Say that the graph with vertices $V$ is bipartite, hence $V=A \cup B$ and $A \cap B = \emptyset$ where all edges connect one vertex in $A$ with one in $B$. Assume there is an element with a "self-loop":
Pick one of those elements, i.e. a node with an edge that connects back to itself, for example $a_0 \in A$.
Because the graph is bipartite, all edges that connect $a_0$ only connect "at the other end" with $b_i \in B$. But we must have $b_i \not= a_0$ for all $i$ since $A \cap B = \emptyset$.
Hence: No, a bipartite graph cannot have an element with a self-loop.
0
Bumbble Comm
On
Presumably, a self-loop is an edge that starts and ends at the same vertex. If so, then you're right: no graph with a "self-loop" is bipartite.
Say that the graph with vertices $V$ is bipartite, hence $V=A \cup B$ and $A \cap B = \emptyset$ where all edges connect one vertex in $A$ with one in $B$. Assume there is an element with a "self-loop":
Pick one of those elements, i.e. a node with an edge that connects back to itself, for example $a_0 \in A$.
Because the graph is bipartite, all edges that connect $a_0$ only connect "at the other end" with $b_i \in B$. But we must have $b_i \not= a_0$ for all $i$ since $A \cap B = \emptyset$.
Hence: No, a bipartite graph cannot have an element with a self-loop.