MHB Solve Exercise 1.6 in Greenberg's Euclidean Geometry: Betweenness and Lying On

[Ackbach]

Exercise 1.6 in Greenberg's Euclidean and Non-Euclidean Geometries: "Betweenness" and "Lying On"

Undefined terms: point, line, lie on, between, and congruent.

Postulate I. For every point $P$ and for every point $A$ not equal to $P$ there exists a unique line $\ell$ that passes through $P$ and $Q$. $\ell=\overset{\longleftrightarrow}{PQ}$.

Definition I (Segment, Endpoints). Given two points $A$ and $B$. The segment $AB$ is the set whose members are the points $A$ and $B$ and all points that lie on the line $\overset{\longleftrightarrow}{AB}$ and are between $A$ and $B$. The two given points $A$ and $B$ are called the endpoints of the segment $AB$.

Postulate II (Segment Extension). For every segment $AB$ and for every segment $CD$ there exists a unique point $E$ on $\overset{\longleftrightarrow}{AB}$ such that $B$ is between $A$ and $E$ and segment $CD$ is congruent to segment $BE$.

Definition II (Circle, Radius, Center). Given two points $O$ and $A$. The set of all points $P$ such that segment $OP$ is congruent to segment $OA$ is called a circle with $O$ as center, and each of the segments $OP$ is called a radius of the circle.

Postulate III (Existence of Circles). For every point $O$ and every point $A$ not equal to $O$ there exists a circle with center $O$ and radius $OA$.

Definition III (Ray, Emanate, Vertex, Part). The ray $\overset{\longrightarrow}{AB}$ is the following set of points lying on the line $\overset{\longleftrightarrow}{AB}$: those points that belong to the segment $AB$ and all points $C$ on $\overset{\longleftrightarrow}{AB}$ such that $B$ is between $A$ and $C$. The ray $\overset{\longrightarrow}{AB}$ is said to emanate from the vertex $A$ and to be part of line $\overset{\longleftrightarrow}{AB}$.

Definition IV (Opposite Rays). Rays $\overset{\longrightarrow}{AB}$ and $\overset{\longrightarrow}{AC}$ are opposite if they are distinct, if they emanate from the same point $A$, and if they are part of the same line $\overset{\longleftrightarrow}{AB}=\overset{\longleftrightarrow}{AC}$.

Definition V (Angle, Vertex, Sides). An angle with vertex $A$ is a point $A$ together with two distinct nonopposite rays $\overset{\longrightarrow}{AB}$ and $\overset{\longrightarrow}{AC}$ (called the sides of the angle) emanating from $A$. Notations: $\sphericalangle A, \; \sphericalangle BAC, \; \sphericalangle CAB$.

Definition VI (Supplementary Angles). If two angles $\sphericalangle BAD$ and $\sphericalangle CAD$ have a common side $\overset{\longrightarrow}{AD}$ and the other two sides $\overset{\longrightarrow}{AB}$ and $\overset{\longrightarrow}{AC}$ form opposite rays, the angles are supplements of each other, or supplementary angles.

Definition VIII (Right Angle). An angle $\sphericalangle BAD$ is a right angle if it has a supplementary angle to which it is congruent.

Postulate IV (Congruence of Right Angles). All right angles are congruent to each other.

Definition VIII (Parallel Lines). Two lines $\ell$ and $m$ are parallel if they do not intersect; i.e., if no point lies on both of them. We denote this by $\ell \| m$.

Euclidean Parallel Postulate (Playfair's Axiom). For every line $\ell$ and for every point $P$ that does not lie on $\ell$ there exists a unique line $m$ through $P$ that is parallel to $\ell$.

Definition (Perpendicular Lines). Two lines $\ell$ and $m$ are perpendicular if they intersect at a point $A$ and if there is a ray $\overset{\longrightarrow}{AB}$ that is a part of $\ell$ and a ray $\overset{\longrightarrow}{AC}$ that is a part of $m$ such that $\sphericalangle BAC$ is a right angle. We denote this by $\ell \perp m$.

So much for the definitions and postulates. The problem statement is the following (part a): Given two points $A$ and $B$ and a third point $C$ between them. Can you think of any way to prove from the postulates that $C$ lies on line $\overset{\longleftrightarrow}{AB}$?

My issue is that the concept of "betweenness", being undefined, does not seem inherently to include the idea of lying on the segment. That is, suppose $C$ is "between" $A$ and $B$. Nowhere in these definitions and postulates does it seem to be implied that $C$ must lie on segment $AB$. I feel like I'm missing something basic here. Any ideas?

If "betweenness" automatically includes the concept of lying on the segment, then I could prove as follows:

By definition, $C$ is on segment $AB$. By Postulate I, there is a unique line $\ell$ on $A$ and $B$, and $\ell=\overset{\longleftrightarrow}{AB}$. By definition, all points in $AB$ lie on $\ell$, hence $C$ lies on $\ell$. Quod Erat Demonstrandum.

Is this all that's being asked?

 

[Deveno]

I am likewise confused. For example, if we were to imagine the real vector space $\Bbb R^2$ as a model for these axioms, it could conceivably be that $C$ being "between" $A$ and $B$ meant that the $x$-coordinate of $C$ lay between the $x$-coordinate of $A$ and the $x$-coordinate of $B$. Of course this makes "vertical segments" all lines, but that doesn't seem inconsistent with the definitions, as given.

One gets a slightly unusual geometry with this interpretation: vertical rays are also lines, and vertical rays can never be "opposite" since every point on the line is "between" any two points on it. I suspect this can be modified to be a model of the projective plane (or perhaps the projective line-a circle, and a cartesian product with the line; that is, a tube), but the exact details escape me.

This interpretation is not *that* far-fetched, being a normal way we actually determine "betweeness" using the injectiveness of a linear function (vertical lines aren't functions).

Perhaps Definition I is meant to imply that the ONLY points between A and B lie on the line segment AB, but it stops somewhat short of saying that.