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In other words, given a segment in the hyperbolic plane is there a straightedge and compass construction of a segment incommensurable with it? In the Euclidean plane one can take the diagonal of the square built on the segment, as Pythagoreans discovered. However, equivalence of this incommensurability and irrationality of $\sqrt{2}$ relies on the Euclidean Pythagorean theorem. Pythagoreans originally believed that any two segments have a common measure, how hard would it have been for them to discover their mistake if we happened to live in a hyperbolic space?

The "straightedge" of course has to be hyperbolic. More precisely, a construction can use all Hilbert's axioms of the hyperbolic plane (including the axiom of Archimedes) except the Cantor's axiom of continuity. Equivalently, the question asks if there is a pair of incommensurable segments in every subset of the hyperbolic plane closed under straightedge and compass constructions, but not necessarily metrically complete.

This may not be as easy as it looks. There are no squares in the hyperbolic plane, and the hypotenuse of an equilateral right triangle can be commensurable with its leg. In fact, it follows from the hyperbolic Pythagorean theorem that any number in $(\sqrt{2},2)$ can be the hypotenuse/leg ratio depending on the size of the triangle. If the ratio is rational for the given segment the Pythagorean construction won't work. One could try doubling/halving the segment multiple times and then taking hypotenuses on various concatenations, but it is conceivable that all of them remain commensurable since there do exist non-rational analytic functions that map rationals into rationals. Perhaps there is a construction more taylored to the hyperbolic plane. But standard constructions of hyperbolic parallels, and therefore of ideal triangles, do use the axiom of continuity.

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Yes. the fundamental theorem is that the constructible angles in the non-Euclidean plane are exactly the constructible angle in the Euclidean plane. Lengths come from the two laws of cosines and the law of sines. In particular, length 1 is not constructible. As i recall, positive length $x$ is constructible if and only if $\sinh x$ is constructible in the Euclidean plane.

See my 1995 Intelligencer article at http://zakuski.utsa.edu/~jagy/bib.html

Let's see, in the fourth edition of his book, Marvin Jay Greenberg included Robin Hartshorne's proof of this, very different sort of language.

EDIT: it seems part of the question was constructible lengths with irrational ratio. $\log 3$ and $\log 2$ should work.

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    $\begingroup$ Any way I look at it, you can’t construct anything with a straightedge and compass unless you are given at least two distinct points. Surely the set of constructible lengths will then depend on the distance between these two points (what if they are already at distance 1?). $\endgroup$
    – Emil Jeřábek
    Commented Sep 2, 2014 at 20:16
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    $\begingroup$ I can’t find a definition of random lengths in your article. Does it mean that you consider $x$ constructible if there exists a construction which always produces a length $x$ starting from two given distinct points, no matter what these points are, or something like that? $\endgroup$
    – Emil Jeřábek
    Commented Sep 2, 2014 at 20:34
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    $\begingroup$ @EmilJeřábek, yes, that sounds perfect. $\endgroup$
    – Will Jagy
    Commented Sep 2, 2014 at 20:36
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    $\begingroup$ Alright, thanks. Now one more thing: the question didn’t ask for a nonconstructible length, but for a pair of constructible lengths with irrational ratio. $\endgroup$
    – Emil Jeřábek
    Commented Sep 2, 2014 at 20:37
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    $\begingroup$ @Conifold: You can’t construct an equilateral triangle from nothing. Straightedge can draw a line through a given pair of distinct points, a compass can draw a circle with a given point as a centre going through another given point, and new points are only constructed as intersections of lines or circles. So, unless you are given two points (or enough lines or circles to get them as intersection points), you are stuck. $\endgroup$
    – Emil Jeřábek
    Commented Sep 2, 2014 at 20:48

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