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Simplify proof of Apostol 1.7.4.

finite-set-exercises
Joshua Potter 2023-05-09 16:28:45 -06:00
parent 22e2e6af2a
commit 123bdbdc20
2 changed files with 53 additions and 75 deletions
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114
Bookshelf/Apostol/Chapter_1_07.tex
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@ -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 $A$.

14
Bookshelf/Apostol/Chapter_1_11.tex
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@ -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}