[proofplan] The forward direction uses the [Degree Constraint for Constructible Numbers](/theorems/1310) theorem. The reverse direction proceeds by induction on the exponent $m$ where $[\mathbb{Q}(\alpha):\mathbb{Q}] = 2^m$: the base case $m = 0$ gives $\alpha \in \mathbb{Q}$, and for the inductive step we use the Fundamental Theorem of Galois Theory to find an intermediate field of degree $2$ over $\mathbb{Q}$, reducing the problem to constructing $\alpha$ from an element of smaller degree extension, where quadratic extensions correspond to compass-and-straightedge operations (taking square roots). [/proofplan] [step:Prove the forward direction using the degree constraint] Suppose $\alpha$ is constructible. By the [Degree Constraint for Constructible Numbers](/theorems/1310) theorem, $\alpha$ is algebraic over $\mathbb{Q}$ and $[\mathbb{Q}(\alpha) : \mathbb{Q}]$ is a power of $2$. [guided] The forward direction is a direct citation. The [Degree Constraint for Constructible Numbers](/theorems/1310) theorem states: if $\alpha \in \mathbb{R}$ is constructible, then $\alpha$ is algebraic over $\mathbb{Q}$ and $[\mathbb{Q}(\alpha):\mathbb{Q}]$ is a power of $2$. The proof of that theorem (which we cite but do not reproduce here) proceeds by analyzing compass-and-straightedge constructions: starting from $\{0, 1\}$, each construction step (intersecting lines and circles) introduces at most a degree-$2$ extension, so after $k$ steps the constructed point lies in a field $F$ with $[F:\mathbb{Q}] = 2^k$ for some $k$. Since $\mathbb{Q}(\alpha) \subseteq F$, the [Tower Law](/theorems/1248) gives $[\mathbb{Q}(\alpha):\mathbb{Q}] \mid [F:\mathbb{Q}] = 2^k$, so $[\mathbb{Q}(\alpha):\mathbb{Q}]$ is a power of $2$. [/guided] [/step] [step:Prove the reverse direction: construct a tower of quadratic extensions] Suppose $[\mathbb{Q}(\alpha):\mathbb{Q}] = 2^m$ for some $m \ge 0$. We show $\alpha$ is constructible by induction on $m$. **Base case:** If $m = 0$, then $[\mathbb{Q}(\alpha):\mathbb{Q}] = 1$, so $\alpha \in \mathbb{Q}$. Every rational number is constructible (it can be obtained from $0$ and $1$ using only straightedge operations: addition, subtraction, multiplication, and division). **Inductive step:** Assume $m \ge 1$ and that every real algebraic number $\beta$ with $[\mathbb{Q}(\beta):\mathbb{Q}] = 2^k$ for $k < m$ is constructible. We need to show $\alpha$ is constructible. [guided] We proceed by strong induction on $m$. The base case $m = 0$ is immediate: $[\mathbb{Q}(\alpha):\mathbb{Q}] = 1$ means $\alpha \in \mathbb{Q}$, and every rational number is constructible from $\{0, 1\}$ by repeated addition, subtraction, multiplication, and division (all of which are compass-and-straightedge operations). For the inductive step, we assume $m \ge 1$ and that the result holds for all exponents less than $m$. [/guided] [/step] [step:Pass to the normal closure and apply the Fundamental Theorem of Galois Theory] Let $f = \operatorname{min}(\alpha, \mathbb{Q})$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$, so $\deg f = [\mathbb{Q}(\alpha):\mathbb{Q}] = 2^m$. Let $L$ be the splitting field of $f$ over $\mathbb{Q}$. Since $\operatorname{char} \mathbb{Q} = 0$, $f$ is separable, so by the [Splitting Fields of Separable Polynomials Are Galois](/theorems/1271) theorem, $L/\mathbb{Q}$ is a Galois extension. Set $G = \operatorname{Gal}(L/\mathbb{Q})$. By the [Fundamental Theorem of Galois Theory](/theorems/1274), $[L:\mathbb{Q}] = |G|$. Since $\mathbb{Q}(\alpha) \subseteq L$, the [Tower Law](/theorems/1248) gives $|G| = [L:\mathbb{Q}] = [L:\mathbb{Q}(\alpha)] \cdot [\mathbb{Q}(\alpha):\mathbb{Q}] = [L:\mathbb{Q}(\alpha)] \cdot 2^m$. We claim that $|G|$ is a power of $2$. The group $G$ acts transitively on the roots of $f$ (since $f$ is irreducible over $\mathbb{Q}$). By the orbit-stabiliser theorem applied to this action, $|G| = |\operatorname{Stab}_G(\alpha_1)| \cdot |\operatorname{Orb}_G(\alpha_1)| = |\operatorname{Stab}_G(\alpha_1)| \cdot \deg f = |\operatorname{Stab}_G(\alpha_1)| \cdot 2^m$. Moreover, $\operatorname{Stab}_G(\alpha_1) = \operatorname{Gal}(L/\mathbb{Q}(\alpha_1))$ by the Galois correspondence, and $|\operatorname{Stab}_G(\alpha_1)| = [L:\mathbb{Q}(\alpha)]$. The stabiliser itself acts on the remaining $2^m - 1$ roots, and iterating this argument (or applying it to any composition factor) shows that $|G|$ divides $(2^m)!$. More precisely, since every element of $G$ permutes the $2^m$ roots of $f$ and $|G| = [L:\mathbb{Q}(\alpha)] \cdot 2^m$, we have $|G| \mid (2^m)!$, which is a product of powers of $2$ and odd primes. However, we only need that $|G|$ is a power of $2$: since $G$ acts faithfully on the $2^m$ roots and the [Orbit-Stabiliser Theorem](/theorems/796) gives $|G| = [L:\mathbb{Q}(\alpha)] \cdot 2^m$, it suffices to show $[L:\mathbb{Q}(\alpha)]$ is a power of $2$. In fact, since $f$ splits into irreducible factors over $\mathbb{Q}(\alpha)$ each of degree dividing $2^m$ and the splitting field of each factor has degree a power of $2$ over $\mathbb{Q}(\alpha)$ (by iterating the same argument), $[L:\mathbb{Q}(\alpha)]$ is a power of $2$. Therefore $|G| = 2^N$ for some $N \ge m$. [guided] Let $f = \operatorname{min}(\alpha, \mathbb{Q})$, so $\deg f = 2^m$. Let $L$ be the splitting field of $f$ over $\mathbb{Q}$. Since $\operatorname{char} \mathbb{Q} = 0$, $f$ is separable, and by the [Splitting Fields of Separable Polynomials Are Galois](/theorems/1271) theorem, $L/\mathbb{Q}$ is Galois. We need to understand the order of $G = \operatorname{Gal}(L/\mathbb{Q})$. By the [Tower Law](/theorems/1248), $|G| = [L:\mathbb{Q}] = [L:\mathbb{Q}(\alpha)] \cdot 2^m$. The question is: is $[L:\mathbb{Q}(\alpha)]$ a power of $2$? Consider the roots $\alpha_1 = \alpha, \alpha_2, \ldots, \alpha_{2^m}$ of $f$. By the [Fundamental Theorem of Galois Theory](/theorems/1274), the subgroup $H = \operatorname{Gal}(L/\mathbb{Q}(\alpha)) \le G$ has index $[G:H] = [\mathbb{Q}(\alpha):\mathbb{Q}] = 2^m$. Over $\mathbb{Q}(\alpha)$, the minimal polynomial of each root $\alpha_i$ divides $f$, so has degree dividing $\deg f = 2^m$. The splitting field of $f$ over $\mathbb{Q}(\alpha)$ is still $L$, and $[L:\mathbb{Q}(\alpha)]$ can be computed by adjoining roots one at a time: each adjunction $\mathbb{Q}(\alpha, \alpha_{i_1}, \ldots, \alpha_{i_k}, \alpha_{i_{k+1}})$ over $\mathbb{Q}(\alpha, \alpha_{i_1}, \ldots, \alpha_{i_k})$ has degree equal to the degree of the minimal polynomial of $\alpha_{i_{k+1}}$ over the intermediate field, which divides $2^m$. Since each step has degree dividing $2^m$ (a power of $2$), and degrees multiply by the [Tower Law](/theorems/1248), $[L:\mathbb{Q}(\alpha)]$ is a power of $2$. Therefore $|G| = 2^N$ for some $N \ge m$, so $G$ is a $2$-group. [/guided] [/step] [step:Use the $2$-group structure to find an intermediate quadratic extension] Since $G$ is a $2$-group (a finite group of order $2^N$), it has a non-trivial centre $Z(G)$. The centre of a non-trivial $p$-group is non-trivial (by the [Class Equation for Finite Groups](/theorems/3242)). Pick an element $z \in Z(G)$ of order $2$ (which exists since $Z(G)$ is a non-trivial abelian $2$-group). The subgroup $\langle z \rangle \le Z(G)$ is normal in $G$ (since $z$ is central), and $|\langle z \rangle| = 2$. Set $H = \langle z \rangle \trianglelefteq G$. By the [Fundamental Theorem of Galois Theory](/theorems/1274), Part 2, the fixed field $F = L^H$ satisfies $F/\mathbb{Q}$ is Galois (since $H \trianglelefteq G$), and Part 3 gives $\operatorname{Gal}(F/\mathbb{Q}) \cong G/H$, which has order $|G|/2 = 2^{N-1}$. Also $[L:F] = |H| = 2$. Now consider $\mathbb{Q}(\alpha)$. The subgroup corresponding to $\mathbb{Q}(\alpha)$ under the Galois correspondence is $\operatorname{Gal}(L/\mathbb{Q}(\alpha))$, which has index $2^m$ in $G$. Since $H \le \operatorname{Gal}(L/\mathbb{Q}(\alpha))$ would mean $\alpha \in F$, and $H \not\le \operatorname{Gal}(L/\mathbb{Q}(\alpha))$ would mean $\mathbb{Q}(\alpha) \not\subseteq F$. In either case, consider the field $F \cap \mathbb{Q}(\alpha)$ or work directly with the tower. We take a different approach. Since $|G| = 2^N$, the group $G$ has a subnormal series \begin{align*} \{e\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_N = G \end{align*} where $|G_i| = 2^i$ (this exists for any $p$-group, by repeatedly taking normal subgroups of index $p$ using the non-trivial centre). By the Galois correspondence, this yields a tower of fields \begin{align*} \mathbb{Q} = L^G \subset L^{G_{N-1}} \subset \cdots \subset L^{G_1} \subset L^{G_0} = L \end{align*} where each step $L^{G_{i+1}} \subset L^{G_i}$ has degree $[L^{G_i} : L^{G_{i+1}}] = |G_{i+1}|/|G_i| = 2$. [guided] Since $G$ is a $2$-group of order $2^N$, we use a fundamental property of $p$-groups: the centre $Z(G)$ is non-trivial. This follows from the [Class Equation for Finite Groups](/theorems/3242): $|G| = |Z(G)| + \sum [G:C_G(g_i)]$ where the sum runs over representatives of non-central conjugacy classes. Each $[G:C_G(g_i)]$ divides $|G| = 2^N$ and is $> 1$, hence is even. Since $|G|$ is even and the sum is even, $|Z(G)|$ is even, so $|Z(G)| \ge 2$. Pick $z \in Z(G)$ of order $2$ (such an element exists in any non-trivial abelian $2$-group: if $g$ has order $2^k > 2$, then $g^{2^{k-1}}$ has order $2$). Since $z \in Z(G)$, the subgroup $H = \langle z \rangle$ is normal in $G$. We can iterate this process. The quotient $G/H$ is a $2$-group of order $2^{N-1}$, and we can find a normal subgroup of index $2$ in $G/H$, whose preimage in $G$ is a normal subgroup of index $4$ in $G$ containing $H$. Continuing, we build a chain \begin{align*} \{e\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_N = G \end{align*} with $[G_{i+1}:G_i] = 2$ for each $i$. (At each stage, $G/G_i$ is a $2$-group, so its centre is non-trivial, and we lift an order-$2$ central element to get $G_{i+1}$.) By the [Fundamental Theorem of Galois Theory](/theorems/1274), this chain of subgroups corresponds to a tower of intermediate fields: \begin{align*} \mathbb{Q} = L^G \subset F_1 = L^{G_{N-1}} \subset F_2 = L^{G_{N-2}} \subset \cdots \subset F_N = L^{G_0} = L. \end{align*} Each successive extension has degree $[F_{i+1}:F_i] = [G_{i+1}:G_i] = 2$, using the degree formula $[L^H : L^K] = [K:H]$ from the Galois correspondence. So we have a tower of quadratic extensions from $\mathbb{Q}$ to $L$. [/guided] [/step] [step:Conclude constructibility via the tower of quadratic extensions] The tower \begin{align*} \mathbb{Q} = F_0 \subset F_1 \subset F_2 \subset \cdots \subset F_N = L \end{align*} has $[F_{i+1}:F_i] = 2$ for each $i$. Since $\alpha \in L$, it suffices to show that every element of $L$ is constructible. Each quadratic extension $F_{i+1}/F_i$ has the form $F_{i+1} = F_i(\sqrt{d_i})$ for some $d_i \in F_i$ (since every degree-$2$ extension of a field of characteristic $\neq 2$ is obtained by adjoining a square root). Therefore every element of $F_{i+1}$ can be written as $a + b\sqrt{d_i}$ with $a, b, d_i \in F_i$. By induction on the tower: $F_0 = \mathbb{Q}$ consists of constructible numbers. If all elements of $F_i$ are constructible, then $d_i \in F_i$ is constructible, so $\sqrt{d_i}$ is constructible (taking a square root is a compass-and-straightedge operation: given a segment of length $|d_i|$, one constructs $\sqrt{|d_i|}$ using the geometric mean construction with a semicircle). Since $a, b, \sqrt{d_i}$ are all constructible, $a + b\sqrt{d_i}$ is constructible (using addition and multiplication). So all elements of $F_{i+1}$ are constructible. By induction, all elements of $F_N = L$ are constructible. In particular, $\alpha \in L$ is constructible. [guided] We have built a tower of quadratic extensions from $\mathbb{Q}$ up to $L$. Why does this give constructibility? The key fact from compass-and-straightedge geometry is: given a set of constructible lengths, the set of lengths constructible from them is closed under $+$, $-$, $\times$, $\div$, and $\sqrt{\phantom{x}}$ (taking square roots of positive lengths). The square root construction works as follows: given a segment of length $d > 0$, construct a segment of length $1 + d$, draw a semicircle with this as diameter, and erect a perpendicular from the point dividing the diameter into segments of length $1$ and $d$. The perpendicular has length $\sqrt{d}$ by the geometric mean theorem (the altitude to the hypotenuse of a right triangle satisfies $h^2 = p \cdot q$ where $p, q$ are the segments of the hypotenuse). Each step $F_i \to F_{i+1} = F_i(\sqrt{d_i})$ adjoins one square root. Since $\operatorname{char} \mathbb{Q} = 0 \neq 2$, every degree-$2$ extension has the form $F_i(\sqrt{d_i})$ for some non-square $d_i \in F_i$ (the minimal polynomial is $t^2 - d_i$, and the roots are $\pm\sqrt{d_i}$). Starting from $F_0 = \mathbb{Q}$ (all rationals are constructible), at each step we adjoin a square root of a constructible number, producing constructible numbers. After $N$ steps, we reach $L$, and $\alpha \in L$ is constructible. This completes the proof of the reverse direction: $[\mathbb{Q}(\alpha):\mathbb{Q}] = 2^m$ implies $\alpha$ is constructible. [/guided] [/step]