**Introduction:**

Ancient Greek mathematicians developed the methodology of “ruler-and-compass” constructions: if one is given only a ruler (without marks) and a compass, what objects can be constructed as a result of a finite set of operations? While they achieved many successes, three problems confounded their efforts: (1) squaring the circle; (2) trisecting an angle; and (3) duplicating a cube (i.e., constructing a cube whose volume is twice that of a given cube). Indeed, countless mathematicians through the ages have attempted to solve these problems, and countless incorrect “proofs” have been offered. We might add that Archimedes discovered a way to trisect an angle, but it did not strictly conform to the rules of ruler-and-compass construction.

The impossibility of squaring the circle was first proved by Lindemann in 1882, who showed that $\pi$ is *transcendental* — it is not the root of any algebraic equation with integer or rational coefficients. The impossibility of trisecting an arbitrary angle was proved earlier, in 1837, by Pierre Wantzel, and the impossibility of duplicating a cube was also proved at about that same time. For additional historical background on the trisection problem, see this Wikipedia article.

In most textbooks, these impossibility proofs are presented only after extensive background in groups, rings, fields, field extensions and Galois theory. Needless to say, very few college-level students, even among those studying majors such as physics or engineering, ever take such coursework. This is typically accepted as an unpleasant but necessary aspect of mathematical pedagogy.

We present here a proof of the impossibility of trisecting an arbitrary angle by ruler-and-compass construction and, as a bonus, the impossibility of cube duplication. We emphasize that this proof is both *elementary* and *complete* — all necessary lemmas and nontrivial facts are proved below. It should be understandable by anyone with a solid high school background in algebra, geometry and trigonometry, although it would help if the reader is familiar with the integers modulo $n$ and algebraic fields. This proof is somewhat longer than others in this series, but this is mainly due to the inclusion of several illustrative examples, as well as proofs of facts, such as the rational root theorem and the formula for cosine of a triple angle, that some might consider “obvious.” The core of the proof (see “Proof of the main theorems” below) is actually the *shortest* of those published so far in this series.

**Gist of the proof:**

We will show that trisecting a 60 degree angle, in particular, is equivalent to constructing the number $\cos (\pi/9)$ (i.e., the cosine of 20 degrees), which is an algebraic number that satisfies an irreducible polynomial of degree 3. Since the only numbers and number fields that can be produced by ruler-and-compass construction have algebraic degrees that are powers of two, this shows that the trisection of a 60-degree angle is impossible.

We first define some basic notions about polynomials, algebraic numbers and field extensions, and prove three basic lemmas (Lemmas 1, 2 and 3). Readers familiar with these lemmas may skip to the “Constructible numbers” section below.

**Definitions: Algebraic numbers, irreducible polynomials, minimal polynomials and degrees.**

By an *algebraic number*, we mean some real or complex number $\alpha$ that is the root of a polynomial $a_0 + a_1 x + a_2 x^2 + \cdots + a_m x^m$ with integer coefficients $a_i$. We will assume here that the $a_i$ have no common factor, since if they do all coefficients can be divided by this factor, and $\alpha$ is still a root. Note also that if $\alpha$ is the root of a polynomial with rational coefficients, then by multiplying all coefficients by the greatest common multiple of the denominators of the $a_i$, one obtains a polynomial with integer coefficients that has $\alpha$ as a root. An *irreducible polynomial* is a polynomial with integer coefficients that cannot be factored into two polynomials of lesser degrees. The *minimal polynomial* of an algebraic number $\alpha$ is the polynomial with integer coefficients of minimal degree (i.e., the irreducible polynomial) that has $\alpha$ as a root, and the *degree* of $\alpha$, denoted ${\rm deg}(\alpha)$, is the degree of its minimal polynomial. Note that if $a_0 + a_1 x + a_2 x^2 + \cdots + a_m x^m$ is the minimal polynomial of $\alpha$, then neither $a_0$ nor $a_m$ can be zero, since otherwise $\alpha$ would satisfy a polynomial of lower degree.

**LEMMA 1 (The rational root theorem)**: If $x = p / q$ is a rational root of a polynomial $a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$, where $a_i$ are integer coefficients with neither $a_n$ nor $a_0$ zero (otherwise the polynomial is equivalent to one of lesser degree), and where $p$ and $q$ are integers without any common factor, then $p$ divides $a_0$ and $q$ divides $a_n$.

**Proof**: By hypothesis, $a_n (p/q)^n + a_{n-1} (p/q)^{n-1} + \cdots + a_1 (p/q) + a_0 = 0.$ After multiplying through by $q^n$, taking the lowest-order (constant) term to the right-hand side and factoring out $p$ from the other terms, we obtain $$p (a_n p^{n-1} + a_{n-1} p^{n-2}q + \cdots + a_1 q^{n-1}) = – a_0 q^n.$$ Since the large expression in parentheses is an integer, $p$ divides the left-hand side and thus also the right-hand side. But since $p$ and $q$ have no common factors, $p$ cannot divide $q^n$. Thus $p$ must divide $a_0$. Similarly, by taking the highest-order term to the right-hand side and factoring out $q$ from the other terms, we can write $$q (a_{n-1} p^{n-1} + a_{n-2} p^{n-2} q + \cdots + a_0 q^{n-1}) = – a_n p^n.$$ But again, since the expression in parentheses is an integer, $q$ divides the left-hand side and thus must also divide the right-hand side. Since $q$ cannot divide $p^n$, it must divide $a_n$.

**Examples**: The constant $1 + \sqrt{2}$ satisfies the polynomial $x^2 – 2 x – 1$, so that its algebraic degree cannot exceed two. By Lemma 1, the only possible rational roots are $x = 1$ and $x = -1$, and neither satisfies $x^2 – 2x – 1 = 0$. Thus the polynomial $x^2 – 2 x – 1$ is irreducible and is the minimal polynomial of $1 + \sqrt{2}$, so that ${\rm deg}(1 + \sqrt{2}) = 2$. As another example, consider $\sqrt[3]{2}$, which satisfies $x^3 – 2 = 0$. If this polynomial factors at all over the rational numbers, it has a linear factor and thus a rational root. But by Lemma 1, the only possible rational roots are $x = \pm 1$ and $x = \pm 2$, and none of these satisfies $x^3 – 2 = 0$. Thus $x^3 – 2$ is irreducible and is the minimal polynomial of $\sqrt[3]{2}$, so that ${\rm deg}(\sqrt[3]{2}) = 3$.

**Definitions: Algebraic fields.**

By an *algebraic field*, we will mean, for our purposes here, a set of numbers $F$ that includes the rational numbers $R$, with the usual four arithmetic operations defined, and with the property that if $x$ and $y$ are in $F$, then so are the sum $x+y$, difference $x-y$, product $x y$, and, if $x \neq 0$, the reciprocal $1/x$. There is a rich theory of algebraic fields, but we will not require anything here more than this definition and two basic lemmas (Lemmas 2 and 3), which we prove here.

**LEMMA 2 (Algebraic extensions are fields)**: Let $R$ be the rational numbers, and let $\alpha$ be algebraic of degree $m \ge 2$, with minimal polynomial $P(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_m x^m$. Then the set $S = \{r_0 + r_1 \alpha + r_2 \alpha^2 + \cdots + r_{m-1} \alpha^{m-1}\}$, where $r_i \in R$, is a field.

**Proof**: If $s = s_0 + s_1 \alpha + s_2 \alpha^2 + \cdots + s_{m-1} \alpha^{m-1}$, where $s_i \in R$, and $t = t_0 + t_1 \alpha + t_2 \alpha^2 + \cdots + t_{m-1} \alpha^{m-1}$, where $t_i \in R$ are elements of $S$, their sum $s + t = (s_0 + t_0) + (s_1 + t_1) \alpha + \cdots + (s_{m-1} + t_{m-1}) \alpha^{m-1}$ is clearly in $S$, as is their difference. We can write their product as: $$st = s_0 t_0 + (s_0 t_1 + s_1 t_0) \alpha + (s_0 t_2 + s_1 t_1 + s_2 t_0) \alpha^2 + \cdots + s_{m-1} t_{m-1} \alpha^{2m-2}.$$ This expression involves terms with $\alpha^k$ for $k \ge m$, but note that the term with $\alpha^m$ can be rewritten by solving the minimal polynomial $P(\alpha) = 0$ to obtain $\alpha^m = -(a_0/a_m) – (a_1/a_m) \alpha – \cdots – (a_{m-1}/a_m) \alpha^{m-1}$; higher powers of $\alpha$ can similarly be rewritten in terms of $\alpha^k$ for $k \le m-1$. Thus the product $st$ is in $S$. The reciprocal of $s = Q(\alpha) = s_0 + s_1 \alpha + s_2 \alpha^2 + \cdots + s_{m-1} \alpha^{m-1}$ can be calculated by applying the extended Euclidean algorithm to the polynomials $P(x)$ and $Q(x)$, where polynomial long division, dropping the remainder, is used instead of ordinary division. In this case, where $P(x)$ and $Q(x)$ have no common factor (since $P(x)$ is irreducible), the extended Euclidean algorithm produces polynomials $A(x)$ and $B(x)$, of lower degree than $P(x)$, such that $A(x) P(x) + B(x) Q(x) = 1$. Substituting $x = \alpha$ and recalling that $P(\alpha) = 0$, we have $B(\alpha) Q(\alpha) = 1$, so that $1/s = 1 / Q(\alpha) = B(\alpha)$, which is in $S$ since ${\rm deg}(B(x)) \le m – 1$. This proves that $S$ is a field.

**The extended Euclidean algorithm**: The extended Euclidean algorithm is best illustrated by using it to find the reciprocal (multiplicative inverse) of an integer $m$ in the field of integers modulo $p$, where $p$ is a prime. By the “field of integers modulo $p$,” we mean the set of integers $\{0, 1, 2, \cdots, p – 1\}$, where multiples of $p$ are added to or subtracted from the results of the operations $+, \, -, \, \times$ to reduce them to the range $(0, 1, 2, \cdots, p – 1)$. We will illustrate the case $m = 100$ and $p = 257$ (a prime). First start with a column vector containing $100$ and $257$, together with a $2 \times 2$ identity matrix. Then operate as follows, where ${\rm int}(\cdot)$ denotes integer part, where $R_1$ and $R_2$ denote the first and second rows of the column vector, and where $C_1$ and $C_2$ denote the first and second columns of the $2 \times 2$ matrix:

\begin{align*}

\left( \begin{array}{r} 100 \\ 257 \end{array} \right) & \quad

\left( \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right) & {\rm int}(257/100) = 2, \; {\rm so} \; R_2 \leftarrow R_2 – 2 \cdot R_1 \; {\rm and} \; C_1 \leftarrow C_1 + 2 \cdot C_2. \\

\left( \begin{array}{r} 100 \\ 57 \end{array} \right) & \quad

\left( \begin{array}{rr} 1 & 0 \\ 2 & 1 \end{array} \right) & {\rm int}(100/57) = 1, \; {\rm so} \; R_1 \leftarrow R_1 – 1 \cdot R_2 \; {\rm and} \; C_2 \leftarrow C_2 + 1 \cdot C_1. \\

\left( \begin{array}{r} 43 \\ 57 \end{array} \right) & \quad

\left( \begin{array}{rr} 1 & 1 \\ 2 & 3 \end{array} \right) & {\rm int}(57/43) = 1, \; {\rm so} \; R_2 \leftarrow R_2 – 1 \cdot R_1 \; {\rm and} \; C_1 \leftarrow C_1 + 1 \cdot C_2. \\

\left( \begin{array}{r} 43 \\ 14 \end{array} \right) & \quad

\left( \begin{array}{rr} 2 & 1 \\ 5 & 3 \end{array} \right) & {\rm int}(43/14) = 3, \; {\rm so} \; R_1 \leftarrow R_1 – 3 \cdot R_2 \; {\rm and} \; C_2 \leftarrow C_2 + 3 \cdot C_1. \\

\left( \begin{array}{r} 1 \\ 14 \end{array} \right) & \quad

\left( \begin{array}{rr} 2 & 7 \\ 5 & 18 \end{array} \right) & {\rm int}(14/1) = 14, \; {\rm so} \; R_2 \leftarrow R_2 – 14 \cdot R_1 \; {\rm and} \; C_1 \leftarrow C_1 + 14 \cdot C_2. \\

\left( \begin{array}{r} 1 \\ 0 \end{array} \right) & \quad

\left( \begin{array}{rr} 100 & 7 \\ 257 & 18 \end{array} \right) & {\rm one \; entry \; of \; vector \; is \; zero, \; so \; process \; ends; \; result \; = \; 18.}

\end{align*}

Thus $100^{-1} \bmod 257 = 18$. Indeed, $18 \cdot 100 – 7 \cdot 257 = 1$. Note that $18$ appears in the final matrix, in the opposite corner from the input value $100$. If the number of iterations performed is odd, the sign of the result is changed (not necessary in above example). The polynomial version of the extended Euclidean algorithm operates in exactly the same way, except: (a) the division operation is replaced by polynomial long division, dropping the remainder; and (b) if at termination the nonzero constant in the column vector is not $1$, then the result must be divided by this constant.

**Example**: Let $\alpha$ be a root of the minimal polynomial $P(x) = 1 + x + x^3$ (note that $P(x)$ is irreducible by Lemma 1), and define $S = \{r_0 + r_1 \alpha + r_2 \alpha^2\}$ for $r_i \in R$. Let $s = 1 + 2 \alpha + \alpha^2$ and $t = 1 + \alpha^2$ be two typical elements of $S$. Clearly their sum $s + t = 2 + 2 \alpha + 2 \alpha^2$ is in $S$, as is their difference. Their product $s t = 1 + 2 \alpha + 2 \alpha^2 + 2 \alpha^3 + \alpha^4$. Since $P(\alpha) = 0$, we can solve the minimal polynomial to write $\alpha^3 = -1 – \alpha$ and $\alpha^4 = – \alpha – \alpha^2$, and rewrite the product as $s t = -1 -\alpha + \alpha^2$, which then is in $S$. To find the reciprocal of $s = 1 + 2 \alpha + \alpha^2$, we apply the extended Euclidean algorithm to the polynomials $P(x) = 1 + x + x^3$ and $Q(x) = 1 + 2 x + x^2$. The algorithm proceeds as follows, where ${\rm quot}(\cdot,\cdot)$ denotes polynomial quotient, dropping remainder:

\begin{align*}

\left( \begin{array}{c} x^2 + 2 x + 1 \\ x^3 + x + 1 \end{array} \right) & \quad \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) & \begin{array}{c} {\rm quot}((x^3 + x + 1), (x^2 + 2 x + 1)) = x – 2, \; {\rm so} \\ R_2 \leftarrow R_2 – (x – 2) \cdot R_1 \; {\rm and} \; C_1 \leftarrow C_1 + (x – 2) \cdot C_2. \end{array} \\

\left( \begin{array}{c} x^2 + 2 x + 1 \\ 4 x + 3 \end{array} \right) & \quad \left( \begin{array}{cc} 1 & 0 \\ x – 2 & 1 \end{array} \right) & \begin{array}{c} {\rm quot}((x^2 + 2 x + 1), (4 x + 3)) = x / 4 + 5/16, \; {\rm so} \\ R_1 \leftarrow R_1 – (x/4 + 5/16) \cdot R_2 \; {\rm and} \; C_2 \leftarrow C_2 + (x/4 + 5/16) \cdot C_1. \end{array} \\

\left( \begin{array}{c} 1/16 \\ 4 x + 3 \end{array} \right) & \quad \left( \begin{array}{cc} 1 & x/4 + 5/16 \\ x – 2 & x^2/4 – 3 x/16 + 3/8 \end{array} \right) & \begin{array}{c} {\rm quot}((4x + 3), 1/16) = 64 x + 48, \; {\rm so} \\ R_1 \leftarrow R_1 – (64 x + 48) \cdot R_2 \; {\rm and} \; C_2 \leftarrow C_2 + (64 x + 48) \cdot C_1. \end{array} \\

\left( \begin{array}{c} 1/16 \\ 0 \end{array} \right) & \quad \left( \begin{array}{cc} 16 + 32 x + 16 x^2 & x / 4 + 5/16 \\ 16 x^3 + 16 x + 16 & x^2/4 – 3 x /16 + 3/8 \end{array} \right) & {\rm process \; ends; \; result} \; = \; x^4 / 4 – 3 x / 16 + 3 / 8. \\

\end{align*}

After dividing the result $(x^4 / 4 – 3 x / 16 + 3 / 8)$ by $1/16$, we obtain $B(x) = 4 x^2 – 3 x + 6$, so that $1/s = B(\alpha) = 6 – 3 \alpha + 4 \alpha^2$. Indeed, $(6 – 3 \alpha + 4 \alpha^2)s = (6 – 3 \alpha + 4 \alpha^2)(1 + 2 \alpha + \alpha^2) = 6 + 9 \alpha + 4 \alpha^2 + 5 \alpha^3 + 4 \alpha^4$, which, after recalling $\alpha^3 = -1 – \alpha$ and $\alpha^4 = -\alpha – \alpha^2$, gives $1$.

**Definitions: Linear independence and degrees.**

Suppose that we have a field $S$ that contains the rational numbers $R$ (i.e., $R \subset S$), such that every $s \in S$ can be written as $s = r_1 \alpha_1 + r_2 \alpha_2 + \cdots + r_m \alpha_m$, with $r_i \in R$, and with the property that the $\alpha_i$ are *linearly independent*: if $r_1 \alpha_1 + r_2 \alpha_2 + \cdots + r_n \alpha_m = 0$, with $r_i \in R$, then each $r_i = 0$. Such an extension is said to be of *degree* $m$, which we will denote here as ${\rm deg}(S / R) = m$ (this is also denoted $[S:R] = m$).

Suppose $\alpha$ is algebraic of degree $m$, with minimal polynomial $a_0 + a_1 x + a_2 x^2 + \cdots + a_m x^m$. Then consider the field $S$ consisting of all $s$ that can be written $s = r_0 + r_1 \alpha + r_2 \alpha^2 + \cdots + r_{m-1} \alpha^{m-1}$, with $r_i \in R$. Since there are $m$ terms, it is clear that ${\rm deg}(S/R) \leq m$. Now suppose that for some choice of $r_0, r_1, \cdots, r_{m-1}$, that $r_0 + r_1 \alpha + r_2 \alpha^2 + \cdots + r_{m-1} \alpha^{m-1} = 0$. Since the minimal polynomial of $\alpha$ has degree $m$, and not degree $m-1$ or less, this can only happen if each $r_i = 0$. Thus ${\rm deg}(S/R) = m$ exactly. In other words, *the notions of algebraic degree and field extension degree coincide when the field is generated by an algebraic number*.

**Example**: Consider the case $S = R(\sqrt[3]{2})$, where $S$ is the set of all $s$ that can be written $s = r_0 + r_1 \sqrt[3]{2} + r_2 (\sqrt[3]{2})^2$, with $r_i \in R$. Clearly $R \subset S$, and, since there are three terms, ${\rm deg}(S/R) \leq 3$. Now suppose that for some choice of $r_0, r_1, r_2$ that $r_0 + r_1 \sqrt[3]{2} + r_2 (\sqrt[3]{2})^2 = 0$. Since $\sqrt[3]{2}$ satisfies the polynomial $x^3 – 2 = 0$, and this polynomial is irreducible, as we showed above, $x^3 – 2$ is the minimal polynomial of $\sqrt[3]{2}$ and ${\rm deg}(\sqrt[3]{2}) = 3$. Since the minimal polynomial of $\sqrt[3]{2}$ is of degree $3$ and not $2$ or less, it follows that $r_0 + r_1 \sqrt[3]{2} + r_2 (\sqrt[3]{2})^2 = 0$ only when $r_0 = r_1 = r_2 = 0$. Thus ${\rm deg}(S/R) = 3$ exactly.

We now prove an important fact about the degrees of field extensions, which will be key to our main impossibility proofs below:

**LEMMA 3 (Multiplication of degrees of field extensions)**: Suppose that we have three fields $R, S, T$ (where $R$ is the field of rational numbers as before) such that $R \subset S \subset T$, and with ${\rm deg}(S/R) = m$ and ${\rm deg}(T/S) = n$. Then ${\rm deg} (T/R) = {\rm deg} (S/R) \cdot {\rm deg} (T/S) = m \cdot n$.

**Proof**: For ease of notation, we will prove this in the specific case $m = 2, n = 3$, i.e., where ${\rm deg} (S/R) = 2$ and ${\rm deg} (T/S) = 3$, although the argument is completely general. By hypothesis, all $s \in S$ can be written as $s = r_1 \alpha_1 + r_2 \alpha_2$, where $r_i \in R$ are rational numbers, and with the property that if $r_1 \alpha_1 + r_2 \alpha_2 = 0$, then each $r_i = 0$. In a similar way, by hypothesis all $t \in T$ can be written as $t = s_1 \beta_1 + s_2 \beta_2 + s_3 \beta_3$, where $s_i \in S$, and with the property that if $s_1 \beta_1 + s_2 \beta_2 + s_3 \beta_3 = 0$, then each $s_i = 0$.

First note that when we write any $t \in T$ as $t = s_1 \beta_1 + s_2 \beta_2 + s_3 \beta_3,$ where $s_1, s_2, s_3 \in S$, each of these $s_i$ can in turn be rewritten in terms of $r_i$ and $\alpha_i$: in particular, $s_1 = a_1 \alpha_1 + a_2 \alpha_2$, for some $a_i \in R$; $s_2 = b_1 \alpha_1 + b_2 \alpha_2$, for some $b_i \in R$; and $s_3 = c_1 \alpha_1 + c_2 \alpha_2$, for some $c_i \in R$. Thus we can write $$t = (a_1 \alpha_1 + a_2 \alpha_2) \beta_1 + (b_1 \alpha_1 + b_2 \alpha_2) \beta_2 + (c_1 \alpha_1 + c_2 \alpha_2) \beta_3 $$ $$= a_1 (\alpha_1 \beta_1) + a_2 (\alpha_2 \beta_1) + b_1 (\alpha_1 \beta_2) + b_2 (\alpha_2 \beta_2) + c_1 (\alpha_1 \beta_3) + c_2 (\alpha_2 \beta_3),$$ where $a_1, a_2, b_1, b_2, c_1, c_2 \in R$. We now see that we can write any $t \in T$ as a linear sum of the six terms shown, with coefficients in $R$. Thus ${\rm deg}(T/R) \leq 6$. Now suppose that some $t \in T$ is zero, so that $$t = (a_1 \alpha_1 + a_2 \alpha_2) \beta_1 + (b_1 \alpha_1 + b_2 \alpha_2) \beta_2 + (c_1 \alpha_1 + c_2 \alpha_2) \beta_3 = 0.$$ By hypothesis, this can only happen if each of the three coefficient expressions $(a_1 \alpha_1 + a_2 \alpha_2)$, $(b_1 \alpha_1 + b_2 \alpha_2)$ and $(c_1 \alpha_1 + c_2 \alpha_2)$ is zero, which in turn is only possible if each $a_1, a_2, b_1, b_2, c_1, c_2$ is zero. Thus the six terms $\alpha_1 \beta_1, \alpha_2 \beta_1, \alpha_1 \beta_2, \alpha_2 \beta_2, \alpha_1 \beta_3$ and $\alpha_2 \beta_3$ are linearly independent, so that ${\rm deg}(T/R) = 6$ exactly.

The more general case where ${\rm deg} (S/R) = m$ and ${\rm deg} (T/S) = n$ is handled in exactly the same way, only with somewhat more complicated notation.

**Definitions: Constructible numbers.**

The next step is to establish what kinds of numbers can be constructed by ruler and compass. Readers familiar with the mathematics of ruler-and-compass constructions may skip to the “Angle trisection” section below.

First, it is clear that addition can be easily done with ruler and compass by marking the two distances on a straight line, then the combined distance is the sum. Subtraction can be done similarly. Multiplication can also be done by ruler and compass, as illustrated in the figure to the right (taken from here), and division is just the inverse of this process.

What’s more, one can compute square roots by ruler and compass. In the second illustration to the right (taken from here), assuming that $AC = 1$, simple properties of right triangles show that the distance $AD = \sqrt{AB}$.

Thus, starting with a distance of unit length, numbers composed of any finite combination of the four arithmetic operations and square roots can be constructed. The converse is also true: Such numbers are the *only* numbers that can be constructed, as we will now show.

**LEMMA 4 (Degrees of constructible numbers)**: If $S$ is the number field resulting from a finite sequence of ruler-and-compass constructions, then ${\rm deg}(S/R) = 2^m$ for some integer $m$.

**Proof**: Ruler-and-compass constructions can only extend the rational number field by a sequence of one of the following operations, each of which has algebraic degree 1 or 2 over the field generated by the previous operation:

- Finding the $(x,y)$ coordinates of the intersection of two lines is equivalent to solving a set of two linear equations in two unknowns, which involves only the four arithmetic operations. Note that this does
*not*involve any field extension. - Finding the $(x,y)$ coordinates of the intersection of a line with a circle is equivalent to solving a system of a linear equation and a quadratic equation, which requires only arithmetic and one square root.
- Finding the $(x,y)$ coordinates of the intersection of two circles is equivalent to solving two simultaneous quadratic equations, which again requires only arithmetic and one square root. See any high school analytic geometry text for details.
- Copying the distance between any two constructed points to the $x$ axis is equivalent to computing the distance between the two points, which requires only arithmetic and one square root.

Thus, by Lemma 3, the algebraic degree over the rationals of the final number field so constructed, after a sequence of $n$ ruler-and-compass operations, is $2^m$ for some integer $m \le n$.

**Definitions: Angle trisection.**

We first must keep in mind that some angles *can* be trisected. For example, a right angle can be trisected, because a 30 degree angle can be constructed, simply by bisecting one of the interior angles of an equilateral triangle. To show the impossibility result, we need only exhibit a single angle that cannot be trisected. We choose a 60 degree angle (i.e., $\pi/3$ radians). In other words, we will show that a 20 degree angle (i.e., $\pi/9$ radians) cannot be constructed. Note that it is sufficient to prove that the number $x = \cos(\pi/9)$ cannot be constructed, because if it could, then by copying this distance to the $x$ axis, with the left end at the origin, then constructing a perpendicular from the right end to the unit circle, the resulting angle will be the desired trisection.

Here we require only an elementary fact of trigonometry, namely the formula for the cosine of a triple angle. For convenience, we will prove this and some related formulas here, based only on elementary geometry and a little algebra. Readers familiar with these formulas may skip to the “Proof of the main theorems” section below.

**LEMMA 5 (Triple angle formula for cosine)**: $\cos(3\alpha) = 4 \cos^3(\alpha) – 3 \cos(\alpha).$

**Proof**: We start with the formula $$\sin (\alpha + \beta) = \sin (\alpha) \cos (\beta) + \cos (\alpha) \sin (\beta).$$ This fact has a simple geometric proof, which is illustrated to the right, where $OP = 1$. First note that $RPQ = \alpha, \, PQ = \sin(\beta)$ and $OQ = \cos (\beta)$. Further, $AQ/OQ = \sin(\alpha)$, so $AQ = \sin(\alpha) \cos(\beta)$, and $PR/PQ = \cos(\alpha)$, so $PR = \cos(\alpha) \sin(\beta)$. Combining these results, $$\sin(\alpha + \beta) = PB = RB + PR = AQ + PR = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta).$$ [This proof and illustration are taken from here.] By applying this formula to $(\pi/2 – \alpha)$ and $(-\beta)$, one obtains the related formula: $$\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) – \sin(\alpha) \sin(\beta).$$ From these formulas one immediately obtains the formulas $\sin(2\alpha) = 2 \sin(\alpha) \cos(\alpha)$ and $\cos(2\alpha) = 2 \cos^2(\alpha) – 1$. Now we can write: $$\cos(3\alpha) = \cos (\alpha + 2\alpha) = \cos(\alpha) \cos(2\alpha) – \sin(\alpha) \sin(2\alpha) = \cos(\alpha) (2 \cos^2(\alpha) – 1) – 2 \sin^2(\alpha)\cos(\alpha)$$ $$= \cos(\alpha) (2 \cos^2(\alpha) – 1) – 2 (1 – \cos^2(\alpha))\cos(\alpha) = 4 \cos^3(\alpha) – 3 \cos(\alpha).$$

**Proof of the main theorems:**

**THEOREM 1**: There is no ruler-and-compass scheme to trisect an arbitrary angle.

**Proof**: As mentioned above, we will show that the angle $\pi/3$ radians or 60 degrees cannot be trisected, or, in other words, that the number $\cos (\pi/9)$ cannot be constructed. From Lemma 5 we can write $$\cos (3(\pi/9)) = \cos (\pi/3) = 1/2 = 4 \cos^3(\pi/9) – 3 \cos(\pi/9),$$ so that $\cos(\pi/9)$ is a root of the polynomial equation $8 x^3 – 6 x – 1 = 0$. If this polynomial factors at all, it must have a linear factor and hence a rational root. But by Lemma 1, the only possible rational roots are $x = \pm 1, \, x = \pm 1/2, \, x = \pm 1/4$ and $x = \pm 1/8$, none of which satisfies $8 x^3 – 6 x – 1 = 0$. Thus $8 x^3 – 6 x – 1$ is irreducible and is the minimal polynomial of $\cos(\pi/9)$, so that $\deg(\cos(\pi/9)) = 3$. Hence any extension field $F$ containing $\cos(\pi/9)$ must have $\deg(F/R) = 3n$ for some integer $n$. However, by Lemma 4 the only possible degrees for number fields composed by ruler-and-compass constructions are powers of two. Since $2^m$ is not divisible by $3$, no finite sequence of ruler-and-compass constructions can compose an angle of $\pi/9$ or 20 degrees, which is the trisection of the angle $\pi/3$ or 60 degrees.

**THEOREM 2**: There is no ruler-and-compass scheme to duplicate an arbitrary cube.

**Proof**: This follows by the same line of reasoning, since duplication of a cube is equivalent to constructing the number $\sqrt[3]{2}$, which is a root of the polynomial equation $x^3 – 2 = 0$. As noted in one of the examples above, if $x^3 – 2$ factors at all, one of the factors must be linear, yielding a rational root. But by Lemma 1, the only possible rational roots are $x = \pm 1$ and $x = \pm 2$, and none of these satisfies $x^3 – 2 = 0$. Thus $x^3 – 2$ is irreducible and is the minimal polynomial of $\sqrt[3]{2}$, so that ${\rm deg}(\sqrt[3]{2}) = 3$. Hence any extension field $F$ containing $\sqrt[3]{2}$ must have $\deg(F/R) = 3n$ for some integer $n$.

For other proofs in this series see the listing at Simple proofs of great theorems.