Confusion about definition of cycle It is said a bipartite graph is such a graph whose set of vertices can be divided into two disjoint parts and for every edge on the graph, a vertex in either set can be found that makes up said edge.
It's also said that a graph being bipartite is equivalent to it containing only even length cycles.
It's not said what exactly constitutes a cycle, though. Well, a set notation is given and intuitively it's understood, aswell. A cycle is what it sounds like: a cycle. You start from somewhere and you end up at the exact same place.
A bipartite graph need not contain any cycle. Do we then say it contains a cycle of length $0$? (even length). 
 A: Here is corresponding information from the classic Graph Theory by F. Harary which might be helpful:

  
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*A bipartite graph $G$ is a graph whose vertex set $V$ can be partitioned into two subsets $V_1$ and $V_2$ such that every edge of $G$ joins $V_1$ with $V_2$.
For example the graph on the left hand side below with vertices
  \begin{align*}
V=V_1\cup V_2\qquad\qquad V_1=\{A_1,A_3,A_5,A_7\},\quad V_2=\{A_2,A_4,A_6,A_8\}
\end{align*} is represented on the right hand side to clearly display the fact that it is a bipartite graph.

                                    

  
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*Theorem: A graph $G$ is bipartite if and only if all its cycles are even.
Proof: If $G$ is bipartite, then its vertex set $V$ can be partitioned into two sets $V_1$ and $V_2$ so that every edge of $G$ joins a vertex of $V_1$ with a vertex of $V_2$. Thus every cycle $A_1A_2\cdots A_nA_1$ in $G$ necessarily has its oddly subscripted vertices in $V_1$, say, and the others in $V_2$, so that its length $n$ is even.
For the converse, we assume, without loss of generality, that $G$ is connected (for otherwise we can consider the components of $G$ separately). Take any node $A_1\in V$, and let $V_1$ consist of $A_1$ and all points at even distance from $A_1$, while $V_2=V\setminus V_1$.
Since all the cycles of $G$ are even, every edge of $G$ joins a vertex of $V_1$ with a vertex of $V_2$. For suppose there is an edge $BC$ joining two vertices $B$ and $C$ of $V_1$. Then the union of the shortest path from $A_1$ to $B$ and the shortest path from $A_1$ to $C$ together with the edge $BC$ contains an odd cycle, a contradiction.
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box$

