Impossibility by field degree

Try

Doubling the volume of a unit cube requires a new side t with t³=2. A numerical approximation such as 1.26 makes a nearly correct cube, but a classical construction asks for an exact length obtained by finitely many compass and unmarked-straightedge operations.

The field-tower theorem from the previous lesson constrains every such coordinate. Starting from rational data, each intersection adds at most a quadratic extension. This converts an apparently unlimited search over constructions into a finite-degree obstruction.

Predict

Predict whether failing to discover a construction proves impossibility. It does not. A proof must rule out every permitted sequence, which is why the algebraic invariant matters.

Predict whether a marked ruler changes the question. It does: allowing simultaneous alignment constraints can introduce cubic equations. An impossibility theorem is always relative to an explicit tool model and initial data.

Experiment

Compare three exact requests.

Request from rational starting data	Required quantity	Obstruction used
Double a unit cube	∛2	Irreducible degree three
Trisect a constructible 60° angle	cos 20°	Irreducible degree three
Square a unit circle	√π	Transcendence of π

The first two rows require only elementary field-degree arguments after polynomial irreducibility is checked. The last relies on a deep external theorem. A table naming that theorem is not a proof of transcendence.

Understand

If a constructible number α lies in a tower of quadratic extensions E over Q, the tower law gives [E:Q]=2ⁿ and [E:Q]=[E:Q(α)][Q(α):Q]. Thus the algebraic degree of α divides a power of two.

For cube duplication, t³−2 is irreducible over Q by Eisenstein at two. Therefore ∛2 has degree three, which cannot divide 2ⁿ. No permitted construction gives the exact side.

For angle trisection, let u=2cos 20°. The triple-angle identity yields u³−3u−1=0. The only possible rational roots are ±1, and neither is a root. A cubic over Q without a rational root is irreducible, so u has degree three. If the angle 20° were constructible from the standard frame, projecting its unit-radius point onto an axis would construct cos 20° and hence u, contradiction. This disproves a universal angle-trisection construction by exhibiting one constructible input that fails.

Apply

Assume the established transcendence theorem for π. A square equal in area to the unit disk would have side √π. Every constructible number is algebraic; squaring an algebraic number remains algebraic. Hence √π cannot be constructible. The argument is complete conditional on the stated transcendence theorem, which this course does not prove.

Repair “Degree a power of two always guarantees constructibility.” The tower law supplies a necessary condition. A general algebraic extension of degree four need not sit inside a sequence of real quadratic adjunctions. For positive construction claims, provide the actual tower or geometric procedure rather than reverse the necessary condition.

For optional depth, regular-polygon constructibility has a stronger classification using cyclotomic fields. We do not derive that theorem here; the current mechanism explains why field structure is relevant.

Continue

For each impossibility proof, distinguish approximate computation, exact construction, initial data and allowed operations. Continue to affine geometry and barycenters to study geometric statements that do not need a length scale at all.