# Big O notation of $O(n/(m-n))$

I'm new to the complexity theory and have a basic question about the big-O notation that I encountered.

I came across a complexity of $$O\big(\frac{n}{m-n}\big)$$, where both $$n$$ and $$m$$ are independent variables in my algorithm (assuming $$m \geq n$$). Since I haven't encountered the subtraction of two variables in the denominator before, I'm curious if it can be simplified further, perhaps to $$O\big(\frac{n}{m}\big)$$ or other nicer forms. I've thought about it for some time, and maybe dividing the fraction by $$n$$ gives $$O\big(\frac{1}{\frac{m}{n}-1}\big) = O\big(\frac{1}{\frac{m}{n}}\big) = O\big(\frac{n}{m}\big)$$, but not sure if this is correct.

Additionally, I'm wondering about the whether containing expressions like $$m-n$$ is common or not. Any help or advice on this would be greatly appreciated :)

• if $$m$$ is almost $$n$$ (not equal to $$n$$, otherwise $$\frac{n}0$$ is not defined), then it is $$\mathcal{O}(n)$$;
• if $$m$$ is big enough compared to $$n$$ (meaning $$m\geqslant (1 + \varepsilon) n$$, with $$\varepsilon > 0$$ being a constant), then it is $$\mathcal{O}(1)$$.
You cannot conclude that it is $$\mathcal{O}\left(\frac{n}{m}\right)$$ in the general case, because you neglected $$1$$ in front of $$\frac{m}{n}$$, but you cannot always do that, for example when $$m = n+1$$.