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Most field theory textbooks will describe the field of constructible numbers, i.e. complex numbers corresponding to points in the Euclidean plane that can be constructed via straightedge and compass. This field is the smallest field of characteristic 0 that is closed under square root (i.e. is Pythagorean) and is closed under conjugation.

I'm interested in know: What is the field of numbers that can be constructed if we disallow compass and use only straightedge?

I have not checked this up, but it seems that this question led Hilbert to formulate his 17th problem, particularly the version involving polynomials with rational coefficients (rather than the real coefficients which Artin proved). I'm also interested in knowing more about this history too.

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    $\begingroup$ How do you get the right angle without a compass? How do you ensure that the perpendicular legs are of equal length? $\endgroup$
    – Ryan Reich
    Commented Apr 8, 2011 at 13:26
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    $\begingroup$ How do you get the right-angle or any angle? Also addition is in my understanding done with the circle. The straightedge has no 'marks'. $\endgroup$
    – user9072
    Commented Apr 8, 2011 at 13:29
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    $\begingroup$ To piggyback on unknown's question, is it even obvious that this is a field? I think that one has to be very careful about the ground rules; for example, are you given the operation of moving / duplicating a segment? If not, then what you can construct seems to me to depend on how your ‘numbers’ are originally positioned. (Of course, switching which tool we're allowed, you doubtless already know about en.wikipedia.org/wiki/Mohr%E2%80%93Mascheroni_theorem.) $\endgroup$
    – LSpice
    Commented Apr 8, 2011 at 13:38
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    $\begingroup$ "Maybe by straightedge I mean at least something that can translate lengths. Not just to connect points." Maybe you need to think about what you mean to ask more precisely before you post the question? $\endgroup$
    – Todd Trimble
    Commented Apr 8, 2011 at 13:56
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    $\begingroup$ Note that moving to a "marked ruler" is a fairly dangerous generalization of the notion of "straightedge". In particular, if you allow yourself a marked straightedge and a compas, then you can trisect and angle: en.wikipedia.org/wiki/Angle_trisection#With_a_marked_ruler $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 8, 2011 at 14:34

4 Answers 4

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In view of the many comments, I will make a (I hope correct) summary of these comments in CW mode; everybody please feel free to edit:

  1. If one starts with a completely 'blank sheet of paper' it seems that one can do almost nothing with a straightedge alone.

  2. However, as mentioned by François Brunault given certain 'initial constellations' one can construct some additional interesting points using a staightedege alone (see here (in French)).

  3. Daniel Briggs suggested to 'add' just one circle with known center (the unit circle). If one does this, then by the Poncelet-Steiner theorem (mentioned by François Brunault) one can already construct everything one can construct with straightedge and compass.

  4. L. Spice mentioned that by the Mohr-Mascheroni theorem the converse situation (only a compass no straightedge) allows also to construct everything one can construct with straightedge and compass.

  5. The book Leçons sur les constructions géométriques by Lebesgue is entirely devoted to the question of geometric constructions using various instruments. The table of contents of this book (in French, again) is available here.

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  • $\begingroup$ I fixed the links, they did not work. $\endgroup$
    – Pieter Naaijkens
    Commented Apr 8, 2011 at 15:44
  • $\begingroup$ I added the reference to Lebesgue's book. $\endgroup$
    – François Brunault
    Commented Apr 8, 2011 at 15:56
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    $\begingroup$ 3: You must add one circle with its center point identified. $\endgroup$
    – Will Jagy
    Commented Apr 8, 2011 at 20:48
  • $\begingroup$ In particular, all classical constructible points can also be constructed with a straightedge and a "rusty" compass (that has a fixed radius, since this obviously gives one circle). For example, a (straight) knife and a fork would work. I think that this formulation is closest to the original question. (A marked straightedge is much more than translating length with a compass.) $\endgroup$
    – user11235
    Commented Apr 8, 2011 at 23:00
  • $\begingroup$ @Will Jagy, yes I know (the 'the unit circle' was meant to convey this, having its center in 'the origin'). Though, on second thought, I agree this should be said explictly; I add it. Thanks for pointing it out. $\endgroup$
    – user9072
    Commented Apr 9, 2011 at 1:05
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The short answer is that nothing is constructible. As is standard, we begin with the two points $(0,0)$ and $(1,0)$. Then we can draw a line between them, and that's it. We can't draw any more lines, and hence we can't construct any new points. The Euclidean rules say that we are only allowed to draw a new line if we are joining two already-constructed points, and a point can only be constructed if it is the intersection of two lines (or, irrelevant to this discussion, two circles or a line and a circle).

However, suppose you begin with a finite collection of points $(x_1,y_1),\ldots, (x_n,y_n)$. Let $C$ be the set of points constructible from this set using only a straightedge (unmarked). If a point $(x,y)$ is in $C$, then $x$ and $y$ can be formed from $x_1,\ldots,x_n,y_1,\ldots,y_n$ using the operations $+$, $-$, $\cdot$, and $\div$ (since new points are created as intersections between lines). However the converse of this is not true (as the two-point example shows). I suppose you could say more about what $C$ looks like, but it would probably be messy.

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It depends on what you regard as a starting point.

For ruler and compass, we start with the points 0 and 1.

For an unmarked ruler, this is not a good start, because an unmarked ruler is good at conserving cross-ratios, but if you start with two (or three including $\infty$) points, there is no cross-ratio yet to conserve.

With a marked ruler, this problem disappears because you obviously get all the integers and then are able to construct parallel lines by building a complete quadrilateral over three equidistant points. So you already get all the rational numbers.

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Bogomolny's site "Cut the Knot" has lots of interesting math... http://www.cut-the-knot.org/impossible/straightedge.shtml

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