Cycle Graph is Bipartite iff Order is Even

Theorem
Let $n \in \N$ be a natural number.

Let $C_n$ be the cycle graph of order $n$.

Then $C_n$ is a bipartite graph $n$ is even.

Proof
Let $V$ be the set of vertices of $C_n$.

Let the elements of $V$ be denoted $v_1, v_2, \ldots, v_n$.

Then by definition of cycle graph, by appropriate selection of subscripts, $C_n$ consists of one cycle $C$ that can be expressed as:
 * $C := \tuple {v_1 v_2 \ldots, v_n v_1}$

Let $v_k \in C_n$.

Then for $1 < k < n$, $v_k$ is adjacent to $v_{k - 1}$ and $v_{k + 1}$ and no other vertices of $C_n$.

We also have that:
 * $v_1$ is adjacent to $v_2$ and $v_n$

and by the same coin:
 * $v_n$ is adjacent to $v_{n - 1}$ and $v_1$.

Necessary Condition
Let $n$ be even.

Then we can partition $V$ into $A$ and $B$ such that:
 * $A = \set {V_k: k = 2 r: r \in \N}$


 * $B = \set {V_k: k = 2 r + 1: r \in \N}$

That is:
 * $A$ contains the vertices with even subscripts
 * $B$ contains the vertices with odd subscripts.

By construction, every edge of $C_n$ is incident to one vertices with even subscript and one vertices with odd subscript.

This also applies to the edge $v_1 v_n$, as $1$ is odd and $n$ is even.

Hence by definition $C_n$ is a bipartite graph.

Sufficient Condition
Let $C_n$ is a bipartite graph.

Let $V$ be partitioned into $A$ and $B$.

, let $v_1 \in A$.

Then:
 * $v_2 \in B$
 * $v_3 \in A$

and so on, such that:
 * $v_k \in A \implies k = 2 r + 1$ for some $r \in \N$
 * $v_k \in B \implies k = 2 r$ for some $r \in \N$

That is:
 * $A$ contains all vertices with odd subscript

and:
 * $B$ contains all vertices with even subscript.

$n$ is odd.

For $1 < k < n$, each pair of adjacent vertices is joined by an edge which is incident to one vertices with even subscript and one vertices with odd subscript.

But the edge $v_1 v_n$ is incident to vertices which both have odd subscripts.

Hence $C_n$ is not bipartite.

It follows by Proof by Contradiction that $n$ is even.