LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingjudge 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.

Ordered-field component bound with its equality case

Statement

For $x, y, r \in R$ with $r \geq 0$ and $r^{2} = x^{2} + y^{2}$ , one has $x \leq r$ . Equality holds exactly when $y = 0$ and $x \geq 0$ .

Statement references

The real numbers form a totally ordered field: The reals form a totally ordered field.

Facts & Assumptions

Given: $x, y, r \in R$ , $r \geq 0$ , and $r^{2} = x^{2} + y^{2}$ .

[F1]

The reals form a totally ordered field: the real numbers have a total order compatible with addition and multiplication; positive elements have positive products.

Proof

Given: [F1] and the hypotheses in the Statement.

1.1F1casesalgebra

By trichotomy, either $y = 0$ , $y > 0$ , or $- y > 0$ . In the first case $y^{2} = 0$ . In either positive case, multiplying the positive element by itself gives a positive square; since $(- y) (- y) = y^{2}$ , this proves $y^{2} \geq 0$ , with strict inequality when $y \neq 0$ .

2.1step 1.1algebra

The hypothesis $r^{2} = x^{2} + y^{2}$ and step 1.1 imply $r^{2} \geq x^{2}$ .

2.2F1step 1.1algebra

If $x = r$ , then $y^{2} = r^{2} - x^{2} = 0$ . Step 1.1 shows a nonzero $y$ would have $y^{2} > 0$ , so $y = 0$ ; and $x = r \geq 0$ .

3.1F1step 2.1algebra

If $x \leq 0$ , then $x \leq r$ because $r \geq 0$ . Otherwise $x > 0$ . If $r < x$ , then $x - r > 0$ and $x + r > 0$ , so $x^{2} - r^{2} = (x - r) (x + r) > 0$ , contradicting step 2.1. Hence $x \leq r$ in all cases.

4.1F1algebra∎

Conversely, if $y = 0$ and $x \geq 0$ , the hypotheses give $r^{2} = x^{2}$ with $r, x \geq 0$ . If $r < x$ , then $(x - r) (x + r) > 0$ , contradicting equality of squares; if $x < r$ , then $(r - x) (r + x) > 0$ , the same contradiction. By trichotomy, $r = x$ . Equality holds exactly in the stated case.

Depends on

The reals form a totally ordered field

Dependency tree · 2 levels

Sources

Princeton University MAT330, Complex Analysis with Applications, Spring 2023 lecture notes