Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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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.

Every odd cycle C2k+1 has Turán density 1/2

Example

For every k1,

χ(C2k+1)=3andπ(C2k+1)=12.

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[F1]

For n3, Cn has the consecutive edges {i,i+1} and the closing edge {n1,0} (Empty and complete graphs, complete bipartite graphs, and the convention that Pn and Cn have n vertices).

[F2]

A proper k-vertex-colouring is a map c:Vk with c(u)c(v) for every edge {u,v}, its fibres are the colour classes, and χ(G)=min{kN:G is k-colourable} (Proper vertex colourings and chromatic number).

[F3]

For every finite graph H with an edge, π(H)=11/(χ(H)1) (The asymptotic extremal density is determined exactly by chromatic number: π(H)=11/(χ(H)1)).

Verification

technique · alternate colours and use odd parity
1.1

In a two-colouring of a cycle, colours must alternate along consecutive vertices. After the odd number 2k+1 of edges, the closing edge would join equal colours, so no proper two-colouring exists.

givenF1F2
2.1

Colour vertices 0,,2k1 alternately with two colours and give vertex 2k a third colour. This is proper, so χ(C2k+1)=3. For k=1 this is the triangle and the same argument applies.

step 1.1givenF1F2
3.1

The density formula gives π(C2k+1)=11/(31)=1/2.

step 2.1givenF3

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 38 results over 16 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