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> Then we form a bijection (x, y) <-> x/y and we see that there are as many pixels as rational numbers. I don't think that works, though. Take for example thes

• HZet0r 12y ago
  You're right - good point. For one way, I guess you could let p_i represent the ith prime number (there are infinitely many) and use p_x / p_y instead. Unsure about the other way.
  • roywiggins 12y ago
    There's a general result that the set of "All pairs of elements from two countable sets" is, itself, countable. (the Cartesian product of two countable sets is countable). For example, the set of all pairs of natural numbers can be counted something like this: 1,1 2,1 2,2 1,2 3,1 3,2 3,3 1,3 2,3 ...
  • function_seven 12y ago
    Nice! I'm convinced now.
• rmidthun 12y ago
  There is a way to map each rational to a specific positive integer. First, you need to define a mapping between positives and integers, this is pretty simple. 0 1 2 3 4 0 -1 1 -2 2 in c code: if(x<0) ? (x * -2)-1 : x * 2; (pretend that's fixed width above) Next, any rational can be broken down into a unique prime factorization. 4/6 = 2^1 * 3^-1, 4/15 = 2^2 * 3^-1 * 5^-1 From the factorization, define a transform where the exponent for each prime is replaced using the mapping above: so (2^2 * 3^-1 * 5^-1) becomes (2^4 * 3^1 * 5^1) = 240. Since the original rational can be negative, you'll need to use the first transformation again, so 4/15 = 240, -4/15 = -240, which work out to 480 and 479 respectively. 0 and 1 are left unchanged by the factorization process so 0 maps to 0, 1 to 2, and -1 to 1.