Why do all additive inverses of natural numbers precede $0_R$ in the integer ring?

Question

${\forall}a^{-1}:{\lnot}(a^{-1}\,{\in}\,\mathbb{N})\,{\land}\,(a^{-1}\,{\in}\,\mathbb{Z})\,{\land}\,({\exists}a\,{\in}\,\mathbb{N}:a*a^{-1}=a^{-1}*a=0):a^{-1}\,{\preceq}\,0$

How to prove the above?

Answer 1

One of the axioms of an ordered ring like $\mathbb Z$ is:

If $R$ is an ordered ring, and $a,b,c\in R$ and $a\leq b$, then $a+c\leq b+c$.

Given any negative integer $-z$, we can start with $0\leq z$ and then add $-z$ to both sides and invoke this axiom: 0+ (-z) $\le$ z+ (-z).