CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-07-31

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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 $O : = {v \in V : \deg_{G} (v) is odd}$ .

[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 $∣ A ∣$ 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 $2 q_{i} + 1$ makes the sum $2 \sum_{i} q_{i} + k$ . 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 $V ∖ O$ 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 · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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