An ordered vector space is the pair $(V , \leq)$ where it satisfies the following:
For all $x,y,z \in V, \lambda \geq 0$,
i) $x \leq y \Rightarrow x+z \leq y+z$
ii) $x \leq y \Rightarrow \lambda x \leq \lambda y$
Question: Solely based on the definition, can we deduce that $x \leq y \Rightarrow \mu x \geq \mu y$ for any $\mu \leq 0, \mu \in \mathbb{R}$? In other words, is it true that for any ordered vector space, the inequality will be flipped over if we multiply both sides by a negative number?
Yes, that is true:
Let $x \le y$ and $\mu \in \mathbb R_-$.
Now $$(-\mu) \cdot x \le (- \mu) \cdot y $$ yields (by adding $\mu \cdot x$ on both sides) $$0 \le (-\mu) \cdot y + \mu \cdot x$$
Adding $\mu \cdot y$ on both sides gives
$$ \mu \cdot y \le \mu \cdot x $$ as desired.
Btw. Note that given a linear ordered $K$-vector space $(V, \le)$, we may define a linear order $(K, \preceq)$ that makes $K$ into an ordered field as follows. Fix $v_0 \in V \setminus \{ \underline 0 \}$. Then for $a,b \in K$
$$ a \preceq b :\Leftrightarrow a \cdot v_0 \le b \cdot v_0 $$