 There are a number of ways to prove Euler's formula. The simplest, for a sufficiently broad definition of simple, relies on certain contraction operations we can perform on a graph G to get a new graph G prime. One proof relies on edge contraction. Suppose an edge joins two vertices U and V in our graph. We can contract the edge to a point, causing the vertices U and V to merge. Note that this will reduce the number of edges by one, and reduce the number of vertices by one. But the number of faces doesn't change. This suggests we can use an induction proof where N is the number of edges, or the number of vertices. Let's try induction on edges first. If we have N equal to one edge, then the number of edges is one. There are two vertices and one face, and so Euler's formula is true for N equals one. Now suppose Euler's formula is true for all graphs with N equals K edges. Let G have K plus one edge. If we contract an edge, we get a new graph with K edges, one fewer vertex, and the same number of faces. Now by our induction assumption, Euler's formula does hold for graphs with N equals K edges. And so we have, which is the same as, since we contracted a graph and merged two vertices, the edges in our new graph are one fewer, and the vertices in our new graph are one fewer. Or rearranging these. So our original graph has V prime plus one vertices, E prime plus one edges, and F prime faces. And so we have, so Euler's formula holds, completing the induction proof. While this proves Euler's formulas, there are aesthetic flaws. We've been working with simple graphs, where no edge joins a vertex to itself, and two vertices are joined by at most one edge. However, edge contraction operations often lead to both situations. So can we find a proof where we always have a simple graph? Now before trying to find a proof, let's motivate why we should. Remember one way to really understand mathematics is that once you prove something, prove it again in a different way. And in this particular case, it helps that there's an objectionable feature about the proof, and we have a guide to how our different proof will be. So let's explore this problem. So an obvious alternative proof is to delete vertices instead. If G prime is produced from G by removing a vertex, then the number of vertices decreases by one, and the number of edges... Well, there's a bit of a problem here because we don't know how many edges we're going to lose. Since we don't know how many edges will be eliminated if we delete a vertex, let's delete the edges first. But if we delete all edges to a vertex, we isolate the point, so we disconnect the graph, and Euler's formula no longer applies. This means we can only delete an edge that isn't a bridge. So let's think about this. Suppose G prime is produced from G by deleting an edge that isn't a bridge. Now we haven't deleted any vertices, so the number of vertices is the same. We've removed one edge, so the number of edges decreases by one, and since we've removed a boundary between two faces, allowing the two faces to merge, then the number of faces decreases by one. Or does it? An edge separates two faces, but could they be parts of the same face? If they were, we could find a path in the ordinary geometric sense, from one side of the edge to the other that doesn't cross the edge. And so removing the edge allows us to complete the path, which gives us a boundary between two parts of the original graph. So the edge removed was a bridge, but we weren't supposed to remove bridges. And so if we produce G prime from G by deleting an edge that isn't a bridge, the number of vertices is the same, the number of edges decreases by one, and the number of faces decreases by one. And so we have. And now, considering the corollaries, this means that the quantity v plus f minus e is the same before and after the removal of the edge. Now, while this proves that v plus f minus e is an invariant of planar graphs, it doesn't tell us what this quantity actually is. Now we could simply find it from any planar graph, but what's the fun of that? Remember, it's the journey, not the destination, and if that's not enough, remember that proof reminds, reveals, and raises new questions. So let's consider instead, if we keep removing edges that aren't bridges, then eventually we'll come to a point where all edges are bridges. But a graph where all edges are bridges is a tree. And so for a tree, remember, the edges and vertices are related by vertices are edges plus one. And if we view a tree as a planar graph, there's one face. And so our quantity v plus f minus e is two, which we can rearrange to get Euler's formula.