LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

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.

Nonnegative real square roots exist uniquely

Statement

In a Dedekind-complete ordered field, every $x \geq 0$ has a unique $y \geq 0$ with $y^{2} = x$ .

Facts & Assumptions

Given: A field satisfying Dedekind-complete ordered real field and $x \geq 0$ .

[A1]

Dedekind-complete ordered real field: every nonempty bounded-above subset has a least upper bound; field operations preserve order.

Proof

1.1givenA1construct

If $x = 0$ , $y = 0$ is a root. If $x > 0$ , let $S = {q \geq 0 : q^{2} \leq x}$ . It contains $0$ . If $q > x + 1$ , then $q^{2} > (x + 1)^{2} > x$ , so $x + 1$ bounds $S$ above. By [A1], $y = \sup S$ exists and $y \geq 0$ .

2.1step 1.1A1algebra

Suppose $y^{2} < x$ . Put $ϵ = \min (1, (x - y^{2}) / (2 y + 1)) > 0$ . Then $2 y ϵ + ϵ^{2} \leq (2 y + 1) ϵ \leq x - y^{2}$ , so $(y + ϵ)^{2} \leq x$ . Thus $y + ϵ \in S$ , contradicting that $y$ is an upper bound. If $y^{2} > x$ , put $ϵ = (y^{2} - x) / (4 y)$ . Since $0 < y^{2} - x \leq y^{2}$ , $0 < ϵ \leq y / 4$ , and $(y - ϵ)^{2} = y^{2} - 2 y ϵ + ϵ^{2} > x$ . Every $q \in S$ therefore satisfies $q < y - ϵ$ , contradicting that $y$ is the least upper bound. Hence $y^{2} = x$ .

3.1step 2.1algebra∎

If $0 \leq u < v$ , then $v^{2} - u^{2} = (v - u) (v + u) > 0$ unless $u = v = 0$ , which is excluded by $u < v$ . Thus two nonnegative roots of $x$ cannot differ; $y$ is unique.

Depends on

Dedekind-complete ordered real field

Used by

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Sources

Steven G. Krantz, Foundations of Analysis, §1.1, Theorem 1.8 (standard reference, not scraped)