Is there a way to define betweeness that doesn't require two cases? ($s$ is between $x,y$ if, if $x<y$, then $x<s<y$, and if $y<x$, then $y<s<x$)

Question

Consider the proof that $(a+b)/2$ is between $a,b \in \mathbb{R}$. The proof requires two cases, according to the definition of in-betweeness: $s$ is in-between $x,y$ if:

1. Case 1: if $x<y$, then $x<s<y$.
2. Case 2: if $y<x$, then $y<s<x$.

Is there a way to define betweeness that is equivalent with the above definition, but wouldn't require two cases to do the proof? I'm thinking that the proof for case #1 and case #2 are so similar (we literally only rename the variables), there must be a way to generalize the definition and do these both in one shot.

Answer 1

You could define "$c$ is between $a$ and $b$" as $$ (c-a)(c-b) < 0. $$

Then the proof that the midpoint is in between is one line.

You would have to show that the usual definition is a theorem.

Answer 2

$s$ is strictly between $x$ and $y$ if and only if $(s-x)(s-y) <0$.

Answer 3

$s$ is between $x$ and $y$ iff $$|x-s|+|y-s|=|x-y|$$