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As the title states, I wonder how I could prove that for some $\epsilon >0 \; \exists k \in \mathbb{N}$ such that $ka - [ka]$ for any irrational $a$. Where $[x]$ would be the floor function of a real number $x$.

MKaissen
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    Should this maybe read "$|ka-[ka]|<\epsilon"$? – lulu Dec 07 '15 at 19:13
  • Wouldn't it be $0 \leq ka-[ka] < 1$. $[x] \leq x < [x]+1 \Rightarrow 0 \leq x - [x] < 1$ – MKaissen Dec 07 '15 at 19:14
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    Your statement doesn't involve $\epsilon$ in any way...I'm trying to guess where it fits in. Also, I expect you mean to fix $a$ from the start...presumably you don't expect a single $k$ to work for all $a$. – lulu Dec 07 '15 at 19:15
  • Yes the $a$ is fixed and it's not for all $\epsilon > 0$ but basically for some given one. – MKaissen Dec 07 '15 at 19:18
  • I believe that what you mean to ask is this "given $a\in \mathbb R - \mathbb Q$, show that, for any $\epsilon>0$ $\exists k\in \mathbb N$ such that $|;ka-[ka]|<\epsilon$. Of course, I might be misunderstanding. – lulu Dec 07 '15 at 19:19
  • Yes, that is what I am asking, also English is not my native language so my wording won't be perfect. – MKaissen Dec 07 '15 at 19:21
  • So, the question, then is that the fractional part of $ka$ should be dense...that is true, and has been asked before on this site, eg, here: http://math.stackexchange.com/questions/903142/for-an-irrational-number-a-the-fractional-part-of-na-for-n-in-mathbb-n-is – lulu Dec 07 '15 at 19:25

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