I was playing around with integer coordinates on a square grid and came up with the following protocol:
Start with the sequence $\{\frac{0}{1},\frac{1}{0}\} = \{0,\infty\}$. Add the numerators and denominators together, and insert the result: $$\left\{\frac{0}{1},\frac{1}{0}\right\} \rightarrow \left\{\frac{0}{1},\frac{0+1}{1+0},\frac{1}{0}\right\}= \{0,1,\infty\}$$ Repeat this process: $$\{0,...,\frac{a}{b},\frac{c}{d},...,\infty\} \rightarrow \{0,...,\frac{a}{b},\frac{a+c}{b+d},\frac{c}{d},...,\infty\}$$
The first few iterations of this process yield: $$\left\{\frac{0}{1}, \frac{1}{1}, \frac{1}{0}\right\}$$ $$\left\{\frac{0}{1}, \frac{1}{2}, \frac{1}{1}, \frac{2}{1}, \frac{1}{0}\right\}$$ $$\left\{\frac{0}{1}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{1}{1}, \frac{3}{2}, \frac{2}{1}, \frac{3}{1}, \frac{1}{0}\right\}$$ $$\left\{\frac{0}{1}, \frac{1}{4}, \frac{1}{3}, \frac{2}{5}, \frac{1}{2}, \frac{3}{5}, \frac{2}{3}, \frac{3}{4}, \frac{1}{1}, \frac{4}{3}, \frac{3}{2}, \frac{5}{3}, \frac{2}{1}, \frac{5}{2}, \frac{3}{1}, \frac{4}{1}, \frac{1}{0}\right\}$$ $$\left\{\frac{0}{1}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{3}{8}, \frac{2}{5}, \frac{3}{7}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{1}{1}, \frac{5}{4}, \frac{4}{3}, \frac{7}{5}, \frac{3}{2}, \frac{8}{5}, \frac{5}{3}, \frac{7}{4}, \frac{2}{1}, \frac{7}{3}, \frac{5}{2}, \frac{8}{3}, \frac{3}{1}, \frac{7}{2}, \frac{4}{1}, \frac{5}{1}, \frac{1}{0}\right\}$$ $$\left\{\frac{0}{1}, \frac{1}{6}, \frac{1}{5}, \frac{2}{9}, \frac{1}{4}, \frac{3}{11}, \frac{2}{7}, \frac{3}{10}, \frac{1}{3}, \frac{4}{11}, \frac{3}{8}, \frac{5}{13}, \frac{2}{5}, \frac{5}{12}, \frac{3}{7}, \frac{4}{9}, \frac{1}{2}, \frac{5}{9}, \frac{4}{7}, \frac{7}{12}, \frac{3}{5}, \frac{8}{13}, \frac{5}{8}, \frac{7}{11}, \frac{2}{3}, \frac{7}{10}, \frac{5}{7}, \frac{8}{11}, \frac{3}{4}, \frac{7}{9}, \frac{4}{5}, \frac{5}{6}, \frac{1}{1}, \frac{6}{5}, \frac{5}{4}, \frac{9}{7}, \frac{4}{3}, \frac{11}{8}, \frac{7}{5}, \frac{10}{7}, \frac{3}{2}, \frac{11}{7}, \frac{8}{5}, \frac{13}{8}, \frac{5}{3}, \frac{12}{7}, \frac{7}{4}, \frac{9}{5}, \frac{2}{1}, \frac{9}{4}, \frac{7}{3}, \frac{12}{5}, \frac{5}{2}, \frac{13}{5}, \frac{8}{3}, \frac{11}{4}, \frac{3}{1}, \frac{10}{3}, \frac{7}{2}, \frac{11}{3}, \frac{4}{1}, \frac{9}{2}, \frac{5}{1}, \frac{6}{1}, \frac{1}{0}\right\}$$
My question is: Does this process eventually create all positive rational numbers? (Is this process well-known?)
My intuition is: There might be some formal definition for "simple" (something like "small numerator and denominator") such that the following statement holds:
If $\frac{a}{b}$ and $\frac{c}{d}$ are positive rational numbers in reduced form with $\frac{a}{b}<\frac{c}{d}$, s.t. for any $x$, $\frac{a}{b}<x<\frac{c}{d}$ $\implies$ $\frac{a}{b}$ and $\frac{c}{d}$ are simpler than $x$, then $\frac{a+c}{b+d}$ is the simplest number that lies between $\frac{a}{b}$ and $\frac{c}{d}$.
Then it would make sense that this process creates all the rational numbers in order of "simplicity."
My attempts: It's pretty straightforward to show that this process creates all positive integers, as well as all rational numbers of the form $\frac{a+1}{a}$ (and their reciprocals).
In general, you can look at a number, and show that all the numbers directly to the left or right have a specific form (e.g. all numbers directly to the left of $\frac 3 2$ have form $\frac{1+3n}{1 + 2n}$).
By hand, I've confirmed that all positive rational numbers with numerator and denominator up to $10$ appear.
I thought maybe I could disprove my conjecture by comparing the process to starting with $\{0,1\}$ and splitting the interval repeatedly in half (which creates all positive rational numbers $\leq 1$ with denominator a power of $2$), but I made no headway.
I wrote a program to show that, at least for the first $20$ iterations, the $\gcd$ of the numerator and the denominator is always $1$, but I haven't been able to prove that this should in general be the case.
This is the Stern-Brocot tree, which is bijective to the positive rationals.