Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)audited 2026-07-24
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The rationals form a field

Statement

(Q,+,,0,1)(\mathbb{Q}, +, \cdot, 0, 1) with the operations of Arithmetic on the rationals is a field: a commutative ring with 101 \ne 0 in which every nonzero element has a multiplicative inverse.

Facts & Assumptions

Given: Rationals with representatives (a,b),(c,d),(e,f)(a,b), (c,d), (e,f), second components nonzero.

[L1]

The operations are independent of representatives (Rational arithmetic is well defined).

[L2]

Z\mathbb{Z} is a commutative ring in which 101 \ne 0 (The integers form a commutative ring; 101 \ne 0 because 101 \ne 0 in N\mathbb{N}).

[L3]

Z\mathbb{Z} has no zero divisors and admits cancellation (The integers have no zero divisors; multiplicative cancellation).

Proof

technique · direct
1.1

By [L1], each axiom may be verified on fixed representatives.

L1suffices
1.2

Associativity of ++: both ((a,b)+(c,d))+(e,f)((a,b)+(c,d))+(e,f) and (a,b)+((c,d)+(e,f))(a,b)+((c,d)+(e,f)) have representative (adf+cbf+ebd,  bdf)(adf + cbf + ebd,\; bdf).

L2
1.3

Commutativity of ++: (ad+cb,bd)(ad+cb,\, bd) and (cb+ad,db)(cb+ad,\, db) are equal pairs.

L2
1.4

Additive identity: (a,b)+(0,1)=(a1+0b,  b1)=(a,b)(a,b) + (0,1) = (a \cdot 1 + 0 \cdot b,\; b \cdot 1) = (a,b).

L2
1.5

Additive inverses: (a,b)+(a,b)=(abab,  b2)=(0,b2)(0,1)(a,b) + (-a,b) = (ab - ab,\; b^2) = (0, b^2) \sim (0,1).

L2
1.6

Associativity and commutativity of \cdot: ((ac)e,(bd)f)=(a(ce),b(df))((ac)e,\, (bd)f) = (a(ce),\, b(df)) and (ac,bd)=(ca,db)(ac,\, bd) = (ca,\, db).

L2
1.7

Multiplicative identity: (a,b)(1,1)=(a,b)(a,b)(1,1) = (a,b); and 101 \ne 0 since 11011 \cdot 1 \ne 0 \cdot 1.

L2
1.8

Distributivity: (a,b)((c,d)+(e,f))(a,b)\bigl((c,d)+(e,f)\bigr) has representative (a(cf+ed),  b(df))(a(cf+ed),\; b(df)), while (a,b)(c,d)+(a,b)(e,f)(a,b)(c,d) + (a,b)(e,f) has representative (acbf+aebd,  b2df)=(b(acf+aed),  b(bdf))(acbf + aebd,\; b^2 df) = (b(acf + aed),\; b(bdf)); cancelling the common nonzero factor bb these are equivalent.

L2L3
1.9

Inverses: [(a,b)]=0[(a,b)] = 0 iff a1=0ba \cdot 1 = 0 \cdot b iff a=0a = 0; for a0a \ne 0, (a,b)(b,a)=(ab,ba)(1,1)(a,b)(b,a) = (ab,\, ba) \sim (1,1).

L2
2.1

All field axioms hold: Q\mathbb{Q} is a field.

step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6step 1.7step 1.8step 1.9

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 27 results over 12 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.

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