(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Show that, for positive integers [itex]a[/itex] and [itex]b[/itex], there exists unique integers [itex]q[/itex] and [itex]r[/itex] such that

[itex]a = bq + r[/itex] and [itex]-b/2 < r \leq b/2[/itex].

2. Relevant equations

We can assume the standard division algorithm holds. That is, for integers [itex]a[/itex] and [itex]b[/itex], there exists unique integers [itex]q[/itex] and [itex]r[/itex] such that

[itex]a = bq + r[/itex] and [itex]0 \leq r < b[/itex].

Let [itex][\cdot][/itex] represent the floor function.

3. The attempt at a solution

Suppose [itex]a[/itex] and [itex]b[/itex] are positive integers. By the division algorithm, there exists unique integers [itex]q[/itex] and [itex]r[/itex] such that

[itex]a = bq + r[/itex] and [itex]0 \leq r < b[/itex].

At this point, I let [itex]r' = [b/2] - r[/itex]. Since [itex]0 \leq r[/itex], we have that [itex]-r \leq 0[/itex], which implies [itex][b/2] - r = r' \leq [b/2] \leq b/2[/itex] by the definition of the floor function. Also, since [itex]r < b[/itex], we see that [itex]-r > -b[/itex], which implies [itex][b/2] - r > [b/2] - b = [-b/2] \geq -b/2[/itex].

This is where I get stuck. I have found an integer [itex]r'[/itex] that gives [itex]-b/2 < r' \leq b/2[/itex], but now I have to find an integer [itex]q'[/itex] to satisfy [itex]a = bq' + r'[/itex]. Any ideas?

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# Homework Help: Proof of Altered Division Algorithm

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