If both $t\_1(n)$ and $t\_2(n)$ are $O(f(n))$, then what is $O(t\_1(n) / t\_2(n))$?

Here is my reasoning... I know that the following property holds:

$$t\_1(n)\cdot t\_2(n) = O(f(n) \cdot f(n)) = O(f(n)^2) $$

But the inverse property does not hold:

$$t\_1(n)/t\_2(n)=O(f(n)/f(n)) = O(1)$$

Since the division property does apparently not hold, does this mean we cannot know $O(t\_1(n)/t\_2(n))$ unless $t\_1(n)$ and $t\_2(n)$ are known?

Bonus points if you can also explain why the division property does not hold.