# Arithmetic progression of rationals

We know that the set of rational numbers is countable. For which $n$ can we order all rational numbers as $a_1,a_2,\dots$ so that every subsequence of length $n$ is not an arithmetic progression?

For $n=1,2$ this is trivially impossible. If $n=3$, then the subsequence $a_1,a_2,2a_2-a_1$ forms an arithmetic progression, so it is also impossible. For large enough $n$ it should be possible, though.

To put in other words (thanks to Greg Martin), for which $n$ does there exist an enumeration $a_1,a_2,\dots$ of all rationals such that any time $a_{j_1},a_{j_2},\dots,a_{j_n}$ forms an arithmetic progression of length $n$, the indices $j_1,\dots,j_n$ are not in increasing order?

• Your question is unclear. When you say "any subsequence" do you mean "some subsequence" or "every subsequence"? – Rob Arthan Oct 16 '16 at 2:26
• In your $n=3$ case, you seem to be assuming that there is an index $i \geq 3$ such that $a_i = 2 a_2 - a_1$? Certainly we can design a sequence such that this fails. – erfink Oct 16 '16 at 2:26
• I mean every subsequence. There must exist $i$ with $a_i=2a_2-a_1$ since all rational numbers appear in the sequence. – pi66 Oct 16 '16 at 2:31
• In other words, the OP wants an enumeration $a_1,a_2,\dots$ of the rationals with the following property: any time $a_{j_1}, a_{j_2}, \dots, a_{j_n}$ forms an arithmetic progression in $\Bbb Q$ of length $n$, then the indices $j_1,\dots,j_n$ must not be in increasing order. – Greg Martin Oct 16 '16 at 3:13