How does one derive $O(n \log{n}) =O(n^2)$? I was studying time complexity where I found that time complexity for sorting is $O(n\log n)=O(n^2)$. Now, I am confused how they found out the right-hand value. According to this $\log n=n$. So, can anyone tell me how they got that value?
Here is the link where I found out the result.

What this equation means is that the class $O(n\log n)$ is included in the class $O(n^2)$. That is, if a sequence is eventually bounded above by a constant times $n \log n$, it will eventually be bounded above by a (possibly different) constant times $n^2$. Can you prove this? The notation is somewhat surprising at first, yes, but you get used to it.
