The only way the statement is false is if there exists a graph G of order 5 or more such that G and G complement is bipartite. So we assume a graph G is bipartite, and show that G complement must have an odd cycle and thus cannot be bipartite. We do not have to prove that if G complement is bipartite then G is not, because the graph G is arbitrary, and G complement relates to G in the same way that G relates to G complement - they are complements of each other.
Lesson proving a bipartite graph has no odd cycles:
Lesson proving a graph with no odd cycles is bipartite:
I hope you find this video helpful, and be sure to ask any questions down in the comments!
+WRATH OF MATH+
◆ Support Wrath of Math on Patreon:
Follow Wrath of Math on...
● Instagram:
● Facebook:
● Twitter:
My Music Channel:
0 Comments