LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24

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.

The rationals are Archimedean

Statement

For every rational $x$ there is a natural number $n$ with $x < n$ . Consequently, for every rational $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ .

Facts & Assumptions

Given: A rational $x = [(a, b)]$ with $b > 0$ .

[L1]

The order and arithmetic of $Q$ (The rationals form a totally ordered field).

[L2]

Integer facts: positive integers are exactly $ι (m)$ with $m \geq 1$ natural; nonnegative integers are the image of $N$ ; the embeddings preserve arithmetic and order (The naturals embed in the integers, The integers embed in the rationals, The integers form a totally ordered ring).

Proof

technique · direct

1.1

Since $b > 0$ , $b$ lies in the image of $N$ and $b \geq 1$ .

1.2

If $a \leq 0$ set $k = 0$ ; otherwise $a$ is a positive integer, so $a = ι (k)$ for some natural $k \geq 1$ . In both cases $a \leq k$ (as integers).

2.1

Then $a < k + 1$ and, since $b \geq 1$ and $k + 1 > 0$ , also $k + 1 \leq (k + 1) b$ .

step 1.1step 1.2L2

3.1

Hence $a < (k + 1) b$ , and dividing by $b > 0$ (order-scaling in the definition of the rational order), $x = [(a, b)] < [((k + 1) b, b)] = k + 1 = : n$ .

step 2.1L1

4.1

For rational $ε > 0$ : apply the above to $x = 1 / ε$ to get $n$ with $1 / ε < n$ , hence $1 / n < ε$ .

step 3.1L1∎

Depends on

Used by

On a closed interval of ℚ there is a continuous unbounded function, a bounded one with no maximum, and one without the intermediate value property Counterexample
The sequence 1/n is null Example
FALSE: the rationals are complete False statement
For a cut A, -A is a cut and A + (-A) = 0^* Lemma
For a positive cut A, the reciprocal A⁻¹ satisfies A · A⁻¹ = 1^* Lemma
The Cauchy-sequence reals are Archimedean Lemma
The Dedekind reals are Archimedean Lemma
Conventions for sequences: indexing, eventually, lim, and rational ε Remark
The reals are complete Theorem

Dependency tree · two levels

24 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

T. Tao, Analysis I, 3rd ed., §4.2 (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)