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

The reals form a field

Statement

R=C/N\mathbb{R} = \mathcal{C}/\mathcal{N} (The real numbers) is a field.

Facts & Assumptions

Given: Classes [(an)],[(bn)]R[(a_n)], [(b_n)] \in \mathbb{R}.

[L1]

N\mathcal{N} is an ideal of C\mathcal{C} (Null sequences form an ideal).

[L2]

C\mathcal{C} is a commutative ring with 11 (Cauchy sequences form a commutative ring).

[L3]

Maximality construction: for non-null (an)(a_n) there is Cauchy (bn)(b_n) with (anbn)1(a_n b_n) - 1 null (The null ideal is maximal).

[L4]

The constant sequence 11 is not a null sequence (its terms stay at 11), so N\mathcal{N} is a proper ideal and 1^0^\hat 1 \ne \hat 0 (Null sequence).

Proof

technique · direct
1.1

Operations on classes via representatives are well defined: for z,wNz, w \in \mathcal{N}, ((a+z)+(b+w))(a+b)=z+wN\bigl((a+z)+(b+w)\bigr) - (a+b) = z + w \in \mathcal{N} and (a+z)(b+w)ab=aw+zb+zwN(a+z)(b+w) - ab = aw + zb + zw \in \mathcal{N}, since ideals absorb products and sums.

L1L2
2.1

The ring axioms descend to the quotient because the operations are well defined and C\mathcal{C} is a ring, verified on representatives; 1^0^\hat 1 \ne \hat 0 since the constant sequence 11 is not null.

step 1.1L1L2L4
2.2

Inverses: a nonzero class has a non-null representative (an)(a_n); taking (bn)(b_n) from the maximality construction, (an)(bn)1N(a_n)(b_n) - 1 \in \mathcal{N}, so [(bn)][(b_n)] is a multiplicative inverse of [(an)][(a_n)].

step 1.1L3
3.1

R\mathbb{R} is a commutative ring with 101 \ne 0 in which every nonzero element is invertible: a field.

step 2.1step 2.2

Depends on

Used by

Cited to discharge well-definedness by The real numbers.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 20 results over 14 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