Proving "3f ≤ 2e" for Planar Graphs

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Homework Statement


Prove that for a simple, connected, planar graph, 3f ≤ 2e where f = faces and e = edges

The Attempt at a Solution


My professor went over this in class, and mentioned something about "double counting," which I do not really understand. I tried saying that it takes 2 counts to define an edge and 3 to define a face, but from here I am not sure what to do as I am unfamiliar with double-counting proofs.

Thanks for the help.
 
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hi kinof! :smile:
kinof said:
My professor went over this in class, and mentioned something about "double counting,"

every edge has exactly two faces, so slice every edge in two, lengthwise …

then you have double the number of edges, and every edge has only one face :wink:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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