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Prove that in any directed acyclic graph (DAG) we have $2 \cdot |E| \leq |V|(|V|−1)$. Use induction on the number of vertices.

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What happens when you have a single vertex? You have no arcs, and so $E = 0$ and $V(V-1) = 0$. When you have two vertices, you can have at most one arc. So $E = 1$, and $V(V-1) = 2$. Now assume true up to $V = K$. On the $K+1$ case, add a vertex and draw arcs from each vertex in the existing graph to the new vertex. You are adding $V$ arcs. So $2E_{k} \leq K(K-1)$ by the Inductive Hypothesis. And so $2E_{k} + 2K \leq K(K-1) + 2K = K(K+1)$.

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  • $\begingroup$ Wouldn't it be $2(E_k + K)$ since you are adding K edges back? $\endgroup$ – sfreeman2494 May 3 '14 at 20:09
  • $\begingroup$ Yep. Thanks for the catch! I've updated the proof. $\endgroup$ – ml0105 May 3 '14 at 20:14

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