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

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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

1.1givenL1

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

2.1step 1.1

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

2.2step 1.1L1L2

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 $- b$ .

3.1step 2.1step 2.2∎

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.

Every rational has a positive-denominator representative

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources