Bipartite proof Let $G$ be a graph of order $5$ or more. Prove that at most one of $G$ and "$G$ complement" is bipartite. 
I'm lost as to what needs to be done. I know that A nontrivial graph $G$ is bipartite if and only if $G$ contains no odd cycles.

Let me give you a hint towards a much, much more elementary solution:
Hint: It is enough to show that if $G$ is bipartite, then $\bar{G}$ (the complement of $G$) is not bipartite.
To that end, suppose that $V(G)=A\cup B$ is a bipartition of $G$.  Then in $E(G)$, there are no edges inside $A$, no edges inside $B$, and there may be some edges between $A$ and $B$.
Based on this, what does $E(\bar{G})$ look like? Since $G$ contains no edges inside $A$ and no edges inside $B$, it must be the case that $\bar{G}$ contains EVERY edge inside $A$ and EVERY edge inside $B$.
So, if $\bar{G}$ WERE bipartite, it couldn't be the case that any two vertices from $A$ or any two vertices from $B$ fell in the same part of the bipartition.
Can you see how to finish it from here, using the assumption that there are at least 5 vertices distributed between $A$ and $B$?
