ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-25

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.

$\sqrt{2}$ exists in every complete ordered field, and is irrational

Example

In any complete ordered field $F$ , the element $2 = 1 + 1$ is positive, so by Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ applied to $a = 2$ it has a unique $s > 0$ with $s^{2} = 2$ : this is $\sqrt{2}$ . Moreover $s$ is not the image of any rational under the embedding $ι : Q \to F$ , because no rational squares to $2$ . Thus every complete ordered field contains $\sqrt{2}$ , the canonical gap that $Q$ lacks, now filled by completeness.

Facts & Assumptions

Given: A complete ordered field $F$ (Complete ordered field (least-upper-bound property)) with unit $1$ ; write $2 : = 1 + 1$ . In any ordered field $1 > 0$ , hence $2 = 1 + 1 > 0$ .

[L1]

Every $a \geq 0$ in $F$ has a unique $s \geq 0$ with $s^{2} = a$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[L2]

There is a unique field homomorphism $ι : Q \to F$ ; it is injective and order-preserving, and satisfies $ι (1) = 1$ (The unique embedding of ℚ into an ordered field).

[L3]

No rational number squares to $2$ (FALSE: some rational number squares to 2).

Verification

technique · direct

1.1

In $F$ we have $2 = 1 + 1 > 0$ , so in particular $2 \geq 0$ .

given

2.1

Apply Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ [L1] with $a = 2$ : there is a unique $s \geq 0$ with $s^{2} = 2$ , and $s \neq 0$ since $s^{2} = 2 > 0$ , so $s > 0$ ; write $\sqrt{2} : = s$ .

L1step 1.1

3.1

The element $s$ is not rational: if $s = ι (q)$ for some $q \in Q$ , then $ι (q^{2}) = ι (q)^{2} = s^{2} = 2 = ι (1) + ι (1) = ι (2)$ , so injectivity of $ι$ [L2] forces $q^{2} = 2$ , which is impossible by [L3]; hence $s$ lies outside $ι (Q)$ .

L2L3step 2.1

4.1

Therefore every complete ordered field contains a unique positive $s = \sqrt{2}$ with $s^{2} = 2$ , and this $s$ is irrational: it is exactly the gap in $Q$ that completeness fills.

step 2.1step 3.1∎

Depends on

Used by

Initial terms of the square-root-two rotation Example

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (Thm 1.21, Cor) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §5.5 (standard reference, not scraped)
University of Colorado analysis notes: The real numbers (standard reference, not scraped)
Elias Zakon, Mathematical Analysis: Irrational numbers (standard reference, not scraped)

22​ exists in every complete ordered field, and is irrational