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let $n=\mid V\mid$

base case: let $n=0$. Thus lemma becomes vacuously true since both bipartitions will contain the empty set thus not a bipartite graph.

Inductive step:

let $k\in\mathbb{N}$, assume $P(k)$ show P(k+1): $\forall G=(V,E), k+1 = \mid V \mid\Rightarrow \mid E\mid\leq\frac{(\mid k+1\mid)^{2}}{4}$, where G is bipartite

I'm having trouble the inductive step, here's my intuition of how I think I should approach. create a subproof to show max possible edges in bipartite graph is when both bipartitions of V has the same size. Then we can assume if a bipartite graph is complete then proposition is satisfied for case 1. I think for Case two, prove for when # of edges are less than the max?

neet
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  • https://math.stackexchange.com/questions/1098587/maximum-number-of-edges-in-a-bipartite-graph – user614287 Apr 11 '19 at 03:29
  • not sure if i'm approaching this correctly but case 1 when $\mid E \mid = \mid V1\mid\cdot\mid V2\mid$, where V1 , V2 are the bipartitions of V. Case 2, when $\mid E \mid < \mid V1\mid\cdot\mid V2\mid$ – neet Apr 11 '19 at 03:44

1 Answers1

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There's no need to consider different cases to do the inductive step.

If for $k+1$ vertices $|E| \geq \frac{(k+1)^2}{4} +1$, remove a vertex with degree $ \leq \lfloor\frac{k + 1}{2}\rfloor$ (why does such a vertex exist?) to get a graph with $k$ vertices and more than $\frac{k^2}{4}$ edges, which contradicts $P(k)$.

PkT
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