I stumbled upon the strange representation of integers where $$8=\langle\langle0,\langle0^{\infty}\rangle,0^{\infty}\rangle,0^{\infty}\rangle$$

I'll try explain the representation in a natural way. What I'm wondering -- as is common -- does this idea have a name? Is it well studied? Is it useful? It's certainly fun to play with.


Naturally all Natural numbers can be represented as their prime decomposition. $$1=2^03^05^0... \quad\quad 2=2^13^05^0... \quad\quad 3=2^03^15^0...$$ $$...$$ $$12=2^23^15^07^0... \quad\quad 13=2^03^05^07^011^013^117^0... \quad\quad 14=2^13^05^07^111^0...$$ $$\text{and so forth...}$$

One could easily use this to represent the integers as endless vectors. The $n$th element corresponds to the power of the $n$th prime in a given prime decomposition. $$1=\langle0^\infty\rangle \quad\quad 2=\langle1,0^\infty\rangle \quad\quad3=\langle0,1,0^\infty\rangle$$ $$...$$ $$12=\langle2,1,0^\infty\rangle \quad\quad 13=\langle0,0,0,0,0,1,0^\infty\rangle \quad\quad 14=\langle1,0,0,1,0^\infty\rangle$$ $$\text{and so forth...}$$

To take it a step further, one could substitute the vector representation of a number wherever that number itself appears in another vector. That is, $$2 \rightarrow \langle1,0^\infty\rangle \rightarrow \langle\langle0^\infty\rangle,0^\infty\rangle$$ $$\text{or}$$ $$8 \rightarrow \langle3, 0^\infty\rangle \rightarrow \langle\langle0,1,0^\infty\rangle, 0^\infty\rangle \rightarrow \langle\langle0,\langle0^\infty\rangle,0^\infty\rangle,0^\infty\rangle$$ This corresponds to the fact that $$8=2^33^05^0...=2^{(2^03^15^0...)}3^05^0...=2^{(2^03^{(2^03^05^0...)}5^0...)}3^05^0...$$

Now all the Natural numbers can be represented with brackets and zeros $$0 = 0 \quad\quad\quad\quad\quad\quad\quad\quad\ \ \ 5 = \langle0,0,\langle0^\infty\rangle,0^\infty\rangle \quad\quad\quad\quad\quad\quad\quad\ $$ $$1 = \langle0^\infty\rangle \quad\quad\quad\quad\quad\quad\quad\ 6 = \langle\langle0^\infty\rangle,\langle0^\infty\rangle,0^\infty\rangle\quad\quad\quad\quad\quad\quad\ \ \ $$ $$2 = \langle\langle0^\infty\rangle,0^\infty\rangle \quad\quad\quad\quad\quad 7 = \langle0,0,0,\langle0^\infty\rangle,0^\infty\rangle\quad\quad\quad\quad\quad\quad\ \ \ $$ $$3 = \langle0,\langle0^\infty\rangle,0^\infty\rangle \quad\quad\quad\quad\ 8 = \langle\langle0,\langle0^\infty\rangle,0^\infty\rangle,0^\infty\rangle\quad\quad\quad\quad\quad\quad$$ $$4 = \langle\langle\langle0^\infty\rangle,0^\infty\rangle,0^\infty\rangle \quad\quad\quad 9 = \langle0,\langle\langle0^\infty\rangle,0^\infty\rangle,0^\infty\rangle\quad\quad\quad\quad\quad\quad$$ I've been calling these things "corporeal" numbers since they remind me of Conway's Surreal numbers. I'll add down below any extra information about these guys that turns up.


$\textbf{Edit}$: These have been rewritten a few times due to new knowledge and the good suggestions of others -- such as https://math.stackexchange.com/q/2356029

The set of corporeal numbers can be defined with set-builder notation $$\mathbb{C}_{\text{orporeal}}=\{ \langle v_1, v_2, v_3, ...\rangle : v_i \in \mathbb{C}_{\text{orporeal}} \}$$

We will also make up some notation for jumping back and forth between the Real numbers and corporeal numbers. $$ [\ ]:\mathbb{R} \rightarrow \mathbb{C}_\text{orporeal} \quad \text{Brackets go from Real to corporeal} $$ $$ \text{Re}: \mathbb{C}_\text{orporeal}\supseteq X \rightarrow \mathbb{R} \quad \text{Re() goes from some corporeals to Real} $$ So for example $$ \text{Re}(\langle\langle0,\langle0^\infty\rangle,0^\infty\rangle,0^\infty\rangle) = 8 \quad\text{and}\quad [8] = \langle\langle0,\langle0^\infty\rangle,0^\infty\rangle,0^\infty\rangle $$ Note that Re() is only defined for $\textit{some}$ of the corporeal numbers because some don't correspond to any Real number. We'll get to those corporeal numbers further down.

$$\bullet\textbf{ Axioms }\bullet$$

Addition exists $$ 1)\quad \text{If } \alpha \text{ and } \beta \text{ are corporeal numbers, then so is } \alpha + \beta$$

Multiplication is element-wise addition $$2)\quad \text{If } \alpha = \langle ...a_i...\rangle \text{ and } \beta = \langle ...b_i...\rangle \text{ then }$$ $$\alpha \cdot \beta = \langle ... (a_i + b_i) ... \rangle$$

There is a corporeal number for each real number. $$3)\quad \text{If } x \in \mathbb{R} \text{ then } [x] \text{ exists and } [x] \in \mathbb{C}_\text{orporeal}$$

Corporeal numbers model prime decomposition $$4)\quad \text{If } x = 2^{a_1}3^{a_2}5^{a_3}7^{a_4}11^{a_5}... \text{ then}$$ $$[x]=\langle [a_1],[a_2],[a_3],[a_4],[a_5],...\rangle$$

Corporeal numbers maintain the structure of addition and multiplication in the Real numbers. $$5)\quad \text{For } a,b \in \mathbb{R} \quad : \quad [a] + [b] = [a+b] \text{ and } [a]\cdot[b] = [ab]$$

$\textbf{Non-Real numbers}$

We can show there are a few corporeal numbers that don't correspond to any Real number.

First is $\omega$ which we'll call a hard-zero because $\omega + [x] = \omega$. Here's why it exists $$\text{The corporeal number }[0] \text{ exists and contains other corporeal numbers}$$ $$\text{so suppose } [0] = \langle\omega,\omega,\omega,...\rangle$$ $$\text{now we make use of how multiplication was defined}$$ $$ [0]\cdot[n] = [0]$$ $$\Downarrow$$ $$\langle...,\omega,...\rangle\cdot\langle ...,[a_i],...\rangle=\langle...(\omega + [a_i]) ... \rangle = \langle ...,\omega,...\rangle$$ $$\Downarrow$$ $$\text{For any } a_i \in \mathbb{R} \text{ if follows that } \omega + [a_i] = \omega$$

The next number is $\zeta$ which corresponds to an imaginary number. It has the property that $$\text{for some Real number -- say } n \quad : \quad n^\zeta = [-1]$$ $$\text{To see why, first note that } [-1] \text{ exists and must contain at least one non-Real element}$$ $$\text{ accordingly we say that } [-1] = \langle 0,...,0,\zeta,0,...\rangle$$ $$\text{ and we make use of how multiplication was defined again}$$ $$[-1]\cdot[-1]=[1]$$ $$\Downarrow$$ $$\langle [0],...,[0],\zeta,[0],...\rangle \cdot \langle [0],...,[0],\zeta,[0],...\rangle$$ $$||$$ $$\langle [0],...,[0],(\zeta + \zeta),[0],...\rangle$$ $$||$$ $$\langle [0],[0],[0],...\rangle$$ $$\Downarrow$$ $$\zeta + \zeta =[0]$$ The axioms of corporeal numbers aren't strong enough to conclude from here that $\zeta = 0$. We did that on purpose. If $\zeta$ was indeed $0$, then all corporeal numbers would be equal as a consequence. And a number system where all numbers are equal is useless.

The explanation here is that $\zeta$ can correspond to multiple values in the complex numbers. The conclusion -- $\zeta + \zeta =[0]$ -- can be satisfied by picking any complex value and it's conjugate for the left and right $\zeta$ respectively.

$\textbf{So what is Pi?}$

There's a sense in which the product of all prime numbers is $4\pi^2$ https://link.springer.com/article/10.1007%2Fs00220-007-0350-z

Accordingly we will say that $$[4\pi^2]=\langle[1]^\infty\rangle \quad\Leftrightarrow\quad [\pi]=\Big\langle[-1]\cdot\Big[\frac{1}{2}\Big],\Big[\frac{1}{2}\Big]^\infty\Big\rangle = \langle[-1]\cdot\langle[-1],[0]^\infty\rangle,\langle[-1],[0]^\infty\rangle^\infty\rangle$$

But this isn't the only sense in which pi can be defined. Euler found that

$$\frac π 4 = \frac 3 4 \cdot \frac 5 4 \cdot \frac 7 8 \cdot \frac {11} {12} \cdot \frac {13} {12} \cdots$$

where the numerators are the prime numbers starting at $3$ and the denominators are the multiples of $4$ closest to the corresponding prime. (See: https://mathoverflow.net/q/137346 and http://mathworld.wolfram.com/PrimeProducts.html [forumla 33])


$$[\pi]=\langle \sum_{}^{\infty} [-2],[1] - \alpha_1,[1]-\alpha_2,[1]-\alpha_3,...\rangle$$

Where each $\alpha_i$ may be -- as far as I know -- finite or infinite.

  • $\begingroup$ I vaguely remember reading about a similar concept: formal prime decompositions, with exponents in the nonnegative integers, and with infinitely many exponents permitted to be positive. Addition was not defined on these. They may have been called "pseudo-integers". $\endgroup$ – Tanner Swett Jul 7 '17 at 23:31
  • $\begingroup$ Sounds like this paper link.springer.com/article/10.1007/BF02760931 ? $\endgroup$ – Christian Woll Jul 7 '17 at 23:52
  • $\begingroup$ Very interesting! $\endgroup$ – diff_math Jul 8 '17 at 22:58
  • 1
    $\begingroup$ My first thought was: "are corporeal numbers well-founded", i.e. when given such number $c_0$ and we have $c_1\in c_0$ and $c_2\in c_1$ and so on, do we have to reach a number without elements after finite such steps? Assuming you follow ZFC set theory, this has to be true. This means you can build $\Bbb C_{\text{oporeal}}$ recursively starting from a no further decomposable number of your choice, maybe $0$, or in your case, $\omega$. Or do you assume that $\omega$ has further elements but we just do not know them yet? $\endgroup$ – M. Winter Jul 12 '17 at 9:05
  • $\begingroup$ I think before worrying about how to represent $\pi$, I would think about how to represent $-1$ or $1/2$ as corporeal numbers. Well you can just define "there are such numbers $c_{-1}$ and $c_{1/2}$ with $[-1]=c_{-1}$ and $[1/2]=c_{1/2}$" without talking about there elements, but I would guess without these definitions your current axiom system is not strong enough to prove their existence. But I may be wrong here. Still an interesting construction. $\endgroup$ – M. Winter Jul 12 '17 at 9:20

This is an answer to the question as it was stated initially. There it seemed as if $\def\corpo{\Bbb C_{\text{orpo}}}\corpo$ should extend the natural numbers $\Bbb N$ in a certain way. Among other things, I assume the addition and multiplication of $\corpo$ to be distributive (the former axiom 4).

I thought a lot about this interesting construct and I found a purely algebraic formulation you are maybe interested in. Note that you cannot use the setbuilder notation in the way you did: you cannot use the defined set inside the braces on the right side. You might run into foundational problems with a too naive approach. So, if we cannot directly construct the corporeals this way, we at least can state (in an algebraic fashion) what we want them to be.

In the natural numbers we have the following kind of prime factorization: if $n\in\Bbb N\setminus\{0\}$ and $p_i$ is the $i$-th prime number, then there are unique $k_i\in\Bbb N$ (only finitely many are non-zero) so that

$$n=\prod_i p_i^{k_i}.$$

It now is natural to extend this in several ways:

  • We want to allow infinitely many non-zero $k_i$, i.e. numbers with infinitely many different prime factors.
  • We want to have exponents not only from $\Bbb N$ but from a more general appropriate space $R$. It would be exceptionally nice if the exponents come from the same space as the one we are currently defining.
  • We do not want to have $0$ as an exception but we want it to have a unique factorization too.

Now let me show you how to express the fact that $\Bbb N$ allows a unique prime factorization in an algebraic way. First note that $(\Bbb N,+,\cdot)$ is a semiring, i.e. has associative, commutative addition with $0$; associative, distributive multiplication with $1$; $0$ is multiplicatively absorbing. The existence of a prime factorization now is equivalent to the existence of a specific isomorphism $\varphi$ between the additive and multiplicative strucuture of $\Bbb N$. More precisely, we want a monoid-isomorphism

$$\varphi :(\Bbb N\setminus\{0\},\cdot) \leftrightarrow (\Bbb N^{\Bbb N}_{\text{fin}},+).$$

Here, $(\Bbb N\setminus\{0\},\cdot)$ is the multiplicative monoid of $\Bbb N\setminus\{0\}$ (just the associative multiplication with $1$). $\Bbb N^{\Bbb N}_{\text{fin}}$ is the set of all sequences of natural numbers with only finitely many non-zero entries. Together with componentwise addition, this will give an additive monoid. This might seem strange at first, but is just your $\langle\cdot\rangle$ notation in disguise. E.g. if $n\in\Bbb N$ with prime factors $p_i^{k_i}$, then

$$n=\langle k_0,k_1,...\rangle=\prod_ip_i^{k_i} \qquad\Longleftrightarrow\qquad \varphi(n)=(k_0,k_1,...).$$

In still other words $\varphi(\langle k_0,k_1,...\rangle)=(k_0,k_1,...)$.

So why do I write this so complicated as an isomorphism? At first, isomorphisms are structure preserving, i.e. $\varphi(n\cdot m)=\varphi(n)+\varphi(m)$. This resembles your former componentwise addition axiom 3. On the other hand isomorphisms are bijective, hence this reflects the fact that a prime factorization is unique, and that any possible prime factorization gives us one unique natural number. All this is packed in only a few words and now one can apply the whole power of abstract algebra to deduce and generalize.

Now lets see where your corporeals can be placed in here. We are looking for a $\Bbb N$-extending semiring $(R,+,\cdot)$ with an even stronger connection between the additive and multiplicative structure, namely we claim the existence of a monoid-isomorphism

$$\varphi:(R,\cdot)\leftrightarrow(R^{\Bbb N},+).$$

Note the following:

  • We have used $R$ instead of $R\setminus\{0\}$ on the left, hence we expect that also the zero $0$ has a unqiue prime factorization.
  • We have removed the $\text{fin}$ from the right side, so we allow a factorization of a number $n\in R$ to contain infinitely many different primes.
  • We have used $R$ on the right instead of only $\Bbb N$, hence our exponents can be all numbers from this newly defined space. This automatically equips our space with an operation of exponentiation.

The question now is, whether such a semiring exists. If one exists we can call it $\corpo$ and take $\Bbb N$ as a specific sub-semiring. Actually you claimed one more thing for $\corpo$ to hold. Above definition says that any number $n\in R$ has a unique prime factorization with exponents in $R$. But we never stated what those primes are. If we specifically want the primes to be the prime numbers from $\Bbb N$, then we have to include the following further axiom: For $p_i\in\Bbb N$ being the $i$-th prime number, we claim


This makes $\varphi(p_i)$ a generating system of $(\corpo^{\Bbb N},+)$. If we do not claim this, it could happen that the former prime numbers are no longer prime in the extended space, but now are composites of new strange corporeal primes. We then cannot be sure that e.g. $\varphi(2)=(1,0,0,...)$. So we claimed it for convenience.

Here are some first observations. Lets begin with the cardinality of $\corpo$. As isomorphisms are bijective, this means that $\corpo$ and $\corpo^{\Bbb N}$ are of the same cardinality. But we have

$$|\Bbb N|^{|\Bbb N|}=|\Bbb R|\qquad \text{but}\qquad |\Bbb R|^{|\Bbb N|}=|\Bbb R|.$$

We therefore see that $\corpo$ must be uncountable. And this is not because it contains representations of all real numbers (you can see that it does not from my other answer), but only generalized integers. This means that it does contain many non-natural numbers.

I have never mentioned your "hard-zero" $\omega$ until now. This is because its existence follows naturally from the fact that we need a prime factorization of $0$. As $0$ is multiplicatively absorbing, the existence of $\varphi$ imples the existence of an additively absorbing element $\omega$ and


  • $\begingroup$ These are wonderful thoughts on the "corporeals". I expect to need a few days to digest the rigor you've offered here. $\endgroup$ – Christian Woll Jul 17 '17 at 17:18

Your structure has big problems when you try to implement additive inverses $[-n]$ for $n\in\Bbb N$ with some reasonable porperties. E.g. using the natural assumption $[-1]\cdot[-1]=[1]$ will lead to a contradiction. Assume a representation

$$[-1]=\langle a_0,a_1,...\rangle.$$

The assumption then tells us that $a_i+a_i=0$. Because of distributivity, this is equivalent to $[2]a_i=0$. Assuming the further representation $a_i=\langle b_0^i,b_1^i,...\rangle$, we can deduce that

$$0=[2]a_i=\langle b_0^i+1,b_1^i,...\rangle\quad\Rightarrow\quad b_0^i+1=\omega,\quad b_j^i=\omega,j>1.$$

This means $a_i=\langle b_0^i,\omega^\infty\rangle$ for some strange corporeal numbers $b_0^i$ with $b_0^i+1=\omega$. But now you see


This means that actually $a_i=\langle\omega^\infty\rangle=0$ and therefore $[-1]=[1]$. This is certainly unwanted. Even worse, as $[-1]$ by definition satisfies $[1]+[-1]=[0]$, we just proved $[2]=[0]$. This is a contradiction.

So what now? You do not seem to have additive inverses with some natural behavior. To implement mutliplicative inverses you need additive inverses. And to implement $[\pi]$ and other real numbers you certainly need one of the above as you already demonstrated.

Actually, the existence of this strange absorbing "hard zero" $\omega$ boggles me. I got the feeling the structure might be much more well behaved if we allowe the corporeal numbers to also be finite sequences and defining $[0]:=\langle\rangle$.


Initial Setup

For the sake of brevity, I'll use $\mathbb{K}$ for the corporeal numbers. You want something like $\mathbb{K}=\left\{ \left\langle v_{1},v_{2},\ldots\right\rangle :v_{i}\in\mathbb{K}\right\}$ . Now, there's a minor set theory quibble about the recursion there, but there are standard workarounds to make that sensible. Except you need a place to start from. Initially, it seems that $\left[0\right]$ is some sort of atomic corporeal, so we can throw that in: $\mathbb{K}=\left\{ \left[0\right]\right\} \cup\left\{ \left\langle v_{1},v_{2},\ldots\right\rangle :v_{i}\in\mathbb{K}\right\}$ . Then you can build $\left[1\right]$ from $\left[0\right]$, and then all the primes, and then all the numbers whose exponents of primes are prime, and then... To be general and unambiguous, I'll write $\left[0\right]^{\omega}$ where the OP would have $0^{\infty}$. As M. Winter mentioned, it seems odd to assume that $0$ is a sequence and there's something even more fundamental like what the OP called $\omega$.

Now, since the (standard) corporeal representations of the positive integers are built out of their prime decompositions, which is related to their multiplicative structure, the natural definition of multiplication on $\mathbb{K}$ presents itself as the OP described: $0\cdot\beta=\alpha\cdot0=0$ and $\left\langle a_{1},a_{2},\ldots\right\rangle \cdot\left\langle b_{1},b_{2},\ldots\right\rangle =\left\langle a_{1}+b_{1},a_{2}+b_{2},\ldots\right\rangle$ .

Thoughts on sums

The problem is to build a definition of addition that agrees with the standard one. Even for sums like $\left[1\right]+\left[1\right]+\left[1\right]+\left[1\right]+\left[1\right]+\left[1\right]$, we either need to use our knowledge of integer arithmetic to find the prime factorization of the numbers $1$ through $6$ to calculate it (which wouldn't generalize to new corporeals), or we need a method of manipulating things like $\left\langle \left[0\right],\left[0\right],\left\langle \left[0\right]^{\omega}\right\rangle ,\left[0\right]^{\omega}\right\rangle$ that amounts to baking in an algorithm for prime factorization. But since prime factorization is a pretty complicated algorithm even with normal representations of numbers, I'm pessimistic that this can be done in a way that's easy to write and generalize. This is especially difficult since the notation seems to know what the primes are, but a prime factorization algorithm probably wouldn't be able to make use of that directly. Now, by factoring out common factors and asserting that multiplication distributes over addition we can reduce the problem to just “how do we add two coprime numbers?” (or at least “how do we add two numbers that don't have any equal prime powers except $p^{0}$?”?), but finding the prime factorization of the sum of two coprime numbers (even with known prime factorizations) doesn't have a tidy answer either.

Putting that thorny issue aside, we can simply assume as the OP does that $\left[n\right]+\left[m\right]=\left[n+m\right]$ for naturals $n$ and $m$ by fiat, and look at what follows from that, even if we don't know how to add any other corporeals. We could multiply any two corporeals corresponding to sequences of naturals (such as $\left\langle \left[1\right]^{\omega}\right\rangle$ or $\left\langle \left[0\right],\left[1\right],\left[2\right],\ldots\right\rangle$ ), but without a general definition of sum (or order!), we wouldn't even know if one of them were equal to $-\left[1\right]$ or another were between $\left[39\right]$ and $\left[40\right]$ (so in the ballpark of $4\pi^{2}$).


Now, if we had $\nu=\left\langle v_{1},v_{2},\ldots\right\rangle =-\left[1\right]$, then we run into problems as M. Winter argued. We'd expect $\nu\cdot\nu=\left[1\right]=\left\langle \left[0\right]^{\omega}\right\rangle$ . So it must be $v_{1}+v_{1}=\left[0\right]$ with $v_{1}\ne0$. So $\left[2\right]v_{1}=\left[0\right]$ and if $v_{1}=\left\langle w_{1},\ldots\right\rangle$ (to avoid $v_{1}=\left[0\right]$), we have $\left\langle w_{1}+\left[1\right],w_{2},\ldots\right\rangle =\left[0\right]$. If corporeals have unique representations, then this is a contradiction immediately (if you keep the OP's $\omega$ then you have to take things a little further as in M. Winter's answer). This shouldn't be too surprising in that you can't really represent a negative number with a product of positive primes.

If we were just building the integers with this sort of prime decomposition representation, we'd need to manually introduce a negative sign. (And as M. Winter said, we'd probably prefer to allow finite sequences instead.) So let's try that: $\mathbb{K}_{\pm}=\left\{ \left[0\right]\right\} \cup\left\{ \left\langle v_{1},v_{2},\ldots\right\rangle \mid v_{i}\in\mathbb{K}_{\pm}\right\} \cup\left\{ -\left\langle v_{1},v_{2},\ldots\right\rangle \mid v_{i}\in\mathbb{K}_{\pm}\right\}$ where the minus sign doesn't distribute; it's just a formal symbol. Then we can write $-\left\langle \left[0\right]^{\omega}\right\rangle$ and declare things like $-\alpha+\alpha=\left[0\right]$ (so that $\left[-1\right]=-\left[1\right]$, etc.). This allows us to do cool things like $\left[\dfrac{1}{2}\right]=\left\langle -\left[1\right],\left[0\right]^{\omega}\right\rangle$ where we know it's reasonable to call it $\dfrac{1}{2}$ since the product with $\left[2\right]$ would be $\left[1\right]$ by the multiplcation definition. This artificially gives us all of the rationals, plus things like $\dfrac{1}{\boldsymbol{1}}=\left\langle \left[-1\right]^{\omega}\right\rangle$ which might or might not have reason to be called $\left[\dfrac{1}{4\pi^{2}}\right]$. My hope would be that we could get all reals with this sort of construction and facts about interpretations about infinite products like $\boldsymbol{1}$, but I really don't know for sure without a reasonable addition (which could give us an order if we could say something like $\alpha<\beta$ if $\beta+\left(-\alpha\right)$ can be written as $\left\langle ...\right\rangle$ without a minus sign in front).

Something Similar?

You mentioned that this reminded you of the surreals, and it makes me think of two vaguely related ideas.

Sign-expansions of surreals

The surreals themselves can be represented as “sequences” with signs via Gonshor's sign expansion, but they could be finite or longer than \omega. We have things like $\left\langle \right\rangle =0$, $\left\langle +\right\rangle =1$, $\left\langle -\right\rangle =-1$, $\left\langle ++\right\rangle =2$, $\left\langle +-\right\rangle =\dfrac{1}{2}$, $\left\langle ++-+-+-\cdots\right\rangle =1+\dfrac{2}{3}$, etc.

The nimber Field

The thing about the surreals is that they have these signs, or a left side and a right side, etc. But what if we did away with that and just had “a nimber is a set of previously created nimbers” (and made some sets equivalent to others so that the sets we cared about are basically the ordinals)? Something like $0=\left\{ \right\}$ , $1=\left\{ 0\right\}$ ,... Then we can add, multiply, subtract and divide, find roots of polynomials, etc. in a fairly natural way. Unfortunately, it's not an ordered field. $1+1=0$, not “$2$”=$\left\{ 0,1\right\}$ like you might hope. But you can still talk about infinite things like “$\omega$”=$\left\{ 0,1,\left\{ 0,1\right\} ,\left\{ 0,1,\left\{ 0,1\right\} \right\} ,\ldots\right\}$ .

  • $\begingroup$ Very fun. These thoughts are all greatly appreciated. I didn't realize how deep the waters of number system construction would be. $\endgroup$ – Christian Woll Jul 17 '17 at 17:21

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