Chapter 4, Problem 51C

Explanation of Solution

Initial considerations:

Assume that the line arrangement “S” be a set of “n” lines in the plane and there is no two lines are parallel and no three lines meet in the same point.

Since the following induction step proves that simple arrangement of “n” lines in ${\text{R}}^{\text{2}}$ has “$\text{n/2}$” vertices “${\text{n}}^{\text{2}}$” edges.

Proof:

Let us first start count the number of edges using induction on “n”. For n=1 there is 12 =1 edge.

If the line is added with one to “n–1” line of arrangement then it is easy to split “n–1” existing edges into two and the “n” new edges are introduced along the newly inserted line