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Advanced/Continued Fractions/main.tex Executable file
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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
shortwarning,
singlenumbering
]{../../resources/ormc_handout}
\usepackage{../../resources/macros}
\usepackage{multicol}
\uptitlel{Advanced 2}
\uptitler{Fall 2023}
\title{Continued Fractions}
\subtitle{
Prepared by \githref{Mark} on \today \\
Based on a handout by Matthew Gherman and Adam Lott
}
\begin{document}
\maketitle
\input{parts/00 euclidean}
\input{parts/01 part A}
\input{parts/02 part B}
\end{document}

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Advanced/Continued Fractions/parts/00 euclidean.tex Executable file
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\section{The Euclidean Algorithm}
\definition{}
The \textit{greatest common divisor} of $a$ and $b$ is the greatest integer that divides both $a$ and $b$. \par
We denote this number with $\gcd(a, b)$. For example, $\gcd(45, 60) = 15$.
\problem{}
Find $\gcd(20, 14)$ by hand.
\begin{solution}
$\gcd(20, 14) = 2$
\end{solution}
\vfill
\theorem{The Division Algorithm}<divalgo>
Given two integers $a, b$, we can find two integers $q, r$, where $0 \leq r < b$ and $a = qb + r$. \par
In other words, we can divide $a$ by $b$ to get $q$ remainder $r$.
\vspace{2mm}
For example, take $14 \div 3$. We can re-write this as $3 \times 4 + 2$. \par
Here, $a$ and $b$ are $14$ and $3$, $q = 4$ and $r = 2$.
\theorem{}<gcd_abc>
For any integers $a, b, c$, \par
$\gcd(ac + b, a) = \gcd(a, b)$
\problem{}
Compute $\gcd(668, 6)$ \hint{$668 = 111 \times 6 + 2$}
Then, compute $\gcd(3 \times 668 + 6, 668)$.
\vfill
\problem{The Euclidean Algorithm}
Using the two theorems above, detail an algorithm for finding $\gcd(a, b)$. \par
Then, compute $\gcd(1610, 207)$ by hand. \par
\begin{solution}
Using \ref{gcd_abc} and the division algorthm,
% Minipage prevents column breaks inside body
\begin{multicols}{2}
\begin{minipage}{\columnwidth}
$\gcd(1610, 207)$ \par
$= \gcd(207, 161)$ \par
$= \gcd(161, 46)$ \par
$= \gcd(46, 23)$ \par
$= \gcd(23, 0) = 23$ \par
\end{minipage}
\columnbreak
\begin{minipage}{\columnwidth}
$1610 = 207 \times 7 + 161$ \par
$207 = 161 \times 1 + 46$ \par
$161 = 46 \times 3 + 23$ \par
$46 = 23 \times 2 + 0$ \par
\end{minipage}
\end{multicols}
\end{solution}
\vfill
\pagebreak

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Advanced/Continued Fractions/parts/01 part A.tex Normal file
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\section{}
\definition{}
A \textit{finite continued fraction} is an expression of the form
\[
a_0 + \cfrac{1}{a_1+\cfrac{1}{a_2 + \cfrac{1}{a_3 + ... + \cfrac{1}{a_{k-1} + \cfrac{1}{a_k}}}}}
\]
where $a_0, a_1, ..., a_k$ are all in $\mathbb{Z}^+$.
We'll denote this as $[a_0, a_1, ..., a_k]$.
\problem{}<num2cf>
Write each of the following as a continued fraction. \par
\hint{Solve for one $a_n$ at a time.}
\begin{itemize}
\item $5/12$
\item $5/3$
\item $33/23$
\item $37/31$
\end{itemize}
\vfill
\problem{}
Write each of the following continued fractions as a regular fraction in lowest terms: \par
\begin{itemize}
\item $[2,3,2]$
\item $[1,4,6,4]$
\item $[2,3,2,3]$
\item $[9,12,21,2]$
\end{itemize}
\vfill
\pagebreak
\problem{}<euclid>
Let $\frac{p}{q}$ be a positive rational number in lowest terms.
Perform the Euclidean algorithm to obtain the following sequence:
\begin{align*}
p \ &= \ q_0 q + r_1 \\
q \ &= \ q_1 r_1 + r_2 \\
r_1 \ &= \ q_2 r_2 + r_3 \\
&\vdots \\
r_{k-1} \ &= \ q_k r_k + 1 \\
r_k \ &= \ q_{k+1}
\end{align*}
We know that we will eventually get $1$ as the remainder because $p$ and $q$ are relatively prime. \par
Show that $p/q = [q_0, q_1, ..., q_{k+1}]$.
\vfill
\problem{}
Repeat \ref{num2cf} using the method outlined in \ref{euclid}.
\vfill
\pagebreak
\definition{}
An \textit{infinite continued fraction} is an expression of the form
\[
a_0 + \cfrac{1}{a_1+\cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{a_4 + ...}}}}
\]
where $a_0, a_1, a_2, ...$ are in $\mathbb{Z}^+$.
To prove that this expression actually makes sense and equals a finite number
is beyond the scope of this worksheet, so we assume it for now.
This is denoted $[a_0, a_1, a_2, ...]$.
\problem{}<irrational>
Using a calculator, compute the first five terms of the
continued fraction expansion of the following numbers.
Do you see any patterns?
\begin{itemize}
\item $\sqrt{2}$
\item $\pi \approx 3.14159...$
\item $\sqrt{5}$
\item $e \approx 2.71828...$
\end{itemize}
\vfill
\problem{}
Show that an $\alpha \in \mathbb{R}^+$ can be written as a finite
continued fraction if and only if $\alpha$ is rational. \par
\hint{For one of the directions, use \ref{euclid}}
\vfill
\pagebreak
\definition{}
The continued fraction $[a_0, a_1, a_2, ...]$ is \textit{periodic} if it ends in a repeating sequence of digits. \par
A few examples are below. We denote the repeating sequence with a line.
\begin{itemize}
\item $[1,2,2,2,...] = [1, \overline{2}]$ is periodic.
\item $[1,2,3,4,5,...]$ is not periodic.
\item $[1,3,7,6,4,3,4,3,4,3,...] = [1,3,7,6,\overline{4,3}]$ is periodic.
\item $[1,2,4,8,16, ...]$ is not periodic.
\end{itemize}
\problem{}
\begin{itemize}
\item Show that $\sqrt{2} = [1, \overline{2}]$.
\item Show that $\sqrt{5} = [1, \overline{4}]$.
\end{itemize}
\hint{use the same strategy as \ref{irrational} but without a calculator.}
\vfill
\problem{Challenge I}
Express the following continued fractions in the form $\frac{a+\sqrt{b}}{c}$ where $a$, $b$, and $c$ are integers: \par
\begin{itemize}
\item $[~\overline{1}~]$
\item $[~\overline{2,5}~]$
\item $[~1, 3, \overline{2,3}~]$
\end{itemize}
\vfill
\problem{Challenge II}
Let $\alpha = [~a_0,~ ...,~ a_r,~ \overline{a_{r+1},~ ...,~ a_{r+p}}~]$ be any periodic continued fraction. \par
Prove that $\alpha$ is of the form $\frac{a+\sqrt{b}}{c}$ for some integers $a,b,c$ where $b$ is not a perfect square.
\vfill
\problem{Challenge III}
Prove that any number of the form $\frac{a+\sqrt{b}}{c}$ where $a,b,c$ are integers
and $b$ is not a perfect square can be written as a periodic continued fraction.
%\begin{rmk}
%Numbers of the form $\frac{a+\sqrt{b}}{c}$ where $a,b,c$ are integers and $b$ is not a perfect square are the ``simplest'' irrational numbers in the following sense. A number is rational if and only if it is the solution to a degree $1$ polynomial equation, $ax+b = 0$. Similarly, a number is of the form $\frac{a+\sqrt{b}}{c}$ if it is the solution to a degree $2$ polynomial equation, $ax^2 + bx + c = 0$ (Bonus exercise: prove this). Such numbers are called \textit{quadratic} irrational numbers or \textit{degree 2} irrational numbers.
%\end{rmk}
%\begin{rmk}
%Notice that the results of this worksheet provide a very clean characterization of continued fraction expansions:
%\begin{itemize}
%\item $\alpha$ is a rational number if and only if it has a finite continued fraction expansion.
%\item $\alpha$ is a degree $2$ irrational number if and only if it has an infinite periodic continued fraction expansion.
%\end{itemize}
%\end{rmk}
\vfill
\pagebreak

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\section{Convergents}
\definition{}
Let $\alpha = [a_0, a_1, a_2, ...]$ be an infinite continued fraction (aka an irrational number). \par
The \emph{$n$th convergent to $\alpha$} is the rational number $[a_0, a_1, ..., a_n]$ and is denoted $C_n(\alpha)$.
\problem{}
Calculate the following convergents and write them in lowest terms: \\
\begin{itemize}
\item $C_3([~ 1,2,3,4, ... ~])$
\item $C_4([~ 0,\overline{2,3} ~])$
\item $C_5([~ \overline{1,5} ~])$
\end{itemize}
\vfill
\problem{}<sqrt5>
Recall from last week that $\sqrt{5} = [2,\overline{4}]$.
Calculate the first five convergents to $\sqrt{5}$ and write them in lowest terms.
Do you notice any patterns? \par
\hint{Look at the numbers $\sqrt{5}-C_j(\sqrt{5})$ for $1 \leq j \leq 5$}
\vfill
\generic{Properties of Convergents}
In this section, we want to show that the $n$th convergent to a real number
$\alpha$ is the best approximation of $\alpha$ with the given denominator.
Let $\alpha = [a_0, a_1, ...]$ be fixed, and we will write $C_n$ instead of
$C_n(\alpha)$ for short. Let $p_n / q_n$ be the expression of $C_n$ as a rational number
in lowest terms. We will eventually prove that $|\alpha-C_n|<\frac{1}{q_n^2}$,
and there is no better rational estimate of $\alpha$ with denominator less than or equal to $q_n$.
First we want the recursive formulas $p_n=a_np_{n-1}+p_{n-2}$ and $q_n=a_nq_{n-1}+q_{n-2}$ given $p_{-1}=1$, $p_0=a_0$, $q_{-1}=0$, and $q_0=1$.
\pagebreak
\problem{}
Verify the recursive formula for $1\leq j\leq 3$ for the convergents $C_j$ of: \par
\begin{itemize}
\item $[~ 1,2,3,4, ... ~]$
\item $[~ 0,\overline{2,3} ~]$
\item $[~ \overline{1,5} ~]$
\end{itemize}
\vfill
\problem{Challenge IV}<rec>
Prove that $p_n = a_np_{n-1} + p_{n-2}$ and $q_n = a_nq_{n-1} + q_{n-2}$ by induction.
\begin{itemize}
\item As the base case, verify the recursive formulas for $n=1$ and $n=2$.
\item Assume the recursive formulas hold for $n\leq m$ and show the formulas hold for $m+1$.
\end{itemize}
\vfill
\problem{}<form1>
Using the recursive formula from \ref{rec},
we will show that $p_n q_{n-1} - p_{n-1}q_n = (-1)^{n-1}$.
\begin{itemize}
\item What is $p_1q_0-p_0q_1$?
\item Substitute $a_np_{n-1}+p_{n-2}$ for $p_n$ and $a_nq_{n-1}+q_{n-2}$ for $q_n$ in $p_n q_{n-1}-p_{n-1}q_n$. Simplify the expression.
\item What happens when $n=2$? Explain why $p_n q_{n-1}-p_{n-1}q_n = (-1)^{n-1}$.
\end{itemize}
\vfill
\problem{Challenge VI}
Similarly derive the formula $p_nq_{n-2}-p_{n-2}q_n = (-1)^{n-2}a_n$.
\vfill
\pagebreak
\problem{}<diff>
Recall $C_n=p_n/q_n$.
Show that $C_n-C_{n-1}=\frac{(-1)^{n-1}}{q_{n-1}q_n}$
and $C_n-C_{n-2}=\frac{(-1)^{n-2}a_n}{q_{n-2}q_n}$.
\hint{Use \ref{form1} and $p_nq_{n-2}-p_{n-2}q_n = (-1)^{n-2}a_n$ respectively}
In \ref{sqrt5}, the value $\alpha-C_n$ alternated between negative and positive
and $|\alpha-C_n|$ got smaller each step. Using the relations in \ref{diff},
we can prove that this is always the case.
Specifically, $\alpha$ is always between $C_n$ and $C_{n+1}$.
\vfill
\problem{}
Let's figure out how well the $n$th convergents estimate $\alpha$.
We will show that $|\alpha-C_n|<\frac{1}{q_n^2}$.
\begin{itemize}
\item Note that $|C_{n+1}-C_n|=\frac{1}{q_nq_{n+1}}$.
\item Why is $|\alpha-C_n|\leq|C_{n+1}-C_n|$?
\item Conclude that $|\alpha-C_n|<\frac{1}{q_n^2}$.
\end{itemize}
We are now ready to prove a fundamental result in the theory of rational approximation.
\problem{Dirichlet's approximation theorem}
Let $\alpha$ be any irrational number.
Prove that there are infinitely many rational numbers $\frac{p}{q}$ such that $|\alpha - \frac{p}{q}| < \frac{1}{q^2}$.
\vfill
\pagebreak
\problem{Challenge VII}
Prove that if $\alpha$ is \emph{rational}, then there are only \emph{finitely} many rational numbers $\frac{p}{q}$
satisfying $|\alpha - \frac{p}{q} | < \frac{1}{q^2}$.
The above result shows that the $n$th convergents estimate $\alpha$ extremely well.
Are there better estimates for $\alpha$ if we want small denominators?
In order to answer this question, we introduce the Farey sequence.
\vfill
\definition{}
The \emph{Farey sequence} of order $n$ is the set of rational numbers between
0 and 1 whose denominators (in lowest terms) are $\leq n$, arranged in increasing order.
\problem{}
List the Farey sequence of order 4. Now figure out the Farey sequence of order 5 by including the relevant rational numbers in the Farey sequence of order 4.
\vfill
\problem{}
Let $\frac{a}{b}$ and $\frac{c}{d}$ be consecutive elements of the Farey sequence of order 5. What does $bc-ad$ equal?
\vfill
\problem{Challenge VIII}<farey>
Prove that $bc-ad=1$ for $\frac{a}{b}$ and $\frac{c}{d}$ consecutive rational numbers in Farey sequence of order $n$.
\begin{itemize}[itemsep=2mm]
\item In the plane, draw the triangle with vertices (0,0), $(b,a)$, $(d,c)$.
Show that the area $A$ of this triangle is $\frac{1}{2}$ using Pick's Theorem.
Recall that Pick's Theorem states $A=\frac{B}{2}+I-1$ where $B$ is the number of
lattice points on the boundary and $I$ is the number of points in the interior. \par
\hint{B=3 and I=0}
\item Show that the area of the triangle is also given by $\frac{1}{2}|ad-bc|$.
\item Why is $bc>ad$?
\item Conclude that $bc-ad=1$.
\end{itemize}
\vfill
\pagebreak
\problem{}
Use the result of \ref{farey} to show that there is no rational number between
$C_{n-1}$ and $C_n$ with denominator less than or equal to $q_n$.
Conclude that if $a/b$ is any rational number with $b \leq q_n$, then
$|\alpha - \frac ab| \geq |\alpha - \frac{p_n}{q_n}|$
%What the above exercise shows is that relative to the size of the denominator, the convergents of the continued fraction expansion of $\a$ are the absolute best rational approximations to $\a$.
\vfill
\problem{Challenge V}
Prove the following strengthening of Dirichlet's approximation theorem.
If $\alpha$ is irrational, then there are infinitely many rational numbers
$\frac{p}{q}$ satisfying $|\alpha - \frac pq| < \frac{1}{2q^2}$.
\begin{itemize}[itemsep = 2mm]
\item Prove that $(x+y)^2 \geq 4xy$ for any real $x,y$.
\item Let $p_n/q_n$ be the $n$th convergent to $\alpha$. Prove that
\[
|\frac{p_n}{q_n} - \frac{p_{n+1}}{q_{n+1}}|^2 \ \geq \ 4 | \frac{p_n}{q_n} - \alpha | | \frac{p_{n+1}}{q_{n+1}} - \alpha |
\]
\hint{$\alpha$ lies in between $\frac{p_n}{q_n}$ and $\frac{p_{n+1}}{q_{n+1}}$}
\item Prove that either $\frac{p_n}{q_n}$ or $\frac{p_{n+1}}{q_{n+1}}$ satisfies the desired inequality (Hint: proof by contradiction).
\item Conclude that there are infinitely many rational numbers $\frac{p}{q}$ satisfying $|\alpha - \frac pq| < \frac{1}{2q^2}$.
\end{itemize}
\vfill
\pagebreak