Section 10.1 267 24. (a) If 0 has a vertex of degree 0, 0 would have the most edges in the case that each pair of the remaining n - 1 vertices were joined with an edge. The graph in all would have at most (n~ 1 ) = (n-l)Jn-2) < m edges. (b) Assume to the contrary that 0 is not connected. Then we can find vertices u and v not joined to each other by a walk. In particular, u and v are not adjacent to any common vertex and they are not adjacent to each other. We conclude that deg u + deg v ~ n - 2. [If deg u = k, then u is adjacent to k vertices, v is not adjacent to these; neither is it adjacent to u or to itself, so deg v ~ n -k - 2.] The remaining n - 2 vertices each have degree at most n - 2 (none can be adjacent to both u and v) so the sum of the degrees of all vertices is at most n -2 + (n -2)(n -2) = n 2 -3n + 2 = (n -l)(n -2). Proposition 9.2.5 then says 1£1 ~ (n-l~n-2), a contradiction. 25. Assume that
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