for a while I have been interested in the details of the construction of the integers from the natural numbers. credit to the software, for as soon as I began writing this, for it drew my attention to a very apposite discussion from a question on MSE about a year ago. i hope to revisit that discussion, as i am still not familiar with some of the references there, but perhaps this exercise will help me to make progress with more abstract ideas. so my question is - is the following method a reasonable way of deriving the integers from the natural numbers, explaining, by the way, a very good reason why a negative times a negative comes out positive. you can actually calculate it, with the $\star$operation introduced here.
the additive monoid $\mathbb{N}$ is also furnished with a multiplicative structure on $\mathbb{N}^+ = \mathbb{N} \setminus\{0\}$ which it pulls back (via the natural bijection) from its own multiplicative monoid of injective endomorphisms.
$\mathbb{N} \times \mathbb{N}$ is an additive monoid with the obvious addition performed elementwise. we define $$(a,b) \star (c,d) = (ac+bd, ad+bc) $$ the $\star$-multiplication is commutative, associative, distributive over addition, and has an identity element $(1,0)$.
$\mathbb{N} \times \mathbb{N}$ is the union of three sets $\mathbb{M}^+,\mathbb{M}^0$ and $\mathbb{M}^-$ which intersect only in $\{0\}$. each is a submonoid. an element $(a,b)$ is in $\mathbb{M}^+,\mathbb{M}^0$ or $\mathbb{M}^-$ as $a \gt b, a=b$ or $a<b$.
with respect to the $\star$-multiplication $\mathbb{M}^+$ and $\mathbb{M}^0$ are closed under this operation. $\mathbb{M}^0$, however has an ideal-like property -it is absorbing. if $m_0 \in \mathbb{M}^0$ then for any other $m \in \mathbb{N} \times \mathbb{N}$ we have $m_0m \in \mathbb{M}^0$. the remaining relations may be summarized by $\mathbb{M}^+\star\mathbb{M}^- \subset \mathbb{M}^-$ and $\mathbb{M}^-\star\mathbb{M}^- \subset \mathbb{M}^+$.
now take two copies of $\mathbb{N}^+$, the second of which we shall call $\mathbb{N}^-$, and define $\mathbb{N}^0=\{0\}$. we define the set $$\mathbb{Z} = \mathbb{N}^+ \cup \mathbb{N}^0 \cup \mathbb{N}^- $$ define $$\psi: \mathbb{N} \times \mathbb{N} \rightarrow (\mathbb{N},0) \cup (0,\mathbb{N}) $$in terms of three auxiliary functions, $\chi^+$ and $\chi^-$ the indicator functions for $\mathbb{M}^+$ and $\mathbb{M}^-$, and $\tau: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ defined by $\tau(a,b)=max(a,b)-min(a,b)$. then we may write $$\psi(m)=(\chi^+(m)\tau(m),\chi^-(m)\tau(m)) $$ although $\psi$ is not quite a morphism, its kernel is $\mathbb{M}^0$, and restricted to either $\mathbb{M}^+$ or $\mathbb{M}^-$ is a morphism of monoids. moreover it preserves the multiplicative structure, subject to the sign permutation already mentioned. so it gives the usual multiplicative structure on $\mathbb{Z}$.
