Extending the Ordinals analogously to the Integers

Question

I have been (recreationally) trying to expand the notion of ordinal numbers in the same way that the natural numbers $\mathbb N$ are extended to the integers $\mathbb Z$.

My objective is to be able to define "Integer Ordinals" from Ordinal Numbers in the same way as the Integers are defined from the Natural Numbers (so that we can have both positive and negative transfinite numbers). In principle this seems pretty straightforward, but the non-commutativity of ordinal addition brings in a few complications.

The Integers, as constructed from the Natural Numbers, can be defined as equivalence classes of ordered pairs of Natural Numbers with an equivalence relation $\sim$. They also have an addition operator $+$. The definitions of both are given as:

$$ [(a,b)] \sim [(c,d)] \Leftrightarrow a+d = b+c, \; \forall a,b,c,d \in \mathbb N $$ $$ [(a,b)] + [(c,d)] = [(a+c,b+d)], \; \forall a,b,c,d \in \mathbb N$$

If we want to extend this notion from the natural numbers to the ordinals, we need to take into account the non-commutativity of ordinal addition. This means we actually end up with 4 possible definitions for $\sim$:

1. $ [(a,b)] \sim [(c,d)] \Leftrightarrow a+d = b+c, \; \forall a,b,c,d \in \mathbb N$
2. $ [(a,b)] \sim [(c,d)] \Leftrightarrow a+d = c+b, \; \forall a,b,c,d \in \mathbb N$
3. $ [(a,b)] \sim [(c,d)] \Leftrightarrow d+a = b+c, \; \forall a,b,c,d \in \mathbb N$
4. $ [(a,b)] \sim [(c,d)] \Leftrightarrow d+a = c+b, \; \forall a,b,c,d \in \mathbb N$

And 4 definitions for $+$:

1. $ [(a,b)] + [(c,d)] = [(a+c,b+d)], \; \forall a,b,c,d \in \mathbb N$
2. $ [(a,b)] + [(c,d)] = [(a+c,d+b)], \; \forall a,b,c,d \in \mathbb N$
3. $ [(a,b)] + [(c,d)] = [(c+a,b+d)], \; \forall a,b,c,d \in \mathbb N$
4. $ [(a,b)] + [(c,d)] = [(c+a,d+b)], \; \forall a,b,c,d \in \mathbb N$

Since for any ordinal $\alpha = [(a,b)]$, $\alpha + 0 = 0 + \alpha = \alpha$ and $0 = [(n,n)]$ for any $n$ it must be true that:

$$ \alpha + 0 \sim \alpha \Rightarrow [(a,b)] + [(n,n)] \sim [(a,b)]$$

And from this, it can be shown that out of the 4 definitions of $\sim$ and $+$ only two pairs are compatible, namely:

1. $ \begin{cases} [(a,b)] \sim [(c,d)] \Leftrightarrow a+d = c+b, \; \forall a,b,c,d \in \mathbb N\\ [(a,b)] + [(c,d)] = [(a+c,d+b)], \; \forall a,b,c,d \in \mathbb N \end{cases} $
2. $ \begin{cases} [(a,b)] \sim [(c,d)] \Leftrightarrow d+a = b+c, \; \forall a,b,c,d \in \mathbb N\\ [(a,b)] + [(c,d)] = [(c+a,b+d)], \; \forall a,b,c,d \in \mathbb N \end{cases} $

So far so good. The problem is that under any of these two sets of definitions, we get that for any ordinal $\alpha$ the following must be true:

$$ \begin{cases} [(0,0)] \sim [(1,1)]\\ [(\alpha,\alpha)] \sim [(0,0)]\\ [(\alpha,\alpha)] \nsim [(1,1)] \end{cases} $$

Which is clearly a contradiction. The only escape from this I can envision is to redefine $\sim$ as: $$ [(a,b)] \sim [(c,d)] \Leftrightarrow (a+d = b+c) \vee (a+d = c+b) , \; \forall a,b,c,d \in \mathbb N $$

And (finally) my question is:

Is this definition good, or will this bring out any contradictions?

I have looked and looked but can't find any, but I cannot prove this.

Thank you very much, and apologies for the long post.

Answer 1

To answer your implicit question of "what definition would be good?", the key is to replace the usual ordinal addition with natural addition. That operation is commutative and cancellative and then the construction for the integers works just fine. These "ordinal integers" are essentially part of the Surreal Numbers.