[Polyhedral Combinatorics] Euler's Formula for Convex Polyhedra using Inductive Proof

If we imagine a light source placed near one of the faces of the polyhedron above a plane, the shadows of the polyhedron edges form a planar graph embedded in the plane. This resulting figure in the plane has \(V\) vertices, \(E\) edges, but \(F-1\) faces, because the face that was closest to the light source becomes the outside face of this figure. In this way, every convex polyhedron can be projected onto the plane without the edges crossing, hence can be represented as a planar graph.

Therefore, by showing that a connected planar graph with V vertices, E edges, and F faces satisfy the equation \(V-E+F =1\), we will have shown that Euler's formula \(V-E+f=2\) holds for a convex polyhedron.

Let statement \(P(n)\): All connected planar graphs with E edges satisfies \(V-E+F =1\).

\((1)\) Base case: \(n=1=E\)

There exist only one planar graph that has one edge. This graph has two vertices, one edge, and no face. Since \(2-1+0 = 1\), it satisfies the formula. So \(P(1)\) is true.

\((2)\) Assume \(P(k)\) is true. This means that every connected planar graph with \(k\) edges satisfies the formula. Now we want to show that \(P(k+1)\) is also true. Consider a connected planar graph \(\Gamma\) with \(k+1\) edges, and say \(\Gamma\) has \(v\) vertices and \(f\) faces. We want to show that \(\Gamma\) satisfies the formula \(v-(k+1)+f =1\). We can do this by removing a carefully chosen edge from \(\Gamma\) to leave a connected planar graph with \(k\) edges, then use \(P(k)\) to show \(P(k+1)\). In the case that \(\Gamma\) has at least one face, that is, \(f\geq 1\), we can remove one edge of this face, and the remaining graph will still be connected with \(k\) edges, \(v\) vertices, and \(f-1\) faces. Since we assumed \(P(k)\) to be true, this remaining graph will satisfy \(v-k+(f-1) = 1\), and this gives \(v-(k+1)+f=1\) as we required. The other case is when \(\Gamma\) has no faces at all, so \(f=0\). This means that there exists at least one vertex that is joined by an edge that connects to only one other vertex. If we remove this edge, our vertex will also be removed. So the resulting graph will have \(v-1\) vertices, \(k\) edges, and no faces. This graph also satisfies the formula by \(P(n)\), which mans that \((v-1) -k+0 =1\), hence \(v-(k+1)+0 =1\). However, in this case the resulting graph is not a connected planar graph anymore, but rather, a disjoint union of two graphs. Therefore, we do not need to include this case.

Therefore, we have shown that \(P(n+1)\) is true whenever \(P(n)\) is true.

Since \(P(n)\) is true for \(n=1\), and \(P(n) \Rightarrow P(n+1)\), by Principle of Mathematical Induction we conclude that every connected planar graph with V vertices, E edges and F faces satisfies \(V-E+F=1\). Since every convex polyhedron can be represented as a planar graph with one less face, we conclude that Euler's formula \(V-E+f=2\) holds for any convex polyhedron.

Note by Tasha Kim
1 week, 3 days ago

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