1.13 Constructions with straight-edge and compass.

The Greeks understood integers and the rational numbers. They were surprised to find that the length of the diagonal of a square of side $1$ , namely, $\sqrt{2}$ , is not rational. They thus realized that they needed to extend their number system. They then hoped that the “constructible” numbers would suffice. Suppose that we are given a length, which we call $1$ , a straight-edge, and a compass (device for drawing circles). A real number (better a length) is constructible if it can be constructed by forming successive intersections of

$\diamond$

lines drawn through two points already constructed, and
$\diamond$

circles with centre a point already constructed and radius a constructed length.

This led them to three famous questions that they were unable to answer: is it possible to duplicate the cube, trisect an angle, or square the circle by straight-edge and compass constructions? We’ll see that the answer to all three is negative.

Let $F$ be a subfield of $\mathbb{R}$ . For a positive $a\in F$ , $\sqrt{a}$ denotes the positive square root of $a$ in $\mathbb{R}{}$ . The $F$ -plane is $F\times F\subset\mathbb{R}\times\mathbb{R}$ . We make the following definitions:

An $F$ -line is a line in $\mathbb{R}{}\times\mathbb{R}{}$ through two points in the $F$ -plane. These are the lines given by equations

$ax+by+c=0,\quad a,b,c\in F.$

An $F$ -circle is a circle in $\mathbb{R}{}\times\mathbb{R}{}$ with centre an $F$ -point and radius an element of $F$ . These are the circles given by equations

$(x-a)^{2}+(y-b)^{2}=c^{2},\quad a,b,c\in F.$

Lemma 1.34

Let $L\neq L^{\prime}$ be $F$ -lines, and let $C\neq C^{\prime}$ be $F$ -circles.

(a)

$L\cap L^{\prime}=\emptyset$ or consists of a single $F$ -point.
(b)

$L\cap C=\emptyset$ or consists of one or two points in the $F[\sqrt{e}]$ -plane, some $e\in F$ , $e>0$ .
(c)

$C\cap C^{\prime}=\emptyset$ or consists of one or two points in the $F[\sqrt{e}]$ -plane, some $e\in F$ , $e>0$ .

Proof.

The points in the intersection are found by solving the simultaneous equations, and hence by solving (at worst) a quadratic equation with coefficients in $F$ . _□

Lemma 1.35

(a) If $c$ and $d$ are constructible, then so also are $c+d$ , $-c$ , $cd$ , and $\frac{c}{d}$ $(d\neq 0)$ .

(b) If $c>0$ is constructible, then so also is $\sqrt{c}$ .

Sketch of proof.

First show that it is possible to construct a line perpendicular to a given line through a given point, and then a line parallel to a given line through a given point. Hence it is possible to construct a triangle similar to a given one on a side with given length. By an astute choice of the triangles, one constructs $cd$ and $c^{-1}$ . For (b), draw a circle of radius $\frac{c+1}{2}$ and centre $(\frac{c+1}{2},0)$ , and draw a vertical line through the point $A=(1,0)$ to meet the circle at $P$ . The length $AP$ is $\sqrt{c}$ . (For more details, see Artin, M., Algebra, 1991, Chapter 13, Section 4.) _□

Theorem 1.36

(a)

The set of constructible numbers is a field.
(b)

A number $\alpha$ is constructible if and only if it is contained in a subfield of $\mathbb{R}{}$ of the form

$\mathbb{Q}{}[\sqrt{a_{1}},\ldots,\sqrt{a_{r}}],\quad a_{i}\in\mathbb{Q}{}[% \sqrt{a_{1}},\ldots,\sqrt{a_{i-1}}],\quad a_{i}>0\text{.}$

Proof.

(a) This restates (a) of Lemma 1.35.

(b) It follows from Lemma 1.34 that every constructible number is contained in such a field $\mathbb{Q}{}[\sqrt{a_{1}},\ldots,\sqrt{a_{r}}]$ . Conversely, if all the elements of $\mathbb{Q}{}[\sqrt{a_{1}},\ldots,\sqrt{a_{i-1}}]$ are constructible, then $\sqrt{a_{i}}$ is constructible (by 1.35b), and so all the elements of $\mathbb{Q}{}[\sqrt{a_{1}},\ldots,\sqrt{a_{i}}]$ are constructible (by (a)). Applying this for $i=0,1,\ldots$ , we find that all the elements of $\mathbb{Q}{}[\sqrt{a_{1}},\ldots,\sqrt{a_{r}}]$ are constructible. _□

Corollary 1.37

If $\alpha$ is constructible, then $\alpha$ is algebraic over $\mathbb{Q}$ , and $[\mathbb{Q}[\alpha]\colon\mathbb{Q}]$ is a power of $2$ .

Proof.

According to Proposition 1.20, $[\mathbb{Q}[\alpha]\colon\mathbb{Q}]$ divides

[\mathbb{Q}[\sqrt{a_{1}}]\cdots[\sqrt{a_{r}}]\colon\mathbb{Q}]

and $[\mathbb{Q}[\sqrt{a_{1}},\ldots,\sqrt{a_{r}}]\colon\mathbb{Q}]$ is a power of $2$ . _□

Corollary 1.38

It is impossible to duplicate the cube by straight-edge and compass constructions.

Proof.

The problem is to construct a cube with volume $2$ . This requires constructing the real root of the polynomial $X^{3}-2$ . But this polynomial is irreducible (by Eisenstein’s criterion 1.16 for example), and so $[\mathbb{Q}[\sqrt[3]{2}]\colon\mathbb{Q}]=3$ . _□

Corollary 1.39

In general, it is impossible to trisect an angle by straight-edge and compass constructions.

Proof.

Knowing an angle is equivalent to knowing the cosine of the angle. Therefore, to trisect $3\alpha$ , we have to construct a solution to

\cos 3\alpha=4\cos^{3}\alpha-3\cos\alpha.

For example, take $3\alpha=60$ degrees. As $\cos 60^{\circ}=\frac{1}{2}$ , to construct $\alpha$ , we have to solve $8x^{3}-6x-1=0$ , which is irreducible (apply 1.11), and so $[\mathbb{Q}{}[\alpha]\colon\mathbb{Q}{}]=3$ . _□

Corollary 1.40

It is impossible to square the circle by straight-edge and compass constructions.

Proof.

A square with the same area as a circle of radius $r$ has side $\sqrt{\pi}r$ . Since $\pi$ is transcendental¹¹¹¹ 11 Proofs of this can be found in many books on number theory, for example, in 11.14 of Hardy, G. H., and Wright, E. M., An Introduction to the Theory of Numbers, Fourth Edition, Oxford, 1960., so also is $\sqrt{\pi}$ . _□

We next consider another problem that goes back to the ancient Greeks: list the integers $n$ such that the regular $n$ -sided polygon can be constructed using only straight-edge and compass. Here we consider the question for a prime $p$ (see LABEL:ag11 for the general case). Note that $X^{p}-1$ is not irreducible; in fact

X^{p}-1=(X-1)(X^{p-1}+X^{p-2}+\cdots+1).

Lemma 1.41

If $p$ is prime, then $X^{p-1}+\cdots+1$ is irreducible; hence $\mathbb{Q}[e^{2\pi i/p}]$ has degree $p-1$ over $\mathbb{Q}.$

Proof.

Let $f(X)=(X^{p}-1)/(X-1)=X^{p-1}+\cdots+1$ ; then

f(X+1)=\frac{(X+1)^{p}-1}{X}=X^{p-1}+\cdots+a_{i}X^{i}+\cdots+p,

with $a_{i}=\tbinom{p}{i+1}$ . We know (1.4) that $p|a_{i}$ for $i=1,...,p-2$ , and so $f(X+1)$ is irreducible by Eisenstein’s criterion 1.16. This implies that $f(X)$ is irreducible. _□

In order to construct a regular $p$ -gon, $p$ an odd prime, we need to construct

\cos\tfrac{2\pi}{p}=\frac{e^{\frac{2\pi i}{p}}+e^{-\frac{2\pi i}{p}}}{2}.

Note that

\mathbb{Q}[e^{\frac{2\pi i}{p}}]\supset\mathbb{Q}[\cos\tfrac{2\pi}{p}]\supset% \mathbb{Q}.

The degree of $\mathbb{Q}[e^{\frac{2\pi i}{p}}]$ over $\mathbb{Q}[\cos\frac{2\pi}{p}]$ is $2$ because the equation

\alpha^{2}-2\cos\tfrac{2\pi}{p}\cdot\alpha+1=0,\quad\alpha=e^{\frac{2\pi i}{p}},

shows that it is at most $2$ , and it is not $1$ because $e^{\frac{2\pi i}{p}}\notin\mathbb{R}{}$ . Hence

[\mathbb{Q}[\cos\tfrac{2\pi}{p}]\colon\mathbb{Q}]=\frac{p-1}{2}.

We deduce that, if the regular $p$ -gon is constructible, then $(p-1)/2$ is a power of $2$ ; later (LABEL:ag11) we’ll prove the converse statement. Thus, the regular $p$ -gon is constructible if and only if $p=2^{r}+1$ for some positive integer $r$ .

A number $2^{r}+1$ can be prime only if $r$ is a power of $2$ : if $t$ is odd, then

Y^{t}+1=(Y+1)(Y^{t-1}-Y^{t-2}+\cdots+1)

and so

2^{st}+1=(2^{s}+1)((2^{s})^{t-1}-(2^{s})^{t-2}+\cdots+1)\text{.}

We conclude that the primes $p$ for which the regular $p$ -gon is constructible are exactly those of the form $2^{2^{r}}+1$ for some $r$ . Such $p$ are called Fermat primes (because Fermat conjectured that all numbers of the form $2^{2^{r}}+1$ are prime). For $r=0,1,2,3,4$ , we have $2^{2^{r}}+1=3,5,17,257,65537$ , which are indeed prime, but Euler showed that $2^{32}+1=(641)(6700417)$ , and we don’t know whether there are any more Fermat primes. Thus, we do not know the list of primes $p$ for which the regular $p$ -gon is constructible. See Wikipedia: Fermat number.

Gauss showed that¹²¹² 12 Or perhaps that $\cos\frac{2\pi}{17}=-\frac{1}{16}+\frac{1}{16}\sqrt{17}+\frac{1}{16}\sqrt{34-2% \sqrt{17}}+\frac{1}{8}\sqrt{17+3\sqrt{17}-2\sqrt{34-2\sqrt{17}}-\sqrt{170-26% \sqrt{17}}}$ — both expressions are correct.

\cos\frac{2\pi}{17}=-\frac{1}{16}+\frac{1}{16}\sqrt{17}+\frac{1}{16}\sqrt{34-2% \sqrt{17}}+\frac{1}{8}\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2% \sqrt{17}}}

when he was 18 years old. This success encouraged him to become a mathematician.