Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The real cube root of two is algebraic but not algebraically constructible

Statement refuted

Every real algebraic number is algebraically constructible.

Facts & Assumptions

Given: The unique positive real number a=23.

[L1]

A constructible real algebraic number has degree over Q equal to a power of 2 (An algebraically constructible real algebraic number has degree over Q equal to a power of two).

[L2]

Eisenstein's criterion proves a primitive integer polynomial irreducible when one prime divides every nonleading coefficient, its square does not divide the constant coefficient, and it does not divide the leading coefficient (Eisenstein criterion over the integers).

[L3]

The degree of a simple algebraic extension equals the degree of the element's minimal polynomial (A simple algebraic extension is its minimal-polynomial quotient and has power basis 1,a,,an1 and degree n).

[L4]

Every nonnegative real has a unique nonnegative n-th root for n1 (Existence and uniqueness of n-th roots: a unique a1/n0 with (a1/n)n=a).

Counterexample

technique · contradiction
1.1

By [L4], the real number a exists and satisfies a3=2, so it is algebraic over Q.

givenL4
2.1

The polynomial t32 is Eisenstein at 2, so [L2] makes it irreducible over Q. Hence [L3] gives [Q(a):Q]=3.

step 1.1L2L3
3.1

Suppose, for contradiction, that a is algebraically constructible. Then [L1] makes its degree a power of 2, contrary to step 2.1 because 3 is odd and greater than 1.

step 2.1L1assume-contra
4.1

Therefore a is algebraic but not algebraically constructible, refuting the universal statement.

step 3.1discharge-contradiction

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 124 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources