Does a norm have to map to $\mathbb R$? Wikipedia defines a norm on a vector space $V$ as a function $p : V \mapsto \mathbb R$. I've seen this defined similarly elsewhere. However, it seems to me that a real codomain isn't always necessary.
Take as an example the vector space over $\mathbb Q$ with vectors $v \in \mathbb Q^n$. If we combine this with the $L^1$ norm, then the codomain is again $\mathbb Q$. 
Am I missing something here? Is my example flawed?
 A: The fact that a norm maps to $\Bbb R$ does not mean that the norm is surjective. It can map to a proper subset of $\Bbb R$.
A: Having the norm map to $\mathbb R$ is just a definition, and the map definitely need not be surjective. There's the example you gave, there's also the example of the $p$-adic metric on $\mathbb Q$ which has image given by all integral powers of $p$. One nice thing about having norms map to $\mathbb R$ is that $\mathbb R$ is complete, and a normed vector space in its own right, although the image of the norm map is just $\mathbb R^{\geq 0}$. Someone can probably add to this.
A: The norm is some measure of size. It is very convenient to measure sizes by real numbers (actually, by non-negative real numbers). In case the vectors you consider all have rational entries then indeed the norm will be a non-negative rational number. Since $\mathbb Q\subseteq \mathbb R$, this case is already included in the more general case. 
For analytic reasons it is very convenient that the codomain for the norm be complete. This is why don't see $\mathbb Q$ is a codomain for a norm. It is possible to consider norms taking values in complete lattices other than $\mathbb R$, depending on the problem at hand. However, usually, when vector spaces are over the field $\mathbb R$ or $\mathbb C$, it is most convenient to consider $\mathbb R$ as the codomain for a norm function. 
