Finish apostol exercises 1.7.

finite-set-exercises
Joshua Potter 2023-05-08 06:52:46 -06:00
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\documentclass{article} \documentclass{article}
\usepackage{amsmath} \usepackage{amsmath}
\usepackage{graphicx}
\usepackage{mathrsfs}
\input{../../preamble} \input{../../preamble}
\graphicspath{{./images/}}
\newcommand{\larea}[2]{\lean{../..}{Bookshelf/Real/Geometry/Area}{#1}{#2}} \newcommand{\larea}[2]{\lean{../..}{Bookshelf/Real/Geometry/Area}{#1}{#2}}
\newcommand{\lrect}[2]{\lean{../..}{Bookshelf/Real/Geometry/Rectangle}{#1}{#2}} \newcommand{\lrect}[2]{\lean{../..}{Bookshelf/Real/Geometry/Rectangle}{#1}{#2}}
@ -9,7 +12,7 @@
\begin{document} \begin{document}
The properties of area in this set of exercises are to be deduced from the 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}% \section{Exercise 1}%
\label{sec:exercise-1} \label{sec:exercise-1}
@ -40,20 +43,20 @@ A set consisting of a finite number of points in a plane.
\begin{proof} \begin{proof}
We show for all $k > 0$, a set consisting of $k$ points in a plane is Define predicate $P(n)$ as "A set consisting of $n$ points in a plane is
measurable with area $0$. measurable with area $0$".
We use induction to prove $P(n)$ holds for all $n > 0$.
\paragraph{Base Case}% \paragraph{Base Case}%
Consider a set $S$ consisting of a single point in a plane. Consider a set $S$ consisting of a single point in a plane.
By \eqref{sub:exercise-1a}, $S$ is measurable with area $0$. By \eqref{sub:exercise-1a}, $S$ is measurable with area $0$.
Thus $P(1)$ holds.
\paragraph{Induction Step}% \paragraph{Induction Step}%
Define our induction hypothesis as, "Let $k > 0$ and assume a set consisting Assume induction hypothesis $P(k)$ holds for some $k > 0$.
of $k$ points in a plane is measurable with area $0$." Let $S_{k+1}$ be a set consisting of $k + 1$ points in a plane.
Consider a set $S_{k+1}$ consisting of $k + 1$ points in a plane.
Pick an arbitrary point of $S_{k+1}$. Pick an arbitrary point of $S_{k+1}$.
Denote the set containing just this point as $T$. Denote the set containing just this point as $T$.
Denote the remaining set of points as $S_k$. 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 \\ & = a(S_k) + a(T) - a(S_k \cap T) \nonumber \\
& = 0 + 0 - a(S_k \cap T). \label{sub:exercise-1b-eq1} & = 0 + 0 - a(S_k \cap T). \label{sub:exercise-1b-eq1}
\end{align} \end{align}
\noindent
There are two cases to consider: There are two cases to consider:
\subparagraph{Case 1}% \subparagraph{Case 1}%
@ -86,12 +87,11 @@ A set consisting of a finite number of points in a plane.
\vspace{8pt} \vspace{8pt}
\noindent \noindent
In both cases, \eqref{sub:exercise-1b-eq1} evaluates to $0$, implying 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}% \paragraph{Conclusion}%
By mathematical induction, it follows for all $n > 0$, a set consisting of By mathematical induction, it follows for all $n > 0$, $P(n)$ is true.
$n$ points in a plane is measurable with area $0$.
\end{proof} \end{proof}
@ -102,8 +102,9 @@ The union of a finite collection of line segments in a plane.
\begin{proof} \begin{proof}
We show for all $k > 0$, a set consisting of $k$ line segments in a plane is Define predicate $P(n)$ as "A set consisting of $n$ line segments in a plane
measurable with area $0$. is measurable with area $0$".
We use induction to prove $P(n)$ holds for all $n > 0$.
\paragraph{Base Case}% \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 By \larea{Choice-of-Scale}{Choice of Scale}, $S$ is measurable with area its
width $w$ times its height $h$. width $w$ times its height $h$.
Therefore $a(S) = wh = 0$. Therefore $a(S) = wh = 0$.
Thus $P(1)$ holds.
\paragraph{Induction Step}% \paragraph{Induction Step}%
Define our induction hypothesis as, "Let $k > 0$ and assume a set consisting Assume induction hypothesis $P(k)$ holds for some $k > 0$.
of $k$ line segments in a plane is measurable with area $0$." Let $S_{k+1}$ be a set consisting of $k + 1$ line segments in a plane.
Consider a set $S_{k+1}$ consisting of $k + 1$ line segments in a plane.
Pick an arbitrary line segment of $S_{k+1}$. Pick an arbitrary line segment of $S_{k+1}$.
Denote the set containing just this line segment as $T$. Denote the set containing just this line segment as $T$.
Denote the remaining set of line segments as $S_k$. 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 \\ & = a(S_k) + a(T) - a(S_k \cap T) \nonumber \\
& = 0 + 0 - a(S_k \cap T). \label{sub:exercise-1c-eq1} & = 0 + 0 - a(S_k \cap T). \label{sub:exercise-1c-eq1}
\end{align} \end{align}
\noindent
There are two cases to consider: There are two cases to consider:
\subparagraph{Case 1}% \subparagraph{Case 1}%
@ -152,12 +150,11 @@ The union of a finite collection of line segments in a plane.
\vspace{8pt} \vspace{8pt}
\noindent \noindent
In both cases, \eqref{sub:exercise-1c-eq1} evaluates to $0$, implying 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}% \paragraph{Conclusion}%
By mathematical induction, it follows for all $n > 0$, a set consisting of By mathematical induction, it follows for all $n > 0$, $P(n)$ is true.
$n$ line segments in a plane is measurable with area $0$.
\end{proof} \end{proof}
@ -165,12 +162,450 @@ The union of a finite collection of line segments in a plane.
\label{sec:exercise-2} \label{sec:exercise-2}
Every right triangular region is measurable because it can be obtained as the Every right triangular region is measurable because it can be obtained as the
intersection of two rectangles. Prove that every triangular region is measurable intersection of two rectangles.
and that its area is one half the product of its base and altitude. Prove that every triangular region is measurable and that its area is one half
the product of its base and altitude.
\begin{proof} \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} \end{proof}

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