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\begin{center}

CONTINUED FRACTIONS

BEST APPROXIMATIONS

\end{center}


$\frac{a}{b}$ is said to be a best approximation to $\alpha\
(\alpha\in Z\ b\in N)$ if

$$\left|\alpha-\frac{p}{q}\right|&lt;\left|\alpha-\frac{a}{b}
\right|\Rightarrow q&gt;b.$$

We now prove that the convergents to an (irrational) number give a
sequence of best approximations.

Note that as in the previous result, we often investigate
$|q\alpha-p|$ rather than $\left|\alpha-\frac{p}{q}\right|$.
Inequalities involving the former are often a bit stronger the
those involving the latter.

%page 21

Theorem

If $|q\alpha-p|&lt;|q_n\alpha-p_n|,\ n&gt;0$ where $\frac{p_n}{q_n}$ is
a convergent of the continued fraction for $\alpha$, then $q&gt;q_n$.

Proof

Assume that $|q\alpha-p|&lt;q_n\alpha-p_n|$ and that $q\leq q_n$. It
follows that $q&lt;q_{n+1}\ (n&gt;0)$. Consider the equations

\begin{eqnarray*}
x.p_n+y.p_{n+1}&amp;=&amp;p\\ x.q_n+y.q_{n+1}=q
\end{eqnarray*}

$p_nq_{n+1}-p_{n+1}q_n=(-1)^n$, so this pair of equations has
integer solutions $x,y$.

Now $y=0\Rightarrow p=xp_n\ q=xq_n,\ x\neq0$ and so
$|q\alpha-p|=|x||q_n\alpha-p_n|\geq|q_n\alpha-p_n|$

If $x=0$ then $y\neq0$ and $q=yq_n$ which contradicts $q\leq q_n$.

So $x$ and $y$ are non-zero.

We now show that $x$ and $y$ are of opposite sign.

$$0&lt;q=xq_n+yq_{n+1}&lt;q_{n+1}$$

$x$ and $y$ cant both be $&lt;0$ as $q&gt;0$

$x$ and $y$ cant both be $&gt;0$ otherwise $&gt;q_{n+1}$

Now $q_n\alpha-p_n$ and $q_{n+1}|alpha-p_{n+1}$ have opposite
signs, since the convergents alternate either side of $\alpha$, so
$x(q_n\alpha-p_n)$ and $y(q_{n+1}\alpha-p_{n+1})$ have the same
sign.

%page 22

Also

$$q\alpha-p=x(q_n\alpha-p_n)+y(q_{n+1}\alpha-p_{n+1})$$

so

\begin{eqnarray*}
|q\alpha-p|&amp;=&amp;|x(q_n\alpha-p_n)|+|y(q_{n+1}\alpha-p_{n+1})|\\
&amp;&gt;&amp;|x(q_n\alpha-p_n)|\geq|q_n\alpha-p_n|
\end{eqnarray*}

This contradiction proves the theorem.

This proof goes back to Legendre, and is quoted in Perron.

Now
$\left|\alpha-\frac{p}{q}\right|&lt;\left|\alpha-\frac{p_n}{q_n}\right|$

and$q\leq q_n$ multiplying the inequalities

$$\Rightarrow |q\alpha-p|&lt;|q_n\alpha-p_n|\Rightarrow q&gt;q_n$$

So the convergents are the best approximations to $\alpha$. But
how good are they?

Now we have already seen the equation

$$\alpha-\frac{p_n}{q_n}=\frac{(-1)^n}{q_n(\alpha_{n+1}q_n+q_{n-1})}$$

so

$$\left|alpha-\frac{p_n}{q_n}\right|=\frac{1}{q_n(\alpha_{n+1}q_n
+q_{n-1})}\leq\frac{1}{q_n(a_{n+1}q_n+q_{n-1})}=\frac{1}{q_nq_{n+1}}
&lt;\frac{1}{q_n^2}$$

We already know from Dirichlet's theorem that an irrati0onal
$\alpha$ has infinitely many rational approximations $\frac{p}{q}$
satisfying $\left|\alpha-\frac{p}{q}\right|&lt;\frac{1}{q^2}$.

The sequence of convergents supplies such a set.

%page 23

This does not give them all however. E.g. consider rational
approximations to $\frac{779}{207}$

The convergents are $$3,\ 4,\ \frac{15}{4},\ \frac{64}{17},\
\frac{143}{38},\ \frac{779}{207}.$$

\begin{eqnarray*}
\frac{779}{207}-\frac{79}{21}&amp;=&amp;\frac{6}{4347}\approx1.38\times10^{-3}\\
\frac{1}{21^2}&amp;\approx&amp;2.27\times10^{-3}
\end{eqnarray*}

However, notice that $\frac{79}{21}=\frac{15+64}{4+17}$

I shall not pursue this, but instead show that if
$\left|\alpha-\frac{p}{q}\right|&lt;\frac{1}{2q^2}\ (p,q)=1$ then
$\frac{p}{q}$ is one of the convergents of the continued fraction
for $\alpha$.

Proof

Suppose not. Then $q_n\leq q\leq q_{n+1}$ determines an integer
$n$, and $|q\alpha-\phi|&lt;|q_n\alpha-p_n|$ is impossible. ( The
earlier theorem can be improved to $q\geq q_{n+1}$)

so $|q_n\alpha-p_n|\leq|q\alpha-p|&lt;\frac{1}{2q}$

i.e $\left|\alpha-\frac{p_n}{q_n}\right|Z&lt;\frac{1}{2qq_n}$

Now

\begin{eqnarray*}
\frac{1}{qq_n}&amp;\leq&amp;\frac{|qp_n-pq_n|}{qq_n}\textrm{ (even if
}q=q_n\ \frac{p}{q}\neq\frac{p_n}{q_n})\\
&amp;=&amp;\left|\frac{p_n}{q_n}-\frac{p}{q}\right|\\
&amp;\leq&amp;\left|\alpha-\frac{p_n}{q_n}\right|+\left|\alpha-\frac{p}{q}
\right|\\ &amp;&lt;&amp;\frac{1}{2qq_n}+\frac{1}{2q^2}\\
\frac{1}{2qq_n}&amp;&lt;&amp;\frac{1}{2q^2}
\end{eqnarray*}

so $q&lt;q_n$ This is a contradiction so the theorem is proved.

%page 24

Now of any two successive convergents, at least one satisfies
$\left|\alpha-\frac{p}{q}\right|&lt;\frac{1}{2q^2}$

Proof

Since the convergents are alternatively greater and less then $x$

$$\left|\frac{p_{n+1}}{q_{n+1}}-\frac{p_n}{q_n} \right|=\left|
\frac{p_{n+1}}{q_{n+1}}-\alpha\right|+\left|\alpha-
\frac{po_n}{q_n} \right|$$

Suppose the result false. Then

\begin{eqnarray*}
\frac{1}{2q_{n+1}^2}+\frac{1}{2q_n^2}&amp;\leq&amp;\left|
\frac{p_{n+1}}{q_{n+1}}-\alpha\right|+\left|\alpha-
\frac{p_n}{q_n}\right|\\
&amp;=&amp;\left|\frac{p_{n+1}}{q_{n+1}}-\frac{p_n}{q_n}\right|\\
&amp;=&amp;\frac{1}{q_nq_{n+1}}
\end{eqnarray*}

i.e. $\left(\frac{1}{q_{n+1}}-\frac{1}{q_n}\right)^2\leq0$ i.e.
$q_{n+1}=q_n$. This is true only if $n=1\ a_1=1\ q_1=q_0=1$.
Otherwise $q_{n+1}&gt;q_n$.

Even in this case

$$0&lt;\frac{p_1}{q_1}-x=1-\frac{1}{1+}\frac{1}{a_2+}&lt;1-
\frac{a_2}{a_2+1}\leq\frac{1}{2}$$

so the theorem is still true.

Further, of any three successive convergents, at least one
satisfies $\left|\alpha-\frac{p}{q}\right|&lt;\frac{1}{q^2\sqrt{5}}$

Proof

%page 25

$$\left|\alpha-\frac{p_n}{q_n}\right|=\frac{1}{q_n(\alpha_{n+1}
q_n+q_{n-1})}=\frac{1}{q_n^2\left(\alpha_{n+1}+\frac{q_{n-1}}{q_n}
\right)}$$

Now suppose that

$$\alpha_i+\frac{q_{i-2}}{q_{i-1}}\leq\sqrt{5}\textrm{ for
}i=n-1,n,n+1$$

Then

$$\alpha_{n-1}=a_{n-1}+\frac{1}{\alpha_n}\textrm{ and
}\frac{q_{n-1}}{q_{n-2}}=a_{n-1}+\frac{q_{n-3}}{q_{n-2}}$$

so

$$\frac{1}{\alpha_n}+\frac{q_{n-1}}{q_{n-2}}=\alpha_{n-1}+
\frac{q_{n-3}}{q_{n-2}}\leq\sqrt{5}$$

by assumption and

\begin{eqnarray*}
1&amp;=&amp;\alpha_n\frac{1}{\alpha_n}\leq\left(\sqrt{5}+\frac{q_{n
-2}}{q_{n-1}}\right)\left(\sqrt{5}-\frac{q_{n-1}}{q_{n-2}}
\right)\\
&amp;=&amp;5+1-\sqrt{5}\left(\frac{q_{n-2}}{q_{n-1}}+\frac{q_{n-1}}{
q_{n-2}}\right)
\end{eqnarray*}

giving
$\frac{q_{n-2}}{q_{n-1}}+\frac{q_{n-1}}{q_{n-2}}\leq\sqrt{5}$

In fact since LHS is rational we have strictly less then, so

\begin{eqnarray*}
\left(\frac{q_{n-1}}{q_{n-1}}\right)^2-\left(\frac{q_{n-2}}{
q_{n-1}}\right)\sqrt{5}+1&amp;&lt;&amp;0\\
\left(\frac{q_{n-2}}{q_{n-1}}-\frac{1}{2}\sqrt{5}\right)^2&amp;&lt;&amp;
\frac{1}{4}\textrm{ i.e.}\\
\frac{q_{n-2}}{q_{n-1}}&amp;&gt;&amp;\frac{1}{2}(\sqrt{5}-1)
\end{eqnarray*}

This has used $i=n-1,n$. Using $1=n,n+1$ gives

$$\frac{q_{n-1}}{q_n}&gt;\frac{1}{2}(\sqrt{5}-1)$$

Now $q_n=a_nq_{n-1}+q_{n-2}$

$$a_n=\frac{q_n}{q_{n-1}}-\frac{q_{n-2}}{q_{n-1}}&lt; \frac{2}{\sqrt{
5}-1}-\frac{1}{2}(\sqrt{5}-1)=1$$

$a_n&lt;1$ is a contradiction.

%page 26

Now let $\alpha=\frac{1}{2}(\sqrt{5}-1)=[0,1,1,1,\ldots]$

Suppose that there are an infinite number of solutions of

$$\left|\alpha-\frac{p}{q}\right|&lt;\frac{1}{Aq^2}\ A&gt;\sqrt{5}$$

$\alpha=\frac{p}{q}+\frac{\delta}{q^2}$ where
$|\delta|&lt;\frac{1}{A}&lt;\frac{1}{\sqrt{5}}$

Then $\frac{\delta}{q}=q\alpha-p$ and

$$\frac{\delta}{q}-\frac{1}{2}q\sqrt{5}=q\left(
\frac{1}{2}\sqrt{5}-1\right)-p-\frac{1}{2}q\sqrt{5}
=-\frac{1}{2}q-p$$

so

$$\left(\frac{\delta}{q}\right)^2-\delta\sqrt{5}+
\frac{5}{4}q^2=\left(\frac{1}{2}q+p\right)^2$$

so

$$\left(\frac{\delta}{q}\right)^2-\delta\sqrt{5}=p^2pq-q^2$$

when $q$ is large, since $|\delta|\sqrt{5}&lt;1$ the LHS is between
$-1$ and +1 whereas RHS is an integer.

So $p^2+pq-q^2=0$ i.e.$(2p+q)^2=5q^2$ which is impossible for
integers $p$ and $q$. So $\frac{1}{\sqrt{5}}$ is the best
possible. This establishes Hurwitz theorem.

%page 27

We now investigate for which numbers $\sqrt{5}$ is best possible.
It turns out that the criterion is that these numbers should end
in an infinite tail of 1's. We generalise this.

Definition

Two irrational numbers $\alpha$ and $\beta$ are equivalent if they
have the same tail to their continued fraction, in the sense that

$\alpha=[a_0;a_1,\ldots,a_k,c_0,c_1c_2\ldots]$

$\beta=[b_0;b_1,\ldots,b_j,c_0,c_1,c_2\ldots]$

Theorem

Two irrational numbers $\alpha$ and $\beta$ are equivalent if and
only if there exist integers $a,b,c,d$ with $ad-bc=\pm1$ such that

$$\alpha=\frac{A\beta+B}{C\beta+D}$$

Lemma

if $x=\frac{P\xi+R}{Q\xi+S}$ where $\xi&gt;1,\ PS-RQ=\pm1$, and
$Q&gt;S&gt;0$ then $\frac{r}{S}$ and $\frac{P}{Q}$ are two consecutive
convergents to the continued function for $x$. If $\frac{R}{S}$ is
the $(n-1)$th, $\frac{P}{Q}$ is the $n$th and $\xi$ is the
$(n-1)$th complete quotient.

Proof

$$\frac{P}{Q}=[a_0,a_1,\ldots,a_n]=\frac{p_n}{q_n}$$

$n$ can be even or odd. Choose it so that $PS-QR=(-1)^{n-1}$

%page 28

$(p,q)=1$ so $P=p_n\ Q=q_n$

so $p_nS-q_nR=(-1)^{n-1}=p_nq_{n-1}-p_{n-1}q_n$

so $p_n(S-q_{n-1})=q_n(R-p_{n-1})$

so $q_n|S-q_{n-1}$ since $(p_n,q_n)=1$.

Now

\begin{eqnarray*}
q_n&amp;=&amp;Q&gt;S&gt;0\\ q_n&amp;\geq&amp;q_{n-1}&gt;0\textrm{ so }\\ q_n&amp;&gt;&amp;| S-q_{n-1}|
\end{eqnarray*}

Hence $S-q_{n-1}=0$ and so $R-p_{n-1}=0$ thus


$\frac{R}{S}=\frac{p_{n-1}}{q_{n-1}}$ and
$x=\frac{p_n\xi+p_{n-1}}{q_n\xi+q_{n-1}}$

i.e.

\begin{eqnarray*}
x&amp;=&amp;[a_0,a_1,\ldots,a_n,\xi]\\ &amp;=&amp;[a_0,a_1,\ldots a_n,c_0,c_1]
\end{eqnarray*}

where$\xi=[c_0;c_1,c_2\ldots]$ and $c_0\neq0$ as $\xi&gt;1$ and so
$\xi$ is the $n+1$th complete quotient.

Proof of theorem

Suppose $\alpha=[a_0,\ldots a_k,c_0,c_1\ldots]=[a_0,\ldots a_k,w]$

$\beta=[b_0,\ldots b_j,c_o,c_1\ldots]=[b_0,\ldots b_j,w]$

then

$$\alpha=\frac{p_kw+p_{k-1}}{q_kw+q_{k-1}}\hspace{1.5cm}p_kq_{k-1}
-p_{k-1}q_k=\pm1$$

$$\beta=\frac{p_j'w+p_{j-1}'}{q_j'w+q_{j-1}'}\hspace{1.5cm}p_j'
q'_{j-1}-p'_{j-1}q;_j=\pm1$$

%page 29

eliminating $w$ will give

$$\alpha=\frac{A\beta+B}{C\beta+D}\textrm{ where }AD-BC=\pm1.$$

Now suppose

$$\alpha=\frac{A\beta+B}{C\beta+D}\ AD-BC=\pm1$$

assume w.l.o.g. $C\beta+D&gt;0$.

Let $\beta=[b_0,\ldots
b_{k-1}\beta_k]=\frac{p_{k-1}\beta_k+p_{k-2}}{q_{k-1}\beta_k+q_{k-
2}}$

substituting fo $\beta$ in $\alpha=\frac{A\beta+B}{C\beta+D}$
gives

$$\alpha=\frac{P\beta_K+R}{q\beta_k+s}$$

where

\begin{eqnarray*}
P&amp;=&amp;Ap_{k-1}+Bq_{k-1}\\ R&amp;=&amp;Ap_{k-2}+Bq_{k-2}\\
Q&amp;=&amp;Cp_{k-1}+Dq_{k-1}\\ S&amp;=&amp;Cp_{k-2}+Dq_{k-2}
\end{eqnarray*}

So $P,Q,R,S\in Z$ and
$$PS_QR=(AD-BC)(p_{k-1}q_{k-2}-p_{k-1}q_{k-1})=\pm1$$

Now
$\left|\beta-\frac{p_{k-1}}{q_{k-1}}\right|&lt;\frac{1}{q_{k-1}^2}$
and
$\left|\beta-\frac{p_{k-2}}{q_{k-2}}\right|&lt;\frac{1}{q_{k-2}^2}$

so

$$p_{k-1}=q_{k-1}\beta+\frac{\varepsilon}{q_{k-1}};p_{k-1}=q_{k-1}
\beta+\frac{\varepsilon'}{q_{k-1}}$$

where $|\varepsilon|&lt;1$ and $|\varepsilon'|&lt;1$.

%page 30

So

\begin{eqnarray*}
Q&amp;=&amp; (C\beta+D) q_{k-1}+ \frac{C\varepsilon}{q_{k-1}}
\\
S&amp;=&amp;(C\beta+D)q_{k-2}+\frac{C\varepsilon'}{q_{k-2}}
\end{eqnarray*}

Now $C\beta+D&gt;0$ and $q_{k-1}&gt;q_{k-1}$ also $q_n\rightarrow\infty$
as $n\rightarrow\infty$. So provided $k$ is sufficiently large,
$Q&gt;S&gt;0$

For such $k,\ \alpha=\frac{P\beta_k+R}{Q\beta_k+S}\ PS-QR=\pm1,
Q&gt;S&gt;0$

so $\beta_k$ is a complete quotient in the continued fraction for
$\alpha$ by the lemma thus $\alpha=[a_0;a_1,\ldots
a_m,b_k,b_{k+1}\ldots]$ i.e. $\alpha$ is equivalent to $\beta$.

We now define the Markov constant of an irrational number $\alpha$
by

$$M(\alpha)=sup\left\{\lambda:\left|\alpha-\frac{p}{q}\right|&lt;
\frac{1}{\lambda q^2}\textrm{ has infinitely many solutions
}\frac{p}{q}\right\}$$

So Huzwitz theorem says

$\forall\alpha\ M(\alpha\geq \sqrt{5}$ and
$M\left(\frac{1+\sqrt{5}}{2}\right)=\sqrt{5}$.

We now extend this :

%page 31

Theorem

If $\alpha$ is equivalent to $\beta$ then $M(\alpha)=M(\beta)$. If
$\alpha$ is not equivalent to $\frac{1+\sqrt{5}}{2}$ then
$M(\alpha)\geq\sqrt{8}$. If $\alpha$ is equivalent to $1+\sqrt{2}$
then $M(\alpha)=\sqrt{8}$.

Proof

Recall that

\begin{eqnarray*}
\left|\alpha-\frac{p_k}{q_k}\right|&amp;=&amp;\frac{1}{q_k(q_k\alpha_{k+1}+q_{k-1})}\\
&amp;=&amp;\frac{1}{q_k^2\left(\alpha_{k+1}+\frac{q_{k-1}}{q_k}\right)}
\end{eqnarray*}

Thus
$$M(\alpha)=\lim_{k\to\infty}sup\left(\alpha_{k+1}+\frac{q_{k-1}}{q_k}\right)$$

Recall from the discussion of symmetric continued fractions that

$$\frac{q_k}{q_{k-1}}=[a_k;a_{k-1},\ldots a_1]$$

so

$$\frac{q_{k-1}}{q_k}=[0;a_ka_{k-1}\ldots a_1]$$

so

$$M(\alpha)=\lim_{k\to\infty}sup\left([0;a_k,a_{k-1},\ldots
a_1]+\alpha_{k+1}\right)$$

Now if $\alpha$ is equivalent to $\beta$ then $\beta_j=\alpha_k$
and $b_j=a_k$ for all sufficiently large $k$ and $j$ for which
$j-k$ has a suitable fixed value $h$.

%page 32

If the convergents of $\beta$ are $\frac{P_j}{Q_j}$ then for $j$
and $k$ differing by $h$, the continued fractions for
$\frac{q_{k-1}}{q_k}$ and $\frac{Q_{j-1}}{Q_j}$ have rhe same
partial quotients at the beginning, and the length of agreement
can be made large by making $j$ and $k$ sufficiently large.

Suppose $\frac{q_{k-1}}{q_k}$ and $\frac{Q_{j-1}}{Q_j}$ agree in
the first $l_1$ partial quotients, and denote the common
convergents by $\frac{r_i}{s_i}\ (i=0,\ldots l)$

so

$$\frac{q_{k-1}}{q_k}=\frac{r_{l-1}x_l+r_{l-2}}{s_{l-1}x_l+s_{l-2}}$$

and

$$\frac{Q_{j-1}}{Q_{j}}=\frac{r_{l-1}y_l+r_{l-2}}{s_{l-1}y_l+s_{l-2}}$$

Then $[x_l]=[y_l]=$ common $l+1$th partial quotient so
$|x_l-y_l|\leq1$.

Then we have

$$\left|\frac{q_{k-1}}{q_k}-\frac{Q_{j-1}}{Q_j}\right|=\frac{|x_l-y_l|}{
(s_{l-1}x_l+s_{l-2})(s_{l-1}y_l+s_{l-1})}\leq\frac{1}{s_{l-1}^2}$$

Now provided $j$ and $k$ are large enough, we have

$$\left|\frac{q_{k-1}}{q_k}-\frac{Q_{j-1}}{Q_j}\right|&lt;\varepsilon$$

since $s_{l-1}\geq(l-1)$th term in Fibonacci sequence.

%page 33

Also for large $j,k\ \alpha_k=\beta_j$, so

$$\left(\alpha_k+\frac{q_{k-1}}{q_k}\right)-\left(\beta_j+\frac{Q_{j-1
}}{Q_j}\right)=\frac{q_{k-1}}{q_k}-\frac{Q_{j-1}}{Q_j}\to0\textrm{
as }j,k\to\infty,\ j-k=h $$

Thus $M(\alpha)=M(\beta)$

If $\alpha$ is not equivalent to $\frac{\sqrt{5}+1}{2}$ then
infinitely many of the $a_k$ are $\geq2$.

If $a_k\geq3$ for infinitely many $k$ then

\begin{eqnarray*}
M(\alpha)&amp;=&amp;\lim
sup\left(\alpha_{k+1}+\frac{q_{k-1}}{q_k}\right)\\ &amp;\geq&amp;\lim
sup(a_{k+1})\geq3
\end{eqnarray*}

So suppose that the $a_k$ contain only 1's and 2's from some point
on.

Case I

$a_k=2$ from some point on. Then $\alpha$ is equivalent to
$1+\sqrt{2}=[2;2,2,\ldots]$

\begin{eqnarray*}
M(\alpha)&amp;=&amp;\lim
sup\left(\alpha_{k+1}+\frac{q_{k-1}}{q_k}\right)\\
\alpha_{k+1}&amp;=&amp;[2;2,\ldots]=1+\sqrt{2}\\
\frac{q_{k-1}}{q_k}&amp;=&amp;[0;\underbrace{2,2,\ldots}_{k \textrm{
times}}]\to\frac{1}{1+\sqrt{2}}\textrm{ as }k\to\infty
\end{eqnarray*}

So $M(\alpha)=1+\sqrt{2}+\frac{1}{1+\sqrt{2}}=\sqrt{8}$

Case II

Suppose thee are infinitely many 1's and 2's.

%page 34

Then there are infinitely many $k$ such that $a_k=1$ and
$a_{k+1}=2$, so

\begin{eqnarray*}
\alpha_{k+1}&amp;=&amp;
2+\frac{1}{a_{k+2}+}\frac{1}{a_{k+3}}\geq2+\frac{1}{2+\frac{1}{1}}=\frac{7}{3}\\
\frac{q_{k-1}}{q_k}&amp;=&amp;
\frac{1}{a_k+}\frac{1}{a_{k-1}+}\ldots\geq\frac{1}{1+\frac{1}{a_{k-1}}}\geq\frac{1}{1+\frac{1}{1}}=\frac{1}{2}
\end{eqnarray*}

So $M(\alpha)\geq\frac{7}{3}+\frac{1}{2}=\frac{17}{6}&gt;\sqrt{8}$

Note: This shows that if $\alpha\not\sim1+\sqrt{2}$ then
$M(\alpha)\geq\frac{17}{6}$.

Theorem

There are uncountably many $\alpha$ with $M(\alpha)=3$

Proof

Let
$\alpha=[\underbrace{1;1,1,\ldots1}_{r_1},2,2,\underbrace{1,1,\ldots1
}_{r_2},2,2,\underbrace{1,1,\ldots1}_{r_3},2,2,1,\ldots]\
r_1&lt;r_2&lt;r_3$

\begin{description}

\item[(i)]
If $a_{k+1}=1$ then $\alpha_{k+1}&lt;2$ and since
$\frac{q_{k-1}}{q_k}&lt;1,\ \alpha_{k+1}+\frac{q_{k-1}}{q_k}&lt;3$.

\item[(ii)]
If $a_{k+1}=2$ and $a_{k+1}=2$ then
$$\alpha_{k+1}+\frac{q_{k-1}}{q_k}=\left(2+\frac{1}{2+}
\frac{1}{1+ }\frac{1}{1+} \ldots\right)+ \left(
\frac{1}{1+}\frac{1}{1+ }\ldots \frac{1}{1} \right)$$

If $k$ is large the sequences of 1's can be made as long as we
like before a 2 appears. So
$\alpha_{k+1}+\frac{q_{k-1}}{q_k}\to2+\frac{1}{2+\frac{\sqrt{5}
-1}{2}}+\frac{\sqrt{5}-1}{2}=3$

\item[(iii)]
If $a_{k+1}=2$ and $a_k=1$ then

$$\alpha_{k+1}+\frac{q_{k-1}}{q_k}=\left(2+\frac{1}{1+}\frac{1}{
1+}\ldots\right)+\left(\frac{1}{2+}\frac{1}{1+}\ldots\frac{1}{1}
\right)\to2+\frac{1}{\frac{\sqrt{5}+1}{2}}+\frac{1}{2+\frac{
\sqrt{5}-1}{2}}=3$$

\end{description}

%page 35

So $M(\alpha)=\lim
sup\left(\alpha_{k+1}+\frac{q_{k-1}}{q_k}\right)=3$

Two such $\alpha$'s are equivalent iff their associated sequences
of $r_i$'s are equivalent in the same sense of having equal tails.
There are uncountably many inequivalent such sequences of $r_i$'s
so uncountably many inequivalent $\alpha$ with $M(\alpha)=3$.


\end{document}
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