[proofplan] The proof uses the successor structure built into the usual order on $\mathbb{N}=\{1,2,3,\dots\}$. The successor $n+1$ is the next element after $n$, so no [natural number](/page/Natural%20Number) can lie strictly between them. The predecessor assertion follows from the fact that every positive natural number except $1$ occurs as the successor of a positive natural number. Once $n=m+1$ is obtained, the same immediate-successor argument applied to $m$ gives the final discreteness statement. [/proofplan] [step:Use the successor property to rule out naturals between $n$ and $n+1$] Fix $n \in \mathbb{N}$. In the usual strict order on the positive natural numbers, the successor $n+1$ is the least element of $\mathbb{N}$ that is strictly greater than $n$. Equivalently, for every $k \in \mathbb{N}$, \begin{align*} n<k \implies n+1 \le k. \end{align*} Suppose, toward a contradiction, that there exists $k \in \mathbb{N}$ such that $n<k<n+1$. From $n<k$, the successor property gives $n+1 \le k$. From $k<n+1$, we have $k<n+1$. Thus \begin{align*} n+1 \le k < n+1, \end{align*} which contradicts the irreflexivity of the strict order, since it implies $n+1<n+1$ after transitivity of $\le$ and $<$. Therefore no such $k$ exists. [guided] Fix $n \in \mathbb{N}$. The key structural fact about the usual order on $\mathbb{N}=\{1,2,3,\dots\}$ is that $n+1$ is not merely some element greater than $n$; it is the immediate successor of $n$. Formally, this means that $n+1$ is the least natural number strictly greater than $n$. Thus, whenever $k \in \mathbb{N}$ satisfies $n<k$, the leastness of $n+1$ gives \begin{align*} n+1 \le k. \end{align*} Now suppose that a natural number $k$ did lie strictly between $n$ and $n+1$. This means that $k \in \mathbb{N}$ and \begin{align*} n<k<n+1. \end{align*} The left inequality $n<k$ forces $n+1 \le k$, because $n+1$ is the least natural number greater than $n$. The right inequality says $k<n+1$. Combining them gives \begin{align*} n+1 \le k < n+1. \end{align*} This is impossible in a strict linear order: if $n+1 \le k$ and $k<n+1$, then transitivity gives $n+1<n+1$, contradicting irreflexivity. Hence no $k \in \mathbb{N}$ satisfies $n<k<n+1$. [/guided] [/step] [step:Express every natural number different from $1$ as a successor] Let $n \in \mathbb{N}$ with $n \ne 1$. Since $\mathbb{N}=\{1,2,3,\dots\}$ is the positive natural numbers ordered by successive application of the successor operation, every element is either the initial element $1$ or the successor of a unique positive natural number. Because $n \ne 1$, there exists $m \in \mathbb{N}$ such that \begin{align*} n=m+1. \end{align*} [/step] [step:Apply immediate-successor discreteness to the predecessor $m$] Let $m \in \mathbb{N}$ be such that $n=m+1$. By the first step applied with $m$ in place of $n$, there is no $k \in \mathbb{N}$ such that \begin{align*} m<k<m+1. \end{align*} Since $m+1=n$, this is exactly the assertion that there is no $k \in \mathbb{N}$ such that \begin{align*} m<k<n. \end{align*} This proves both the successor discreteness statement and the predecessor discreteness statement. [/step]