Proof of the impossibility of doubling the cube

Builds on Constructible number Doubling the cube Eisenstein's criterion Minimal polynomial of an algebraic element The minimal polynomial divides every polynomial vanishing at the element Degree of a simple extension is the degree of the minimal polynomial The degree of a constructible number is a power of two The constructible numbers form a field closed under square roots Complex conjugate Modulus of a complex number A complex number times its conjugate

Proof

Let $c = \sqrt[3]{2}$ , the real cube root of $2$ , and let $\mathcal{C}$ be the set of the constructible numbers.

The polynomial $f = x^3 - 2$ satisfies Eisenstein's criterion with $p = 2$ : $2$ divides $0, 0, -2$ , $2$ does not divide $1$ , and $4$ does not divide $-2$ . So $f$ is irreducible in $\Q[x]$ .
The minimal polynomial of $c$ over $\Q$ divides $f$ , since $f(c) = 0$ . As $f$ is irreducible and both are monic, the minimal polynomial is $f$ .
By the degree of a simple extension, $[\Q(c) : \Q] = 3$ .
If $c$ were constructible, then $[\Q(c) : \Q]$ would be a power of $2$ . But $3$ is not a power of $2$ . So $c \notin \mathcal{C}$ .

Now we turn to the construction problem. Take the edge of the given cube as the segment from $0$ to $1$ . A doubling of the cube would construct two points $P, Q$ with $\abs{P - Q} = c$ , since the doubled cube has volume $2 = c^3$ .

By the field theorem, $\mathcal{C}$ is a subfield of $\C$ closed under square roots, and it is the smallest such subfield.
Let $\overline{\mathcal{C}} = \set{\bar z : z \in \mathcal{C}}$ , using the complex conjugate. Conjugation preserves sums, products and inverses, so $\overline{\mathcal{C}}$ is a subfield of $\C$ .
If $w = \bar z$ with $z \in \mathcal{C}$ , and $s^2 = w$ , then $\bar s^2 = z$ . So $\bar s \in \mathcal{C}$ and $s \in \overline{\mathcal{C}}$ . Thus $\overline{\mathcal{C}}$ is closed under square roots.
By minimality, $\mathcal{C} \subseteq \overline{\mathcal{C}}$ . So for $z \in \mathcal{C}$ we have $z = \bar u$ with $u \in \mathcal{C}$ , and $\bar z = u \in \mathcal{C}$ .
Let $w = P - Q$ . Then $w \in \mathcal{C}$ , so $\bar w \in \mathcal{C}$ by step 8, and $w \bar w = \abs{w}^2$ lies in $\mathcal{C}$ .
The modulus $\abs{w} = c$ is a square root of $\abs{w}^2$ , so $c \in \mathcal{C}$ by step 5. This contradicts step 4.
So $\sqrt[3]{2}$ is not constructible, and the cube cannot be doubled with straightedge and compass.

Idea

The number $\sqrt[3]{2}$ has degree $3$ over $\Q$ , while constructible numbers have degree a power of $2$ . A construction of the doubled edge would make its length constructible, because the constructible numbers are closed under conjugation, products and square roots.

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Proof of the impossibility of doubling the cube proof

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Tag	`1GGWHERXSZ`
URI	`m2://1GGWHERXSZ`
Kind	proof
Writing	W0 stub
MSC	12F05
Proves	Impossibility of doubling the cube
Uses	Constructible number, Doubling the cube, Eisenstein's criterion, Minimal polynomial of an algebraic element, The minimal polynomial divides every polynomial vanishing at the element, Degree of a simple extension is the degree of the minimal polynomial, The degree of a constructible number is a power of two, The constructible numbers form a field closed under square roots, Complex conjugate, Modulus of a complex number, A complex number times its conjugate
Changed	4 November 2026

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