00:01
E as a simple graph with at least 11 vertices, g bar is complement.
00:06
Prove that at least one of g and g bar is not planar.
00:10
We're going to prove by contradiction.
00:11
So assume g and g bar are both cleaner.
00:27
Okay, we're going to let e1 be the number of edges in g.
00:43
And let e2 be the number of edges in the complement.
01:00
Okay, we have a theorem about the relationship between the number of edges and the number of vertices.
01:06
Then e1 is at most 3v minus 6.
01:13
But the number of vertices in the graph is 11.
01:17
So that's going to be 3 times 11 minus 6 is 33 minus 6.
01:23
Is 27.
01:26
E2 is also at most 3v minus 6.
01:30
The complement is on the same vertex set us the original graph, so that's still 3 times 11 minus 6, so that's also 27.
01:41
Thus, e1 plus e2 is at most 54, the sum of 227s.
01:48
Okay.
01:51
Okay, now g and g bar together, are the complete graph on 11 vertices.
02:19
And so they have a total of 11 choose two edges.
02:34
And the complete graph means that every vertex is connected with every other vertex.
02:40
So how many ways can i do that? well, i just choose two vertices from the 11.
02:46
11 choose 2 is 11 times 10 over 2.
02:51
That's going to be 11 times 5 with just 55.
02:53
Okay, so e1 plus e2 is at most 54, but i'm also saying that they have a total of 55 edges.
03:03
That's a contradiction.
03:12
Right.
03:14
This would be e1 plus e2.
03:16
That's a contradiction.
03:17
It can't be the sum 55 and also be at most 54.
03:21
Okay, so we can conclude one of g or g bar is not planar.
03:40
Now we want to find a simple graph g in eight vertices.
03:43
Such that g and g bar are both cleaner.
03:47
I'll be one, two, three, four, five, six, seven, eight.
03:57
One, two, three, four, five, six, seven, eight.
04:05
Okay, i'm gonna start by making g kind of like as connected as possible while still being planer.
04:12
I'm just going to connect.
04:23
Okay, one can also kind of kind of...