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Let $F \subset \mathbb R$ be the set of lengths of line segments that one can construct, starting from the points $(0,0)$ and $(1,0)$, using a straightedge, compass, and an Archimedean spiral - the curve defined in polar coordinates by the equation $r=\theta$.

To clarify, if we can construct a line or circle, then we can construct all the intersection points of that line or circle with the Archimedean spiral. We cannot move the spiral. Cartesian and polar coordinates are lined up normally, so we can construct the line $\theta=0$.

By classical arguments from the first two constructions, $F$ is a field and is closed under square roots. Due to the third construction, it is closed under taking roots of Chebyshev polynomials, and includes $\pi$.

What is $F$? What is its transcendence degree? Is $F(i)$ algebraically closed? Can we exhibit a number that is not in $F$?

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Any noncomputable number, such as Chaitin's constant, will not be in $F$. But I suspect you want a better answer. –  Alex Becker Dec 20 '12 at 22:26
    
$F$ is countable, so a diagonalisation argument will give you an explicit computable number not in $F$. It seems obvious that the transcendence degree is infinite, though I do not have a proof at the moment. –  Goldstern Dec 20 '12 at 22:33
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On your second question: note that if $r\in F$ so too is $\sin(r)$ and $\cos(r).$ Hence, $F(i)$ contains $e^{ir}$ for any $r\in F.$ As an algebraic extension of $\mathbb{Q}$ of infinite degree, e.g. the constructable numbers, it follows by the Lindemann–Weierstrass theorem that $F(i)$ and hence $F$ has infinite transcendence degree. –  JSpecter Dec 21 '12 at 2:38
    
The second sentence should read: As F contains an algebraic extension... –  JSpecter Dec 21 '12 at 18:02

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