Representation theory of the additive group of the rationals? What do the finite-dimensional continuous complex representations of the additive group $\mathbb{Q}$ with the usual topology look like?  With the discrete topology?  Which representations are indecomposable?  Irreducible?
The only ones I can think of are of the form $t \mapsto e^{tA}$ for some $A \in \mathcal{M}_n(\mathbb{C})$.  I would be willing to believe that these are the only ones in the first case, but I'm less sure in the second case.
 A: Taking Pontrjagin duals of the short exact sequence of discrete abelian groups 
$0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z} \rightarrow 0$
gives
$0 \rightarrow \hat{\mathbb{Z}} \rightarrow \mathbb{Q}^{\vee} \rightarrow S^1 \rightarrow 0$, 
-- here $\hat{\mathbb{Z}}$ is the profinite completion of $\mathbb{Z}$ -- so this classifies the one-dimensional "unitary" continuous representations of $\mathbb{Q}$.  I believe that the exact sequence above is split; of all places, this came up at a dinner party I attended last week.  (Added: no, it is not split as a sequence of topological groups or even as a sequence of groups: see Matt E's comment below.)
N.B.: If you ask a number theorist (broadly construed) what the "usual topology" on $\mathbb{Q}$ is, she will say that it is the discrete topology.  This is the topology it inherits from the topology on the adele ring $A_{\mathbb{Q}}$.  
A: One way to think of $\mathbb Q$ is the direct limit over positive integers $n$ of $\frac{1}{n} \mathbb Z$.  Thus giving a character of $\mathbb Q$ is the same
as giving an element in the projective limit of the character groups of
$\frac{1}{n}\mathbb Z$.  In particular, if we restrict to unitary characters,
we find that $\mathbb Q^{\vee}$ is the projective limit of circle groups $S^1$ under the $n$th power maps.  This object is (I think) called a solenoid; to number theorists it is better known as the adele class group $\mathbb A/\mathbb Q$.  (Here and throughout I am using the discrete topology; if one instead considers the
induced topology from $\mathbb R$, then, as Robin explains, one just gets 
characters of $\mathbb R$.)
The exact sequence $0 \to \hat{\mathbb Z} \to \mathbb Q^{\vee} \to S^1 \to 0$
in Pete's answer arises from the map taking the solenoid to the base $S^1$; the fibres of this map are copies of $\hat{\mathbb Z}$. 
If we wanted not necessarily unitary characters, we would instead get
the projective limit of copies of $\mathbb C^{\times}$ under the $n$th power maps.  Since $\mathbb C^{\times} = \mathbb R_{> 0} \times S^1$, and since
$\mathbb R_{> 0}$ is uniquely divisible, this projective limit is simply 
$\mathbb R_{> 0}$ times the solenoid.
On a slightly tangential note, let me remark that
the relationship with the adeles is important (e.g. it is the first step in Tate's thesis):
Since the adeles are the (restricted) product of $\mathbb R$ and each $\mathbb Q_p$, and since these are all self-dual, it is easy to see that $\mathbb A$ is self-dual.
One then has the exact sequence
$$0 \to \mathbb Q \to \mathbb A \to \mathbb A/\mathbb Q \to 0$$
which is again self-dual (the duality swaps $\mathbb Q$ and the solenoid
$\mathbb A/\mathbb Q$).
One should compare this with the exact sequence
$$0 \to \mathbb Z \to \mathbb R \to \mathbb R/\mathbb Z = S^1 \to 0.$$
This is again self-dual ($\mathbb R$ is self-dual,
and duality swaps the integers and the circle).
This brings out the important intuition that the adeles are to $\mathbb Q$ as
$\mathbb R$ is to $\mathbb Z$.
A: In the usual topology, each continuous
representation of $(\mathbb{Q},{+})$ will extend to a continuous
representation of $(\mathbb{R},{+})$ so you'll only get the obvious ones.
In the discrete topology there are non-obvious representations even
in dimension one. Consider the map
$\alpha:\mathbb{Q}\to \mathbb{Q}_p\to\mathbb{Q}_p/\mathbb{Z}_p
\cong \mathbb{Z}[1/p]/\mathbb{Z}$. Then $t\mapsto\exp(2\pi i\alpha(t))$
is a character of $\mathbb{Q}$, but not of the form $t\mapsto e^{at}$.
