I am not sure exactly what question is being debated. At least under some rules, many aspect ratios of Moebius strips are foldable. The tri-hexa-flexagon, for instance, is a Moebius strip. On Sun, Oct 25, 2015 at 6:59 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Martin Gardner offered an argument (due to ?) that isn't quite valid:
Take a 100 x 1 rectangle of paper and pleat it 1001 times in the 100 direction (i.e., make 1000 folds back and forth, 100/1001 apart from each other).
You get a very thin rectangle that is 100/1001 x 1. Now bend it around with a half-twist and glue the pleated edges together to get a Moebius band. (Because there were an even number of folds, they match up.)
This may "work" with paper, but not isometrically with a genuine flat 2-manifold in R^3.
—Dan
On Oct 25, 2015, at 1:55 PM, James Propp <jamespropp@gmail.com> wrote:
Regarding Mobius bands, Dan's remark
(If differentiability is assumed to be only C^1, then the answer is inf{h}
= 0, because of the Nash-Kuiper isometric embedding theorem. It's very hard to imagine just how this works.)
makes me wonder what the situation is for PL embeddings. Can a clever origamist turn ANY rectangle into a Mobius band?
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