prove that $X'(K_{m, n}) = max({m, n})$

Question

I started by assuming that $\Delta(G) > X'(G)$, and then i took the vertex with the highest degree, meaning this vertex must have n-1 adjacent edges (assuming the number of vertices in the graph is n), but it would be impossible to have $n-1 > X'(G)$, and the only option is for $X'(G) = \Delta(G) = n-1.$

But I'm not sure how to complete it, nor if this is the right approach, I'd appreciate some help.

Answer 1

Vertices in the $n$ partition have degree $m$ and those in the $m$ partition have degree $n$ so $\chi'(K_{m,n}) \ge \max(m,n)$.

By simply rotating the edge colors for successive vertices in the smaller (higher-degree) partition, we can ensure a proper coloring at the vertices in the other partition also. So $\chi'(K_{m,n}) = \max(m,n)$.