TITLE: What happens when we (incorrectly) make improper fractions proper again?
QUESTION [164 upvotes]: Many folks avoid the "mixed number" notation such as $4\frac{2}{3}$ due to its ambiguity. The example could mean "$4$ and two thirds", i.e. $4+\frac{2}{3}$, but one may also be tempted to multiply, resulting in $\frac{8}{3}$.
My questions pertain to what happens when we iterate this process -- alternating between changing a fraction
to a mixed number, then "incorrectly" multiplying the mixed
fraction. The iteration terminates when you arrive at a proper
fraction (numerator $\leq$ denominator) or an integer. I'll "define" this process via sufficiently-complicated example:
$$\frac{14}{3} \rightarrow 4 \frac{2}{3} \rightarrow \frac{8}{3} \rightarrow 2 \frac{2}{3} \rightarrow \frac{4}{3} \rightarrow 1\frac{1}{3}\rightarrow \frac{1}{3}.$$

Does this process always terminate?

For which $(p,q)\in\mathbb{N}\times(\mathbb{N}\setminus\{0\})$ does this process, with initial iterate $\frac{p}{q}$, terminate at $\frac{p \mod q}{q}$?

REPLY [8 votes]: Does this process always terminate?


Yes. The process described is simply the evaluation of the recurrence relation...
$$a_{n+1}=\lfloor a_n\rfloor(a_n-\lfloor a_n\rfloor);\quad a_0=\frac{p}{q}$$
...where $\lfloor\cdot\rfloor$ is the floor function, while $a_n>0$.
To show that the process terminates, it suffices to show that there is some $n$ such that $a_n=0$.
A proof-sketch follows:
Suppose that there is some $n$ such that $a_n$ is an integer. Then $a_{n+1}=0$ and we are done.
Suppose that there is no $n$ such that $a_n$ is an integer. Let $c=\max\{a_n-\lfloor a_n\rfloor:n\in\Bbb{N}\}$ (note that $c<1$). It follows that...
$$a_{n+1}\le\lfloor a_n\rfloor c\le a_nc$$
Because $a_{n+1}=a_nc$ has closed form $a_n=a_0c^n$, we know that $a_n\le a_0c^n$. Since $\lim_{n\to\infty}a_0c^n=0$ and $0\le a_n$ for all $n$, it follows that $\lim_{n\to\infty}a_n=0$. By definition of the limit of a sequence, there must exist some $n$ such that $a_n-0<1$. Let $k$ be the least such $n$, then $a_{k+1}=0$ and we are done.
The convention employed @GeoffreyTrang can be used for dealing with negative rationals.


For which $(p,q)\in\mathbb{N}\times(\mathbb{N}\setminus\{0\})$ does this process, with initial iterate $\frac{p}{q}$, terminate at $\frac{p \mod q}{q}$?


It is easier to consider the pairs for which the process doesn't terminate at $\frac{p\mod q}{q}$.
To start with, note let $h(p,q)$ be the last value prior to termination, as described above. It is worth noting that $h(p,q)=h(np,nq)$ for any $n\in\Bbb{Z}^+$ - so it is only necessary to consider the proportion between $p$ and $q$. We can identify each such proportion with a line through $\Bbb{N}\times\Bbb{Z}^+$.
(more to follow)
(see @Paul 's answer for solution)