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I know that for unmarked ruler and compass a number (distance) is constructible from $\mathbb Q$ iff it lies in a finite tower of field extensions $\mathbb Q=K_{0}...K_{n}$, where $[K_{i} :K_{i+1}] =2$ for all $i$. And an angle $x$ is constructible iff $\cos x$ meets the same conditions. My question is, for marked ruler and compass, a number is constructible iff what? Must we merely impose the condition that $[K_{i} :K_{i+1}] = 2$ or $3$? Can you give an explanation and proof please?

Edit: To be more specific, the ruler is marked with only points that may be constructed with compass and unmarked straightedge, and points that may be constructed with ruler and compass. This may also be relevant: we only allow verging between lines, not lines and curves or curves and curves.

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  • $\begingroup$ To be more specific, the ruler is marked with only points that may be constructed with compass and unmarked straightedge, and points that may be constructed with ruler and compass. This may also be relevant: we only allow verging between lines, not lines and curves or curves and curves. $\endgroup$ – Phillip Feldman Apr 27 at 7:46
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The important question is: What kind of markings does your ruler have. Suppose, that you have the perfect ruler, which has a mark for each number in $ \Bbb{R} $. In this case you can construct each point in the plane. Because your ruler has each real number as a marking you can construct the real line - just lie your ruler through $ 0 $ and $ 1 $ than you can make a point for each real number. Therefore $ \Bbb{R} $ is constructible. Now we have $ [\Bbb{C} : \Bbb{R}] = 2 $. Now the rule you have pointed out does work for any extension $ K \subset \Bbb{C} $ of $ \Bbb{Q}$:

A point $ x $ of $ \Bbb{C} $ is constructible out of $ K $, iff $ x $ lies in an iterated quadratic extension of $ K $

Thus also $ \Bbb{C} $ is constructible. Which just means, that you can construct every point in the plane.

Remark: You don't really have to worry about marking off uncountably many points of your ruler. Because if you want to construct an explicit point it is easy to see, that you only have to use finitely many marked off points.

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  • $\begingroup$ This is a good point. Even if the ruler had only 3 marks then $\pi$ might be constructible as the marks might be at $0$, $1$, and $\pi$. Just 2 marks might be the most interesting case or marks at just the integers or rationals. $\endgroup$ – badjohn Apr 27 at 7:33
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It depends on how it is marked, but a single notch does not not really help, so the least number of notches that's interesting is two. Already Archimedes proved that one can do trisection with that. If we only use it to do trisection than $[K_{i} :K_{i+1}] = 2, 3$ is right. This is proved by Gleason in Angle Trisection, the Heptagon, and the Triskaidecagon. This exactly covers what ancient Greeks called "solid constructions", i.e. those doable by using conic sections. As a result, the regular heptagon becomes constructible (although Archimedes's construction of it is based on a different kind of neusis). But the twice-notched straightedge is more powerful. Here is Theorem 5.1 from Baragar's Constructions Using a Compass and Twice-Notched Straightedge:

Suppose $α\in\mathbb{C}$ is constructible using a compass and twice-notched straightedge. Then $α$ belongs to a field $K$ that lies in a tower of fields $\mathbb{Q}=K_0⊂K_1⊂K_2⊂···⊂K_n=K$ for which the index $[K_j:K_{j−1}]$ at each step is $2,3,5$, or $6$. In particular, if $N=[K:\mathbb{Q}]$, then the only primes dividing $N$ are $2,3$, and $5$.

This has consequences for constructing regular polygons as well, not even twice-notched straightedge (and compass) can construct the regular $23$-gon. As of Baragar's writing in 2002, it was not known if the regular $11$-gon can be so constructed. For a more general discussion of neusis see Lyle Ramshaw's answer on Math Overflow.

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  • $\begingroup$ So I want my ruler to be marked by every constructible number. I also want it to be marked by any new number I construct by angle trisection and cube rooting. I don't want to be able to construct the regular 11 gon (if that's possible). Can you explain where 2,3,5,6 come from? $\endgroup$ – Phillip Feldman Apr 27 at 8:56
  • $\begingroup$ @PhillipFeldman I do not think adding notches at constructible distances makes a difference, you can probably simulate that with the original straightedge. Baragar shows that one can construct points of intersection of a circle and a conchoid of Nicomedes using twice-notched straightedge, that is where 5 and 6 come from. By the way, the paper I linked is freely accessible, see pp. 160-161. $\endgroup$ – Conifold Apr 27 at 10:07

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