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Archytas of Tarentum

Archytas, son of Hestiaeus, of Taras (Ἀρχύτας ὁ Ταραντῖνος)

A course in two strands: the life of the Pythagorean general who governed a Greek city in southern Italy and rescued Plato from a tyrant, and the mathematics he left behind: the three means, the first rigorous theorem of number theory applied to music, and a solution of the doubling of the cube in three dimensions.

Born c. 435–410 BC, Tarentum (Taras) Died c. 360–350 BC Signature result two mean proportionals from a cylinder, a torus and a cone Level advanced undergraduate
Archytas’ three surfaces. The half-cylinder on the circle AC (grey) meets the surface swept by a semicircle turning about A in a curve (red). The cone about AC (green) cuts that curve at P. With AC = 2 and the cone’s angle set by a chord of length 1, the foot M of P gives AM = ∛2.
Lesson 00

How to use this course

Archytas is the earliest Greek mathematician from whom we have both genuine fragments in his own words and a detailed proof reported by a reliable witness. He was also a working politician: for seven years running his city elected him general. Plato knew him, Eudoxus learned geometry from him, and Aristotle wrote books about him that are now lost.

Much more has survived under his name than he wrote. Ancient forgers produced treatises in a Doric dialect and passed them off as his, and the forgeries outnumber the genuine fragments many times over. This course follows Carl Huffman’s edition (2005), which accepts four fragments (B1–B4 in the standard numbering of Diels and Kranz) and a set of reliable testimonia (the “A” texts). For each idea, the lessons say which ancient author reports it.

Prerequisites for Part II
  • Fractions, ratios and the idea of a greatest common divisor.
  • Elementary solid geometry: cylinders, cones, spheres and their equations in coordinates.
  • Logarithms (for the measurement of musical intervals in cents).
  • No musical training is needed; Lesson 06 explains the intervals from scratch.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proof sketches and exercises with folded solutions. The companion courses on Plato, Eudoxus and Euclid treat the same circle of mathematicians from other sides; references such as VIII.8 are to Euclid’s Elements in Heath’s numbering.

Reference

Timeline

Archytas’ dates are inferred, not recorded. Eudemus calls him a contemporary of Plato (born 428/427 BC), and he was apparently dead before Tarentum began to hire foreign generals in the 340s BC. All dates BC below are approximate unless stated.

c. 706 BCTraditional date for the founding of Taras (Tarentum) by colonists from Sparta.
c. 435–410 BCArchytas born at Tarentum, son of Hestiaeus (or, in one report, Mnesagoras).
late 5th c. BCProbably taught by the Pythagorean Philolaus of Croton (Cicero).
c. 388/387 BCPlato’s first journey to Italy and Sicily; he very likely meets Archytas.
380s–370s BCEudoxus of Cnidus studies geometry with him.
c. 370s–360s BCElected general of Tarentum seven years in succession; said never to have been defeated.
361–360 BCPlato’s third visit to Syracuse. Archytas sends a thirty-oared ship and the envoy Lamiscus to secure his release.
c. 360–350 BCDeath of Archytas; place and manner unknown.
4th c. BCAristoxenus of Tarentum writes a Life of Archytas; Aristotle writes on his philosophy; Eudemus records his solution of the Delian problem and his argument for an unlimited universe.
c. 300 BCEuclid’s Elements Book VIII, whose theory of continued proportion may draw on Archytas.
23 BCHorace publishes Odes I–III; Ode I.28 imagines the tomb of Archytas on the Apulian shore.
2nd c. ADPtolemy’s Harmonics reports his divisions of the tetrachord; Aulus Gellius tells of his wooden dove.
3rd c. ADPorphyry quotes fragments B1 and B2 in his commentary on Ptolemy; Diogenes Laertius writes his short Life.
6th c. ADBoethius preserves his proof on superparticular ratios; Eutocius copies Eudemus’ account of the cube duplication; Simplicius preserves the staff argument.
2005 ADCarl Huffman’s edition and commentary separates the genuine fragments from the forgeries.
Part I · Formation
Lesson 01

Tarentum and the sources

Taras, called Tarentum by the Romans and Taranto today, lies on a sheltered bay in the instep of the Italian boot. Tradition says colonists from Sparta founded it around 706 BC. By the fourth century BC it was the richest Greek city of southern Italy, with a large fleet, a strong cavalry and, according to Aristotle and Strabo, a democratic constitution. Its neighbours were the Italic peoples of the interior, the Messapians and Lucanians, with whom it fought repeatedly.

Archytas was born there sometime between about 435 and 410 BC. Diogenes Laertius gives his father as Mnesagoras or, on the authority of Aristoxenus, Hestiaeus. Aristoxenus was himself a Tarentine whose father had probably known Archytas, so his version is usually preferred.

The sources and their reliability

  • Fragments B1–B4, quoted by Porphyry (3rd century AD), Iamblichus and Stobaeus. B1 and B2 come from a work on harmonics, B3 from a work On Sciences, B4 from Discourses.
  • Aristoxenus, the music theorist and pupil of Aristotle, wrote a Life of Archytas, now lost. Athenaeus and Diogenes quote it.
  • Eudemus of Rhodes, another pupil of Aristotle, wrote histories of geometry and astronomy. Through Eutocius and Simplicius he preserves the cube duplication and the staff argument.
  • Ptolemy’s Harmonics (2nd century AD) reports Archytas’ divisions of the tetrachord, and Boethius (6th century AD) reports his proof about superparticular ratios.
  • Diogenes Laertius, Lives VIII.79–83, gives a short biography mixing good information with a spurious letter to Plato.
  • The pseudo-Pythagorean writings: treatises such as On the Ten Categories or On Law and Justice, composed centuries later. Huffman treats them as forgeries, and this course does not use them.

Exercise 1.1Rank these three reports by reliability and give a reason for each: (a) Eutocius’ account of the cube duplication, copied from Eudemus; (b) a treatise On the Ten Categories under Archytas’ name; (c) Aulus Gellius’ story of the flying wooden dove.

Show solution
(a) is best: Eudemus wrote within a generation of Archytas, belonged to Aristotle’s school with its interest in accurate doxography, and Eutocius says he is copying him. (c) is weaker: Gellius wrote about five centuries later and cites only his teacher Favorinus, with no earlier authority. (b) is worst: the work uses Aristotle’s doctrine of categories, which postdates Archytas, and belongs to the class of Hellenistic forgeries written to make the Pythagoreans look like the source of Plato and Aristotle.
Lesson 02

A Pythagorean education

The Pythagorean communities of southern Italy had been violently broken up in the mid-fifth century BC, and survivors scattered to Greece and to cities such as Tarentum and Rhegium. Cicero says Archytas was a pupil of Philolaus of Croton, the first Pythagorean to publish a book. The claim is plausible on dates and doctrine; Huffman accepts it as likely but unproven. Philolaus held that everything known has number, and his musical scale, with tones of ratio 9:8, reappears in Plato’s Timaeus.

Archytas’ mathematics builds on the work of Hippocrates of Chios, who in the later fifth century BC compiled the first Elements of geometry and reduced the doubling of the cube to finding two mean proportionals. There is no evidence that the two men met, but Archytas’ solution (Lesson 09) begins exactly where Hippocrates stopped.

Learning and finding out

A fragment from On Sciences (B3) describes the two ways of coming to know: learning from another, or finding out for oneself. Finding out without a teacher requires inquiry, and inquiry is hard for one who does not know how to inquire. The same fragment praises calculation (logismos): once discovered, it stops civil strife and increases concord, because where there is calculation there is equality, and rich and poor can see how to deal fairly with each other.

Fragment B4 makes a bolder claim. Logistic, the science of number, surpasses the other sciences in wisdom and is clearer than geometry, and it completes demonstrations where geometry fails. Historians read this as a programme: proofs about ratios and proportion should rest on numbers. Lessons 05 and 07 show Archytas carrying it out.

Exercise 2.1Fragment B3 says calculation produces fairness between rich and poor. Give a concrete example of a political decision that the arithmetic, geometric and harmonic means (Lesson 04) would settle differently, and say which kind of “equality” each embodies.

Show solution
Suppose two citizens contribute 2 and 8 talents to a common fund, and a single honour or share must be fixed for a third party between them. The arithmetic mean gives 5, equal differences (absolute equality); the geometric mean gives 4, equal ratios (proportional equality, the kind Plato and Aristotle later call “geometric equality” and favour in distributive justice); the harmonic mean gives 3.2, which leans toward the smaller value. Greek political theory after Archytas used exactly this contrast between arithmetic and geometric equality.
Lesson 03

General and statesman

Aristoxenus, as reported by Diogenes, says the Tarentines elected Archytas general (strategos) seven times in succession, although a law forbade holding the office twice running, and that he was never defeated. When envy forced him out of office, the army was beaten at once. The story comes from a Tarentine admirer, but the core, a long series of commands, is accepted by modern historians. His campaigns were most likely against the Messapians and Lucanians. He may have held the title autokrator, giving him freedom in diplomacy and war, within a democratic constitution.

Later writers made him a model of self-control. Aristoxenus and Cicero tell that he refused to punish a slave while angry; others that he wrote a curse on a wall rather than say it aloud. Athenaeus preserves from Aristoxenus a debate in which Polyarchus, an envoy of Dionysius II, praises bodily pleasure and Archytas answers that reason is the best part of us. Cicero reworks a related argument: no one in the grip of the greatest pleasure can think. Aristotle praised a remark of his that an arbitrator and an altar are alike, since the wronged take refuge in both.

Some scholars see in Archytas one model for Plato’s philosopher-king: a mathematician who actually governed a city well. Plato never names him in the dialogues, though the Republic quotes his saying that astronomy and harmonics are sister sciences (Lesson 08).

Exercise 3.1Name two features of the generalship story that make a historian cautious, and one feature that supports its core.

Show solution
Cautions: the source, Aristoxenus, was a fellow Tarentine with a family connection to Archytas and wrote an admiring Life; and “never defeated, and the city lost as soon as he left” is a stock motif of praise. Support: the detail that a law against successive terms was suspended for him is specific and checkable by contemporaries, and Archytas’ ability to send a state ship and embassy to Syracuse in 361 BC (Lesson 11) shows he had real authority.
Part II · The mathematics
Lesson 04

The three means

For the general reader. Between 6 and 12 one can put 9, halfway in difference; or about 8.49, halfway in ratio; or 8, which divides the octave on a lyre into a fifth and a fourth. Archytas gave the first surviving definitions of these three “means” and named the third one harmonic. Each corresponds to a different notion of “in between”, and all three are still in daily use.

Goals
  • State Archytas’ definitions of the arithmetic, geometric and harmonic means.
  • Derive the formula for the harmonic mean and the identity \(AH=G^2\).
  • Explain why the harmonic mean is the musical one.

Fragment B2

Porphyry quotes Archytas as saying that there are three means in music. In paraphrase, for three terms \(a>b>c\):

Definitions (Archytas, fragment B2)
  • Arithmetic: the first exceeds the second by the same amount as the second exceeds the third, \(a-b=b-c\). The ratio of the greater terms is then smaller than the ratio of the lesser terms.
  • Geometric: the first is to the second as the second is to the third, \(a:b=b:c\). The greater and lesser terms make the same ratio.
  • Subcontrary, “which we call harmonic”: by whatever part of itself the first exceeds the second, by that part of the third the middle exceeds the third. The ratio of the greater terms is then larger.

In symbols the harmonic condition says \(a-b=\frac{a}{k}\) and \(b-c=\frac{c}{k}\) for the same \(k\), so \(\dfrac{a-b}{b-c}=\dfrac ac\). Iamblichus says that Archytas and Hippasus renamed the old “subcontrary” mean harmonic, because it produces the concordant intervals.

Proposition

For positive \(a>c\) the three means are \(A=\frac{a+c}2\), \(G=\sqrt{ac}\) and \(H=\frac{2ac}{a+c}\). They satisfy \(A>G>H\), \(AH=G^2\), and \(\frac1H=\frac12\big(\frac1a+\frac1c\big)\).

Proof. From \((a-H)c=(H-c)a\) we get \(H(a+c)=2ac\). Then \(AH=\frac{a+c}2\cdot\frac{2ac}{a+c}=ac=G^2\), and \(A-G=\frac12(\sqrt a-\sqrt c)^2>0\). Since \(G^2=AH\) and \(A>G\), we get \(H<G\). The last identity says the harmonic mean is the reciprocal of the arithmetic mean of reciprocals. \(\square\)

Why “harmonic”

Take string lengths 12 and 6, an octave apart. The harmonic mean is \(\frac{2\cdot72}{18}=8\) and the arithmetic mean is 9. Then \(12:8=3:2\) is a fifth, \(8:6=4:3\) a fourth, and \(12:9=4:3\), \(9:6=3:2\) reverse them. The interval between the two means is \(9:8\), the whole tone. The four numbers 12, 9, 8, 6 contain every concordance the Greeks recognized, which is why later writers such as Nicomachus called this the most perfect proportion. Because \(A\cdot H=G^2\), the two means sit symmetrically about the geometric centre of the octave, a fact the Greeks expressed as \(a:A=H:c\) (here \(12:9=8:6\)).

Exercise 4.1Find the harmonic and arithmetic means of 4 and 3 (a fourth) and of 3 and 2 (a fifth). Which intervals do they produce?

Show solution
For 4 and 3: \(A=\frac72\), \(H=\frac{24}{7}\). Then \(4:\frac{24}7=7:6\) and \(\frac{24}7:3=8:7\); the arithmetic mean gives \(8:7\) and \(7:6\) in the other order. For 3 and 2: \(A=\frac52\), \(H=\frac{12}5\), giving \(5:4\) and \(6:5\), the major and minor thirds. Archytas’ interval 8:7 appears in his diatonic and 5:4 in his enharmonic (Lesson 06): splitting concords by means is one way he may have found his ratios.

Exercise 4.2Prove the ratio statements in B2: in the arithmetic proportion \(\frac ab<\frac bc\), and in the harmonic proportion \(\frac ab>\frac bc\).

Show solution
Arithmetic: \(b^2=\big(\frac{a+c}2\big)^2>ac\) when \(a\ne c\), so \(\frac bc>\frac ab\). Harmonic: \(b=H<G\), so \(b^2<ac\), hence \(\frac ab>\frac bc\). The geometric mean is the borderline case \(b^2=ac\).

Exercise 4.3A car drives 60 km at 30 km/h and 60 km at 60 km/h. Show that the average speed is the harmonic mean of the two speeds, and explain why.

Show solution
Total time is \(2+1=3\) hours for 120 km, so 40 km/h, and \(H(30,60)=\frac{2\cdot1800}{90}=40\). Equal distances mean we average the times per unit distance, \(\frac1v\), so we take the arithmetic mean of reciprocals and invert.
Lesson 05

Superparticular ratios have no mean

For the general reader. Can the whole tone, the step from one note to the next, be split exactly in half? Archytas proved that it cannot, and more: no interval whose ratio has the form “one more than a number to that number”, such as 2:1, 3:2, 4:3 or 9:8, can be divided into two equal parts. It is the earliest surviving proof in Greek number theory, and it is a statement about irrationality in musical dress.

Goals
  • State the theorem and know its source.
  • Follow Archytas’ proof using numbers in lowest terms.
  • Translate it into the modern statement that \(\sqrt{(n+1)/n}\) is irrational.

A ratio \(m:n\) of whole numbers is superparticular (Greek epimoric) if \(m=n+1\), or more generally if \(m\) exceeds \(n\) by a part (unit fraction) of \(n\). The basic musical intervals are of this kind: octave 2:1, fifth 3:2, fourth 4:3, tone 9:8. A mean proportional between \(m\) and \(n\) is a whole number \(k\) with \(m:k=k:n\).

Theorem (Archytas; Boethius, De institutione musica III.11)

There is no whole number that is a mean proportional between two numbers in superparticular ratio.

Archytas’ proof, as reported by Boethius

  1. Let \(A:B\) be superparticular. Take \(C\) and \(D+E\) to be the smallest numbers in the same ratio, with \(D+E\) the larger, \(D=C\).
  2. Since \(D+E\) exceeds \(C\) by a part of itself and of \(C\), and these are the least numbers in the ratio, the excess \(E\) is a part of both and must be the unit: \(E=1\).
  3. If a mean fell between \(A\) and \(B\), then a mean would also fall between the smallest numbers in the same ratio, \(C\) and \(D+E\) (this is the content of Elements VIII.8 in the case of one mean).
  4. But \(D+E\) exceeds \(C\) by a unit, and no whole number lies strictly between two consecutive whole numbers. Contradiction.

Euclid’s Elements gives the results in step 2 and step 3 in Books VII and VIII, and the Sectio canonis ascribed to Euclid (Proposition 3) repeats the proof. The word “unit” matters: Archytas works within whole numbers and their ratios, as fragment B4 recommends.

Modern form

For every integer \(n\ge1\), the number \(\sqrt{\frac{n+1}{n}}\) is irrational. Equivalently, \(n(n+1)\) is never a perfect square.

Proof. \(\gcd(n,n+1)=1\). If \(n(n+1)=k^2\), unique factorization forces both \(n\) and \(n+1\) to be squares, since they share no prime. But two positive squares differ by at least 3. \(\square\) An integer mean \(k\) between \(m\) and \(n\) in any ratio means \(mn=k^2\); the theorem says this never happens for the ratio \(n+1:n\) scaled by any common factor.

Exercise 5.1Show directly that there is no whole number \(k\) with \(18:k=k:16\), and conclude that the tone 9:8 cannot be divided into two equal tones.

Show solution
We would need \(k^2=288\), but \(16^2=256\) and \(17^2=289\). In general, if \(9t:k=k:8t\) then \(k^2=72t^2\), so \(72=(k/t)^2\) would be the square of a rational, and \(72=36\cdot2\) is not. So no ratio of whole numbers is exactly half a tone. The Pythagorean “semitone” 256:243 is smaller than half a tone (90 cents against about 102), and the remainder 2187:2048 is larger.

Exercise 5.2Prove step 2 in modern terms: if \(m:n\) is superparticular and \(m,n\) are in lowest terms, then \(m=n+1\).

Show solution
Superparticular means \(m-n=\frac nq\) for some whole \(q\), so \(m-n\) divides \(n\), hence also \(m=n+(m-n)\). Thus \(m-n\) divides \(\gcd(m,n)=1\), so \(m-n=1\).

Exercise 5.3The octave is 2:1. What does the theorem say about the equal-tempered tritone, and why did Greek theorists who followed Archytas avoid it?

Show solution
The tritone of equal temperament is exactly half an octave, ratio \(\sqrt2:1\). The theorem with \(n=1\) says no ratio of whole numbers has this value. Theorists who measured intervals by ratios of string lengths, like Archytas, therefore had no “exact half-octave”. Aristoxenus, who treated intervals as continuous magnitudes, did divide the tone and the octave into equal parts, and the two schools argued about this for centuries.
Lesson 06

Dividing the tetrachord

For the general reader. Greek scales were built from tetrachords, four notes spanning a fourth, whose two outer notes were fixed and whose two inner notes moved. Different placements gave three “genera”: diatonic, chromatic and enharmonic. Archytas gave a precise ratio for every step in each genus. His numbers, preserved by Ptolemy, look as if they were designed to match the music actually played in his day.

Goals
  • Know Archytas’ three tetrachord divisions and check that each composes to 4:3.
  • Measure intervals in cents and compare with Philolaus and Plato.
  • Understand the role of the shared 28:27 step.

Intervals compose by multiplying ratios. To compare sizes, musicians use the cent: an interval of ratio \(r\) measures \(1200\log_2 r\) cents, so the octave is 1200 and the equal-tempered semitone 100.

Archytas’ divisions (Ptolemy, Harmonics I.13)

Each tetrachord descends from the top note through three steps whose product is the fourth, 4:3.

  • Enharmonic: \(\frac54\cdot\frac{36}{35}\cdot\frac{28}{27}=\frac43\) (386 + 49 + 63 cents).
  • Chromatic: \(\frac{32}{27}\cdot\frac{243}{224}\cdot\frac{28}{27}=\frac43\) (294 + 141 + 63 cents).
  • Diatonic: \(\frac98\cdot\frac87\cdot\frac{28}{27}=\frac43\) (204 + 231 + 63 cents).

The lowest step is 28:27 in all three genera, so the second-lowest note is the same in each. For comparison, the diatonic of Philolaus, used in Plato’s Timaeus, is \(\frac98\cdot\frac98\cdot\frac{256}{243}\): two equal tones and a remainder. Archytas’ diatonic uses the larger step 8:7 and the prime 7, which no Pythagorean scale before him used. Ptolemy praised him as the Pythagorean most concerned with music, but complained that Archytas did not respect perceived consonance in places. He also claimed Archytas used only superparticular ratios, which the chromatic 32:27 and 243:224 contradict. Huffman argues that Archytas started from the scales musicians tuned by ear and then found the simplest ratios that fit them.

How the numbers fit together

The chromatic step 32:27 is a Pythagorean minor third, the interval of a fourth less a tone: \(\frac43\big/\frac98=\frac{32}{27}\). The enharmonic 5:4 is the major third. With 28:27 fixed at the bottom, the middle step is whatever remains: \(\frac43\big/\big(\frac54\cdot\frac{28}{27}\big)=\frac{36}{35}\), and \(\frac43\big/\big(\frac{32}{27}\cdot\frac{28}{27}\big)=\frac{243}{224}\).

Exercise 6.1Verify that each of the three products equals 4:3, and compute the middle step of the diatonic from the other two.

Show solution
Enharmonic: \(\frac{5\cdot36\cdot28}{4\cdot35\cdot27}=\frac{5040}{3780}=\frac43\). Chromatic: \(\frac{32\cdot243\cdot28}{27\cdot224\cdot27}=\frac{217728}{163296}=\frac43\). Diatonic: \(\frac{9\cdot8\cdot28}{8\cdot7\cdot27}=\frac{2016}{1512}=\frac43\). Middle diatonic step: \(\frac43\big/\big(\frac98\cdot\frac{28}{27}\big)=\frac43\cdot\frac{216}{252}=\frac{288}{252}=\frac87\).

Exercise 6.2Compute in cents the Philolaus–Plato diatonic tetrachord and compare its lowest step with Archytas’ 28:27.

Show solution
\(1200\log_2\frac98\approx203.9\), twice gives 407.8, and \(1200\log_2\frac{256}{243}\approx90.2\); total 498.0, the fourth. Archytas’ 28:27 is about 63.0 cents, much smaller. His lowest step is narrower, and his upper steps (204 and 231) are wider and unequal, a more uneven tetrachord than the Pythagorean one.

Exercise 6.3Show that 36:35 and 28:27 are superparticular but 243:224 is not, and check whether 243:224 is in lowest terms.

Show solution
\(36=35+1\) and \(28=27+1\). For 243:224 the difference 19 does not divide 224, so it is not superparticular. \(243=3^5\) and \(224=2^5\cdot7\) share no prime, so the ratio is in lowest terms. Ptolemy’s claim that all of Archytas’ ratios are superparticular is therefore inaccurate.
Lesson 07

Continued proportion and Elements VIII

For the general reader. A list of numbers in continued proportion, such as 8, 12, 18, 27, grows by the same ratio at each step. Book VIII of Euclid’s Elements studies when such lists can be inserted between two given numbers. Its theorems are what Archytas needed for his proof about musical intervals, and since the work of Bartel van der Waerden many historians have attributed the core of Book VIII to Archytas. The attribution rests on style and content, not on an ancient statement.

Goals
  • State VIII.8, VIII.11 and VIII.12.
  • Connect the number-theoretic “two means between cubes” with the geometric doubling of the cube.
  • Weigh the evidence for Archytas’ authorship.
Elements VIII.8

If between two numbers \(a,b\) there fall \(k\) numbers in continued proportion, then between any two numbers in the same ratio as \(a:b\) there also fall \(k\) numbers in continued proportion.

Elements VIII.11 and VIII.12

Between two square numbers there is one mean proportional number, and the square has to the square the duplicate ratio of the side to the side. Between two cube numbers there are two mean proportional numbers, and the cube has to the cube the triplicate ratio of the side to the side.

For squares \(p^2,q^2\) the mean is \(pq\). For cubes \(p^3,q^3\) the two means are \(p^2q\) and \(pq^2\):

\[ p^3 : p^2q = p^2q : pq^2 = pq^2 : q^3 = p:q. \]

Two means turn a ratio into its cube. This is the arithmetic shadow of Hippocrates’ reduction of the Delian problem: if \(a:x=x:y=y:b\) then \(\left(\frac ax\right)^3=\frac ab\), so with \(b=2a\) the cube on \(x\) is double the cube on \(a\). For numbers, two means between \(1\) and \(2\) do not exist, because 2 is not the cube of a ratio of whole numbers; for lines, Archytas showed how to construct them (Lesson 09).

Proof idea of VIII.8

Reduce \(a:b\) to lowest terms \(a_0:b_0\). If \(a,x_1,\dots,x_k,b\) is in continued proportion, the least numbers with that ratio pattern are \(u^{k+1},u^kv,\dots,v^{k+1}\) with \(\gcd(u,v)=1\) (VIII.2–3), and these have \(u^{k+1}:v^{k+1}=a_0:b_0\). Any \(c:d\) equal to \(a:b\) is \(tu^{k+1}:tv^{k+1}\) for some \(t\), and \(tu^{k+1-j}v^j\) are the required means.

Exercise 7.1Insert two mean proportionals between 8 and 27, and between 16 and 54.

Show solution
\(8=2^3\), \(27=3^3\): means \(2^2\cdot3=12\) and \(2\cdot3^2=18\), giving 8, 12, 18, 27 with ratio 2:3. Since \(16:54=8:27\), VIII.8 gives means \(2\cdot12=24\) and \(2\cdot18=36\): 16, 24, 36, 54.

Exercise 7.2Use VIII.8 and VIII.12 to show that there are no two whole numbers in continued proportion between \(n\) and \(2n\).

Show solution
If there were, then by VIII.8 two means would fall between the least numbers in the ratio, 1 and 2. From the proof idea, the least numbers with two means are \(u^3\) and \(v^3\) with \(\gcd(u,v)=1\), so \(1=u^3\) and \(2=v^3\), impossible. This is why the doubling of the cube could not be solved in numbers and had to become a problem about lines.
Lesson 08

Pitch, sound and the sister sciences

For the general reader. Fragment B1 is the opening of Archytas’ book on harmonics. It praises the mathematical sciences as one family, and then gives a theory of sound: sounds are produced by impacts, and fast impacts sound high, slow ones low. The theory is wrong in an instructive way, and its first sentence named the group of four sciences that the Middle Ages called the quadrivium.

Goals
  • Summarize the content of B1 and its influence on Plato.
  • Compare Archytas’ “speed” theory of pitch with the modern frequency theory.
  • Relate string lengths to pitch ratios.

The four sciences

B1, quoted by Porphyry, says that the mathematicians have given clear knowledge of the speed of the stars, their risings and settings, of geometry, numbers and sphaeric, and of music, and that these sciences seem to be sisters. Plato’s Republic (530d) repeats the saying about astronomy and harmonics as sister sciences, attributing it to “the Pythagoreans”. Through Nicomachus and Boethius, the grouping of arithmetic, geometry, music and astronomy became the medieval quadrivium.

The theory of pitch

B1 continues: there can be no sound unless objects strike each other; many impacts are too weak to be heard; sounds that reach us quickly and powerfully seem high, those that come slowly and weakly seem low. A stick moved fast gives a higher note than one moved slowly; in a pipe, breath driven hard gives a higher note. Plato (Timaeus 67b, 80a) and Aristotle adopted versions of this theory.

Modern statement

Perceived pitch is determined by the frequency \(f\) of the periodic pressure wave. In a given medium all frequencies travel at the same speed \(v\) (about 343 m/s in air at 20 °C). For an ideal string of length \(L\), tension \(T\) and mass per unit length \(\mu\), the fundamental frequency is \(f=\dfrac{1}{2L}\sqrt{\dfrac T\mu}\).

Archytas’ intuition was half right: a faster stroke does produce more vibrations per second. His mistake was to place the speed in the travel of the sound rather than in the rate of its pulses. The formula shows why the Greek ratios work: at fixed tension, frequency is inversely proportional to string length, so a string of length 12 and one of length 6 sound an octave apart, as in Lesson 04.

Exercise 8.1A monochord string of length 120 cm sounds a note. Where must the bridge be placed to sound the fourth, fifth and octave above it, and the tone 9:8?

Show solution
Higher pitch means shorter length in the inverse ratio: fourth at \(120\cdot\frac34=90\) cm, fifth at \(120\cdot\frac23=80\) cm, octave at 60 cm, tone at \(120\cdot\frac89\approx106.7\) cm. The lengths 120, 90, 80, 60 are 12, 9, 8, 6 scaled by 10.

Exercise 8.2Give an everyday observation that refutes the claim that high notes travel faster than low notes.

Show solution
Music heard from far away, such as a band across a field, still sounds in tune: chords arrive together, with high and low notes simultaneous. If pitch depended on travel speed, high notes would arrive first and the timing would smear with distance. Thunder is another example: its low rumble and sharp crack start at intervals set by the lightning path, not by pitch.
Lesson 09

Doubling the cube

For the general reader. The Delian problem asks for the side of a cube with twice the volume of a given cube. Hippocrates had shown it is enough to find two lines in continued proportion between the given side and its double. Archytas found them by imagining three surfaces in space, a half-cylinder, a surface swept by a turning semicircle and a cone, and taking the point where all three meet. It is the first known solution, and a striking piece of three-dimensional geometry for the early fourth century BC.

Goals
  • Know the source of the solution and Hippocrates’ reduction.
  • Follow Archytas’ construction and prove it in coordinates.
  • Compare it with the later solution of Menaechmus.

The source

Eutocius of Ascalon, in his sixth-century commentary on Archimedes’ On the Sphere and Cylinder, collected a dozen solutions of the problem. For Archytas he says he copies the account of Eudemus. Eratosthenes, in a letter that Eutocius also preserves, warns readers not to attempt the cumbersome “cylinders of Archytas”. Diogenes Laertius says Archytas was the first to use mechanical motion in a geometrical construction, seeking two mean proportionals by a section of a half-cylinder. Some scholars doubt that Eutocius reproduces Eudemus word for word, but the construction itself is not in dispute.

Problem (two mean proportionals)

Given lines \(a>b\), construct \(x,y\) with \(a:x=x:y=y:b\). Then \(x^3=a^2b\) and \(y^3=ab^2\); with \(a=2b\), \(y^3=2b^3\) and \(y\) is the side of the doubled cube.

The construction

In a horizontal plane draw a circle with diameter \(AC=a\) and a chord \(AB=b\). Three surfaces are built on this figure:

  1. the half-cylinder standing vertically on the semicircle \(ABC\);
  2. the surface swept by a vertical semicircle on \(AC\) as it turns about the vertical line through \(A\) (a torus with zero inner radius);
  3. the cone obtained by turning the triangle \(ACD\) about \(AC\), where \(D\) is the point where \(AB\) extended meets the tangent at \(C\). Its generators make with \(AC\) the angle whose cosine is \(b/a\).

The first two surfaces meet in a curve on the cylinder; the cone cuts this curve at a point \(P\). Let \(M\) be the foot of the perpendicular from \(P\) to the horizontal plane; \(M\) lies on the circle. Then \(AP\) and \(AM\) are the required means.

Theorem (Archytas)

With the construction above, \(AC:AP=AP:AM=AM:AB\).

Proof in coordinates

Put \(A\) at the origin, \(C=(a,0,0)\), the \(z\)-axis vertical. Write \(r=|AP|\) and \(\rho=|AM|=\sqrt{x^2+y^2}\) for the point \(P=(x,y,z)\). The three surfaces are

\[ \begin{aligned} \text{cylinder:}\quad & x^2+y^2=ax &&\iff \rho^2=ax,\\ \text{torus:}\quad & x^2+y^2+z^2=a\sqrt{x^2+y^2} &&\iff r^2=a\rho,\\ \text{cone:}\quad & x^2+y^2+z^2=\tfrac{a^2}{b^2}x^2 &&\iff r=\tfrac ab\,x. \end{aligned} \]

From the cone, \(x=\frac{b r}{a}\). Substituting in the cylinder gives \(\rho^2=br\). With the torus, \(r^2=a\rho\). Hence

\[ \frac ar=\frac r\rho \quad\text{and}\quad \frac r\rho=\frac\rho b, \qquad\text{so}\qquad a:r=r:\rho=\rho:b. \]

Archytas argued with similar right triangles instead of equations. Let \(C'\) be the turned position of \(C\). The triangle \(APC'\) is right-angled at \(P\), since \(P\) lies on the turned semicircle, and \(M\) is the foot of its altitude; this gives \(r^2=a\rho\). The triangle \(AMC\) is right-angled at \(M\), since \(M\) lies on the base circle; this gives \(\rho^2=a\,x\). The cone fixes the angle between \(AP\) and \(AC\), so that \(x:r=b:a\). Combining the three similar-triangle relations gives the same chain of proportions.

The curve

On the cylinder use the angle \(\theta\) between \(AM\) and \(AC\). Then \(\rho=a\cos\theta\), \(x=a\cos^2\theta\), and the torus gives \(z=a\sqrt{\cos\theta(1-\cos\theta)}\). This space curve is sometimes called the curve of Archytas. The cone meets it where \(\cos^3\theta=b^2/a^2\).

Exercise 9.1Take \(a=2\), \(b=1\). Find \(\cos\theta\), \(AM\), \(AP\) and the height of \(P\), and check the proportion numerically.

Show solution
\(\cos^3\theta=\frac14\), so \(\cos\theta=4^{-1/3}\approx0.630\). Then \(AM=\rho=2\cos\theta=2^{1/3}\approx1.260\), \(AP=r=\sqrt{a\rho}=\sqrt{2\cdot2^{1/3}}=2^{2/3}\approx1.587\), and \(z=\sqrt{r^2-\rho^2}=\sqrt{2^{4/3}-2^{2/3}}\approx0.966\). Check: \(2/1.587\approx1.260\), \(1.587/1.260\approx1.260\), \(1.260/1\approx1.260\). The cube on \(AM\) is \(2\), double the cube on \(AB=1\).

Exercise 9.2Show that the triangle \(APC'\), where \(C'\) is the far end of the turned semicircle, is right-angled at \(P\), and deduce \(r^2=a\rho\) geometrically.

Show solution
The turned semicircle has diameter \(AC'\) of length \(a\), lying in the vertical plane through \(A\) and \(M\); by Thales’ theorem the angle at \(P\) is right. In a right triangle the leg is the mean proportional between the hypotenuse and its projection on the hypotenuse. The projection of \(P\) onto the line \(AC'\) is \(M\), because \(PM\) is vertical and \(AC'\) horizontal in that plane. So \(AP^2=AC'\cdot AM\), that is \(r^2=a\rho\).

Exercise 9.3Menaechmus later found the two means as the intersection of the parabolas \(x^2=ay\) and \(y^2=bx\). Verify this and compare the two solutions.

Show solution
From \(x^2=ay\) we get \(a:x=x:y\), and from \(y^2=bx\) we get \(x:y=y:b\). So any intersection other than the origin gives the two means, and \(x^4=a^2y^2=a^2bx\) yields \(x^3=a^2b\). Menaechmus works in a plane with two curves of a new kind, conic sections; Archytas uses only familiar solids (cylinder, cone, the surface of a turning semicircle) but needs three dimensions. Both show that the problem cannot be done by the circle-and-line constructions of the Elements, which was proved only in 1837 by Pierre Wantzel.
Lesson 10

Mechanics and the edge of the cosmos

For the general reader. Ancient writers called Archytas the founder of mechanics and told of a wooden dove that flew. Most of these stories are late and hard to check. One argument of his is well attested, and philosophers have discussed it ever since: if you stood at the edge of the universe, could you stretch out your hand?

Goals
  • Separate attested from legendary claims about Archytas’ mechanics.
  • State the law of the lever as the Peripatetic Mechanical Problems derives it.
  • Analyse the staff argument for an unlimited universe.

What the sources say

  • Diogenes Laertius (VIII.83): Archytas was the first to treat mechanics systematically by means of mathematical principles. A book on mechanics circulated under his name, though some ascribed it to a namesake.
  • Vitruvius (On Architecture VII, preface) lists Archytas among writers on machines.
  • Plutarch (Life of Marcellus 14; Table Talk VIII.2) says Eudoxus and Archytas began the use of instruments in geometry and that Plato objected. Huffman judges this a later story, since the cube construction uses no machine.
  • Aristotle (Politics VIII.6) mentions the “rattle of Archytas”, a toy that keeps small children busy so they break nothing in the house.
  • Aulus Gellius (Attic Nights X.12), citing Favorinus, says Archytas made a wooden model of a dove that flew, apparently balanced by weights and moved by air enclosed within it. Gellius himself treats the story with caution.

The Mechanical Problems in the Aristotelian corpus is the oldest surviving Greek text on mechanics. It was probably written in Aristotle’s school; a few modern scholars have suggested Archytas as author, but there is no ancient support. It reduces the lever, the balance, the pulley and the wedge to the properties of the circle.

Law of the lever (Mechanical Problems; proved by Archimedes)

Weights \(W_1,W_2\) at distances \(d_1,d_2\) from the fulcrum balance when \(W_1d_1=W_2d_2\), that is, when the weights are inversely proportional to their distances.

The Mechanical Problems argues that a point farther from the centre of a rotating rod moves faster, covering an arc proportional to its radius, and so a small force at a long arm is equivalent to a larger force at a short one. Archimedes later gave an axiomatic proof.

The staff argument

Eudemus, as quoted by Simplicius, reports Archytas’ question. If I were at the outermost edge of the heaven, could I stretch my hand or a staff outside it, or not? It would be absurd if I could not. If I can, then what lies outside is either body or place. The same question can be asked again at the new edge, and so on without end. So whatever there is beyond, whether body or place, is unlimited.

Structure of the argument
  1. Suppose the cosmos has an outermost boundary.
  2. At any boundary point one can extend a staff outward (nothing could prevent it without being itself something beyond).
  3. So there is something (body or place) beyond the boundary, and the supposed edge is not an edge.
  4. The argument applies at every proposed edge; hence there is no edge.

Aristotle denied premise 2 by denying that there is any place outside the cosmos: “outside” has no meaning there. Lucretius retold the argument with a javelin thrown from the edge. In modern terms Archytas confused unboundedness with infinitude, a distinction that became visible only with non-Euclidean geometry: the surface of a sphere, or a closed three-dimensional universe, is finite but has no edge.

Exercise 10.1A 600 N stone rests 0.2 m from the fulcrum of a 2.2 m crowbar. What force at the far end lifts it, and through what distance must the hand move to raise the stone 1 cm?

Show solution
The far end is 2.0 m from the fulcrum, so \(F\cdot2.0=600\cdot0.2\) and \(F=60\) N. By similar arcs the hand moves \(2.0/0.2=10\) times as far, 10 cm. The product force × distance (6 J) is the same on both sides, the conservation principle hidden in the Mechanical Problems’ appeal to the circle.

Exercise 10.2An ant lives on the surface of a sphere and repeats Archytas’ argument. Which premise holds and which conclusion fails? What does this show about the argument?

Show solution
Premise 2 holds: at every point the ant can always step forward; the surface has no edge. But the conclusion that the surface is unlimited in extent fails: its area is finite. The argument proves that space has no boundary, not that it is infinite. Archytas and his ancient readers assumed Euclidean space, where the two coincide; Bernhard Riemann’s 1854 lecture made the distinction explicit.
Part III · Life and legacy
Lesson 11

Plato, Dionysius and the last years

Plato visited southern Italy and Sicily around 388/387 BC, aged about forty, and very likely met Archytas then. The relationship that followed is told mainly in Plato’s Seventh Letter, whose authenticity is disputed but which most historians use as evidence for events.

According to the letter, when Plato returned to Syracuse in 367/366 BC to advise the young tyrant Dionysius II, he brought about friendly ties between Dionysius and Archytas and the Tarentines. In 361 BC Dionysius pressed Plato to come a third time; letters from Archytas and his friends, assuring Plato that the tyrant had become serious about philosophy, helped persuade him. The visit went badly. Dionysius seized the property of Plato’s friend Dion, and Plato found himself confined and in danger. He wrote to Archytas. The letter says Archytas and the Tarentines sent a thirty-oared ship with an envoy, Lamiscus, under the pretext of an embassy from the city. Lamiscus pleaded with Dionysius, who let Plato go in 360 BC and even paid for the journey.

Diogenes Laertius quotes an exchange of letters between Archytas and Plato. These are generally considered forgeries, as is a letter from Archytas to Dionysius.

Death

Nothing reliable is known about the date or manner of Archytas’ death. He was probably dead by about 350 BC. Horace’s Ode I.28 begins by addressing Archytas, measurer of land, sea and countless sand, now held by a little gift of dust near the Matine shore in Apulia. Older readers took this as evidence that he drowned in a shipwreck. The speaker of the poem is in fact an unburied drowned sailor, and modern scholars, Huffman among them, do not use the ode as evidence for Archytas’ death.

Exercise 11.1The rescue story depends on the Seventh Letter. Suppose the letter is not by Plato. Which parts of the story would you still accept, and why?

Show solution
Even a forger writing in the fourth century BC, close to the events, would need a story his readers found credible; the letter shows detailed knowledge of Syracusan politics. The friendship between Plato and Archytas is supported independently by the Eudemian list in Proclus, which places Archytas among Plato’s associates, by Plato’s use of Archytas’ saying in the Republic, and by Aristoxenus’ Life, which featured an envoy of Dionysius. So the connection between Plato, Archytas and the Syracusan court is secure; the specific ship, envoy and date are as good as the letter.
Lesson 12

Legacy: from Eudoxus to the quadrivium

Mathematics

Eudoxus learned geometry from Archytas, and took up both of his problems: the theory of means and the doubling of the cube. Proclus, following Eudemus, says Eudoxus added three further means to the three of Archytas. Menaechmus, Eudoxus’ pupil, found the conic-section solution of the Delian problem. Archytas’ number theory of proportions probably survives, in Euclid’s systematic form, in Book VIII of the Elements. His proof on superparticular ratios became Proposition 3 of the Sectio canonis and, through Boethius, part of every medieval music curriculum.

Music and philosophy

Ptolemy treated Archytas as the most serious of the Pythagorean theorists and reported his tetrachords in detail. His theory of pitch, adapted by Plato and Aristotle, governed acoustics until the frequency theory of the seventeenth century. Aristotle wrote a lost work in several books on his philosophy, and Diogenes’ catalogue lists a separate summary of Plato’s Timaeus and the writings of Archytas. The staff argument was taken up by the Stoics and Epicureans and was cited in the early modern debate on infinite space.

Reputation and honours

His fame as a wise ruler produced a flood of forgeries in the Hellenistic period, written in his name to make Pythagoreanism the source of Platonic and Aristotelian philosophy. Separating the genuine Archytas from these texts was the work of nineteenth- and twentieth-century scholarship, culminating in Huffman’s edition. A lunar crater, Archytas, and the nearby rille Rima Archytas are named after him, as is the asteroid 14995 Archytas.

Exercise 12.1List the three problems of Archytas that Eudoxus and his school developed further, and say what each became in Euclid or later Greek mathematics.

Show solution
(1) The means: Eudoxus added three more, and later writers (Nicomachus, Pappus) list ten. (2) The doubling of the cube: Eudoxus used curves now lost, Menaechmus used conic sections, and Eutocius collected a dozen later solutions. (3) The arithmetic of ratios: the theory of numbers in continued proportion in Elements VIII, while Eudoxus extended proportion from numbers to arbitrary magnitudes in Book V.

Final projectChoose one of Lessons 05, 06 or 09. Write two pages for a strong undergraduate that set out what the ancient source reports, reconstruct the mathematics in modern notation, and explain what Archytas’ result tells us about the relation between numbers and magnitudes in his time.

Reference

Connections

People linked to Archytas of Tarentum across this collection. A name without a link has no page here yet. See Archytas of Tarentum in the network of all 337 lives →

Teachers and advisors

  • Philolaus of Croton

Students

Influenced by

  • Hippocrates of Chios

Influenced

Friends

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Ancient sources

  1. Diogenes Laertius, Lives of the Eminent Philosophers VIII.79–83, transl. C. D. Yonge (Wikisource).
  2. Plato, Seventh Letter 338c–339d and 350a–b; Greek text and translation of 350a.
  3. Plato, Republic VII, 530d; Timaeus 67b, 80a.
  4. Aristotle, Politics VIII.6; Metaphysics H.2; Rhetoric III.11.
  5. Eutocius, Commentary on Archimedes’ On the Sphere and Cylinder II, with the account of Eudemus; transl. in T. L. Heath, A History of Greek Mathematics, vol. 1, Oxford, 1921, pp. 246–249.
  6. Ptolemy, Harmonics I.13–14; Porphyry, Commentary on Ptolemy’s Harmonics.
  7. Boethius, De institutione musica III.11.
  8. Aulus Gellius, Attic Nights X.12; Horace, Odes I.28.

Modern studies

  1. C. A. Huffman, Archytas, Stanford Encyclopedia of Philosophy.
  2. C. A. Huffman, Archytas of Tarentum: Pythagorean, Philosopher and Mathematician King, Cambridge University Press, 2005.
  3. J. J. O’Connor and E. F. Robertson, Archytas of Tarentum, MacTutor History of Mathematics.
  4. Archytas, Wikipedia.
  5. A. Barker, Greek Musical Writings, vol. 2: Harmonic and Acoustic Theory, Cambridge University Press, 1989.
  6. W. R. Knorr, The Ancient Tradition of Geometric Problems, Birkhäuser, 1986.
  7. B. L. van der Waerden, Science Awakening, Noordhoff, 1954.
  8. T. L. Heath, The Thirteen Books of Euclid’s Elements, 2nd ed., Cambridge, 1926, vol. 2 (Book VIII).