Is it possible to construct a strictly monotonic sequence of all rational numbers? I know that the set of all rational numbers is countable, and can be enumerated by a sequence, say $\{a_n\}$.  But can we construct a monotonic $\{a_n\}_{n=1}^{\infty}$, e.g. with $a_k<a_{k+1}$?  It doesn't seem plausible to me, because then $a_1$ would be the smallest rational number, which clearly can't be any finite number.  Am I mistaken?

To parallel Thomas's answer with a more obscure one:
If $a_k < a_{k+1}$ and $a_k = \frac{b_k}{c_k}$ then the mediant $x:= \frac{b_k+b_{k+1}}{c_k+c_{k+1}}$ is a rational number between those two, so no. 
($b_k$ and $c_k$ are relatively prime/are in lowest terms) 
It is elementary but not obvious that the mediant $x$ is between the two (exercise for the reader). The reason this answer is even a thing (Thomas gave a much simpler and easier to understand answer) is because using the mediant we can construct an ordering on the rationals that is countable, and in addition gives all and only the rationals. 
The usual proof that the (positive) rationals have equal cardinality to the naturals is by 'dovetailing'. counting along antidiagonals of pairs and then ignoring rationals that have already been seen (ignoring if gcd $\neq 1$). This is somewhat unsatisfying because it doesn't give an explicit bijection with the naturals. To get the implied rational-natural bijection from gcd, see the Stern-Brocot tree for details.
