Finish apostol exercises 1.7.
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\documentclass{article}
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\usepackage{amsmath}
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\usepackage{graphicx}
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\usepackage{mathrsfs}
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\input{../../preamble}
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\graphicspath{{./images/}}
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\newcommand{\larea}[2]{\lean{../..}{Bookshelf/Real/Geometry/Area}{#1}{#2}}
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\newcommand{\lrect}[2]{\lean{../..}{Bookshelf/Real/Geometry/Rectangle}{#1}{#2}}
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@ -9,7 +12,7 @@
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\begin{document}
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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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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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@ -40,20 +43,20 @@ A set consisting of a finite number of points in a plane.
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\begin{proof}
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We show for all $k > 0$, a set consisting of $k$ points in a plane is
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measurable with area $0$.
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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 \eqref{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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Define our induction hypothesis as, "Let $k > 0$ and assume a set consisting
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of $k$ points in a plane is measurable with area $0$."
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Consider a set $S_{k+1}$ consisting of $k + 1$ points in a plane.
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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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@ -68,8 +71,6 @@ A set consisting of a finite number of points in a plane.
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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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\noindent
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There are two cases to consider:
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\subparagraph{Case 1}%
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@ -86,12 +87,11 @@ A set consisting of a finite number of points in a plane.
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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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$a(S_{k+1}) = 0$ as expected.
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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 $n > 0$, a set consisting of
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$n$ points in a plane is measurable with area $0$.
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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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@ -102,8 +102,9 @@ The union of a finite collection of line segments in a plane.
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\begin{proof}
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We show for all $k > 0$, a set consisting of $k$ line segments in a plane is
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measurable with area $0$.
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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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@ -113,13 +114,12 @@ The union of a finite collection of line segments in a plane.
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By \larea{Choice-of-Scale}{Choice of Scale}, $S$ is measurable with area its
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width $w$ 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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Define our induction hypothesis as, "Let $k > 0$ and assume a set consisting
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of $k$ line segments in a plane is measurable with area $0$."
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Consider a set $S_{k+1}$ consisting of $k + 1$ line segments in a plane.
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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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@ -134,8 +134,6 @@ The union of a finite collection of line segments in a plane.
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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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\noindent
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There are two cases to consider:
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\subparagraph{Case 1}%
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@ -152,12 +150,11 @@ The union of a finite collection of line segments in a plane.
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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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$a(S_{k+1}) = 0$ as expected.
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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 $n > 0$, a set consisting of
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$n$ line segments in a plane is measurable with area $0$.
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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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@ -165,12 +162,450 @@ The union of a finite collection of line segments in a plane.
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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. Prove that every triangular region is measurable
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and that its area is one half the product of its base and altitude.
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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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TODO
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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 \larea{Choice-of-Scale}{Choice of 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 \larea{Additive-Property}{Additive Property}, $R \cup S$ and $R \cap S$
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are both 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 \larea{Invariance-Under-Congruence}{Invariance Under Congruence},
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$a(T') = a(T)$, concluding our proof.
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\end{proof}
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\section{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 \larea{Choice-of-Scale}{Choice of Scale}, $R$ is measurable.
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By \eqref{sec:exercise-2}, $T$ is measurable.
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By the \larea{Additive-Property}{Additive Property}, $R \cup T$ and $R \cap T$
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are both 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 & \eqref{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 \eqref{sec:exercise-2}, $T$ is measurable.
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By Case 1, $R$ is measurable.
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By the \larea{Additive-Property}{Additive Property}, $R \cup T$ and $R \cap T$
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are both 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) & \eqref{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 \larea{Invariance-Under-Congruence}{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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\vspace{4pt}
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\hrule
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\vspace{10pt}
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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 \larea{Invariance-Under-Congruence}{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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A point $(x, y)$ is called a \textit{lattice point} if both coordinates $x$ and
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$y$ are integers.
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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{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 width $w$, height $h$, and lattice points for
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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 \larea{Choice-of-Scale}{Choice of Scale}, $P$ is measurable with area
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$a(P) = wh$.
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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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$a(P)$:
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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{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 $T'$ be a right triangle with width $w$ and height $h$.
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Let $T$ be the triangle $T'$ translated, rotated, and reflected such that the
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its vertices are $(0, 0)$, $(0, w)$, and $(w, h)$, where $w \leq h$.
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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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There are two cases to consider:
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\paragraph{Case 1}%
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Suppose $h / w$ is an integral value.
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Then there exist $w + 1$ lattice points on $T$'s hypotenuse.
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The number of interior lattices points of $T$ is
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\begin{align*}
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I
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& = \frac{1}{2}\left[ (w - 1)(h - 1) - (w - 1) \right] \\
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& = \frac{1}{2}\left[ wh - 2w - h + 2 \right].
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\end{align*}
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The number of boundary lattice points of $T$ is
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\begin{align*}
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B
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& = (w + 1) + h + (w - 1) \\
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& = 2w + h.
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\end{align*}
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Thus
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\begin{align*}
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I + \frac{1}{2}B - 1
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& = \frac{wh - 2w - h + 2}{2} + \frac{2w + h}{2} - 1 \\
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& = \frac{wh - 2w - h + 2 + 2w + h - 2}{2} \\
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& = \frac{wh}{2}.
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\end{align*}
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\paragraph{Case 2}%
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Suppose $h / w$ is not an integral value.
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Then there exist exactly 2 lattice points on $T$'s hypotenuse.
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The number of interior lattice points of $T$ is
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\begin{align*}
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I
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& = \frac{1}{2}\left[ (w - 1)(h - 1) \right] \\
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& = \frac{1}{2}\left[ wh - w - h + 1 \right].
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\end{align*}
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The number of boundary lattice points of $T$ is
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\begin{align*}
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B
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& = (w + 1) + h \\
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& = w + h + 1.
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\end{align*}
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Thus
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\begin{align*}
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I + \frac{1}{2}B - 1
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& = \frac{wh - w - h + 1}{2} + \frac{w + h + 1}{2} - 1 \\
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& = \frac{wh - w - h + 1 + w + h + 1 - 2}{2} \\
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& = \frac{wh}{2}.
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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 $$a(T) = I + \frac{1}{2}B - 1.$$
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We do not prove this formula is valid for parallelograms here.
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Instead, refer to \eqref{sub:exercise-4c} below.
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\end{proof}
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\subsection{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 \eqref{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 $P$.
|
||||
Let $B_P$ be the number of boundary lattice points of $P$.
|
||||
Similarly, let $I_S$, $I_T$, $B_S$, and $B_T$ be the number of interior
|
||||
and boundary lattice points of $S$ and $T$.
|
||||
Let $c$ denote the number of boundary points shared between $S$ and $T$.
|
||||
|
||||
By our induction hypothesis, $a(S) = I_S + \frac{1}{2}B_S - 1$.
|
||||
By our base case, $a(T) = I_T + \frac{1}{2}B_T - 1$.
|
||||
By construction, it follows:
|
||||
\begin{align*}
|
||||
I_P & = I_S + I_T + c - 2 \\
|
||||
B_P & = B_S + B_T - (c - 2) - c \\
|
||||
& = B_S + B_T - 2c + 2.
|
||||
\end{align*}
|
||||
Applying the lattice point area formula to $P$ yields the following:
|
||||
\begin{align*}
|
||||
& I_P + \frac{1}{2}B_P - 1 \\
|
||||
& = (I_S + I_T + c - 2) + \frac{1}{2}(B_S + B_T - 2c + 2) - 1 \\
|
||||
& = I_S + I_T + c - 2 + \frac{1}{2}B_S + \frac{1}{2}B_T - c + 1 - 1 \\
|
||||
& = (I_S + \frac{1}{2}B_S - 1) + (I_T + \frac{1}{2}B_T - 1) \\
|
||||
& = a(S) + (I_T + \frac{1}{2}B_T - 1) & \text{induction hypothesis} \\
|
||||
& = a(S) + a(T). & \text{base case}
|
||||
\end{align*}
|
||||
By the \larea{Additive-Property}{Additive Property}, $S \cup T$ is
|
||||
measurable, $S \cap T$ is measurable, and
|
||||
\begin{align*}
|
||||
a(P)
|
||||
& = a(S \cup T) \\
|
||||
& = a(S) + a(T) - a(S \cap T) \\
|
||||
& = a(S) + a(T). & \text{by construction}
|
||||
\end{align*}
|
||||
This shows the lattice point area formula is in agreement with our axiomatic
|
||||
definition of area.
|
||||
Thus $P(k + 1)$ holds.
|
||||
|
||||
\paragraph{Conclusion}%
|
||||
|
||||
By mathematical induction, it follows for all $n \geq 3$, $P(n)$ is true.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{Exercise 5}%
|
||||
\label{sub:exercise-5}
|
||||
|
||||
Prove that a triangle whose vertices are lattice points cannot be equilateral.
|
||||
|
||||
[\textit{Hint:} Assume there is such a triangle and compute its area in two
|
||||
ways, using Exercises 2 and 4.]
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Proceed by contradiction.
|
||||
Let $T$ be an equilateral triangle whose vertices are lattice points.
|
||||
Assume each side of $T$ has length $a$.
|
||||
Then $T$ has height $h = (a\sqrt{3}) / 2$.
|
||||
By \eqref{sec:exercise-2},
|
||||
\begin{equation}
|
||||
\label{sub:exercise-5-eq1}
|
||||
\tag{5.1}
|
||||
a(T) = \frac{1}{2}ah = \frac{a^2\sqrt{3}}{4}.
|
||||
\end{equation}
|
||||
Let $I$ and $B$ denote the number of interior and boundary lattice points of
|
||||
$T$ respectively.
|
||||
By \eqref{sec:exercise-4},
|
||||
\begin{equation}
|
||||
\label{sub:exercise-5-eq2}
|
||||
\tag{5.2}
|
||||
a(T) = I + \frac{1}{2}B - 1.
|
||||
\end{equation}
|
||||
But \eqref{sub:exercise-5-eq1} is irrational whereas
|
||||
\eqref{sub:exercise-5-eq2} is not.
|
||||
This is a contradiction.
|
||||
Thus, there is \textit{no} equilateral triangle whose vertices are lattice
|
||||
points.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{Exercise 6}%
|
||||
\label{sub:exercise-6}
|
||||
|
||||
Let $A = \{1, 2, 3, 4, 5\}$, and let $\mathscr{M}$ denote the class of all
|
||||
subsets of $A$.
|
||||
(There are 32 altogether, counting $A$ itself and the empty set $\emptyset$.)
|
||||
For each set $S$ in $\mathscr{M}$, let $n(S)$ denote the number of distinct
|
||||
elements in $S$.
|
||||
If $S = \{1, 2, 3, 4\}$ and $T = \{3, 4, 5\}$, compute $n(S \cup T)$,
|
||||
$n(S \cap T)$, $n(S - T)$, and $n(T - S)$.
|
||||
Prove that the set function $n$ satisfies the first three axioms for area.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
Let $S = \{1, 2, 3, 4\}$ and $T = \{3, 4, 5\}$.
|
||||
Then
|
||||
\begin{align*}
|
||||
n(S \cup T)
|
||||
& = n(\{1, 2, 3, 4\} \cup \{3, 4, 5\}) \\
|
||||
& = n(\{1, 2, 3, 4, 5\}) \\
|
||||
& = 5. \\
|
||||
n(S \cap T)
|
||||
& = n(\{1, 2, 3, 4\} \cap \{3, 4, 5\}) \\
|
||||
& = n(\{3, 4\}) \\
|
||||
& = 2. \\
|
||||
n(S - T)
|
||||
& = n(\{1, 2, 3, 4\} - \{3, 4, 5\}) \\
|
||||
& = n(\{1, 2\}) \\
|
||||
& = 2. \\
|
||||
n(T - S)
|
||||
& = n(\{3, 4, 5\} - \{1, 2, 3, 4\}) \\
|
||||
& = n(\{5\}) \\
|
||||
& = 1.
|
||||
\end{align*}
|
||||
We now prove $n$ satisfies the first three axioms for area.
|
||||
|
||||
\paragraph{Nonnegative Property}%
|
||||
|
||||
$n$ returns the length of some member of $\mathscr{M}$.
|
||||
By hypothesis, the smallest possible input to $n$ is $\emptyset$.
|
||||
Since $n(\emptyset) = 0$, it follows $n(S) \geq 0$ for all $S \subset A$.
|
||||
|
||||
\paragraph{Additive Property}%
|
||||
|
||||
Let $S$ and $T$ be members of $\mathscr{M}$.
|
||||
It trivially follows that both $S \cup T$ and $S \cap T$ are in
|
||||
$\mathscr{M}$.
|
||||
Consider the value of $n(S \cup T)$.
|
||||
There are two cases to consider:
|
||||
|
||||
\subparagraph{Case 1}%
|
||||
|
||||
Suppose $S \cap T = \emptyset$.
|
||||
That is, there is no common element shared between $S$ and $T$.
|
||||
Thus
|
||||
\begin{align*}
|
||||
n(S \cup T)
|
||||
& = n(S) + n(T) \\
|
||||
& = n(S) + n(T) - 0 \\
|
||||
& = n(S) + n(T) - n(S \cap T).
|
||||
\end{align*}
|
||||
|
||||
\subparagraph{Case 2}%
|
||||
|
||||
Suppose $S \cap T \neq \emptyset$.
|
||||
Then $n(S) + n(T)$ counts each element of $S \cap T$ twice.
|
||||
Therefore $n(S \cup T) = n(S) + n(T) - n(S \cap T)$.
|
||||
|
||||
\subparagraph{Conclusion}%
|
||||
|
||||
These cases are exhaustive and in agreement with one another.
|
||||
Thus $n(S \cup T) = n(S) + n(T) - n(S \cap T)$.
|
||||
|
||||
\paragraph{Difference Property}%
|
||||
|
||||
Suppose $S, T \in \mathscr{M}$ such that $S \subseteq T$.
|
||||
That is, every member of $S$ is a member of $T$.
|
||||
By definition, $T - S$ consists of members in $T$ but not in $S$.
|
||||
Thus $n(T - S) = n(T) - n(S)$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
|
|
|
|||
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