A complete Euclidean coordinate model

Try

For P=(1,−1) and Q=(4,3), the displacement is (3,4) and the distance is five. The same pair determines the line P+t(3,4). Values 0<t<1 give its open segment. These familiar formulas can be organized into a model rather than left as unrelated computational rules.

Define points to be R², lines to be sets P+Rv with v≠0, distance to be the Euclidean norm of a difference, and strict betweenness by an affine parameter between zero and one. Angle between nonzero vectors is the unique value in [0,π] with cosine given by their normalized dot product.

Predict

Predict which assumptions guarantee that the angle formula is meaningful. Cauchy–Schwarz places its cosine in [−1,1], and the vectors must be nonzero so the denominator is nonzero.

Predict whether verifying one line formula proves every possible synthetic geometry is coordinatized this way. It does not. This lesson verifies properties of one model; uniqueness of all complete models up to isomorphism is a separate theorem.

Experiment

Audit how the same displacement controls three structures.

Input	Calculation	Geometric interpretation
P=(1,−1), Q=(4,3)	Q−P=(3,4)	Direction
t=1/4	P+t(Q−P)=(7/4,0)	Interior segment point
Dot product	3²+4²=25	Squared length
Normal n=(4,−3)	n·(3,4)=0	Line equation 4x−3y=7

Change t to −1/2 and explain which properties remain: the new point is still collinear but is no longer between P and Q. Replace the normal by twice itself; the equation changes scale but the geometric line does not.

Understand

Distinct P,Q have a unique joining line. Their difference is nonzero, so P+R(Q−P) is a line containing both. Any affine line containing them must have a direction proportional to Q−P and therefore has exactly the same points.

Distance positivity and symmetry follow from the norm. For the triangle inequality, write u=Q−P and v=R−Q. Cauchy–Schwarz gives ||u+v||²=||u||²+2u·v+||v||²≤(||u||+||v||)². Taking nonnegative square roots proves d(P,R)≤d(P,Q)+d(Q,R). With nonzero displacements, equality holds precisely when u and v point in the same direction.

Polarization reconstructs dot products from distances: 2u·v=||u||²+||v||²−||u−v||². This identity will make the isometry classification a consequence of distance preservation rather than an assumption about the shape of transformations.

Verify a declared planar axiom package

A model assertion needs a finite, explicit checklist. Here the package is: unique joins and noncollinear points; strict linear betweenness with extension and Pasch; segment equivalence, unique transport on a ray and adjacent-segment addition; unsigned-angle equivalence and unique transport into a specified open half-plane; SAS; the Archimedean and line-cut completeness principles. These are planar incidence/order/congruence/continuity axioms in a Hilbert-style formulation. We additionally verify Playfair. We do not claim to have checked an unspecified alternative formulation, a three-dimensional incidence system, or the hyperbolic model in this paragraph.

1. Incidence. The unique-join proof above applies to every pair of distinct points. Each line P+Rv has at least P and P+v. Points (0,0),(1,0),(0,1) are noncollinear because the first two determine y=0 and the third is not on it.
2. Order and extension. On P+Rv use the real parameter t. Strict betweenness is ordinary real betweenness, so endpoints may be reversed, the points are distinct, exactly one of three distinct collinear points lies between the other two, and extensions exist by taking parameters outside the closed interval. This is independent of the choice of parameter origin or direction because another parameter is an affine monotone or antimonotone change. The Pasch sign proof verifies the open-side crossing axiom for every line avoiding the triangle’s vertices; equal strict signs cannot acquire a zero along a segment.
3. Segment congruence. Define it by equality of nonnegative norm lengths. Equality gives reflexivity, symmetry and transitivity. On a specified ray P+t v with t≥0 and v≠0, the unique point at distance a>0 from P is P+(a/||v||)v. Along adjacent collinear segments with a common direction, norm lengths add because their displacement vectors are nonnegative scalar multiples; therefore congruent adjacent pieces have congruent unions. Degenerate zero segments transport to P separately.
4. Angle transport. For a ray with unit direction e, choose the unit normal n pointing into the specified open half-plane. Given an angle θ strictly between 0 andπ, the unique transported unit direction in that half-plane is cosθ e+sinθ n. Its dot product with e is cosθ, and its normal component is positive. Conversely a unit direction there has unique coefficients of this form. Equality of angles supplies equivalence. Straight and zero angles lie on the boundary rays and are treated separately, not as rays inside the open half-plane.
5. SAS. Translate the vertex to the origin and orthogonally rotate/reflect the specified side onto the positive horizontal axis. These operations preserve all lengths and unsigned angles. Two triangles with matching sides b,c and included angle θ on the same chosen side then both have vertices 0,(c,0),(b cosθ,b sinθ). Hence their third sides and other angles agree. This proves the model’s SAS assertion without assuming that every abstract model already has coordinates.
6. Archimedean segments and line cuts. Repeating a positive length a along a ray passes any prescribed distance b because a natural n satisfies na>b. For a Dedekind cut of a ruler line into nonempty lower and upper classes, with every lower point preceding every upper point, use the supremum of the lower coordinates. It determines the cut point; depending on the cut convention that point lies in one class or is separated out. Its uniqueness follows because two distinct candidate coordinates have a midpoint between them. These are consequences of completeness of R, not finite numerical approximations.
7. Euclidean parallel choice. The normal-vector proof constructs exactly one disjoint line through any point off a given line. Thus this model satisfies the additional Euclidean choice, rather than deriving it from the neutral blocks.

This verifies every named block of this chosen package. The direction is real-number model ⇒ axioms. It supplies the Euclidean side of a relative-consistency argument if real arithmetic is consistent; independence of Playfair additionally requires checking the same neutral package in a model where Playfair fails. The next model verification supplies that second step; a geodesic picture alone would not.

Change the metric and find the exact failed block

Retain real affine points, lines and ordinary Euclidean angles, but replace distance by d₁(P,Q)=|Δx|+|Δy|. Compare triangles with vertices O=(0,0), A=(1,0), B=(0,1) and O, C=(1/2,1/2), D=(−1/2,1/2). Are the two sides at O and their included angles the same? What happens to the third side?

Feedback: a countermodel needs a failed axiom, not a visual objection

All four incident sides have d₁ length 1, and both Euclidean included angles are 90°. But d₁(A,B)=2 whereas d₁(C,D)=1. Thus this combined distance/angle structure violates SAS. It still has unique affine joins and the same betweenness/Pasch structure. Losing one congruence axiom is enough to prevent calling it a model of the complete chosen package; it is not a counterexample to a theorem whose hypotheses include SAS.

Primary axiom comparison: Birkhoff’s plane postulates in the University of Kentucky geometry course, inspected 30 October 2026. That is a distinct short metric/angle axiom formulation, explicitly not a license to interchange all versions. The verification above states its own planar checklist so every claim has an identifiable target.

Verify the same neutral package in a non-Euclidean model

To turn the preceding consistency comparison into an independence argument, we must verify the same incidence, order, congruence and continuity blocks, not replace them with a different convenient list. Use the upper hyperboloid H: p₀²−p₁²−p₂²=1, p₀>0, with Lorentz form L(p,q)=−p₀q₀+p₁q₁+p₂q₂. Lines are its intersections with two-dimensional vector planes through zero that meet H; angles use the positive-definite tangent restriction of L. Betweenness will be order along each complete geodesic. The hyperboloid/Klein lesson proves tangent positivity and gives the projection to the open unit disk.

First establish the distance, rather than assuming it. Spatial rotations preserve L. After such a rotation any p has form (cosh a,sinh a,0). The linear map with first two coordinate block [[cosh a,−sinh a],[−sinh a,cosh a]] and third coordinate unchanged sends p to o=(1,0,0), preserves L, and preserves the positive sheet (the new first coordinate is positive because p₀>|p₁| and cosh a>|sinh a|). Thus these maps act transitively and preserve tangent lengths. Polar coordinates about o are (cosh r,sinh r cosθ,sinh r sinθ). Differentiation gives ds²=dr²+sinh²r dθ². Every piecewise smooth path from o to radial coordinate r has length at least r, since speed is at least |r′|; if a path meets o, apply the estimate on intervals outside an arbitrarily small radial disk and pass to its boundary limit. The radial path has length r. Consequently the path distance satisfies cosh d(p,q)=−L(p,q), by sending p to o. Equality in the radial estimate requires constant angular direction wherever r>0, so the minimizing segment is unique. The complete unit-speed lines are γ(t)=cosh t·p+sinh t·v for unit tangent v at p, and d(γ(s),γ(t))=|s−t|. Central projection k=(p₁/p₀,p₂/p₀) sends these lines bijectively to entire open-disk chords and sends their ordered segments to ordinary chord segments.

1. Incidence and order/Pasch. Two distinct Klein points determine exactly one affine line and therefore one chord; the disk has noncollinear triples. Chord betweenness is affine betweenness. Every segment extends in both directions within its chord; in the geodesic parameter this is the unrestricted real line. For a noncollinear triangle, its straight sides and interior lie in the convex disk. An affine line cutting an open side and avoiding vertices cuts precisely one other open side by the same sign proof used above, and that second crossing is still in the disk. Thus the chord model satisfies every declared order axiom, including Pasch.
2. Segments and completeness. Congruence is equality of d. On a chosen ray γ(t), t≥0, there is exactly one point of prescribed length a, namely γ(a). Adjacent lengths add by the displayed distance identity. Each line is an ordered isometric copy of R, so both repeated-length Archimedean extension and Dedekind cut completeness follow exactly as in the Euclidean verification. Euclidean chord endpoints on the disk boundary are not missing finite-distance points: they correspond to t tending to positive or negative infinity.
3. Angles and SAS. At any p the unit tangent directions form the ordinary circle for a positive-definite two-dimensional inner product. A chosen half-plane selects one semicircle, so for each angle strictly between zero and π there is precisely one transported ray, by the same cosine/sine formula in an orthonormal tangent basis. To check SAS, send the first vertex to o using the explicit boost and rotation. Every orthogonal map on the two spatial coordinates extends to an L-preserving map fixing o, so the first side can be put in a fixed direction and the second in a chosen half-plane. Vertices of the two sides then have the unique expressions (cosh b,sinh b,0) and (cosh c,sinh c cosθ,sinh c sinθ). Matching b,c,θ therefore gives identical normalized triangles. Undoing the normalizations gives a length- and angle-preserving map between them. This verifies SAS, not just a formula for the opposite side. Angle equivalence is equality, hence reflexive, symmetric and transitive.
4. Playfair fails. In the Klein disk take the line y=0 and P=(0,1/2). Both chords on y=1/2 and y=1/2+x/4 pass through P and avoid y=0 inside the disk: the second affine line meets y=0 only at x=−2, outside the disk. They are distinct. Thus there is more than one disjoint model line through P.

We have now supplied a real-number model of every neutral block in the declared package, in which Playfair fails, and a real-number model of that same package in which it holds. Relative to real arithmetic, neither Playfair nor its negation follows from that neutral package. This conclusion is not the reverse coordinatization theorem: it classifies neither every model nor every possible version of Hilbert’s axioms. In particular, three-dimensional incidence axioms and Hilbert’s maximality formulation of completeness are not silently substituted for the planar line-cut formulation used here.

Audit a changed choice of angles. Could the Klein picture use its ordinary Euclidean chord angles instead? No. For example, the radial line element from the hyperboloid calculation is dr², whereas tangential motion at the same r has coefficient sinh²r; after the radial change ρ=tanh r, the metric is dρ²/(1−ρ²)²+ρ²dθ²/(1−ρ²). Away from the center its radial and tangential scaling factors differ. Therefore the central projection is not conformal, and Euclidean chord angles do not implement the transported tangent-angle definition. A correct incidence diagram alone does not verify SAS for an altered angle rule.

Apply

Repair “Any invertible linear map preserves this geometry.” Invertibility preserves incidence and affine ratios, but a shear changes distances and angles. The chosen structure matters: an automorphism of affine geometry need not be an isometry of Euclidean geometry.

For optional depth, the real field provides square roots and continuity needed for familiar circle constructions and geometric transport. The declared planar package above now supplies systematic angle, order, congruence and continuity checks. A full coordinatization theorem in the reverse direction is not proved by the calculations here.

Continue

When claiming that a map preserves geometry, list what geometry means: incidence, order, distance, oriented area, or some combination. Continue to Euclidean isometries for a theorem identifying every distance-preserving map of this model.