The Elements: axioms and constructions

In the last chapter we handled the plane like an old friend: rotated it, stretched it, measured lengths and added angles. And never once asked how we know all this. Draw a triangle, measure its angles with a protractor and add them up. You’ll get something around $180^\circ$, perhaps $179^\circ$ or $181^\circ$. The protractor won’t tell you the sum is exactly $180^\circ$. Still less will it tell you anything about the triangles nobody has ever drawn.

How can you learn something about all figures at once, when you can only draw a few? Around 300 BC Euclid of Alexandria answered this question with a book. It’s called the Elements, it has thirteen parts, and it rests on a simple idea: accept without proof a very small number of obvious statements, and derive everything else from them. Euclid has two tools, compass and straightedge. The rules are strict, the moves few, and the problems sometimes demand real ingenuity. It makes a game, and in this chapter we’ll play it.

The rules of the game

Euclid’s straightedge has no markings: you can’t measure anything with it, only draw a line through two points already on the drawing. With the compass you can draw a circle centred at one marked point and passing through another. New points appear only as intersections of lines and circles. Nothing else is allowed: no “by eye”, no “roughly here”, no set square, no protractor.

Euclid’s compass “collapses”: lift it off the paper and the setting is lost, so you can’t carry a length to a new place. That seems a serious restriction. But the Elements’ very second proposition shows how to lay off from any point a segment equal to a given one using only collapsing moves. So a collapsing compass is no worse than an ordinary one; it just takes longer to work with.

The book’s first problem goes like this: on a given segment, construct an equilateral triangle. Before you read the solution, try it yourself. Tap points on the board: with the compass, first the centre, then the point the circle will pass through; with the straightedge, two points of the line. The dashed outline shows what you need to get. The counter counts every circle and every line, and par is the smallest number of moves the problem can be solved in.

The solution takes four moves: two circles of radius $AB$ centred at $A$ and $B$, and two segments from the ends of $AB$ to the point $C$ where the circles meet. More interesting is why the triangle comes out equilateral. That’s the next section.

Five postulates and five common notions

Book I of the Elements opens with twenty-three definitions. “A point is that which has no part.” “A line is breadthless length.” The first of them are descriptions rather than definitions: no proof relies on them. Next come five statements that Euclid asks us to accept without proof. In Greek they’re called aitemata, requests; in Latin translations, postulates.

The first three are the very rules of our game: the straightedge, extending with the straightedge, the compass. The fourth sounds odd: can right angles be unequal? It says the plane is the same everywhere: a right angle in one place is the same as in any other. The fifth is longer than all the others put together, and the longest story in geometry is tied to it. We’ll come back to it.

After the postulates come five common notions: statements not about figures but about magnitudes in general.

Now we can write out the solution of the first problem the way Euclid does: every step with a reference to what permits it.

Every step is a reference to a postulate, a common notion or a definition. We take nothing from the picture.

Draw the circle with centre $A$ and radius $AB$: postulate 3 permits it. Draw the circle with centre $B$ and radius $BA$: postulate 3 again. Let $C$ be the point where the circles meet. Draw the segments $CA$ and $CB$: postulate 1 permits it. The point $C$ lies on the first circle, so $AC = AB$: all radii of a circle are equal, which is how Euclid defines a circle. The point $C$ also lies on the second circle, so $BC = BA$. The segments $AC$ and $BC$ are equal to the same segment $AB$, and so equal to one another (common notion 1). All three sides of triangle $ABC$ are equal.

Euclid ends constructions with the words “which was to be done”, and theorems with “which was to be proved”. Latin editions shortened these to Q.E.F. and Q.E.D.; the second abbreviation is alive to this day.

Notice what the proof doesn’t contain. There are no words “as can be seen from the figure”. The picture helps you understand, but the argument is only references to postulates and common notions. A little later we’ll see that this proof does have a hole after all, and understand why nobody noticed it for two thousand years.

First moves

The first problem hides a technique that comes in handy almost everywhere. Two circles of the same radius centred at $A$ and $B$ meet at two points, and each of them is the same distance from $A$ as from $B$. The line through these points is special.

The perpendicular bisector immediately gives the midpoint of a segment. And looking at the same picture from the other side, it gives the perpendicular to a line from a point not on it: it’s enough to find the point symmetric to the given one in the line. For an angle there’s a technique of its own.

All four problems below take three or four moves. If you get stuck, remember that a rhombus has all its sides equal, and its diagonals are perpendicular and bisect its angles. And a rhombus made of radii takes the compass just two touches.

In Euclid these problems come in a different order: first he bisects an angle, and only with its help a segment. The order in the Elements is determined not by convenience but by what has been proved up to that point: a proposition can be used only after it has been proved.

On a segment $AB$ of length $6$, two circles of radius $AB$ are drawn with centres $A$ and $B$. Find the distance between their two points of intersection. You can answer exactly (for example, 2sqrt(3)) or approximately, to two decimal places.

Each point of intersection forms, together with $A$ and $B$, an equilateral triangle with side $6$. Its height bisects the base, and by Pythagoras’ theorem (it has chapter 18 to itself) it equals $\sqrt{6^2 - 3^2} = \sqrt{27} = 3\sqrt3$. The points lie on the perpendicular bisector on opposite sides of $AB$, so the distance between them is $2 \cdot 3\sqrt3 = 6\sqrt3 \approx 10.39$.

A book copied for twenty-three centuries

Very little is known about Euclid himself: he taught mathematics in Alexandria, most likely under King Ptolemy I, around 300 BC. Everything else is legend. The best known is told by Proclus, a fifth-century philosopher who wrote a commentary on Book I: the king asked whether there was a shorter way to geometry than the Elements, and Euclid replied that there is no royal road to geometry.

Much of the Elements wasn’t discovered by Euclid. The theory of proportion in Book V is usually linked to Eudoxus, the theory of regular polyhedra to Theaetetus. Euclid’s achievement lies elsewhere: he arranged about 465 propositions into a single chain in which each rests only on those before it. Books I–IV are plane geometry: triangles, parallels, areas, Pythagoras’ theorem, the circle, inscribed polygons. Book V is proportion, VI similarity. Books VII–IX are about numbers; they’re home to the Euclidean algorithm and the proof that there are infinitely many primes. Book X is about incommensurable segments, Books XI–XIII about solids. It all ends with the five regular polyhedra and a proof that there’s no sixth.

The Egyptians and Babylonians knew many facts about figures long before the Greeks. What was new in the Elements was not the content but the structure: knowledge became a building with a foundation and floors. Below is the whole of Book I: 48 propositions on five postulates. Each proposition stands one floor above the highest of those its proof refers to.

Pythagoras’ theorem (I.47) rests on 28 of the 46 propositions before it and on all five postulates. And proposition I.4, the congruence of triangles with two sides and the angle between them equal, rests on nothing at all: Euclid proves it by mentally laying one triangle on top of the other. There’s no such move among the postulates, and that’s another hole we’ll come back to.

The fifth postulate

The fifth postulate looks like a theorem someone forgot to prove. Take two lines and a third one crossing them. If the interior angles on one side of the third line add up to less than $180^\circ$, the lines lean towards each other and will meet somewhere on that side. It sounds obvious. But “somewhere” can be very far away: if the angles add up to $179.99^\circ$ and the lines cross the third one a metre apart, they meet almost six kilometres away. No drawing will show that.

Before turning to the fifth postulate, let’s see how far we can get without it. Euclid does the same: the first 28 propositions of Book I manage with the first four postulates alone. The main tool there is the exterior angle theorem.

Double a median, and one of the interior angles gets “carried” inside the exterior angle.

Extend the side $BC$ of triangle $ABC$ beyond $C$ to a point $D$. Let’s prove that the exterior angle $ACD$ is greater than angle $A$. Bisect the side $AC$ at the point $E$ (constructing the midpoint of a segment, I.10). Draw the segment $BE$ and extend it beyond $E$ by the same length: $EF = BE$ (postulate 2 and laying off a segment, I.2–I.3). The angles $AEB$ and $CEF$ are equal as vertical angles. So triangles $ABE$ and $CFE$ are congruent by two sides and the angle between them: $AE = CE$, $BE = FE$. This congruence test is proposition I.4; there’s more about it in chapter 17. Congruent triangles have equal corresponding angles: $\angle ECF = \angle EAB$, that is, angle $ACF$ equals angle $A$. The ray $CF$ runs inside the angle $ACD$: the point $F$ lies on the same side of $BD$ as $A$, and on the other side of $AC$ from $B$. So $\angle ACD > \angle ACF = \angle A$. For angle $B$ the argument is the same, only the midpoint is taken on the side $BC$, and the comparison is with the angle vertical to $ACD$.

Note the last step: “the point lies on that side” is knowledge about the order of points, which Euclid takes from the drawing. We’ll come back to this subtlety in the section on the holes in the Elements. Two consequences follow at once from the exterior angle theorem.

It follows that at least one parallel exists even without the fifth postulate: through a point off the line, it’s enough to draw a line making the same alternate angle with a transversal. Euclid does this in proposition I.31. Try it yourself: it’s the next level of the game.

But proving that there’s no other parallel through the same point can’t be done without the fifth postulate. What’s more, forbidding a second parallel is the fifth postulate itself, just said more briefly.

Compare any line through the point $P$ with the one line whose alternate angles are equal.

Let a line $l$ and a point $P$ off it be given. Draw a transversal $PQ$ and a line $m$ through $P$ making the same alternate angle with $PQ$ as $l$ does. By the test for parallels (I.27) the line $m$ is parallel to $l$; the fifth postulate isn’t needed here. Take any other line $m'$ through $P$. It’s turned relative to $m$, so on one side of the transversal the interior angles that $m'$ and $l$ make with it add up to less than $180^\circ$, short by exactly the angle between $m$ and $m'$. By the fifth postulate $m'$ and $l$ meet on that side. So no line except $m$ is parallel to $l$: Playfair’s axiom follows from the fifth postulate. Conversely, accept Playfair’s axiom, and let a line $m'$ make with the transversal and the line $l$ interior angles on one side adding up to less than $180^\circ$. The line $m'$ is different from $m$, and there’s only one parallel through $P$, so $m'$ meets $l$ at some point $X$. The point $X$ can’t lie on the other side of the transversal: there the interior angles add up to more than $180^\circ$, while in triangle $PQX$ any two angles together are less than $180^\circ$ (I.17). So $X$ lies on the side where the sum is smaller, which is what the fifth postulate says.

Euclid seems to have sensed the postulate’s special status: he puts it off to the last and first uses it only in proposition I.29. Go back to the map of Book I and tap “Postulate 5”. The part of the building that rests on it lights up; everything else would stand without it.

How many propositions of Book I manage without the fifth postulate? The answer can be found on the map above.

The fifth postulate is needed by proposition I.29 and everything that rests on it, and also by proposition I.44, which cites the postulate directly. Together that’s 19 propositions: I.29, I.30 and I.32–I.48. That leaves $48 - 19 = 29$: the first 28 propositions in a row and the construction of the parallel, I.31. The famous “first 28 propositions of Euclid, proved without the fifth postulate” are one solid block at the start of the book.

For two thousand years mathematicians tried to derive the fifth postulate from the other four. Proclus writes that Ptolemy already tried, and offers a proof of his own; later Omar Khayyam and the Italian Jesuit Giovanni Saccheri worked on it. Every proof turned out to be wrong: somewhere in each one hid a statement equivalent to the postulate itself. In the nineteenth century it became clear why. There’s a geometry in which the first four postulates hold and the fifth doesn’t. The journey there is in chapter 58.

Which of these statements can’t be proved without the fifth postulate?

The angles of a triangle add up to two right angles. Through a point not on a line, a line parallel to it can be drawn. Vertical angles are equal. In an isosceles triangle the base angles are equal.

The angle sum is proposition I.32, which rests on I.29, the first use of the fifth postulate. The existence of a parallel (I.31), vertical angles (I.15) and the base angles (I.5) are proved without it. And on a sphere, where there are no parallels at all, the angles of a triangle add up to more than $180^\circ$.

The hexagon and the square

Compass and straightedge treat regular polygons differently. We’ve already constructed the equilateral triangle. Euclid inscribes the regular hexagon in a circle in Book IV (IV.15), and it’s the shortest problem in the chapter once you notice one thing about its side. The square on a segment is proposition I.46, the second-to-last in Book I. Euclid builds it with a perpendicular and parallels, but we’ll manage with less.

In the proof for the square we say that an isosceles right triangle gives an angle of $45^\circ$. That follows from the angle sum of a triangle, that is, from the fifth postulate. It couldn’t have been otherwise: as chapter 58 will show, in a geometry where the first four postulates hold and the fifth doesn’t, there are no squares at all: a quadrilateral with equal sides and four right angles is impossible there.

Doubling the number of sides with bisectors, from the square we get the octagon, the 16-gon and so on, and from the hexagon the dodecagon. Euclid could also construct the regular pentagon (IV.11) and the 15-gon (IV.16). The next step was taken only by the eighteen-year-old Gauss: in 1796 he found a construction for the regular 17-gon. But the heptagon can’t be constructed with compass and straightedge. Exactly which polygons can be is explained in chapter 43.

A regular hexagon is inscribed in a circle of radius $7$. Find its perimeter.

The centre and two neighbouring vertices form a triangle two of whose sides are radii, with the angle $360^\circ : 6 = 60^\circ$ between them. Such a triangle is isosceles with a $60^\circ$ angle at the apex, so its other two angles are $60^\circ$ too: it’s equilateral. So the side of the hexagon equals the radius, $7$, and the perimeter is $6 \cdot 7 = 42$.

All triangles are isosceles

Here’s a proof that every triangle has two equal sides. Each of its steps refers to a true statement, and the conclusion is obviously false.

This sophism has long wandered from one collection of mathematical recreations to another. Try to find the mistake while the picture looks the way it’s usually drawn freehand. Then switch to the exact drawing, or drag a vertex.

Steps one to four are right wherever the point $O$ is. The mistake is in the last one: the equation $AB = AE + EB$ assumes that $E$ lies between $A$ and $B$. In the exact drawing the point $O$ always ends up outside the triangle, on the circumcircle, at the midpoint of the arc $BC$. Then one foot of a perpendicular lies on a side and the other on its extension, and instead of a sum we get a difference: for example, $AB = AE - EB$, but $AC = AF + FC$.

The lesson of this sophism is more serious than it seems. The Elements has no axiom at all about what it means for a point to lie between two others. Euclid takes such things from the drawing, and his drawings are always right. But a drawing isn’t an argument, and a wrong drawing gives a wrong theorem.

Holes in the Elements

Let’s go back to the very first proof. Why do the two circles in proposition I.1 meet at all? The picture shows it, but no postulate guarantees it. You can set up a “plane” in which all five postulates hold and the circles pass right through each other without meeting.

Keep in the plane only the points with rational coordinates. The line through two such points consists of rational points, and so do circles, and all five postulates still hold. Let $A = (0, 0)$ and $B = (1, 0)$. The circles of radius $1$ centred at $A$ and $B$ are given by the equations

$$x^2 + y^2 = 1, \qquad (x - 1)^2 + y^2 = 1.$$

Subtract the second from the first: $2x - 1 = 0$, that is, $x = \frac12$, and then $y^2 = \frac34$, $y = \pm\frac{\sqrt3}{2}$. But $\sqrt3$ is irrational (why is in chapter 6), so the rational plane has no such points. The circles cross at a “hole”. For proposition I.1 to be true, an axiom of continuity is needed: the plane has no holes.

The second hole is the superposition in I.4. Euclid moves the triangle around as if it were cut out of cardboard, although the postulates say nothing about moving figures. The third is the order of points on a line, which is what tripped up the sophism. Moritz Pasch noticed it and in 1882 added to geometry an axiom that Euclid didn’t have.

From it and the other axioms of order much of what Euclid took from the drawing can be derived, for example that in the proof of the exterior angle theorem the point $F$ really does lie inside the angle $ACD$.

With Hilbert’s axioms, Euclid’s theorems are proved without a single glance at the drawing. Euclid himself didn’t get a single theorem wrong: the holes were in the justification, not in the results. But only after Hilbert did geometry become what it had wanted to be from the start: knowledge that rests on logic, not on pictures.

Three problems that won’t give in

There were three problems the ancient Greeks never managed to solve with compass and straightedge. The first is doubling the cube: construct the edge of a cube whose volume is twice that of a given one. Legend has it that the people of the island of Delos asked the oracle how to end a plague and were told to double a cube-shaped altar. The second is trisecting an angle.

The third problem is squaring the circle: construct a square with the same area as a given circle. People struggled with these problems for two thousand years, and the answer came from an unexpected direction: algebra.

The idea of the proof, and why the whole of it doesn’t fit here

Take the given segment as the unit and introduce coordinates. A line through two points is given by an equation of degree one, a circle by one of degree two, so every new intersection point comes from solving a linear or quadratic equation. So compass and straightedge can get from known lengths their sums, differences, products, quotients and square roots, and nothing more.

Doubling the cube needs the length $\sqrt[3]{2}$, a root of $x^3 = 2$, and trisecting $60^\circ$ needs the number $\cos 20^\circ$, a root of $8x^3 - 6x - 1 = 0$. In 1837 Pierre Wantzel proved that the roots of such cubic equations can’t be expressed with square roots. Squaring the circle needs the length $\sqrt\pi$, and in 1882 Ferdinand von Lindemann proved that $\pi$ can’t be a root of any polynomial with integer coefficients, still less be obtained from one by square roots.

Making this into a rigorous proof needs the language of field extensions and degrees of polynomials, which is the subject of chapter 43. Lindemann’s proof is longer still: it relies on analysis and takes several pages even in modern textbooks.

Squaring the circle is linked to big question 2: can a disc be cut into pieces and reassembled into a square of the same area? Such a square can’t be constructed with compass and straightedge, and scissors cutting along straight lines won’t do it either: that’s chapter 20. The final and very strange answer awaits in chapter 54.

Which of these angles can be divided into three equal parts with compass and straightedge?

$90^\circ$ $60^\circ$ $120^\circ$ $30^\circ$

A third of a right angle is $30^\circ$, which is easy to construct: it’s half the angle of an equilateral triangle. A third of $60^\circ$ is $20^\circ$, a third of $120^\circ$ is $40^\circ$, a third of $30^\circ$ is $10^\circ$, and none of these can be constructed. “Trisection is impossible” means “not every angle has a construction”, not “no angle does”.

The impossibility here is a property of the rules, not of the angle. Loosen the rules a little, and the problem can be solved. A construction attributed to Archimedes divides any angle into three with a straightedge carrying two marks. Such a ruler may be slid around until the marks fall on the right lines; the Greeks called this technique neusis, “verging”.

Let the vertex of angle $AOB$ be the centre of a circle of radius $r$, with $A$ and $B$ on the circle. Put the ruler against the point $B$ and slide it so that one mark runs along the line $OA$ beyond the centre (that’s the point $P$) and the other lands on the circle at the point $Q$. The distance between the marks equals the radius, so $PQ = QO = OB = r$.

The angle $\alpha$ between the ruler and the line $OA$: the angle at the mark $P$. The angle $2\alpha$ at the vertex $B$ between the ruler and the radius $OB$. The given angle $AOB$ that we’re dividing into three. Example: for a $75^\circ$ angle the ruler settles at $25^\circ$ to the line $OA$; for $60^\circ$ at $20^\circ$, which can’t be constructed with compass and straightedge.

Two isosceles triangles, and twice over the exterior angle theorem in its exact form: an exterior angle equals the sum of the two interior angles not adjacent to it (chapter 17).

The distance between the marks equals the radius, so $PQ = QO = OB = r$: $QO$ and $OB$ are radii. Triangle $PQO$ is isosceles ($PQ = QO$), so the angles at its base $PO$ are equal: $\angle QOP = \angle QPO = \alpha$. The angle $OQB$ is the exterior angle of triangle $PQO$ at the vertex $Q$. It equals the sum of the other two angles of that triangle: $\alpha + \alpha = 2\alpha$. Triangle $OQB$ is isosceles ($OQ = OB$), so $\angle OBQ = \angle OQB = 2\alpha$. The angle $AOB$ is the exterior angle of triangle $POB$ at the vertex $O$, and it equals the sum of the angles at $P$ and at $B$: $\alpha + 2\alpha = 3\alpha$. So the angle at $P$ is exactly a third of angle $AOB$. The exact form of the exterior angle theorem rests on the angle sum of a triangle, and so on the fifth postulate.

And if we take away the straightedge?

The rules can be tightened as well as loosened. In 1672 the Dane Georg Mohr, and in 1797, independently, the Italian Lorenzo Mascheroni proved that everything that can be constructed with compass and straightedge can be constructed with the compass alone. Of course the compass can’t draw a line, but any points the straightedge would have given can be found without it. The idea of the proof: it’s enough to learn to find with the compass alone the intersection of two lines and the intersections of a line with a circle, when each line is given by two points. The second is done by reflecting the centre of the circle in the line; the first is harder, by inversion, which is covered in chapter 23. And in 1833 Jakob Steiner showed that the straightedge alone is enough too, provided one circle with its centre marked has already been drawn in the plane.

Where next

Triangles do most of the work in the Elements: the equilateral one in I.1, pairs of congruent triangles in almost every proof, the triangle made of three segments in I.22. That’s no accident. Any polygon can be cut into triangles by diagonals, so once we understand the triangle, we understand any figure with straight sides. The first question about it seems simple. A triangle has three sides and three angles: six numbers. How many of them, and which ones, do you need to know for the triangle to be determined uniquely? Three numbers aren’t always enough, and in one case three numbers describe two different triangles at once. That’s chapter 17.