Constructible Numbers and Classical Problems

We must show: (a) $K$ is closed under $+,-,\times$, and division by nonzero elements; (b) $K \supseteq \mathbb{Q}$; (c) $K$ is closed under positive square roots. All constructions below use only straightedge and compass and are classical Euclidean constructions, which we verify explicitly.

Coordinates suffice. First we record the standard reduction: a point $(x,y) \in \mathbb{R}^2$ is constructible from $\{O,I\}$ if and only if both coordinates $x, y$ are constructible numbers in the sense of the definition (i.e. lie in $K$). Indeed, given a constructible point $(x,y)$, drop a perpendicular from $(x,y)$ to the $x$-axis (a standard straightedge-compass construction: erect a perpendicular through a point to a line, both constructible from existing points) to get $(x,0)$, showing $x$ is constructible; a similar rotation/perpendicular argument gives $y$. Conversely if $x,y \in K$, mark $(x,0)$ and $(y,0)$ on the axis, transfer the length $y$ perpendicular to the $x$-axis at the origin using a compass to get $(0,y)$, and intersect the perpendiculars through $(x,0)$ and $(0,y)$ to obtain $(x,y)$. We take this standard reduction as given and work with lengths on the number line.

(b) $\mathbb{Q} \subseteq K$. Since $0,1 \in K$ by definition, and $K$ is closed under the field operations (proved next), every rational number is obtained from $0,1$ by finitely many additions, subtractions, multiplications, and divisions, hence lies in $K$.

(a) Closure under $+,-$. Given constructible lengths $a,b > 0$ marked as points $(a,0), (b,0)$, construct $a+b$ by using the compass to transfer the segment of length $b$ starting at $(a,0)$ along the axis (draw the circle centered at $(a,0)$ of radius $b$; it meets the axis at $(a+b,0)$ and $(a-b,0)$, giving both $a+b$ and $a-b$ as constructible). Negation is immediate by symmetry through $O$.

(a) Closure under multiplication and division. Given $a, b > 0$ constructible, we construct $ab$ using similar triangles. Mark points $A=(a,0)$, and on a second line through $O$ at a different angle mark $B$ at distance $b$ from $O$ and $U$ at distance $1$ from $O$ (all constructible by transferring compass lengths along a newly drawn line through $O$, which is itself constructible since $O$ is a marked point and any second point determines a line). Draw the line through $U$ and $A$. Draw the line through $B$ parallel to line $UA$ (constructing a parallel through a point is a standard straightedge-compass construction: copy the angle at $U$ to the point $B$). This parallel meets the axis through $O$ and $A$ at a point $C$; by similar triangles $OC/OB = OA/OU$, i.e. $OC/b = a/1$, so $OC = ab$. Hence $ab \in K$. For division, run the same similar-triangle construction with the roles of $U$ and $B$ swapped: constructing $C'$ with $OC' = a/b$. Thus $K$ is closed under multiplication and division by nonzero elements, and combined with additive closure this makes $K$ a field.

(c) Closure under square roots. Given $a > 0$ constructible, we construct $\sqrt{a}$ using Thales' theorem (angle in a semicircle is a right angle). Mark points $P = (-1, 0)$ and $Q = (a, 0)$ on the axis (both constructible since $-1, a \in K$ using parts already shown), and let $M$ be the midpoint of segment $PQ$, which is constructible (bisect a segment with straightedge and compass). Draw the circle with center $M$ through $P$ (hence also through $Q$, since $M$ is the midpoint); this circle has diameter $PQ$ of length $a+1$. Erect the perpendicular to the axis at $O$ (constructible, since $O$ lies on segment $PQ$ between $P$ and $Q$ as $-1 < 0 < a$); let it meet the circle at a point $H = (0, h)$ with $h > 0$. By Thales' theorem, since $PQ$ is a diameter, $\angle PHQ = 90^\circ$. The altitude from the right angle $H$ to the hypotenuse $PQ$ has foot $O$, and by the geometric mean relation for right triangles (altitude-on-hypotenuse theorem),

so $h = \sqrt{a}$. Since $H$ is an intersection of two constructed objects (a line and a circle), $H$ is a constructible point, so $h = \sqrt{a} \in K$. This proves $K$ is a subfield of $\mathbb{R}$ containing $\mathbb{Q}$ and closed under positive square roots. $\blacksquare$

(Main theorem, using the lemma. Proof technique: induction on the number of construction steps, via the <span class="wiki-link-missing">Tower Law</span>.) We must show every $\alpha \in K$ lies in a field $F_n$ with $[F_n:\mathbb{Q}] = 2^m$ for some $m$, since then $\mathbb{Q}(\alpha) \subseteq F_n$ forces $[\mathbb{Q}(\alpha):\mathbb{Q}]$ to divide $2^m$ by the Tower Law applied to $\mathbb{Q} \subseteq \mathbb{Q}(\alpha) \subseteq F_n$, and any divisor of a power of $2$ is itself a power of $2$.

Step 1: coordinates of constructed points lie in iterated quadratic extensions. We prove, by induction on $n$, the following stronger statement: for every point $P_n = (x_n,y_n)$ constructible in $n$ steps from $\{O,I\}$, there is a tower of fields

with each $[F_i : F_{i-1}] \in \{1,2\}$, such that $x_n, y_n \in F_n$. Base case ($n=0$): the only points available are $O=(0,0)$ and $I=(1,0)$, both with coordinates in $\mathbb{Q} = F_0$. Inductive step: suppose the claim holds for all points constructible in fewer than $n$ steps, so that all coordinates of $P_1,\dots,P_{n-1}$ lie in some field $F_{n-1}$ obtained from $\mathbb{Q}$ by a tower of extensions each of degree $1$ or $2$ (take $F_{n-1}$ to be the compositum of the towers for each of the finitely many prior points; a compositum of finitely many towers of quadratic/trivial steps is again such a tower, since adjoining a square root of an element of $F_{n-1}$ that is already a square in $F_{n-1}$ contributes a trivial step, and otherwise contributes a genuine quadratic step — in either case each step has degree $1$ or $2$, and the Tower Law lets us concatenate). Now $P_n$ is the intersection of two objects (lines or circles) determined by points among $\{O,I,P_1,\dots,P_{n-1}\}$, whose coordinates all lie in $F_{n-1}$. We check each of the three cases:

• Line meets line. A line through two points with coordinates in $F_{n-1}$ has an equation $ax+by=c$ with $a,b,c \in F_{n-1}$ (indeed $a = y_2-y_1$, $b=x_1-x_2$, $c = x_1y_2-x_2y_1$ for points $(x_1,y_1),(x_2,y_2)$, all in $F_{n-1}$). Two such lines, generically nonparallel, intersect at a point found by solving a linear system with coefficients in $F_{n-1}$; by Cramer's rule the solution $(x_n,y_n)$ has coordinates in $F_{n-1}$ itself (a field is closed under $+,-,\times,\div$). So we may take $F_n = F_{n-1}$, a trivial (degree $1$) extension.

• Line meets circle. A circle centered at a point of $F_{n-1}$-coordinates with radius equal to a distance between two such points has equation $(x-p)^2+(y-q)^2 = r^2$ with $p,q,r^2 \in F_{n-1}$ (note $r^2$, the squared radius, is a sum of squares of differences of $F_{n-1}$-elements, hence lies in $F_{n-1}$, even though $r$ itself might not). Substituting the linear relation $y = \frac{c-ax}{b}$ (or $x$ constant, if the line is vertical) from the line's equation into the circle's equation yields a quadratic equation in $x$ with coefficients in $F_{n-1}$. Its roots lie in $F_{n-1}(\sqrt{D})$ where $D \in F_{n-1}$ is the discriminant, by the quadratic formula. Set $F_n = F_{n-1}(\sqrt{D})$; then $[F_n:F_{n-1}] \in \{1,2\}$ (degree $1$ if $D$ is already a square in $F_{n-1}$, degree $2$ otherwise, since $x^2 - D$ is then irreducible over $F_{n-1}$). Having found $x_n \in F_n$, we get $y_n = \frac{c-a x_n}{b} \in F_n$ as well, since $F_n$ is a field containing $a,b,c$.

• Circle meets circle. Two circles $(x-p_1)^2+(y-q_1)^2=r_1^2$ and $(x-p_2)^2+(y-q_2)^2=r_2^2$ with all $p_i,q_i,r_i^2 \in F_{n-1}$: subtracting the two equations eliminates the quadratic terms $x^2, y^2$, producing a linear equation in $x,y$ with coefficients in $F_{n-1}$ (the radical axis of the two circles). Any intersection point of the two circles also satisfies this linear equation together with one of the original circle equations, so we are reduced to the line-meets-circle case just handled. Again $F_n = F_{n-1}(\sqrt{D})$ for the appropriate discriminant $D \in F_{n-1}$, with $[F_n:F_{n-1}] \in \{1,2\}$.

In every case, $x_n,y_n \in F_n$ for a field $F_n \supseteq F_{n-1}$ with $[F_n:F_{n-1}] \in \{1,2\}$, completing the induction.

Step 2: conclude the degree is a power of $2$. Let $\alpha \in K$, so $(\alpha,0)$ is constructible in some finite number $n$ of steps, hence by Step 1 there is a tower $\mathbb{Q}=F_0 \subseteq \cdots \subseteq F_n$ with each step of degree $1$ or $2$ and $\alpha \in F_n$. By the Tower Law,

for some $m \le n$ (namely $m$ is the number of genuinely quadratic steps). In particular $[F_n:\mathbb{Q}]$ is finite, so $F_n$ (and hence $\alpha$) is algebraic over $\mathbb{Q}$. Applying the Tower Law again to $\mathbb{Q} \subseteq \mathbb{Q}(\alpha) \subseteq F_n$,

so $[\mathbb{Q}(\alpha):\mathbb{Q}]$ divides $2^m$. A positive divisor of a power of $2$ is itself a power of $2$ (unique factorization in $\mathbb{Z}$: any prime dividing $[\mathbb{Q}(\alpha):\mathbb{Q}]$ must divide $2^m$, forcing that prime to be $2$). Hence $[\mathbb{Q}(\alpha):\mathbb{Q}] = 2^k$ for some $k \ge 0$. $\blacksquare$

Both statements are the contrapositive of the theorem. If $[\mathbb{Q}(\alpha):\mathbb{Q}]$ is finite but not a power of $2$, constructibility of $\alpha$ would force $[\mathbb{Q}(\alpha):\mathbb{Q}] = 2^k$ by the theorem, a contradiction. If $\alpha$ is transcendental, $[\mathbb{Q}(\alpha):\mathbb{Q}] = \infty$, so $\alpha$ cannot be algebraic of degree $2^k$; the theorem again gives a contradiction if $\alpha$ were constructible. $\blacksquare$