Let $G$ be a graph obtained from the complete graph $K_n$ by deleting any one edge.Prove that $χ(G) =n−1$
My work: $C=\{1,2,3,4.....n\}$ be a set of $n$ colors. Suppose that the edge removed was $e=(u,v)$ where $u,v ∈ V$ and $V$ is the set of $n$ vertices of $K_n$.
Color all vertices except $u$ and $v$ using $n-2$ colors from the set $C$. This leaves us with two colors in the set $C$ that are not used. Use one of these colors to color the vertex $u$. Since $u$ and $v$ are not connected, the same color can be used to color $v$. Hence we have colored $G$ using $n-1$ colors so $χ(G) =n−1$. Is this proof correct?
The proof shows that $\chi(G) \leq n-1$. Because $G$ has $K_{n-1}$ as subgraph, $\chi(G) \geq n-1$. Hence $\chi(G)=n-1$. Alternatively, $\chi(G)\leq n-1$ by brooks' theorem.