In Wikipedia, the third order axiom of Hilbert's axioms states that

"Of any three points situated on a line, there is no more than one which lies between the other two.

Note: The existence part ("there is at least one") is a theorem.

So if there are three points on a line, then there should be one and only one point that is between the other two. However the axiom states only that there can't be more than one such point, and does not tell anything about the existence of the middle point. How do you prove the existence of the point?

The whole order axioms are

  1. If a point B lies between points A and C, B is also between C and A, and there exists a line containing the distinct points A, B, C.

  2. If A and C are two points, then there exists at least one point B on the line AC such that C lies between A and B.

  3. Of any three points situated on a line, there is no more than one which lies between the other two.

  4. Pasch's Axiom: Let A, B, C be three points not lying in the same line and let a be a line lying in the plane ABC and not passing through any of the points A, B, C. Then, if the line a passes through a point of the segment AB, it will also pass through either a point of the segment BC or a point of the segment AC.

I do think the proof can be done with only these axioms (and of course with the obvious incidence axioms) without concerning congruency and continuity axioms, however I can't come up with one.

  • $\begingroup$ The best proof I suggest is this : if $(x_1,y_1),(x_2,y_2)$ and $(x_3y_3)$ are the co-linear points, then $x_2 = \frac{mx_3 + nx_1}{m+n} $ and $y_2 = \frac{my_3 + ny_1}{m+n} $ if $(x_2,y_2)$ is between the rest. $\endgroup$
    – Spectre
    Sep 20 '20 at 4:57
  • $\begingroup$ i think there should be a proof that uses the axioms hilbert suggests, with the pasch's axiom but i cant come up with one $\endgroup$
    – JHL
    Sep 20 '20 at 5:01
  • $\begingroup$ I don't know Hilbert's axioms, but of course there should be something in coordinate geometry that will help. $\endgroup$
    – Spectre
    Sep 20 '20 at 5:02
  • $\begingroup$ @Spectre I don't know for sure, but I believe/guess that at this point of development of the theory there are no coordinates, so you cannot do coordinate arithmetic. It is a game played with axioms only. In particular, the meaning of between cannot be visualized. $\endgroup$ Sep 20 '20 at 6:11
  • 2
    $\begingroup$ @JyrkiLahtonen Betweenness is just a primitive ternary relation satisfying the four axioms above, that's all we know about it. (But segment, in Pasch's axiom, is defined in terms of betweenness: segment $AB$ consists of all points between $A$ and $B$.) $\endgroup$ Sep 20 '20 at 6:18

Let $A,B,$ and $C$ be three distinct points on a line $\ell$, and suppose neither $A$ nor $C$ is between the other two points. We will show that $B$ must be between $A$ and $C$.

Take a point $E$ which is not on $\ell$, and take a point $D$ such that $E$ is between $D$ and $B$. Then line $CE$ passes through segment $BD$, and so by Pasch's axiom it must pass through segment $AB$ or $AD$. But it cannot pass through segment $AB$ since $C$ is not between $A$ and $B$, so it meets segment $AD$ at some point $F$. Similarly, line $AE$ meets segment $CD$ at some point $G$.

Now line $CE$ meets line $AD$ at $F$, line $DG$ at $C$, and line $AG$ at $E$. Note that $F$ is between $A$ and $D$, but $C$ is not between $D$ and $G$ (since $G$ is between $C$ and $D$). So, by Pasch's axiom (for line $CE$ and triangle $ADG$), $E$ must be between $A$ and $G$.

Finally, consider line $BD$ and triangle $ACG$. Line $BD$ meets segment $AG$ at $E$, and meets line $CG$ at $D$ which is not on segment $CG$ (since $G$ is between $C$ and $D$). Thus by Pasch's axiom, line $BD$ passes through segment $AC$. But the intersection of line $BD$ and line $AC$ is $B$, so $B$ is between $A$ and $C$, as desired.

(In each application of Pasch's axiom, I have omitted verifying that the line does not pass through any of the vertices of the triangle, since these verifications are rather tedious. They mostly boil down to concluding that all the points would be on the same line, contradicting our choice of $D$ as not being on line $\ell$.)

  • $\begingroup$ "~ AB or BD." -> "~AB or AD" in line 4 i guess? $\endgroup$
    – JHL
    Sep 20 '20 at 6:34
  • $\begingroup$ great! i kept trying to construct points outside of the triangle rather than inside. i also didnt mention that the axioms didn't explicitly say that you can pick a point between two distinct points, yet i do have proof for that one. $\endgroup$
    – JHL
    Sep 20 '20 at 6:44
  • $\begingroup$ Oh, for the latter, I misread axiom 2 and thought that's what it said. But yeah, you can prove that by Pasch's axiom by just constructing a triangle with those two points as the vertices and a line which passes through exactly one of the other sides. $\endgroup$ Sep 20 '20 at 13:58
  • 1
    $\begingroup$ Or just pick $E$ first, then $D$. $\endgroup$ Sep 20 '20 at 14:58

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