Continued Fractions edits
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@@ -78,7 +78,7 @@ An \textit{infinite continued fraction} is an expression of the form
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a_0 + \cfrac{1}{a_1+\cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{a_4 + ...}}}}
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\]
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where $a_0, a_1, a_2, ...$ are in $\mathbb{Z}^+_0$.
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To prove that this expression actually makes sense and equals a finite number
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Showing that this expression converges to a finite number
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is beyond the scope of this worksheet, so we assume it for now.
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This is denoted $[a_0, a_1, a_2, ...]$.
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@@ -133,7 +133,7 @@ A few examples are below. We denote the repeating sequence with a line.
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\problem{}
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\begin{itemize}
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\item Show that $\sqrt{2} = [1, \overline{2}]$.
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\item Show that $\sqrt{5} = [1, \overline{4}]$.
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\item Show that $\sqrt{5} = [2, \overline{4}]$.
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\end{itemize}
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\hint{use the same strategy as \ref{irrational} but without a calculator.}
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@@ -159,7 +159,7 @@ Express the following continued fractions in the form $\frac{a+\sqrt{b}}{c}$ whe
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\problem{Challenge II}
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Let $\alpha = [~a_0,~ ...,~ a_r,~ \overline{a_{r+1},~ ...,~ a_{r+p}}~]$ be any periodic continued fraction. \par
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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.
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Show that $\alpha$ is of the form $\frac{a+\sqrt{b}}{c}$ for some integers $a,b,c$ where $b$ is not a perfect square.
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@@ -168,7 +168,7 @@ Prove that $\alpha$ is of the form $\frac{a+\sqrt{b}}{c}$ for some integers $a,b
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\problem{Challenge III}
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Prove that any number of the form $\frac{a+\sqrt{b}}{c}$ where $a,b,c$ are integers
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Show that any number of the form $\frac{a+\sqrt{b}}{c}$ where $a,b,c$ are integers
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and $b$ is not a perfect square can be written as a periodic continued fraction.
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@@ -65,7 +65,7 @@ Verify the recursive formula for $1\leq j\leq 3$ for the convergents $C_j$ of: \
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\problem{Challenge IV}<rec>
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Prove that $p_n = a_np_{n-1} + p_{n-2}$ and $q_n = a_nq_{n-1} + q_{n-2}$ by induction.
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Show that $p_n = a_np_{n-1} + p_{n-2}$ and $q_n = a_nq_{n-1} + q_{n-2}$ by induction.
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\begin{itemize}
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\item As the base case, verify the recursive formulas for $n=1$ and $n=2$.
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\item Assume the recursive formulas hold for $n\leq m$ and show the formulas hold for $m+1$.
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@@ -97,7 +97,7 @@ we will show that $p_n q_{n-1} - p_{n-1}q_n = (-1)^{n-1}$.
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\problem{Challenge VI}
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\problem{Challenge V}
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Similarly derive the formula $p_nq_{n-2}-p_{n-2}q_n = (-1)^{n-2}a_n$.
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@@ -141,7 +141,7 @@ We will show that $|\alpha-C_n|<\frac{1}{q_n^2}$.
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We are now ready to prove a fundamental result in the theory of rational approximation.
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\problem{Dirichlet's approximation theorem}
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Let $\alpha$ be any irrational number.
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Prove that there are infinitely many rational numbers $\frac{p}{q}$ such that $|\alpha - \frac{p}{q}| < \frac{1}{q^2}$.
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Show that there are infinitely many rational numbers $\frac{p}{q}$ such that $|\alpha - \frac{p}{q}| < \frac{1}{q^2}$.
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@@ -154,8 +154,8 @@ Prove that there are infinitely many rational numbers $\frac{p}{q}$ such that $|
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\problem{Challenge VII}
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Prove that if $\alpha$ is \emph{rational}, then there are only \emph{finitely} many rational numbers $\frac{p}{q}$
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\problem{Challenge VI}
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Show that if $\alpha$ is \emph{rational}, then there are only \emph{finitely} many rational numbers $\frac{p}{q}$
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satisfying $|\alpha - \frac{p}{q} | < \frac{1}{q^2}$.
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@@ -195,8 +195,8 @@ Let $\frac{a}{b}$ and $\frac{c}{d}$ be consecutive elements of the Farey sequenc
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\problem{Challenge VIII}<farey>
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Prove that $bc-ad=1$ for $\frac{a}{b}$ and $\frac{c}{d}$ consecutive rational numbers in Farey sequence of order $n$.
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\problem{Challenge VII}<farey>
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Show that $bc-ad=1$ for $\frac{a}{b}$ and $\frac{c}{d}$ consecutive rational numbers in Farey sequence of order $n$.
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\begin{itemize}[itemsep=2mm]
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\item In the plane, draw the triangle with vertices (0,0), $(b,a)$, $(d,c)$.
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@@ -237,7 +237,7 @@ $|\alpha - \frac ab| \geq |\alpha - \frac{p_n}{q_n}|$
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\problem{Challenge X}
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\problem{Challenge VIII}
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Prove the following strengthening of Dirichlet's approximation theorem.
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If $\alpha$ is irrational, then there are infinitely many rational numbers
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$\frac{p}{q}$ satisfying $|\alpha - \frac pq| < \frac{1}{2q^2}$.
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