Compass and Straightedge Constructions
and Constructible Numbers
This outline will bring closure to our discussion of compass and straightedge constructions by providing
a framework within which these constructions can be analyzed algebraically. The basic idea is to
provide a coordinate system for the Euclidean plane, by means of which points acquire a numerical
description. The points which can be constructed have coordinates that must lie in a particular field
extension of the rational numbers. These constructible numbers are in turn characterized in terms of the
equations they satisfy; as it turns out, polynomial equations. Ultimately we obtain a means for showing
that specific numbers are not constructible.
As you work through this outline, you will see that many of the developments are stated without proof.
You should fill in the missing proofs as you go along. Space has been left for you to write in your
proofs, but you will probably find it best to work out a careful argument on scratch paper first, and only
transcribe it to the outline when you are sure it is correct.
Basics. Begin with an initial point, the origin. Through that point draw a line, and then draw a
perpendicular line. Label these as the x and y axes in the usual way, with the positive and negative
directions designated in the customary manner. With the compass point at the origin, place the pencil at
an arbitrary point on either axis, and swing the pencil to mark points on the other three half-lines making
up the axes. From now on, the radius of this circle will be considered to have length 1. All other
numbers will be defined in terms of this designated unit. This is the basic configuration. The first part
of the outline defines constructible numbers, and shows they form a subfield of the reals and an
extension field of the rationals.
1. Definitions: A point in the plane is constructible if and only if it can be constructed from the basic
configuration through a finite number of compass and straightedge steps. A real number r will be called
constructible if and only if the point (r,0) is constructible.
2. Lemma: A real number r is constructible if and only if the point (0,r) is constructible.
Proof: [Hint: describe the construction steps necessary to construct (0,r) once (r,0) has been
constructed, and vice versa.]
3. Lemma: The point (x,y) is constructible if and only if both (x,0) and (0,y) are constructible.
Proof: [Hint: if (x,y) has been constructed, explain how to construct (x,0) and (0,y), and vice versa.]
4. Theorem: The point (x,y) is constructible if and only if both x and y are constructible numbers.
Proof:
5. Lemma: The set of constructible numbers is nonempty and closed under subtraction.
Proof: Note first that 0 and 1 are constructible (why?).
6. Lemma: The set of constructible numbers is closed under multiplication.
Proof: Let a and b be constructible numbers, so that the points (a,0), (0,b), and (0,1) can be constructed.
Since r is constructible if and only if –r is (Why?), we may assume that a and b are both non-negative.
If either of them is 0, then ab =0 is constructible. Otherwise, neither one is 0, so we can assume that a
and b are both positive. Now draw the line L through (a,0) and (0,1). Let M be the line through (0,b)
parallel to L. Let (x,0) be the intersection of M with the x axis. Since x is clearly constructible, the proof
will be completed by showing that x = ab. (Do that below).
7. Lemma: The set of constructible numbers is a subring of the reals.
Proof: (Hint: use a theorem in section 3.2).
8. Theorem: The constructible numbers are a subfield of the reals and contain the rationals.
Proof: (Hints: To show that reciprocals of constructible numbers are constructible, use a similar
approach as the proof of closure under multiplication above. To show the rationals are all constructible,
argue that every subfield of the reals contains the rationals.)
Fields. Next we will associate with each geometric construction a sequence of field extensions of the
rationals. First, note that any time we make use of the intersection of two lines or of a line and a circle,
or of two circles, we have constructed a point. In carrying out a finite number of construction steps, we
will produce a list of constructed points. This list of points, in the order in which they were constructed,
defines the point sequence for the overall construction. Now suppose that a construction has point
sequence {(x1,y1),( x2,y2), … ,( xn,yn)}. We define the corresponding fields as follows:
F0 = Q (the rationals)
F 1 = F 0 (x1,y1)
F 2 = F 1 (x2,y2)

F n = F n-1 (xn,yn)
That is, F 1 is the smallest field that contains the rationals and both coordinates of the first constructed
point. Then, F 2 is the smallest field that contains F 1 and both coordinates of the second constructed
point. At each stage, the extension field that is defined is a subfield of the constructible numbers,
because that is a field containing Q and the coordinates of all the constructed points.
Note that in some cases, successive fields in this arrangement might be identical. For example, if both
coordinates of the first constructed point are rational, then F 1 = F 0. More generally, if the coordinates
of the kth constructed point are both included in the preceding field F k-1, then F k = F k-1. The goal for
the next section of this outline is to see what happens when this is not the case, that is, when F k  F k-1.
We will see that this only happens when the kth constructed point is an intersection of a line with a circle
or of two circles, and that in these cases, [F k : F k-1] = 2.
Before proceeding to that discussion, let’s see what it implies about constructible numbers. If x is
constructible, it arises as a coordinate of a constructed point after a finite number of steps. That means it
is in an extension field Fk for some k. By showing that each such field is either equal to the preceding
field or is of degree 2 over that preceding field, we also establish that [F k : Q] = 2j for some j (which is
at most equal to k). In particular, every element of Fk is algebraic over Q, and satisfies a minimal
polynomial whose degree is a divisor of 2j, hence a power of 2. This shows that a number can only be
constructible if the degree of its minimal polynomial is a power of 2. Put another way, for any
irreducible polynomial in Q[x] whose degree is not a power of 2, none of the roots is constructible.
Constructed Points. New points arise in a construction as intersections of lines and circles. How are
these lines and circles defined? Lines are drawn by specifying two existing points. In Euclidean
geometry, two such points determine a unique line, and we visualize drawing this line by placing the
straightedge so that it touches each point. Therefore, to describe each such line with an equation, we
should begin with two previously constructed points. Similarly, a circle is created by placing the point
of the compass at one previously constructed point and the pencil on another. In both of these situations,
previously constructed refers to the steps of the construction up to, but not including, the production of
the new point under consideration. For concreteness, if we are constructing the kth point, then the
necessary lines or circles come from the first k-1 points. That means that the coordinates of these
previously constructed points lie in Fk-1. To simplify the notation, let us refer to this field as F, without
repeating the subscript of k-1 at each instance. Similarly, let us agree to refer to the kth constructed
point as P.
We proceed to determine equations for the lines and circles involved in defining P. First, consider a line
through previously constructed points (a,b) and (c,d). It is particularly convenient to express these lines
in parametric form, as
x = a + (c-a)t and y = b + (d-b)t
where a, b, c, and d are all members of F. This shows that we can always express such a line in the form
x = a + et
y = b + ft
where again, all of the coefficients are elements of F. This parametric form will be useful when we
contemplate the intersection of a line with a circle. But first, to deal with the intersection of two lines, it
is more convenient if we eliminate t to define a single equation in x and y. Multiply the first equation by
f and the second by e, then subtract to find:
fx - ey = fa – eb.
This gives us a second general form for the equation of our line:
fx - ey = g
again with all coefficients in F.
If the point P is at the intersection of two such lines, its coordinates define a solution to a system of two
linear equations in two unknowns. We know from linear algebra that such solutions are rational
combinations of the coefficients. That is, they are found by adding, subtracting, multiplying and
dividing the coefficients. This shows that the solutions are themselves elements of F. Indeed, we can
display these elements explicitly using Cramer’s rule. If the two equations are
f1x – e1y = g1
f2x – e2y = g2
then the solutions are given by
x 
 g1 e2  g 2 e1
 f 1 e2  f 2 e1
y 
 f 1 g 2  f 2 g1
 f 1 e2  f 2 e1
and we can be sure that the common denominator is not 0 because the lines which are being intersected
could not be parallel or coincident. However, these formulas for x and y are not really needed. We
simply need to see that the coordinates x and y are in the field F, given that the coefficients of the two
equations are in that field. Therefore, if the point P is defined as the intersection of two constructed
lines, we can conclude that the extension field Fk is the same as the preceding field F.
Next we consider the intersection of two circles. First let us again work out a general formula for the
equation of a single circle. If the point of the compass is put at (a,b), then that defines the center of the
circle. If the pencil is at the point (c,d), then the radius r of the circle is the distance between the two
points, and hence satisfies
r2 = (c-a) 2 + (d-b) 2.
That leads to the equation of the circle in the form
(x-a) 2 + (y-b) 2= (c-a) 2 + (d-b) 2.
The right-hand side of this equation is an element of the field F, so let us represent it with a single
parameter u. This gives a standard form for the equation of a circle available for the construction of P as
(x-a) 2 + (y-b) 2= u
where a, b, and u are elements of F.
With two such circles, we can find intersection points (if any) by solving the respective equations
simultaneously. Write the equations as
(x-a1) 2 + (y-b1) 2= u1
(x-a2) 2 + (y-b2) 2= u2
Now as a first step in solving this system, let us expand the squared terms of each equation. We then get
x 2 – 2a1 x + a1 2 + y 2 – 2b1 y + b1 2= u1
x 2 – 2a2 x + a2 2 + y 2 – 2b2 y + b2 2= u2
and subtracting then leads to
2(a2 – a1)x + 2(b2 – b1)y = u1 – u2 + a2 2 – a1 2 + b2 2 – b1 2
Now make two observations. First, this is the equation of a line with coefficients in F. Second, if the
circles intersect, both points must lie on this line. So we see that points of intersection between two
circles with coefficients in F, can be reduced to the case of points of intersection of a line and a circle,
each with coefficients in F. We will look at that case next.
We must now consider the intersection of a line with a circle. We have already seen that the line can be
expressed in parametric equations of the form
x = a + et
y = b + ft
To find the points where this line intersects a given circle, we will substitute a + et for x and b + ft for y
in the equation of the circle. That will result in a single equation in t. The solution(s) of this equation
can then be used with the parametric equations above to find the coordinates x and y for the point of
intersection. Let the equation of the circle be
(x-v) 2 + (y-w) 2= u
and keep in mind that u, v, w, a, b, e, and f are all in F. In the space on the next page, make the
substitutions for x and y into the equation of the circle, and then rearrange the resulting equation as a
quadratic equation in the variable t. Your result should be in the form
At2 + Bt + C = 0
where the coefficients A, B, and C are combinations of u, v, w, a, b, e, and f.
Do your work here, and record your expressions for A, B, and C below.
Your expressions for A, B, and C should convince you that these coefficients are members of F. That
means the quadratic is a polynomial in F[x] and one of two cases holds: Either the quadratic splits or it
is irreducible over F. In the first case, the solutions for t are in F, and so the coordinates x = a + et and y
= b + ft are also in F. In this case, the construction of the new point P does not require extending the
field, and Fk = F. In the second case, when the quadratic is irreducible, the roots are not in F. Let t* be
one such root, and create the extension field F(t*). Then we know that [F(t*):F] = 2, and that both roots
of the quadratic are included in this extension field. Consequently, the coordinates of P will be given by
x = a + et and y = b + ft with t given by either t* or the other root of the irreducible quadratic. In either
case, the coordinates of P are also in F(t*). This shows that F(t*) is a field that contains F and both
coordinates of the newly constructed point P. But Fk is defined as the smallest such field, so we
conclude that Fk is contained in F(t*). On the other hand, since both x and y are in Fk we can solve for t
in either x = a + et or y = b + ft to deduce that the corresponding value of t* must also be in Fk. (To do
this we have to know that at least one of e and f is nonzero. How can you justify that?) That shows that
Fk contains F(t*). Combining the two inclusions shows that Fk = F(t*). Therefore, we have shown that
[Fk:F] = 2.
Summarizing all of the above, we have seen that P, the kth construction point, occurs either as an
intersection of two lines or of a circle with a line, and that in either case, the coordinates of P are either
already in the field F, or they are in an extension field of degree 2 over F. We formalize this in the
following lemma.
9. Lemma: Suppose a finite number of compass and straightedge steps are performed, generating (in
order) the sequence of constructed points {(x1,y1),( x2,y2), … ,( xn,yn)}. Define a sequence of field
extensions of the rationals according to
F0 = Q (the rationals)
F 1 = F 0 (x1,y1)
F 2 = F 1 (x2,y2)

F n = F n-1 (xn,yn)
Then:
1. For each 1 < k < n, [Fk:Fk-1] is either 1 or 2.
2. [Fn:Q] = 2m for some nonnegative integer m.
From this lemma we can now easily deduce a theorem which tells us which algebraic numbers are
constructible.
10. Theorem: Every constructible number is algebraic over Q, with minimal polynomial of degree 2m
for some nonnegative integer m.
Proof: (This is one for you to prove.)
The following corollary is a somewhat more convenient restatement of the preceding theorem.
11. Corollary: Let the real number  be algebraic over Q, and suppose the minimal polynomial for 
has degree d which is not a power of 2. Then  is not constructible.
Proof: (You prove this one too. Try proof by contradiction.)
Exercises
1. Let  be any root of the polynomial 4x3 – 3x – ½ . Show that  is not constructible.
2. Use the identities below to express cos3 in terms of cos .
cos 2cos2sin2sin 2cossincos ( + ) coscossinsin

(Hint: let  = 2).
3. Show that cos 20 is a root of the polynomial in exercise 1.
4. Show that it is possible to construct an angle of 60 with vertex at the origin, the positive x axis
for one side and a line in the first quadrant as the other side. (Hint: make an equilateral triangle.)
5. Show that it is not possible to construct an angle of 20 with vertex at the origin, the positive x
axis for one side and a line in the first quadrant as the other side. (Hint: argue by contradiction.
If such a line were possible, it would also be possible to construct the intersection of that line
with a circle of radius one centered at the origin. What are the coordinates of that point? Are
these coordinates constructible?
6. What do 4 and 5 tell you about the existence of a compass and straightedge construction for
trisecting angles? Why?