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In what position does the number $\frac{14}{15}$ appear in the bijection of the positive rational numbers with the natural numbers?

The first few terms of the bijection are:

$\frac 11$, $\frac12$, $\frac21$, $\frac13$, $\frac31$, $\frac14$, $\frac23$, $\frac32$, $\frac41$, $\frac51$, $\frac16$, $\frac25$, $\frac34$, $\frac43$, $\frac52$, $\frac61$, $\frac17$, $\frac35$,...

user327936
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    Obviously, it depends on the bijection. What bijection are you considering? You have listed some positive rational numbers, but it is not clear what is the pattern (for example, why $1/5$ does not appear?). – Crostul Apr 02 '16 at 09:11
  • It's unlikely that there is a faster way to find out, than to write out the whole sequence... – Karolis Juodelė Apr 02 '16 at 09:13
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    Note that for each $n$, the number of reduced fractions $a/b$ with $a+b=n$ is $\phi(n)$ (that's the Euler phi-function). So compute the sum of $\phi(n)$ up to $n=28$, then figure out how far along in the part of the sequence with $a+b=29$ you have to go to get to $14/15$. – Gerry Myerson Apr 02 '16 at 10:28
  • Are you still here? – Gerry Myerson Apr 03 '16 at 11:56
  • @GerryMyerson My apologies for the late reply. I couldn't quite understand your method however, I ended up using Sterns Diatomic series to find the position of 14/15 as outlined here: http://math.stackexchange.com/questions/7643/produce-an-explicit-bijection-between-rationals-and-naturals – user327936 Apr 04 '16 at 02:12
  • There are a number of answers at that URL, and I'm not sure exactly which one you are using, but are you sure it gives you the same ordering of the rationals as the one that you start in your question? – Gerry Myerson Apr 04 '16 at 02:44

1 Answers1

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Option 1

Writing them out gives:$${1, \frac{1}{2}, 2, \frac{1}{3}, 3, \frac{1}{4}, \frac{2}{3}, \frac{3}{2}, 4, \frac{1}{5}, 5, \frac{1}{6}, \frac{2}{5}, \frac{3}{4}, \frac{4}{3}, \frac{5}{2}, 6, \frac{1}{7}, \frac{3}{5}, \frac{5}{3}, 7, \frac{1}{8}, \frac{2}{7}, \frac{4}{5}, \frac{5}{4}, \frac{7}{2}, 8, \frac{1}{9}, \frac{3}{7}, \frac{7}{3}, 9, \frac{1}{10}, \frac{2}{9}, \frac{3}{8}, \frac{4}{7}, \frac{5}{6}, \frac{6}{5}, \frac{7}{4}, \frac{8}{3}, \frac{9}{2}, 10, \frac{1}{11}, \frac{5}{7}, \frac{7}{5}, 11, \frac{1}{12}, \frac{2}{11}, \frac{3}{10}, \frac{4}{9}, \frac{5}{8}, \frac{6}{7}, \frac{7}{6}, \frac{8}{5}, \frac{9}{4}, \frac{10}{3}, \frac{11}{2}, 12, \frac{1}{13}, \frac{3}{11}, \frac{5}{9}, \frac{9}{5}, \frac{11}{3}, 13, \frac{1}{14}, \frac{2}{13}, \frac{4}{11}, \frac{7}{8}, \frac{8}{7}, \frac{11}{4}, \frac{13}{2}, 14, \frac{1}{15}, \frac{3}{13}, \frac{5}{11}, \frac{7}{9}, \frac{9}{7}, \frac{11}{5}, \frac{13}{3}, 15, \frac{1}{16}, \frac{2}{15}, \frac{3}{14}, \frac{4}{13}, \frac{5}{12}, \frac{6}{11}, \frac{7}{10}, \frac{8}{9}, \frac{9}{8}, \frac{10}{7}, \frac{11}{6}, \frac{12}{5}, \frac{13}{4}, \frac{14}{3}, \frac{15}{2}, 16, \frac{1}{17}, \frac{5}{13}, \frac{7}{11}, \frac{11}{7}, \frac{13}{5}, 17, \frac{1}{18}, \frac{2}{17}, \frac{3}{16}, \frac{4}{15}, \frac{5}{14}, \frac{6}{13}, \frac{7}{12}, \frac{8}{11}, \frac{9}{10}, \frac{10}{9}, \frac{11}{8}, \frac{12}{7}, \frac{13}{6}, \frac{14}{5}, \frac{15}{4}, \frac{16}{3}, \frac{17}{2}, 18, \frac{1}{19}, \frac{3}{17}, \frac{7}{13}, \frac{9}{11}, \frac{11}{9}, \frac{13}{7}, \frac{17}{3}, 19, \frac{1}{20}, \frac{2}{19}, \frac{4}{17}, \frac{5}{16}, \frac{8}{13}, \frac{10}{11}, \frac{11}{10}, \frac{13}{8}, \frac{16}{5}, \frac{17}{4}, \frac{19}{2}, 20, \frac{1}{21}, \frac{3}{19}, \frac{5}{17}, \frac{7}{15}, \frac{9}{13}, \frac{13}{9}, \frac{15}{7}, \frac{17}{5}, \frac{19}{3}, 21, \frac{1}{22}, \frac{2}{21}, \frac{3}{20}, \frac{4}{19}, \frac{5}{18}, \frac{6}{17}, \frac{7}{16}, \frac{8}{15}, \frac{9}{14}, \frac{10}{13}, \frac{11}{12}, \frac{12}{11}, \frac{13}{10}, \frac{14}{9}, \frac{15}{8}, \frac{16}{7}, \frac{17}{6}, \frac{18}{5}, \frac{19}{4}, \frac{20}{3}, \frac{21}{2}, 22, \frac{1}{23}, \frac{5}{19}, \frac{7}{17}, \frac{11}{13}, \frac{13}{11}, \frac{17}{7}, \frac{19}{5}, 23, \frac{1}{24}, \frac{2}{23}, \frac{3}{22}, \frac{4}{21}, \frac{6}{19}, \frac{7}{18}, \frac{8}{17}, \frac{9}{16}, \frac{11}{14}, \frac{12}{13}, \frac{13}{12}, \frac{14}{11}, \frac{16}{9}, \frac{17}{8}, \frac{18}{7}, \frac{19}{6}, \frac{21}{4}, \frac{22}{3}, \frac{23}{2}, 24, \frac{1}{25}, \frac{3}{23}, \frac{5}{21}, \frac{7}{19}, \frac{9}{17}, \frac{11}{15}, \frac{15}{11}, \frac{17}{9}, \frac{19}{7}, \frac{21}{5}, \frac{23}{3}, 25, \frac{1}{26}, \frac{2}{25}, \frac{4}{23}, \frac{5}{22}, \frac{7}{20}, \frac{8}{19}, \frac{10}{17}, \frac{11}{16}, \frac{13}{14}, \frac{14}{13}, \frac{16}{11}, \frac{17}{10}, \frac{19}{8}, \frac{20}{7}, \frac{22}{5}, \frac{23}{4}, \frac{25}{2}, 26, \frac{1}{27}, \frac{3}{25}, \frac{5}{23}, \frac{9}{19}, \frac{11}{17}, \frac{13}{15}, \frac{15}{13}, \frac{17}{11}, \frac{19}{9}, \frac{23}{5}, \frac{25}{3}, 27, \frac{1}{28}, \frac{2}{27}, \frac{3}{26}, \frac{4}{25}, \frac{5}{24}, \frac{6}{23}, \frac{7}{22}, \frac{8}{21}, \frac{9}{20}, \frac{10}{19}, \frac{11}{18}, \frac{12}{17}, \frac{13}{16}, \frac{14}{15}, \frac{15}{14}, \frac{16}{13}, \frac{17}{12}, \frac{18}{11}, \frac{19}{10}, \frac{20}{9}, \frac{21}{8}, \frac{22}{7}, \frac{23}{6}, \frac{24}{5}, \frac{25}{4}, \frac{26}{3}, \frac{27}{2}, 28}$$

Counting then gives it in position 255.

Option 2

Alternatively you can add up terms in Euler Totient function until the 28th then do the case for 29 by hand. The first 28 add up to 241. The 29th term is 28 so we know that no terms with numerator plus denominator equals 29 cancel out. So adding 14 to 241 gives 255.

Ian Miller
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