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CONTINUED FRACTIONS

IRRATIONAL NUMBERS

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Irrational numbers

Terminology:

If $\alpha=a_0+\frac{1}{a_1+}\frac{1}{a_2+}\ldots$ then the $a_i$
are called partial quotients.

$$a_0+\frac{1}{a_1+}\frac{1}{a_2+}\ldots\frac{1}{a_n}=
\frac{p_n}{q_n}$$

is called the $n$-th convergent

If we write

$$\alpha=a_0+\frac{1}{a_1+}\frac{1}{a_2+}\ldots\frac{1}{a_{n-1}+}
\frac{1}{\alpha_n}$$

$\alpha_n$ is called a complete quotient.

To develop a number $\alpha\in R-Q$ as a continued fraction, we
use the following recursive scheme (we take $\alpha>0$)

$\alpha=a_0+\frac{1}{\alpha_1}$ where $a_0=[\alpha]$ and so
$\alpha_1>1$

$\alpha+n=a_n+\frac{1}{\alpha_{n+1}}$ where $a_n=[\alpha_n]$ and
so $\alpha_n>1$.

We need to give a meaning to the infinite expression

$$[a_0;a_1,a_2\ldots]$$

and to associate it with $\alpha$.

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Suppose we take this development of an irrational and truncate it
so

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

and

$$\alpha=[a_0;a_1\ldots a_n\alpha_{n+1}]$$

\begin{eqnarray*}
\alpha-\frac{p_n}{q_n}&=&\frac{\alpha_{n+1}p_n+p_{n-1}}{
\alpha_{n+1}q_n+q_{n-1}}-\frac{p_n}{q_n}\\
&=&\frac{(-1)^n}{q_n(\alpha_{n+1}q_n+q_{n-1})}\to0\textrm{ as
}n\to\infty
\end{eqnarray*}

So every number has a continued fraction expression - finite if
rational and infinite if irrational. It is unique apart from the
choice at the end of a finite continued fraction.

Further more every continued fraction does converge. We have seen
that the even convergents form a sequence bounded above, and the
odd convergents form a sequence bounded below.

Also $\frac{p_n}{q_n}-\frac{p_{n-1}}{q_{n-1}}=\frac{(-1)^{n-1}}{
q_nq_{n-1}}\to 0$ as $n\to\infty$ as $q_n\to\infty$ as
$n\to\infty$

Hence there is a 1-1 correspondence between sequences of natural
numbers and irrational numbers. The Cantor diagonal argument is
even easier in this case than with decimals, to prove that the
irrationals are uncountable.

List irrationals $\alpha_1,\ldots\alpha_n\ldots$

Let $\alpha$ be the irrational whose continued fraction has its
$n$-th partial quotient 1 more than that of $\alpha_n$, for each
$n$.

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We can say a bit more about convergence to $\alpha$, as follows:

Again writing $\alpha=[a_0;a_1\ldots a_n\alpha_{n+1}]$

$$|q_n\alpha-p_m|=\frac{1}{\alpha_{n+1}q_n+q_{n-1}}$$

Now $\alpha_{n+1}q_n+q_{n-1}>a_{n+1}q_n+q_{n-1}=q_{n+1}$

However
$\alpha_{n+1}q_n+q_{n-1}<(a_{n+1}+1)q_n+q_{n-1}=q_{n+1}+q_n
<q_{n+2}$ so

$$\frac{1}{q_{n+2}}<|q_n\alpha-p_n|<\frac{1}{q_{n+1}}$$

$$\frac{1}{q_{n+3}}<|q_{n+1}\alpha-p_{n+1}|<\frac{1}{q_{n+2}}$$

so $|q_{n+1}\alpha-p_{n+1}|<|q_n\alpha-p_n|$

Divide LHS by $q_{n+1}$ and RHS by $q_n$ ($q_{n+1}>q_n$) to give

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

So the convergents get successively nearer to $\alpha$,
alternating either side of $\alpha$.

An aside:

Consider now the set of all real numbers whose first $(n+1)$
partial quotients are fixed.

$$\alpha=[a_0;\ldots a_n,\alpha_{n+1}]\textrm{ so
}1\leq\alpha_{n+1}<\infty,\ 0<\frac{1}{\alpha_{n+1}} \leq1$$

Consider $\alpha_x=[a_o;\ldots a_n,x]\ \alpha_y=[a_0, \ldots
a_n,y]$

\begin{eqnarray*}
\alpha_x-\alpha_y&=&\frac{xp_n+p_{n-1}}{xq_n+q_{n-1}}-
\frac{yp_n+p_{n-1}}{yq_n+q_{n-1}}\\
&=&\frac{(xp_n++p_{n-1})(yq_n+q_{n-1})-(yp_n+p_{n-1})
(xq_n+q_{n-1})}{(xq_n+q_{n-1})(yq_n+q_{n-1})}\\
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&=&\frac{(x-y)(p_nq_{n-1}-p_{n-1}q_n)}{(xq_n+q_{n-1}
)(yq_n+q_{n-1})}\\
&=&\frac{(x-y)(-1)^{n-1}}{(xq_n+q_{n-1})(yq_n+q_{n-1})}
\end{eqnarray*}

so if $n$ is even $\alpha_x$ is a decreasing function of $x$ and
if $n$ is odd $\alpha_x$ is an increasing function of $x$.

So with $0<x\leq1,\ \alpha_x$ occupies an interval, with one end
point (included)

$$[a_0;\ldots,a_n,1]=\frac{p_n+q_{n-1}}{q_n+q_{n-1}}$$

and the other end point (excluded)

$$[a_0;\ldots a_n]=\frac{p_n}{q_n}$$

whose length is

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


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