CorollaryStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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Every finite graph has an even number of vertices of odd degree

Statement

In every finite simple graph, the set of vertices having odd degree has even cardinality.

Facts & Assumptions

Given: A finite simple graph $G=(V,E)$ and the finite set $O:=\{v\in V:\deg_G(v)\text{ is odd}\}$ .

[L1]

The handshake lemma gives $\sum_{v\in V}\deg_G(v)=2|E|$ , an even natural number (Handshake lemma: the sum of the vertex degrees is twice the number of edges).

[F1]

The cardinality $|O|$ is a natural number because $O$ is a subset of the finite set $V$ (The cardinality $\lvert A\rvert$ of a finite set).

Proof

technique · direct

1.1

Split the degree sum into the vertices in $O$ and those outside $O$ . The latter summands are even, so their sum is even.

given

1.2

A finite sum of $k$ odd natural numbers has the same parity as $k$ : writing each summand as $2q_i+1$ makes the sum $2\sum_iq_i+k$ . Hence $\sum_{v\in O}\deg_G(v)$ has the same parity as $|O|$ .

F1algebra

2.1

The total degree sum is even by [L1], and subtracting the even contribution from $V\setminus O$ leaves an even sum over $O$ . Step 1.2 therefore makes $|O|$ even.

step 1.1step 1.2L1∎

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: 33 results over 17 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

Cornell Mathematics, Graph Theory notes (standard reference, not scraped)