# Some arithmetic of fractional part of a irrational number

This is a very simple question, but i don't have a clue how to prove this..

Let $[x+y]$ denote $x+y$ and $(x)$ be a fractional part of $x$

Suppose $x$ is an irrational number and $n$ is an integer.

Let $\delta \in (0,\frac{1}{2}\min\{(nx),1-(nx)\})$

Here, how do i prove that $(n[x-\delta])=(nx)-(n\delta)$ and $(nx)<(n[x+\delta])$?

$[x\pm\delta]$ is an integer, so $n[x\pm\delta]$ is an integer, so $(n[x\pm\delta]) = 0$, so both claims seem to be false, unless I've misunderstood. –  Billy Dec 20 '12 at 2:46
Because $[\cdot]$ only returns integers. –  Billy Dec 20 '12 at 2:51
This claim still seems false.. If $\pi$ for $x$ and $0.06$ for $\delta$ and $n=8$, the equality doesn't hold. There must be a typo in my book, but cannot figure out what that would be... I'll check and edit this post soon. –  Katlus Dec 20 '12 at 3:17