Five Color Theorem

Theorem
Any planar graph $$G \ $$ has the property that every vertex can be assigned a number $$\left\{{0,1,2,3,4}\right\} \ $$, called "colors," such that no edge has the same color at both ends.

Proof
It is obvious the theorem is true for a graph with only one vertex. We will induct on the number of vertices.

Since $$G \ $$ is planar, $$\chi(G) = 2 \ $$ by the Euler polyhedron formula.

Suppose every vertex of $$G \ $$ has 6 edges or more. Then $$2 = \chi(G) = v - e + f \leq v - (3v) + f \ $$, so $$f \geq 2+2v \ $$. But each face is bounded by at least 3 edges, and each edge bounds at most 2 faces, so so $$f \leq \tfrac{3e}{2} \ $$ and hence $$3e \geq 4+4v \ $$

Therefore, $$G \ $$ has at least one vertex with at most 5 edges.

Remove that vertex $$x \ $$ from $$G \ $$ to create another graph, $$G ' \ $$. By induction hypothesis, $$G' \ $$ is five-colorable. If all five colors were not connected to $$x \ $$, then we can give x the missing color and thus five-color $$G \ $$. If all five colors were connected to $$x \ $$, we examine the five vertices $$x \ $$ was attached to, call them $$y_1, y_2, y_3, y_4, y_5 \ $$.