Counterexample to big-O claim looks wrong
The counterexample is fine. Any function that is $O(n)$ is also $O(n^2)$&nbsp;&ndash; if a function's "smaller than" $n$, it's also "smaller than" $n^2$.

And note that even weaker statements are false.  In fact, there is no function&nbsp;$g$ such that $T_1,T_2=O(f)$ implies that $T_1(n)/T_2(n)=O(g(n))$.  For example, take $T_1(n)=1$ and $T_2(n)=1/g(n)^2$. So, now, $T_1(n)/T_2(n) = g(n)^2\neq O(g(n))$.