Let the sequence of the two numbers be $(x_n,y_n)=\left(\frac{p_n}{q_n},\frac{r_n}{s_n}\right)$ and $(\vec{u_n},\vec{v_n})=((p_n,q_n),(r_n,s_n))$, for $n\geq 0$, where $y_n$ is the new number obtained by simplifying $\frac{p_{n-1}+r_{n-1}}{q_{n-1}+s_{n-1}}$, (or equivalently taking $\vec{v_n}$ to be the smallest multiple of $\vec{u_{n-1}}+\vec{v_{n-1}}$ which belongs to $\mathbb{Z}^2$) and $x_n$ Is either $x_{n-1}$ or $y_{n-1}$. Also define $d_n=\det (\vec{u_n},\vec{v_n})=p_ns_n-q_nr_n$. Now we note that for all $n$ $d_{n+1}|d_n$. This is because $d_{n+1}=\det(\vec{u_{n+1}},\vec{v_{n+1}})=\det(\vec{u_n},\vec{v_{n+1}})|\det(\vec{u_n},\vec{u_n}+\vec{v_n})=\det(\vec{u_n},\vec{v_n})=d_n$ (we wlog chose $\vec{u_{n+1}}=\vec{u_n}$, but the other case is the same thing), where the divisibility is because $\vec{u_n}+\vec{v_n}=k\vec{v_{n+1}}$ for some $k\in \mathbb{Z}$. Also if $|\vec{v_{n+1}}|<|\vec{u_n}+\vec{v_n}|$ (i.e. we had to reduce the fraction) we must have the strict inequality $|d_{n+1}|<|d_n|$. However this can't happen infinitely many times, since $d_0$ has only finitely many integer divisors, so it follows that we had to reduce the fraction only finitely many times (i.e. $d_n$ is eventually constant).