Aggregate Apostol LaTeX into single file.
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\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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\label{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*{\unverified{Exercise 1a}}%
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\label{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*{\unverified{Exercise 1b}}%
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\label{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*{\unverified{Exercise 1c}}%
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\label{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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\label{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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\label{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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\label{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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\label{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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\label{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 + 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[ wh \right] & \text{\nameref{sub:exercise-4a}}.
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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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\label{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.
|
||||
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 \nameref{A:sec: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}
|
||||
|
||||
\section*{\unverified{Exercise 5}}%
|
||||
\label{sec: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 \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}}%
|
||||
\label{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}
|
|
@ -1,471 +0,0 @@
|
|||
\documentclass{article}
|
||||
|
||||
\input{../../preamble}
|
||||
|
||||
\externaldocument[C:1:07:]{Chapter_1_07}[Chapter_1_07.pdf]
|
||||
|
||||
\newcommand{\lean}[1]{\leanref
|
||||
{./Chapter\_1\_11.html\#Apostol.Chapter\_1\_11.#1}
|
||||
{Apostol.Chapter\_1\_11.#1}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\header{Exercises 1.11}{Tom M. Apostol}
|
||||
|
||||
\section*{Exercise 4}%
|
||||
\label{sec:exercise-4}
|
||||
|
||||
Prove that the greatest-integer function has the properties indicated:
|
||||
|
||||
\subsection*{\verified{Exercise 4a}}%
|
||||
\label{sub:exercise-4a}
|
||||
|
||||
$\floor{x + n} = \floor{x} + n$ for every integer $n$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{exercise\_4a}
|
||||
|
||||
\divider
|
||||
|
||||
Let $x$ be a real number and $n$ an integer.
|
||||
Let $m = \floor{x + n}$.
|
||||
By definition of the floor function, $m$ is the unique integer such that
|
||||
$m \leq x + n < m + 1$.
|
||||
Then $m - n \leq x < (m - n) + 1$.
|
||||
That is, $m - n = \floor{x}$.
|
||||
Rearranging terms we see that $m = \floor{x} + n$ as expected.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection*{\verified{Exercise 4b}}%
|
||||
\label{sub:exercise-4b}
|
||||
|
||||
$\floor{-x} =
|
||||
\begin{cases}
|
||||
-\floor{x} & \text{if } x \text{ is an integer}, \\
|
||||
-\floor{x} - 1 & \text{otherwise}.
|
||||
\end{cases}$
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\ \vspace{6pt}
|
||||
|
||||
\lean{exercise\_4b\_1}
|
||||
|
||||
\lean{exercise\_4b\_2}
|
||||
|
||||
\divider
|
||||
|
||||
There are two cases to consider:
|
||||
|
||||
\paragraph{Case 1}%
|
||||
|
||||
Suppose $x$ is an integer.
|
||||
Then $x = \floor{x}$ and $-x = \floor{-x}$.
|
||||
It immediately follows that $$\floor{-x} = -x = -\floor{x}.$$
|
||||
|
||||
\paragraph{Case 2}%
|
||||
|
||||
Suppose $x$ is not an integer.
|
||||
Let $m = \floor{-x}$.
|
||||
By definition of the floor function, $m$ is the unique integer such that
|
||||
$m \leq -x < m + 1$.
|
||||
Equivalently, $-m - 1 < x \leq -m$.
|
||||
Since $x$ is not an integer, it follows $-m - 1 \leq x < -m$.
|
||||
Then, by definition of the floor function, $\floor{x} = -m - 1$.
|
||||
Solving for $m$ yields $$\floor{-x} = m = -\floor{x} - 1.$$
|
||||
|
||||
\paragraph{Conclusion}%
|
||||
|
||||
The above two cases are exhaustive. Thus
|
||||
$$\floor{-x} =
|
||||
\begin{cases}
|
||||
-\floor{x} & \text{if } x \text{ is an integer}, \\
|
||||
-\floor{x} - 1 & \text{otherwise}.
|
||||
\end{cases}$$
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection*{\verified{Exercise 4c}}%
|
||||
\label{sub:exercise-4c}
|
||||
|
||||
$\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{exercise\_4c}
|
||||
|
||||
\divider
|
||||
|
||||
Rewrite $x$ and $y$ as the sum of their floor and fractional components:
|
||||
$x = \floor{x} + \{x\}$ and $y = \floor{y} + \{y\}$.
|
||||
Now
|
||||
\begin{align}
|
||||
\floor{x + y}
|
||||
& = \floor{\floor{x} + \{x\} + \floor{y} + \{y\}} \nonumber \\
|
||||
& = \floor{\floor{x} + \floor{y} + \{x\} + \{y\}} \nonumber \\
|
||||
& = \floor{x} + \floor{y} + \floor{\{x\} + \{y\}}
|
||||
& \text{\nameref{sub:exercise-4a}} \label{sub:exercise-4c-eq1}
|
||||
\end{align}
|
||||
There are two cases to consider:
|
||||
|
||||
\paragraph{Case 1}%
|
||||
|
||||
Suppose $\{x\} + \{y\} < 1$.
|
||||
Then $\floor{\{x\} + \{y\}} = 0$.
|
||||
Substituting this value into \eqref{sub:exercise-4c-eq1} yields
|
||||
$$\floor{x + y} = \floor{x} + \floor{y}.$$
|
||||
|
||||
\paragraph{Case 2}%
|
||||
|
||||
Suppose $\{x\} + \{y\} \geq 1$.
|
||||
Because $\{x\}$ and $\{y\}$ are both less than $1$, $\{x\} + \{y\} < 2$.
|
||||
Thus $\floor{\{x\} + \{y\}} = 1$.
|
||||
Substituting this value into \eqref{sub:exercise-4c-eq1} yields
|
||||
$$\floor{x + y} = \floor{x} + \floor{y} + 1.$$
|
||||
|
||||
\paragraph{Conclusion}%
|
||||
|
||||
Since the above two cases are exhaustive, it follows
|
||||
$\floor{x + y} = \floor{x} + \floor{y}$ or $\floor{x} + \floor{y} + 1$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection*{\partial{Exercise 4d}}%
|
||||
\label{sub:exercise-4d}
|
||||
|
||||
$\floor{2x} = \floor{x} + \floor{x + \frac{1}{2}}.$
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{exercise\_4d}
|
||||
|
||||
\divider
|
||||
|
||||
This is immediately proven by applying Hermite's Identity as shown in
|
||||
\nameref{sec:exercise-5}.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection*{\partial{Exercise 4e}}%
|
||||
\label{sub:exercise-4e}
|
||||
|
||||
$\floor{3x} = \floor{x} + \floor{x + \frac{1}{3}} + \floor{x + \frac{2}{3}}.$
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{exercise\_4e}
|
||||
|
||||
\divider
|
||||
|
||||
This is immediately proven by applying Hermite's Identity as shown in
|
||||
\nameref{sec:exercise-5}.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\partial{Exercise 5}}%
|
||||
\label{sec:exercise-5}
|
||||
|
||||
The formulas in Exercises 4(d) and 4(e) suggest a generalization for
|
||||
$\floor{nx}$.
|
||||
State and prove such a generalization.
|
||||
|
||||
\note{The stated generalization is known as "Hermite's Identity."}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{exercise\_5}
|
||||
|
||||
\divider
|
||||
|
||||
We prove that for all natural numbers $n$ and real numbers $x$, the following
|
||||
identity holds:
|
||||
\begin{equation}
|
||||
\label{sec:exercise-5-eq1}
|
||||
\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
|
||||
$r \in \ico{0}{1}$.
|
||||
Define $S$ as the partition of non-overlapping subintervals
|
||||
$$\ico{0}{\frac{1}{n}}, \ico{\frac{1}{n}}{\frac{2}{n}}, \ldots,
|
||||
\ico{\frac{n-1}{n}}{1}.$$
|
||||
By construction, $\cup\; S = \ico{0}{1}$.
|
||||
Therefore there exists some $j \in \mathbb{N}$ such that
|
||||
\begin{equation}
|
||||
\label{sec:exercise-5-eq2}
|
||||
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
|
||||
of \eqref{sec:exercise-5-eq1} evaluate to the same number.
|
||||
|
||||
\paragraph{Left-Hand Side}%
|
||||
|
||||
Consider the left-hand side of identity \eqref{sec:exercise-5-eq1}.
|
||||
By \eqref{sec:exercise-5-eq2}, $nr \in \ico{j}{j + 1}$.
|
||||
Therefore $\floor{nr} = j$.
|
||||
Thus
|
||||
\begin{align}
|
||||
\floor{nx}
|
||||
& = \floor{n(\floor{x} + r)} \nonumber \\
|
||||
& = \floor{n\floor{x} + nr} \nonumber \\
|
||||
& = \floor{n\floor{x}} + \floor{nr}. \nonumber
|
||||
& \text{\nameref{sub:exercise-4a}} \\
|
||||
& = \floor{n\floor{x}} + j \nonumber \\
|
||||
& = n\floor{x} + j. \label{sec:exercise-5-eq3}
|
||||
\end{align}
|
||||
|
||||
\paragraph{Right-Hand Side}%
|
||||
|
||||
Now consider the right-hand side of identity \eqref{sec:exercise-5-eq1}.
|
||||
We note each summand, by construction, is the floor of $x$ added to a
|
||||
nonnegative number less than one.
|
||||
Therefore each summand contributes either $\floor{x}$ or $\floor{x} + 1$ to
|
||||
the total.
|
||||
Letting $z$ denote the number of summands that contribute $\floor{x} + 1$,
|
||||
we have
|
||||
\begin{equation}
|
||||
\label{sec:exercise-5-eq4}
|
||||
\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
|
||||
$$\frac{i}{n} \geq 1 - r.$$
|
||||
By \eqref{sec:exercise-5-eq2}, it follows
|
||||
\begin{align*}
|
||||
1 - r
|
||||
& \in \ioc{1 - \frac{j+1}{n}}{1-\frac{j}{n}} \\
|
||||
& = \ioc{\frac{n - j - 1}{n}}{\frac{n - j}{n}}.
|
||||
\end{align*}
|
||||
Thus we can determine the value of $z$ by instead counting the number of
|
||||
indices $i$ that satisfy $$\frac{i}{n} \geq \frac{n - j}{n}.$$
|
||||
Rearranging terms, we see that $i \geq n - j$ holds for
|
||||
$z = (n - 1) - (n - j) + 1 = j$ of the $n$ summands.
|
||||
Substituting the value of $z$ into \eqref{sec:exercise-5-eq4} yields
|
||||
\begin{equation}
|
||||
\label{sec:exercise-5-eq5}
|
||||
\sum_{i=0}^{n-1} \floor{x + \frac{i}{n}} = n\floor{x} + j.
|
||||
\end{equation}
|
||||
|
||||
\paragraph{Conclusion}%
|
||||
|
||||
Since \eqref{sec:exercise-5-eq3} and \eqref{sec:exercise-5-eq5} agree with
|
||||
one another, it follows identity \eqref{sec:exercise-5-eq1} holds.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\unverified{Exercise 6}}%
|
||||
\label{sec:exercise-6}
|
||||
|
||||
Recall that a lattice point $(x, y)$ in the plane is one whose coordinates are
|
||||
integers.
|
||||
Let $f$ be a nonnegative function whose domain is the interval $[a, b]$, where
|
||||
$a$ and $b$ are integers, $a < b$.
|
||||
Let $S$ denote the set of points $(x, y)$ satisfying $a \leq x \leq b$,
|
||||
$0 < y \leq f(x)$.
|
||||
Prove that the number of lattice points in $S$ is equal to the sum
|
||||
$$\sum_{n=a}^b \floor{f(n)}.$$
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Let $i = a, \ldots, b$ and define $S_i = \mathbb{N} \cap \ioc{0}{f(i)}$.
|
||||
By construction, the number of lattice points in $S$ is
|
||||
\begin{equation}
|
||||
\label{sec:exercise-6-eq1}
|
||||
\sum_{n = a}^b \abs{S_n}.
|
||||
\end{equation}
|
||||
All that remains is to show $\abs{S_i} = \floor{f(i)}$.
|
||||
There are two cases to consider:
|
||||
|
||||
\paragraph{Case 1}%
|
||||
|
||||
Suppose $f(i)$ is an integer.
|
||||
Then the number of integers in $\ioc{0}{f(i)}$ is $f(i) = \floor{f(i)}$.
|
||||
|
||||
\paragraph{Case 2}%
|
||||
|
||||
Suppose $f(i)$ is not an integer.
|
||||
Then the number of integers in $\ioc{0}{f(i)}$ is the same as that of
|
||||
$\ioc{0}{\floor{f(i)}}$.
|
||||
Once again, that number is $\floor{f(i)}$.
|
||||
|
||||
\paragraph{Conclusion}%
|
||||
|
||||
By cases 1 and 2, $\abs{S_i} = \floor{f(i)}$.
|
||||
Substituting this identity into \eqref{sec:exercise-6-eq1} finishes the
|
||||
proof.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{Exercise 7}%
|
||||
\label{sec:exercise-7}
|
||||
|
||||
If $a$ and $b$ are positive integers with no common factor, we have the formula
|
||||
$$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$
|
||||
When $b = 1$, the sum on the left is understood to be $0$.
|
||||
|
||||
\note{When $b = 1$, the proofs of (a) and (b) are trivial. We continue under the
|
||||
assumption $b > 1$.}
|
||||
|
||||
\subsection*{\unverified{Exercise 7a}}%
|
||||
\label{sub:exercise-7a}
|
||||
|
||||
Derive this result by a geometric argument, counting lattice points in a right
|
||||
triangle.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Let $f \colon [1, b - 1] \rightarrow \mathbb{R}$ be given by $f(x) = ax / b$.
|
||||
Let $S$ denote the set of points $(x, y)$ satisfying $1 \leq x \leq b - 1$,
|
||||
$0 < y \leq f(x)$.
|
||||
By \nameref{sec:exercise-6}, the number of lattice points of $S$ is equal to
|
||||
the sum
|
||||
\begin{equation}
|
||||
\label{sub:exercise-7a-eq1}
|
||||
\sum_{n=1}^{b-1} \floor{f(n)} = \sum_{n=1}^{b-1} \floor{\frac{na}{b}}.
|
||||
\end{equation}
|
||||
Define $T$ to be the triangle of width $w = b$ and height $h = f(b) = a$
|
||||
as $$T = \{ (x, y) : 0 < x < b, 0 < y \leq f(x) \}.$$
|
||||
By construction, $T$ does not introduce any additional lattice points.
|
||||
Thus $S$ and $T$ have the same number of lattice points.
|
||||
Let $H_L$ denote the number of boundary points on $T$'s hypotenuse.
|
||||
We prove that (i) $H_L = 2$ and (ii) that $T$ has $\frac{(a - 1)(b - 1)}{2}$
|
||||
lattice points.
|
||||
|
||||
\paragraph{(i)}%
|
||||
\label{par:exercise-7a-i}
|
||||
|
||||
Consider the line $L$ overlapping the hypotenuse of $T$.
|
||||
By construction, $T$'s hypotenuse has endpoints $(0, 0)$ and $(b, a)$.
|
||||
By hypothesis, $a$ and $b$ are positive, excluding the possibility of $L$
|
||||
being vertical.
|
||||
Define the slope of $L$ as $$m = \frac{a}{b}.$$
|
||||
$H_L$ coincides with the number of indices $i = 0, \ldots, b$ such that
|
||||
$(i, i * m)$ is a lattice point.
|
||||
But $a$ and $b$ are coprime by hypothesis and $i \leq b$.
|
||||
Thus $i * m$ is an integer if and only if $i = 0$ or $i = b$.
|
||||
Thus $H_L = 2$.
|
||||
|
||||
\paragraph{(ii)}%
|
||||
|
||||
Next we count the number of lattice points in $T$.
|
||||
Let $R$ be the overlapping retangle of width $w$ and height $h$, situated
|
||||
with bottom-left corner at $(0, 0)$.
|
||||
Let $I_R$ denote the number of interior lattice points of $R$.
|
||||
Let $I_T$ and $B_T$ denote the interior and boundary lattice points of $T$
|
||||
respectively.
|
||||
By \nameref{C:1:07:sub:exercise-4b-eq2},
|
||||
\begin{align}
|
||||
I_T
|
||||
& = \frac{1}{2}(I_R - (H_L - 2)) \nonumber \\
|
||||
& = \frac{1}{2}(I_R - (2 - 2))
|
||||
& \text{\nameref{par:exercise-7a-i}} \nonumber \\
|
||||
& = \frac{1}{2}I_R. & \label{sub:exercise-7a-eq2}
|
||||
\end{align}
|
||||
Furthermore, since both the adjacent and opposite side of $T$ are not
|
||||
included in $T$ and there exist no lattice points on $T$'s hypotenuse
|
||||
besides the endpoints, it follows
|
||||
\begin{equation}
|
||||
\label{sub:exercise-7a-eq3}
|
||||
B_T = 0.
|
||||
\end{equation}
|
||||
Thus the number of lattice points of $T$ equals
|
||||
\begin{align}
|
||||
I_T + B_T
|
||||
& = I_T & \eqref{sub:exercise-7a-eq3} \nonumber \\
|
||||
& = \frac{1}{2}I_R & \eqref{sub:exercise-7a-eq2} \nonumber \\
|
||||
& = \frac{(b - 1)(a - 1)}{2}.
|
||||
& \text{\nameref{C:1:07:sub:exercise-4a}}
|
||||
\label{sub:exercise-7a-eq4}
|
||||
\end{align}
|
||||
|
||||
\paragraph{Conclusion}%
|
||||
|
||||
By \eqref{sub:exercise-7a-eq1} the number of lattice points of $S$ is equal
|
||||
to the sum $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}}.$$
|
||||
But the number of lattice points of $S$ is the same as that of $T$.
|
||||
By \eqref{sub:exercise-7a-eq4}, the number of lattice points in $T$ is equal
|
||||
to $$\frac{(b - 1)(a - 1)}{2}.$$
|
||||
Thus $$\sum_{n=1}^{b-1} \floor{\frac{na}{b}} = \frac{(a - 1)(b - 1)}{2}.$$
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection*{\partial{Exercise 7b}}%
|
||||
\label{sub:exercise-7b}
|
||||
|
||||
Derive the result analytically as follows:
|
||||
By changing the index of summation, note that
|
||||
$\sum_{n=1}^{b-1} \floor{na / b} = \sum_{n=1}^{b-1} \floor{a(b - n) / b}$.
|
||||
Now apply Exercises 4(a) and (b) to the bracket on the right.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{exercise\_7b}
|
||||
|
||||
\divider
|
||||
|
||||
Let $n = 1, \ldots, b - 1$.
|
||||
By hypothesis, $a$ and $b$ are coprime.
|
||||
Furthermore, $n < b$ for all values of $n$.
|
||||
Thus $an / b$ is not an integer.
|
||||
By \nameref{sub:exercise-4b},
|
||||
\begin{equation}
|
||||
\label{sub:exercise-7b-eq1}
|
||||
\floor{-\frac{an}{b}} = -\floor{\frac{an}{b}} - 1.
|
||||
\end{equation}
|
||||
Consider the following:
|
||||
\begin{align*}
|
||||
\sum_{n=1}^{b-1} \floor{\frac{na}{b}}
|
||||
& = \sum_{n=1}^{b-1} \floor{\frac{a(b - n)}{b}} \\
|
||||
& = \sum_{n=1}^{b-1} \floor{\frac{ab - an}{b}} \\
|
||||
& = \sum_{n=1}^{b-1} \floor{-\frac{an}{b} + a} \\
|
||||
& = \sum_{n=1}^{b-1} \floor{-\frac{an}{b}} + a.
|
||||
& \text{\nameref{sub:exercise-4a}} \\
|
||||
& = \sum_{n=1}^{b-1} -\floor{\frac{an}{b}} - 1 + a
|
||||
& \eqref{sub:exercise-7b-eq1} \\
|
||||
& = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - \sum_{n=1}^{b-1} 1 +
|
||||
\sum_{n=1}^{b-1} a \\
|
||||
& = -\sum_{n=1}^{b-1} \floor{\frac{an}{b}} - (b - 1) + a(b - 1).
|
||||
\end{align*}
|
||||
Rearranging the above yields
|
||||
$$2\sum_{n=1}^{b-1} \floor{\frac{an}{b}} = (a - 1)(b - 1).$$
|
||||
Dividing both sides of the above identity concludes the proof.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\partial{Exercise 8}}%
|
||||
\label{sec:exercise-8}
|
||||
|
||||
Let $S$ be a set of points on the real line.
|
||||
The \textit{characteristic function} of $S$ is, by definition, the function
|
||||
$\mathcal{X}_S$ such that $\mathcal{X}_S(x) = 1$ for every $x$ in $S$, and
|
||||
$\mathcal{X}_S(x) = 0$ for those $x$ not in $S$.
|
||||
Let $f$ be a step function which takes the constant value $c_k$ on the $k$th
|
||||
open subinterval $I_k$ of some partition of an interval $[a, b]$.
|
||||
Prove that for each $x$ in the union $I_1 \cup I_2 \cup \cdots \cup I_n$ we have
|
||||
$$f(x) = \sum_{k=1}^n c_k\mathcal{X}_{I_k}(x).$$
|
||||
This property is described by saying that every step function is a linear
|
||||
combination of characteristic functions of intervals.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Let $x \in I_1 \cup I_2 \cup \cdots \cup I_n$ and $N = \{1, \ldots, n\}$.
|
||||
Let $k \in N$ such that $x \in I_k$.
|
||||
Consider an arbitrary $j \in N - \{k\}$.
|
||||
By definition of a partition, $I_j \cap I_k = \emptyset$.
|
||||
That is, $I_j$ and $I_k$ are disjoint for all $j \in N - \{k\}$.
|
||||
Therefore, by definition of the characteristic function,
|
||||
$\mathcal{X}_{I_k}(x) = 1$ and $\mathcal{X}_{I_j}(x) = 0$ for all
|
||||
$j \in N - \{k\}$.
|
||||
Thus
|
||||
\begin{align*}
|
||||
f(x)
|
||||
& = c_k \\
|
||||
& = (c_k)(1) + \sum\nolimits_{j \in N - \{k\}} (c_j)(0) \\
|
||||
& = c_k\mathcal{X}_{I_k}(x) +
|
||||
\sum\nolimits_{j \in N - \{k\}} c_j\mathcal{X}_{I_j}(x) \\
|
||||
& = \sum_{k=1}^n c_k\mathcal{X}_{I_k}(x).
|
||||
\end{align*}
|
||||
|
||||
\end{proof}
|
||||
|
||||
\end{document}
|
|
@ -1,405 +0,0 @@
|
|||
\documentclass{article}
|
||||
|
||||
\input{../../preamble}
|
||||
|
||||
\newcommand{\lean}[1]{\leanref
|
||||
{./Chapter\_I\_03.html\#Apostol.Chapter\_I\_03.#1}
|
||||
{Apostol.Chapter\_I\_03.#1}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\header{A Set of Axioms for the Real-Number System}{Tom M. Apostol}
|
||||
|
||||
\section*{\verified{Lemma 1}}%
|
||||
\label{sec:lemma-1}
|
||||
|
||||
Nonempty set $S$ has supremum $L$ if and only if set $-S$ has infimum $-L$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{is\_lub\_neg\_set\_iff\_is\_glb\_set\_neg}
|
||||
|
||||
\divider
|
||||
|
||||
Suppose $L = \sup{S}$ and fix $x \in S$.
|
||||
By definition of the supremum, $x \leq L$ and $L$ is the smallest value
|
||||
satisfying this inequality.
|
||||
Negating both sides of the inequality yields $-x \geq -L$.
|
||||
Furthermore, $-L$ must be the largest value satisfying this inequality.
|
||||
Therefore $-L = \inf{-S}$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\verified{Theorem I.27}}%
|
||||
\label{sec:theorem-i.27}
|
||||
|
||||
Every nonempty set $S$ that is bounded below has a greatest lower bound; that
|
||||
is, there is a real number $L$ such that $L = \inf{S}$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{exists\_isGLB}
|
||||
|
||||
\divider
|
||||
|
||||
Let $S$ be a nonempty set bounded below by $x$.
|
||||
Then $-S$ is nonempty and bounded above by $x$.
|
||||
By the completeness axiom, there exists a supremum $L$ of $-S$.
|
||||
By \nameref{sec:lemma-1}, $L$ is a supremum of $-S$ if and only if $-L$ is an
|
||||
infimum of $S$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\verified{Theorem I.29}}%
|
||||
\label{sec:theorem-i.29}
|
||||
|
||||
For every real $x$ there exists a positive integer $n$ such that $n > x$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{exists\_pnat\_geq\_self}
|
||||
|
||||
\divider
|
||||
|
||||
Let $n = \abs{\ceil{x}} + 1$.
|
||||
It is trivial to see $n$ is a positive integer satisfying $n \geq 1$.
|
||||
Thus all that remains to be shown is that $n > x$.
|
||||
If $x$ is nonpositive, $n > x$ immediately follows from $n \geq 1$.
|
||||
If $x$ is positive,
|
||||
$$x = \abs{x} \leq \abs{\ceil{x}} < \abs{\ceil{x}} + 1 = n.$$
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\verified{Theorem I.30}}%
|
||||
\label{sec:theorem-i.30}
|
||||
|
||||
If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive
|
||||
integer $n$ such that $nx > y$.
|
||||
|
||||
\note{This is known as the "Archimedean Property of the Reals."}
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{exists\_pnat\_mul\_self\_geq\_of\_pos}
|
||||
|
||||
\divider
|
||||
|
||||
Let $x > 0$ and $y$ be an arbitrary real number.
|
||||
By \nameref{sec:theorem-i.29}, there exists a positive integer $n$ such that
|
||||
$n > y / x$.
|
||||
Multiplying both sides of the inequality yields $nx > y$ as expected.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\verified{Theorem I.31}}%
|
||||
\label{sec:theorem-i.31}
|
||||
|
||||
If three real numbers $a$, $x$, and $y$ satisfy the inequalities
|
||||
$$a \leq x \leq a + \frac{y}{n}$$ for every integer $n \geq 1$, then $x = a$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{forall\_pnat\_leq\_self\_leq\_frac\_imp\_eq}
|
||||
|
||||
\divider
|
||||
|
||||
By the trichotomy of the reals, there are three cases to consider:
|
||||
|
||||
\paragraph{Case 1}%
|
||||
|
||||
Suppose $x = a$.
|
||||
Then we are immediately finished.
|
||||
|
||||
\paragraph{Case 2}%
|
||||
|
||||
Suppose $x < a$.
|
||||
But by hypothesis, $a \leq x$.
|
||||
Thus $a < a$, a contradiction.
|
||||
|
||||
\paragraph{Case 3}%
|
||||
|
||||
Suppose $x > a$.
|
||||
Then there exists some $c > 0$ such that $a + c = x$.
|
||||
By \nameref{sec:theorem-i.30}, there exists an integer $n > 0$ such that
|
||||
$nc > y$.
|
||||
Rearranging terms, we see $y / n < c$.
|
||||
Therefore $a + y / n < a + c = x$.
|
||||
But by hypothesis, $x \leq a + y / n$.
|
||||
Thus $a + y / n < a + y / n$, a contradiction.
|
||||
|
||||
\paragraph{Conclusion}%
|
||||
|
||||
Since these cases are exhaustive and both case 2 and 3 lead to
|
||||
contradictions, $x = a$ is the only possibility.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\verified{Lemma 2}}%
|
||||
\label{sec:lemma-2}
|
||||
|
||||
If three real numbers $a$, $x$, and $y$ satisfy the inequalities
|
||||
$$a - y / n \leq x \leq a$$ for every integer $n \geq 1$, then $x = a$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{forall\_pnat\_frac\_leq\_self\_leq\_imp\_eq}
|
||||
|
||||
\divider
|
||||
|
||||
By the trichotomy of the reals, there are three cases to consider:
|
||||
|
||||
\paragraph{Case 1}%
|
||||
|
||||
Suppose $x = a$.
|
||||
Then we are immediately finished.
|
||||
|
||||
\paragraph{Case 2}%
|
||||
|
||||
Suppose $x < a$.
|
||||
Then there exists some $c > 0$ such that $x = a - c$.
|
||||
By \nameref{sec:theorem-i.30}, there exists an integer $n > 0$ such that
|
||||
$nc > y$.
|
||||
Rearranging terms, we see that $y / n < c$.
|
||||
Therefore $a - y / n > a - c = x$.
|
||||
But by hypothesis, $x \geq a - y / n$.
|
||||
Thus $a - y / n < a - y / n$, a contradiction.
|
||||
|
||||
\paragraph{Case 3}%
|
||||
|
||||
Suppose $x > a$.
|
||||
But by hypothesis $x \leq a$.
|
||||
Thus $a < a$, a contradiction.
|
||||
|
||||
\paragraph{Conclusion}%
|
||||
|
||||
Since these cases are exhaustive and both case 2 and 3 lead to
|
||||
contradictions, $x = a$ is the only possibility.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{Theorem I.32}%
|
||||
\label{sec:theorem-i.32}
|
||||
|
||||
Let $h$ be a given positive number and let $S$ be a set of real numbers.
|
||||
|
||||
\subsection*{\verified{Theorem I.32a}}%
|
||||
\label{sub:theorem-i.32a}
|
||||
|
||||
If $S$ has a supremum, then for some $x$ in $S$ we have $x > \sup{S} - h$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{sup\_imp\_exists\_gt\_sup\_sub\_delta}
|
||||
|
||||
\divider
|
||||
|
||||
By definition of a supremum, $\sup{S}$ is the least upper bound of $S$.
|
||||
For the sake of contradiction, suppose for all $x \in S$,
|
||||
$x \leq \sup{S} - h$.
|
||||
This immediately implies $\sup{S} - h$ is an upper bound of $S$.
|
||||
But $\sup{S} - h < \sup{S}$, contradicting $\sup{S}$ being the \textit{least}
|
||||
upper bound.
|
||||
Therefore our original hypothesis was wrong.
|
||||
That is, there exists some $x \in S$ such that $x > \sup{S} - h$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection*{\verified{Theorem I.32b}}%
|
||||
\label{sub:theorem-i.32b}
|
||||
|
||||
If $S$ has an infimum, then for some $x$ in $S$ we have $x < \inf{S} + h$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{inf\_imp\_exists\_lt\_inf\_add\_delta}
|
||||
|
||||
\divider
|
||||
|
||||
By definition of an infimum, $\inf{S}$ is the greatest lower bound of $S$.
|
||||
For the sake of contradiction, suppose for all $x \in S$,
|
||||
$x \geq \inf{S} + h$.
|
||||
This immediately implies $\inf{S} + h$ is a lower bound of $S$.
|
||||
But $\inf{S} + h > \inf{S}$, contradicting $\inf{S}$ being the
|
||||
\textit{greatest} lower bound.
|
||||
Therefore our original hypothesis was wrong.
|
||||
That is, there exists some $x \in S$ such that $x < \inf{S} + h$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{Theorem I.33}%
|
||||
\label{sec:theorem-i.33}
|
||||
|
||||
Given nonempty subsets $A$ and $B$ of $\mathbb{R}$, let $C$ denote the set
|
||||
$$C = \{a + b : a \in A, b \in B\}.$$
|
||||
|
||||
\note{This is known as the "Additive Property."}
|
||||
|
||||
\subsection*{\verified{Theorem I.33a}}%
|
||||
\label{sub:theorem-i.33a}
|
||||
|
||||
If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and
|
||||
$$\sup{C} = \sup{A} + \sup{B}.$$
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{sup\_minkowski\_sum\_eq\_sup\_add\_sup}
|
||||
|
||||
\divider
|
||||
|
||||
We prove (i) $\sup{A} + \sup{B}$ is an upper bound of $C$ and (ii)
|
||||
$\sup{A} + \sup{B}$ is the \textit{least} upper bound of $C$.
|
||||
|
||||
\paragraph{(i)}%
|
||||
\label{par:theorem-i.33a-i}
|
||||
|
||||
Let $x \in C$.
|
||||
By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such
|
||||
that $x = a' + b'$.
|
||||
By definition of a supremum, $a' \leq \sup{A}$.
|
||||
Likewise, $b' \leq \sup{B}$.
|
||||
Therefore $a' + b' \leq \sup{A} + \sup{B}$.
|
||||
Since $x = a' + b'$ was arbitrarily chosen, it follows $\sup{A} + \sup{B}$
|
||||
is an upper bound of $C$.
|
||||
|
||||
\paragraph{(ii)}%
|
||||
|
||||
Since $A$ and $B$ have supremums, $C$ is nonempty.
|
||||
By \nameref{par:theorem-i.33a-i}, $C$ is bounded above.
|
||||
Therefore the completeness axiom tells us $C$ has a supremum.
|
||||
Let $n > 0$ be an integer.
|
||||
We now prove that
|
||||
\begin{equation}
|
||||
\label{par:theorem-i.33a-ii-eq1}
|
||||
\sup{C} \leq \sup{A} + \sup{B} \leq \sup{C} + 1 / n.
|
||||
\end{equation}
|
||||
|
||||
\subparagraph{Left-Hand Side}%
|
||||
|
||||
First consider the left-hand side of \eqref{par:theorem-i.33a-ii-eq1}.
|
||||
By \nameref{par:theorem-i.33a-i}, $\sup{A} + \sup{B}$ is an upper bound of
|
||||
$C$.
|
||||
Since $\sup{C}$ is the \textit{least} upper bound of $C$, it follows
|
||||
$\sup{C} \leq \sup{A} + \sup{B}$.
|
||||
|
||||
\subparagraph{Right-Hand Side}%
|
||||
|
||||
Next consider the right-hand side of \eqref{par:theorem-i.33a-ii-eq1}.
|
||||
By \nameref{sub:theorem-i.32a}, there exists some $a' \in A$ such that
|
||||
$\sup{A} < a' + 1 / (2n)$.
|
||||
Likewise, there exists some $b' \in B$ such that
|
||||
$\sup{B} < b' + 1 / (2n)$.
|
||||
Adding these two inequalities together shows
|
||||
\begin{align*}
|
||||
\sup{A} + \sup{B}
|
||||
& < a' + b' + 1 / n \\
|
||||
& \leq \sup{C} + 1 / n.
|
||||
\end{align*}
|
||||
|
||||
\subparagraph{Conclusion}%
|
||||
|
||||
Applying \nameref{sec:theorem-i.31} to \eqref{par:theorem-i.33a-ii-eq1}
|
||||
proves $\sup{C} = \sup{A} + \sup{B}$ as expected.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection*{\verified{Theorem I.33b}}%
|
||||
\label{sub:theorem-i.33b}
|
||||
|
||||
If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and
|
||||
$$\inf{C} = \inf{A} + \inf{B}.$$
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{inf\_minkowski\_sum\_eq\_inf\_add\_inf}
|
||||
|
||||
\divider
|
||||
|
||||
We prove (i) $\inf{A} + \inf{B}$ is a lower bound of $C$ and (ii)
|
||||
$\inf{A} + \inf{B}$ is the \textit{greatest} lower bound of $C$.
|
||||
|
||||
\paragraph{(i)}%
|
||||
\label{par:theorem-i.33b-i}
|
||||
|
||||
Let $x \in C$.
|
||||
By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such
|
||||
that $x = a' + b'$.
|
||||
By definition of an infimum, $a' \geq \inf{A}$.
|
||||
Likewise, $b' \geq \inf{B}$.
|
||||
Therefore $a' + b' \geq \inf{A} + \inf{B}$.
|
||||
Since $x = a' + b'$ was arbitrarily chosen, it follows $\inf{A} + \inf{B}$
|
||||
is a lower bound of $C$.
|
||||
|
||||
\paragraph{(ii)}%
|
||||
|
||||
Since $A$ and $B$ have infimums, $C$ is nonempty.
|
||||
By \nameref{par:theorem-i.33b-i}, $C$ is bounded below.
|
||||
Therefore \nameref{sec:theorem-i.27} tells us $C$ has an infimum.
|
||||
Let $n > 0$ be an integer.
|
||||
We now prove that
|
||||
\begin{equation}
|
||||
\label{par:theorem-i.33b-ii-eq1}
|
||||
\inf{C} - 1 / n \leq \inf{A} + \inf{B} \leq \inf{C}.
|
||||
\end{equation}
|
||||
|
||||
\subparagraph{Right-Hand Side}%
|
||||
|
||||
First consider the right-hand side of \eqref{par:theorem-i.33b-ii-eq1}.
|
||||
By \nameref{par:theorem-i.33b-i}, $\inf{A} + \inf{B}$ is a lower bound of
|
||||
$C$.
|
||||
Since $\inf{C}$ is the \textit{greatest} upper bound of $C$, it follows
|
||||
$\inf{C} \geq \inf{A} + \inf{B}$.
|
||||
|
||||
\subparagraph{Left-Hand Side}%
|
||||
|
||||
Next consider the left-hand side of \eqref{par:theorem-i.33b-ii-eq1}.
|
||||
By \nameref{sub:theorem-i.32b}, there exists some $a' \in A$ such that
|
||||
$\inf{A} > a' - 1 / (2n)$.
|
||||
Likewise, there exists some $b' \in B$ such that
|
||||
$\inf{B} > b' - 1 / (2n)$.
|
||||
Adding these two inequalities together shows
|
||||
\begin{align*}
|
||||
\inf{A} + \inf{B}
|
||||
& > a' + b' - 1 / n \\
|
||||
& \geq \inf{C} - 1 / n.
|
||||
\end{align*}
|
||||
|
||||
\subparagraph{Conclusion}%
|
||||
|
||||
Applying \nameref{sec:lemma-2} to \eqref{par:theorem-i.33b-ii-eq1}
|
||||
proves $\inf{C} = \inf{A} + \inf{B}$ as expected.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\verified{Theorem I.34}}%
|
||||
\label{sec:theorem-i.34}
|
||||
|
||||
Given two nonempty subsets $S$ and $T$ of $\mathbb{R}$ such that $$s \leq t$$
|
||||
for every $s$ in $S$ and every $t$ in $T$. Then $S$ has a supremum, and $T$
|
||||
has an infimum, and they satisfy the inequality $$\sup{S} \leq \inf{T}.$$
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\lean{forall\_mem\_le\_forall\_mem\_imp\_sup\_le\_inf}
|
||||
|
||||
\divider
|
||||
|
||||
By hypothesis, $S$ and $T$ are nonempty sets.
|
||||
Let $s \in S$ and $t \in T$.
|
||||
Then $t$ is an upper bound of $S$ and $s$ is a lower bound of $T$.
|
||||
By the completeness axiom, $S$ has a supremum.
|
||||
By \nameref{sec:theorem-i.27}, $T$ has an infimum.
|
||||
All that remains is showing $\sup{S} \leq \inf{T}$.
|
||||
|
||||
For the sake of contradiction, suppose $\sup{S} > \inf{T}$.
|
||||
Then there exists some $c > 0$ such that $\sup{S} = \inf{T} + c$.
|
||||
Therefore $\inf{T} < \sup{S} - c / 2$.
|
||||
By \nameref{sub:theorem-i.32a}, there exists some $x \in S$ such that
|
||||
$\sup{S} - c / 2 < x$.
|
||||
Thus $$\inf{T} < \sup{S} - c / 2 < x.$$
|
||||
But by hypothesis, $x \in S$ is a lower bound of $T$ meaning $x \leq \inf{T}$.
|
||||
Therefore $x < x$, a contradiction.
|
||||
Out original assumption is incorrect; that is, $\sup{S} \leq \inf{T}$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\end{document}
|
|
@ -1,57 +0,0 @@
|
|||
\documentclass{article}
|
||||
|
||||
\input{../../preamble}
|
||||
|
||||
\newcommand{\lean}[2]{\leanref{../../#1.html\##2}{#2}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\tableofcontents
|
||||
|
||||
\section{The Concepts of Integral Calculus}%
|
||||
\label{sec:concepts-integral-calculus}
|
||||
|
||||
\subsection{\defined{Partition}}%
|
||||
\label{sub:partition}
|
||||
|
||||
Let $[a, b]$ be a closed interval decomposed into $n$ subintervals by inserting
|
||||
$n - 1$ points of subdivision, say $x_1$, $x_2$, $\ldots$, $x_{n-1}$, subject
|
||||
only to the restriction
|
||||
\begin{equation}
|
||||
\label{sec:partition-eq1}
|
||||
a < x_1 < x_2 < \cdots < x_{n-1} < b.
|
||||
\end{equation}
|
||||
It is convenient to denote the point $a$ itself by $x_0$ and the point $b$ by
|
||||
$x_n$.
|
||||
A collection of points satisfying \eqref{sec:partition-eq1} is called a
|
||||
\textbf{partition} $P$ of $[a, b]$, and we use the symbol
|
||||
$$P = \{x_0, x_1, \ldots, x_n\}$$ to designate this partition.
|
||||
|
||||
\begin{definition}
|
||||
|
||||
\lean{Common/Set/Intervals/Partition}{Set.Intervals.Partition}
|
||||
|
||||
\end{definition}
|
||||
|
||||
\subsection{\defined{Step Function}}%
|
||||
\label{sub:step-function}
|
||||
|
||||
A function $s$, whose domain is a closed interval $[a, b]$, is called a step
|
||||
function if there is a \nameref{sub:partition} $P = \{x_0, x_1, \ldots, x_n\}$
|
||||
of $[a b]$ such that $s$ is constant on each open subinterval of $P$.
|
||||
That is to say, for each $k = 1, 2, \ldots, n$, there is a real number $s_k$
|
||||
such that $$s(x) = s_k \quad\text{if}\quad x_{k-1} < x < x_k.$$
|
||||
Step functions are sometimes called piecewise constant functions.
|
||||
|
||||
\vspace{8pt}
|
||||
\noindent
|
||||
\textit{Note:} At each of the endpoints $x_{k-1}$ and $x_k$ the function must
|
||||
have some well-defined value, but this need not be the same as $s_k$.
|
||||
|
||||
\begin{definition}
|
||||
|
||||
\lean{Common/Set/Intervals/StepFunction}{Set.Intervals.StepFunction}
|
||||
|
||||
\end{definition}
|
||||
|
||||
\end{document}
|
|
@ -1,93 +0,0 @@
|
|||
\documentclass{article}
|
||||
|
||||
\input{../../../preamble}
|
||||
|
||||
\newcommand{\lean}[2]{\leanref{./Area.html\##1}{#2}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\header{Axiomatic Framework of Area}{Tom M. Apostol}
|
||||
|
||||
We assume there exists a class $\mathscr{M}$ of measurable sets in the plane and
|
||||
a set function $a$, whose domain is $\mathscr{M}$, with the following
|
||||
properties:
|
||||
|
||||
\section*{\defined{Nonnegative Property}}%
|
||||
\label{sec:nonnegative-property}
|
||||
|
||||
For each set $S$ in $\mathscr{M}$, we have $a(S) \geq 0$.
|
||||
|
||||
\begin{axiom}
|
||||
|
||||
\lean{Nonnegative-Property}{Nonnegative Property}
|
||||
|
||||
\end{axiom}
|
||||
|
||||
\section*{\defined{Additive Property}}%
|
||||
\label{sec:additive-property}
|
||||
|
||||
If $S$ and $T$ are in $\mathscr{M}$, then $S \cup T$ and $S \cap T$ are in
|
||||
$\mathscr{M}$, and we have $a(S \cup T) = a(S) + a(T) - a(S \cap T)$.
|
||||
|
||||
\begin{axiom}
|
||||
|
||||
\lean{Additive-Property}{Additive Property}
|
||||
|
||||
\end{axiom}
|
||||
|
||||
\section*{\defined{Difference Property}}%
|
||||
\label{sec:difference-property}
|
||||
|
||||
If $S$ and $T$ are in $\mathscr{M}$ with $S \subseteq T$, then $T - S$ is in
|
||||
$\mathscr{M}$, and we have $a(T - S) = a(T) - a(S)$.
|
||||
|
||||
\begin{axiom}
|
||||
|
||||
\lean{Difference-Property}{Difference Property}
|
||||
|
||||
\end{axiom}
|
||||
|
||||
\section*{\defined{Invariance Under Congruence}}%
|
||||
\label{sec:invariance-under-congruence}
|
||||
|
||||
If a set $S$ is in $\mathscr{M}$ and if $T$ is congruent to $S$, then $T$ is
|
||||
also in $\mathscr{M}$ and we have $a(S) = a(T)$.
|
||||
|
||||
\begin{axiom}
|
||||
|
||||
\lean{Invariant-Under-Congruence}{Invariance Under Congruence}
|
||||
|
||||
\end{axiom}
|
||||
|
||||
\section*{\defined{Choice of Scale}}%
|
||||
\label{sec:choice-scale}
|
||||
|
||||
Every rectangle $R$ is in $\mathscr{M}$.
|
||||
If the edges of $R$ have lengths $h$ and $k$, then $a(R) = hk$.
|
||||
|
||||
\begin{axiom}
|
||||
|
||||
\lean{Choice-of-Scale}{Choice of Scale}
|
||||
|
||||
\end{axiom}
|
||||
|
||||
\section*{\partial{Exhaustion Property}}%
|
||||
\label{sec:exhaustion-property}
|
||||
|
||||
Let $Q$ be a set that can be enclosed between two step regions $S$ and $T$, so
|
||||
that
|
||||
\begin{equation}
|
||||
\label{sec:exhaustion-property-eq1}
|
||||
S \subseteq Q \subseteq T.
|
||||
\end{equation}
|
||||
If there is one and only one number $c$ which satisfies the inequalities
|
||||
$$a(S) \leq c \leq a(T)$$ for all step regions $S$ and $T$ satisfying (1.1),
|
||||
then $Q$ is measurable and $a(Q) = c$.
|
||||
|
||||
\begin{axiom}
|
||||
|
||||
\lean{Exhaustion-Property}{Exhaustion Property}
|
||||
|
||||
\end{axiom}
|
||||
|
||||
\end{document}
|
|
@ -47,14 +47,14 @@ def index : BaseHtmlM Html := do templateExtends (baseHtml "Index") <|
|
|||
LaTeX and Lean.
|
||||
</li>
|
||||
<li>
|
||||
<span style="color:magenta">Magenta statements </span> are reserved
|
||||
<span style="color:fuchsia">Fuchsia statements </span> are reserved
|
||||
for definitions, axioms, statements, theorems, lemmas, etc. that
|
||||
have been proven or encoded in LaTeX but not yet proven or encoded
|
||||
in Lean.
|
||||
</li>
|
||||
<li>
|
||||
<span style="color:red">Red </span> serves as a catch-all for all
|
||||
statements that don't fit the above categorizations. Incomplete
|
||||
<span style="color:maroon">Maroon </span> serves as a catch-all for
|
||||
all statements that don't fit the above categorizations. Incomplete
|
||||
definitions, statements without proof, etc. belong here.
|
||||
</li>
|
||||
</ul>
|
||||
|
|
|
@ -1,11 +1,12 @@
|
|||
\usepackage{amsfonts, amsmath, amsthm}
|
||||
\usepackage{comment}
|
||||
\usepackage[shortlabels]{enumitem}
|
||||
\usepackage{environ}
|
||||
\usepackage{fancybox}
|
||||
\usepackage{fontawesome5}
|
||||
\usepackage{mathrsfs}
|
||||
\usepackage{soul}
|
||||
\usepackage{xcolor}
|
||||
\usepackage[usenames,dvipsnames]{xcolor}
|
||||
% `hyperref` comes after `xr-hyper`.
|
||||
\usepackage{xr-hyper}
|
||||
\usepackage{hyperref}
|
||||
|
@ -16,6 +17,7 @@
|
|||
|
||||
\hypersetup{colorlinks=true, linkcolor=blue, urlcolor=blue}
|
||||
\newcommand{\leanref}[2]{\textcolor{blue}{$\pmb{\exists}\;{-}\;$}\href{#1}{#2}}
|
||||
\newcommand{\textref}[1]{\text{\nameref{#1}}}
|
||||
|
||||
% ========================================
|
||||
% Environments
|
||||
|
@ -55,9 +57,9 @@
|
|||
\DeclareRobustCommand{\verified}[1]{%
|
||||
\texorpdfstring{\color{teal}\faCheckCircle\ #1}{#1}}
|
||||
\DeclareRobustCommand{\partial}[1]{%
|
||||
\texorpdfstring{\color{magenta}\faPencil*\ #1}{#1}}
|
||||
\texorpdfstring{\color{Fuchsia}\faPencil*\ #1}{#1}}
|
||||
\DeclareRobustCommand{\unverified}[1]{%
|
||||
\texorpdfstring{\color{red}\faExclamationCircle\ #1}{#1}}
|
||||
\texorpdfstring{\color{Maroon}\faExclamationCircle\ #1}{#1}}
|
||||
|
||||
% ========================================
|
||||
% Math
|
||||
|
|
Loading…
Reference in New Issue