Extending the proof of Mantel’s Theorem given in class (see lecture slides),prove that for each n ≥ 1, the only n-vertex triangle-free simple graphwith the maximum number of edges is K⌊n/2⌋,⌈n/2⌉. (Other proofs do notcount.)

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The proof of Mantel's Theorem given in class shows that any triangle-free graph on n vertices has at most ⌊n^2/4⌋ edges. The graph K⌊n/2⌋,⌈n/2⌉ is a bipartite graph with ⌊n/2⌋ vertices in one part and ⌈n/2⌉ vertices in the other part. Each vertex in the first part is connected to every vertex in the second part, so the total number of edges is ⌊n/2⌋ * ⌈n/2⌉, which equals ⌊n^2/4⌋ when n is even and (n^2-1)/4 when n is odd.

Now, we need to prove that no other n-vertex triangle-free graph can have more edges than K⌊n/2⌋,⌈n/2⌉. Suppose for contradiction that there is a triangle-free graph G on n vertices with more than ⌊n^2/4⌋ edges. Let v be a vertex of G with maximum degree. Since G is triangle-free, the neighbors of v form an independent set. Therefore, the degree of v is at most ⌊n/2⌋, and the total number of edges in G is at most ⌊n/2⌋ * (n-1) = ⌊n^2/2⌋ - ⌊n/2⌋. But this is less than or equal to ⌊n^2/4⌋ for all n ≥ 1, a contradiction. Therefore, the only n-vertex triangle-free graph with the maximum number of edges is K⌊n/2⌋,⌈n/2⌉.