Theorem
Let
$\frac{p_n}{q_n}$
be the nth convergent of the infinite continued fraction representing the irrational number
$x$
, and let
$\frac{a}{b}$
be any rational number with
$0 \lt b \lt q_{n+1}$
, Then for
$n \gt 1$
,
$\| q_n x - p_n \| \le \| nx-a \|$
with equality holding only of
$a=p_n, \; b=q_n$
.
Proof
Let
$\frac{a}{b}$
be a rational number with
$b \lt q_{n+1}$
and suppose it is not the case that
$a=p_n, \; b=q_n$
, for which the inequality holds.
Consider the system
$a=rp_n+sp_{n+1}$
(1)
$b=rq_n+sq_{n+1}$
(2)
$q_n \times (1)-p_n \times (2)$
gives
$aq_n-bp_n=s(sp_{n+1}q_n-p_nq_{n+1})$
and using Properties of Convergents of Finite Continued Fractions we write
$s=(-1)^{n+1} (aq_n-bp_n)$
.
Similarly
$r=(-1)^{n+1}(bp_{n+1}-aq_{n+1})$
&.
$r, \; s \in \mathbb{N}$
and neither is zero (if
$r=0$
then
$aq_{n+1}=bp_{n+1}$
and Euclid's Lemma forces
$q_{n+1}$
to divide
$b$
since
$gcd(p_{n+1}, \; q_{n+1})=1$
$0 \lt b \lt q_{n+1}$
. If
$s=0$
then
$\frac{a}{b} = \frac{p_n}{q_n}$
which has been excluded).
$0 \lt b \lt rq_n+sq_{n+1} \lt q_{n+1}$
so the non zero integers
$r, \; s$
must have opposite sign.
$x- \frac{p_n}{q_n}, \; x- \frac{p_{n+1}}{q_{n+1}}$
also have opposite signs because the convergents
$\frac{p_n}{q_n}, \; \frac{p_{n+1}}{q_{n+1}}$
are alternately less than and greater than
$x$
. It follows that
$r(q_nx-p_n)$
and
$s(q_{n+1}x-p_{n+1})$
have the same sign.
\begin{aligned} \| bx-a \| &= \| (rq_n-sq_{n+1})x -(rp_n+sp_{n+1} ) \| \\ &= \| r(q_nx-p_n ) +s(q_{n+1} x-p_{n+1}) \| \\ &= \| r \| \| q_nx-p_n \| + \| s \| \| q_{n+1} x -p_{n+1} \| \\ & \gt \| r \| \| q_nx-p_n \| \ge \| s \| \| q_n x -p_n \| \end{aligned}