00:02
So this question, we'll try that if g is connected, then it is possible to remove vertices to disconnect g if and only if g is not complete.
00:12
So one way to approach this problem is to show that if g is complete, then we cannot remove vertices to disconnect g.
00:30
And alternatively, if g is not complete, then we can remove vertices to disconnect g, right? doing so, essentially showing that these two parts are both true will give us information needed to draw the conclusion of problem 51.
01:00
So first off, let's assume that g is a connected graph on n vertices, right? so if we remove one vertex, then we'll get a connected graph on n -minus -1 vertices, right? because if we remove one vertex, then the degree of all the remaining vertices goes down by exactly one.
01:25
So all m -minus -1 vertices will have degree m -minus 2, because the edge is connecting, connecting all pairs, all possible pairs of these n minus one remaining vertices will still be intact.
01:42
There are still be edges.
01:43
So basically, if we remove, removing j vertices then, right, gives us a complete graph on n minus j on all n minus j remaining vertices.
01:59
So this goes on forever.
02:01
And so we cannot, in conclusion, we cannot remove, cannot disconnect g by removing vertices because no matter how many we remove, we will just get a complete graph on the remaining vertices in the graph.
02:21
So the first part of the statement is complete, right? we've shown that if g is complete then we no matter what we what vertices remove, g will always remain connected...