Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11
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K5 and K3,3 are nonplanar

Statement

The complete graph K5 and complete bipartite graph K3,3 of Empty and complete graphs, complete bipartite graphs, and the convention that Pn and Cn have n vertices are nonplanar.

Facts & Assumptions

Given: The two finite simple graphs K5 and K3,3; their degrees may also be counted with Handshake lemma: the sum of the vertex degrees is twice the number of edges.

[L1]

The complete graph KV on n vertices has exactly (n2) edges (The complete graph on an n-element vertex set has (n2) edges).

[L2]

Every simple planar graph with n≥3 vertices has at most 3n−6 edges (Every simple planar graph with n≥3 vertices has at most 3n−6 edges, with equality for every plane triangulation).

[L3]

Every triangle-free simple planar graph with n≥3 vertices has at most 2n−4 edges (Every triangle-free simple planar graph with n≥3 vertices has at most 2n−4 edges).

Proof

technique · direct
1.1

By [L1], K5 has (52)=10 edges, but [L2] would permit at most 3⋅5−6=9 in a planar graph. Hence K5 is nonplanar.

L1L2algebra
2.1

The graph K3,3 has six vertices and nine edges. It is triangle-free because a closed walk alternates between its two parts and therefore has even length. A planar embedding would contradict [L3], whose bound is 2⋅6−4=8.

step 1.1L3algebra∎

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Dependency tree · two levels

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Sources