Betweenness, separation and Pasch

Try

On the real coordinate line, B=1 lies between A=−2 and C=4 because B=(1/2)A+(1/2)C. More generally, for distinct plane points A,C, define strict betweenness by B=(1−t)A+tC with 0<t<1. This definition includes collinearity and excludes endpoints.

Incidence alone does not select which of three collinear points is in the middle. A synthetic order system takes betweenness as primitive and imposes axioms: endpoint symmetry, distinctness, extension, compatible order along a line, and a plane-crossing principle such as Pasch. We will verify representative order arguments in the real model rather than claim one displayed formula is a complete axiom list.

Predict

Predict whether B between A and C implies A between B and C. It does not: replacing the middle point changes the claim. But reversing the endpoints preserves it, since 1−t remains between zero and one.

Predict whether a line entering a triangle through an open side must meet one other open side. The answer requires exclusions: the line must not pass through a vertex and must not coincide with a side. These are hypotheses, not minor drawing conventions.

Experiment

Use the triangle A=(0,0), B=(4,0), C=(0,3). Evaluate L(x,y)=x+y−2.

Vertex	L value	Side of the cutting line
A	−2	Negative
B	2	Positive
C	1	Positive

The line L=0 crosses AB and AC, but not BC. Solve along AB to obtain (2,0), and along AC to obtain (0,2). Now change the constant from 2 to 3. The line passes through C; explain why a statement about two open-side intersections must be revised. The table records exact affine signs, so no apparent overlap in a drawing determines the result.

Understand

Here is the coordinate proof of the relevant Pasch principle. Suppose L(A)<0<L(B) and L(C)≠0. Along any segment PQ, L((1−t)P+tQ)=(1−t)L(P)+tL(Q). Opposite endpoint signs give a unique zero with 0<t<1; equal strict signs give no zero.

If L(C)>0, AC has opposite signs and BC has positive signs. If L(C)<0, BC has opposite signs and AC has negative signs. Hence a line meeting open AB and avoiding the vertices meets exactly one of open AC and open BC.

The argument also explains half-plane convexity: two points with positive L have a whole connecting segment with positive L. The negative half-plane is convex by the same calculation. In a synthetic treatment, separation properties are proved from order axioms; they must not be imported merely because a line appears to divide paper.

Apply

Repair “The crossing is obvious from the figure.” Name the open segments and supply either the order theorem or the sign calculation. If a line passes through a vertex, report that endpoint explicitly instead of counting it as an open-side intersection.

For optional depth, a finite totally ordered line cannot satisfy indefinite extension. A finite line containing at least two points has a greatest point; extending beyond it contradicts maximality. This explains precisely why the finite incidence models from the previous lesson are not counterexamples to theorems requiring ordered-line extension.

Continue

In a geometric proof, annotate each use of “inside,” “same side,” “between” or “beyond.” Those words usually invoke order information. Continue to congruence and SAS, where equality of segments and angles introduces another independent layer.