Natural Numbers are Non-Negative Integers

Theorem
Let $m \in \Z$. Then:


 * $(1) \quad m \in \N \iff 0 \le m$;
 * $(2) \quad m \in \N^* \iff 0 < m$;
 * $(3) \quad m \notin \N \iff -m \in \N^*$.

That is, the natural numbers are precisely those integers which are greater than or equal to zero.

Proof

 * Let $m \in \N$.

Then by definition of $0$, $0 \le m$.

Conversely, let $m \in \Z: 0 \le m$.

Then $\exists x, y \in \N: m = x - y$.

Thus $y \le m + y = x$.

By Naturally Ordered Semigroup: NO 3, $\exists z \in \N: z + y = x = m + y$.

Hence, $m = z \in \N^*$ as $y$ is cancellable from Naturally Ordered Semigroup: NO 2.

Thus $(1)$ holds.


 * $(2)$ follows from $(1)$.


 * We infer from $(1)$ that $m \notin \N \iff m < 0$.

We infer from $(2)$ that $-m > 0 \iff -m \in \N^*$.

But by Ordering of Inverses, $m < 0 \iff -m > 0$.

Therefore $(3)$ also holds.

Comment
From a strictly purist point of view it is inaccurate to say that the natural numbers are the non-negative integers, as an integer is technically an element of an equivalence class composed of pairs of elements of $\N$, constructed as detailed in Construction of Inverse Completion.

However, because an Inverse Completion is Unique, it follows that the natural numbers can be considered to be a substructure of the integers from the Inverse Completion Theorem.

Therefore the theorem holds.