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CONTINUED FRACTIONS

RATIONAL NUMBERS

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We now insist that $a_n\in N$ for all $n$. Every finite continued
fraction represents a rational number.

Furthermore every rational number can be represented as a finite
continued fraction, from the Euclidean algorithm.

Now suppose that

$$x=a_0+\frac{1}{a_1+}\ldots\frac{1}{a_N}\hspace{1cm}x\in Q$$

If we insist that the last $a_N$ is $>1$ then this is unique.
Otherwise the only other representation is

$$x=a_0+\frac{1}{a_1+}+\ldots\frac{1}{a_{N-1}+}\frac{1}{
(a_{N}-1)+}\frac{1}{1}$$

Proof

Let $x\in Q$ and suppose $x=[a_0;a_i\ldots a_N]=[b_0;b_1\ldots
b_M]$

Let $a_n'=[a_n;a_{n+1}\ldots a_N]\ b_n'[n_n;n_{n+1}\ldots b_M]$

So

\begin{eqnarray*}
x&=&[a_0;a_1\ldots a_{n-1},a_n']\ 1\leq n\leq N\\
&=&[b_0;b_1\ldots b_{n-1},b_n']\ 1\leq n\leq M
\end{eqnarray*}

For $n<N$

$$a_n'=a_n+\frac{1}{a_{n+1}'},\ (a_{n+1},>1)$$

so that we must have $a_n=[a_n']$

Similarly for $n<M\ b_n=[b_n']$

Now we must have $a_0=b_0=[x]$, so suppose $a_1=b_1,\ldots
a_{n-1}=b_{n-1}$

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so $\frac{p_{n-1}}{q_{n-1}}=[a_0;\ldots,a_{n-1}]=[b_0;\ldots
b_{n-1}]$

Thus

$$x=\frac{a_n'p_{n-1}+p_{n-2}}{a_n'q_{n-1}+q_{n-2}}=\frac{b_n'
p_{n-1}+p_{n-2}}{b_n'q_{n-1}+q_{n-2}}$$

Cross multiplying gives

$$b_n'(p_{n-2}q_{n-1}-p_{n-1}q_{n-2})=a_n'(p_{n-2}q_{n-1}-
p_{n-1}q_{n-2})$$

so $a_n'=b_n'$ since the term in the bracket =$(-1)^n\neq0$.

So, since $a_n=[a_n']$ and $b_n=[b_n']\ a_n=b_n$.

So $a_n=b_n$ for all $n$, and when one expansion terminates so
does the other.

Note that $a_n=[a_n']$ requires $a_n+1>1$. If $a_{n+1}'=1$ then
$a_n=[a_n']-1$ and $a_{n+1}=1$, and the expansion terminates. All
previous $a_i'$ are $>1$, otherwise the expansion would have
terminated earlier.

Linear Diophantine Equations.

Because of the ambiguity at the end of the expansion of a rational
number, we can always an expansion $[a_o;\ldots a_N]$ with $N$
odd.

Let $x=\frac{a}{b}$. So $\frac{a}{b}=\frac{p_n}{q_n}$

$$p_Nq_{N-1}-q_Np_{N-1}=(-1)^{N-1}=1$$

i.e.

$$aq_{N-1}-bp_{N-1}=1$$

So $x=p_{N-1}\ y=q_{N-1}$ is a positive integer solution of the
equation $ax-by=1$

Notice that this shows $(P_{N-1},q_{N-1})=1$

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In fact since $p_nq_{n-1}-q_np_{n-1}=(-1)^{n-1}, (p_n,q_n)=1$ for
any $n$

So when we \lq\lq add up'' a continued fraction we always obtain
the rational answer in it's lowest terms.

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The following result concerns approximation.

Let $\alpha$ be a real algebraic number of degree $n$. So

$$f(\alpha)=a_n\alpha_n+a_{n-1}\alpha^{n-1}+\ldots+a_0=0,\
a_n\neq0,\ a_i\in Z$$

Now $\exists M, \forall x\in (\alpha-1,\alpha+1) |f'(x)|<M$
Suppose $\frac{p}{q}\neq alpha$ is an approximation to $\alpha$,
close enough to be in $(\alpha-1,\alpha+1)$, and nearer to
$\alpha$ than any other root of $f(x)=0$, so
$f\left(\frac{p}{q}\right)\neq0$.

$$\left|f\left(\frac{p}{q}\right)\right|=\frac{|a_0p^n
+a_1p^{n-1}q+\ldots| }{q^n}\geq\frac{1}{q^n}$$

since the numerator is a positive integer.

$$F\left(\frac{p}{q}\right)=f\left(\frac{p}{q}\right)-f(\alpha)=
\left(\frac{p}{q}-\alpha\right)f'(\xi)\ MUT$$

with $\xi$ between $\frac{p}{q}$ and $\alpha$ so
$$\left|\frac{p}{q}-\alpha\right|=\left|\frac{f\left(\frac{p}{q}
\right)}{f'\left(\xi\right)}\right|>\frac{1}{Mq^n}=\frac{K}{q^n}$$

We have already shown that if $\alpha$ is irrational then there
are infinitely many solutions of

$$\left|\frac{p}{q}-\alpha\right|<\frac{1}{q^2}$$

$\alpha$ is said to be approximable to order 2.

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Now let $\alpha=0.110001000\ldots=\frac{1}{10}+\frac{1}{10^{2!}}+
\frac{1}{10^{3!}}+\ldots$

Let $n_0\in N$ and let $n>n_0$. Let
$$\alpha_n=\frac{p}{10^{n!}}=\frac{p}{q}$$

be the sum of the first $n$ terms of the series

$$0<\alpha-\frac{p}{q}=\frac{1}{10^{(n+1)!}}+
\frac{1}{10^{(n+2)!}}+\ldots$$

$$<\frac{2}{10^{(n+1)!}}<\frac{2}{q^{n+1}}< \frac{2}{q^{n_0}}$$

so $\alpha$ is approximable to order $n_0$, for any $n_0$.

Hence $\alpha$ is transcendental.

The theorem above shows that an algebraic number of degree $n$ is
not approximable to order $\nu=n$

\begin{tabular}{ccc}
A Thue&(1901)& improved this to $\nu=\frac{n}{2}+1$\\ C L Siegel
&(1921)&improved this to $\nu>min_{1\leq s\leq n-1, s\in
N}\left(\frac{n}{s+1}+s\right)$\\ F J Dyson &(1947)& improved this
to $\nu>2\sqrt{n}$\\ K F Roth &(1955)& improved this to
$\nu=2+\varepsilon$ for all $\varepsilon>0$
\end{tabular}

This result is best possible.


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