Nice, Alex. So, the only question is whether we get _all_ the nonnegative rational numbers.

This is one of those questions that's not good to attack directly. It's better to sneak up on it gradually, by noticing all the patterns here:

$$\frac{1}{1},$$

$$\frac{1}{2}, \frac{2}{1},$$

$$\frac{1}{3}, \frac{3}{2}, \frac{2}{3}, \frac{3}{1},$$

$$\frac{1}{4}, \frac{4}{3}, \frac{3}{5}, \frac{5}{2}, \frac{2}{5}, \frac{5}{3}, \frac{3}{4}, \frac{4}{1},$$

$$\frac{1}{5}, \frac{5}{4}, \frac{4}{7}, \frac{7}{3}, \frac{3}{8}, \frac{8}{5}, \frac{5}{7}, \frac{7}{2}, \frac{2}{7}, \frac{7}{5}, \frac{5}{8}, \frac{8}{3}, \frac{3}{7}, \frac{7}{4}, \frac{4}{5}, \frac{5}{1},$$

$$\frac{1}{6}, \frac{6}{5}, \frac{5}{9}, \frac{9}{4}, \frac{4}{11}, \frac{11}{7}, \frac{7}{10}, \frac{10}{3}, \frac{3}{11}, \frac{11}{8}, \frac{8}{13}, \frac{13}{5}, \frac{5}{12}, \frac{12}{7}, \frac{7}{9}, \frac{9}{2}, \frac{2}{9}, \frac{9}{7}, \frac{7}{12}, \frac{12}{5}, \frac{5}{13}, \frac{13}{8}, \frac{8}{11}, \frac{11}{3}, \frac{3}{10}, \frac{10}{7}, \frac{7}{11}, \frac{11}{4}, \frac{4}{9}, \frac{9}{5}, \frac{5}{6}, \frac{6}{1},$$

etcetera, and figuring out why these patterns work as they do.