Constructible coordinates and quadratic extensions

Try

To construct √6 from a unit segment and a segment of length six, draw a semicircle with diameter seven. Split the diameter into pieces of lengths one and six, and erect a perpendicular at the split. Its positive intersection height h satisfies h²=6, so h=√6.

The argument uses a geometric relation, not a ruler reading. Compass and unmarked straightedge constructions produce exact points through permitted intersections. Their coordinate fields reveal which exact lengths can arise after finitely many operations.

Predict

Predict whether repeated construction steps can produce arbitrary algebraic roots. Line–line intersections solve linear equations. A line substituted into a circle gives an equation of degree at most two. This suggests a tower of square roots, not unrestricted polynomial solving.

Predict whether two circles introduce degree four because both equations are quadratic. Their equal quadratic parts cancel when subtracted, reducing the problem to a line–circle intersection.

Experiment

Track the algebraic cost of each operation, assuming existing coordinates lie in a field K contained in R.

Operation	Coordinate equation	Possible extension
Two nonparallel lines	Linear system	None beyond K
Line and circle	Quadratic in line parameter	Degree at most two
Distinct-center circles	Radical axis, then quadratic	Degree at most two

Parallel lines may have no intersection; tangent cases may have one; disjoint circles may produce none. A construction must justify existence of the selected real intersection, not merely write a formal quadratic formula.

Understand

Parametrize a line with coordinates in K by P+tv. Substitution into a circle equation gives a quadratic in t with coefficients in K. Any real solution lies in K or in K(√Δ), where Δ is the discriminant. Coordinates of the resulting point use the same field operations. For two circles with different centers, subtract their equations to obtain the nonzero linear equation of their radical axis, then apply the previous argument.

Inducting over a finite construction yields a field tower K₀⊆K₁⊆…⊆Kₙ with each extension degree one or two. Starting with a rational coordinate frame, K₀=Q. The total extension degree is a power of two.

Conversely, geometric constructions implement addition and subtraction by segment transfer, multiplication and division by similar triangles, and positive square roots by the semicircle construction. Thus real numbers obtainable through a finite tower of real quadratic extensions are constructible, provided the starting data are included in the initial field.

Apply

Justify the opening height formula. The point on the semicircle forms a right triangle with the diameter endpoints by Thales’ theorem. The altitude to its hypotenuse splits it into two similar right triangles, giving h²=1·6. Every invoked theorem is Euclidean; this is not a neutral or hyperbolic construction claim.

Repair “Any root of a polynomial of degree four is constructible because four is a power of two.” The tower condition is stronger than a bare degree condition. Degree being a power of two is necessary, but not sufficient in general. The next lesson uses the necessary condition for impossibility proofs without claiming it is a complete test.

Continue

Record starting coordinates, permitted tools, intersection existence and the successive field extensions in a construction portfolio. Continue to constructibility obstructions for proofs that no finite sequence of these operations can solve certain classical requests.