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