Can a 2 dimensional shape have more vertices than its edges, if there are no curved edges? For example, a hexagon has 6 edges, and 6 vertices. However, is there any shape, open or closed, where the number of vertices exceeds the number of edges?
Just draw a shape by its vertices. (I assumed a shape is a polygon here. Otherwise this is not true. For instance this does not hold when we draw a square with its diagonals.)
- The first vertex will connect the first two edges. We now have one vertex and two edges.
- Adding a new vertex will now add one edge (because one edge was already accounted for by the previous vertex). So for $m$ vertices we now have $m+1$ edges.
- This will hold untill we draw the last edge. The last edge will connect two edges which are already acconted for.
Thus in the end we must have $n$ vertices and $n$ edges, for some $n$, by construction.