Non-reduced, non-crystallographic root systems Let $V$ be a $n$-dimensional real vector space with standard inner product $(\cdot,\cdot)$. For any $\alpha \neq 0 \in V$, set $\alpha^\vee := \frac{2}{(\alpha,\alpha)}\alpha$. For $\alpha \neq 0,\beta \in V$ set $n_{\alpha}(\beta) := (\beta,\alpha^\vee)$ and $s_\alpha(\beta) := \beta - n_{\alpha}(\beta)\cdot\alpha$.
A root system in $V$ is a finite set $\Phi$ of non-zero vectors in $V$ satisfying:


*

*$\Phi$ spans $V$;

*$s_{\alpha}(\Phi) = \Phi$ for any $\alpha \in \Phi$;

*$\mathrm{Span}\{\alpha\} \cap \Phi = \{\alpha,-\alpha\}$ for any $\alpha \in \Phi$;

*$n_{\alpha}(\beta) \in \mathbb{Z}$ for any $\alpha,\beta \in \Phi$.


This is by now a standard definition and there is a very satisfying classification theory for root systems based on Dynkin diagrams (the so-called Cartan-Killing classification). 
Nevertheless, sometimes small modifications to this definition are considered. For instance, sometimes condition 4 above is omitted and root systems satisfying condition 4 are called crystallographic. However, considering non-crystallographic root systems doesn't change much: there are only a few more families of irreducible non-crystallographic root systems.
Similarly, sometimes condition 3 is omitted and root systems satisfying condition 3 are called reduced. Once again this does not change the structure theory so much: from 4 it follows that for any $\alpha \in \Phi$ we have $\mathrm{Span}\{\alpha\} \cap \Phi \subseteq \{2\alpha,\alpha,-\alpha\,-2\alpha\}$, and I think then it is not hard to show that any irreducible non-reduced root system is of the form $A \cup B \cup 2A$ where $A\cup B$ and $2A \cup B$ are irreducible reduced root systems (see Proposition 13, Section 1.4, Chapter VI of Bourbaki's "Lie Groups and Lie Algebras").
I wonder if anyone has ever considered what happens when we eliminate both 3 and 4 from the above. Now things get a bit worse: even in rank one (i.e. $n=1$) there are infinitely many different root systems- any symmetric set of finite vectors in $\mathbb{R}^1$ is a root system by this definition. From what I can gather from some quick searches on the internet, nobody ever tries to remove both 3 and 4 from the definition of root systems, and maybe that's because the resulting theory is horrible. However, I wonder if this is really the case: is there a nice structural classification of non-reduced, non-crystallographic root systems or not?
 A: Let's stick to the OP's definition of a root system.
Let $\Phi_0$ be the set of normalized roots $\frac{\alpha}{||\alpha||}$, $\alpha\in\Phi$. This is a root system satisfying 1, 2 and 3. Thus it is in the list of not necessarily crystallographic root systems. To reconstruct $\Phi$ we need for every $\alpha_0\in\Phi_0$ the length spectrum $L(\alpha_0)=\{||\alpha||\mid\alpha\in\mathbb{R}_{>0}\alpha_0\}\subseteq\mathbb{R}_{>0}$. These spectra can be chosen arbitrarily, subject to the condition that
$$
L(w\alpha_0)=L(\alpha_0)\text{ for all $w$ in the Weyl group $W$ of $\Phi_0$}
$$
Thus root systems with 1 and 2 are classified by a non-crystallographic root system and a finite set of positive numbers for each $W$-conjugacy class of roots.
This can be made more concrete. For this, let $S$ be a set of simple roots of $\Phi_0$. Attached to it is a labeled graph where the label $n_{\alpha\beta}$ above the edge between $\alpha$ and $\beta$ indicates the order of $s_\alpha s_\beta$. Each root is conjugate to a simple root. So it suffices to know $L(\alpha)$ for $\alpha\in S$. If the $n_{\alpha\beta}$ is odd then $\alpha$ is conjugate to $\beta$ (easy exercise). Thus we need
$$
\alpha,\beta\in S\text{ with }n_{\alpha\beta}\text{ odd}\Rightarrow L(\alpha)=L(\beta).
$$
A not so easy theorem on Coxeter groups (somewhere in Bourbaki) implies that this condition above is also sufficient.
