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:
$$\frac{14}{3} \rightarrow 4 \frac{2}{3} \rightarrow \frac{8}{3} \rightarrow 2 \frac{2}{3} \rightarrow \frac{4}{3} \rightarrow 1\frac{1}{3}\rightarrow \frac{1}{3}.$$
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Does this process always terminate?
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For which $(p,q)\in\mathbb{N}\times(\mathbb{N}\setminus\{0\})$ does this process, with initial iterate $\frac{p}{q}$, terminate at $\frac{p \mod q}{q}$?
Best Answer
Yes, the process does always terminate.
Here's why:
Consider the mixed number $a\frac{b}{c}$, where $0 \le b < c$ and $a > 0$. Then, it is clear that $ab < ac+b$, and so the process always continues to lead to smaller and smaller fractions with the same denominator $c$ until the numerator finally becomes smaller than $c$.
In case of a negative mixed number $-a\frac{b}{c}$, remember that this means "$-(a+\frac{b}{c})$", not "$(-a)+\frac{b}{c}$". But one can easily ignore the negative sign, so without loss of generality, one can consider positive mixed numbers only.