cardinality at most $n/q$. Let $F$ be the smallest component in ${\overline{G}}$ and let $D = V(F)$. Then in $G$ every vertex in $V \setminus
D$ is adjacent to all vertices in $D$, and so $D$ is a FD-set in $G$. Thus, ${{\rm fd}}(G) \le |D| \le n/q \le n/2$. [$\Box$]{}

Next we consider the fair domination number of the line graph, $L(G)$, of a graph $G$.

Let $G$ be a graph of size $\mG$ and let $L(G)$ denote the line graph of $G$. If $H$ is a spanning $r$-regular subgraph of $G$, where $r
> 0$ and where $H$ is not necessarily induced, of size $\mH$, then $${{\rm fd}}(L(G)) \le \mH = \left( \frac{\mH}{\mG} \right) |V(L(G)))|.$$ \[l:line1\]

**Proof.** Let $D$ be the set of vertices in the line graph $L(G)$ of $G$ corresponding to the edges in $H$, and let $v \in
V(L(G)) \setminus D$. Thus the vertex $v$ corresponds to an edge $e
\in E(G) \setminus E(H)$, and so both ends of $e$ are incident with precisely $r$ edges of $H$. Hence in $L(G)$, the vertex $v$ is adjacent to exactly $2r$ vertices in $D$. Thus in $L(G)$, we have that $|N(v) \cap D| = 2r$ for every 