I don't understand why the $1^{st}$ is false but I think I see why the $2^{nd}$ is true.

  1. If $f(n) = O(n^2)$ and $g(n) = O(n^2)$, then $f(n) = O(g(n))$.

  2. If $f(n) = O(g(n))$ and $g(n) = O(n^2)$, then $f(n) = O(n^2)$.

I understand why the second is true but not the $1^{st}$. For case 1, if

  • $f(n) < c_1n^2$ for some $n > n_1$ and
  • $g(n) < c_2n^2$ for some $n > n_2$

by using constants instead of big-O notation , can't we find $c_3$ such that $f(n) < c_3g(n)$ for some $n > n_3$ ?


1 Answer 1


As a counterexample you can take $f(n)=n$ and $g(n)=\sqrt{n}$. You can think that $f(n)=O(n^2)$ means that an upper bound for $f(n)$ is $n^2$ (of course, without considering multiplicative constant), so the fact that both $f$ and $g$ are bounded by $n^2$ does not implies nothing about the relative behavior of $f$ and $g$.

  • $\begingroup$ Thanks. I see that with $g(n) = \frac{1}{n}$, then this doesn't $f(n) \neq O(g(n))$ $\endgroup$ Apr 17, 2021 at 18:11

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