Sorry for the wall-of-text in advance, hope some of you make it till the end ;)
When constructing the integers from the natural numbers, I've been using the following definitions. Let $\sim$ be an equicalence relation, which for all $a,b,c,d\in\mathbb{N}$ is defined as \begin{align*} (a,b)\sim (c,d):\Leftrightarrow a+d=b+c. \end{align*} The set of integers is then defined as the set \begin{align*} \mathbb{Z}_{1}:=\{[(a,b)]_{\sim}\mid (a,b)\in\mathbb{N}\times\mathbb{N}\}. \end{align*} Note that every equivalence class $[(a,b)]_{\sim}$ can be written either as $[(n,0)]_{\sim}$ or as $[(0,n)]_{\sim}$, so we define for all $\hat{n}\in\mathbb{Z}_{1}$ \begin{align*} \hat{n}&:=[(n,0)]_{\sim} \\ -\hat{n}&:=[(0,n)]_{\sim}. \end{align*}
Now, addition $\oplus$, multiplication $\odot$, and total order $\preceq$ on $\mathbb{Z}_{1}$ are defined as \begin{align*} [(a,b)]_{\sim}\oplus[(c,d)]_{\sim}&:=[(a+c,b+d)]_{\sim} \\ [(a,b)]_{\sim}\odot[(c,d)]_{\sim}&:=[(ac+bd,ad+bc)]_{\sim} \\ [(a,b)]_{\sim}\preceq [(c,d)]_{\sim}&:\Leftrightarrow a+d\leq b+c. \end{align*}
When starting with the real numbers, we can define the natural numbers as intersection of all inductive sets. With the definition \begin{align*} -\mathbb{N}:=\{x\in\mathbb{R}\mid \exists n\in\mathbb{N}:x=-n\} \end{align*} we then can define the integers as \begin{align*} \mathbb{Z}_{2}:=-\mathbb{N}\cup \mathbb{N}. \end{align*}
Since we started with the real numbers, addition $+$, multiplication $\cdot$ and total order $\leq$ on the set $\mathbb{Z}_{2}$ are given by the axioms of the real numbers.
I would like to prove that the sets $\mathbb{Z}_{1}$ and $\mathbb{Z}_{2}$ are (at least up to isomorphism) identical. My idea was to use a map $f:(-\mathbb{N}\cup\mathbb{N})\rightarrow (\mathbb{N}\times\mathbb{N})$, defined as \begin{align*} f(n)=\left\{\begin{array}{ll} [(n,0)]_{\sim}&\text{for $n\in\mathbb{N}$} \\ [(0,n)]_{\sim}&\text{for $n\in -\mathbb{N}$}. \end{array}\right. \end{align*} So I have to show the following:
- $f$ is bijective.
- For all $x,y\in\mathbb{Z}_{2}$ we have $f(x+y)=f(x)\oplus f(y)$.
- For all $x,y\in\mathbb{Z}_{2}$ we have $f(x\cdot y)=f(x)\odot f(y)$.
- For all $x,y\in\mathbb{Z}_{2}$ we have $x\leq y\Leftrightarrow f(x)\preceq f(y)$.
I've proven the first property but I'm struggling with the second one. My approach to prove the second property was to distinguish between the following four cases:
Let $(x\in\mathbb{N})$ and $(y\in\mathbb{N})$. This case works out just fine; We have $f(x)=[(x,0)]_{\sim}$ and $f(y)=[(0,y)]_{\sim}$. Because $x+y\in\mathbb{N}$ we also have $f(x+y)=[(x+y,0)]_{\sim}$. Using the definition of $\oplus$ we also get $f(x)\oplus f(y)=[(x+y,0)]_{\sim}$. Therefore we have \begin{align*} f(x+y)=f(x)\oplus f(y). \end{align*}
Let $x\in\mathbb{N}$ and $y\in -\mathbb{N}$. Then we have $f(x)=[(x,0)]_{\sim}$ and $f(y)=[(0,y)]_{\sim}$ and, as a consequence \begin{align*} f(x)\oplus f(y)=[(x,y)]_{\sim}. \end{align*} Next it could be that $x+y\leq 0$ or $0\leq x+y$, so we have to consider two or three subcases. And this is where my trouble starts, because now we have the cases 1) $x+y \leq 0$. Then $f(x+y)=[(0,x+y)]_{\sim}$, and 2) $0\leq x+y$. Then $f(x+y)=[(x+y,0)]_{\sim}$. Note, that in the case $x+y=0$, we have $f(x+y)=[(x+y,x+y)]_{\sim}$.
Somehow I have to find a way to show that for $0\leq x+y$ we have \begin{align*} [(x+y,0)]_{\sim}=[(x,y)]_{\sim} \end{align*} and for $x+y\leq 0$ we have \begin{align*} [(0,x+y)]_{\sim}=[(x,y)]_{\sim} \end{align*} Does anyone have an idea or is my approach totally wrong? Proving the next two cases shouldn't be a big issue after this.Let $x\in-\mathbb{N}$ and $y\in\mathbb{N}$.
Let $x,y\in-\mathbb{N}$.
The proof of $f(x\cdot y)=f(x)\odot f(y)$ and $x\leq y\Leftrightarrow f(x)\preceq f(y)$ will probably face the same issue.
It would be great if anyone could help me with this proof. Or is there a better way to proof the sets $\mathbb{Z}_{1}$ and $\mathbb{Z}_{2}$ are the same?