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{\bf Math 431 Exam III On Approximation of Irrational Numbers}

\vspace{0.1in}
{by  Pit-Mann
Wong}
\end{center}

\medskip
\noindent
{\bf Problem I.}\; Find a simple continued fraction expansion of
$1729/625$.

\vskip 4in

\medskip
\noindent
{\bf Problem II.}\; The simple continue fraction $p/q = [a_0, a_1, ...,
a_n]$ of a rational number is unique if $a_n = 1$. If $a_n \ne 1$ show
that $p/q = [a_0, a_1, ...,
a_n-1, 1]$.

\newpage

\medskip
By definition the $k$-th convergent of a simple continue fraction 
$p/q = [a_0, a_1, ...,
a_n]$ is
given inductively by
$$
C_k = \frac{p_k}{q_k} = \frac{a_k p_{k-1} + p_{k-2}}{a_k q_{k-1} +
q_{k-2}}
$$
with $p/q = C_n$.

\medskip
\noindent
{\bf Proposition 1.}\;{\it Let $[a_0, a_1, ..., a_n]$ be a simple
continued fraction. Write the
$k$-th convergent
$C_k = p_k/q_k$ where $p_k, q_k$ are relatively prime then $p_0 = a_0,
q_0 = 1$ and for $k \ge 1$
$$
\begin{pmatrix}
p_k &q_k\\
p_{k-1} &q_{k-1}
\end{pmatrix} = \begin{pmatrix}
a_k &1\\
1 &0
\end{pmatrix} \begin{pmatrix}
p_{k-1} &q_{k-1}\\
p_{k-2} &q_{k-2}
\end{pmatrix}.
$$
Moreover, we have
$$
\begin{cases}
p_kq_{k-1} - q_kp_{k-1} = (-1)^{k-1},\\
p_kq_{k-2} - q_kp_{k-2} = (-1)^k a_k.
\end{cases}
$$}

\medskip
The preceding proposition may be used to solve the Diophantine equations:
$$
Ax - By = \pm 1
$$
provide that $A$ and $B$ are relatively prime.

\medskip
\noindent
{\bf Example 2.}\; To solve the equation $257 x - 97 y = -1$, first find
the ssimple continue fraction expansion of $257/97$:
$$
\frac{257}{97} = [2, 1, 1, 1, 5, 1, 4]
$$
and its convergents  $2, 3, 5/2, 8/3, 45/17, 53/20, 257/97$. Applying
proposition 1 with $k = 6$ yields $$257 \cdot 20 - 97 \cdot 53 =
(-1)^{6-1} = - 1$$
so $x = 20$ and $y = 53$ is a solution. If we ant to solve the equation
$257 x - 97 y = 1$ then we have to replace the expansion by (see Problem
II):
$$
\frac{257}{97} = [2, 1, 1, 1, 5, 1, 3, 1]
$$
with convergents $2, 3, 5/2, 8/3, 45/17, 53/20, 204/77, 257/97$.
proposition 1 with $k = 7$ yields:
$$257 \cdot 77 - 97 \cdot 204 =
(-1)^{7-1} = 1$$
thus $x = 77, y = 204$ is a solution.

\medskip
\noindent
{\bf Problem III.}\; Solve the equations (1) 417 x - 172 y = 1$, (2) 1163
x - 815 y = 1$.

\newpage

\medskip
\noindent
{\bf Example 3.}\; Let $a/b = [a_0, a_1, ...,
a_n, a_n, ..., a_1, a_0], a_i > 0$ for all $i$. A continue fraction of
this form is said to be symmetric. Let
$C_k = p_k/q_k$ be the $k$-th convergent how that $\{p_k\}, \{q_k\}$ are
increasing sequences and
$$
\begin{cases}
p_n/p_{n-1} = [a_n, ..., a_1, a_0],\\
q_n/p_{n-1} = [a_n, ..., a_1].
\end{cases}
$$

\medskip
\noindent
{\it Proof.}\;By Proposition 1, $p_k = a_k p_{k-1} + p_{k-2}$ which
implies that $p_k$ increases as each term on the right is positive.
Moreover, we get
$$
\frac{p_n}{p_{n-1}} = a_n + \frac{p_{n-2}}{p_{n-1}}
$$
hence the first term of the continue fraction expansion for $p_n/p_{n-1}$
is $a_n$ with fractional part $p_{n-2}/p_{n-1}$. The next term of the
expansion is obtained from $p_{n-1} = a_{n-1} p_{n-2} + p_{n-3}$,
dividing through by $p_{n-2}$ yields
$$
\frac{p_{n-1}}{p_{n-2}} = a_{n-1} + \frac{p_{n-3}}{p_{n-2}}
$$
which shows that the second term of the continue fraction expansion 
for $p_n/p_{n-1}$
is $a_{n-1}$ with fractional part $p_{n-3}/p_{n-2}$. It is clear that
this process may be continued until we arrive at $p_0 = a_0$ with no
fractional part. The proof is entirely analogous for $q_k$.\; QED

\medskip
\noindent
{\bf Example 4.}\; If the convergents of a simple continue fraction are
given by $p_k/q_k$ then
$$
(p_k^2 - q_k^2)(p_{k-1}^2 - q_{k-1}^2) = (p_kp_{k-1} - q_kq_{k-1})^2 - 1.
$$

\medskip
\noindent
{\it Proof.}\;Squaring the right hamd side above yields
$$
RHS = p_k^2p_{k-1}^2 + q_k^2q_{k-1}^2 - 2p_kp_{k-1}q_kq_{k-1} - 1.
$$
On the other hand, the left hand side above equals
$$
LHS = p_k^2p_{k-1}^2 + q_k^2q_{k-1}^2 -  p_k^2q_{k-1}^2 -
q_k^2p_{k-1}^2.$$ By the second part of Proposition 1,
$$p_kq_{k-1} - q_kp_{k-1} = (-1)^{k-1}$$
and squaring yields
$$
p_k^2q_{k-1}^2 +  q_k^2p_{k-1}^2 - 2p_kp_{k-1}q_kq_{k-1} = 1
$$
which implies that $RHS = LHS$.\; QED

\medskip
\noindent
{\bf Example 5.}\; Let $a/b = [a_0, a_1, ...,
a_n, a_n, ..., a_1, a_0] = p_{2n+1}/{q_{2n+1}}, a_i > 0$ for all $i$.
Show that $p_{2n+1} = p_n^2 + p_{n-1}^2$ and $q_{2n+1} = p_nq_n + p_{n-1}
q_{n-1}
$.

\medskip
\noindent
{\it Proof.}\; Let $x = [a_n, ..., a_1, a_0]$ then $$\frac{a}{b} = [a_0,
a_1, ..., a_n, x] = (n+1)-\text{st convergent of}\;[a_0,
a_1, ..., a_n, x] = \frac{xp_n + p_{n-1}}{xq_n + q_{n-1}}.$$ By example 3
$p_n/p_{n-1} = x$ hence the preceding can also be expressed as
$$
\frac{p_{2n+1}}{q_{2n+1}} = \frac{a}{b} = \frac{p_n^2 +
p_{n-1}^2}{p_{n}q_n + p_{n-1}q_{n-1}}.
$$
QED

\medskip
\noindent
{\bf Example 6.}\; Let $a/b = [a_0, a_1, ...,
a_n, a_n, ..., a_1, a_0]$ be a symmetric continue fraction. Show that
$p_{2n} = b$ and $a | b^2 + 1$.

\medskip
\noindent
{\it Proof.}\; Since there are $2n+2$ terms in the simple fraction $a/b =
C_{2n+1} = p_{2n+1}q_{2n+1}$ hence $p_{2n+1} = a, q_{2n+1} = b$. By
Example 3 and the condition that the continue fraction is symeetric we
onclude that
$$
\frac{p_{2n+1}}{p_{2n}} = [a_0, a_1, ...,
a_n, a_n, ..., a_1, a_0] = \frac{a}{b}
$$
hence $p_{2n} = b$. By the second part of Proposition 1,
$$p_{2n+1}q_{2n} - q_{2n+1}p_{2n} = (-1)^{2n+1-1} = 1$$
hence $aq_{2n} - b^2 = 1$.\;QED

\medskip
\noindent
{\bf Problem IV.}\;Let $a/b = [a_0, a_1, ...,
a_{n-1}, a_n, a_{n-1} ..., a_1, a_0]$ show that $a
| b^2 - 1$.
\vskip 3in

\medskip
\noindent
{\bf Problem V.}\; Recall that the Fibonacci numbers are defined by the
conditions $f_1 = f_2 = 1$ and $f_n = f_{n-1} + f_{n-2}$ for $n \ge 3$.
Show that

\medskip
(a)\; The simple continue fraction expansion of $f_{n+1}/f_n = [1, ...,
1]$ ($n$ ones).

\medskip
(b)\; What identities can you derive for the Fibonaci sequences from
Examples 5, 6 and problem IV?

\medskip
(c)\;Conclude that any thre consecutive Fibonacci numbers have no common
factor.

\newpage

\medskip
\noindent
{\bf Definition 6.}\; A rational number $p/q, p. q$ relatively prime is
said to be a good (resp. best) approximation of an irrational number
$\alpha$ if
$$
|\alpha - \frac{p}{q}| < |\alpha - \frac{p^{'}}{q^{'}}|\;\; {\rm
(resp.}\;\;|q\alpha -
p| < |q^{'} \alpha - p^{'}|)
$$
for all rational number $p^{'}/q^{'}$ with $p^{'}, q^{'}$ relatively
prime and $q^{'} < q$.

\medskip
\noindent
{\bf Proposition 7.}\;{\it If $p/q$ is a best rational approximation of
an irrational number $\alpha$ then it is also a good approximation.}

\medskip
\noindent
{\bf Theorem 8.}\;{\it A raional number is a best
approximation of an irrational number $\alpha$ if and only if it is a
convergent of the simple continuous frction expansion of $\alpha$.}

\medskip
\noindent
{\bf Example 9.}\;  Rational numbers with respective
denominators $1, 2, 3, 4, 5, 6$ approximating $\pi \sim 3.1415927 $ are
respectively
$3/1, 7/2, 10/3, 13/4, 16/5, 19/6$. The approximation $3$ is a best
approximation, $7/2, 10/3$ are not good approximations, $13/4, 16/5,
19/6$ are good approximations but not best.  Thus the converse of
Proposition 1 is not valid. The simple continuous fraction expansion of
$\pi$ is given by
$[3, 7, 15, 1, 292, 1, ...]$. The convergents are given by
\begin{align*}
C_0 &= 3,\\
C_1 &= 3 + \frac{1}{7} = \frac{22}{7},\\
C_2 &= 3 + \frac{1}{7+\frac{1}{15}} = \frac{333}{106},\\
C_3 &= 3 + \frac{1}{7+\frac{1}{15+ \frac{1}{1}}} = \frac{355}{113},\\
C_4 &= 3 + \frac{1}{7+\frac{1}{15+ \frac{1}{1}}} = \frac{355}{113},\\
C_5 &= 3 + \frac{1}{7+\frac{1}{15+ \frac{1}{1+\frac{1}{292}}}} =
\frac{111613}{37204}.
\end{align*}

\medskip
\noindent
{\bf Corollary 10.}\;{\it If a rational number $p/q, p, q$ relatively
prime satisfyies
$$
|\alpha - \frac{p}{q}| < \frac{1}{2q^2}
$$
then $p/q$ is a convergent of $\alpha$.}

\medskip
\noindent
{\bf Corollary 11.}\;{\it A number $\alpha$ is irrational if and only
if the inequality
$$
|\alpha - \frac{p}{q}| < \frac{1}{q^2}
$$
admits ininitely many rational solutions $p/q, p, q$ relatively prime.}

\medskip
\noindent
{\bf Problem VI.}\; In Example 9 verify that The approximation $3$ is a best
approximation, $7/2, 10/3$ are not good approximations, $13/4, 16/5,
19/6$ are good approximations but not best.



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