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

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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.
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The reciprocal on the rationals is well-defined

Statement

On $Q ∖ {0}$ the reciprocal $[(a, b)]^{- 1} : = [(b, a)]$ (Arithmetic on the rationals) is independent of the chosen representative: if $(a, b) \sim (a^{'}, b^{'})$ with $a, a^{'}, b, b^{'} \neq 0$ , then $(b, a) \sim (b^{'}, a^{'})$ . Hence the reciprocal is a well-defined function on $Q ∖ {0}$ .

Facts & Assumptions

Given: Nonzero rationals represented by integer pairs $(a, b)$ and $(a^{'}, b^{'})$ with $a, a^{'}, b, b^{'} \neq 0$ , where $(x, y) \sim (z, w) ⟺ x w = z y$ in $Z$ (The rationals as equivalence classes of pairs of integers).

[L1]

Multiplication in $Z$ is commutative (The integers form a commutative ring).

Proof

technique · direct

1.1

By hypothesis $(a, b) \sim (a^{'}, b^{'})$ , that is $a b^{'} = a^{'} b$ in $Z$ .

given

2.1

Commuting each product by [L1], $b^{'} a = b a^{'}$ , hence $b a^{'} = b^{'} a$ .

step 1.1L1

3.1

The equation $b a^{'} = b^{'} a$ is exactly the defining relation $(b, a) \sim (b^{'}, a^{'})$ , and since $a, a^{'} \neq 0$ the pairs $(b, a), (b^{'}, a^{'})$ are legal rational representatives; therefore $[(a, b)]^{- 1} = [(b, a)] = [(b^{'}, a^{'})] = [(a^{'}, b^{'})]^{- 1}$ , so the reciprocal is well-defined on $Q ∖ {0}$ .

step 2.1given∎

Depends on

Used by

Nothing in the library uses this result yet.

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

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

T. Tao, Analysis I, 3rd ed., §4.2 (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)