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What Straightedge and Compass Cannot Do

Turn ruler-and-compass geometry into field theory: every constructible number lives in a tower of quadratic extensions, so its degree is a power of two. That single constraint settles three problems the Greeks could not.

Geometry becomes degree

Start with two points, call their distance 1, and place them at 0 and 1 in the plane. A point is constructible if you can reach it by a finite sequence of the only two legal moves: draw a line through two known points, draw a circle centered at a known point through another. Every new point is an intersection of two such lines/circles — and solving line-line, line-circle, or circle-circle intersections involves only field operations and at worst a single square root.

Translate that to algebra. The constructible numbers form a field (closed under +, −, ×, ÷), and crucially closed under taking square roots of positive elements. Each construction step adjoins at most one square root, so a constructible number α sits at the top of a tower ℚ = K₀ ⊆ K₁ ⊆ … ⊆ K_n ∋ α where each [K_{i+1} : K_i] = 1 or 2.

Three impossibilities

Now the famous problems fall in a line. Doubling the cube: building an edge for a cube of twice the volume means constructing α = 2^(1/3). Its minimal polynomial is x³ − 2 (Eisenstein at 2), so [ℚ(2^(1/3)):ℚ] = 3. Three is not a power of 2 — impossible. Trisecting a general angle: a 60° angle is constructible, but trisecting it requires cos 20°, which satisfies the irreducible 8x³ − 6x − 1, giving degree 3 again — impossible.

Trisecting 60 degrees is impossible.

Triple-angle identity:  cos(3t) = 4 cos^3(t) - 3 cos(t).
Put 3t = 60, so cos(3t) = 1/2, and let c = cos(20):
  1/2 = 4 c^3 - 3 c
  8 c^3 - 6 c - 1 = 0.

Let f(x) = 8x^3 - 6x - 1.  Test for rational roots p/q with p | 1, q | 8:
  candidates +-1, +-1/2, +-1/4, +-1/8 -> none give 0 (check x=1/2: 1 - 3 - 1 = -3, etc.)
A cubic with no rational root is irreducible over Q, so [Q(c):Q] = 3.

3 is NOT a power of 2  =>  cos(20) is not constructible  =>  60 deg cannot be trisected.

(Same template kills doubling the cube: x^3 - 2 irreducible, degree 3.)
The triple-angle identity converts trisection into an irreducible cubic of degree 3.

Squaring the circle: a square with the same area as a unit circle needs a side of length √π. But π is transcendental (Lindemann, 1882), so √π is too — it satisfies *no* polynomial over ℚ at all, let alone one of 2-power degree. This is the deepest of the three: it rests not on a degree count but on transcendence, a result well outside our toolkit, and honestly so.

And one thing it can do

The degree test is necessary but not by itself sufficient — but for the regular n-gon the full story is beautiful and is where Gauss enters. The regular 17-gon *is* constructible: cos(2π/17) lives in the cyclotomic extension ℚ(ζ₁₇), whose Galois degree is φ(17) = 16 = 2⁴, a tower of quadratics. The complete theorem (Gauss–Wantzel): a regular n-gon is constructible iff n is a power of 2 times distinct Fermat primes. The teenage Gauss constructing the 17-gon is the historical moment field theory and geometry shook hands.