Representing a sequence via a 1-1 enumeration of the rational numbers

Let {r_k} be a 1-1 enumeration of the rational numbers and {x_n} be an arbitrary sequence consisting of rational numbers. There exists three permutations \pi_1, \pi_2, \pi_3 of the natural numbers such that x_n = r_{\pi_1(n)} + r_{\pi_2(n)} + r_{\pi_3(n)} for every n = 1, 2, 3, ... These three permutations can be found as follows: 
   - First, pick \pi_1(1) = 1 and choose \pi_2(1), \pi_3(1) arbitrarily such that r_{\pi_2(1)} + r_{\pi_3(1)} = x_1 - r_1. 
   - Choose \pi_2(2) to be the smallest positive integer that are not in the set {\pi_2(1)} and choose \pi_1(2), \pi_3(2) such that \pi_1(2) is not in the set {\pi_1(1)}, \pi_3(2) is not in the set {\pi_3(1)} and r_{\pi_1(2)} + r_{\pi_3(2)} = x_2 - r_{\pi_2(2)}. This can be done since there are infinitely many pairs of rational numbers summing to x_2 - r_{\pi_2(2)} and we only need to avoid finitely many of them. 
   - Similarly, choose \pi_3(3) to be the smallest positive integer that are not in the set {\pi_3(1), \pi_3(2)} and choose \pi_1(3), \pi_2(3) such that \pi_1(3) is not in the set {\pi_1(1), \pi_1(2)}, \pi_2(3) is not in the set {\pi_2(1), \pi_2(2)} and r_{\pi_1(3)} + r_{\pi_2(3)} = x_3 - r_{\pi_3(3)}. Again, this is possible since there are infinitely many pairs of rational numbers summing to x_3 - r_{\pi_3(3)}. 
   - Choose \pi_1(4) to be the smallest positive integer that are not in the set {\pi_1(1), \pi_1(2), \pi_1(3)} and choose \pi_2(4), \pi_3(4) such that \pi_2(4) is not in the set {\pi_2(1), \pi_2(2), \pi_2(3)}, \pi_3(4) is not in the set {\pi_3(1), \pi_3(2), \pi_3(3)} and r_{\pi_2(4)} + r_{\pi_3(4)} = x_4 - r_{\pi_1(4)}. 
   - We continue this infinite process periodically. 
However, if only two permutations are allowed, the task might be impossible. That is, it is not always true that there exists two permutations \pi_1, \pi_2 of the natural numbers such that x_n = r_{\pi_1(n)} + r_{\pi_2(n)} for every n = 1, 2, 3, ... For example, if we pick the sequence x_1 = 0 and x_2 = x_3 = ... = 1, assume that we have successfully selected \pi_1(1) and \pi_2(1). Then if r_{\pi_2(n)} = r_{\pi_2(1)} + 1 for some n >= 2, then r_{\pi_1(n)} = 1 - r_{\pi_2(n)} = 1 - r_{\pi_2(1)} - 1 = -r_{\pi_2(1)} = r_{\pi_1(1)}, which is a contradiction. This contradiction shows that the task is impossible. 

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