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Because the nonnegative rationals $\mathbb{Q}_+$ are countable, I can enumerate them as a sequence $\{q_i\}_{i\in\mathbb{N}}$. For any pair of rationals $q$ and $p$, I can perform the comparison $q\leq p$ according to the ordering of the real numbers. Is it possible to enumerate the rationals as an increasing sequence, i.e., $q_i\leq q_{i+1}$ for all $i\in\mathbb{N}$? More generally, given an ordering on $\mathbb{Q}_+$, can I enumerate $\mathbb{Q}_+$ as an increasing sequence under this ordering?

I do not think this is true, but I cannot think of a good approach to proving it false.

EDIT: I meant to only ask about the nonnegative rationals, so I am correcting my question to fix that mistake. The original version of the question asked about enumerating the rationals.

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Thanks to Brian M. Scott's and Old John's answers, I now know the answer to my own question, and Brian has encouraged me to post my own answer to the question.

Let $\{q_i\}_{i\in\mathbb{N}}$ be an ordered enumeration of the nonnegative rationals. Clearly, $q_1=0$. By definition, $q_2\neq0$, and $q_2\leq q$ for all rational $q>0$, but $q_2/2$ is rational and $q_2/2 \leq q_2$. However, $q_2 \leq q_2/2$. This implies that $q_2=2q_2$, i.e., $q_2=0$, which is a contradiction. Hence, no such enumeration exists.

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    Excellent; upvoted.2012-09-26
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No, it is not possible, since if you had such an ordering then by considering two consecutive (distinct) rationals in your ordering, there are (in the real line) an infinite number of rationals between them, but not in your ordering.

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    Yes, that is even clearer!2012-09-26
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Note: This answer was for the original version of the question.

No, you can’t: what could possibly be the first rational in your list? No matter what $q$ you choose, $q-1$ is a smaller rational that ought to precede $q$ in your list.

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    Even if you restrict to rationals in any (non-trivial) interval, either open or closed, a small variation of Brian's answer or mine would still give you a contradiction.2012-09-26