Explanation:
We can use Euler's formula to approach this problem:
For a connected planar graph with V vertices, E edges, and F faces (including the outer face), Euler's formula states that:
V - E + F = 2
We know that our graph has 7 vertices and 16 edges. Assume, for the sake of contradiction, that this graph is planar. Then it must have a certain number of faces.
We know that each face of a planar graph must have at least 3 edges, and each edge is adjacent to two faces. Thus, the total number of edges in all the faces is at least 3F. On the other hand, each edge in the graph is adjacent to two vertices, so the total number of edges is 2V. Therefore:
3F ≤ 2V (1)
Using the fact that the graph is simple, each face of the graph is bounded by at least three edges. Therefore, the total number of edges in the graph is at least 3F/2. On the other hand, we know that the graph has 16 edges. Thus:
3F/2 ≤ 16 (2)
Combining inequalities (1) and (2) gives:
3F/2 ≤ 2V
3F/2 ≤ 14 (since V = 7)
F ≤ 9/2
This means that the number of faces in the graph is less than or equal to 9/2, which is impossible since the number of faces must be a non-negative integer. Therefore, our assumption that the graph is planar must be false, and we have proven that any simple connected graph with 7 vertices and 16 edges is not planar.