Proofs that K5 and K3,3 are not planar
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This proof is not correct, because you don't distinguish the "outer face". So for K5 the result is that the number of edges should be bigger then 3*6/2=9. And here you don't have a contradiction because 10>9.
Tiranodino 1 week ago
@Tiranodino Hi. You must consider the "outer face" as that is part of the Euler Theorem. Hence it is 3*7/2=10.5 giving the contradiction. I'm not making this up. This is the standard quoted proof.
rfbarrow 1 week ago
Hi. Consider K5. Between each two nodes (vertices) there is only one arc (edge). So a region cannot be bounded by connections to two nodes, since this is only one arc. So a third node is required, in which case we have three arcs is required. (Think of a triangle).
In K3,3 there cannot be any triangles because this would mean that two nodes on the same 'side' are connected (not allowed). Hence at least four nodes involved in each bounded region, and so at least four arcs are needed.
rfbarrow 1 month ago
If I may, I find quite annoying that the structure of the proofs (the fact that they are reductio ad absurdum) is that much stressed. What's really going on here isn't proof by contradiction. The key actor is Euler's formula and it would be easy to write the proof differently so that it isn't by contradiction.
More generally, I think that insisting on the arguments is almost always better than insisting on the structure. If you just remember “Euler's formula” you know a proof of this theorem.
JeanPierrePompin 4 months ago 2
@JeanPierrePompin A good point, but the video is not aimed at people like yourself who clearly understand the process. It is aimed at people who have difficulties in following any mathematical proof. Hence the slight overemphasis on a particular structure (not, as you say, the only one). For my target audience, remembering Euler's formula relatively easy. Remembering what to do with it is not. But thanks for your comment.
rfbarrow 4 months ago