In Harry Gonshor's An Introduction to the Theory of Surreal Numbers, on page 50, Gonshor points to a method for intuitively guessing what the square root of the countable infinity is in his construction -- I have a generalization of this intuitive guess method that I am curious about.

For some context, in Gonshor's construction:

A surreal number is a function $x:\alpha\rightarrow\{+,-\}$ from an ordinal $\alpha\in O_n$ into a two element set $\{+,-\}$, where we refer to $\alpha$ as the length of $x$ and write it as $\ell_x$ in general. The class $\mathbb{R}_\infty$ of all such functions is called the Surreal numbers, and it is ordered lexicographically by $$-<0<+,$$ where by convention $x(\ell_x)=0$ for all $x$; that is, a surreal number takes on a value of $0$ at the first ordinal for which it is undefined, its length.

I believe the above is sufficient for the discussion at hand, although I will add a definition of multiplication if it is deemed appropriate. The positive natural number $n$ is given by the string of $n$ $+$'s, and the negative integer $-n$ is given by a string of $n$ $-$'s, and in general we can obtain a positive or negative ordinal $\alpha$ or $-\alpha$ as a string of $\alpha$ $+$'s or $-$'s respectively.

At the first successor position where the sign changes we begin averaging, so $$++=2$$ $$+++=3$$ $$+++-=\frac{2+3}{2}=2.5$$ $$+++--=\frac{2+2.5}{2}=2.25$$ $$+++-+=\frac{2.5+3}{2}=2.75$$ $$\dots$$ in very similar fashion to binary expansions of real numbers with $1\equiv+$ and $0\equiv-$ (note that we would have to pick a new symbol for the thing ordered between $+$ and $-$).

If we look at a string of $\omega$ $+$'s followed by $\omega$ $-$'s, we end up with $\frac{\omega}{2}$: $$\frac{\omega}{2}=+++\dots---\dots,$$ and I believe in general we have that for $\gamma$-numbers $\alpha$ a string of $\alpha$ $+$'s followed by $\alpha$ $-$'s is $\frac{\alpha}{2}$, and strings of $+$'s and $-$'s which switch at limit ordinals all generate nice ordinal fractions in this manner as long as the strings are of additively comparable length, much like an iterated version of the averaging process above.

More interestingly, if we take a string of $\omega$ $+$'s and then attach a string of $\omega^2$ $-$'s we get the square root of $\omega$; that is, $$\sqrt{\omega}=+++\dots---\dots---\dots---\dots.,$$ where $\dots.$ indicates that we repeat this countable string of $-$'s a countable number of times. I suspect we then have that $\omega$ $+$'s followed by a string of $\omega^3$ $-$'s is the cube root of $\omega$: $$\sqrt[3]{\omega}=+++\dots---\dots.---\dots.,$$ and we can obtain higher finite roots in the expected fashion. My question is this:

Is a string of $\omega$ $+$'s followed by $\omega^\omega$ $-$'s the $\omega^{th}$ root of $\omega$? That is, does $$\sqrt[\omega]{\omega}=+++\dots---\dots..$$ hold, where $\dots..$ indicates that we repeat the countable string of countable strings of countable strings of countable strings of... of $-'s$ a countable number of times?

If yes, my followup questions are:

  1. Does this behavior generalize in the expected fashion to higher transfinite roots, or to higher transfinite $\gamma$-numbers?

  2. Does this imply that there is a canonical 'infinite product' structure on $\mathbb{R}_\infty$, seeing as we have canonical infinite-th roots?

Any assistance is greatly appreciated.

  • 1
    $\begingroup$ There is an infinite product structure on No (lifting infinite sums with the exponential) but it does not allow one to multiply the same number by itself an ordinal amount of times (except for $1$), because the corresponding family is not "Hahn-summable". $\endgroup$
    – nombre
    Mar 24, 2018 at 13:03

1 Answer 1


Chapter 3 of ONAG (see parts "$\omega$-map" and "Sign-expansions and normal forms") contains a way to obtain $\omega^x$ for any number $x$ (including transfinite cardinals). UPD (replacing a false statement): according to the reference, for $x$ having sign expansion $x_0 x_1 \ldots$ (indexed by ordinals up to some $\alpha$), the sign expansion of $\omega^x$ is $x_0^{e_0 + 1} x_1^{e_1 + 1} \ldots$, where $e_\beta$ is the (ordinal) number of $+$'s among $x_{\delta}$ for $\delta < \beta$. According to this, the number $+^{\omega}-^{\omega^3}$ is $\sqrt[\omega]{\omega}$ rather than $\sqrt[3]{\omega} = +^{\omega}-^{\omega^2}+^{\omega^2}-^{\omega^3}+^{\omega^3}\ldots$ ($\omega$ terms).

In particular case when $x$ is a limit ordinal we can see that $x = +^x$ and $1/x = + -^x$. It follows that $\sqrt[x]{\omega} = +^{\omega} -^{\omega^2 x}$, and $+^{\omega} -^{\omega^{\omega}} = \sqrt[\omega^{\omega}]{\omega}$.

  • $\begingroup$ Very nice! I think this settles all questions except 2, which (I believe) must be an implicit yes? $\endgroup$
    – Alec Rhea
    Oct 26, 2017 at 21:52
  • $\begingroup$ @AlecRhea I am not sure what kind of "canonical infinite products" you want to consider. Do you have a specific result in mind that you would like to extend/mimic in surreal case? If so, please state this original result. $\endgroup$ Oct 26, 2017 at 22:04
  • $\begingroup$ I use the word 'canonical' in a somewhat loose sense here -- it has seemed to me for some time that the surreals 'should' support an intuitively nice kind of infinite product structure, perhaps as a limiting process over partial products of some ordinal length. The existence of transfinite roots suggests to me that there is such a product structure, under which we would compose $\sqrt[\omega^\omega]{\omega}$ with itself $\omega^\omega$ times and obtain $\omega$. All of this is driving at the idea that we can nicely generalize finite/countable notions over the reals to higher infinities in No. $\endgroup$
    – Alec Rhea
    Oct 26, 2017 at 22:22
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    $\begingroup$ Note that $\omega$-map is not the same as "analytic exponentiation", so usual exponentiation properties do not necessarily apply (just as well as ordinal exponentiation has its caveats). The notation $\sqrt[x]{\omega}$ may be tempting to get ideas about, but really is just a formal symbol for $\omega$-map of $1 / x$. There is an elusive note in ONAG about Martin Kruskal's definition of "proper" exponentiation that allegedly has nice properties (see discussion here). $\endgroup$ Oct 26, 2017 at 22:46
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    $\begingroup$ The only thing I can say about two exponentiation notions is that they do not generally agree, so you may have trouble introducing sound infinite products with properties of both. $\endgroup$ Oct 26, 2017 at 22:51

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