Natural Numbers are Non-Negative Integers

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


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

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.