Prove that G or bar G must be nonplanar if G has 11 vertices. The exact problem is the following: "Let G be a graph with 11 vertices. Prove that G or bar G must be nonplanar. I went to my professor to discuss the proof, but she said it didn't quite work, that something was "off" about it. The following is the proof I presented: Let G be a graph with 11 vertices. It is clear that |E(G)|+|E(bar G)|=|E(K11)|=55. Assume, for the sake of contradiction, that both G and bar G are nonplanar. Then |E(G)|+|E(bar G)|<=(3|V(G)|−6)+(3|V(bar G)|−6)=54.Rightarrow Leftarrow Therefore, either G or bar G is nonplanar. So, where exactly does this proof fail? Any tips on the aesthetics of the proof would also be greatly appreciated.

Melissa Walker

Melissa Walker

Answered question

2022-11-17

Prove that G or G ¯ must be nonplanar if G has 11 vertices.
The exact problem is the following: "Let G be a graph with 11 vertices. Prove that G or G ¯ must be nonplanar.
I went to my professor to discuss the proof, but she said it didn't quite work, that something was "off" about it. The following is the proof I presented:
Let G be a graph with 11 vertices. It is clear that | E ( G ¯ ) |
Assume, for the sake of contradiction, that both G and G ¯ are nonplanar.
Then
| E ( G ) | + | E ( G ¯ ) | ( 3 | V ( G ) | 6 ) + ( 3 | V ( G ¯ ) | 6 ) = 54.
Therefore, either G or G ¯ is nonplanar.
So, where exactly does this proof fail? Any tips on the aesthetics of the proof would also be greatly appreciated.

Answer & Explanation

Maffei2el

Maffei2el

Beginner2022-11-18Added 20 answers

In an arithmetic series we can find common difference between two terms a m and a n by using formula d = a n - a m n - m , if common difference is same throughout, we can say this is an arithmetic sequence.
Here we can get three values of d
1. a 5 - a 3 5 - 3 = 27 - 21 2 = 6 2 = 3
2. a 12 - a 5 12 - 5 = 48 - 27 7 = 21 7 = 3 and
3. a 12 - a 3 12 - 3 = 48 - 21 9 = 27 3 = 3
As we get d as constantly same, it is an arithmetic series.

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