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Apr 13, 2017 at 12:57 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
May 26, 2011 at 14:19 vote accept James Propp
May 24, 2011 at 23:40 answer added George Lowther timeline score: 15
May 24, 2011 at 18:16 comment added Pete L. Clark Nice question. Just a comment: the standard proof works in an ordered field $R$ such that (i) Cauchy sequences converge and (ii) For all $0 < r < 1$, $r^n \rightarrow 0$. But the point is that these properties force $R$ to be complete Archimedean, i.e., $R \cong \mathbb{R}$.
May 24, 2011 at 17:14 history edited James Propp CC BY-SA 3.0
added 491 characters in body
May 24, 2011 at 17:04 history asked James Propp CC BY-SA 3.0