Prove Euler formula by mathematical induction;
(1) When R= 2, from the description of 1, these two regions can be imagined as two hemispheres with the equator as the boundary, and there are two "vertices" on the equator that divide the equator into two "boundaries", that is, R= 2, V= 2 and E = 2；; So R+ V- E= 2, euler theorem is established.
(2) Let R= m(m≥2) and euler theorem is established. It is proved that euler theorem also holds when R= m+ 1.
It can be seen from Note 2 that if we select an area X on the map with R= m+ 1, then X must have an area Y adjacent to it, so after removing the unique boundary between X and Y, there are only m areas on the map; After removing the boundaries of X and Y, if the vertices at both ends of the original boundary are still vertices of three or more boundaries, the vertices will remain, while the number of other boundaries will remain unchanged. If the vertex at one or both ends of the original boundary is now the vertex of two boundaries, the vertex is deleted, and the two boundaries on both sides of the vertex become one boundary. Therefore, when removing the unique boundary between x and y, there are only three situations:
(1) Reduce an area and a boundary.
② Reduce an area, a vertex and two boundaries.
③ Reduce an area, two vertices and three boundaries.
That is to say, when the boundary between x and y is removed, there must be "the number of reduced regions+the number of reduced vertices = the number of reduced boundaries" anyway. We reverse the above process and turn it into a graph of R= m+ 1 again. In this process, it must be "increased number of regions+increased number of vertices = increased number of boundaries". Therefore, if euler theorem holds when R= m (m≥2), euler theorem holds when R= m+ 1. According to (1) and (2), euler theorem holds for any positive integer R≥2.
Refer to the above content: Baidu Encyclopedia-Euler Formula