diff --git a/Bookshelf/Apostol/Chapter_1_07.tex b/Bookshelf/Apostol/Chapter_1_07.tex index a057f13..e798ebf 100644 --- a/Bookshelf/Apostol/Chapter_1_07.tex +++ b/Bookshelf/Apostol/Chapter_1_07.tex @@ -320,8 +320,6 @@ Prove that every trapezoid and every parallelogram is measurable and derive the \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. @@ -334,16 +332,17 @@ Prove that the formula is valid for rectangles with sides parallel to the \begin{proof} - Let $P$ be a rectangle with width $w$, height $h$, and lattice points for - vertices. + 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 \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 - $a(P)$: + the expected result: \begin{align*} I + \frac{1}{2}B - 1 & = (w - 1)(h - 1) + \frac{1}{2}\left[ 2(w + h) \right] - 1 \\ @@ -361,67 +360,50 @@ 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 $P$ 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 $(0, 0)$. - There are two cases to consider: + Let $I_R$ and $B_R$ be the number of interior and boundary points + of $R$ respectively. - \paragraph{Case 1}% + 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 + 2 + B_T - 2 \right] \\ + & = \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[ (w - 1)(h - 1) + \frac{1}{2}(2(w + h)) - 1 \right] + & \eqref{sub:exercise-4a} \\ + & = \frac{1}{2}\left[ (w - 1)(h - 1) + w + h - 1 \right] \\ + & = \frac{1}{2}\left[ wh - w - h + 1 + w + h - 1 \right] \\ + & = \frac{wh}{2}. + \end{align*} - 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. + We do not prove this formula is valid for parallelograms here. + Instead, refer to \eqref{sub:exercise-4c} below. \end{proof} @@ -491,8 +473,8 @@ Use induction on the number of edges to construct a proof for general polygons. \end{proof} -\subsection{Exercise 5}% -\label{sub:exercise-5} +\section{Exercise 5}% +\label{sec:exercise-5} Prove that a triangle whose vertices are lattice points cannot be equilateral. @@ -527,8 +509,8 @@ ways, using Exercises 2 and 4.] \end{proof} -\subsection{Exercise 6}% -\label{sub:exercise-6} +\section{Exercise 6}% +\label{sec:exercise-6} Let $A = \{1, 2, 3, 4, 5\}$, and let $\mathscr{M}$ denote the class of all subsets of $A$. diff --git a/Bookshelf/Apostol/Chapter_1_11.tex b/Bookshelf/Apostol/Chapter_1_11.tex index 593ee86..8dfd3e2 100644 --- a/Bookshelf/Apostol/Chapter_1_11.tex +++ b/Bookshelf/Apostol/Chapter_1_11.tex @@ -97,15 +97,14 @@ State and prove such a generalization. \link{exercise\_5} - \vspace{6pt} + \vspace{10pt} \hrule - \vspace{6pt} + \vspace{10pt} We prove that for all natural numbers $n$ and real numbers $x$, the following identity holds: \begin{equation} \label{sec:exercise-5-eq1} - \tag{5.1} \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 @@ -117,7 +116,6 @@ State and prove such a generalization. Therefore there exists some $j \in \mathbb{N}$ such that \begin{equation} \label{sec:exercise-5-eq2} - \tag{5.2} 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 @@ -129,15 +127,15 @@ State and prove such a generalization. By \eqref{sec:exercise-5-eq2}, $nr \in \ico{j}{j + 1}$. Therefore $\floor{nr} = j$. Thus - \begin{align*} + \begin{align} \floor{nx} & = \floor{n(\floor{x} + r)} \nonumber \\ & = \floor{n\floor{x} + nr} \nonumber \\ & = \floor{n\floor{x}} + \floor{nr}. \nonumber & \eqref{sub:exercise-4a} \\ & = \floor{n\floor{x}} + j \nonumber \\ - & = n\floor{x} + j. \label{sec:exercise-5-eq3} \tag{5.3} - \end{align*} + & = n\floor{x} + j. \label{sec:exercise-5-eq3} + \end{align} \paragraph{Right-Hand Side}% @@ -150,7 +148,6 @@ State and prove such a generalization. we have \begin{equation} \label{sec:exercise-5-eq4} - \tag{5.4} \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 @@ -168,7 +165,6 @@ State and prove such a generalization. Substituting the value of $z$ into \eqref{sec:exercise-5-eq4} yields \begin{equation} \label{sec:exercise-5-eq5} - \tag{5.5} \sum_{i=0}^{n-1} \floor{x + \frac{i}{n}} = n\floor{x} + j. \end{equation}