Each edge borders 2 faces, and every face has at least 3 edges -> 2E ≥ 3F
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Each edge borders 2 faces, and every face has at least 3 edges -> 2E ≥ 3F

[From: ] [author: ] [Date: 11-05-20] [Hit: ]
since each edge borders two different faces. Thus we have actually shown that 2E ≥ 3F.......
This is a particular part of a proof that K5 (complete graph of 5 vertices) is not planar. And I don't see how to go from the left side of '->' to the right side. Can anyone explain this?

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Let's try to count the number of edges. Each face has at least 3 edges, so if we count three edges per face we get E ≥ 3F. However, this counts each edge twice, since each edge borders two different faces. Thus we have actually shown that 2E ≥ 3F.
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