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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}

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.

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    I forgot to say, $c$ and $d$ should also be integers. – 2017-02-02
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    have you tried solving it for some examples and try to get some intuition why this is true? – 2017-02-02
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    To tell you the truth, no. I tried to prove it from the start. – 2017-02-02
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    I would start there. Choose some two integers $a,b$ and check which integers you can write as $a-cb$. Look on these integers on the line $\mathbb{R}$ and you should see why this result must be true. – 2017-02-02
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    I tried it for $a=5$, $b=3$ and got $c=2$ and $d=-1$. So $d$ is not some kind of remainder. – 2017-02-02

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

If $d'

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'

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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.

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    I know, but to show that I used the well ordering of the naturals. And looks like I can't use it here – 2017-02-02
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    @Amontillado You don't have to show it from the start again. Just modify the $x$ and $y$ in such a way that you get $c,d$. – 2017-02-02