Well, the Calkin-Wilf tree is still the best, of course. But given the grid, really I suppose I should say the ordered pairs of positive integers than the rationals, the usual way to traverse it is by diagonals, that leads to a mapping like (n,k) -> (n+k choose 2) + k - 1, or something along those lines. At least, that's what I always seem to see in Cantor and infinity talks and books out there. Today I had a 6th grader suggest that we should show that it's a countable set with the mapping (n,k) -> (2k-1) * 2^(n-1). Probably this is old hat to people who have thought about this sort of thing, but I had never thought of it that way before. It seems much more "natural" because usually when I'm telling people about this stuff, we first use the mapping n -> n+1 to show we can add 1, then (1,k) -> 2k-1 and (2,k) -> 2k to show that we can double things, and then we need to find a whole new solution for "infinity squared". But we can just iterate the same trick we used for "2 times infinity" instead! First apply it to rows 1 and 2, then 2 and 3, then 3 and 4, and so on. [She doesn't know any algebra, so that's the way she explained it, and I showed the formula in terms of a power of 2 times an odd factor.] Another nice thing about her method is that it's quicker and easier to show that it's 1-1 because the inverse function is so simple to explain. Is there a relevant OEIS entry? Ah, yes, http://oeis.org/A135764 . There it is. Does it need a comment about this being a 1-1 mapping from ordered pairs of positive integers to the positive integers, or is that so obvious that it goes without saying? --Joshua Zucker