Linear ordering $\leq$ on $\mathbb{Z}$ in ZFC

Question

Given a set of natural numbers $\mathbb{N}$ in ZFC, we define $\mathbb{Z}$ by first defining an equivalence relation $\simeq$ on $\mathbb{N}\times\mathbb{N}$: $(n,m) \simeq (n',m') \Longleftrightarrow n + m' = n' + m$. Then we consider the set $(\mathbb{N}\times\mathbb{N})/{\simeq}$ of equivalence classes of $\mathbb{N}\times\mathbb{N}$ with respect to $\simeq$ as the set of integers $\mathbb{Z}$.

However, we then need to define a linear ordering $\leq$ on our newfound $\mathbb{Z}$. The idea is to set $[(n,m)] \leq [(n',m')]$ whenever $n + m' \leq n' + m$. I'm having a trouble in showing that this definition doesn't depend on the choie of $(n,m) \leq (n',m')$. That is, that for any natural numbers $n_1,m_1,n_2,m_2,n_3,m_3,n_4,m_4$ if we have $$n_1 + m_3 \leq n_3 + m_1,$$ $$n_1 + m_2 = n_2 + m_1,$$ $$n_3 + m_4 = n_4 + m_3,$$ then we have $$n_2 + m_4 \leq n_4 + m_2.$$ Any help would be appreciated.

Answer 1

Well we can go and do a straightforward verification using the definition of the equivalence relation.

Suppose $(n,m)\simeq(n',m')$ and $(i,j)\simeq(i',j')$, and suppose that $[(i,j)]\leq[(n,m)]$. Namely, $i+m\leq n+j$. We want to prove that $i'+m'\leq n'+j'$.

What do we know? We know that $i+j'=i'+j$ and $n+m'=n'+m$.

$$\begin{align*} i+m&\leq n+j\\ i+m+n'+j'&\leq n+j+n'+j'\\ i+j'+m+n'&\leq n+n'+j+j'\\ i'+j+m'+n&\leq n+n'+j+j'\\ i'+m'+j+n&\leq n'+j'+j+ n\\ i'+m'&\leq n'+j' \end{align*}$$ Which is what we wanted.

Answer 2

My preferred way is to show that every integer has a unique representative that is either $[(0, m)]$ or $[(m, 0)]$. Once you've done that, then the order's uniqueness follows immediately.

Answer 3

$$n_1 + m_3 \leq n_3 + m_1 \implies m_3-n_3\le m_1 - n_1 $$

$$n_1 + m_2 = n_2 + m_1\implies m_1 - n_1=m_2 - n_2 $$

$$n_3 + m_4 = n_4 + m_3 \implies m_4-n_4=m_3-n_3 $$

Thus $$ m_4-n_4=m_3-n_3\le m_1 - n_1=m_2 - n_2$$ Which implies

$$n_2 + m_4 \leq n_4 + m_2$$.