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I am stuck on how to evaluate whether the following condition is true:

Let $\{k\}$ be the fractional part of a real number such that $\{k\} = k - \lfloor{k}\rfloor$.

if $\{\frac{x}{2}\} < \frac{1}{2} + \frac{\{x\}}{2}$, does it then follow that:

$$\{\frac{x+1}{2}\} \ge \frac{1}{2} + \frac{\{x+1\}}{2}$$

I can see that:

$\{x+1\} = \{x\}$

if $\{\frac{x}{2}\} < \frac{1}{2}$, $\{\frac{x+1}{2}\} = \{\frac{x}{2}\} + \frac{1}{2}$

$\frac{\{x\}}{2} < \frac{1}{2}$

So, it follows that $\frac{1}{2} + \frac{\{x\}}{2} < 1$

I can that it is true in the case where $\{x\} = \frac{1}{2}$ since:

$\{\frac{1}{4}\} < \frac{1}{2} + \frac{\{\frac{1}{2}\}}{2}$ and $\{\frac{3}{4}\} \ge \frac{1}{2} + \frac{\{\frac{1}{2} + 1\}}{2}$

Can anyone help me to evaluate this condition is always true?



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up vote 3 down vote accepted

Let $x = n + r,$ where $n = \lfloor x \rfloor$ and $r = \{x\}.$ If $n$ is odd, then $\left\{\frac{n+r}{2}\right\} = \frac{1}{2} + \frac{r}{2},$ so the inequality fails. Hence $n$ is even. Then $n + 1$ is odd, whence $\left\{\frac{n+1+r}{2}\right\} = \frac{1}{2} + \frac{r}{2} = \frac{1}{2} + \frac{\{n+1+r\}}{2},$ so in fact, it is simply an equality.

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