{R}$, let $C$ denote the set + $$C = \{a + b : a \in A, b \in B\}.$$ + +\note{This is known as the "Additive Property."} + +\subsection{\verified{Theorem I.33a}}% +\label{sub:theorem-i.33a} + +If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and + $$\sup{C} = \sup{A} + \sup{B}.$$ + +\begin{proof} + + \lean{Chapter\_I\_03} + {Apostol.Chapter\_I\_03.sup\_minkowski\_sum\_eq\_sup\_add\_sup} + + \divider + + We prove (i) $\sup{A} + \sup{B}$ is an upper bound of $C$ and (ii) + $\sup{A} + \sup{B}$ is the \textit{least} upper bound of $C$. + + \paragraph{(i)}% + \label{par:theorem-i.33a-i} + + Let $x \in C$. + By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such + that $x = a' + b'$. + By definition of a supremum, $a' \leq \sup{A}$. + Likewise, $b' \leq \sup{B}$. + Therefore $a' + b' \leq \sup{A} + \sup{B}$. + Since $x = a' + b'$ was arbitrarily chosen, it follows $\sup{A} + \sup{B}$ + is an upper bound of $C$. + + \paragraph{(ii)}% + + Since $A$ and $B$ have supremums, $C$ is nonempty. + By \nameref{par:theorem-i.33a-i}, $C$ is bounded above. + Therefore the completeness axiom tells us $C$ has a supremum. + Let $n > 0$ be an integer. + We now prove that + \begin{equation} + \label{par:theorem-i.33a-ii-eq1} + \sup{C} \leq \sup{A} + \sup{B} \leq \sup{C} + 1 / n. + \end{equation} + + \subparagraph{Left-Hand Side}% + + First consider the left-hand side of \eqref{par:theorem-i.33a-ii-eq1}. + By \nameref{par:theorem-i.33a-i}, $\sup{A} + \sup{B}$ is an upper bound of + $C$. + Since $\sup{C}$ is the \textit{least} upper bound of $C$, it follows + $\sup{C} \leq \sup{A} + \sup{B}$. + + \subparagraph{Right-Hand Side}% + + Next consider the right-hand side of \eqref{par:theorem-i.33a-ii-eq1}. + By \nameref{sub:theorem-i.32a}, there exists some $a' \in A$ such that + $\sup{A} < a' + 1 / (2n)$. + Likewise, there exists some $b' \in B$ such that + $\sup{B} < b' + 1 / (2n)$. + Adding these two inequalities together shows + \begin{align*} + \sup{A} + \sup{B} + & < a' + b' + 1 / n \\ + & \leq \sup{C} + 1 / n. + \end{align*} + + \subparagraph{Conclusion}% + + Applying \nameref{sec:theorem-i.31} to \eqref{par:theorem-i.33a-ii-eq1} + proves $\sup{C} = \sup{A} + \sup{B}$ as expected. + +\end{proof} + +\subsection{\verified{Theorem I.33b}}% +\label{sub:theorem-i.33b} + +If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and + $$\inf{C} = \inf{A} + \inf{B}.$$ + +\begin{proof} + + \lean{Chapter\_I\_03} + {Apostol.Chapter\_I\_03.inf\_minkowski\_sum\_eq\_inf\_add\_inf} + + \divider + + We prove (i) $\inf{A} + \inf{B}$ is a lower bound of $C$ and (ii) + $\inf{A} + \inf{B}$ is the \textit{greatest} lower bound of $C$. + + \paragraph{(i)}% + \label{par:theorem-i.33b-i} + + Let $x \in C$. + By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such + that $x = a' + b'$. + By definition of an infimum, $a' \geq \inf{A}$. + Likewise, $b' \geq \inf{B}$. + Therefore $a' + b' \geq \inf{A} + \inf{B}$. + Since $x = a' + b'$ was arbitrarily chosen, it follows $\inf{A} + \inf{B}$ + is a lower bound of $C$. + + \paragraph{(ii)}% + + Since $A$ and $B$ have infimums, $C$ is nonempty. + By \nameref{par:theorem-i.33b-i}, $C$ is bounded below. + Therefore \nameref{sec:theorem-i.27} tells us $C$ has an infimum. + Let $n > 0$ be an integer. + We now prove that + \begin{equation} + \label{par:theorem-i.33b-ii-eq1} + \inf{C} - 1 / n \leq \inf{A} + \inf{B} \leq \inf{C}. + \end{equation} + + \subparagraph{Right-Hand Side}% + + First consider the right-hand side of \eqref{par:theorem-i.33b-ii-eq1}. + By \nameref{par:theorem-i.33b-i}, $\inf{A} + \inf{B}$ is a lower bound of + $C$. + Since $\inf{C}$ is the \textit{greatest} upper bound of $C$, it follows + $\inf{C} \geq \inf{A} + \inf{B}$. + + \subparagraph{Left-Hand Side}% + + Next consider the left-hand side of \eqref{par:theorem-i.33b-ii-eq1}. + By \nameref{sub:theorem-i.32b}, there exists some $a' \in A$ such that + $\inf{A} > a' - 1 / (2n)$. + Likewise, there exists some $b' \in B$ such that + $\inf{B} > b' - 1 / (2n)$. + Adding these two inequalities together shows + \begin{align*} + \inf{A} + \inf{B} + & > a' + b' - 1 / n \\ + & \geq \inf{C} - 1 / n. + \end{align*} + + \subparagraph{Conclusion}% + + Applying \nameref{sec:lemma-2} to \eqref{par:theorem-i.33b-ii-eq1} + proves $\inf{C} = \inf{A} + \inf{B}$ as expected. + +\end{proof} + +\section{\verified{Theorem I.34}}% +\label{sec:theorem-i.34} + +Given two nonempty subsets $S$ and $T$ of $\mathbb{R}$ such that $$s \leq t$$ + for every $s$ in $S$ and every $t$ in $T$. Then $S$ has a supremum, and $T$ + has an infimum, and they satisfy the inequality $$\sup{S} \leq \inf{T}.$$ + +\begin{proof} + + \lean{Chapter\_I\_03} + {Apostol.Chapter\_I\_03.forall\_mem\_le\_forall\_mem\_imp\_sup\_le\_inf} + + \divider + + By hypothesis, $S$ and $T$ are nonempty sets. + Let $s \in S$ and $t \in T$. + Then $t$ is an upper bound of $S$ and $s$ is a lower bound of $T$. + By the completeness axiom, $S$ has a supremum. + By \nameref{sec:theorem-i.27}, $T$ has an infimum. + All that remains is showing $\sup{S} \leq \inf{T}$. + + For the sake of contradiction, suppose $\sup{S} > \inf{T}$. + Then there exists some $c > 0$ such that $\sup{S} = \inf{T} + c$. + Therefore $\inf{T} < \sup{S} - c / 2$. + By \nameref{sub:theorem-i.32a}, there exists some $x \in S$ such that + $\sup{S} - c / 2 < x$. + Thus $$\inf{T} < \sup{S} - c / 2 < x.$$ + But by hypothesis, $x \in S$ is a lower bound of $T$ meaning $x \leq \inf{T}$. + Therefore $x < x$, a contradiction. + Out original assumption is incorrect; that is, $\sup{S} \leq \inf{T}$. + +\end{proof} + +\chapter{The Concept of Area as a Set Function}% +\label{chap:concept-area-set-function} + +We assume there exists a class $\mathscr{M}$ of measurable sets in the plane and + a set function $a$, whose domain is $\mathscr{M}$, with the following + properties: + +\section{\defined{Nonnegative Property}}% +\label{sec:nonnegative-property} + +For each set $S$ in $\mathscr{M}$, we have $a(S) \geq 0$. + +\begin{axiom} + + \lean{Common/Real/Geometry/Area}{Nonnegative-Property} + +\end{axiom} + +\section{\defined{Additive Property}}% +\label{sec:additive-property} + +If $S$ and $T$ are in $\mathscr{M}$, then $S \cup T$ and $S \cap T$ are in + $\mathscr{M}$, and we have $a(S \cup T) = a(S) + a(T) - a(S \cap T)$. + +\begin{axiom} + + \lean{Common/Real/Geometry/Area}{Additive-Property} + +\end{axiom} + +\section{\defined{Difference Property}}% +\label{sec:difference-property} + +If $S$ and $T$ are in $\mathscr{M}$ with $S \subseteq T$, then $T - S$ is in + $\mathscr{M}$, and we have $a(T - S) = a(T) - a(S)$. + +\begin{axiom} + + \lean{Common/Real/Geometry/Area}{Difference-Property} + +\end{axiom} + +\section{\defined{Invariance Under Congruence}}% +\label{sec:invariance-under-congruence} + +If a set $S$ is in $\mathscr{M}$ and if $T$ is congruent to $S$, then $T$ is + also in $\mathscr{M}$ and we have $a(S) = a(T)$. + +\begin{axiom} + + \lean{Common/Real/Geometry/Area}{Invariance-Under-Congruence} + +\end{axiom} + +\section{\defined{Choice of Scale}}% +\label{sec:choice-scale} + +Every rectangle $R$ is in $\mathscr{M}$. +If the edges of $R$ have lengths $h$ and $k$, then $a(R) = hk$. + +\begin{axiom} + + \lean{Common/Real/Geometry/Area}{Choice-of-Scale} + +\end{axiom} + +\section{\partial{Exhaustion Property}}% +\label{sec:exhaustion-property} + +Let $Q$ be a set that can be enclosed between two step regions $S$ and $T$, so + that + \begin{equation} + \label{sec:exhaustion-property-eq1} + S \subseteq Q \subseteq T. + \end{equation} +If there is one and only one number $c$ which satisfies the inequalities + $$a(S) \leq c \leq a(T)$$ for all step regions $S$ and $T$ satisfying (1.1), + then $Q$ is measurable and $a(Q) = c$. + +\begin{axiom} + + \lean{Common/Real/Geometry/Area}{Exhaustion-Property} + +\end{axiom} + +\chapter{Exercises 1.7}% +\label{chap:exercises-1.7} + +\section{Exercise 1.7.1}% +\label{sec:exercise-1.7.1} + +Prove that each of the following sets is measurable and has zero area: + +\subsection{\unverified{Exercise 1.7.1a}}% +\label{sub:exercise-1.7.1a} + +A set consisting of a single point. + +\begin{proof} + + Let $S$ be a set consisting of a single point. + By definition of a Point, $S$ is a rectangle in which all vertices coincide. + By \nameref{sec:choice-scale}, $S$ is measurable with area its width times + its height. + The width and height of $S$ is trivially zero. + Therefore $a(S) = (0)(0) = 0$. + +\end{proof} + +\subsection{\unverified{Exercise 1.7.1b}}% +\label{sub:exercise-1.7.1b} + +A set consisting of a finite number of points in a plane. + +\begin{proof} + + Define predicate $P(n)$ as "A set consisting of $n$ points in a plane is + measurable with area $0$". + We use induction to prove $P(n)$ holds for all $n > 0$. + + \paragraph{Base Case}% + + Consider a set $S$ consisting of a single point in a plane. + By \nameref{sub:exercise-1.7.1a}, $S$ is measurable with area $0$. + Thus $P(1)$ holds. + + \paragraph{Induction Step}% + + Assume induction hypothesis $P(k)$ holds for some $k > 0$. + Let $S_{k+1}$ be a set consisting of $k + 1$ points in a plane. + Pick an arbitrary point of $S_{k+1}$. + Denote the set containing just this point as $T$. + Denote the remaining set of points as $S_k$. + By construction, $S_{k+1} = S_k \cup T$. + By the induction hypothesis, $S_k$ is measurable with area $0$. + By \nameref{sub:exercise-1.7.1a}, $T$ is measurable with area $0$. + By the \nameref{sec:additive-property}, $S_k \cup T$ is + measurable, $S_k \cap T$ is measurable, and + \begin{align} + a(S_{k+1}) + & = a(S_k \cup T) \nonumber \\ + & = a(S_k) + a(T) - a(S_k \cap T) \nonumber \\ + & = 0 + 0 - a(S_k \cap T). \label{sub:exercise-1.7.1b-eq1} + \end{align} + There are two cases to consider: + + \subparagraph{Case 1}% + + $S_k \cap T = \emptyset$. + Then it trivially follows that $a(S_k \cap T) = 0$. + + \subparagraph{Case 2}% + + $S_k \cap T \neq \emptyset$. + Since $T$ consists of a single point, $S_k \cap T = T$. + By \nameref{sub:exercise-1.7.1a}, $a(S_k \cap T) = a(T) = 0$. + + \vspace{8pt} + \noindent + In both cases, \eqref{sub:exercise-1.7.1b-eq1} evaluates to $0$, implying + $P(k + 1)$ as expected. + + \paragraph{Conclusion}% + + By mathematical induction, it follows for all $n > 0$, $P(n)$ is true. + +\end{proof} + +\subsection{\unverified{Exercise 1.7.1c}}% +\label{sub:exercise-1.7.1c} + +The union of a finite collection of line segments in a plane. + +\begin{proof} + + Define predicate $P(n)$ as "A set consisting of $n$ line segments in a plane + is measurable with area $0$". + We use induction to prove $P(n)$ holds for all $n > 0$. + + \paragraph{Base Case}% + + Consider a set $S$ consisting of a single line segment in a plane. + By definition of a Line Segment, $S$ is a rectangle in which one side has + dimension $0$. + By \nameref{sec:choice-scale}, $S$ is measurable with area its width $w$ + times its height $h$. + Therefore $a(S) = wh = 0$. + Thus $P(1)$ holds. + + \paragraph{Induction Step}% + + Assume induction hypothesis $P(k)$ holds for some $k > 0$. + Let $S_{k+1}$ be a set consisting of $k + 1$ line segments in a plane. + Pick an arbitrary line segment of $S_{k+1}$. + Denote the set containing just this line segment as $T$. + Denote the remaining set of line segments as $S_k$. + By construction, $S_{k+1} = S_k \cup T$. + By the induction hypothesis, $S_k$ is measurable with area $0$. + By the base case, $T$ is measurable with area $0$. + By the \nameref{sec:additive-property}, $S_k \cup T$ is measurable, + $S_k \cap T$ is measurable, and + \begin{align} + a(S_{k+1}) + & = a(S_k \cup T) \nonumber \\ + & = a(S_k) + a(T) - a(S_k \cap T) \nonumber \\ + & = 0 + 0 - a(S_k \cap T). \label{sub:exercise-1.7.1c-eq1} + \end{align} + There are two cases to consider: + + \subparagraph{Case 1}% + + $S_k \cap T = \emptyset$. + Then it trivially follows that $a(S_k \cap T) = 0$. + + \subparagraph{Case 2}% + + $S_k \cap T \neq \emptyset$. + Since $T$ consists of a single point, $S_k \cap T = T$. + By the base case, $a(S_k \cap T) = a(T) = 0$. + + \vspace{8pt} + \noindent + In both cases, \eqref{sub:exercise-1.7.1c-eq1} evaluates to $0$, implying + $P(k + 1)$ as expected. + + \paragraph{Conclusion}% + + By mathematical induction, it follows for all $n > 0$, $P(n)$ is true. + +\end{proof} + +\section{\unverified{Exercise 1.7.2}}% +\label{sec:exercise-1.7.2} + +Every right triangular region is measurable because it can be obtained as the + intersection of two rectangles. +Prove that every triangular region is measurable and that its area is one half + the product of its base and altitude. + +\begin{proof} + + Let $T'$ be a triangular region with base of length $a$, height of length $b$, + and hypotenuse of length $c$. + Consider the translation and rotation of $T'$, say $T$, such that its + hypotenuse is entirely within quadrant I and the vertex opposite the + hypotenuse is situated at point $(0, 0)$. + + Let $R$ be a rectangle of width $a$, height $b$, and bottom-left corner at + $(0, 0)$. + By construction, $R$ covers all of $T$. + Let $S$ be a rectangle of width $c$ and height $a\sin{\theta}$, where $\theta$ + is the acute angle measured from the bottom-right corner of $T$ relative + to the $x$-axis. + As an example, consider the image below of triangle $T$ with width $4$ and + height $3$: + + \begin{figure}[h] + \includegraphics{right-triangle} + \centering + \end{figure} + + By \nameref{sec:choice-scale}, both $R$ and $S$ are measurable. + By this same axiom, $a(R) = ab$ and $a(S) = ca\sin{\theta}$. + By the \nameref{sec:additive-property}, $R \cup S$ and $R \cap S$ are both + measurable. + $a(R \cap S) = a(T)$ and $a(R \cup S)$ can be determined by noting that + $R$'s construction implies identity $a(R) = 2a(T)$. + Therefore + \begin{align*} + a(T) + & = a(R \cap S) \\ + & = a(R) + a(S) - a(R \cup S) \\ + & = ab + ca\sin{\theta} - a(R \cup S) \\ + & = ab + ca\sin{\theta} - (ca\sin{\theta} + \frac{1}{2}a(R)) \\ + & = ab + ca\sin{\theta} - ca\sin{\theta} - a(T). + \end{align*} + Solving for $a(T)$ gives the desired identity: $$a(T) = \frac{1}{2}ab.$$ + By \nameref{sec:invariance-under-congruence}, $a(T') = a(T)$, concluding our + proof. + +\end{proof} + +\section{\unverified{Exercise 1.7.3}}% +\label{sec:exercise-1.7.3} + +Prove that every trapezoid and every parallelogram is measurable and derive the + usual formulas for their areas. + +\begin{proof} + + We begin by proving the formula for a trapezoid. + Let $S$ be a trapezoid with height $h$ and bases $b_1$ and $b_2$, $b_1 < b_2$. + There are three cases to consider: + + \begin{figure}[h] + \includegraphics[width=\textwidth]{trapezoid-cases} + \centering + \end{figure} + + \paragraph{Case 1}% + + Suppose $S$ is a right trapezoid. + Then $S$ is the union of non-overlapping rectangle $R$ of width $b_1$ and + height $h$ with right triangle $T$ of base $b_2 - b_1$ and height $h$. + By \nameref{sec:choice-scale}, $R$ is measurable. + By \nameref{sec:exercise-1.7.2}, $T$ is measurable. + By the \nameref{sec:additive-property}, $R \cup T$ and $R \cap T$ are both + measurable and + \begin{align*} + a(S) + & = a(R \cup T) \\ + & = a(R) + a(T) - a(R \cap T) \\ + & = a(R) + a(T) & \text{by construction} \\ + & = b_1h + a(T) & \text{Choice of Scale} \\ + & = b_1h + \frac{1}{2}(b_2 - b_1)h & \textref{sec:exercise-1.7.2} \\ + & = \frac{b_1 + b_2}{2}h. + \end{align*} + + \paragraph{Case 2}% + + Suppose $S$ is an acute trapezoid. + Then $S$ is the union of non-overlapping triangle $T$ and right trapezoid $R$. + Let $c$ denote the length of base $T$. + Then $R$ has longer base edge of length $b_2 - c$. + By \nameref{sec:exercise-1.7.2}, $T$ is measurable. + By Case 1, $R$ is measurable. + By the \nameref{sec:additive-property}, $R \cup T$ and $R \cap T$ are both + measurable and + \begin{align*} + a(S) + & = a(T) + a(R) - a(R \cap T) \\ + & = a(T) + a(R) & \text{by construction} \\ + & = \frac{1}{2}ch + a(R) & \textref{sec:exercise-1.7.2} \\ + & = \frac{1}{2}ch + \frac{b_1 + b_2 - c}{2}h & \text{Case 1} \\ + & = \frac{b_1 + b_2}{2}h. + \end{align*} + + \paragraph{Case 3}% + + Suppose $S$ is an obtuse trapezoid. + Then $S$ is the union of non-overlapping triangle $T$ and right trapezoid $R$. + Let $c$ denote the length of base $T$. + Reflect $T$ vertically to form another right triangle, say $T'$. + Then $T' \cup R$ is an acute trapezoid. + By \nameref{sec:invariance-under-congruence}, + \begin{equation} + \label{sec:exercise-1.7.3-eq1} + \tag{3.1} + a(T' \cup R) = a(T \cup R). + \end{equation} + By construction, $T' \cup R$ has height $h$ and bases $b_1 - c$ and $b_2 + c$ + meaning + \begin{align*} + a(T \cup R) + & = a(T' \cup R) & \eqref{sec:exercise-1.7.3-eq1} \\ + & = \frac{b_1 - c + b_2 + c}{2}h & \text{Case 2} \\ + & = \frac{b_1 + b_2}{2}h. + \end{align*} + + \paragraph{Conclusion}% + + These cases are exhaustive and in agreement with one another. + Thus $S$ is measurable and $$a(S) = \frac{b_1 + b_2}{2}h.$$ + + \divider + + Let $P$ be a parallelogram with base $b$ and height $h$. + Then $P$ is the union of non-overlapping triangle $T$ and right trapezoid $R$. + Let $c$ denote the length of base $T$. + Reflect $T$ vertically to form another right triangle, say $T'$. + Then $T' \cup R$ is an acute trapezoid. + By \nameref{sec:invariance-under-congruence}, + \begin{equation} + \label{sec:exercise-1.7.3-eq2} + \tag{3.2} + a(T' \cup R) = a(T \cup R). + \end{equation} + By construction, $T' \cup R$ has height $h$ and bases $b - c$ and $b + c$ + meaning + \begin{align*} + a(T \cup R) + & = a(T' \cup R) & \eqref{sec:exercise-1.7.3-eq2} \\ + & = \frac{b - c + b + c}{2}h & \text{Area of Trapezoid} \\ + & = bh. + \end{align*} + +\end{proof} + +\section{Exercise 1.7.4}% +\label{sec:exercise-1.7.4} + +Let $P$ be a polygon whose vertices are lattice points. +The area of $P$ is $I + \frac{1}{2}B - 1$, where $I$ denotes the number of + lattice points inside the polygon and $B$ denotes the number on the boundary. + +\subsection{\unverified{Exercise 1.7.4a}}% +\label{sub:exercise-1.7.4a} + +Prove that the formula is valid for rectangles with sides parallel to the + coordinate axes. + +\begin{proof} + + Let $P$ be a rectangle with sides parallel to the coordinate axes, with width + $w$, height $h$, and lattice points for vertices. + We assume $P$ has three non-collinear points, ruling out any instances of + points or line segments. + + By \nameref{sec:choice-scale}, $P$ is measurable with area $a(P) = wh$. + By construction, $P$ has $I = (w - 1)(h - 1)$ interior lattice points and + $B = 2(w + h)$ lattice points on its boundary. + The following shows the lattice point area formula is in agreement with + the expected result: + \begin{align*} + I + \frac{1}{2}B - 1 + & = (w - 1)(h - 1) + \frac{1}{2}\left[ 2(w + h) \right] - 1 \\ + & = (wh - w - h + 1) + \frac{1}{2}\left[ 2(w + h) \right] - 1 \\ + & = (wh - w - h + 1) + (w + h) - 1 \\ + & = wh. + \end{align*} + +\end{proof} + +\subsection{\unverified{Exercise 1.7.4b}}% +\label{sub:exercise-1.7.4b} + +Prove that the formula is valid for right triangles and parallelograms. + +\begin{proof} + + Let $P$ be a right triangle with width $w > 0$, height $h > 0$, and lattice + points for vertices. + Let $T$ be the triangle $P$ translated, rotated, and reflected such that the + its vertices are $(0, 0)$, $(0, w)$, and $(w, h)$. + Let $I_T$ and $B_T$ be the number of interior and boundary points of $T$ + respectively. + Let $H_L$ denote the number of lattice points on $T$'s hypotenuse. + + Let $R$ be the overlapping rectangle of width $w$ and height $h$, situated + with bottom-left corner at $(0, 0)$. + Let $I_R$ and $B_R$ be the number of interior and boundary points + of $R$ respectively. + + By construction, $T$ shares two sides with $R$. + Therefore + \begin{equation} + \label{sub:exercise-1.7.4b-eq1} + B_T = \frac{1}{2}B_R - 1 + H_L. + \end{equation} + Likewise, + \begin{equation} + \label{sub:exercise-1.7.4b-eq2} + I_T = \frac{1}{2}(I_R - (H_L - 2)). + \end{equation} + The following shows the lattice point area formula is in agreement with + the expected result: + \begin{align*} + I_T + \frac{1}{2}B_T - 1 + & = \frac{1}{2}(I_R - (H_L - 2)) + \frac{1}{2}B_T - 1 + & \eqref{sub:exercise-1.7.4b-eq2} \\ + & = \frac{1}{2}\left[ I_R - H_L + B_T \right] \\ + & = \frac{1}{2}\left[ I_R - H_L + \frac{1}{2}B_R - 1 + H_L \right] + & \eqref{sub:exercise-1.7.4b-eq1} \\ + & = \frac{1}{2}\left[ I_R + \frac{1}{2}B_R - 1 \right] \\ + & = \frac{1}{2}\left[ wh \right] & \textref{sub:exercise-1.7.4a}. + \end{align*} + + We do not prove this formula is valid for parallelograms here. + Instead, refer to \nameref{sub:exercise-1.7.4c} below. + +\end{proof} + +\subsection{\unverified{Exercise 1.7.4c}}% +\label{sub:exercise-1.7.4c} + +Use induction on the number of edges to construct a proof for general polygons. + +\begin{proof} + + Define predicate $P(n)$ as "An $n$-polygon with vertices on lattice points has + area $I + \frac{1}{2}B - 1$." + We use induction to prove $P(n)$ holds for all $n \geq 3$. + + \paragraph{Base Case}% + + A $3$-polygon is a triangle. + By \nameref{sub:exercise-1.7.4b}, the lattice point area formula holds. + Thus $P(3)$ holds. + + \paragraph{Induction Step}% + + Assume induction hypothesis $P(k)$ holds for some $k \geq 3$. + Let $P$ be a $(k + 1)$-polygon with vertices on lattice points. + Such a polygon is equivalent to the union of a $k$-polygon $S$ with a + triangle $T$. + That is, $P = S \cup T$. + + Let $I_P$ be the number of interior lattice points of $P$. + Let $B_P$ be the number of boundary lattice points of $P$. + Similarly, let $I_S$, $I_T$, $B_S$, and $B_T$ be the number of interior + and boundary lattice points of $S$ and $T$. + Let $c$ denote the number of boundary points shared between $S$ and $T$. + + By our induction hypothesis, $a(S) = I_S + \frac{1}{2}B_S - 1$. + By our base case, $a(T) = I_T + \frac{1}{2}B_T - 1$. + By construction, it follows: + \begin{align*} + I_P & = I_S + I_T + c - 2 \\ + B_P & = B_S + B_T - (c - 2) - c \\ + & = B_S + B_T - 2c + 2. + \end{align*} + Applying the lattice point area formula to $P$ yields the following: + \begin{align*} + & I_P + \frac{1}{2}B_P - 1 \\ + & = (I_S + I_T + c - 2) + \frac{1}{2}(B_S + B_T - 2c + 2) - 1 \\ + & = I_S + I_T + c - 2 + \frac{1}{2}B_S + \frac{1}{2}B_T - c + 1 - 1 \\ + & = (I_S + \frac{1}{2}B_S - 1) + (I_T + \frac{1}{2}B_T - 1) \\ + & = a(S) + (I_T + \frac{1}{2}B_T - 1) & \text{induction hypothesis} \\ + & = a(S) + a(T). & \text{base case} + \end{align*} + By the \nameref{sec:additive-property}, $S \cup T$ is measurable, + $S \cap T$ is measurable, and + \begin{align*} + a(P) + & = a(S \cup T) \\ + & = a(S) + a(T) - a(S \cap T) \\ + & = a(S) + a(T). & \text{by construction} + \end{align*} + This shows the lattice point area formula is in agreement with our axiomatic + definition of area. + Thus $P(k + 1)$ holds. + + \paragraph{Conclusion}% + + By mathematical induction, it follows for all $n \geq 3$, $P(n)$ is true. + +\end{proof} + +\section{\unverified{Exercise 1.7.5}}% +\label{sec:exercise-1.7.5} + +Prove that a triangle whose vertices are lattice points cannot be equilateral. + +[\textit{Hint:} Assume there is such a triangle and compute its area in two +ways, using Exercises 2 and 4.] + +\begin{proof} + + Proceed by contradiction. + Let $T$ be an equilateral triangle whose vertices are lattice points. + Assume each side of $T$ has length $a$. + Then $T$ has height $h = (a\sqrt{3}) / 2$. + By \nameref{sec:exercise-1.7.2}, + \begin{equation} + \label{sub:exercise-1.7.5-eq1} + \tag{5.1} + a(T) = \frac{1}{2}ah = \frac{a^2\sqrt{3}}{4}. + \end{equation} + Let $I$ and $B$ denote the number of interior and boundary lattice points of + $T$ respectively. + By \nameref{sec:exercise-1.7.4}, + \begin{equation} + \label{sub:exercise-1.7.5-eq2} + \tag{5.2} + a(T) = I + \frac{1}{2}B - 1. + \end{equation} + But \eqref{sub:exercise-1.7.5-eq1} is irrational whereas + \eqref{sub:exercise-1.7.5-eq2} is not. + This is a contradiction. + Thus, there is \textit{no} equilateral triangle whose vertices are lattice + points. + +\end{proof} + +\section{\unverified{Exercise 1.7.6}}% +\label{sec:exercise-1.7.6} + +Let $A = \{1, 2, 3, 4, 5\}$, and let $\mathscr{M}$ denote the class of all + subsets of $A$. +(There are 32 altogether, counting $A$ itself and the empty set $\emptyset$.) +For each set $S$ in $\mathscr{M}$, let $n(S)$ denote the number of distinct + elements in $S$. +If $S = \{1, 2, 3, 4\}$ and $T = \{3, 4, 5\}$, compute $n(S \cup T)$, + $n(S \cap T)$, $n(S - T)$, and $n(T - S)$. +Prove that the set function $n$ satisfies the first three axioms for area. + +\begin{proof} + + Let $S = \{1, 2, 3, 4\}$ and $T = \{3, 4, 5\}$. + Then + \begin{align*} + n(S \cup T) + & = n(\{1, 2, 3, 4\} \cup \{3, 4, 5\}) \\ + & = n(\{1, 2, 3, 4, 5\}) \\ + & = 5. \\ + n(S \cap T) + & = n(\{1, 2, 3, 4\} \cap \{3, 4, 5\}) \\ + & = n(\{3, 4\}) \\ + & = 2. \\ + n(S - T) + & = n(\{1, 2, 3, 4\} - \{3, 4, 5\}) \\ + & = n(\{1, 2\}) \\ + & = 2. \\ + n(T - S) + & = n(\{3, 4, 5\} - \{1, 2, 3, 4\}) \\ + & = n(\{5\}) \\ + & = 1. + \end{align*} + We now prove $n$ satisfies the first three axioms for area. + + \paragraph{Nonnegative Property}% + + $n$ returns the length of some member of $\mathscr{M}$. + By hypothesis, the smallest possible input to $n$ is $\emptyset$. + Since $n(\emptyset) = 0$, it follows $n(S) \geq 0$ for all $S \subset A$. + + \paragraph{Additive Property}% + + Let $S$ and $T$ be members of $\mathscr{M}$. + It trivially follows that both $S \cup T$ and $S \cap T$ are in + $\mathscr{M}$. + Consider the value of $n(S \cup T)$. + There are two cases to consider: + + \subparagraph{Case 1}% + + Suppose $S \cap T = \emptyset$. + That is, there is no common element shared between $S$ and $T$. + Thus + \begin{align*} + n(S \cup T) + & = n(S) + n(T) \\ + & = n(S) + n(T) - 0 \\ + & = n(S) + n(T) - n(S \cap T). + \end{align*} + + \subparagraph{Case 2}% + + Suppose $S \cap T \neq \emptyset$. + Then $n(S) + n(T)$ counts each element of $S \cap T$ twice. + Therefore $n(S \cup T) = n(S) + n(T) - n(S \cap T)$. + + \subparagraph{Conclusion}% + + These cases are exhaustive and in agreement with one another. + Thus $n(S \cup T) = n(S) + n(T) - n(S \cap T)$. + + \paragraph{Difference Property}% + + Suppose $S, T \in \mathscr{M}$ such that $S \subseteq T$. + That is, every member of $S$ is a member of $T$. + By definition, $T - S$ consists of members in $T$ but not in $S$. + Thus $n(T - S) = n(T) - n(S)$. + +\end{proof} + +\chapter{Partitions and Step Functions}% +\label{chap:partitions-step-functions} + +\section{\defined{Partition}}% +\label{sec:partition} + +Let $[a, b]$ be a closed interval decomposed into $n$ subintervals by inserting + $n - 1$ points of subdivision, say $x_1$, $x_2$, $\ldots$, $x_{n-1}$, subject + only to the restriction + \begin{equation} + \label{sec:partition-eq1} + a < x_1 < x_2 < \cdots < x_{n-1} < b. + \end{equation} +It is convenient to denote the point $a$ itself by $x_0$ and the point $b$ by + $x_n$. +A collection of points satisfying \eqref{sec:partition-eq1} is called a + \textbf{partition} $P$ of $[a, b]$, and we use the symbol + $$P = \{x_0, x_1, \ldots, x_n\}$$ to designate this partition. + +\begin{definition} + + \lean{Common/Set/Intervals/Partition}{Set.Intervals.Partition} + +\end{definition} + +\section{\defined{Step Function}}% +\label{sec:step-function} + +A function $s$, whose domain is a closed interval $[a, b]$, is called a step + function if there is a \nameref{sec:partition} $P = \{x_0, x_1, \ldots, x_n\}$ + of $[a b]$ such that $s$ is constant on each open subinterval of $P$. +That is to say, for each $k = 1, 2, \ldots, n$, there is a real number $s_k$ + such that $$s(x) = s_k \quad\text{if}\quad x_{k-1} < x < x_k.$$ +Step functions are sometimes called piecewise constant functions. + +\vspace{8pt} +\noindent +\textit{Note:} At each of the endpoints $x_{k-1}$ and $x_k$ the function must + have some well-defined value, but this need not be the same as $s_k$. + +\begin{definition} + + \lean{Common/Set/Intervals/StepFunction}{Set.Intervals.StepFunction} + +\end{definition} + +\chapter{Exercises 1.11}% +\label{chap:exercises-1-11} + +\section{Exercise 1.11.4}% +\label{sec:exercise-1.11.4} + +Prove that the greatest-integer function has the properties indicated: + +\subsection{\verified{Exercise 1.11.4a}}% +\label{sub:exercise-1.11.4a} + +$\floor{x + n} = \floor{x} + n$ for every integer $n$. + +\begin{proof} + + \lean{Bookshelf/Apostol/Chapter\_1\_11} + {Apostol.Chapter\_1\_11.exercise\_4a} + + \divider + + Let $x$ be a real number and $n$ an integer. + Let $m = \floor{x + n}$. + By definition of the floor function, $m$ is the unique integer such that + $m \leq x + n < m + 1$. + Then $m - n \leq x < (m - n) + 1$. + That is, $m - n = \floor{x}$. + Rearranging terms we see that $m = \floor{x} + n$ as expected. + +\end{proof} + +\subsection{\verified{Exercise 1.11.4b}}% +\label{sub:exercise-1.11.4b} + +$\floor{-x} = + \begin{cases} + -\floor{x} & \text{if } x \text{ is an integer}, \\ + -\floor{x} - 1 & \text{otherwise}. + \end{cases}$ + +\begin{proof} + + \ \vspace{6pt} + + \lean{Bookshelf/Apostol/Chapter\_1\_11} + {Apostol.Chapter\_1\_11.exercise\_4b\_1} + + \lean{Bookshelf/Apostol/Chapter\_1\_11} + {Apostol.Chapter\_1\_11.exercise\_4b\_2} + + \divider + + There are two cases to consider: + + \paragraph{Case 1}% + + Suppose $x$ is an integer. + Then $x = \floor{x}$ and $-x = \floor{-x}$. + It immediately follows that $$\floor{-x} = -x = -\floor{x}.$$ + + \paragraph{Case 2}% + + Suppose $x$ is not an integer. + Let $m = \floor{-x}$. + By definition of the floor function, $m$ is the unique integer such that + $m \leq -x < m + 1$. + Equivalently, $-m - 1 < x \leq -m$. + Since $x$ is not an integer, it follows $-m - 1 \leq x < -m$. + Then, by definition of the floor function, $\floor{x} = -m - 1$. + Solving for $m$ yields $$\floor{-x} = m = -\floor{x} - 1.$$ + + \paragraph{Conclusion}% + + The above two cases are exhaustive. Thus + $$\floor{-x} = + \begin{cases} + -\floor{x} & \text{if } x \text{ is an integer}, \\ + -\floor{x} - 1 & \text{otherwise}. + \end{cases}$$ + +\end{proof} + +\subsection{\verified{Exercise 1.11.4c}}% +\label{sub:exercise-1.11.4c} + +$\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$. + +\begin{proof} + + \lean{Bookshelf/Apostol/Chapter\_1\_11} + {Apostol.Chapter\_1\_11.exercise\_4c} + + \divider + + Rewrite $x$ and $y$ as the sum of their floor and fractional components: + $x = \floor{x} + \{x\}$ and $y = \floor{y} + \{y\}$. + Now + \begin{align} + \floor{x + y} + & = \floor{\floor{x} + \{x\} + \floor{y} + \{y\}} \nonumber \\ + & = \floor{\floor{x} + \floor{y} + \{x\} + \{y\}} \nonumber \\ + & = \floor{x} + \floor{y} + \floor{\{x\} + \{y\}} + & \textref{sub:exercise-1.11.4a} \label{sub:exercise-1.11.4c-eq1} + \end{align} + There are two cases to consider: + + \paragraph{Case 1}% + + Suppose $\{x\} + \{y\} < 1$. + Then $\floor{\{x\} + \{y\}} = 0$. + Substituting this value into \eqref{sub:exercise-1.11.4c-eq1} yields + $$\floor{x + y} = \floor{x} + \floor{y}.$$ + + \paragraph{Case 2}% + + Suppose $\{x\} + \{y\} \geq 1$. + Because $\{x\}$ and $\{y\}$ are both less than $1$, $\{x\} + \{y\} < 2$. + Thus $\floor{\{x\} + \{y\}} = 1$. + Substituting this value into \eqref{sub:exercise-1.11.4c-eq1} yields + $$\floor{x + y} = \floor{x} + \floor{y} + 1.$$ + + \paragraph{Conclusion}% + + Since the above two cases are exhaustive, it follows + $\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$. + +\end{proof} + +\subsection{\partial{Exercise 1.11.4d}}% +\label{sub:exercise-1.11.4d} + +$\floor{2x} = \floor{x} + \floor{x + \frac{1}{2}}.$ + +\begin{proof} + + \lean{Bookshelf/Apostol/Chapter\_1\_11} + {Apostol.Chapter\_1\_11.exercise\_4d} + + \divider + + This is immediately proven by applying Hermite's Identity as shown in + \nameref{sec:exercise-1.11.5}. + +\end{proof} + +\subsection{\partial{Exercise 1.11.4e}}% +\label{sub:exercise-1.11.4e} + +$\floor{3x} = \floor{x} + \floor{x + \frac{1}{3}} + \floor{x + \frac{2}{3}}.$ + +\begin{proof} + + \lean{Bookshelf/Apostol/Chapter\_1\_11} + {Apostol.Chapter\_1\_11.exercise\_4e} + + \divider + + This is immediately proven by applying Hermite's Identity as shown in + \nameref{sec:exercise-1.11.5}. + +\end{proof} + +\section{\partial{Exercise 1.11.5}}% +\label{sec:exercise-1.11.5} + +The formulas in Exercises 4(d) and 4(e) suggest a generalization for + $\floor{nx}$. +State and prove such a generalization. + +\note{The stated generalization is known as "Hermite's Identity."} + +\begin{proof} + + \lean{Bookshelf/Apostol/Chapter\_1\_11} + {Apostol.Chapter\_1\_11.exercise\_5} + + \divider + + We prove that for all natural numbers $n$ and real numbers $x$, the following + identity holds: + \begin{equation} + \label{sec:exercise-1.11.5-eq1} + \floor{nx} = \sum_{i=0}^{n-1} \floor{x + \frac{i}{n}} + \end{equation} + By definition of the floor function, $x = \floor{x} + r$ for some + $r \in \ico{0}{1}$. + Define $S$ as the partition of non-overlapping subintervals + $$\ico{0}{\frac{1}{n}}, \ico{\frac{1}{n}}{\frac{2}{n}}, \ldots, + \ico{\frac{n-1}{n}}{1}.$$ + By construction, $\cup\; S = \ico{0}{1}$. + Therefore there exists some $j \in \mathbb{N}$ such that + \begin{equation} + \label{sec:exercise-1.11.5-eq2} + r \in \ico{\frac{j}{n}}{\frac{j+1}{n}}. + \end{equation} + With these definitions established, we now show the left- and right-hand sides + of \eqref{sec:exercise-1.11.5-eq1} evaluate to the same number. + + \paragraph{Left-Hand Side}% + + Consider the left-hand side of identity \eqref{sec:exercise-1.11.5-eq1}. + By \eqref{sec:exercise-1.11.5-eq2}, $nr \in \ico{j}{j + 1}$. + Therefore $\floor{nr} = j$. + Thus + \begin{align} + \floor{nx} + & = \floor{n(\floor{x} + r)} \nonumber \\ + & = \floor{n\floor{x} + nr} \nonumber \\ + & = \floor{n\floor{x}} + \floor{nr}. \nonumber + & \textref{sub:exercise-1.11.4a} \\ + & = \floor{n\floor{x}} + j \nonumber \\ + & = n\floor{x} + j. \label{sec:exercise-1.11.5-eq3} + \end{align} + + \paragraph{Right-Hand Side}% + + Now consider the right-hand side of identity + \eqref{sec:exercise-1.11.5-eq1}. + We note each summand, by construction, is the floor of $x$ added to a + nonnegative number less than one. + Therefore each summand contributes either $\floor{x}$ or $\floor{x} + 1$ to + the total. + Letting $z$ denote the number of summands that contribute $\floor{x} + 1$, + we have + \begin{equation} + \label{sec:exercise-1.11.5-eq4} + \sum_{i=0}^{n-1} \floor{x + \frac{i}{n}} = n\floor{x} + z. + \end{equation} + The value of $z$ corresponds to the number of indices $i$ that satisfy + $$\frac{i}{n} \geq 1 - r.$$ + By \eqref{sec:exercise-1.11.5-eq2}, it follows + \begin{align*} + 1 - r + & \in \ioc{1 - \frac{j+1}{n}}{1-\frac{j}{n}} \\ + & = \ioc{\frac{n - j - 1}{n}}{\frac{n - j}{n}}. + \end{align*} + Thus we can determine the value of $z$ by instead counting the number of + indices $i$ that satisfy $$\frac{i}{n} \geq \frac{n - j}{n}.$$ + Rearranging terms, we see that $i \geq n - j$ holds for + $z = (n - 1) - (n - j) + 1 = j$ of the $n$ summands. + Substituting the value of $z$ into \eqref{sec:exercise-1.11.5-eq4} yields + \begin{equation} + \label{sec:exercise-1.11.5-eq5} + \sum_{i=0}^{n-1} \floor{x + \frac{i}{n}} = n\floor{x} + j. + \end{equation} + + \paragraph{Conclusion}% + + Since \eqref{sec:exercise-1.11.5-eq3} and \eqref{sec:exercise-1.11.5-eq5} + agree with one another, it follows identity + \eqref{sec:exercise-1.11.5-eq1} holds. + +\end{proof} + +\section{\unverified{Exercise 1.11.6}}% +\label{sec:exercise-1.11.6} + +Recall that a lattice point $(x, y)$ in the plane is one whose coordinates are + integers. +Let $f$ be a nonnegative function whose domain is the interval $[a, b]$, where + $a$ and $b$ are integers, $a < b$. +Let $S$ denote the set of points $(x, y)$ satisfying $a \leq x \leq b$, + $0 < y \leq f(x)$. +Prove that the number of lattice points in $S$ is equal to the sum + $$\sum_{n=a}^b \floor{f(n)}.$$ + +\begin{proof} + + Let $i = a, \ldots, b$ and define $S_i = \mathbb{N} \cap \ioc{0}{f(i)}$. + By construction, the number of lattice points in $S$ is + \begin{equation} + \label{sec:exercise-1.11.6-eq1} + \sum_{n = a}^b \abs{S_n}. + \end{equation} + All that remains is to show $\abs{S_i} = \floor{f(i)}$. + There are two cases to consider: + + \paragraph{Case 1}% + + Suppose $f(i)$ is an integer. + Then the number of integers in $\ioc{0}{f(i)}$ is $f(i) = \floor{f(i)}$. + + \paragraph{Case 2}% + + Suppose $f(i)$ is not an integer. + Then the number of integers in $\ioc{0}{f(i)}$ is the same as that of + $\ioc{0}{\floor{f(i)}}$. + Once again, that number is $\floor{f(i)}$. + + \paragraph{Conclusion}% + + By cases 1 and 2, $\abs{S_i} = \floor{f(i)}$. + Substituting this identity into \eqref{sec:exercise-1.11.6-eq1} finishes the + proof. + +\end{proof} + +\section{Exercise 1.11.7}% +\label{sec:exercise-1.11.7} + +If $a$ and $b$ are positive integers with no common factor, we have the formula + $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$ +When $b = 1$, the sum on the left is understood to be $0$. + +\note{When $b = 1$, the proofs of (a) and (b) are trivial. We continue under the + assumption $b > 1$.} + +\subsection{\unverified{Exercise 1.11.7a}}% +\label{sub:exercise-1.11.7a} + +Derive this result by a geometric argument, counting lattice points in a right + triangle. + +\begin{proof} + + Let $f \colon [1, b - 1] \rightarrow \mathbb{R}$ be given by $f(x) = ax / b$. + Let $S$ denote the set of points $(x, y)$ satisfying $1 \leq x \leq b - 1$, + $0 < y \leq f(x)$. + By \nameref{sec:exercise-1.11.6}, the number of lattice points of $S$ is equal + to the sum + \begin{equation} + \label{sub:exercise-1.11.7a-eq1} + \sum_{n=1}^{b-1} \floor{f(n)} = \sum_{n=1}^{b-1} \floor{\frac{na}{b}}. + \end{equation} + Define $T$ to be the triangle of width $w = b$ and height $h = f(b) = a$ + as $$T = \{ (x, y) : 0 < x < b, 0 < y \leq f(x) \}.$$ + By construction, $T$ does not introduce any additional lattice points. + Thus $S$ and $T$ have the same number of lattice points. + Let $H_L$ denote the number of boundary points on $T$'s hypotenuse. + We prove that (i) $H_L = 2$ and (ii) that $T$ has $\frac{(a - 1)(b - 1)}{2}$ + lattice points. + + \paragraph{(i)}% + \label{par:exercise-1.11.7a-i} + + Consider the line $L$ overlapping the hypotenuse of $T$. + By construction, $T$'s hypotenuse has endpoints $(0, 0)$ and $(b, a)$. + By hypothesis, $a$ and $b$ are positive, excluding the possibility of $L$ + being vertical. + Define the slope of $L$ as $$m = \frac{a}{b}.$$ + $H_L$ coincides with the number of indices $i = 0, \ldots, b$ such that + $(i, i * m)$ is a lattice point. + But $a$ and $b$ are coprime by hypothesis and $i \leq b$. + Thus $i * m$ is an integer if and only if $i = 0$ or $i = b$. + Thus $H_L = 2$. + + \paragraph{(ii)}% + + Next we count the number of lattice points in $T$. + Let $R$ be the overlapping retangle of width $w$ and height $h$, situated + with bottom-left corner at $(0, 0)$. + Let $I_R$ denote the number of interior lattice points of $R$. + Let $I_T$ and $B_T$ denote the interior and boundary lattice points of $T$ + respectively. + By \nameref{sub:exercise-1.7.4b-eq2}, + \begin{align} + I_T + & = \frac{1}{2}(I_R - (H_L - 2)) \nonumber \\ + & = \frac{1}{2}(I_R - (2 - 2)) + & \textref{par:exercise-1.11.7a-i} \nonumber \\ + & = \frac{1}{2}I_R. & \label{sub:exercise-1.11.7a-eq2} + \end{align} + Furthermore, since both the adjacent and opposite side of $T$ are not + included in $T$ and there exist no lattice points on $T$'s hypotenuse + besides the endpoints, it follows + \begin{equation} + \label{sub:exercise-1.11.7a-eq3} + B_T = 0. + \end{equation} + Thus the number of lattice points of $T$ equals + \begin{align} + I_T + B_T + & = I_T & \eqref{sub:exercise-1.11.7a-eq3} \nonumber \\ + & = \frac{1}{2}I_R & \eqref{sub:exercise-1.11.7a-eq2} \nonumber \\ + & = \frac{(b - 1)(a - 1)}{2}. + & \textref{sub:exercise-1.7.4a} \label{sub:exercise-1.11.7a-eq4} + \end{align} + + \paragraph{Conclusion}% + + By \eqref{sub:exercise-1.11.7a-eq1} the number of lattice points of $S$ is + equal to the sum $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}}.$$ + But the number of lattice points of $S$ is the same as that of $T$. + By \eqref{sub:exercise-1.11.7a-eq4}, the number of lattice points in $T$ is + equal to $$\frac{(b - 1)(a - 1)}{2}.$$ + Thus $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$ + +\end{proof} + +\subsection{\partial{Exercise 1.11.7b}}% +\label{sub:exercise-1.11.7b} + +Derive the result analytically as follows: +By changing the index of summation, note that + $\sum_{n=1}^{b-1} \floor{na / b} = \sum_{n=1}^{b-1} \floor{a(b - n) / b}$. +Now apply Exercises 4(a) and (b) to the bracket on the right. + +\begin{proof} + + \lean{Bookshelf/Apostol/Chapter\_1\_11} + {Apostol.Chapter\_1\_11.exercise\_7b} + + \divider + + Let $n = 1, \ldots, b - 1$. + By hypothesis, $a$ and $b$ are coprime. + Furthermore, $n < b$ for all values of $n$. + Thus $an / b$ is not an integer. + By \nameref{sub:exercise-1.11.4b}, + \begin{equation} + \label{sub:exercise-1.11.7b-eq1} + \floor{-\frac{an}{b}} = -\floor{\frac{an}{b}} - 1. + \end{equation} + Consider the following: + \begin{align*} + \sum_{n=1}^{b-1} \floor{\frac{na}{b}} + & = \sum_{n=1}^{b-1} \floor{\frac{a(b - n)}{b}} \\ + & = \sum_{n=1}^{b-1} \floor{\frac{ab - an}{b}} \\ + & = \sum_{n=1}^{b-1} \floor{-\frac{an}{b} + a} \\ + & = \sum_{n=1}^{b-1} \floor{-\frac{an}{b}} + a. + & \textref{sub:exercise-1.11.4a} \\ + & = \sum_{n=1}^{b-1} -\floor{\frac{an}{b}} - 1 + a + & \eqref{sub:exercise-1.11.7b-eq1} \\ + & = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - \sum_{n=1}^{b-1} 1 + + \sum_{n=1}^{b-1} a \\ + & = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - (b - 1) + a(b - 1). + \end{align*} + Rearranging the above yields + $$2\sum_{n=1}^{b-1} \floor{\frac{an}{b}} = (a - 1)(b - 1).$$ + Dividing both sides of the above identity concludes the proof. + +\end{proof} + +\section{\partial{Exercise 1.11.8}}% +\label{sec:exercise-1.11.8} + +Let $S$ be a set of points on the real line. +The \textit{characteristic function} of $S$ is, by definition, the function + $\mathcal{X}_S$ such that $\mathcal{X}_S(x) = 1$ for every $x$ in $S$, and + $\mathcal{X}_S(x) = 0$ for those $x$ not in $S$. +Let $f$ be a step function which takes the constant value $c_k$ on the $k$th + open subinterval $I_k$ of some partition of an interval $[a, b]$. +Prove that for each $x$ in the union $I_1 \cup I_2 \cup \cdots \cup I_n$ we have + $$f(x) = \sum_{k=1}^n c_k\mathcal{X}_{I_k}(x).$$ +This property is described by saying that every step function is a linear + combination of characteristic functions of intervals. + +\begin{proof} + + Let $x \in I_1 \cup I_2 \cup \cdots \cup I_n$ and $N = \{1, \ldots, n\}$. + Let $k \in N$ such that $x \in I_k$. + Consider an arbitrary $j \in N - \{k\}$. + By definition of a partition, $I_j \cap I_k = \emptyset$. + That is, $I_j$ and $I_k$ are disjoint for all $j \in N - \{k\}$. + Therefore, by definition of the characteristic function, + $\mathcal{X}_{I_k}(x) = 1$ and $\mathcal{X}_{I_j}(x) = 0$ for all + $j \in N - \{k\}$. + Thus + \begin{align*} + f(x) + & = c_k \\ + & = (c_k)(1) + \sum\nolimits_{j \in N - \{k\}} (c_j)(0) \\ + & = c_k\mathcal{X}_{I_k}(x) + + \sum\nolimits_{j \in N - \{k\}} c_j\mathcal{X}_{I_j}(x) \\ + & = \sum_{k=1}^n c_k\mathcal{X}_{I_k}(x). + \end{align*} + +\end{proof} + +\end{document} diff --git a/Bookshelf/Apostol/Chapter_1_07.tex b/Bookshelf/Apostol/Chapter_1_07.tex deleted file mode 100644 index 6a02590..0000000 --- a/Bookshelf/Apostol/Chapter_1_07.tex +++ /dev/null @@ -1,589 +0,0 @@ -\documentclass{article} - -\usepackage{graphicx} - -\input{../../preamble} - -\graphicspath{{./images/}} - -\externaldocument[A:] - {../../Common/Real/Geometry/Area} - [../../Common/Real/Geometry/Area.pdf] - -\begin{document} - -\header{Exercises 1.7}{Tom M. Apostol} - -The properties of area in this set of exercises are to be deduced from the - axioms for area stated in the foregoing section. - -\section*{Exercise 1}% -\label{sec:exercise-1} - -Prove that each of the following sets is measurable and has zero area: - -\subsection*{\unverified{Exercise 1a}}% -\label{sub:exercise-1a} - -A set consisting of a single point. - -\begin{proof} - - Let $S$ be a set consisting of a single point. - By definition of a Point, $S$ is a rectangle in which all vertices coincide. - By \nameref{A:sec:choice-scale}, $S$ is measurable with area its width times - its height. - The width and height of $S$ is trivially zero. - Therefore $a(S) = (0)(0) = 0$. - -\end{proof} - -\subsection*{\unverified{Exercise 1b}}% -\label{sub:exercise-1b} - -A set consisting of a finite number of points in a plane. - -\begin{proof} - - Define predicate $P(n)$ as "A set consisting of $n$ points in a plane is - measurable with area $0$". - We use induction to prove $P(n)$ holds for all $n > 0$. - - \paragraph{Base Case}% - - Consider a set $S$ consisting of a single point in a plane. - By \nameref{sub:exercise-1a}, $S$ is measurable with area $0$. - Thus $P(1)$ holds. - - \paragraph{Induction Step}% - - Assume induction hypothesis $P(k)$ holds for some $k > 0$. - Let $S_{k+1}$ be a set consisting of $k + 1$ points in a plane. - Pick an arbitrary point of $S_{k+1}$. - Denote the set containing just this point as $T$. - Denote the remaining set of points as $S_k$. - By construction, $S_{k+1} = S_k \cup T$. - By the induction hypothesis, $S_k$ is measurable with area $0$. - By \nameref{sub:exercise-1a}, $T$ is measurable with area $0$. - By the \nameref{A:sec:additive-property}, $S_k \cup T$ is - measurable, $S_k \cap T$ is measurable, and - \begin{align} - a(S_{k+1}) - & = a(S_k \cup T) \nonumber \\ - & = a(S_k) + a(T) - a(S_k \cap T) \nonumber \\ - & = 0 + 0 - a(S_k \cap T). \label{sub:exercise-1b-eq1} - \end{align} - There are two cases to consider: - - \subparagraph{Case 1}% - - $S_k \cap T = \emptyset$. - Then it trivially follows that $a(S_k \cap T) = 0$. - - \subparagraph{Case 2}% - - $S_k \cap T \neq \emptyset$. - Since $T$ consists of a single point, $S_k \cap T = T$. - By \nameref{sub:exercise-1a}, $a(S_k \cap T) = a(T) = 0$. - - \vspace{8pt} - \noindent - In both cases, \eqref{sub:exercise-1b-eq1} evaluates to $0$, implying - $P(k + 1)$ as expected. - - \paragraph{Conclusion}% - - By mathematical induction, it follows for all $n > 0$, $P(n)$ is true. - -\end{proof} - -\subsection*{\unverified{Exercise 1c}}% -\label{sub:exercise-1c} - -The union of a finite collection of line segments in a plane. - -\begin{proof} - - Define predicate $P(n)$ as "A set consisting of $n$ line segments in a plane - is measurable with area $0$". - We use induction to prove $P(n)$ holds for all $n > 0$. - - \paragraph{Base Case}% - - Consider a set $S$ consisting of a single line segment in a plane. - By definition of a Line Segment, $S$ is a rectangle in which one side has - dimension $0$. - By \nameref{A:sec:choice-scale}, $S$ is measurable with area its width $w$ - times its height $h$. - Therefore $a(S) = wh = 0$. - Thus $P(1)$ holds. - - \paragraph{Induction Step}% - - Assume induction hypothesis $P(k)$ holds for some $k > 0$. - Let $S_{k+1}$ be a set consisting of $k + 1$ line segments in a plane. - Pick an arbitrary line segment of $S_{k+1}$. - Denote the set containing just this line segment as $T$. - Denote the remaining set of line segments as $S_k$. - By construction, $S_{k+1} = S_k \cup T$. - By the induction hypothesis, $S_k$ is measurable with area $0$. - By the base case, $T$ is measurable with area $0$. - By the \nameref{A:sec:additive-property}, $S_k \cup T$ is measurable, - $S_k \cap T$ is measurable, and - \begin{align} - a(S_{k+1}) - & = a(S_k \cup T) \nonumber \\ - & = a(S_k) + a(T) - a(S_k \cap T) \nonumber \\ - & = 0 + 0 - a(S_k \cap T). \label{sub:exercise-1c-eq1} - \end{align} - There are two cases to consider: - - \subparagraph{Case 1}% - - $S_k \cap T = \emptyset$. - Then it trivially follows that $a(S_k \cap T) = 0$. - - \subparagraph{Case 2}% - - $S_k \cap T \neq \emptyset$. - Since $T$ consists of a single point, $S_k \cap T = T$. - By the base case, $a(S_k \cap T) = a(T) = 0$. - - \vspace{8pt} - \noindent - In both cases, \eqref{sub:exercise-1c-eq1} evaluates to $0$, implying - $P(k + 1)$ as expected. - - \paragraph{Conclusion}% - - By mathematical induction, it follows for all $n > 0$, $P(n)$ is true. - -\end{proof} - -\section*{\unverified{Exercise 2}}% -\label{sec:exercise-2} - -Every right triangular region is measurable because it can be obtained as the - intersection of two rectangles. -Prove that every triangular region is measurable and that its area is one half - the product of its base and altitude. - -\begin{proof} - - Let $T'$ be a triangular region with base of length $a$, height of length $b$, - and hypotenuse of length $c$. - Consider the translation and rotation of $T'$, say $T$, such that its - hypotenuse is entirely within quadrant I and the vertex opposite the - hypotenuse is situated at point $(0, 0)$. - - Let $R$ be a rectangle of width $a$, height $b$, and bottom-left corner at - $(0, 0)$. - By construction, $R$ covers all of $T$. - Let $S$ be a rectangle of width $c$ and height $a\sin{\theta}$, where $\theta$ - is the acute angle measured from the bottom-right corner of $T$ relative - to the $x$-axis. - As an example, consider the image below of triangle $T$ with width $4$ and - height $3$: - - \begin{figure}[h] - \includegraphics{right-triangle} - \centering - \end{figure} - - By \nameref{A:sec:choice-scale}, both $R$ and $S$ are measurable. - By this same axiom, $a(R) = ab$ and $a(S) = ca\sin{\theta}$. - By the \nameref{A:sec:additive-property}, $R \cup S$ and $R \cap S$ are both - measurable. - $a(R \cap S) = a(T)$ and $a(R \cup S)$ can be determined by noting that - $R$'s construction implies identity $a(R) = 2a(T)$. - Therefore - \begin{align*} - a(T) - & = a(R \cap S) \\ - & = a(R) + a(S) - a(R \cup S) \\ - & = ab + ca\sin{\theta} - a(R \cup S) \\ - & = ab + ca\sin{\theta} - (ca\sin{\theta} + \frac{1}{2}a(R)) \\ - & = ab + ca\sin{\theta} - ca\sin{\theta} - a(T). - \end{align*} - Solving for $a(T)$ gives the desired identity: $$a(T) = \frac{1}{2}ab.$$ - By \nameref{A:sec:invariance-under-congruence}, $a(T') = a(T)$, concluding our - proof. - -\end{proof} - -\section*{\unverified{Exercise 3}}% -\label{sec:exercise-3} - -Prove that every trapezoid and every parallelogram is measurable and derive the - usual formulas for their areas. - -\begin{proof} - - We begin by proving the formula for a trapezoid. - Let $S$ be a trapezoid with height $h$ and bases $b_1$ and $b_2$, $b_1 < b_2$. - There are three cases to consider: - - \begin{figure}[h] - \includegraphics[width=\textwidth]{trapezoid-cases} - \centering - \end{figure} - - \paragraph{Case 1}% - - Suppose $S$ is a right trapezoid. - Then $S$ is the union of non-overlapping rectangle $R$ of width $b_1$ and - height $h$ with right triangle $T$ of base $b_2 - b_1$ and height $h$. - By \nameref{A:sec:choice-scale}, $R$ is measurable. - By \nameref{sec:exercise-2}, $T$ is measurable. - By the \nameref{A:sec:additive-property}, $R \cup T$ and $R \cap T$ are both - measurable and - \begin{align*} - a(S) - & = a(R \cup T) \\ - & = a(R) + a(T) - a(R \cap T) \\ - & = a(R) + a(T) & \text{by construction} \\ - & = b_1h + a(T) & \text{Choice of Scale} \\ - & = b_1h + \frac{1}{2}(b_2 - b_1)h - & \text{\nameref{sec:exercise-2}} \\ - & = \frac{b_1 + b_2}{2}h. - \end{align*} - - \paragraph{Case 2}% - - Suppose $S$ is an acute trapezoid. - Then $S$ is the union of non-overlapping triangle $T$ and right trapezoid $R$. - Let $c$ denote the length of base $T$. - Then $R$ has longer base edge of length $b_2 - c$. - By \nameref{sec:exercise-2}, $T$ is measurable. - By Case 1, $R$ is measurable. - By the \nameref{A:sec:additive-property}, $R \cup T$ and $R \cap T$ are both - measurable and - \begin{align*} - a(S) - & = a(T) + a(R) - a(R \cap T) \\ - & = a(T) + a(R) & \text{by construction} \\ - & = \frac{1}{2}ch + a(R) & \text{\nameref{sec:exercise-2}} \\ - & = \frac{1}{2}ch + \frac{b_1 + b_2 - c}{2}h & \text{Case 1} \\ - & = \frac{b_1 + b_2}{2}h. - \end{align*} - - \paragraph{Case 3}% - - Suppose $S$ is an obtuse trapezoid. - Then $S$ is the union of non-overlapping triangle $T$ and right trapezoid $R$. - Let $c$ denote the length of base $T$. - Reflect $T$ vertically to form another right triangle, say $T'$. - Then $T' \cup R$ is an acute trapezoid. - By \nameref{A:sec:invariance-under-congruence}, - \begin{equation} - \label{par:exercise-3-case-3-eq1} - \tag{3.1} - a(T' \cup R) = a(T \cup R). - \end{equation} - By construction, $T' \cup R$ has height $h$ and bases $b_1 - c$ and $b_2 + c$ - meaning - \begin{align*} - a(T \cup R) - & = a(T' \cup R) & \eqref{par:exercise-3-case-3-eq1} \\ - & = \frac{b_1 - c + b_2 + c}{2}h & \text{Case 2} \\ - & = \frac{b_1 + b_2}{2}h. - \end{align*} - - \paragraph{Conclusion}% - - These cases are exhaustive and in agreement with one another. - Thus $S$ is measurable and $$a(S) = \frac{b_1 + b_2}{2}h.$$ - - \divider - - Let $P$ be a parallelogram with base $b$ and height $h$. - Then $P$ is the union of non-overlapping triangle $T$ and right trapezoid $R$. - Let $c$ denote the length of base $T$. - Reflect $T$ vertically to form another right triangle, say $T'$. - Then $T' \cup R$ is an acute trapezoid. - By \nameref{A:sec:invariance-under-congruence}, - \begin{equation} - \label{par:exercise-3-eq2} - \tag{3.2} - a(T' \cup R) = a(T \cup R). - \end{equation} - By construction, $T' \cup R$ has height $h$ and bases $b - c$ and $b + c$ - meaning - \begin{align*} - a(T \cup R) - & = a(T' \cup R) & \eqref{par:exercise-3-eq2} \\ - & = \frac{b - c + b + c}{2}h & \text{Area of Trapezoid} \\ - & = bh. - \end{align*} - -\end{proof} - -\section*{Exercise 4}% -\label{sec:exercise-4} - -Let $P$ be a polygon whose vertices are lattice points. -The area of $P$ is $I + \frac{1}{2}B - 1$, where $I$ denotes the number of - lattice points inside the polygon and $B$ denotes the number on the boundary. - -\subsection*{\unverified{Exercise 4a}}% -\label{sub:exercise-4a} - -Prove that the formula is valid for rectangles with sides parallel to the - coordinate axes. - -\begin{proof} - - Let $P$ be a rectangle with sides parallel to the coordinate axes, with width - $w$, height $h$, and lattice points for vertices. - We assume $P$ has three non-collinear points, ruling out any instances of - points or line segments. - - By \nameref{A:sec:choice-scale}, $P$ is measurable with area $a(P) = wh$. - By construction, $P$ has $I = (w - 1)(h - 1)$ interior lattice points and - $B = 2(w + h)$ lattice points on its boundary. - The following shows the lattice point area formula is in agreement with - the expected result: - \begin{align*} - I + \frac{1}{2}B - 1 - & = (w - 1)(h - 1) + \frac{1}{2}\left[ 2(w + h) \right] - 1 \\ - & = (wh - w - h + 1) + \frac{1}{2}\left[ 2(w + h) \right] - 1 \\ - & = (wh - w - h + 1) + (w + h) - 1 \\ - & = wh. - \end{align*} - -\end{proof} - -\subsection*{\unverified{Exercise 4b}}% -\label{sub:exercise-4b} - -Prove that the formula is valid for right triangles and parallelograms. - -\begin{proof} - - Let $P$ be a right triangle with width $w > 0$, height $h > 0$, and lattice - points for vertices. - Let $T$ be the triangle $P$ translated, rotated, and reflected such that the - its vertices are $(0, 0)$, $(0, w)$, and $(w, h)$. - Let $I_T$ and $B_T$ be the number of interior and boundary points of $T$ - respectively. - Let $H_L$ denote the number of lattice points on $T$'s hypotenuse. - - Let $R$ be the overlapping rectangle of width $w$ and height $h$, situated - with bottom-left corner at $(0, 0)$. - Let $I_R$ and $B_R$ be the number of interior and boundary points - of $R$ respectively. - - By construction, $T$ shares two sides with $R$. - Therefore - \begin{equation} - \label{sub:exercise-4b-eq1} - B_T = \frac{1}{2}B_R - 1 + H_L. - \end{equation} - Likewise, - \begin{equation} - \label{sub:exercise-4b-eq2} - I_T = \frac{1}{2}(I_R - (H_L - 2)). - \end{equation} - The following shows the lattice point area formula is in agreement with - the expected result: - \begin{align*} - I_T + \frac{1}{2}B_T - 1 - & = \frac{1}{2}(I_R - (H_L - 2)) + \frac{1}{2}B_T - 1 - & \eqref{sub:exercise-4b-eq2} \\ - & = \frac{1}{2}\left[ I_R - H_L + B_T \right] \\ - & = \frac{1}{2}\left[ I_R - H_L + \frac{1}{2}B_R - 1 + H_L \right] - & \eqref{sub:exercise-4b-eq1} \\ - & = \frac{1}{2}\left[ I_R + \frac{1}{2}B_R - 1 \right] \\ - & = \frac{1}{2}\left[ wh \right] & \text{\nameref{sub:exercise-4a}}. - \end{align*} - - We do not prove this formula is valid for parallelograms here. - Instead, refer to \nameref{sub:exercise-4c} below. - -\end{proof} - -\subsection*{\unverified{Exercise 4c}}% -\label{sub:exercise-4c} - -Use induction on the number of edges to construct a proof for general polygons. - -\begin{proof} - - Define predicate $P(n)$ as "An $n$-polygon with vertices on lattice points has - area $I + \frac{1}{2}B - 1$." - We use induction to prove $P(n)$ holds for all $n \geq 3$. - - \paragraph{Base Case}% - - A $3$-polygon is a triangle. - By \nameref{sub:exercise-4b}, the lattice point area formula holds. - Thus $P(3)$ holds. - - \paragraph{Induction Step}% - - Assume induction hypothesis $P(k)$ holds for some $k \geq 3$. - Let $P$ be a $(k + 1)$-polygon with vertices on lattice points. - Such a polygon is equivalent to the union of a $k$-polygon $S$ with a - triangle $T$. - That is, $P = S \cup T$. - - Let $I_P$ be the number of interior lattice points of $P$. - Let $B_P$ be the number of boundary lattice points of $P$. - Similarly, let $I_S$, $I_T$, $B_S$, and $B_T$ be the number of interior - and boundary lattice points of $S$ and $T$. - Let $c$ denote the number of boundary points shared between $S$ and $T$. - - By our induction hypothesis, $a(S) = I_S + \frac{1}{2}B_S - 1$. - By our base case, $a(T) = I_T + \frac{1}{2}B_T - 1$. - By construction, it follows: - \begin{align*} - I_P & = I_S + I_T + c - 2 \\ - B_P & = B_S + B_T - (c - 2) - c \\ - & = B_S + B_T - 2c + 2. - \end{align*} - Applying the lattice point area formula to $P$ yields the following: - \begin{align*} - & I_P + \frac{1}{2}B_P - 1 \\ - & = (I_S + I_T + c - 2) + \frac{1}{2}(B_S + B_T - 2c + 2) - 1 \\ - & = I_S + I_T + c - 2 + \frac{1}{2}B_S + \frac{1}{2}B_T - c + 1 - 1 \\ - & = (I_S + \frac{1}{2}B_S - 1) + (I_T + \frac{1}{2}B_T - 1) \\ - & = a(S) + (I_T + \frac{1}{2}B_T - 1) & \text{induction hypothesis} \\ - & = a(S) + a(T). & \text{base case} - \end{align*} - By the \nameref{A:sec:additive-property}, $S \cup T$ is measurable, - $S \cap T$ is measurable, and - \begin{align*} - a(P) - & = a(S \cup T) \\ - & = a(S) + a(T) - a(S \cap T) \\ - & = a(S) + a(T). & \text{by construction} - \end{align*} - This shows the lattice point area formula is in agreement with our axiomatic - definition of area. - Thus $P(k + 1)$ holds. - - \paragraph{Conclusion}% - - By mathematical induction, it follows for all $n \geq 3$, $P(n)$ is true. - -\end{proof} - -\section*{\unverified{Exercise 5}}% -\label{sec:exercise-5} - -Prove that a triangle whose vertices are lattice points cannot be equilateral. - -[\textit{Hint:} Assume there is such a triangle and compute its area in two -ways, using Exercises 2 and 4.] - -\begin{proof} - - Proceed by contradiction. - Let $T$ be an equilateral triangle whose vertices are lattice points. - Assume each side of $T$ has length $a$. - Then $T$ has height $h = (a\sqrt{3}) / 2$. - By \nameref{sec:exercise-2}, - \begin{equation} - \label{sub:exercise-5-eq1} - \tag{5.1} - a(T) = \frac{1}{2}ah = \frac{a^2\sqrt{3}}{4}. - \end{equation} - Let $I$ and $B$ denote the number of interior and boundary lattice points of - $T$ respectively. - By \nameref{sec:exercise-4}, - \begin{equation} - \label{sub:exercise-5-eq2} - \tag{5.2} - a(T) = I + \frac{1}{2}B - 1. - \end{equation} - But \eqref{sub:exercise-5-eq1} is irrational whereas - \eqref{sub:exercise-5-eq2} is not. - This is a contradiction. - Thus, there is \textit{no} equilateral triangle whose vertices are lattice - points. - -\end{proof} - -\section*{\unverified{Exercise 6}}% -\label{sec:exercise-6} - -Let $A = \{1, 2, 3, 4, 5\}$, and let $\mathscr{M}$ denote the class of all - subsets of $A$. -(There are 32 altogether, counting $A$ itself and the empty set $\emptyset$.) -For each set $S$ in $\mathscr{M}$, let $n(S)$ denote the number of distinct - elements in $S$. -If $S = \{1, 2, 3, 4\}$ and $T = \{3, 4, 5\}$, compute $n(S \cup T)$, - $n(S \cap T)$, $n(S - T)$, and $n(T - S)$. -Prove that the set function $n$ satisfies the first three axioms for area. - -\begin{proof} - - Let $S = \{1, 2, 3, 4\}$ and $T = \{3, 4, 5\}$. - Then - \begin{align*} - n(S \cup T) - & = n(\{1, 2, 3, 4\} \cup \{3, 4, 5\}) \\ - & = n(\{1, 2, 3, 4, 5\}) \\ - & = 5. \\ - n(S \cap T) - & = n(\{1, 2, 3, 4\} \cap \{3, 4, 5\}) \\ - & = n(\{3, 4\}) \\ - & = 2. \\ - n(S - T) - & = n(\{1, 2, 3, 4\} - \{3, 4, 5\}) \\ - & = n(\{1, 2\}) \\ - & = 2. \\ - n(T - S) - & = n(\{3, 4, 5\} - \{1, 2, 3, 4\}) \\ - & = n(\{5\}) \\ - & = 1. - \end{align*} - We now prove $n$ satisfies the first three axioms for area. - - \paragraph{Nonnegative Property}% - - $n$ returns the length of some member of $\mathscr{M}$. - By hypothesis, the smallest possible input to $n$ is $\emptyset$. - Since $n(\emptyset) = 0$, it follows $n(S) \geq 0$ for all $S \subset A$. - - \paragraph{Additive Property}% - - Let $S$ and $T$ be members of $\mathscr{M}$. - It trivially follows that both $S \cup T$ and $S \cap T$ are in - $\mathscr{M}$. - Consider the value of $n(S \cup T)$. - There are two cases to consider: - - \subparagraph{Case 1}% - - Suppose $S \cap T = \emptyset$. - That is, there is no common element shared between $S$ and $T$. - Thus - \begin{align*} - n(S \cup T) - & = n(S) + n(T) \\ - & = n(S) + n(T) - 0 \\ - & = n(S) + n(T) - n(S \cap T). - \end{align*} - - \subparagraph{Case 2}% - - Suppose $S \cap T \neq \emptyset$. - Then $n(S) + n(T)$ counts each element of $S \cap T$ twice. - Therefore $n(S \cup T) = n(S) + n(T) - n(S \cap T)$. - - \subparagraph{Conclusion}% - - These cases are exhaustive and in agreement with one another. - Thus $n(S \cup T) = n(S) + n(T) - n(S \cap T)$. - - \paragraph{Difference Property}% - - Suppose $S, T \in \mathscr{M}$ such that $S \subseteq T$. - That is, every member of $S$ is a member of $T$. - By definition, $T - S$ consists of members in $T$ but not in $S$. - Thus $n(T - S) = n(T) - n(S)$. - -\end{proof} - -\end{document} diff --git a/Bookshelf/Apostol/Chapter_1_11.tex b/Bookshelf/Apostol/Chapter_1_11.tex deleted file mode 100644 index 53120b6..0000000 --- a/Bookshelf/Apostol/Chapter_1_11.tex +++ /dev/null @@ -1,471 +0,0 @@ -\documentclass{article} - -\input{../../preamble} - -\externaldocument[C:1:07:]{Chapter_1_07}[Chapter_1_07.pdf] - -\newcommand{\lean}[1]{\leanref - {./Chapter\_1\_11.html\#Apostol.Chapter\_1\_11.#1} - {Apostol.Chapter\_1\_11.#1}} - -\begin{document} - -\header{Exercises 1.11}{Tom M. Apostol} - -\section*{Exercise 4}% -\label{sec:exercise-4} - -Prove that the greatest-integer function has the properties indicated: - -\subsection*{\verified{Exercise 4a}}% -\label{sub:exercise-4a} - -$\floor{x + n} = \floor{x} + n$ for every integer $n$. - -\begin{proof} - - \lean{exercise\_4a} - - \divider - - Let $x$ be a real number and $n$ an integer. - Let $m = \floor{x + n}$. - By definition of the floor function, $m$ is the unique integer such that - $m \leq x + n < m + 1$. - Then $m - n \leq x < (m - n) + 1$. - That is, $m - n = \floor{x}$. - Rearranging terms we see that $m = \floor{x} + n$ as expected. - -\end{proof} - -\subsection*{\verified{Exercise 4b}}% -\label{sub:exercise-4b} - -$\floor{-x} = - \begin{cases} - -\floor{x} & \text{if } x \text{ is an integer}, \\ - -\floor{x} - 1 & \text{otherwise}. - \end{cases}$ - -\begin{proof} - - \ \vspace{6pt} - - \lean{exercise\_4b\_1} - - \lean{exercise\_4b\_2} - - \divider - - There are two cases to consider: - - \paragraph{Case 1}% - - Suppose $x$ is an integer. - Then $x = \floor{x}$ and $-x = \floor{-x}$. - It immediately follows that $$\floor{-x} = -x = -\floor{x}.$$ - - \paragraph{Case 2}% - - Suppose $x$ is not an integer. - Let $m = \floor{-x}$. - By definition of the floor function, $m$ is the unique integer such that - $m \leq -x < m + 1$. - Equivalently, $-m - 1 < x \leq -m$. - Since $x$ is not an integer, it follows $-m - 1 \leq x < -m$. - Then, by definition of the floor function, $\floor{x} = -m - 1$. - Solving for $m$ yields $$\floor{-x} = m = -\floor{x} - 1.$$ - - \paragraph{Conclusion}% - - The above two cases are exhaustive. Thus - $$\floor{-x} = - \begin{cases} - -\floor{x} & \text{if } x \text{ is an integer}, \\ - -\floor{x} - 1 & \text{otherwise}. - \end{cases}$$ - -\end{proof} - -\subsection*{\verified{Exercise 4c}}% -\label{sub:exercise-4c} - -$\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$. - -\begin{proof} - - \lean{exercise\_4c} - - \divider - - Rewrite $x$ and $y$ as the sum of their floor and fractional components: - $x = \floor{x} + \{x\}$ and $y = \floor{y} + \{y\}$. - Now - \begin{align} - \floor{x + y} - & = \floor{\floor{x} + \{x\} + \floor{y} + \{y\}} \nonumber \\ - & = \floor{\floor{x} + \floor{y} + \{x\} + \{y\}} \nonumber \\ - & = \floor{x} + \floor{y} + \floor{\{x\} + \{y\}} - & \text{\nameref{sub:exercise-4a}} \label{sub:exercise-4c-eq1} - \end{align} - There are two cases to consider: - - \paragraph{Case 1}% - - Suppose $\{x\} + \{y\} < 1$. - Then $\floor{\{x\} + \{y\}} = 0$. - Substituting this value into \eqref{sub:exercise-4c-eq1} yields - $$\floor{x + y} = \floor{x} + \floor{y}.$$ - - \paragraph{Case 2}% - - Suppose $\{x\} + \{y\} \geq 1$. - Because $\{x\}$ and $\{y\}$ are both less than $1$, $\{x\} + \{y\} < 2$. - Thus $\floor{\{x\} + \{y\}} = 1$. - Substituting this value into \eqref{sub:exercise-4c-eq1} yields - $$\floor{x + y} = \floor{x} + \floor{y} + 1.$$ - - \paragraph{Conclusion}% - - Since the above two cases are exhaustive, it follows - $\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$. - -\end{proof} - -\subsection*{\partial{Exercise 4d}}% -\label{sub:exercise-4d} - -$\floor{2x} = \floor{x} + \floor{x + \frac{1}{2}}.$ - -\begin{proof} - - \lean{exercise\_4d} - - \divider - - This is immediately proven by applying Hermite's Identity as shown in - \nameref{sec:exercise-5}. - -\end{proof} - -\subsection*{\partial{Exercise 4e}}% -\label{sub:exercise-4e} - -$\floor{3x} = \floor{x} + \floor{x + \frac{1}{3}} + \floor{x + \frac{2}{3}}.$ - -\begin{proof} - - \lean{exercise\_4e} - - \divider - - This is immediately proven by applying Hermite's Identity as shown in - \nameref{sec:exercise-5}. - -\end{proof} - -\section*{\partial{Exercise 5}}% -\label{sec:exercise-5} - -The formulas in Exercises 4(d) and 4(e) suggest a generalization for - $\floor{nx}$. -State and prove such a generalization. - -\note{The stated generalization is known as "Hermite's Identity."} - -\begin{proof} - - \lean{exercise\_5} - - \divider - - We prove that for all natural numbers $n$ and real numbers $x$, the following - identity holds: - \begin{equation} - \label{sec:exercise-5-eq1} - \floor{nx} = \sum_{i=0}^{n-1} \floor{x + \frac{i}{n}} - \end{equation} - By definition of the floor function, $x = \floor{x} + r$ for some - $r \in \ico{0}{1}$. - Define $S$ as the partition of non-overlapping subintervals - $$\ico{0}{\frac{1}{n}}, \ico{\frac{1}{n}}{\frac{2}{n}}, \ldots, - \ico{\frac{n-1}{n}}{1}.$$ - By construction, $\cup\; S = \ico{0}{1}$. - Therefore there exists some $j \in \mathbb{N}$ such that - \begin{equation} - \label{sec:exercise-5-eq2} - r \in \ico{\frac{j}{n}}{\frac{j+1}{n}}. - \end{equation} - With these definitions established, we now show the left- and right-hand sides - of \eqref{sec:exercise-5-eq1} evaluate to the same number. - - \paragraph{Left-Hand Side}% - - Consider the left-hand side of identity \eqref{sec:exercise-5-eq1}. - By \eqref{sec:exercise-5-eq2}, $nr \in \ico{j}{j + 1}$. - Therefore $\floor{nr} = j$. - Thus - \begin{align} - \floor{nx} - & = \floor{n(\floor{x} + r)} \nonumber \\ - & = \floor{n\floor{x} + nr} \nonumber \\ - & = \floor{n\floor{x}} + \floor{nr}. \nonumber - & \text{\nameref{sub:exercise-4a}} \\ - & = \floor{n\floor{x}} + j \nonumber \\ - & = n\floor{x} + j. \label{sec:exercise-5-eq3} - \end{align} - - \paragraph{Right-Hand Side}% - - Now consider the right-hand side of identity \eqref{sec:exercise-5-eq1}. - We note each summand, by construction, is the floor of $x$ added to a - nonnegative number less than one. - Therefore each summand contributes either $\floor{x}$ or $\floor{x} + 1$ to - the total. - Letting $z$ denote the number of summands that contribute $\floor{x} + 1$, - we have - \begin{equation} - \label{sec:exercise-5-eq4} - \sum_{i=0}^{n-1} \floor{x + \frac{i}{n}} = n\floor{x} + z. - \end{equation} - The value of $z$ corresponds to the number of indices $i$ that satisfy - $$\frac{i}{n} \geq 1 - r.$$ - By \eqref{sec:exercise-5-eq2}, it follows - \begin{align*} - 1 - r - & \in \ioc{1 - \frac{j+1}{n}}{1-\frac{j}{n}} \\ - & = \ioc{\frac{n - j - 1}{n}}{\frac{n - j}{n}}. - \end{align*} - Thus we can determine the value of $z$ by instead counting the number of - indices $i$ that satisfy $$\frac{i}{n} \geq \frac{n - j}{n}.$$ - Rearranging terms, we see that $i \geq n - j$ holds for - $z = (n - 1) - (n - j) + 1 = j$ of the $n$ summands. - Substituting the value of $z$ into \eqref{sec:exercise-5-eq4} yields - \begin{equation} - \label{sec:exercise-5-eq5} - \sum_{i=0}^{n-1} \floor{x + \frac{i}{n}} = n\floor{x} + j. - \end{equation} - - \paragraph{Conclusion}% - - Since \eqref{sec:exercise-5-eq3} and \eqref{sec:exercise-5-eq5} agree with - one another, it follows identity \eqref{sec:exercise-5-eq1} holds. - -\end{proof} - -\section*{\unverified{Exercise 6}}% -\label{sec:exercise-6} - -Recall that a lattice point $(x, y)$ in the plane is one whose coordinates are - integers. -Let $f$ be a nonnegative function whose domain is the interval $[a, b]$, where - $a$ and $b$ are integers, $a < b$. -Let $S$ denote the set of points $(x, y)$ satisfying $a \leq x \leq b$, - $0 < y \leq f(x)$. -Prove that the number of lattice points in $S$ is equal to the sum - $$\sum_{n=a}^b \floor{f(n)}.$$ - -\begin{proof} - - Let $i = a, \ldots, b$ and define $S_i = \mathbb{N} \cap \ioc{0}{f(i)}$. - By construction, the number of lattice points in $S$ is - \begin{equation} - \label{sec:exercise-6-eq1} - \sum_{n = a}^b \abs{S_n}. - \end{equation} - All that remains is to show $\abs{S_i} = \floor{f(i)}$. - There are two cases to consider: - - \paragraph{Case 1}% - - Suppose $f(i)$ is an integer. - Then the number of integers in $\ioc{0}{f(i)}$ is $f(i) = \floor{f(i)}$. - - \paragraph{Case 2}% - - Suppose $f(i)$ is not an integer. - Then the number of integers in $\ioc{0}{f(i)}$ is the same as that of - $\ioc{0}{\floor{f(i)}}$. - Once again, that number is $\floor{f(i)}$. - - \paragraph{Conclusion}% - - By cases 1 and 2, $\abs{S_i} = \floor{f(i)}$. - Substituting this identity into \eqref{sec:exercise-6-eq1} finishes the - proof. - -\end{proof} - -\section*{Exercise 7}% -\label{sec:exercise-7} - -If $a$ and $b$ are positive integers with no common factor, we have the formula - $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$ -When $b = 1$, the sum on the left is understood to be $0$. - -\note{When $b = 1$, the proofs of (a) and (b) are trivial. We continue under the - assumption $b > 1$.} - -\subsection*{\unverified{Exercise 7a}}% -\label{sub:exercise-7a} - -Derive this result by a geometric argument, counting lattice points in a right - triangle. - -\begin{proof} - - Let $f \colon [1, b - 1] \rightarrow \mathbb{R}$ be given by $f(x) = ax / b$. - Let $S$ denote the set of points $(x, y)$ satisfying $1 \leq x \leq b - 1$, - $0 < y \leq f(x)$. - By \nameref{sec:exercise-6}, the number of lattice points of $S$ is equal to - the sum - \begin{equation} - \label{sub:exercise-7a-eq1} - \sum_{n=1}^{b-1} \floor{f(n)} = \sum_{n=1}^{b-1} \floor{\frac{na}{b}}. - \end{equation} - Define $T$ to be the triangle of width $w = b$ and height $h = f(b) = a$ - as $$T = \{ (x, y) : 0 < x < b, 0 < y \leq f(x) \}.$$ - By construction, $T$ does not introduce any additional lattice points. - Thus $S$ and $T$ have the same number of lattice points. - Let $H_L$ denote the number of boundary points on $T$'s hypotenuse. - We prove that (i) $H_L = 2$ and (ii) that $T$ has $\frac{(a - 1)(b - 1)}{2}$ - lattice points. - - \paragraph{(i)}% - \label{par:exercise-7a-i} - - Consider the line $L$ overlapping the hypotenuse of $T$. - By construction, $T$'s hypotenuse has endpoints $(0, 0)$ and $(b, a)$. - By hypothesis, $a$ and $b$ are positive, excluding the possibility of $L$ - being vertical. - Define the slope of $L$ as $$m = \frac{a}{b}.$$ - $H_L$ coincides with the number of indices $i = 0, \ldots, b$ such that - $(i, i * m)$ is a lattice point. - But $a$ and $b$ are coprime by hypothesis and $i \leq b$. - Thus $i * m$ is an integer if and only if $i = 0$ or $i = b$. - Thus $H_L = 2$. - - \paragraph{(ii)}% - - Next we count the number of lattice points in $T$. - Let $R$ be the overlapping retangle of width $w$ and height $h$, situated - with bottom-left corner at $(0, 0)$. - Let $I_R$ denote the number of interior lattice points of $R$. - Let $I_T$ and $B_T$ denote the interior and boundary lattice points of $T$ - respectively. - By \nameref{C:1:07:sub:exercise-4b-eq2}, - \begin{align} - I_T - & = \frac{1}{2}(I_R - (H_L - 2)) \nonumber \\ - & = \frac{1}{2}(I_R - (2 - 2)) - & \text{\nameref{par:exercise-7a-i}} \nonumber \\ - & = \frac{1}{2}I_R. & \label{sub:exercise-7a-eq2} - \end{align} - Furthermore, since both the adjacent and opposite side of $T$ are not - included in $T$ and there exist no lattice points on $T$'s hypotenuse - besides the endpoints, it follows - \begin{equation} - \label{sub:exercise-7a-eq3} - B_T = 0. - \end{equation} - Thus the number of lattice points of $T$ equals - \begin{align} - I_T + B_T - & = I_T & \eqref{sub:exercise-7a-eq3} \nonumber \\ - & = \frac{1}{2}I_R & \eqref{sub:exercise-7a-eq2} \nonumber \\ - & = \frac{(b - 1)(a - 1)}{2}. - & \text{\nameref{C:1:07:sub:exercise-4a}} - \label{sub:exercise-7a-eq4} - \end{align} - - \paragraph{Conclusion}% - - By \eqref{sub:exercise-7a-eq1} the number of lattice points of $S$ is equal - to the sum $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}}.$$ - But the number of lattice points of $S$ is the same as that of $T$. - By \eqref{sub:exercise-7a-eq4}, the number of lattice points in $T$ is equal - to $$\frac{(b - 1)(a - 1)}{2}.$$ - Thus $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$ - -\end{proof} - -\subsection*{\partial{Exercise 7b}}% -\label{sub:exercise-7b} - -Derive the result analytically as follows: -By changing the index of summation, note that - $\sum_{n=1}^{b-1} \floor{na / b} = \sum_{n=1}^{b-1} \floor{a(b - n) / b}$. -Now apply Exercises 4(a) and (b) to the bracket on the right. - -\begin{proof} - - \lean{exercise\_7b} - - \divider - - Let $n = 1, \ldots, b - 1$. - By hypothesis, $a$ and $b$ are coprime. - Furthermore, $n < b$ for all values of $n$. - Thus $an / b$ is not an integer. - By \nameref{sub:exercise-4b}, - \begin{equation} - \label{sub:exercise-7b-eq1} - \floor{-\frac{an}{b}} = -\floor{\frac{an}{b}} - 1. - \end{equation} - Consider the following: - \begin{align*} - \sum_{n=1}^{b-1} \floor{\frac{na}{b}} - & = \sum_{n=1}^{b-1} \floor{\frac{a(b - n)}{b}} \\ - & = \sum_{n=1}^{b-1} \floor{\frac{ab - an}{b}} \\ - & = \sum_{n=1}^{b-1} \floor{-\frac{an}{b} + a} \\ - & = \sum_{n=1}^{b-1} \floor{-\frac{an}{b}} + a. - & \text{\nameref{sub:exercise-4a}} \\ - & = \sum_{n=1}^{b-1} -\floor{\frac{an}{b}} - 1 + a - & \eqref{sub:exercise-7b-eq1} \\ - & = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - \sum_{n=1}^{b-1} 1 + - \sum_{n=1}^{b-1} a \\ - & = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - (b - 1) + a(b - 1). - \end{align*} - Rearranging the above yields - $$2\sum_{n=1}^{b-1} \floor{\frac{an}{b}} = (a - 1)(b - 1).$$ - Dividing both sides of the above identity concludes the proof. - -\end{proof} - -\section*{\partial{Exercise 8}}% -\label{sec:exercise-8} - -Let $S$ be a set of points on the real line. -The \textit{characteristic function} of $S$ is, by definition, the function - $\mathcal{X}_S$ such that $\mathcal{X}_S(x) = 1$ for every $x$ in $S$, and - $\mathcal{X}_S(x) = 0$ for those $x$ not in $S$. -Let $f$ be a step function which takes the constant value $c_k$ on the $k$th - open subinterval $I_k$ of some partition of an interval $[a, b]$. -Prove that for each $x$ in the union $I_1 \cup I_2 \cup \cdots \cup I_n$ we have - $$f(x) = \sum_{k=1}^n c_k\mathcal{X}_{I_k}(x).$$ -This property is described by saying that every step function is a linear - combination of characteristic functions of intervals. - -\begin{proof} - - Let $x \in I_1 \cup I_2 \cup \cdots \cup I_n$ and $N = \{1, \ldots, n\}$. - Let $k \in N$ such that $x \in I_k$. - Consider an arbitrary $j \in N - \{k\}$. - By definition of a partition, $I_j \cap I_k = \emptyset$. - That is, $I_j$ and $I_k$ are disjoint for all $j \in N - \{k\}$. - Therefore, by definition of the characteristic function, - $\mathcal{X}_{I_k}(x) = 1$ and $\mathcal{X}_{I_j}(x) = 0$ for all - $j \in N - \{k\}$. - Thus - \begin{align*} - f(x) - & = c_k \\ - & = (c_k)(1) + \sum\nolimits_{j \in N - \{k\}} (c_j)(0) \\ - & = c_k\mathcal{X}_{I_k}(x) + - \sum\nolimits_{j \in N - \{k\}} c_j\mathcal{X}_{I_j}(x) \\ - & = \sum_{k=1}^n c_k\mathcal{X}_{I_k}(x). - \end{align*} - -\end{proof} - -\end{document} diff --git a/Bookshelf/Apostol/Chapter_I_03.tex b/Bookshelf/Apostol/Chapter_I_03.tex deleted file mode 100644 index c4838b7..0000000 --- a/Bookshelf/Apostol/Chapter_I_03.tex +++ /dev/null @@ -1,405 +0,0 @@ -\documentclass{article} - -\input{../../preamble} - -\newcommand{\lean}[1]{\leanref - {./Chapter\_I\_03.html\#Apostol.Chapter\_I\_03.#1} - {Apostol.Chapter\_I\_03.#1}} - -\begin{document} - -\header{A Set of Axioms for the Real-Number System}{Tom M. Apostol} - -\section*{\verified{Lemma 1}}% -\label{sec:lemma-1} - -Nonempty set $S$ has supremum $L$ if and only if set $-S$ has infimum $-L$. - -\begin{proof} - - \lean{is\_lub\_neg\_set\_iff\_is\_glb\_set\_neg} - - \divider - - Suppose $L = \sup{S}$ and fix $x \in S$. - By definition of the supremum, $x \leq L$ and $L$ is the smallest value - satisfying this inequality. - Negating both sides of the inequality yields $-x \geq -L$. - Furthermore, $-L$ must be the largest value satisfying this inequality. - Therefore $-L = \inf{-S}$. - -\end{proof} - -\section*{\verified{Theorem I.27}}% -\label{sec:theorem-i.27} - -Every nonempty set $S$ that is bounded below has a greatest lower bound; that - is, there is a real number $L$ such that $L = \inf{S}$. - -\begin{proof} - - \lean{exists\_isGLB} - - \divider - - Let $S$ be a nonempty set bounded below by $x$. - Then $-S$ is nonempty and bounded above by $x$. - By the completeness axiom, there exists a supremum $L$ of $-S$. - By \nameref{sec:lemma-1}, $L$ is a supremum of $-S$ if and only if $-L$ is an - infimum of $S$. - -\end{proof} - -\section*{\verified{Theorem I.29}}% -\label{sec:theorem-i.29} - -For every real $x$ there exists a positive integer $n$ such that $n > x$. - -\begin{proof} - - \lean{exists\_pnat\_geq\_self} - - \divider - - Let $n = \abs{\ceil{x}} + 1$. - It is trivial to see $n$ is a positive integer satisfying $n \geq 1$. - Thus all that remains to be shown is that $n > x$. - If $x$ is nonpositive, $n > x$ immediately follows from $n \geq 1$. - If $x$ is positive, - $$x = \abs{x} \leq \abs{\ceil{x}} < \abs{\ceil{x}} + 1 = n.$$ - -\end{proof} - -\section*{\verified{Theorem I.30}}% -\label{sec:theorem-i.30} - -If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive - integer $n$ such that $nx > y$. - -\note{This is known as the "Archimedean Property of the Reals."} - -\begin{proof} - - \lean{exists\_pnat\_mul\_self\_geq\_of\_pos} - - \divider - - Let $x > 0$ and $y$ be an arbitrary real number. - By \nameref{sec:theorem-i.29}, there exists a positive integer $n$ such that - $n > y / x$. - Multiplying both sides of the inequality yields $nx > y$ as expected. - -\end{proof} - -\section*{\verified{Theorem I.31}}% -\label{sec:theorem-i.31} - -If three real numbers $a$, $x$, and $y$ satisfy the inequalities - $$a \leq x \leq a + \frac{y}{n}$$ for every integer $n \geq 1$, then $x = a$. - -\begin{proof} - - \lean{forall\_pnat\_leq\_self\_leq\_frac\_imp\_eq} - - \divider - - By the trichotomy of the reals, there are three cases to consider: - - \paragraph{Case 1}% - - Suppose $x = a$. - Then we are immediately finished. - - \paragraph{Case 2}% - - Suppose $x < a$. - But by hypothesis, $a \leq x$. - Thus $a < a$, a contradiction. - - \paragraph{Case 3}% - - Suppose $x > a$. - Then there exists some $c > 0$ such that $a + c = x$. - By \nameref{sec:theorem-i.30}, there exists an integer $n > 0$ such that - $nc > y$. - Rearranging terms, we see $y / n < c$. - Therefore $a + y / n < a + c = x$. - But by hypothesis, $x \leq a + y / n$. - Thus $a + y / n < a + y / n$, a contradiction. - - \paragraph{Conclusion}% - - Since these cases are exhaustive and both case 2 and 3 lead to - contradictions, $x = a$ is the only possibility. - -\end{proof} - -\section*{\verified{Lemma 2}}% -\label{sec:lemma-2} - -If three real numbers $a$, $x$, and $y$ satisfy the inequalities - $$a - y / n \leq x \leq a$$ for every integer $n \geq 1$, then $x = a$. - -\begin{proof} - - \lean{forall\_pnat\_frac\_leq\_self\_leq\_imp\_eq} - - \divider - - By the trichotomy of the reals, there are three cases to consider: - - \paragraph{Case 1}% - - Suppose $x = a$. - Then we are immediately finished. - - \paragraph{Case 2}% - - Suppose $x < a$. - Then there exists some $c > 0$ such that $x = a - c$. - By \nameref{sec:theorem-i.30}, there exists an integer $n > 0$ such that - $nc > y$. - Rearranging terms, we see that $y / n < c$. - Therefore $a - y / n > a - c = x$. - But by hypothesis, $x \geq a - y / n$. - Thus $a - y / n < a - y / n$, a contradiction. - - \paragraph{Case 3}% - - Suppose $x > a$. - But by hypothesis $x \leq a$. - Thus $a < a$, a contradiction. - - \paragraph{Conclusion}% - - Since these cases are exhaustive and both case 2 and 3 lead to - contradictions, $x = a$ is the only possibility. - -\end{proof} - -\section*{Theorem I.32}% -\label{sec:theorem-i.32} - -Let $h$ be a given positive number and let $S$ be a set of real numbers. - -\subsection*{\verified{Theorem I.32a}}% -\label{sub:theorem-i.32a} - -If $S$ has a supremum, then for some $x$ in $S$ we have $x > \sup{S} - h$. - -\begin{proof} - - \lean{sup\_imp\_exists\_gt\_sup\_sub\_delta} - - \divider - - By definition of a supremum, $\sup{S}$ is the least upper bound of $S$. - For the sake of contradiction, suppose for all $x \in S$, - $x \leq \sup{S} - h$. - This immediately implies $\sup{S} - h$ is an upper bound of $S$. - But $\sup{S} - h < \sup{S}$, contradicting $\sup{S}$ being the \textit{least} - upper bound. - Therefore our original hypothesis was wrong. - That is, there exists some $x \in S$ such that $x > \sup{S} - h$. - -\end{proof} - -\subsection*{\verified{Theorem I.32b}}% -\label{sub:theorem-i.32b} - -If $S$ has an infimum, then for some $x$ in $S$ we have $x < \inf{S} + h$. - -\begin{proof} - - \lean{inf\_imp\_exists\_lt\_inf\_add\_delta} - - \divider - - By definition of an infimum, $\inf{S}$ is the greatest lower bound of $S$. - For the sake of contradiction, suppose for all $x \in S$, - $x \geq \inf{S} + h$. - This immediately implies $\inf{S} + h$ is a lower bound of $S$. - But $\inf{S} + h > \inf{S}$, contradicting $\inf{S}$ being the - \textit{greatest} lower bound. - Therefore our original hypothesis was wrong. - That is, there exists some $x \in S$ such that $x < \inf{S} + h$. - -\end{proof} - -\section*{Theorem I.33}% -\label{sec:theorem-i.33} - -Given nonempty subsets $A$ and $B$ of $\mathbb{R}$, let $C$ denote the set - $$C = \{a + b : a \in A, b \in B\}.$$ - -\note{This is known as the "Additive Property."} - -\subsection*{\verified{Theorem I.33a}}% -\label{sub:theorem-i.33a} - -If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and - $$\sup{C} = \sup{A} + \sup{B}.$$ - -\begin{proof} - - \lean{sup\_minkowski\_sum\_eq\_sup\_add\_sup} - - \divider - - We prove (i) $\sup{A} + \sup{B}$ is an upper bound of $C$ and (ii) - $\sup{A} + \sup{B}$ is the \textit{least} upper bound of $C$. - - \paragraph{(i)}% - \label{par:theorem-i.33a-i} - - Let $x \in C$. - By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such - that $x = a' + b'$. - By definition of a supremum, $a' \leq \sup{A}$. - Likewise, $b' \leq \sup{B}$. - Therefore $a' + b' \leq \sup{A} + \sup{B}$. - Since $x = a' + b'$ was arbitrarily chosen, it follows $\sup{A} + \sup{B}$ - is an upper bound of $C$. - - \paragraph{(ii)}% - - Since $A$ and $B$ have supremums, $C$ is nonempty. - By \nameref{par:theorem-i.33a-i}, $C$ is bounded above. - Therefore the completeness axiom tells us $C$ has a supremum. - Let $n > 0$ be an integer. - We now prove that - \begin{equation} - \label{par:theorem-i.33a-ii-eq1} - \sup{C} \leq \sup{A} + \sup{B} \leq \sup{C} + 1 / n. - \end{equation} - - \subparagraph{Left-Hand Side}% - - First consider the left-hand side of \eqref{par:theorem-i.33a-ii-eq1}. - By \nameref{par:theorem-i.33a-i}, $\sup{A} + \sup{B}$ is an upper bound of - $C$. - Since $\sup{C}$ is the \textit{least} upper bound of $C$, it follows - $\sup{C} \leq \sup{A} + \sup{B}$. - - \subparagraph{Right-Hand Side}% - - Next consider the right-hand side of \eqref{par:theorem-i.33a-ii-eq1}. - By \nameref{sub:theorem-i.32a}, there exists some $a' \in A$ such that - $\sup{A} < a' + 1 / (2n)$. - Likewise, there exists some $b' \in B$ such that - $\sup{B} < b' + 1 / (2n)$. - Adding these two inequalities together shows - \begin{align*} - \sup{A} + \sup{B} - & < a' + b' + 1 / n \\ - & \leq \sup{C} + 1 / n. - \end{align*} - - \subparagraph{Conclusion}% - - Applying \nameref{sec:theorem-i.31} to \eqref{par:theorem-i.33a-ii-eq1} - proves $\sup{C} = \sup{A} + \sup{B}$ as expected. - -\end{proof} - -\subsection*{\verified{Theorem I.33b}}% -\label{sub:theorem-i.33b} - -If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and - $$\inf{C} = \inf{A} + \inf{B}.$$ - -\begin{proof} - - \lean{inf\_minkowski\_sum\_eq\_inf\_add\_inf} - - \divider - - We prove (i) $\inf{A} + \inf{B}$ is a lower bound of $C$ and (ii) - $\inf{A} + \inf{B}$ is the \textit{greatest} lower bound of $C$. - - \paragraph{(i)}% - \label{par:theorem-i.33b-i} - - Let $x \in C$. - By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such - that $x = a' + b'$. - By definition of an infimum, $a' \geq \inf{A}$. - Likewise, $b' \geq \inf{B}$. - Therefore $a' + b' \geq \inf{A} + \inf{B}$. - Since $x = a' + b'$ was arbitrarily chosen, it follows $\inf{A} + \inf{B}$ - is a lower bound of $C$. - - \paragraph{(ii)}% - - Since $A$ and $B$ have infimums, $C$ is nonempty. - By \nameref{par:theorem-i.33b-i}, $C$ is bounded below. - Therefore \nameref{sec:theorem-i.27} tells us $C$ has an infimum. - Let $n > 0$ be an integer. - We now prove that - \begin{equation} - \label{par:theorem-i.33b-ii-eq1} - \inf{C} - 1 / n \leq \inf{A} + \inf{B} \leq \inf{C}. - \end{equation} - - \subparagraph{Right-Hand Side}% - - First consider the right-hand side of \eqref{par:theorem-i.33b-ii-eq1}. - By \nameref{par:theorem-i.33b-i}, $\inf{A} + \inf{B}$ is a lower bound of - $C$. - Since $\inf{C}$ is the \textit{greatest} upper bound of $C$, it follows - $\inf{C} \geq \inf{A} + \inf{B}$. - - \subparagraph{Left-Hand Side}% - - Next consider the left-hand side of \eqref{par:theorem-i.33b-ii-eq1}. - By \nameref{sub:theorem-i.32b}, there exists some $a' \in A$ such that - $\inf{A} > a' - 1 / (2n)$. - Likewise, there exists some $b' \in B$ such that - $\inf{B} > b' - 1 / (2n)$. - Adding these two inequalities together shows - \begin{align*} - \inf{A} + \inf{B} - & > a' + b' - 1 / n \\ - & \geq \inf{C} - 1 / n. - \end{align*} - - \subparagraph{Conclusion}% - - Applying \nameref{sec:lemma-2} to \eqref{par:theorem-i.33b-ii-eq1} - proves $\inf{C} = \inf{A} + \inf{B}$ as expected. - -\end{proof} - -\section*{\verified{Theorem I.34}}% -\label{sec:theorem-i.34} - -Given two nonempty subsets $S$ and $T$ of $\mathbb{R}$ such that $$s \leq t$$ - for every $s$ in $S$ and every $t$ in $T$. Then $S$ has a supremum, and $T$ - has an infimum, and they satisfy the inequality $$\sup{S} \leq \inf{T}.$$ - -\begin{proof} - - \lean{forall\_mem\_le\_forall\_mem\_imp\_sup\_le\_inf} - - \divider - - By hypothesis, $S$ and $T$ are nonempty sets. - Let $s \in S$ and $t \in T$. - Then $t$ is an upper bound of $S$ and $s$ is a lower bound of $T$. - By the completeness axiom, $S$ has a supremum. - By \nameref{sec:theorem-i.27}, $T$ has an infimum. - All that remains is showing $\sup{S} \leq \inf{T}$. - - For the sake of contradiction, suppose $\sup{S} > \inf{T}$. - Then there exists some $c > 0$ such that $\sup{S} = \inf{T} + c$. - Therefore $\inf{T} < \sup{S} - c / 2$. - By \nameref{sub:theorem-i.32a}, there exists some $x \in S$ such that - $\sup{S} - c / 2 < x$. - Thus $$\inf{T} < \sup{S} - c / 2 < x.$$ - But by hypothesis, $x \in S$ is a lower bound of $T$ meaning $x \leq \inf{T}$. - Therefore $x < x$, a contradiction. - Out original assumption is incorrect; that is, $\sup{S} \leq \inf{T}$. - -\end{proof} - -\end{document} diff --git a/Bookshelf/Apostol/Glossary.tex b/Bookshelf/Apostol/Glossary.tex deleted file mode 100644 index 69ec110..0000000 --- a/Bookshelf/Apostol/Glossary.tex +++ /dev/null @@ -1,57 +0,0 @@ -\documentclass{article} - -\input{../../preamble} - -\newcommand{\lean}[2]{\leanref{../../#1.html\##2}{#2}} - -\begin{document} - -\tableofcontents - -\section{The Concepts of Integral Calculus}% -\label{sec:concepts-integral-calculus} - -\subsection{\defined{Partition}}% -\label{sub:partition} - -Let $[a, b]$ be a closed interval decomposed into $n$ subintervals by inserting - $n - 1$ points of subdivision, say $x_1$, $x_2$, $\ldots$, $x_{n-1}$, subject - only to the restriction - \begin{equation} - \label{sec:partition-eq1} - a < x_1 < x_2 < \cdots < x_{n-1} < b. - \end{equation} -It is convenient to denote the point $a$ itself by $x_0$ and the point $b$ by - $x_n$. -A collection of points satisfying \eqref{sec:partition-eq1} is called a - \textbf{partition} $P$ of $[a, b]$, and we use the symbol - $$P = \{x_0, x_1, \ldots, x_n\}$$ to designate this partition. - -\begin{definition} - - \lean{Common/Set/Intervals/Partition}{Set.Intervals.Partition} - -\end{definition} - -\subsection{\defined{Step Function}}% -\label{sub:step-function} - -A function $s$, whose domain is a closed interval $[a, b]$, is called a step - function if there is a \nameref{sub:partition} $P = \{x_0, x_1, \ldots, x_n\}$ - of $[a b]$ such that $s$ is constant on each open subinterval of $P$. -That is to say, for each $k = 1, 2, \ldots, n$, there is a real number $s_k$ - such that $$s(x) = s_k \quad\text{if}\quad x_{k-1} < x < x_k.$$ -Step functions are sometimes called piecewise constant functions. - -\vspace{8pt} -\noindent -\textit{Note:} At each of the endpoints $x_{k-1}$ and $x_k$ the function must - have some well-defined value, but this need not be the same as $s_k$. - -\begin{definition} - - \lean{Common/Set/Intervals/StepFunction}{Set.Intervals.StepFunction} - -\end{definition} - -\end{document} diff --git a/Common/Real/Geometry/Area.tex b/Common/Real/Geometry/Area.tex deleted file mode 100644 index 60a9aeb..0000000 --- a/Common/Real/Geometry/Area.tex +++ /dev/null @@ -1,93 +0,0 @@ -\documentclass{article} - -\input{../../../preamble} - -\newcommand{\lean}[2]{\leanref{./Area.html\##1}{#2}} - -\begin{document} - -\header{Axiomatic Framework of Area}{Tom M. Apostol} - -We assume there exists a class $\mathscr{M}$ of measurable sets in the plane and - a set function $a$, whose domain is $\mathscr{M}$, with the following - properties: - -\section*{\defined{Nonnegative Property}}% -\label{sec:nonnegative-property} - -For each set $S$ in $\mathscr{M}$, we have $a(S) \geq 0$. - -\begin{axiom} - - \lean{Nonnegative-Property}{Nonnegative Property} - -\end{axiom} - -\section*{\defined{Additive Property}}% -\label{sec:additive-property} - -If $S$ and $T$ are in $\mathscr{M}$, then $S \cup T$ and $S \cap T$ are in - $\mathscr{M}$, and we have $a(S \cup T) = a(S) + a(T) - a(S \cap T)$. - -\begin{axiom} - - \lean{Additive-Property}{Additive Property} - -\end{axiom} - -\section*{\defined{Difference Property}}% -\label{sec:difference-property} - -If $S$ and $T$ are in $\mathscr{M}$ with $S \subseteq T$, then $T - S$ is in - $\mathscr{M}$, and we have $a(T - S) = a(T) - a(S)$. - -\begin{axiom} - - \lean{Difference-Property}{Difference Property} - -\end{axiom} - -\section*{\defined{Invariance Under Congruence}}% -\label{sec:invariance-under-congruence} - -If a set $S$ is in $\mathscr{M}$ and if $T$ is congruent to $S$, then $T$ is - also in $\mathscr{M}$ and we have $a(S) = a(T)$. - -\begin{axiom} - - \lean{Invariant-Under-Congruence}{Invariance Under Congruence} - -\end{axiom} - -\section*{\defined{Choice of Scale}}% -\label{sec:choice-scale} - -Every rectangle $R$ is in $\mathscr{M}$. -If the edges of $R$ have lengths $h$ and $k$, then $a(R) = hk$. - -\begin{axiom} - - \lean{Choice-of-Scale}{Choice of Scale} - -\end{axiom} - -\section*{\partial{Exhaustion Property}}% -\label{sec:exhaustion-property} - -Let $Q$ be a set that can be enclosed between two step regions $S$ and $T$, so - that - \begin{equation} - \label{sec:exhaustion-property-eq1} - S \subseteq Q \subseteq T. - \end{equation} -If there is one and only one number $c$ which satisfies the inequalities - $$a(S) \leq c \leq a(T)$$ for all step regions $S$ and $T$ satisfying (1.1), - then $Q$ is measurable and $a(Q) = c$. - -\begin{axiom} - - \lean{Exhaustion-Property}{Exhaustion Property} - -\end{axiom} - -\end{document} diff --git a/DocGen4/Output/Index.lean b/DocGen4/Output/Index.lean index 8bdf79d..fd9b16e 100644 --- a/DocGen4/Output/Index.lean +++ b/DocGen4/Output/Index.lean @@ -47,14 +47,14 @@ def index : BaseHtmlM Html := do templateExtends (baseHtml "Index") <| LaTeX and Lean.
  • - Magenta statements are reserved + Fuchsia statements are reserved for definitions, axioms, statements, theorems, lemmas, etc. that have been proven or encoded in LaTeX but not yet proven or encoded in Lean.
  • - Red serves as a catch-all for all - statements that don't fit the above categorizations. Incomplete + Maroon serves as a catch-all for + all statements that don't fit the above categorizations. Incomplete definitions, statements without proof, etc. belong here.
    • diff --git a/preamble.tex b/preamble.tex index da9f2cf..476ab0a 100644 --- a/preamble.tex +++ b/preamble.tex @@ -1,11 +1,12 @@ \usepackage{amsfonts, amsmath, amsthm} +\usepackage{comment} \usepackage[shortlabels]{enumitem} \usepackage{environ} \usepackage{fancybox} \usepackage{fontawesome5} \usepackage{mathrsfs} \usepackage{soul} -\usepackage{xcolor} +\usepackage[usenames,dvipsnames]{xcolor} % `hyperref` comes after `xr-hyper`. \usepackage{xr-hyper} \usepackage{hyperref} @@ -16,6 +17,7 @@ \hypersetup{colorlinks=true, linkcolor=blue, urlcolor=blue} \newcommand{\leanref}[2]{\textcolor{blue}{$\pmb{\exists}\;{-}\;$}\href{#1}{#2}} +\newcommand{\textref}[1]{\text{\nameref{#1}}} % ======================================== % Environments @@ -55,9 +57,9 @@ \DeclareRobustCommand{\verified}[1]{% \texorpdfstring{\color{teal}\faCheckCircle\ #1}{#1}} \DeclareRobustCommand{\partial}[1]{% - \texorpdfstring{\color{magenta}\faPencil*\ #1}{#1}} + \texorpdfstring{\color{Fuchsia}\faPencil*\ #1}{#1}} \DeclareRobustCommand{\unverified}[1]{% - \texorpdfstring{\color{red}\faExclamationCircle\ #1}{#1}} + \texorpdfstring{\color{Maroon}\faExclamationCircle\ #1}{#1}} % ======================================== % Math