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I am wondering why is $\{\frac{y}{n} + k\} = \{\frac{y}{n}\}$ where $k$ is an integer. $\{x\}$ is a fractional part function where $\{x\} = x - \lfloor x\rfloor$. I know it makes sense logically, but when I try to prove it I can not seem to get rid of $k$.

$\{\frac{y}{n} + k\} = \frac{y}{n} + k - \lfloor\frac{y}{n} + k\rfloor$. How does $k$ disappear from this equation?

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  • $\begingroup$ It would help if you stated any particular restrictions on the variables $n,k,y$ since some of the properties of the floor, ceiling, etc., functions are tied to such restrictions. I imagine the result doesn't old true, for example, for all $n,k,y$ in the reals. $\endgroup$ – Eevee Trainer Feb 14 '19 at 9:22
  • $\begingroup$ Edited, $k$ is an integer. $\endgroup$ – Michael Munta Feb 14 '19 at 9:29
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First of all, in general, $\{\frac yn+k\}\neq \{\frac yn\}$. For example, $y=0,n=1,k=\frac12$ are a counterexample. But yes, the equality holds if $k$ is an integer. Then, it can even be written more simply as $\{x+k\}=\{x\}$, where $x$ can be any real number.

To actually prove that equality, you can use the following hint:

The function $\lfloor x\rfloor$ has the property that $\lfloor x+n\rfloor = \lfloor x \rfloor + n$ for all $n\in\mathbb N$.

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  • $\begingroup$ I know of this property, but $n$ is still positive so it can not cancel out. $\endgroup$ – Michael Munta Feb 14 '19 at 10:01
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    $\begingroup$ @MichaelMunta $\{x + k\} = x+k - \lfloor x+k\rfloor = x+k-(\lfloor x\rfloor + k) = \dots$. $\endgroup$ – 5xum Feb 14 '19 at 10:10

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