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LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)audited 2026-07-24
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The rational-defining relation is an equivalence relation

Statement

The relation (a,b)(c,d)    ad=cb(a,b) \sim (c,d) \iff ad = cb on pairs of integers with nonzero second component (The rationals as equivalence classes of pairs of integers) is an equivalence relation.

Facts & Assumptions

Given: Pairs (a,b),(c,d),(e,f)(a,b), (c,d), (e,f) of integers with b,d,f0b, d, f \ne 0.

[L1]

Z\mathbb{Z} is a commutative ring (The integers form a commutative ring).

[L2]

Multiplicative cancellation in Z\mathbb{Z}: ud=vdud = vd with d0d \ne 0 implies u=vu = v (The integers have no zero divisors; multiplicative cancellation).

Proof

technique · direct
1.1

Reflexivity: ab=baab = ba, so (a,b)(a,b)(a,b) \sim (a,b).

L1
1.2

Symmetry: if ad=cbad = cb then cb=adcb = ad, which is the defining equation for (c,d)(a,b)(c,d) \sim (a,b).

L1
1.3

Suppose (a,b)(c,d)(a,b) \sim (c,d) and (c,d)(e,f)(c,d) \sim (e,f), i.e. ad=cbad = cb and cf=edcf = ed.

given
2.1

Multiplying the first equation by ff and the second by bb: adf=cbfadf = cbf and cfb=edbcfb = edb.

step 1.3L1
3.1

Chaining: (af)d=adf=cbf=cfb=edb=(eb)d(af)d = adf = cbf = cfb = edb = (eb)d.

step 2.1L1
4.1

Cancelling the nonzero dd: af=ebaf = eb, so (a,b)(e,f)(a,b) \sim (e,f); the relation is transitive.

step 3.1L2
5.1

The relation is reflexive, symmetric, and transitive, hence an equivalence relation.

step 1.1step 1.2step 4.1

Depends on

Used by

Nothing in the library uses this result yet.

Cited to discharge well-definedness by The rationals as equivalence classes of pairs of integers.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 25 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources