Talk:Euler's Theorem for Planar Graphs

My undergraduate text, "John M. Harris, Jeffry L. Hirst, Michael J. Mossinghoff Combinatorics and Graph Theory", on page 78 gives a proof by induction on the number of edges. It has two cases (plus a base case), and seems to be more elegant than the proof given here. Rough outline of the proof [in my own words] now follows. A connected planar graph with no edges has one face; $1 - 0 + 1 = 2$ satisfies the base case. A connected planar graph either has a cycle, or does not have a cycle (i.e., it is a tree). If the graph is a tree with $V$ number of vertices, then by Finite Connected Graph is Tree iff Size is One Less than Order, it must have $V-1$ edges. $V-(V-1)+1=2$ satisfies this case. If our graph was instead not a tree, then it must have at least one cycle. We can remove one edge from a cycle to merge two faces of our graph into one face; \begin{align*} & V-(E-1)+(F-1) && = 2 \\ \leadsto & V-E+F && = 2 \end{align*} (this is similar to the reasoning of the "loop" case on the page). By our induction hypothesis, we are done. I'll write up a more formal version of this when I get some free time. --michael macleod (talk) 23:13, 07 May 2020 (PST)


 * Please do. It's what this site is for. Plenty of activity at the moment, the more the merrier. --prime mover (talk) 02:21, 8 May 2020 (EDT)