not in $D$ are dominated by the same number of vertices from $D$; that is, $|N(u)
\cap D| = |N(v) \cap D| > 0$ for every two vertices $u, v \in V
\setminus D$. We remark that if $G \ne {{\overline{K}}}_n$, then $G$ contains a vertex $v$ that is not isolated in $G$ and the set $V \setminus \{v\}$ is a FD-set in $G$. Hence every graph that is not empty has a FD-set of cardinality strictly less than its order. The *fair domination number*, denoted by ${{\rm fd}}(G)$, of a graph $G$ that is not the empty graph is the minimum cardinality of a FD-set in $G$. By convention, if $G = {{\overline{K}}}_n$, we define ${{\rm fd}}(G) =
n$. Hence if $G$ is not the empty graph, then ${{\rm fd}}(G) = \min \{
{{\rm fd}_k}(G) \}$, where the minimum is taken over all integers $k$ where $1 \le k \le |V|-1$. A FD-set of $G$ of cardinality ${{\rm fd}}(G)$ is called a ${{\rm fd}}(G)$-set. Every FD-set in a graph $G$ is a dominating set in $G$. Hence we have the following observation.

Let $G$ be a graph of order $n$. Then the following holds.\
$\gamma(G) \le {{\rm fd}}(G)$.\
${{\rm 