595 lines
20 KiB
TeX
595 lines
20 KiB
TeX
\documentclass{article}
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\usepackage{graphicx}
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\input{../../preamble}
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\graphicspath{{./images/}}
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\externaldocument[A:]
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{../../Common/Real/Geometry/Area}
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[../../Common/Real/Geometry/Area.pdf]
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\begin{document}
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\header{Exercises 1.7}{Tom M. Apostol}
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The properties of area in this set of exercises are to be deduced from the
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axioms for area stated in the foregoing section.
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\section*{Exercise 1}%
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\hyperlabel{sec:exercise-1}%
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Prove that each of the following sets is measurable and has zero area:
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\subsection*{\proceeding{Exercise 1a}}%
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\hyperlabel{sub:exercise-1a}%
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A set consisting of a single point.
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\begin{proof}
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Let $S$ be a set consisting of a single point.
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By definition of a Point, $S$ is a rectangle in which all vertices coincide.
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By \nameref{A:sec:choice-scale}, $S$ is measurable with area its width times
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its height.
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The width and height of $S$ is trivially zero.
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Therefore $a(S) = (0)(0) = 0$.
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\end{proof}
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\subsection*{\proceeding{Exercise 1b}}%
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\hyperlabel{sub:exercise-1b}%
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A set consisting of a finite number of points in a plane.
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\begin{proof}
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Define predicate $P(n)$ as "A set consisting of $n$ points in a plane is
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measurable with area $0$".
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We use induction to prove $P(n)$ holds for all $n > 0$.
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\paragraph{Base Case}%
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Consider a set $S$ consisting of a single point in a plane.
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By \nameref{sub:exercise-1a}, $S$ is measurable with area $0$.
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Thus $P(1)$ holds.
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\paragraph{Induction Step}%
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Assume induction hypothesis $P(k)$ holds for some $k > 0$.
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Let $S_{k+1}$ be a set consisting of $k + 1$ points in a plane.
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Pick an arbitrary point of $S_{k+1}$.
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Denote the set containing just this point as $T$.
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Denote the remaining set of points as $S_k$.
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By construction, $S_{k+1} = S_k \cup T$.
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By the induction hypothesis, $S_k$ is measurable with area $0$.
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By \nameref{sub:exercise-1a}, $T$ is measurable with area $0$.
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By the \nameref{A:sec:additive-property}, $S_k \cup T$ is
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measurable, $S_k \cap T$ is measurable, and
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\begin{align}
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a(S_{k+1})
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& = a(S_k \cup T) \nonumber \\
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& = a(S_k) + a(T) - a(S_k \cap T) \nonumber \\
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& = 0 + 0 - a(S_k \cap T). \label{sub:exercise-1b-eq1}
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\end{align}
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There are two cases to consider:
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\subparagraph{Case 1}%
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$S_k \cap T = \emptyset$.
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Then it trivially follows that $a(S_k \cap T) = 0$.
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\subparagraph{Case 2}%
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$S_k \cap T \neq \emptyset$.
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Since $T$ consists of a single point, $S_k \cap T = T$.
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By \nameref{sub:exercise-1a}, $a(S_k \cap T) = a(T) = 0$.
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\vspace{8pt}
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\noindent
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In both cases, \eqref{sub:exercise-1b-eq1} evaluates to $0$, implying
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$P(k + 1)$ as expected.
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\paragraph{Conclusion}%
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By mathematical induction, it follows for all $n > 0$, $P(n)$ is true.
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\end{proof}
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\subsection*{\proceeding{Exercise 1c}}%
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\hyperlabel{sub:exercise-1c}%
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The union of a finite collection of line segments in a plane.
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\begin{proof}
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Define predicate $P(n)$ as "A set consisting of $n$ line segments in a plane
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is measurable with area $0$".
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We use induction to prove $P(n)$ holds for all $n > 0$.
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\paragraph{Base Case}%
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Consider a set $S$ consisting of a single line segment in a plane.
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By definition of a Line Segment, $S$ is a rectangle in which one side has
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dimension $0$.
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By \nameref{A:sec:choice-scale}, $S$ is measurable with area its width $w$
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times its height $h$.
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Therefore $a(S) = wh = 0$.
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Thus $P(1)$ holds.
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\paragraph{Induction Step}%
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Assume induction hypothesis $P(k)$ holds for some $k > 0$.
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Let $S_{k+1}$ be a set consisting of $k + 1$ line segments in a plane.
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Pick an arbitrary line segment of $S_{k+1}$.
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Denote the set containing just this line segment as $T$.
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Denote the remaining set of line segments as $S_k$.
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By construction, $S_{k+1} = S_k \cup T$.
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By the induction hypothesis, $S_k$ is measurable with area $0$.
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By the base case, $T$ is measurable with area $0$.
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By the \nameref{A:sec:additive-property}, $S_k \cup T$ is measurable,
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$S_k \cap T$ is measurable, and
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\begin{align}
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a(S_{k+1})
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& = a(S_k \cup T) \nonumber \\
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& = a(S_k) + a(T) - a(S_k \cap T) \nonumber \\
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& = 0 + 0 - a(S_k \cap T). \label{sub:exercise-1c-eq1}
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\end{align}
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There are two cases to consider:
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\subparagraph{Case 1}%
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$S_k \cap T = \emptyset$.
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Then it trivially follows that $a(S_k \cap T) = 0$.
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\subparagraph{Case 2}%
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$S_k \cap T \neq \emptyset$.
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Since $T$ consists of a single point, $S_k \cap T = T$.
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By the base case, $a(S_k \cap T) = a(T) = 0$.
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\vspace{8pt}
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\noindent
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In both cases, \eqref{sub:exercise-1c-eq1} evaluates to $0$, implying
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$P(k + 1)$ as expected.
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\paragraph{Conclusion}%
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By mathematical induction, it follows for all $n > 0$, $P(n)$ is true.
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\end{proof}
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\section*{\unverified{Exercise 2}}%
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\hyperlabel{sec:exercise-2}%
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Every right triangular region is measurable because it can be obtained as the
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intersection of two rectangles.
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Prove that every triangular region is measurable and that its area is one half
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the product of its base and altitude.
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\begin{proof}
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Let $T'$ be a triangular region with base of length $a$, height of length $b$,
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and hypotenuse of length $c$.
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Consider the translation and rotation of $T'$, say $T$, such that its
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hypotenuse is entirely within quadrant I and the vertex opposite the
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hypotenuse is situated at point $(0, 0)$.
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Let $R$ be a rectangle of width $a$, height $b$, and bottom-left corner at
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$(0, 0)$.
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By construction, $R$ covers all of $T$.
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Let $S$ be a rectangle of width $c$ and height $a\sin{\theta}$, where $\theta$
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is the acute angle measured from the bottom-right corner of $T$ relative
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to the $x$-axis.
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As an example, consider the image below of triangle $T$ with width $4$ and
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height $3$:
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\begin{figure}[h]
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\includegraphics{right-triangle}
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\centering
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\end{figure}
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By \nameref{A:sec:choice-scale}, both $R$ and $S$ are measurable.
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By this same axiom, $a(R) = ab$ and $a(S) = ca\sin{\theta}$.
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By the \nameref{A:sec:additive-property}, $R \cup S$ and $R \cap S$ are both
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measurable.
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$a(R \cap S) = a(T)$ and $a(R \cup S)$ can be determined by noting that
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$R$'s construction implies identity $a(R) = 2a(T)$.
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Therefore
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\begin{align*}
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a(T)
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& = a(R \cap S) \\
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& = a(R) + a(S) - a(R \cup S) \\
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& = ab + ca\sin{\theta} - a(R \cup S) \\
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& = ab + ca\sin{\theta} - (ca\sin{\theta} + \frac{1}{2}a(R)) \\
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& = ab + ca\sin{\theta} - ca\sin{\theta} - a(T).
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\end{align*}
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Solving for $a(T)$ gives the desired identity: $$a(T) = \frac{1}{2}ab.$$
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By \nameref{A:sec:invariance-under-congruence}, $a(T') = a(T)$, concluding our
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proof.
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\end{proof}
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\section*{\unverified{Exercise 3}}%
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\hyperlabel{sec:exercise-3}%
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Prove that every trapezoid and every parallelogram is measurable and derive the
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usual formulas for their areas.
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\begin{proof}
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We begin by proving the formula for a trapezoid.
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Let $S$ be a trapezoid with height $h$ and bases $b_1$ and $b_2$, $b_1 < b_2$.
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There are three cases to consider:
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\begin{figure}[h]
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\includegraphics[width=\textwidth]{trapezoid-cases}
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\centering
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\end{figure}
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\paragraph{Case 1}%
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Suppose $S$ is a right trapezoid.
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Then $S$ is the union of non-overlapping rectangle $R$ of width $b_1$ and
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height $h$ with right triangle $T$ of base $b_2 - b_1$ and height $h$.
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By \nameref{A:sec:choice-scale}, $R$ is measurable.
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By \nameref{sec:exercise-2}, $T$ is measurable.
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By the \nameref{A:sec:additive-property}, $R \cup T$ and $R \cap T$ are both
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measurable and
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\begin{align*}
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a(S)
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& = a(R \cup T) \\
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& = a(R) + a(T) - a(R \cap T) \\
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& = a(R) + a(T) & \text{by construction} \\
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& = b_1h + a(T) & \text{Choice of Scale} \\
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& = b_1h + \frac{1}{2}(b_2 - b_1)h
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& \text{\nameref{sec:exercise-2}} \\
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& = \frac{b_1 + b_2}{2}h.
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\end{align*}
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\paragraph{Case 2}%
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Suppose $S$ is an acute trapezoid.
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Then $S$ is the union of non-overlapping triangle $T$ and right trapezoid $R$.
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Let $c$ denote the length of base $T$.
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Then $R$ has longer base edge of length $b_2 - c$.
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By \nameref{sec:exercise-2}, $T$ is measurable.
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By Case 1, $R$ is measurable.
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By the \nameref{A:sec:additive-property}, $R \cup T$ and $R \cap T$ are both
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measurable and
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\begin{align*}
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a(S)
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& = a(T) + a(R) - a(R \cap T) \\
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& = a(T) + a(R) & \text{by construction} \\
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& = \frac{1}{2}ch + a(R) & \text{\nameref{sec:exercise-2}} \\
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& = \frac{1}{2}ch + \frac{b_1 + b_2 - c}{2}h & \text{Case 1} \\
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& = \frac{b_1 + b_2}{2}h.
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\end{align*}
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\paragraph{Case 3}%
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Suppose $S$ is an obtuse trapezoid.
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Then $S$ is the union of non-overlapping triangle $T$ and right trapezoid $R$.
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Let $c$ denote the length of base $T$.
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Reflect $T$ vertically to form another right triangle, say $T'$.
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Then $T' \cup R$ is an acute trapezoid.
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By \nameref{A:sec:invariance-under-congruence},
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\begin{equation}
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\label{par:exercise-3-case-3-eq1}
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\tag{3.1}
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a(T' \cup R) = a(T \cup R).
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\end{equation}
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By construction, $T' \cup R$ has height $h$ and bases $b_1 - c$ and $b_2 + c$
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meaning
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\begin{align*}
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a(T \cup R)
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& = a(T' \cup R) & \eqref{par:exercise-3-case-3-eq1} \\
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& = \frac{b_1 - c + b_2 + c}{2}h & \text{Case 2} \\
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& = \frac{b_1 + b_2}{2}h.
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\end{align*}
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\paragraph{Conclusion}%
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These cases are exhaustive and in agreement with one another.
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Thus $S$ is measurable and $$a(S) = \frac{b_1 + b_2}{2}h.$$
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\divider
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Let $P$ be a parallelogram with base $b$ and height $h$.
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Then $P$ is the union of non-overlapping triangle $T$ and right trapezoid $R$.
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Let $c$ denote the length of base $T$.
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Reflect $T$ vertically to form another right triangle, say $T'$.
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Then $T' \cup R$ is an acute trapezoid.
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By \nameref{A:sec:invariance-under-congruence},
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\begin{equation}
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\label{par:exercise-3-eq2}
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\tag{3.2}
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a(T' \cup R) = a(T \cup R).
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\end{equation}
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By construction, $T' \cup R$ has height $h$ and bases $b - c$ and $b + c$
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meaning
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\begin{align*}
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a(T \cup R)
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& = a(T' \cup R) & \eqref{par:exercise-3-eq2} \\
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& = \frac{b - c + b + c}{2}h & \text{Area of Trapezoid} \\
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& = bh.
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\end{align*}
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\end{proof}
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\section*{Exercise 4}%
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\hyperlabel{sec:exercise-4}%
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Let $P$ be a polygon whose vertices are lattice points.
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The area of $P$ is $I + \frac{1}{2}B - 1$, where $I$ denotes the number of
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lattice points inside the polygon and $B$ denotes the number on the boundary.
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\subsection*{\unverified{Exercise 4a}}%
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\hyperlabel{sub:exercise-4a}%
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Prove that the formula is valid for rectangles with sides parallel to the
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coordinate axes.
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\begin{proof}
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Let $P$ be a rectangle with sides parallel to the coordinate axes, with width
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$w$, height $h$, and lattice points for vertices.
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We assume $P$ has three non-collinear points, ruling out any instances of
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points or line segments.
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By \nameref{A:sec:choice-scale}, $P$ is measurable with area $a(P) = wh$.
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By construction, $P$ has $I = (w - 1)(h - 1)$ interior lattice points and
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$B = 2(w + h)$ lattice points on its boundary.
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The following shows the lattice point area formula is in agreement with
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the expected result:
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\begin{align*}
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I + \frac{1}{2}B - 1
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& = (w - 1)(h - 1) + \frac{1}{2}\left[ 2(w + h) \right] - 1 \\
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& = (wh - w - h + 1) + \frac{1}{2}\left[ 2(w + h) \right] - 1 \\
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& = (wh - w - h + 1) + (w + h) - 1 \\
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& = wh.
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\end{align*}
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\end{proof}
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\subsection*{\unverified{Exercise 4b}}%
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\hyperlabel{sub:exercise-4b}%
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Prove that the formula is valid for right triangles and parallelograms.
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\begin{proof}
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Let $P$ be a right triangle with width $w > 0$, height $h > 0$, and lattice
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points for vertices.
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Let $T$ be the triangle $P$ translated, rotated, and reflected such that the
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its vertices are $(0, 0)$, $(0, w)$, and $(w, h)$.
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Let $I_T$ and $B_T$ be the number of interior and boundary points of $T$
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respectively.
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Let $H_L$ denote the number of lattice points on $T$'s hypotenuse.
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Let $R$ be the overlapping rectangle of width $w$ and height $h$, situated
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with bottom-left corner at $(0, 0)$.
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Let $I_R$ and $B_R$ be the number of interior and boundary points
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of $R$ respectively.
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By construction, $T$ shares two sides with $R$.
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Therefore
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\begin{equation}
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\label{sub:exercise-4b-eq1}
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B_T = \frac{1}{2}B_R - 1 + H_L.
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\end{equation}
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Likewise,
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\begin{equation}
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\label{sub:exercise-4b-eq2}
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I_T = \frac{1}{2}(I_R - H_L + 2).
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\end{equation}
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The following shows the lattice point area formula is in agreement with
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the expected result:
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\begin{align*}
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I_T + \frac{1}{2}B_T - 1
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& = \frac{1}{2}(I_R - H_L + 2) + \frac{1}{2}B_T - 1
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& \eqref{sub:exercise-4b-eq2} \\
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& = \frac{1}{2}\left[ I_R - H_L + 2 + B_T - 2 \right] \\
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& = \frac{1}{2}\left[ I_R - H_L + B_T \right] \\
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& = \frac{1}{2}\left[ I_R - H_L + \frac{1}{2}B_R - 1 + H_L \right]
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& \eqref{sub:exercise-4b-eq1} \\
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& = \frac{1}{2}\left[ I_R + \frac{1}{2}B_R - 1 \right] \\
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& = \frac{1}{2}\left[ (w - 1)(h - 1) + \frac{1}{2}(2(w + h)) - 1 \right]
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& \text{\nameref{sub:exercise-4a}} \\
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& = \frac{1}{2}\left[ (w - 1)(h - 1) + w + h - 1 \right] \\
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& = \frac{1}{2}\left[ wh - w - h + 1 + w + h - 1 \right] \\
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& = \frac{wh}{2}.
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\end{align*}
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We do not prove this formula is valid for parallelograms here.
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Instead, refer to \nameref{sub:exercise-4c} below.
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\end{proof}
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\subsection*{\unverified{Exercise 4c}}%
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\hyperlabel{sub:exercise-4c}%
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Use induction on the number of edges to construct a proof for general polygons.
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\begin{proof}
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Define predicate $P(n)$ as "An $n$-polygon with vertices on lattice points has
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area $I + \frac{1}{2}B - 1$."
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We use induction to prove $P(n)$ holds for all $n \geq 3$.
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\paragraph{Base Case}%
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A $3$-polygon is a triangle.
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By \nameref{sub:exercise-4b}, the lattice point area formula holds.
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Thus $P(3)$ holds.
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\paragraph{Induction Step}%
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Assume induction hypothesis $P(k)$ holds for some $k \geq 3$.
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Let $P$ be a $(k + 1)$-polygon with vertices on lattice points.
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Such a polygon is equivalent to the union of a $k$-polygon $S$ with a
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triangle $T$.
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That is, $P = S \cup T$.
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Let $I_P$ be the number of interior lattice points of $P$.
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Let $B_P$ be the number of boundary lattice points of $P$.
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Similarly, let $I_S$, $I_T$, $B_S$, and $B_T$ be the number of interior
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and boundary lattice points of $S$ and $T$.
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Let $c$ denote the number of boundary points shared between $S$ and $T$.
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By our induction hypothesis, $a(S) = I_S + \frac{1}{2}B_S - 1$.
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By our base case, $a(T) = I_T + \frac{1}{2}B_T - 1$.
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By construction, it follows:
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\begin{align*}
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I_P & = I_S + I_T + c - 2 \\
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B_P & = B_S + B_T - (c - 2) - c \\
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& = B_S + B_T - 2c + 2.
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\end{align*}
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Applying the lattice point area formula to $P$ yields the following:
|
|
\begin{align*}
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|
& I_P + \frac{1}{2}B_P - 1 \\
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|
& = (I_S + I_T + c - 2) + \frac{1}{2}(B_S + B_T - 2c + 2) - 1 \\
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|
& = I_S + I_T + c - 2 + \frac{1}{2}B_S + \frac{1}{2}B_T - c + 1 - 1 \\
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& = (I_S + \frac{1}{2}B_S - 1) + (I_T + \frac{1}{2}B_T - 1) \\
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& = a(S) + (I_T + \frac{1}{2}B_T - 1) & \text{induction hypothesis} \\
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|
& = a(S) + a(T). & \text{base case}
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|
\end{align*}
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|
By the \nameref{A:sec:additive-property}, $S \cup T$ is measurable,
|
|
$S \cap T$ is measurable, and
|
|
\begin{align*}
|
|
a(P)
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|
& = a(S \cup T) \\
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|
& = a(S) + a(T) - a(S \cap T) \\
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& = a(S) + a(T). & \text{by construction}
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|
\end{align*}
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|
This shows the lattice point area formula is in agreement with our axiomatic
|
|
definition of area.
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|
Thus $P(k + 1)$ holds.
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|
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|
\paragraph{Conclusion}%
|
|
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|
By mathematical induction, it follows for all $n \geq 3$, $P(n)$ is true.
|
|
|
|
\end{proof}
|
|
|
|
\section*{\unverified{Exercise 5}}%
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|
\hyperlabel{sec:exercise-5}%
|
|
|
|
Prove that a triangle whose vertices are lattice points cannot be equilateral.
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|
|
|
[\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.
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|
Assume each side of $T$ has length $a$.
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|
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}}%
|
|
\hyperlabel{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}
|