Division with remainder property:

Let $a,b \in \mathbb{Z}$ with $b > 0$. Then there exist unique $q,r \in \mathbb{Z}$ such that $a = bq + r$ and $0 \le r < b$.

Generalized division with remainder property:

Let $a,b \in \mathbb{Z}$ with $b > 0$, and let $x \in \mathbb{R}$. Then there exist unique $q, r \in \mathbb{Z}$ such that $a = bq + r$ and $r \in [x,\ x + b)$.

Show that the generalized division with remainder property also holds for the interval $(x,\ x + b]$. Does it hold in general for the intervals $[x,\ x + b]$ or $(x,\ x + b)$ ?

Consider the interval $(x,\ x + b]$, which contains exactly $b$ integers, namely $\lfloor x \rfloor + 1,\ ... , \lfloor x \rfloor + b$. Applying the division with remainder property with $a - (\lfloor x \rfloor + 1)$ in place of $a$:

$$\begin{aligned} a - (\lfloor x \rfloor + 1) &= bq + r \\ a &= bq + r + \lfloor x \rfloor + 1 \\ \end{aligned}$$

Let $r' = r + \lfloor x \rfloor + 1$. Given $r \in [0,\ b)$,

$$\begin{aligned} r &\in \{0,\ ...,\ b - 1\} \\ r' &\in \{\lfloor x \rfloor + 1,\ ..., \lfloor x \rfloor + 1 + b - 1\} = \{\lfloor x \rfloor + 1,\ ..., \lfloor x \rfloor + b\} \\ r' &\in (x,\ x + b] \end{aligned}$$

In other words, the generalized division with remainder property holds for the interval $(x,\ x + b]$.

Consider the interval $[x,\ x + b]$, which contains integers:

$$\begin{aligned} \{ \lceil x \rceil,\ ..., \lceil x \rceil + b - 1, \lfloor x \rfloor + b\} \end{aligned}$$

In particular, when $x$ is an integer, this set will contain exactly $b + 1$ integers, and therefore the interval $[x,\ x + b]$ does not hold in general. (It would hold, however, in the special case when x is not an integer, since then the set would contain exactly $b$ integers. Do you see why ?)

Similarly, consider the interval $(x,\ x + b)$, which contains integers:

$$\begin{aligned} \{ \lfloor x \rfloor + 1,\ ..., \lfloor x \rfloor + b - 1, \lceil x \rceil + b - 1 \} \end{aligned}$$

In particular, when $x$ is an integer, this set will contain exactly $b - 1$ integers, and therefore the interval $(x,\ x + b)$ does not hold in general. (It would hold, however, in the special case when x is not an integer, since then the set would contain exactly $b$ integers. Do you see why ?)

$\Box$

# posted by rot13(Unafba Pune) @ 10:58 PM