From $q'=\frac{q-1}{2q-1}$ we get $q+q'=2qq'+1$. Starting from $q=0$, we can change $q$ adding $\pm 1$ or flip to $q'$ and again do the same. Let us consider the substitutions $u=q-1/2, u'=q'-1/2$. Quick calculation gets $uu'=-1/4$. Thus, we reduce the problem as: starting from $u=1/2$, we add $\pm 1$ to $u$ or flip to $u'$. Finally, let $v=2u,v'=2u'$, so $vv'=-1$ and the problem boils down to following rule: We start from $v=-1$, we can make consecutively one of the two following steps: 1) Add $\pm 2$ to $v$ 2) Flip to $v'$ satisfying: $vv '=-1$ Let $v=\frac{r}{s}$, where $r,s$ are integers. Consider pair $(r,s)$. We transform that pair to either $(r\pm 2s,s)$ or $(-s,r)$. Starting from $(-1,1)$, we can obviously get only $(\mbox{odd }, \mbox{ odd})$ pairs. But $\frac{1}{2018}$ corresponds to $v=2\left(\frac{1}{2018}-\frac{1}{2}\right)=-\frac{1008}{1009}$, which corresponds to $(r,s)=(-1008, 1009)$. It means we can never get to this pair following the above rules. That answers to the a) For b), to get to $(-1008,1009)$, I suppose we can start from any pair $(r,s)$ with g.c.d=1, which is not $(\mbox{ odd }, \mbox{ odd})$.

Actually for a) one may notice that the fraction is always in its lowest terms. Thus, the denominator residue $\mod 2$ must be $0$. However. notice that $a_2=\pm 1$ and there is no point going back to $0$. Thus we cannot reach $\dfrac{1}{2n}$

dgrozev wrote: But $\frac{1}{2018}$ corresponds to $v=2\left(\frac{1}{2018}-\frac{1}{2}\right)=-\frac{1008}{2009}$, which corresponds to $(r,s)=(-1008, 2009)$. Actually, $v=2\left(\frac{1}{2018}-\frac{1}{2}\right)=-\frac{1008}{1009}$. But this doesn't change the vality of the proof.

@above: Thanks, edited.