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