For rational numbers $a, b$, what is the range of $b$ such that $\lceil a + b \rfloor = \lceil a \rfloor$ holds? For rational numbers $a, b$, what is the range of $b$ such that  $\lceil a + b \rfloor = \lceil a \rfloor$  holds?
Clearly, b=0 gives us the result.
What are the lower and upper bounds of $b$?
$\lceil \cdot \rfloor$ : is a rounding function that rounds a rational number to the nearest integer
The range of $a$ is $[-\frac{x}{2}, \frac{x}{2})$ for some positive integer $x$.
 A: For positive $a$, $\lceil a \rfloor = \lfloor a + \frac 12 \rfloor$, being the floor function. If $a = a_q+a_r$ where $a_q$ is the whole number part and $a_r$ is the fractional part, $\lceil a \rfloor = a_q + \lfloor a_r + \frac 12 \rfloor$. Similarly, if $b$ (also positive) is $b_q + b_r$ then
\begin{align}
\lceil a + b \rfloor  & = \big \lfloor a_q+a_r + b_q + b_r + \frac 12 \big \rfloor \\
& = a_q + b_q + \big \lfloor a_r + b_r + \frac 12 \big \rfloor
\end{align}
If this is to equal $\lceil a \rfloor$ then we must have $b_q = 0$ and we want to find when $\lfloor a_r + b_r + \frac 12 \rfloor  = \lfloor a_r + \frac 12 \rfloor$.
This is best broken into parts: $a_r < \frac 12$ and $a_r \ge \frac 12$. For the first of these we find that $b_r \in [0, \frac 12 - a_r)$, while for the second $b_r \in [0, \frac 32 - a_r)$.
For negative $a$, it is easier to think of $a_q< 0$ and $a_r \ge 0$, thus $-5.6 = -6 + 0.4$. The above arguments are true and we still split it into two: when $a_r< \frac 12, b\in [0, \frac 12 - a_r)$ and when $a_r \ge \frac 12, b \in [0, \frac 32 - a_r)$.
