Let $\mathbb{Z}$ be equipped with its usual ordered ring structure, and let $\mathbb{N} := \{n \in \mathbb{Z} : n > 0\} = \{1,2,3,\dots\}$ denote the positive integers inside $\mathbb{Z}$. Then $1$ is the least positive integer: if $n \in \mathbb{Z}$ and $n > 0$, then $1 \le n$.
paragraph
admin
Moreover, if $S \subset \mathbb{Z}$ is nonempty and bounded below in $\mathbb{Z}$, meaning there exists $b \in \mathbb{Z}$ such that $b \le s$ for every $s \in S$, then there exists $\ell \in S$ such that $\ell \le s$ for every $s \in S$.