LemmaStatement: AI-adaptedProof: 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.

Every rational has a positive-denominator representative

Statement

Every rational has a representative $(a, b)$ with $b > 0$ : for a class $[(a, b)]$ (where $b \neq 0$ ), if $b > 0$ take $(a, b)$ itself, and if $b < 0$ then $(a, b) \sim (- a, - b)$ with $- b > 0$ . Consequently the order on $Q$ (Order on the rationals), which is stated on positive-denominator representatives, is defined on all of $Q$ .

Facts & Assumptions

Given: A rational represented by $(a, b)$ with $a, b \in Z$ , $b \neq 0$ , and the relation $(x, y) \sim (z, w) ⟺ x w = z y$ (The rationals as equivalence classes of pairs of integers).

[L1]

Trichotomy in $Z$ : each nonzero integer is either $> 0$ or $< 0$ , and $b < 0$ iff $- b > 0$ (The integers form a totally ordered ring).

[L2]

In the commutative ring $Z$ , $a (- b) = - (a b) = (- a) b$ (both products are the additive inverse of $a b$ , by distributivity) (The integers form a commutative ring).

Proof

technique · direct

1.1

Since $b \neq 0$ , by trichotomy [L1] either $b > 0$ or $b < 0$ .

givenL1

2.1

If $b > 0$ , the representative $(a, b)$ already has positive denominator.

step 1.1

2.2

If $b < 0$ , then $- b > 0$ by [L1], and $a (- b) = (- a) b$ by [L2] is exactly the defining relation $(a, b) \sim (- a, - b)$ ; so $(- a, - b)$ represents the same class and has positive denominator $- b$ .

step 1.1L1L2

3.1

In either case the class has a representative with positive denominator; hence the order Order on the rationals, stated on such representatives, is defined for every rational.

step 2.1step 2.2∎

Depends on

Used by

Rational powers aʳ of a positive base Definition
Monotonicity of r ↦ aʳ and of a ↦ aʳ Lemma
ℚ is countably infinite Theorem
The number e is irrational Theorem
The reals are complete Theorem
Weighted AM-GM inequality with rational weights Theorem

Cited to discharge well-definedness by Order on the rationals.

Dependency tree · two levels

21 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)
Rational number (Wikipedia) (standard reference, not scraped)
L. S. Krapp, Constructions of the real numbers: a set theoretical approach (Oxford, 2014) (standard reference, not scraped)