Prove that in $\Bbb{Z}$, $\forall a,b\in\Bbb{Z}$ such that $a = bq +r$, we can find $r$ such that $-\frac{1}{2}b \leq r \leq \frac{1}{2}b$

Question

In $\Bbb{Z}$, we know that for all $a, b \in \Bbb{Z}$, we can express $a = bq + r$ such that $|r| < |b|$. However, I read from this post

Prove that the Gaussian Integer's ring is a Euclidean domain

that it is also true that we can find r such that $-\frac{1}{2}b \leq r \leq \frac{1}{2}b$. I am able to prove this by considering four case, when $r$ is less than half of $b$,more than half of $b$, $r$ is half of $2b$ or $r$ is twice of $2b$. However, i find this proof to be very ugly. Can someone provide me with a more elegant proof of this fact. Thanks.

Answer 1

Among any $b$ consecutive integers you can find a multiple of $b$ (because there's only $b$ different possible remainders). $(a- \lfloor \frac{b}{2} \rfloor), ..., (a+ \lfloor \frac{b}{2} \rfloor) $ is a list of at least $b$ consecutive integers, so there is a number $qb$ among them. Then $r=a-qb$ will suffice.

Answer 2

Just apply the division algorithm to y = a + floor(b/2) to get y = qb + r'.

Set r = r' - floor(b/2).