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"Universal" means that the construction steps are independent of the length of the given segment. In the Euclidean plane one can take the diagonal of a square built on it. Without the "universal" the answer is affirmative. However, that is due to the fact that the hyperbolic plane has a natural unit of length, and one takes advantage of comparing it to the given length, which is excluded here.

Define the Schweikart's length as the length of a perpendicular to a line such that the asymptotic parallel to it through the endpoint of the perpendicular makes $\pi/4$ angle with the perpendicular. A construction that goes back to Bolyai (the converse to the better known one, see Will Jagy's paper) allows one to construct a segment of Schweikart's length from "nothing", i.e. from a segment of arbitrary length. So one can proceed as follows: construct a Schweikart's segment, if it is incommensurable with the given one we are done, if it is commensurable then construct a segment incommensurable with the Schweikart's (which is easy).

However, this is clearly not "universal", and in fact is similar in spirit to putting marks on the straightedge and measuring. In Euclidean plane at least marked straightedge is a strictly stronger tool. One can think of a universal construction as implementing a continuous (even real analytic) function $f(l)$ of a given length $l$. Since it is supposed to produce an incommensurable for every positive $l$ we have that $f(l)/l$ never takes rational values, i.e. it is a constant. Therefore, if such $f(l)$ exists it should be $f(l)=\alpha l$ with irrational $\alpha$. This is exactly what happens with the diagonal of the square in the Euclidean plane.

My intuition is that the answer is "no", nontrivial elementary hyperbolic constructions transform lengths nonlinearly, but I am not sure. Is there a hyperbolic straightedge and compass construction that does that?

EDIT: Another way to put it is that I am looking for an algorithm that is guaranteed to halt after finitely many steps with an incommensurable for any input. Or rather proving that, as I suspect, none exists. Comparison to the natural length would involve Euclidean algorithm presumably, to test for commensurability, and that would never halt if the input is in fact incommensurable. By laying off and bisecting one can implement $l\mapsto\frac{m}{2^n}l$ with hyperbolic straightedge and compass, but I suspect that these are the only linear transformations possible. Even trisecting a segment $l\mapsto\frac{1}{3}l$ is impossible, Euclidean trisection relies on the intercept theorem, and hence on the parallel postulate.

The length here refers to the standard assignment via binary fractions as in chapter 4 of Greenberg's text for example, which works uniformly in both hyperbolic and Euclidean geometries, not the specialized multiplicative length of Hartshorne. However, that in the hyperbolic case the lengths themselves do not form (the positive part of) a field, but rather their exponentiations do, is perhaps the reason why no such algorithm exists. Since $x=e^l$ an equivalent question would be if $x\mapsto x^\alpha$ with some irrational $\alpha$ can be done with Euclidean straightedge and compass for any length $x$, probably not, but I don't know how to prove that either.

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  • $\begingroup$ not sure what you are getting at, but the natural thing to consider is angles, between pairs of lines or two circles that meet or a circle and a line. The fundamental theorem is that the constructible angles in the hyperbolic plane are exactly the same as the constructible angles in the Euclidean plane. $\endgroup$
    – Will Jagy
    Mar 26, 2015 at 22:47
  • $\begingroup$ This is probably why Hartshorne switched to a "multiplicative length" for his (book) axiomatic treatment of the hyperbolic plane. For a length most of us would call $x,$ his multiplicative length is just $e^x,$ which is then in the "constructible field." $\endgroup$
    – Will Jagy
    Mar 26, 2015 at 22:52
  • $\begingroup$ @Will Jagy I tried to explain in the edit, but not sure if this addresses your question. Could you explain more how to use angles? $\endgroup$
    – Conifold
    Mar 27, 2015 at 18:02
  • $\begingroup$ It appears that you do understand that a length $x$ in the hyperbolic plane, take curvature as $-1,$ is constructible if and only if $e^x,$ $\cosh x,$ $\sinh x,$ $\tanh x$ are in the field of lengths constructible in the Euclidean plane. As that does not answer your question, I don't think I can help you at a distance. Commensurability is not a reasonable condition in the hyperbolic plane, there are explicit transcendental functions involved. $\endgroup$
    – Will Jagy
    Mar 27, 2015 at 18:12
  • $\begingroup$ Anyway, in case what you want is to be given an unknown segment and produce something else, how about taking the given length and making that two legs of a right triangle? The hypotenuse can be solved for explicitly. $\endgroup$
    – Will Jagy
    Mar 27, 2015 at 18:28

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