Eudoxus' Method: Commentary

Eudoxus of Cnidos

Commentary on the text

1. Euclid of Alexandria (325?--265?) is one of the most important mathematicians of all time, so it is surprising that we know virtually nothing of his life. He is the author of the Elements, far and away the most important piece of mathematical literature ever written. The Elements is divided into 13 books, the first six on plane geometry (points, lines, parallels, polygons, and circles), the next three on numbers and divisibility (primes, greatest common divisors, geometric progressions), a book on irrational quantities, and three books on solid geometry.

The content of the book was not original with Euclid; he assembled it as a kind of Encyclopedia Mathematica, a synthesis of the mathematical knowledge of the day. But his editing skills, and more importantly, his understanding of the logical relationships amongst the ideas he included was brilliant. It is from the Elements that mathematicians learned the spare definition-axiom-theorem-proof style of writing that has characterized mathematical literature to this day. It is this far reaching legacy of the Elements that makes it so important. For centuries and around the world, unabridged editions of Books i-vi and xi-xiii have been used as geometry textbooks for generations of students. The following assessment by B. L. van der Waerden comes from an extensive article on Euclid in the Encyclopedia Britannica Online:

Almost from the time of its writing and lasting almost to the present, the Elements has exerted a continuous and major influence on human affairs. It was the primary source of geometric reasoning, theorems, and methods at least until the advent of non-Euclidean geometry in the 19th century. It is sometimes said that, next to the Bible, the Elements may be the most translated, published, and studied of all the books produced in the Western world.

The excerpt we read here comes from the beginning of Book xii where Euclid discusses the area problem for the circle.

2. The proposition is claiming that

If we let

then this can be abbreviated to

or equivalently,

In this last form, it says that the ratio of the area of a circle to the square of its diameter is the same for any two circles. If we let k represent this constant ratio, then for any circle with area A and diameter d we have

Finally, since the radius r of the circle satisfies d = 2r, this gives the formula

for the area of a circle. In other words, k = p/4. Consequently the proposition proved here by Eudoxus is a Greek version of the familiar result for the area of a circle.

3. The squares are not drawn in the diagram. As an aside, it is worth noting that nearly all modern editions of Euclid include diagrams with the text as does the one we follow here. But we cannot know if Euclid had exactly these diagrams in mind, for we do not have original texts. The oldest nearly complete edition dates only to the tenth century (more than 1000 years after Euclid!) and the oldest texts of the Elements known are just fragments from the third century (still 500 years after Euclid).

4. Here is the introduction of the double reductio argument. Eudoxus takes the position that some area S must satisfy the proportion

and he wants (area S) = (area of circle EFGH). To conclude this, he aims to show that neither (area S) < (area of circle EFGH) nor (area S) > (area of circle EFGH) can be true.

5. In this paragraph he is interested in showing only that the square EFGH has area more than half that of the circle EFGH. This may seem obvious from the diagram, but Eudoxus is careful about not making any leaps of faith. The need for this conclusion will be addressed later in note 7.
To show that (area of square EFGH) > (1/2)(area of circle EFGH), he introduces the square UVXY circumscribed about the circle with sides tangent at points E, F, G, H.

The conclusion follows from combining the observations that (area of square EFGH) = (1/2)(area of square UVXY) and (area of square UVXY) > (area of circle EFGH).

6. This part of the argument is similar in spirit to the last (note 5); the aim is to show that the triangles that are being added on to square EFGH to obtain polygon EKFLGMHN are each larger than half the corresponding segments of the circle that contain them. The proof he gives for triangle EKF (which follows the diagram in the text) will work for all the other triangles.
The parallelogram on EF whose opposite side passes through K is actually a rectangle, but this fact is irrelevant. What is important is that (area of triangle EKF) = (1/2)(area of the parallelogram) and (area of the parallelogram) > (area of circular segment on EF). Thus, (area of triangle EKF) > (1/2)(area of circular segment on EF).

7. The next stage of the approximation is only sketched: continue to bisect the arcs of the circle between the vertices of the polygon we have so far constructed. These points determine thin triangles with the neighboring points of the polygon, and each of these triangles (as before) has an area greater than half the area of the segment in which it lies.
The reason for requiring that the additional areas be more than half of the remaining portion of the circle is finally revealed here, and Euclid refers the reader to Proposition x.i where he has supplied the necessary justification for this. Here is a more detailed explanation.
Recall that this part of the double reductio begins with the assumption that (area S) < (area of circle EFGH). Therefore, we can identify the excess

Next, let A = area of circle EFGH . Remove first the square EFGH which, since what we remove is more than half of A, leaves four segments of the circle whose total area is < (1/2)A. At the next stage of the process, we remove from these four segments four triangles each of which is greater than half the size of the corresponding segment in which it lies. So the remaining eight segments of the circle (see the diagram in the text) represent an area

We repeat this at the next stage to produce a slightly larger polygon, with twice as many sides as before, so that

Since the quantities (1/2)A, (1/4)A, (1/8)A, ... are necessarily shrinking in size, we must eventually come to a stage where we can ensure that

no matter how small e might have been to start out with. This is what Eudoxus/Euclid means when he writes "by doing this continually, we shall leave some segments of the circle which will be less than the excess by which circle EFGH exceeds area S". And it is this process which "exhausts" the circle by filling it up more and more with a larger and larger polygon.

8. For the purpose of the argument, the exhaustion is assumed to have been completed at the second stage (see previous note). It may, however, require many more.

9. Since

it follows that (area of polygon) > (area S).

10. The bracketed text is not part of Euclid; it is supplied by the editor (here, Sir Thomas Heath, the nineteenth century British historian of mathematics) to point the reader to those earlier propositions in the Elements that justify the steps that Euclid is taking here. At this point we are referred to Proposition xii.1, the result that immediately precedes the text we are reading here. It allows us to conclude that the similar polygons produced by the method of exhaustion that have been inscribed in the two circles (ABCD and EFGH) have areas whose ratio is the same as the ratio of the squares on the diameters of the corresponding circles:

11. See note 4.

12. Proposition v.11 says that given the proportions A/B = C/D and C/D = E/F, then A/B = E/F. Applied here, since

(note 4) and

(note 10), then

13. In Euclid, given a proportion A/B = C/D, the alternate proportion is A/C = B/D. Proposition v.16 justifies the validity of the alternate proportion. A modern reader will understand this by the simple algebra step of multiplying through the original proportion by B/C. Applied to the present argument, we deduce

14. Since the ratio on the left side of the proportion in note 13 is greater than 1, so must the ratio on the right be. But from note 9 the argument implies that the ratio on the right side is less than 1. Here is the contradiction. This means that the assumption made at the start, that (area S) < (area of circle EFGH), cannot be true (see note 4). Eudoxus thus completes the first of the two reductio arguments that make up the proof.

15. If he applies the same argument he has just completed to the situation obtained by switching the roles of the two circles, then he would be arguing that if area S is chosen to satisfy

then it is impossible for (area S) < (area of circle ABCD). Compare with note 4. This observation will allow him us to simplify the second half of the double reductio considerably.

16. This begins the second half of the double reductio. See note 4. Eudoxus now assumes that

where (area S) > (area of circle EFGH). The goal is to derive a contradiction as before.

17. In Euclid, given a proportion A/B = C/D, the inverse proportion is B/A = D/C. So from the proportion stated in note 16, he derives

18. See note 15.

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