# Natural Numbers are Non-Negative Integers

## Theorem

Let $m \in \Z$. Then:

- $(1): \quad m \in \N \iff 0 \le m$
- $(2): \quad m \in \N_{> 0} \iff 0 < m$
- $(3): \quad m \notin \N \iff -m \in \N_{> 0}$

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

## Proof 1

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:

- $\exists z \in \N: z + y = x = m + y$

From Naturally Ordered Semigroup: NO 2, $y$ is cancellable.

Hence: $m = z \in \N_{> 0}$

Thus $(1)$ holds.

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

We infer from $(1)$ that:

- $m \notin \N \iff m < 0$

We infer from $(2)$ that:

- $-m > 0 \iff -m \in \N_{> 0}$

But by Ordering of Inverses in Ordered Monoid:

- $m < 0 \iff -m > 0$

Therefore $(3)$ also holds.

$\blacksquare$

## Proof 2

Consider the formal definition of the integers: $x = \eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers.

Let $x \in \Z_{>0}$ be a (strictly) positive integer.

Thus by definition:

- $x > 0$

This is equivalent to the condition that $a > b$.

Hence:

- $x = \eqclass {b + u, b} {}$

for some $u \in \N_{>0}$.

It is immediate that:

- $\forall c \in \N: \eqclass {b + u, b} {} = \eqclass {c + u, c} {}$

Consider the mapping $\phi: \N_{>0} \to \Z_{>0}$ defined as:

- $\forall u \in \N_{>0}: \map \phi u = u'$

where $u' \in \Z$ be the (strictly) positive integer $\eqclass {b + u, b} {}$.

From the above we have that $\phi$ is well-defined.

Let $\eqclass {b + u, b} {} = \eqclass {c + v, c} {}$.

Then:

- $b + u + c = b + c + v$

and so:

- $u = v$

Hence $\phi$ is an injection.

Next note that all $u' \in \Z_{>0}$ can be expressed in the form:

- $u' = \eqclass {b + u, b} {}$

for arbitrary $b$.

Hence $\phi$ is a surjection.

Hence $\phi$ is an isomorphism.

By defining $\map \phi 0 = \eqclass {b, b} {}$ we see that $\N$ and $\Z_{\ge 0}$ are isomorphic.

$\blacksquare$

## Warning

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.

With that caveat in mind, the theorem holds.

## Sources

- 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers