Question: Just a d\*mb question on Lie algebras:

Given a Dynkin diagram of a root system (or a Cartan Matrix), how do I know which combination of simple roots are roots?

Eg. Consider the root system of G\_2, let a be the short root and b be the long one, it is clear that a, b, b+a, b+2a, b+3a are positive roots. But it is not clear to me that 2b+3a is a root just from the Dynkin diagram.

Answer: 
Looking at Fulton and Harris like Mariano suggests is a good idea, but here's another answer which might be helpful to think about.

In the simply-laced case, put an orientation on the edges. Then the positive roots are in bijection with indecomposable representations (over the complex numbers) of the corresponding quiver. More precisely, once we pick a way to label the nodes with simple roots, the dimensions of an indecomposable representation give the coefficients of a linear combination for a positive root. So for example, in type A, a positive linear combination of simple roots is a root only if its support is connected (otherwise you can write it as a direct sum of the two pieces). This also tells you that the coefficients have to be 1 if you play around with it.

For the non simply-laced case, we can reduce to the simply laced case via folding. So for example, $G\_2$ is the folding of $D\_4$ by the order 3 automorphism $\sigma$ which spins it around. The right replacement (for our purposes) for representations of a "$G\_2$ quiver" are representations of $D\_4$ which are invariant under $\sigma$. And then again positive roots correspond to the dimensions of the indecomposable $\sigma$-invariant representations (indecomposable considered as an object in the category of $\sigma$-invariant representations). See Hubery's paper <http://www.ams.org/mathscinet-getitem?mr=2025328> for details.

So for example, the short root corresponds to the middle node in $D\_4$, while the long root corresponds to the orbit of 3 nodes. Orient the edges of $D\_4$ inward, and put ${\bf C}^3$ in the middle node with basis $e\_1, e\_2, e\_3$. Set the outer nodes to be two dimensional, and set their images to be the subspaces $\langle e\_1, e\_2 \rangle$, $\langle e\_2, e\_3 \rangle$, and $\langle e\_3, e\_1 \rangle$ respectively. Then this representation is $\sigma$-invariant and has no $\sigma$-invariant summands (though it is decomposable as a $D\_4$-representation. This corresponds to the root $3a+2b$.

This also works for non-Dynkin diagrams.
Just a related point for hashing out arguments quickly:

If you ever find yourself in the situation where you want to rule out some vector being a root in a hurry, you can check it against the other roots you do know to see if you get a Cartan integer out of their dot products.