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I have to decide, wether the following theorem is right or wrong. There are functions, that satisfy the following conditions:

$ f(n) \in \mathcal{O}(h(n)) $ and $ g(n) \in \mathcal{O}(h(n))$

Now it should hold: $$ \frac{f(n)}{g(n)} =\mathcal{O}(1) $$

By definition I get: $f(n) \leq c_1 \cdot h(n) \ \forall n\geq N$ and $g(n) \leq c_2 \cdot h(n) \ \forall n\geq N'$

So, $$ \frac{f(n)}{g(n)} = \frac{c_1}{c_2} \cdot 1 \ \forall n\geq max\{N,N'\}$$

I'm not sure. g(n) could be 0. What do you think?

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    $\begingroup$ I don't understand why you went from saying f and g are elements of O(something) to f/g being equal to O(1). O(1) is a set, and f/g is a function, so surely the question is whether it is an element of O(1), right? Was this a typo? Can you explain the question more clearly? $\endgroup$ – Eric Lippert Apr 26 at 15:13
  • $\begingroup$ @EricLippert It is extremely common in big O notation to mix and match the element symbol and equality symbol. They both mean element of. Yes it's very confusing. $\endgroup$ – Brady Gilg May 1 at 18:35
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You can't go from $f \leqslant c_1 h$ and $g \leqslant c_2 h$ to $\frac{f}{g} = \frac{c_1}{c_2}$.

And the initial claim is false. Take, for example, $f(n) = h(n) = n^2$, $g(n) = n$.

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  • $\begingroup$ Why can't I go to $ \frac{f}{g}$ $\endgroup$ – Leon1998 Apr 26 at 11:01
  • $\begingroup$ Because why you would? Even with just numbers, no functions: $1 < 2$, $3 < 4$ but $\frac{1}{3} \neq \frac{2}{4}$. May be you wanted to write $\frac{f}{g} \leqslant \frac{c_1}{c_2}$? It's still not true: you have $\frac{1}{g} \geqslant \frac{1}{c_2 h}$, but you can multiply inequalities only with same direction. $\endgroup$ – mihaild Apr 26 at 11:06
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    $\begingroup$ Ok thank you. Now I see my mistake:) $\endgroup$ – Leon1998 Apr 26 at 11:34
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Consider $f(n)=h(n)=1$ and $g(n)=1/n$.

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