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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.

user3563894
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  • Such an $n$ always exists. Do you need to find the exact value of $n$, or just prove that it exists? – Calvin Lin Apr 17 '20 at 20:37
  • Find a value for $n$ – user3563894 Apr 17 '20 at 20:37
  • This is ambiguous. What, precisely, would count as a disproof of your more general statement? For contrast, we can easily do better and better with a real number such as $\sqrt 2,$ forever, by a simple recurrence. Probably no algorithm for real number $\pi$ – Will Jagy Apr 17 '20 at 20:53
  • @HallaSurvivor, It sure answers on exitance. I would glad to have an upper bound on $n$ – user3563894 Apr 17 '20 at 20:56
  • @WillJagy disprove : there is an $\epsilon_0>0$ and $x_0$ s.t for each $n$ we have that the above remainder is larger than $\epsilon_0$. However, I think that this was proved. – user3563894 Apr 17 '20 at 21:00
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    The topic you want is called continued fractions. https://en.wikipedia.org/wiki/Continued_fraction You mention an upper bound on $n,$ that will be possible for very well behaved irrational numbers, but not at all nice for general $x.$ Suggest you take $x = \pi$ and try some modest $\epsilon$ such as $1/1000,$ find an $n$ that works. – Will Jagy Apr 17 '20 at 21:07
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    There is a section in the article Will Jagy cited called best rational approximations. It goes the other direction from what you ask. Given $n$ as a maximum denominator of a rational approximating $x$, find the rational. A good approach for you is to check the convergents to $x$ against your desired $\epsilon$. The first that comes within $\epsilon$ is a candidate. Then use the technique in the article to find a few more rationals to try that will have smaller $n$ at the price of greater error. Maybe one of those will fit your $\epsilon$. – Ross Millikan Apr 17 '20 at 21:15

2 Answers2

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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.

user3563894
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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.

IanF1
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