Arithmetic, Order, Discreteness and Intervals of the Integers

Adopt the notation of the definition of the integers: $\mathbb{R}$ is the real numbers with the order $\le$ of its ordered field structure, $\mathbb{N}$ is the set of natural numbers, $\iota:\mathbb{N}\to\mathbb{R}$ is the canonical map, and $\mathbb{Z}$ is the set of integers. For $s,t\in\mathbb{R}$ write $s<t$ to mean $s\le t$ and $s\ne t$, and write $\iota(\mathbb{N})=\{\iota(n):n\in\mathbb{N}\}$ and $-\iota(\mathbb{N})=\{-\iota(n):n\in\mathbb{N}\}$. Then the following hold.

1. (Sign trichotomy.)¶ The sets $\{0\}$, $\iota(\mathbb{N})$ and $-\iota(\mathbb{N})$ are pairwise disjoint and their union is $\mathbb{Z}$. Moreover

$$\iota(\mathbb{N})=\{x\in\mathbb{Z}:0<x\},\qquad-\iota(\mathbb{N})=\{x\in\mathbb{Z}:x<0\}.$$

2. (Closure under the ring operations.)¶ $0\in\mathbb{Z}$ and $1\in\mathbb{Z}$, and for all $x,y\in\mathbb{Z}$ the real numbers $-x$, $x+y$, $x-y$ and $xy$ belong to $\mathbb{Z}$.

3. (Discreteness.)¶ If $x\in\mathbb{Z}$ and $0<x$, then $1\le x$. Consequently there is no $x\in\mathbb{Z}$ with $0<x<1$, and if $x,y\in\mathbb{Z}$ satisfy $x<y$ then $x+1\le y$.

4. (Intervals.)¶ Let $a,b\in\mathbb{Z}$ with $a\le b$. Then there is exactly one $p\in\mathbb{N}$ with $\iota(p)=b-a+1$, and the set $\{x\in\mathbb{Z}:a\le x\le b\}$ is finite and has $p$ elements.