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This picture from Wikipedia's article on Algebraic numbers shows a visualization of Algebraic numbers coloured by degree.

I'm wondering if this is a fractal?

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The points in that picture are sized based on how small the integer coefficient of the minimal polynomial are (so, e.g., the point at zero is huge). Naturally, the smaller the points get, the more of them there are, which gives an illusion of self-similar structure--but you could get something similar by looking at the rationals and putting an interval of size, say, $5^{-max(\pm p, q)}$ about each rational $p/q$. The resulting set would have positive Lebesgue measure, hence fractal dimension 1, but its boundary might be interesting. –  Charles Staats Jul 13 '10 at 1:50
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The similarity of this particular picture to the Mandelbrot set is perhaps misleading. Consider that the algebraic numbers include all possible $ r + s i$ where $r,s$ are rational numbers and $i^2 = -1.$ So, while the field has many self-similarity properties built in, it is probably best to think of the algebraic numbers as a sort of fog that is roughly the same everywhere. –  Will Jagy Jul 13 '10 at 1:55
    
Fair enough. Thanks for the clarification. –  M.S. Jul 14 '10 at 17:19
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John Baez has a page on a similar picture of Dan Christensen, and some feathery patterns lend additional credibility to this: math.ucr.edu/home/baez/roots There are some references at the bottom. –  Robert Haraway Jan 3 '11 at 3:14
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The notion of having the radius proportional to the degree feels somewhat strange, and does not easily compare to other methods of creating fractals. However, it would be the boundary that is the interesting set here, as it will constitute of a union of circles. Such fractals are not unheard of, see for example the Apollonian gasket, and it does not feel arcane to think that there is some non-obvious self-similarity going on there. Odds are that the boundary has some non-integer Haussdorff dimension. –  Per Alexandersson Aug 30 '11 at 9:22

2 Answers 2

If you consider the set of roots of polynomials whose coefficients are entirely $1$ or $-1$, and take the topological closure of that set, you get a fractal pattern closely related to the Dragon curve.

see: http://math.ucr.edu/home/baez/roots/beauty.pdf

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The algebraic numbers are countable hence $\dim_{H}A=0$ for each subset. But one defines a fractal by non-intger Hausdorff dimenson.

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Not quite. Everyone agrees that the boundary of the Mandelbrot set is a fractal, but it has Hausdorff dimension 2. Of course its topological dimension is 1, so a better definition would be the one of Mandelbrot, namely Hausdorff dimension strictly bigger than topological dimension. But even this definition is not universally agreed upon... –  Wolfgang Loehr Sep 10 '12 at 16:31
    
Many folks I know actually use Minkowski dimension rather than Hausdorff dimension for this. For which countability does not imply dimension zero. –  BSteinhurst Sep 11 '12 at 2:12
    
Dear Wolfgang, you are right it would be better to put it this way, but also by your definition a contbale set is not a fractal. –  Jörg Neunhäuserer Sep 20 '12 at 16:40
    
Dear Bsteinhurst, to use Minkowski dimension is know to be problematic. You would have to say for instance that $\{0\}\cup\{1/n|n\ge 1\}$ is a fractal. –  Jörg Neunhäuserer Sep 20 '12 at 16:44
    
Dear Jorg, in some sense it is. But one can create a non-trivial self-similar fractal with integer Hausdorff dimension as well. So dimensionality alone isn't a good definition for what a fractal is. –  BSteinhurst Oct 8 '12 at 22:40

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