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$A-B-C$ means that there is a function $f$ on the line containing $A,B,C$ such that $f(A)<f(B)<f(C)$. I think this is called the betweenness axiom in some geometry books.

My professor said that we need to show there is another function on the line containing $A,B,C$ that maps $A,B,C$ to a codomain such that $f(C) < f(B) < f(A)$. I'm kind of confused, can I just define that other function to be whatever I want? It seems like making the function $g(x) = -x$ would accomplish that. Am I allowed to just that function however I want?

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It depends on what kind of functions $f\colon\{A,B,C\}\to X$ you allow. From your usage of the symbol $<$, I infer that $X$ is in fact a (totally?) ordered set. If in additon to that $X$ happens to be a totally ordered abelian group, and the subset of allowed functions $\{A,B,C\}\to X$ carries enough structure (e.g. as subgroup of te abelian group $X^{\{A,B,C\}}$), then your idea of simply letting $g(x)=-f(x)$ is fine.

Alternatively, if $X$ is just some ordered set and you are allowed to switch (sufficietly arbitrarily) the codomain of your functions, then just replacing $X$ with $X^{\operatorname{op}}$ would be ok as well (where $X^{\operatorname{op}}$ is the set $X$ with order reversed, i.e. $a<^{\operatorname{op}}b\iff b<a$).

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There must be some information missing from the question, because as stated, the condition doesn't really tell us anything about the relation between $A$, $B$, and $C$. If you pick any random $A$, $B$, and $C$, as long as they are all different, you can define the function $f$ by: $$ f(p) = \begin{cases} 0 &\text{if }p=A \\ 1 &\text{if }p=B \\ 2 &\text{for all other }p\end{cases}$$ and this $f$ will then prove "$A-B-C$".

This clearly doesn't succeed in defining any kind of "betweenness", so I suspect there are additional conditions on $f$ that you haven't mentioned in the question -- for example something like "$f(p)$ must be an affine function of the coordinates of $p$".

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