Is it possible to multiple a real number by integer, so that the factional part is low as possible Given a real number $x$, and a real number $\epsilon>0$ arbitrary small as possible, can we find an integer $n$ such that $|nx-\left \lfloor{xn}\right \rfloor |<\epsilon$ or or $|nx-\left \lceil{xn}\right \rceil|<\epsilon$. ? and if so- can we find either an exact number $n=f(x,\epsilon)$ or even an upper bound over $n$?
Edit: For $x\in\mathbb R\setminus\mathbb Q$, the set $\{nx-\lfloor nx\rfloor: n\in \mathbb{N}\}$ is dense on $[0,1)$ proved that such $n$ exists. However, I wish to know any upper bound\exact formula for $n$ :) 
I know that the above is true by replacing the above constrain by $|nx-\left \lfloor{xn}\right \rfloor |<\epsilon\cdot n$. This follows simply as for each irrational $x$ number there is a series of rational numbers converging to $x$. However, I wish to prove\disprove a more generalized statement.
 A: I found a really simply upper bound using Dirichlet's approximation theorem:
https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem
By using the upper bound $n\leq\left \lceil{1/\epsilon}\right \rceil$.
Note: It doesn't really matter for me if it is $|nx-\left \lfloor{xn}\right \rfloor |<\epsilon$ or $|nx-\left \lceil{xn}\right \rceil|<\epsilon$. I will update my question above.
A: I think I found a simple (but naive and inefficient) way to find a multiplier to generate a product with fractional part less than 10^-n.
Multiply the real number by 10^n different numbers and look at the first n digits of the fractional part of each. If two of them are the same then the difference in the two multipliers will work.
If not then there are 10^n different values of this first n digits, which means every combination is covered including n zeroes, which again does the trick.
Apologies (a) for lack of mathjax (I'm on a phone and in a rush), (b) if I have a out-by-1 error (then 10^(n+1) will do), and (c) if i turn out to be completely wrong.
