Condition on Approximation to Infinite Continued Fraction

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  
\[gcd(p_{n+1}, \; q_{n+1})=1\]
  and this contradicts  
\[0 \lt b \lt q_{n+1}\]
. 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.
\[\begin{equation} \begin{aligned} \| bx-a \| &= \| (rq_n-sq_{n+1})x -(rp_n+sp_{n+1} ) \| \\ &= \| r(q_nx-p_n ) +s(q_{n+1} x-p_{n+1}) \| \\ &= \| r \| \| q_nx-p_n \| + \| s \| \| q_{n+1} x -p_{n+1} \| \\ & \gt \| r \| \| q_nx-p_n \| \ge \| s \| \| q_n x -p_n \| \end{aligned} \end{equation}\]

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