1. ## Graph Theory

every planar connected simple graph G has at least 3 vertices of degree no more than 5.

every planar connected simple graph G with 11 or fewer vertices has at least 1 vertex of degree no more than 5.

Question: show that every planar simple graph with 11 or fewer vertices is 5 colourable.

~~~~~~~~~~~#
im not sure how to prove the first statment for the 3 vertices but I need to know how to prove the last statement.

Thanks.

2. For the first question we work by contradiction.

Assume there are at most 2 vertices of degrees less or equal than 5. Then $
e = \tfrac{1}
{2} \cdot \left( {\sum\limits_{x \in V\left( G \right)} {\deg \left( x \right)} } \right) \geqslant \tfrac{1}
{2} \cdot \left( {\deg(u)+\deg(w) + 6 \cdot \left( {v - 2} \right)} \right)=
\tfrac{{\deg \left( u \right) + \deg \left( w \right)}}
{2}
+3(v-2)
$
(1) - u and w are the vertices that may have degree less or equal than 5 -

However, if G is planar and $v=|V(G)|\geq{3}$ we have: $
3v - 6 \geqslant e
$
(2) NOw note that (1) and (2) cannot be true simultaneously. -since G is connected-

3. how can I prove it's 5 - colourable?

4. Just use induction to the number of vertices. Remove a vertice with 5 or less edges, make a 5-coloring and then inspect the colors. if the node had less than 5 edges you are done, if the vertice had 5 edges of less than 5 colors ur done. If the vertice had 5 edges with 5 colors the fact that the K_5 isn't planar saves you, you can smelt two non-connected nodes together, make a 5-coloring, then remove the nodes again and color the removed node in the remaining color.

That's how the 5 coloring proof goes, if you need a proper one with mathematical notation etc. feel free to ask and I'll cook one up.