Worked solution: Wantzel's algebraic impossibility proof for doubling the cube (1837)

Frequently asked questions

\sqrt[3]{2}\approx1.26

itself an unusual or exotic number?

Not at all — it is a perfectly ordinary algebraic number, easy to approximate numerically or mark on a ruler with a scale. What straightedge and compass cannot do is produce it from only an initial unit length using lines and circles; the restriction is on the tools, not on the number itself.

Source

Pierre Laurent Wantzel, 1837

Recherches sur les moyens de reconnaître si un problème de géométrie peut se résoudre avec la règle et le compas

Step 4 of 5: Conclusion:

3

is not a power of

2

, so the cube cannot be doubled

Previous step Next step

In plain words

Two facts now collide. Step 2 says any constructible number's degree has to be $1, 2, 4, 8, 16,\ldots$ — always a power of two. Step 3 computed that $\sqrt[3]{2}$ 's degree is exactly $3$ , and $3$ never shows up on that list.

So $\sqrt[3]{2}$ is simply out of reach for straightedge and compass. No matter how cleverly one tries — no sequence of ruler-and-compass moves will ever land exactly on this length, because its algebraic 'signature' is the wrong shape from the very start.

[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3 \ne 2^m \implies \sqrt[3]{2} \text{ is not constructible}

Detailed analysis

Step 2 proved every constructible real number has degree $2^m$ over $\mathbb{Q}$ for some integer $m\ge0$ , and Step 3 computed $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$ . Since $3$ is odd and greater than $1$ , it cannot equal $2^m$ for any $m$ . Therefore $\sqrt[3]{2}$ is not constructible with straightedge and compass, and since the edge of a cube with twice the volume of a unit cube must equal $\sqrt[3]{2}$ , doubling the cube is impossible with straightedge and compass alone (Wantzel 1837, §III).

This closes a problem that had resisted construction attempts since at least the 5th century BCE. It also illustrates the strength of Wantzel's method: rather than exhausting the (infinite) space of possible constructions, the proof shows a single algebraic obstruction that rules them all out simultaneously.

As with the trisection problem, this impossibility is specific to the classical unmarked straightedge and compass; more powerful tools solve the problem directly, a point taken up in the closing step.