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Let $L$ be a semisimple Lie algebra over $\mathbb{C}$; $H$ maximal abelian subalgebra. So $L$ has decomposition $$L=H\oplus (\oplus_{\alpha\in\Phi}L_{\alpha}).$$ The set $\Phi$ is root system of $L$ w.r.t. $H$; it is spanning subset of $H^*=Hom_{\mathbb{C}}(H,\mathbb{C})$.

If $\{\alpha_1,\cdots,\alpha_l\}=\Delta\subset\Phi$ is a basis of $H^*$, then every $\alpha\in\Phi$ is $\mathbb{Q}$-linear combination of $\alpha_i$'s.

Therefore $\mathbb{Q}$-span of $\Delta$ is $\mathbb{Q}$-vector space of dimension $l$.

After above description of decomposition of semisimple Lie algebra, we move to investigate geometric properties of $\Phi$; which in turn reflect structure of $L$.

A vector space over $\mathbb{Q}$ has also an inner product; then why it is necessary to extend scalars from $\mathbb{Q}$ to $\mathbb{R}$, where we investigate properties of root system $\Phi$?

Beginner
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I have asked myself that question sometimes. The best answer I have come up with is that in the more advanced study of root systems, one makes good use of the geometry and topology of $\Bbb R^n$ as an ambient space. The first instance of this that comes to my mind is: For a fixed Weyl chamber $C$ and a given element $v\in V$, there is exactly one element in the Weyl orbit of $v$ that lies in the closure $\bar C$. Whereas this feels almost self-evident in $\Bbb R^n$ (or at least in dimensions 2 and 3 where one can visualise it), I assume it would be quite awkward to state, let alone prove a variant of this if one worked just in the $\Bbb Q$-span of the roots.

Of course one could avoid explicit mention of the real numbers if one really wanted to (I'm quite sure that everything of interest stays at least algebraic over $\Bbb Q$; e.g. eigenvalues of the Weyl group elements), but I think it is actually one of the motivations for considering root systems that they are a very basic geometric object that can be visualised best with our intuition of $n$-dimensional Euclidean (in particular, real) space. I've written more about this in the answer to Picture of Root System of $\mathfrak{sl}_{3}(\mathbb{C})$.

Torsten Schoeneberg
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    Torsten, you are one of the users whose taste in math I trust. I don't know if you have heard of the Pearl Dive project, and what you think about it. But I (and a few others) have decided to seriously promote the good stuff. If you see a particularly sleek answer, or a good/interesting question that takes a bit of extra effort to answer, consider endorsing it in that chat room. I will happily sponsor a bit of extra visibility :-) – Jyrki Lahtonen Feb 24 '20 at 19:09