For a rational number $x \in [-a/4, a/4)$ with a positive integer $a$, is it possible to separate it into its integral and fractional parts?

Question

For a rational number $x \in [-a/4, a/4)$ for a positive integer $a$, is it possible to separate it into its integral and fractional parts? Namely, can we represent $x$ as $x = b + c$ where $b \in \mathbb{Z}_a, c \in [-1/4, 1/4)$?

Answer 1

We see that the interval length for $c$ is $\frac12$ so we can multiply by $2$ to get closer to the ordinary fractional part definition.
Multypliing by $2$ gives
$$-\frac{a}2\le 2x<\frac{a}2,\,\\ 2x=2b+2c,\,\\ -\frac12\le 2c <\frac12,\hbox{so}$$ $0\le 2c+\frac12<1$ and $\left\{2c+\frac12\right\}=\left\{2x+\frac12\right\}$ so we can $c=\frac12\left(\left\{2x+\frac12\right\}-\frac12\right)$ and $b=x-c$.
However, this implies $2b=\lfloor2x+\frac12\rfloor$ to be integer, not $b$ itself. Let's take some $x$ such as $2b=1$, e.g. $x=\frac12$ and there will be no such $c,-\frac14\le c<\frac14$ for $x=b+c$ for some $b\in\mathbb{Z}$.