Let $\frac{p}{q}$ be such a rational number. Define $a_0=\frac{p}{q}$ and inductively $a_n \in (0,1)$ by $a_n=2a_{n-1}$ if $a_{n-1}<\frac{1}{2}$ (call this step A) and $a_n=\frac{1}{a_{n-1}}-1$ if $a_{n-1}>\frac{1}{2}$ (call this step B). Stop if $a_{n-1}=\frac{1}{2}$. By construction, we are left to show that $\frac{1}{2}$ eventually occurs in our sequence, whatever the values of $p$ and $q$. Let $a_n=\frac{p_n}{q_n}$ be the representation in minimal terms. Clearly, the value of $q_n$ decreases when applying step $B$ and it does not decrease when we apply step $A$. Supposing that $\frac{1}{2}$ never occurs in our sequence, it must be infinite, so $q_n$ must be eventually constant. But this means that we must eventually only apply step $A$ which is of course impossible since at some point we would obtain a number greater than $\frac{1}{2}$.