From 5bfc41b1716e1df43244d9044121747db0936577 Mon Sep 17 00:00:00 2001 From: Joshua Potter Date: Mon, 8 May 2023 06:52:46 -0600 Subject: [PATCH] Finish apostol exercises 1.7. --- Exercises/Apostol/Exercises_1_7.tex | 487 ++++++++++++++++++- Exercises/Apostol/images/right-triangle.png | Bin 0 -> 14119 bytes Exercises/Apostol/images/trapezoid-cases.png | Bin 0 -> 9116 bytes 3 files changed, 461 insertions(+), 26 deletions(-) create mode 100644 Exercises/Apostol/images/right-triangle.png create mode 100644 Exercises/Apostol/images/trapezoid-cases.png diff --git a/Exercises/Apostol/Exercises_1_7.tex b/Exercises/Apostol/Exercises_1_7.tex index 51268f3..4b0b4a5 100644 --- a/Exercises/Apostol/Exercises_1_7.tex +++ b/Exercises/Apostol/Exercises_1_7.tex @@ -1,7 +1,10 @@ \documentclass{article} \usepackage{amsmath} +\usepackage{graphicx} +\usepackage{mathrsfs} \input{../../preamble} +\graphicspath{{./images/}} \newcommand{\larea}[2]{\lean{../..}{Bookshelf/Real/Geometry/Area}{#1}{#2}} \newcommand{\lrect}[2]{\lean{../..}{Bookshelf/Real/Geometry/Rectangle}{#1}{#2}} @@ -9,7 +12,7 @@ \begin{document} The properties of area in this set of exercises are to be deduced from the -axioms for area stated in the foregoing section. + axioms for area stated in the foregoing section. \section{Exercise 1}% \label{sec:exercise-1} @@ -40,20 +43,20 @@ A set consisting of a finite number of points in a plane. \begin{proof} - We show for all $k > 0$, a set consisting of $k$ points in a plane is - measurable with area $0$. + 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 \eqref{sub:exercise-1a}, $S$ is measurable with area $0$. + Thus $P(1)$ holds. \paragraph{Induction Step}% - Define our induction hypothesis as, "Let $k > 0$ and assume a set consisting - of $k$ points in a plane is measurable with area $0$." - - Consider a set $S_{k+1}$ consisting of $k + 1$ points in a plane. + 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 $S_k$. @@ -68,8 +71,6 @@ A set consisting of a finite number of points in a plane. & = 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} - - \noindent There are two cases to consider: \subparagraph{Case 1}% @@ -86,12 +87,11 @@ A set consisting of a finite number of points in a plane. \vspace{8pt} \noindent In both cases, \eqref{sub:exercise-1b-eq1} evaluates to $0$, implying - $a(S_{k+1}) = 0$ as expected. + $P(k + 1)$ as expected. \paragraph{Conclusion}% - By mathematical induction, it follows for all $n > 0$, a set consisting of - $n$ points in a plane is measurable with area $0$. + By mathematical induction, it follows for all $n > 0$, $P(n)$ is true. \end{proof} @@ -102,8 +102,9 @@ The union of a finite collection of line segments in a plane. \begin{proof} - We show for all $k > 0$, a set consisting of $k$ line segments in a plane is - measurable with area $0$. + 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}% @@ -113,13 +114,12 @@ The union of a finite collection of line segments in a plane. By \larea{Choice-of-Scale}{Choice of 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}% - Define our induction hypothesis as, "Let $k > 0$ and assume a set consisting - of $k$ line segments in a plane is measurable with area $0$." - - Consider a set $S_{k+1}$ consisting of $k + 1$ line segments in a plane. + 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 $S_k$. @@ -134,8 +134,6 @@ The union of a finite collection of line segments in a plane. & = 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} - - \noindent There are two cases to consider: \subparagraph{Case 1}% @@ -152,12 +150,11 @@ The union of a finite collection of line segments in a plane. \vspace{8pt} \noindent In both cases, \eqref{sub:exercise-1c-eq1} evaluates to $0$, implying - $a(S_{k+1}) = 0$ as expected. + $P(k + 1)$ as expected. \paragraph{Conclusion}% - By mathematical induction, it follows for all $n > 0$, a set consisting of - $n$ line segments in a plane is measurable with area $0$. + By mathematical induction, it follows for all $n > 0$, $P(n)$ is true. \end{proof} @@ -165,12 +162,450 @@ The union of a finite collection of line segments in a plane. \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. + 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} - TODO + 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 \larea{Choice-of-Scale}{Choice of Scale}, both $R$ and $S$ are measurable. + By this same axiom, $a(R) = ab$ and $a(S) = ca\sin{\theta}$. + By the \larea{Additive-Property}{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 \larea{Invariance-Under-Congruence}{Invariance Under Congruence}, + $a(T') = a(T)$, concluding our proof. + +\end{proof} + +\section{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 \larea{Choice-of-Scale}{Choice of Scale}, $R$ is measurable. + By \eqref{sec:exercise-2}, $T$ is measurable. + By the \larea{Additive-Property}{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 & \eqref{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 \eqref{sec:exercise-2}, $T$ is measurable. + By Case 1, $R$ is measurable. + By the \larea{Additive-Property}{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) & \eqref{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 \larea{Invariance-Under-Congruence}{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 $$a(S) = \frac{b_1 + b_2}{2}h.$$ + + \vspace{4pt} + \hrule + \vspace{10pt} + + 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 \larea{Invariance-Under-Congruence}{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} + +A point $(x, y)$ is called a \textit{lattice point} if both coordinates $x$ and + $y$ are integers. +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{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 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 \larea{Choice-of-Scale}{Choice of 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 + $a(P)$: + \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{Exercise 4b}% +\label{sub:exercise-4b} + +Prove that the formula is valid for right triangles and parallelograms. + +\begin{proof} + + Let $T'$ be a right triangle with width $w$ and height $h$. + Let $T$ be the triangle $T'$ translated, rotated, and reflected such that the + its vertices are $(0, 0)$, $(0, w)$, and $(w, h)$, where $w \leq h$. + Let $R$ be the overlapping rectangle of width $w$ and height $h$, situated + with bottom-left corner at $(0, 0)$. + There are two cases to consider: + + \paragraph{Case 1}% + + Suppose $h / w$ is an integral value. + Then there exist $w + 1$ lattice points on $T$'s hypotenuse. + The number of interior lattices points of $T$ is + \begin{align*} + I + & = \frac{1}{2}\left[ (w - 1)(h - 1) - (w - 1) \right] \\ + & = \frac{1}{2}\left[ wh - 2w - h + 2 \right]. + \end{align*} + The number of boundary lattice points of $T$ is + \begin{align*} + B + & = (w + 1) + h + (w - 1) \\ + & = 2w + h. + \end{align*} + Thus + \begin{align*} + I + \frac{1}{2}B - 1 + & = \frac{wh - 2w - h + 2}{2} + \frac{2w + h}{2} - 1 \\ + & = \frac{wh - 2w - h + 2 + 2w + h - 2}{2} \\ + & = \frac{wh}{2}. + \end{align*} + + \paragraph{Case 2}% + + Suppose $h / w$ is not an integral value. + Then there exist exactly 2 lattice points on $T$'s hypotenuse. + The number of interior lattice points of $T$ is + \begin{align*} + I + & = \frac{1}{2}\left[ (w - 1)(h - 1) \right] \\ + & = \frac{1}{2}\left[ wh - w - h + 1 \right]. + \end{align*} + The number of boundary lattice points of $T$ is + \begin{align*} + B + & = (w + 1) + h \\ + & = w + h + 1. + \end{align*} + Thus + \begin{align*} + I + \frac{1}{2}B - 1 + & = \frac{wh - w - h + 1}{2} + \frac{w + h + 1}{2} - 1 \\ + & = \frac{wh - w - h + 1 + w + h + 1 - 2}{2} \\ + & = \frac{wh}{2}. + \end{align*} + + \paragraph{Conclusion}% + + These cases are exhaustive and in agreement with one another. + Thus $$a(T) = I + \frac{1}{2}B - 1.$$ + We do not prove this formula is valid for parallelograms here. + Instead, refer to \eqref{sub:exercise-4c} below. + +\end{proof} + +\subsection{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 \eqref{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 \larea{Additive-Property}{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} + +\subsection{Exercise 5}% +\label{sub: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 \eqref{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 \eqref{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} + +\subsection{Exercise 6}% +\label{sub: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} diff --git a/Exercises/Apostol/images/right-triangle.png b/Exercises/Apostol/images/right-triangle.png new file mode 100644 index 0000000000000000000000000000000000000000..280106c8f5a2d75b9907ec88221071e5b87fd750 GIT binary patch literal 14119 zcmbWegk7(;@;Rn^kP#uTCOV2lm{;WYwO zTtw9)eSg7SH(~wu*YU9Pkl|t8r`NBazNA`tCuL^U&pfDBWSPpuB|n(J+gilSE;Tfu 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