# How did the author get at $z = -(|a|+1)$ in the proof of division theorem?

Theorem Let $a,n \in Z$ with $n \neq 0$. There are $q,r \in Z$ with $a = nq+r$ and $0 \leq r < |n|$, and both $q$ and $r$ are unique with these properties.

Proof The author intends to prove the existence of $q$ and $r$ using well-ordering principle. So it is required to prove that $r \geq 0$ with $r= a-qn$. Consider $S = \{a-zn|z \in Z\}$ and show that the intesection of $S$ and $N$ (set of natural numbers) is non-empty. If $z = -(|a|+1)$ then $a-zn=a+(|a|+1)n \geq a + |a| + n \geq n > 0$. Consequently intersection of $S$ and $N$ is non-empty. The proof continues further.

Question

1. How was the author able to choose the value of $z$ as $-(|a|+1)$?
2. How does proving $n >0$ allow us to conclude that the intersection of $S$ and $N$ is non-empty?

The argument is erroneous as stated: the step $a+(|a|+1)n\ge a+|a|+n$ requires that $n$ be positive, which was not assumed. It can be corrected by letting $e=-\frac{|n|}{n}$, so that $en=-|n|$. Let $S=\{a-zn:z\in\Bbb Z\}$ as before, but take $z_0=e(|a|+1)$. Then
$$a-z_0n=a-e(|a|+1)n=a+|n|(|a|+1)\ge a+|a|+1\ge 1\;,$$
since $|n|>0$ and $a+|a|\ge 0$. This shows that $S$ has at least one positive element, namely, $a-z_0n$. The argument then proceeds by using the well-ordering principle to take the smallest member of $S\cap\Bbb Z^+$, which we can do because we now know that $S\cap\Bbb Z^+$ is a non-empty set of positive integers.
Coming up with $z_0=e(|a|+1)$ in the first place is just a matter of figuring out what sort of $z$ you need in order to be sure that $a-zn>0$. You need $z<\frac{a}n$. Making $z$ a very negative number will certainly do the job, but how negative is ‘very negative’? Clearly $\left|\frac{a}n\right|\le|a|$, since $|n|$ can be no smaller than $1$. Thus, $\frac{a}n\ge-\left|\frac{a}n\right|\ge-|a|$. If $n>0$, we can take $z=-|a|-1$. If $n<0$, however, multiplying by $n$ changes the sign, so we need $z=|a|+1$. The trick with $e$ just avoids having to consider the two cases separately.