most recent 30 from http://mathoverflow.net 2013-05-21T23:42:34Z http://mathoverflow.net/feeds/question/24891 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24891/the-root-of-a-line Max Muller 2010-05-16T15:22:53Z 2010-05-17T15:53:04Z

<p>Ok, imagine having a finite line segment from point (a) to point (b) in $R^2$ . I'm not familiar with mathematical terminology of this kind, but let me state that the line we began with is the geometric interpretation of $A^1$. The geometric interpretation of $A^2$ is a square with sides $A^1$. We could go on by saying that $ A^3$ is a cube with edge length $ A^{1}$ again. </p> <p>I wonder what the geometric interpration of $ \sqrt[2]{A}=A^{1/2}$ would look like. Is it a (straight) line? Is it constructible? Of course, we could extend this question by asking ourselves what $A^{k} $ would look like, in which $k \in \mathbb{R}$ or even $\mathbb{C}$ . When $k>2$, $A^k$ probably isn't constructible anymore on a sheet of paper, but one can still think about how these constructions of $A$ would 'look like'.</p> <p>Thanks in advance, </p> <p>Max Muller</p> <p>P.S. I realize I ask more than one question now, which is an indictation I don't know a lot about this (yet) and I'd like to know more about this subject. Should this be a community wiki? Feel free to modify the Tags, I don't know how to classify this question exactly.</p>

http://mathoverflow.net/questions/24891/the-root-of-a-line/24899#24899 Henno Brandsma 2010-05-16T16:30:48Z 2010-05-16T16:30:48Z

<p>From a topology standpoint, $\sqrt{\mathbb{R}}$ would be a space $Y$ such that $Y x Y$ is homeomorphic to $\mathbb{R}$. Such a space cannot exist as $Y$ would be connected (as an image of a connected space) hence a singleton or an interval. And an interval times an interval has non-cutpoints (removing such a point leaves the space connected) while $\mathbb{R}$ has none. One can also show that no odd power of $\mathbb{R}$ has such a square root, IIRC.</p>

http://mathoverflow.net/questions/24891/the-root-of-a-line/24924#24924 André Henriques 2010-05-16T20:00:29Z 2010-05-16T20:00:29Z

<p>Here's my proposal for the square root of a line segment.</p> <p>It's the Cantor set obtaining by repeatedly splitting the intervals into four, and removing the two middle pieces. When you take the cartesian square of that space, you obtain something whose projection is exactly an interval:</p> <p><img src="http://www.staff.science.uu.nl/~henri105/drawing.pdf" alt="alt text"></p>

http://mathoverflow.net/questions/24891/the-root-of-a-line/24981#24981 Tom Boardman 2010-05-17T08:56:37Z 2010-05-17T15:53:04Z

<p>The answers above seem to suggest there is a homomorphism from this interesting little line group to $\mathbb{R}$ sending the cartesian product to multiplication. But we have $A^1 \times A^2 =A^3$ when the answer we want is surely $A^2$. Cartesian product is additive on dimensions so perhaps eg. Andre has found a value for $ \frac12 A$. My guess as to how to get a multiplicative structure would be $X*Y:=hom(X,Y)$ in some suitable category, for example if $A^i=\mathbb{R}^i$ then linear maps would do the job. But now $\sqrt{\mathbb{R}}=[Y, hom(Y,Y)=\mathbb{R}] =\mathbb{R}$ and the real (and probably impossible) fun comes in finding '$\sqrt{\mathbb{R}^2}$' </p> <p>[Edit: re-read question- didn't realise those were powers of $A$- thought it was just an index! Oh well- will leave this here as I think it's an interesting reformulation...]</p> <p>Edit2: Bit of a fiddle but you can make this work in the category of $\mathbb{Q}$ Vector spaces with continuous $\mathbb{Q}$-linear maps where $\sqrt{\mathbb{Q}^2}=\mathbb{Q}[\sqrt{m}]$ for any non-square m and $\sqrt{\mathbb{Q}^n}=\mathbb{Q}[^n\sqrt{m},^n\sqrt{m}^2...]$ for any non-nth power m, we've lost the opportunity for fractions and a cube root seems unlikely, but it's sort of neat...</p>