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I was thinking about $log(\omega)$ which appears to be $\{\mathbb{N}|\omega^{\frac{1}{n}}\}_{n\in\mathbb{N}}\stackrel{?}{=}\omega^\frac{1}{\omega}$. Intuitively, there's the idea that, if the highest power of $\omega$ is greater for one surreal number, then it is greater, but this seems incorrect for two reasons: $$n<\omega^\frac{1}{\omega}\rightarrow n^\omega <\omega$$ The statement on the right is not true, hence the question. Additionally, from a real analysis perspective, $$\lim_{n\to\infty}{n^{\frac{1}{n}}}=1$$

Asa Kaplan
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Every surreal number has a unique representation as a Hahn series of $\omega$ with coefficients being real numbers and exponents being themselves surreal numbers, and they are compared lexicographically as such Hahn series. (This form is known as a "Conway normal form". See, e.g., Norman Alling, Foundations of Analysis over Surreal Number Fields (North-Holland Math Studies 141 (1987)), esp., around §6.50, although pretty much the whole book really.) So $\omega^x$ is greater than all real numbers for any $x>0$, including $1/\omega$.

Gro-Tsen
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  • Thank you for the response. With this in mind, why does exponentiating by $\omega$ fail? Is exponentiating no longer order preserving? – Asa Kaplan Apr 21 '16 at 19:00
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    There simply isn't a single (if any) concept of exponentiation on the surreal numbers like there is on the ordinals. Have a look at http://mathoverflow.net/questions/88927/surreal-exponentiation-are-the-varying-definitions-equivalent-if-not-is-the — insofar as there is a concept of exponentiation, it is not the one being used when writing $\omega^x$. – Gro-Tsen Apr 21 '16 at 19:22
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    Also, it is correct that $\log(\omega) = \omega^{1/\omega}$ (for the Gonshor exponential/logarithm). If you were to compute $x^{1/x}$ as $\exp(\log(x)/x$ for $x=\omega$ (not what is written $\omega^{1/\omega}$), you would get $1+\omega^{-1+1/\omega}+\frac{1}{2}\omega^{-2+2/\omega}+\cdots$ which is, indeed, infinitesimally above $1$. – Gro-Tsen Apr 21 '16 at 20:03