A student claims to have found a ruler-and-compass construction that trisects a 60° angle, achieved through an unusually long sequence of steps. Why must this claim be false?
ATrisecting any angle requires at least one step that cannot be performed with a compass
Bcos(20°) has degree 3 over ℚ, and no tower of degree-2 extensions can reach a number of odd degree
CThe construction would require more precision than any physical compass can provide
DAngle trisection was proven impossible by exhaustive computational search
Trisecting 60° requires constructing cos(20°), which is a root of the irreducible polynomial 8x³ - 6x - 1 over ℚ. This means [ℚ(cos 20°) : ℚ] = 3. Since every ruler-and-compass step adds at most a degree-2 extension, the degree of any constructible number over ℚ must be a power of 2. Since 3 is not a power of 2, no sequence of ruler-and-compass steps — however long or clever — can construct cos(20°). The impossibility is algebraic, not practical.
Question 2 Multiple Choice
Squaring the circle (constructing a line segment of length √π from a unit segment) is impossible for a fundamentally different reason than doubling the cube. What is that reason?
Aπ is irrational, while ∛2 is rational, so the degree argument applies differently
Bπ is transcendental — it satisfies no polynomial over ℚ — so it lies outside the entire algebraic hierarchy, not merely in a wrong-degree extension
CThe circle has infinite area, making it geometrically incomparable to a square
D√π has degree 4 over ℚ, which is a power of 2 but still not constructible
Doubling the cube fails because ∛2 is algebraic of degree 3 — it's inside the algebraic hierarchy but in the wrong-degree extension. Squaring the circle fails for a deeper reason: π is transcendental, meaning it satisfies no polynomial equation with rational coefficients at all. It cannot even be reached by any algebraic extension of ℚ, making the degree argument irrelevant. The impossibility is of a completely different kind.
Question 3 True / False
Each ruler-and-compass step corresponds algebraically to solving at most a quadratic equation, which is why the degree of any constructible number over ℚ must be a power of 2.
TTrue
FFalse
Answer: True
True. Drawing a line between two points or finding intersections solves linear equations (degree 1, staying in the same field). Finding circle-circle or line-circle intersections solves quadratic equations, adjoining a square root and creating a degree-2 extension. By the tower law, a chain of n such steps produces [Fₙ : ℚ] = 2ⁿ. Since every constructible number lives in such a tower, its degree over ℚ divides some power of 2 — and if it is a minimal polynomial, its degree must be a power of 2.
Question 4 True / False
If [ℚ(α) : ℚ] = 4, then α is constructible by ruler and compass.
TTrue
FFalse
Answer: False
False. Degree being a power of 2 is necessary but not sufficient for constructibility. The number must lie in a tower of degree-2 extensions Q ⊂ F₁ ⊂ ... ⊂ Fₙ where each step has degree 2. An element of degree 4 over ℚ could lie in an extension that is a degree-4 extension not decomposable into two quadratic steps. Constructibility requires the correct tower structure, not merely the correct degree.
Question 5 Short Answer
Explain why the impossibility of doubling the cube is not a matter of insufficient geometric ingenuity but of algebraic necessity.
Think about your answer, then reveal below.
Model answer: Doubling the cube requires constructing ∛2, which has degree 3 over ℚ. Every ruler-and-compass step can only adjoin a square root, creating a degree-2 extension. Any tower of such steps produces a field of degree 2ⁿ over ℚ. Since 3 does not divide any power of 2, ∛2 cannot live in any such tower — no matter how many steps are used.
The key shift is from geometry to algebra: ruler-and-compass operations are exactly the operations that produce towers of quadratic extensions. The algebraic constraint on degrees is absolute. A more creative sequence of steps doesn't help because every step is still a linear or quadratic equation. The impossibility proof converts an ancient geometric question into an algebra problem that has a clean answer.