My attempt is that G have at least one cut vertex and I know that I have to divide the edges of v in two groups but I don't know how to show that the edge connectivity have to be less or equal than r/2. Thanks in advance.
2026-03-26 06:22:23.1774506143
If $G$ is a $r-regular$ graph and $K(G) = 1$ (vertex connectivity) show that the edge connectivity λ(G) $<=$ $r/2$
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Let $v$ be the cut-vertex of $G$. Then $G-v$ has at least two connected components. Since the number of edges incident with $v$ is $r$ then there exists a component for which there are at most $r/2$ edges that connect $v$ to that component. Removing those edges will disconnect $G$ into exactly two components.