If $d\mid a$ and $a\mathrm{\ne }0$ then $d\mathrm{\ne }0$ and $\mathrm{\mid }d\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }$; hence the set of divisors of a nonzero integer is bounded above by $\mathrm{\mid }a\mathrm{\mid }$

Write $a=dq$. If $d=0$ then $a=0\cdot q=0$, and if $q=0$ then $a=d\cdot 0=0$; both contradict $a\mathrm{\ne }0$, so $d\mathrm{\ne }0$ and $q\mathrm{\ne }0$.

If $0\le u$ and $0\le v$ then $0\le uv$: if $u=0$ or $v=0$ then $uv=0$, and otherwise $0<u$ and $0<v$, so $0<uv$ because positives are closed under multiplication.

Discreteness of $\mathbb{Z}$: if $0<x$ then $1\le x$. Indeed $x\ge 0$, so $x=\iota (k)$ for some $k\in \mathbb{N}$; $k\mathrm{\ne }0$ because $\iota (0)=0\mathrm{\ne }x$; hence $0<k$ in $\mathbb{N}$, so $1=\sigma (0)\le k$, and applying $\iota$, which preserves the order, gives $1=\iota (1)\le \iota (k)=x$.

$\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }d\mathrm{\mid }\mathrm{\mid }q\mathrm{\mid }$, and $\mathrm{\mid }q\mathrm{\mid }>0$ because $\mathrm{\mid }q\mathrm{\mid }\ge 0$ and $\mathrm{\mid }q\mathrm{\mid }\mathrm{\ne }0$, the latter since $q\mathrm{\ne }0$.

Hence $1\le \mathrm{\mid }q\mathrm{\mid }$, so $0\le \mathrm{\mid }q\mathrm{\mid }-1$ by compatibility of the order with addition.

Since $\mathrm{\mid }d\mathrm{\mid }\ge 0$ and $0\le \mathrm{\mid }q\mathrm{\mid }-1$, the product $\mathrm{\mid }d\mathrm{\mid }(\mathrm{\mid }q\mathrm{\mid }-1)$ is nonnegative, and it equals $\mathrm{\mid }d\mathrm{\mid }\mathrm{\mid }q\mathrm{\mid }-\mathrm{\mid }d\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }-\mathrm{\mid }d\mathrm{\mid }$ by distributivity; adding $\mathrm{\mid }d\mathrm{\mid }$ gives $\mathrm{\mid }d\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }$.

Finally, every divisor $d$ of $a$ satisfies $d\le \mathrm{\mid }d\mathrm{\mid }$ and $\mathrm{\mid }d\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }$, hence $d\le \mathrm{\mid }a\mathrm{\mid }$ by transitivity: $\mathrm{\mid }a\mathrm{\mid }$ is an upper bound for the set of divisors of $a$.