**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. However, as we will see below, the impossibility of trisection or cube duplication, at least, can be proven much more simply.

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 has previously been introduced to the notion of an algebraic field. 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 two basic lemmas (Lemma 1 and Lemma 2). Readers familiar with these two lemmas may skip to the “Constructible numbers” section below.

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

By an *algebraic number*, we mean some real or complex number $\alpha$ that is the root of a polynomial equation with integer coefficients. The *minimal polynomial* of $\alpha$, namely $P(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_m x^m$, is the polynomial with integer coefficients of minimal degree that has $\alpha$ as a root, and the *degree* of $\alpha$, denoted ${\rm deg}(\alpha)$, is the degree $m$ of its minimal polynomial. We will assume here that the integer coefficients $a_i$ have no common factor, since if they do all coefficients can be divided by this factor.

**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**: $\sqrt{3}$ satisfies the equation $x^2 – 3 = 0$, so that its minimal polynomial is of degree not greater than two. By Lemma 1, any possible rational root $x = p/q$, where $p$ and $q$ have no common factor, would need $p$ dividing 3 and $q$ dividing one. Since there are no such integers $p$ and $q$ that satisfy this requirement and $(p/q) ^2 – 3 = 0$, the polynomial $x^2 – 3$ is irreducible. Thus $x^2 – 3$ is the minimal polynomial of $\sqrt{3}$, and hence $\sqrt{3}$ is irrational. Similarly, $1 + \sqrt{2}$ satisfies the polynomial $x^2 – 2 x – 1$, so that its algebraic degree cannot exceed two, and since this polynomial is *irreducible*, its degree cannot be less than two. Irreducibility can be seen in this case by checking the various possibilities for integers $a, b, c, d$ such that $x^2 – 2 x – 1 = (ax + b) (cx + d) = ac x^2 + (ad + bc) x + bd$, where the coefficients of $x^2, \, x$ and the constant term must match term by term on both sides of the equation. By Lemma 1, the only possibilities for $a, b, c, d$ are $1$ or $-1$, and none of these combinations give $-2x$ for the middle term. Thus $x^2 – 2x – 1$ is irreducible, and its roots are irrational.

**Definitions: Algebraic fields and extensions.**

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 is 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 simple definition, the notions of linear independence and degrees, which we define below, and a basic fact (Lemma 2), which we prove below.

**Examples**: The simplest examples of fields are extensions of the rationals generated by one or more algebraic numbers. Suppose, for example, that $\alpha$ is algebraic of degree $m \ge 2$, satisfying a minimal polynomial $P(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_m x^m$. Let the set $S$ be the set of all $s$ that can be written $s = r_0 + r_1 \alpha + r_2 \alpha^2 + \cdots + r_{m-1} \alpha^{m-1}$, where $r_i \in R$ are rational. It is fairly easy to show that $S$ is closed under addition, subtraction and multiplication, and that the reciprocal of any nonzero value is also in $S$. Note, for instance, that if $\alpha$ is algebraic of degree $m$, then the constant term $a_0$ of its minimal polynomial $P(x)$ cannot be zero (otherwise its degree would be $m-1$), so that we can solve the equation $P(\alpha) = a_0 + a_1 \alpha + a_2 \alpha^2 + \cdots + a_m \alpha^m = 0$ to obtain $1/\alpha = -(a_1/a_0) – (a_2/a_0) \alpha – (a_3/a_0) \alpha^2 – \cdots – (a_m/a_0) \alpha^{m-1}$, which is in the set $S$. Thus $S$ is a field.

**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*: This means that there are no linear relations among the $\alpha_i$, so that in particular, 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 $[S:R] = m$).

**Examples**: For example, suppose $\alpha$ is algebraic of degree $m$, with minimal polynomial $a_0 + a_1 \alpha + a_2 \alpha^2 + \cdots + a_m \alpha^m$. 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*.

As a specific example, let $R$ be the rational numbers, and consider the extension $S = R(\sqrt{2}) = \{s = r_0 \cdot 1 + r_1 \cdot \sqrt{2}\}$ with $r_0, r_1 \in R$. If $s, t \in S$, then clearly the sum and difference are also in $S$. So is their product: note, for instance, if $s = 2 + 3 \sqrt{2}$ and $t = 1 – 2 \sqrt{2}$, that the product $st = (2 + 3 \sqrt{2})(1 – 2 \sqrt{2}) = 2 – 4 \sqrt{3} + 3 \sqrt{3} – 12 = -10 – \sqrt{2}$, which is in $S$. Further, by “rationalizing the denominator,” we can write $1/s = 1/(2 + 3 \sqrt{2}) = (2 – 3 \sqrt{2})/((2 + 3 \sqrt{2})(2 -3 \sqrt{2})) = -1/7 + 3/14 \sqrt{2}$, which is also in $S$. Now note that the numbers $\{1, \sqrt{2}\}$ are linearly independent, since if $r_0 + r_1 \sqrt{2} = 0$ and $r_1 \ne 0$, this means that $\sqrt{2} = – r_0 / r_1$, which clearly cannot happen because $\sqrt{2}$ is irrational (this also follows by Lemma 1). Thus $r_0 = 0$ and $r_1 = 0$, so that ${\rm deg} (S/R) = 2$.

Similarly, consider the case where $S = R(\sqrt[3]{2})$, 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 by Lemma 1, $\sqrt[3]{2}$ cannot satisfy any polynomial with coefficients in $R$ with degree 2 or less. Thus it follows that $r_0 + r_1 \sqrt[3]{2} + r_2 (\sqrt[3]{2})^2 = 0$ only when $r_0 = 0, \, r_1 = 0, \, r_2 = 0$ and thus ${\rm deg}(S/R) = 3$ exactly. This also proves rigorously that $\sqrt[3]{2}$ is irrational.

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

**LEMMA 2 (Multiplication of degrees of field extensions)**: Suppose that we have three fields $R, S, T$ (where $R$ is the 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 of the form $2/3 + \sqrt{1 + \sqrt{5/2}}$, or any finite combination of operations involving 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 3 (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 numbers by a sequence of square root operations, each of which has algebraic degree 1 or 2:

- The intersection of two lines is algebraically the solution of two sets of linear equations, which involves only addition, subtraction, multiplication and division. Note that this does
*not*involve any field extension. - The intersection of a line with a circle involves the solution of a system of a linear equation and a quadratic equation, which involves only arithmetic and square roots.
- The intersection of two circles involves the solution of two simultaneous quadratic equations, which again involves only arithmetic and square roots. See any high school analytic geometry text for details.

**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 4 (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 $OP = 1$. Note that angle $RPQ = \alpha$, because $OQA = \pi/2 – \alpha$, so that $RQO = \alpha$, and $RPQ = \pi/2 – RQP = \pi/2 – (\pi/2 – RQO) = RQO = \alpha$. Then note that $PQ = \sin(\beta), \, OQ = \cos (\beta)$. 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, we have $$\sin(\alpha + \beta) = PB = RB + PR = AQ + PR = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta).$$ [This proof and illustration are taken from here.] By applying this formula to $(\pi/2 – \alpha)$ and $-\beta$, we obtain the formula: $$\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) – \sin(\alpha) \sin(\beta).$$ From these formulas we immediately derive the double-angle 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**: From Lemma 4 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 $8 x^3 – 6 x – 1$. A quick check of possible integer coefficients $a, b, c, d, e$ for the equality $8 x^3 – 6 x – 1 = (ax + b) (cx^2 + dx + e) = ac x^3 + (ad + bc)x^2 + (ae + bd)x + be$, attempting to match coefficients term by term on each side, and applying Lemma 1, shows that the polynomial $8 x^3 – 6 x – 1$ is irreducible. In other words, $\cos(\pi/9)$ is an algebraic number whose minimal (irreducible) polynomial is $8 x^3 – 6 x – 1$, so that it lies in an extension field $F$ of the rationals $R$ with $\deg(F/R) = 3$ or some multiple of $3$. However, by Lemma 3 the only possible degrees for number fields constructed by ruler and compass 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 $\sqrt[3]{2}$ is clearly of algebraic degree 3. As noted above, the irreducibility of the associated polynomial $x^3 – 2$ follows by Lemma 1. In particular, if $x^3 – 2$ factors at all, one of the factors must be linear, yielding a rational root. But there is no rational root, since by Lemma 1 any rational root $p/q$, where $p$ and $q$ are integers having no common factor, must have $p$ dividing $3$ and $q$ dividing $1$, and none of the possibilities $(3,1), \, (3, -1), \, (-3, 1)$ or $(-3,-1)$ work.

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