Representation theory of the additive group of the rationals?

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 $\mathbb Z$.

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 $t\mapsto e^{at}$.

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 $A_{\mathbb{Q}}$.