Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck pass
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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.

Alternating polynomial signs force the Haar zero-count contradiction

Statement

Let n≥0, let p be a nonzero real polynomial of degree at most n, and let x0<⋯<xn+1 be real numbers. The values p(xi) cannot all be nonzero with alternating signs.

Facts & Assumptions

Given: n≥0, a nonzero real polynomial p of degree at most n, and x0<⋯<xn+1 with each p(xi)≠0.

[F2]

Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on [a,b] takes every value between f(a) and f(b): if a continuous real function has endpoint values of opposite signs on a closed interval, it has a zero in the open interval.

[F3]

A nonzero real polynomial of degree n has no more than n distinct real roots: a nonzero real polynomial of degree at most n has at most n distinct real roots.

Proof

1.1givenF1F2construct

Suppose that the nonzero values p(xi) alternate in sign. By [F1], p is continuous on each [xi,xi+1]. Since the endpoint values are nonzero and have opposite signs, [F2] gives a root zi∈(xi,xi+1) for every 0≤i≤n. These open intervals are pairwise disjoint, so the zi are n+1 distinct real roots.

2.1givenF3step 1.1algebra∎

The n+1 distinct roots from step 1.1 contradict [F3], because p is nonzero and has degree at most n. Therefore the assumed alternating nonzero values cannot occur.

Depends on

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