An exercise in number theory and divisibility

Question

The following was given as an exersise to me and I 'm stuck.

If $a$ and $b$ are positive integers and $b \ne 0$, then show that there are unique $c$ and $d$, integers, so that $a = cb + d$ and $-{b \over 2}<d \le {b\over2}$.

If $b$ is even then by setting $S= \{ a-kb + {b\over2}:k\in \Bbb Z, a-kb + {b\over2} < 0\} $ it's not hard to show it by using the well ordering principle.

But if $b$ is odd then $S$ is a set of rationals and I can't use the well ordering principle. Can you help me? I'm new to number theory, sorry if this is trivial or already answered.

Answer 1

Apply division algorithm first.So u get unique c' and d' such that a=c'b+d' where $0\leq d'<b$

If $d'<b/2$ then we can take d = d' and c = c'.

If not then b>d'>b/2 => b/2>(d'-b)>-b/2 so,a=(c'+1)b+(d-b')

take c =c'+1 and d =d'-b

Uniqueness follows from uniqueness of c' and d'

Answer 2

Hint:

We know that there exists a unique pair of integers $x,y$ such that $$a=xb + y$$

and $0\leq y < b$.

This statement is very close to the one you want to prove.