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For questions about the surreal numbers, which are a real-closed ordered proper-class-sized field that contains both the real numbers and the ordinal numbers. Thus they contain both infinite numbers (including the ordinals, but also infinite numbers like ω-1 and sqrt(ω)) and infinitesimal numbers (like 1/ω). They can also be identified with a subclass of two-player partisan games.
15
votes
Accepted
In surreal numbers, what is $\ln \omega$?
In general this is taken to mean the value of Gonshor's logarithm at $\omega$. This was defined in the tenth chapter of his 1986 book An introduction to the theory of surreal numbers where you can fin …
1
vote
Growth of the hyperexponential
To partially answer the question, exponential levels are convex classes, and they are not intervals. This is because for instance an exponential level is closed under adding or substracting $1$, which …
15
votes
Are there any interesting surreal constants?
$\DeclareMathOperator{\Noo}{\mathbf{No}}$When seen as a big ordered field, $\Noo$ hasn't much to offer in terms of constants besides real numbers; indeed every other surreal can be sent to pretty much …
8
votes
Is the inverse of surreal numbers actually well-defined?
I think that you could take this as $y$ admitting the definition
$y:= \left\{\left.0,\frac{1+(x^R-x)Y^L}{x^R},\frac{1+(x^L-x)Y^R}{x^L}\right|\frac{1+(x^L-x)Y^L}{x^L},\frac{1+(x^R-x)Y^R}{x^R}\right\}$ …
2
votes
Accepted
Smallest ring whose field of fractions includes all the reals (subring of omnific integers?)
Assume for contradiction that such a ring $R$ exists.
Consider the ring $R_{\omega}:=\mathbb{Z} \oplus \omega\mathbb{R}[\omega]$ which is contained in $\mathbf{Oz}$. We have $\mathbb{R} \not\subseteq …
5
votes
Accepted
In surreal numbers, do the automorphisms allow us to define $\omega_2=\partial(\omega_1)$?
It is not the case that $0<\alpha < \operatorname{e}^{\omega} \Longrightarrow \partial(\alpha)< \alpha$. For instance $\partial(\operatorname{e}^{\omega}-\omega^{-1})= \operatorname{e}^{\omega}+\omega …
4
votes
A "surnatural numbers" as a largest model of the natural numbers
I think it should be said that the first order theory of the semi-ring $\mathbb{N}$ of non-negative integers is much more difficult to work with than that of the real ordered field.
The existence of c …
9
votes
Is the surreal number $\omega(\sqrt{2}+1)+1$ a prime?
This is not an answer but it is too long for a comment. I think that the question is still open, and that in general there are no known prime elements with finite support besides primes in $\mathbb{Z} …
2
votes
Accepted
Confusion regarding $\ln \omega$
The exponentiation in $\omega^{\frac{1}{\omega}} = \ln \omega$ is not that given by the exponential function. It is the value at $\frac{1}{\omega}$ of Conway's $\omega$-map.
2
votes
What makes the surreals special among other surreal-like fields?
I don't know if this answers the question, but one specificity of surreal numbers which would probably be difficult to translate into properties of the omega-map is that they have this compatible simp …
7
votes
Accepted
The surreal version of $e$
$\DeclareMathOperator{\ee}{e}$If $\varepsilon$ is an infinitesimal surreal, the quantity $\log(1+\varepsilon)$ is actually equal to the formal sum à la Hahn series $\sum \limits_{n \in \mathbb{N}} \fr …
4
votes
Surreal numbers and the Axiom of Choice
This is not an answer but just a set of comments:
A cousin of 1. is a yet unanswered question of J. D. Hamkins.
Notice that if an ordered field embeds into $\mathbf{No}$, then so does, as an ordered …
3
votes
Accepted
'Smallest' subfield of the Surreals which is isomorphic to the Surreals as an ordered group
$\DeclareMathOperator{\Noo}{\mathbf{No}}$This might actually be a dead end.
This is because if $F$ is isomorphic as an ordered group to $\Noo$, then their value classes under natural ordered group va …
18
votes
What's wrong with the surreals?
I take advantage of this question to propose an answer in the vein of that of Philip Ehrlich, and to advertise a little bit for $\mathbf{No}$.
I feel like the surreal numbers should be attractive for …
8
votes
Accepted
Transfinitely iterating the Levi-Civita, Hahn or Puiseux constructions
Let us work in NBG set theory with global choice. There is, up to non unique isomorphism, a unique real-closed field that is $\kappa$-saturated for all infinite cardinals $\kappa$. Let's denote it by …