 Given a planar graph with e edges and v vertices, we proved that e has to be less than or equal to 3v minus 6. Intuitively, this means a planar graph can't have too many edges. This suggests that the minimum degree can't be too high either. Let's think about that. Suppose g is planar with minimum degree equal to d. If it has v vertices, then the degree sum will be at least d times v. Now, remember the handshake theorem tells us the sum of the degrees of all vertices in a graph is twice the number of edges. And so by the handshake theorem, we have 2e must be greater than or equal to dv. Consequently, there are greater than or equal to dv over two edges. Now again, if g is planar, e has to be less than or equal to 3v minus 6, and we know that e itself is greater than or equal to dv halves, so we can replace and do some algebra and Consequently, d is strictly less than 6. Consequently, in a planar graph, the minimum degree is at most 5. This leads to an important result known as the six-color theorem. While we can prove it immediately, remember it's the journey, not the destination. What are some consequences of the fact the minimum degree of a planar graph has to be 5 or less? Let's think about that. Let di be the number of vertices of degree i. So remember the degree sum and the number of edges are related by twice the number of edges is the degree sum, and we can express that degree sum this way, i times the number of vertices of degree i. Since 5 seems to be an important number, let's split our sum. So we're looking at the sum of the vertices with degree 5 or less and all the rest. And notice that this first sum, because we're multiplying the number of vertices by the degree of the vertices, that's going to necessarily be greater than just the sum of the vertices. Likewise, for the other sum, since every degree here is 6 or more, we can replace it with a smaller value and get the inequality and factor out the constant, which gives us the inequality. But now this first sum is just the number of vertices of degree 5 or less. We'll call that number k, and this second sum is just the number of vertices of degree 6 or more. And since there are v vertices altogether and k have degree 5 or less, there will be v minus k vertices of degree 6 or more. And putting this all together, we find that, but again in a planar graph, e has to be less than or equal to 3v minus 6, and so this gives us the inequality. And with a little bit of algebra we find, consequently, k must be greater than or equal to 12 fifths. Now since k is the number of vertices of degree 5 or less, and k has to be greater than or equal to this amount, this means that the planar graph has at least three vertices with degree 5 or less. In fact, we could obtain even more detail. While we use the inequality, that the first part of the sum was greater than or equal to the sum of the number of vertices of degree 5 or less, it's a finite sum with a definite cap, so we can actually write it out explicitly. And remember k was the number of vertices of degree 5 or less, well, that's just the sum d1 through d5. And so with a little bit of algebra we find, let di be the number of vertices in our graph with degree i. If g is planar, then this sum must be greater than or equal to 12. And in some sense this is a remarkable result because it means we don't have to know what the graph looks like to be able to say something about the degrees of sum of its vertices. For example, suppose we see a portion of a planar graph, and we know that no vertex has degree 1 and there are no other degree 5 vertices, what do we know about the rest of the graph? Now since we know we have a particular sum and the number of vertices of degree 1 is zero and there's only one degree 5 vertex, then we have, and at this point we can list all feasible combinations of degree 2, 3, and 4 vertices. Since we need the inequality to be satisfied, one solution is to have d2 equal to 3, and this gives us 3 degree 2 vertices. If d2 equals 2, in other words we have 2 degree 2 vertices, then in order for our inequality to be satisfied, we need d3 greater than or equal to 1, or d4 greater than or equal to 2. And so that's 2 degree 2 vertices and 1 degree 3 vertex, or 2 degree 2 vertices and 2 degree 4 vertices. And there are other possibilities. And now we're ready to prove our first major result. If g is a planar graph, the chromatic number is less than or equal to 6. In other words, we can color the graph using 6 or fewer colors. We'll prove this by induction. Clearly the theorem is true if we only have 1, 2, 3, 4, 5, or 6 vertices, so suppose the theorem is true for graphs with n equals k vertices. Let g be a planar graph with k plus 1 vertices. Now, we know that the minimum degree of g is less than or equal to 5, so there is guaranteed to be a vertex in g with degree 5 or less. So remove the vertex and begin a graph with k vertices, and we know by our induction assumption that the chromatic number is less than or equal to 6. So now we'll put phi back, and since phi has degree 5 or less, it has 5 or fewer neighbors, and so if we have 6 colors, we can definitely assign it to one of the existing 6 colors. And so the chromatic number of g is less than or equal to 6, and so any planar graph has a vertex coloring with 6 or fewer colors.