How to use this course
Niccolò Tartaglia is usually met as the wronged party in the story of the cubic equation: the man who found the rule, told Gerolamo Cardano under an oath of secrecy, and saw it printed in Cardano’s Ars Magna. That story is real, and this course tells it carefully. But Tartaglia was also the first to treat the flight of a cannonball as a problem in mathematics, the first to print Euclid in a modern European language, an editor of Archimedes, and the author of a six-part encyclopedia of practical mathematics that contains the arithmetic triangle, a volume formula for the tetrahedron, and a critique of an early rule for dividing stakes.
- Algebra of polynomials, including completing the square.
- Basic trigonometry and the kinematics of uniform acceleration, for Lesson 06.
- Vectors in three dimensions and determinants, for Lesson 09.
- Elementary probability, for Lesson 10.
Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Tartaglia wrote in Venetian Italian and stated his rules in words and worked numbers. We use modern notation and say where his own formulation differs. Much of what we know of his early life comes from an autobiographical passage in his Quesiti et inventioni diverse (1546, Book VI, question 8), and much of the cubic story comes from Book IX of the same work, written by an angry man. We flag claims that rest on his word alone.
This page is a companion to the Cardano course. The two accounts of the cubic dispute agree in their facts and differ in where they stand.
Timeline
| 1499/1500 | Born Niccolò Fontana in Brescia, in the Venetian Republic. His father Michele is a mounted courier. |
| c. 1506 | His father dies, murdered on the road by most accounts, leaving the family poor. |
| 1512 | French troops sack Brescia. Sheltering in the cathedral, he receives sabre wounds to the jaw and palate that leave him with a stammer. |
| 1516–18 | Moves to Verona as a teacher of the abacus, probably in this period. |
| 1534 | Moves to Venice as a teacher of mathematics. |
| 1535 | 13 February: finds the rule for the cubic \(x^3+px=q\) and defeats Antonio Maria Fior in a public contest. |
| 1537 | Nova scientia, the first book on the mathematics of gunnery. |
| 1539 | 25 March: in Milan, gives Cardano the cubic rule in a poem after Cardano swears secrecy. August: refuses to help Cardano with the irreducible case. |
| 1543 | Euclide Megarense, the first printed translation of the Elements into a modern European language; a Latin edition of works of Archimedes. |
| 1545 | Cardano’s Ars Magna publishes the cubic rule, crediting Scipione del Ferro and Tartaglia. |
| 1546 | Quesiti et inventioni diverse; Book IX accuses Cardano of breaking his oath. |
| 1547–48 | Exchange of printed challenges (cartelli) with Lodovico Ferrari, 10 February 1547 to 24 July 1548. |
| 1548 | 10 August: public disputation with Ferrari in Milan. Tartaglia leaves after the first day. Takes up a lectureship in Brescia. |
| 1549 | Returns to Venice after about eighteen months, his salary unpaid. |
| 1551 | Travagliata inventione, on raising sunken ships, with the Ragionamenti on Archimedes’ floating bodies. |
| 1556 | Parts I–III of the General trattato di numeri et misure. |
| 1557 | Will dated 10 December. Dies in Venice on 13 December, near the Rialto. |
| 1560 | Parts IV–VI of the General trattato published posthumously by Curtio Troiano. |
Brescia, 1499–1516
Niccolò Fontana was born in Brescia in 1499 or 1500. Brescia was then a city of the Venetian Republic. His father, Michele Fontana, carried post on horseback between Brescia and the towns around it. Around 1506 Michele died, murdered on the road according to most accounts. His widow was left with three children and almost nothing.
The sack of 1512
In 1512, during the War of the League of Cambrai, French troops under Gaston de Foix retook Brescia after the city had revolted, and sacked it. The family took refuge in the cathedral. A soldier cut the boy across the face with a sabre, splitting his jaw and palate, and left him for dead. The Dictionary of Scientific Biography counts five serious head wounds. His mother could not afford a doctor and nursed him herself. He recovered, but with a lasting difficulty in speech. Other children called him Tartaglia, “stammerer”, and he adopted the name. He wore a beard for the rest of his life to hide the scars.
Teaching himself
Tartaglia’s own account, in the Quesiti, says that at about fourteen he went to a writing master to learn the alphabet, and stopped at the letter k when the money ran out. From then on he taught himself, from whatever books he could find. MacTutor reports that a patron, Ludovico Balbisonio, took him to Padua for a period of study. By his late teens he knew enough mathematics to earn a living teaching it.
Abacus master in Verona and Venice, 1516–1535
Probably between 1516 and 1518 Tartaglia moved to Verona as a maestro d’abaco, a teacher of commercial arithmetic. The abacus schools of Italian cities taught merchants’ sons to calculate with Hindu–Arabic numerals, convert currencies, compute interest and partnerships, and solve problems that led to linear and quadratic equations. Their masters lived by fees and by reputation, and reputation was won in public. A master challenged a rival with a list of problems; the loser lost pupils.
He married in Verona and had children, and stayed poor. He also advised gunners and architects, an unusual sideline for an abacus master that led to his first book. Wikipedia records that he once sued a client who had paid him with a worn-out cloak.
In 1534 he moved to Venice, the great printing centre of Europe, and taught mathematics there. In Venice the questions put to him by gunners, merchants and fellow masters fill the Quesiti of 1546. One correspondent, the Brescian teacher Zuanne da Coi, sent him problems that led to cubic equations. Tartaglia later said that this had set him working on cubics of the type “cube and squares equal to a number”, \(x^3+ax^2=b\), for which he claimed a method by the early 1530s.
The contest, the oath and the quarrel, 1535–1548
Luca Pacioli’s Summa (1494) had suggested that cubic equations could not be solved by a general rule. Around the 1510s Scipione del Ferro, a professor at Bologna, found how to solve \(x^3+px=q\). He kept it secret and, before he died in 1526, passed it to his student Antonio Maria Fior.
The contest of 1535
Fior, a weaker mathematician, challenged Tartaglia to a public contest in Venice. Each man deposited thirty problems with a notary, and each had forty or fifty days to solve the other’s. Tartaglia set a varied list. Fior set thirty problems all of the one type he could solve, “a cube and things equal to a number”. Tartaglia knew a rumour that Fior had a secret rule. Shortly before the deadline, in the early hours of 13 February 1535, he found the rule for himself, and he then solved all thirty of Fior’s problems in about two hours. Fior was beaten decisively. Tartaglia declined the small prize.
Milan, 1539
News of the contest reached Cardano in Milan, who was writing a large arithmetic and wanted the rule. Through an intermediary, Zuan Antonio da Bassano, he asked for it in January 1539. Tartaglia refused: he intended to publish it in a book of his own. Letters followed, some angry. Cardano hinted that he could introduce Tartaglia to Alfonso d’Avalos, the Marquis del Vasto, governor of Milan, a useful patron for a man who wrote on artillery. Tartaglia came to Milan. On 25 March 1539, in Cardano’s house, after Cardano swore not to publish it, Tartaglia gave him the rule in the form of a poem of three stanzas. He did not wait to meet the marquis. In August he refused to help Cardano with a cubic the rule could not handle (Lesson 05), and accused him of bad faith.
Ars Magna and the quarrel
Cardano and his pupil Lodovico Ferrari extended the rule to every type of cubic, and Ferrari reduced the quartic to a cubic. In 1543 they visited Bologna, where Annibale della Nave, del Ferro’s son-in-law and successor, showed them del Ferro’s notebook. Cardano decided that the oath did not bind him to suppress a discovery that was not Tartaglia’s alone. Ars Magna appeared in 1545. It credits del Ferro with the first discovery and Tartaglia with an independent one made in competition.
Tartaglia printed his version in Book IX of the Quesiti (1546), with the letters, the oath and the poem, and with abuse of Cardano. Ferrari answered for his master. Between 10 February 1547 and 24 July 1548 the two exchanged six printed challenges each, the cartelli and Tartaglia’s risposte, each with problems for the other. The quarrel ended in a public disputation on 10 August 1548 in the church of Santa Maria del Giardino in Milan, presided over by the governor, Ferrante Gonzaga. Each side had posed sixty-two problems. By the end of the first day Ferrari was clearly ahead, and Tartaglia left Milan that night.
DiscussionTartaglia kept secret a rule that del Ferro had found first; Cardano broke an oath but printed the rule with credit to both. Del Ferro and Fior kept the rule as a weapon for contests. What did secrecy protect in a world of abacus schools and public challenges, and what did it cost mathematics?
A cube and things: the rule of 1535
For the general reader. On a February night in 1535 Tartaglia found how to solve “a cube and things equal to a number”, an equation like \(x^3+3x=14\). The idea is to write the unknown as a difference of two cube roots, chosen so that the cubic falls apart into a quadratic. With that one idea he answered thirty problems in two hours.
- Derive the rule for \(x^3+px=q\) with \(p,q>0\).
- See why every one of Fior’s problems fell to it.
- Recognize simple roots hidden in nested radicals.
Let \(p,q>0\). The equation \(x^3+px=q\) has exactly one real root, and it is positive:
\[ x=\sqrt[3]{\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}+\frac q2}\;-\;\sqrt[3]{\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}-\frac q2}. \]Derivation
Try \(x=u-v\). For any \(u,v\),
\[ (u-v)^3+3uv\,(u-v)=u^3-v^3. \]So \(x=u-v\) is a root as soon as \(3uv=p\) and \(u^3-v^3=q\). Write \(U=u^3\), \(V=v^3\). Then \(U-V=q\) and \(UV=(p/3)^3\). Two numbers with known difference and product are found by a quadratic: \(U\) and \(-V\) are the roots of \(t^2-qt-(p/3)^3=0\), so
\[ U=\frac q2+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}},\qquad V=\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}-\frac q2, \]both positive. Taking real cube roots gives the formula. The function \(x^3+px\) is strictly increasing, so the root is the only real one.
Tartaglia’s words
The first stanza of the poem of 1539 says this in verse: when the cube and the things together equal a number, find two other numbers that differ by that number and whose product is the cube of a third of the things; then the difference of their cube roots is the thing sought. In the notation above: \(U-V=q\), \(UV=(p/3)^3\), \(x=\sqrt[3]U-\sqrt[3]V\). No symbols, no proof, but a complete algorithm.
Why Fior lost
Fior’s thirty problems were word problems that each reduced to one equation of the form \(x^3+px=q\) with positive \(p,q\). A single rule therefore solved all of them. Tartaglia’s problems were of many kinds, including ones that needed geometry, and Fior had no general method to fall back on.
Exercise 4.1Solve \(x^3+3x=14\) by the rule, then show that the answer simplifies to an integer.
Show solution
Exercise 4.2A typical contest problem: find a number which, added to its cube, gives 6. Write the root by the rule and evaluate it to four decimal places.
Show solution
Exercise 4.3Show that \(U\) and \(-V\) are the roots of \(t^2-qt-p^3/27=0\), and explain why \(U>0>-V\) whenever \(p>0\).
Show solution
The poem and the other cubics
For the general reader. Without negative numbers, a cubic could be written in several genuinely different ways: “a cube equal to things and a number”, “a cube and a number equal to things”, “a cube and squares equal to a number”. Tartaglia’s poem covers three of these. One of them hides a trap: for some equations with three ordinary roots the rule asks for the square root of a negative number. Cardano hit the trap in 1539, and Tartaglia refused to help.
- Derive the rule for \(x^3=px+q\) and connect it to \(x^3+q=px\).
- Recognize the irreducible case from the coefficients.
- Reduce “cube and squares” to a depressed cubic.
The second and third stanzas
The second stanza treats the cube “left alone”, \(x^3=px+q\). The same substitution with a sum, \(x=u+v\), \(3uv=p\), \(u^3+v^3=q\), gives
Let \(p,q>0\) and suppose \(q^2/4\ge p^3/27\). Then a positive root of \(x^3=px+q\) is
\[ x=\sqrt[3]{\frac q2+\sqrt{\frac{q^2}{4}-\frac{p^3}{27}}}+\sqrt[3]{\frac q2-\sqrt{\frac{q^2}{4}-\frac{p^3}{27}}}. \]The third stanza says that \(x^3+q=px\) is handled through the second case. In modern terms, \(x_0\) solves \(x^3=px+q\) if and only if \(-x_0\) solves \(x^3+q=px\), and the remaining roots come from a quadratic.
The irreducible case
Let \(p,q>0\). If \(q^2/4<p^3/27\), then \(x^3=px+q\) has three distinct real roots, and the square root in the rule is the square root of a negative number.
Proof. Let \(f(x)=x^3-px-q\). Its critical points are \(\pm\sqrt{p/3}\), and \(f(-\sqrt{p/3})=\tfrac{2p}{3}\sqrt{p/3}-q\). This is positive exactly when \(q^2<\tfrac{4p^3}{27}\). Then \(f\) goes from \(-\infty\) up to a positive local maximum, down to a negative local minimum (since \(f(\sqrt{p/3})<-q<0\)), and up to \(+\infty\): three sign changes, three real roots.
Later in 1539 Cardano asked Tartaglia about the equation “a cube equal to nine things plus ten”, \(x^3=9x+10\). Here \(q^2/4=25<27=p^3/27\). In his reply of August 1539 Tartaglia did not answer the question and complained that Cardano had broken faith. Neither man could resolve the case. Rafael Bombelli showed in 1572 how to compute through it with complex numbers.
Cube and squares
The type Tartaglia had studied first, \(x^3+ax^2=b\), reduces to the others. The shift \(x=y-a/3\) removes the square term:
\[ x^3+ax^2-b=y^3-\frac{a^2}{3}\,y+\Big(\frac{2a^3}{27}-b\Big). \]Exercise 5.1Show that \(x^3=6x+40\) falls under the rule, and simplify the answer.
Show solution
Exercise 5.2Find all three roots of Cardano’s \(x^3=9x+10\) without the rule.
Show solution
Exercise 5.3Reduce \(x^3+3x^2=5\) to a depressed cubic, solve it by the rule, and give the root to three decimals.
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Nova scientia: the new science of gunnery
For the general reader. Gunners in the 1530s aimed by experience. Tartaglia’s Nova scientia (Venice, 1537) tried to turn their craft into geometry. His physics was still Aristotle’s and his picture of the path was wrong in detail, but he asked the right questions, insisted that a shot’s path is curved throughout, and stated the correct answer to the gunner’s main question: the greatest range comes at an elevation of 45 degrees.
- Describe Tartaglia’s model of a trajectory and what was new in it.
- Prove the 45° theorem in the vacuum model.
- Use the range formula to aim at a target.
The book
Nova scientia was dedicated to Francesco Maria della Rovere, Duke of Urbino and captain-general of the Venetian forces. It was the first book to apply mathematics to the paths of projectiles. In it, and later in the Quesiti, Tartaglia described instruments for measuring inaccessible heights and distances, introduced firing tables, and analysed a trajectory in three parts: a straight “violent” motion, a circular arc, and a straight “natural” fall. In the Quesiti he went further and said that the path is curved everywhere, since gravity acts on the ball from the start.
His argument for 45° was not sound by later standards. The result is still called Tartaglia’s theorem, and it is exactly correct in the absence of air resistance. Galileo Galilei, who owned and annotated copies of Tartaglia’s ballistic works, proved in his Two New Sciences (1638) that the path in a vacuum is a parabola, and derived the 45° result from it.
A projectile launched from level ground with speed \(v\) at elevation \(\theta\in(0,\pi/2)\), under constant gravity \(g\) and with no air resistance, lands at distance
\[ R(\theta)=\frac{v^2\sin2\theta}{g}. \]The range is greatest, equal to \(v^2/g\), at \(\theta=45^\circ\), and elevations \(45^\circ\pm\alpha\) give equal ranges.
Proof. The position at time \(t\) is \(x=vt\cos\theta\), \(y=vt\sin\theta-\tfrac12gt^2\). The shot lands when \(y=0\), at \(t=2v\sin\theta/g\), so \(R=2v^2\sin\theta\cos\theta/g=v^2\sin2\theta/g\). The factor \(\sin2\theta\) is largest at \(2\theta=90^\circ\), and \(\sin(90^\circ+2\alpha)=\sin(90^\circ-2\alpha)\). Eliminating \(t\) shows the path is the parabola \(y=x\tan\theta-\dfrac{g\,x^2}{2v^2\cos^2\theta}\), curved at every point, as Tartaglia had said.
With air resistance the best elevation is lower than 45°, and much lower for fast, light shot. Tartaglia’s firing practice and his theory therefore did not agree perfectly, and later gunners kept relying on trial.
Exercise 6.1In the vacuum model a gun has a greatest range of 1000 m. At what elevations will it hit a target 800 m away on level ground?
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Exercise 6.2Show that the highest point of the vacuum trajectory is at height \(\tfrac{v^2\sin^2\theta}{2g}\), and that at \(\theta=45^\circ\) it equals a quarter of the range.
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Exercise 6.3Tartaglia’s model contains a straight initial segment. Using the parabola, compute the angle between the direction of motion and the launch direction after the shot has covered one tenth of the range at \(45^\circ\). Is the path “nearly straight” there?
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Euclid, Archimedes and floating bodies
For the general reader. In 1543 Tartaglia published two editions that changed what Italian readers could study. His Euclide Megarense was the first translation of Euclid’s Elements into a modern European language, and it got right the subtlest part of the book, the theory of proportion in Book V. His Latin Archimedes, and his later Italian version of part of On Floating Bodies, put the founding texts of statics and hydrostatics into circulation.
- State Euclid’s definition of equal ratios and see why it is subtle.
- State Archimedes’ principle and use it to analyse an alloy.
- Understand the editorial controversies around Tartaglia’s editions.
Book V
The Elements had circulated in Latin in the medieval version of Campanus of Novara, in which the definitions of Book V were garbled. Tartaglia’s Italian translation, with commentary, rendered them correctly. The key definition, due to Eudoxus, defines when two ratios are equal without assuming that ratios are numbers, which is how it copes with incommensurable magnitudes.
Magnitudes \(a,b\) (of one kind) and \(c,d\) (of one kind) are in the same ratio, \(a:b=c:d\), if for all positive integers \(m,n\):
\[ ma>nb\iff mc>nd,\qquad ma=nb\iff mc=nd,\qquad ma<nb\iff mc<nd. \]In modern terms, \(a:b\) is determined by the set of rationals \(n/m\) below it and the set above it. This is in essence a cut in the sense of Richard Dedekind.
Archimedes
Tartaglia’s Opera Archimedis (1543) printed Latin texts on the parabola, the circle, centres of gravity and floating bodies. Critics then and since pointed out that the text was essentially William of Moerbeke’s thirteenth-century translation, and that Tartaglia did not say so. In Travagliata inventione (1551), a book on raising sunken ships, its Ragionamento primo gave an Italian translation, with commentary, of part of Book I of On Floating Bodies, and the book discussed specific weights.
A solid immersed in a fluid is buoyed up by a force equal to the weight of fluid it displaces. A solid lighter than the fluid floats with so much of it submerged that the displaced fluid weighs as much as the whole solid.
Exercise 7.1Using the definition, show that the diagonal and side of a square, \(d:s\), are in the same ratio as \(2s:d\).
Show solution
Exercise 7.2A crown weighs 1000 g in air and 940 g when fully immersed in water. Assuming it is gold (density 19.3 g/cm³) alloyed with silver (10.5 g/cm³), how much gold does it contain?
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The triangle and the extraction of roots
For the general reader. Italians still call the arithmetic triangle il triangolo di Tartaglia. It appears in Part II of his General trattato di numeri et misure (1556), about a century before Blaise Pascal’s treatise. Tartaglia did not invent it; versions appear earlier in Chinese, Islamic and European sources. He used it for a practical purpose: to extract roots of any degree, digit by digit.
- Prove the addition rule and the binomial theorem from it.
- Use binomial coefficients to extract a cube root by hand.
- Place Tartaglia’s triangle in its long history.
Define \(\binom{n}{k}\) by \(\binom{n}{0}=\binom{n}{n}=1\) and \(\binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}\) for \(1\le k\le n\). Then for all \(a,b\) and \(n\ge0\),
\[ (a+b)^n=\sum_{k=0}^{n}\binom nk a^{n-k}b^k. \]Proof. Induction on \(n\). Multiply the expansion of \((a+b)^n\) by \(a+b\). The coefficient of \(a^{n+1-k}b^k\) collects \(\binom nk\) from multiplying by \(a\) and \(\binom n{k-1}\) from multiplying by \(b\), and their sum is \(\binom{n+1}{k}\) by the addition rule.
Tartaglia’s text states the addition rule by example: two neighbouring entries such as 15 and 20 in one row add to give the 35 below them. Those are \(\binom62=15,\ \binom63=20,\ \binom73=35\).
Roots digit by digit
To extract a cube root, write the root as \(10a+b\) where \(a\) is already known and \(b\) is the next digit. Then
\[ (10a+b)^3=1000a^3+3\cdot100a^2\,b+3\cdot10a\,b^2+b^3, \]and the coefficients \(1,3,3,1\) come from the triangle. Subtract \(1000a^3\) and choose the largest digit \(b\) with \(300a^2b+30ab^2+b^3\) not exceeding the remainder. Row \(n\) of the triangle gives the same procedure for \(n\)th roots. The General trattato also treats the rationalization of denominators, the companion skill for computing with the roots in Lessons 04 and 05.
Exercise 8.1Extract the cube root of 12167 by the digit method.
Show solution
Exercise 8.2Prove that the entries of row \(n\) sum to \(2^n\) and that their alternating sum is 0 for \(n\ge1\). Then show \(\binom nk=\dfrac{n!}{k!\,(n-k)!}\) satisfies the addition rule.
Show solution
The volume of a tetrahedron
For the general reader. Heron of Alexandria found the area of a triangle from its three sides. Part IV of Tartaglia’s General trattato does the same for a pyramid on a triangular base: given all six edges, find the height and hence the volume. He worked by example, using the law of cosines without proof, and the method is sound. The modern formula, a determinant in the squared edge lengths, is named after Arthur Cayley and Karl Menger, and is sometimes called Tartaglia’s formula.
- Compute the height of a tetrahedron from its edges, as Tartaglia did.
- State the Cayley–Menger determinant formula.
- Check it on regular and irregular examples.
Tartaglia’s procedure
Let the base be triangle \(ABC\) and the apex \(D\). Put \(B\) at the origin and \(C\) on the first axis. The law of cosines locates \(A\) in the plane. The foot \(H\) of the altitude from \(D\) is found by the same reasoning: the differences \(DB^2-DC^2\) and \(DB^2-DA^2\) equal the corresponding differences \(HB^2-HC^2\) and \(HB^2-HA^2\), which are linear in the coordinates of \(H\). Then the height is \(h=\sqrt{DB^2-HB^2}\) and the volume is one third of base area times height.
Let a tetrahedron have vertices \(P_1,\dots,P_4\) and let \(d_{ij}=|P_iP_j|\). Its volume \(V\) satisfies
\[ 288\,V^2=\det\begin{pmatrix}0&1&1&1&1\\ 1&0&d_{12}^2&d_{13}^2&d_{14}^2\\ 1&d_{12}^2&0&d_{23}^2&d_{24}^2\\ 1&d_{13}^2&d_{23}^2&0&d_{34}^2\\ 1&d_{14}^2&d_{24}^2&d_{34}^2&0\end{pmatrix}. \]Proof sketch. With \(P_4\) at the origin and \(u_i=P_i\), \(6V=|\det(u_1,u_2,u_3)|\), so \(36V^2=\det G\) where \(G_{ij}=u_i\cdot u_j\) is the Gram matrix. By the law of cosines \(u_i\cdot u_j=\tfrac12(d_{i4}^2+d_{j4}^2-d_{ij}^2)\). Row and column operations on the bordered matrix above reduce it to \(\pm8\det G\), with the sign making the right side positive, and \(8\cdot36=288\). The analogous \(4\times4\) determinant gives \(16T^2\) for a triangle, which is Heron’s formula.
Exercise 9.1Find the volume of a regular tetrahedron with edge \(a\), first by Tartaglia’s method (height over the centroid) and then by checking the determinant.
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Exercise 9.2Tartaglia’s worked example uses edges 13, 14, 15, 20, 18, 16. Take the base \(AB=13\), \(BC=14\), \(CA=15\) and lateral edges \(DA=20\), \(DB=18\), \(DC=16\). Find the height and the volume.
Show solution
The same part of the General trattato poses the problem of inscribing three circles in a triangle, now known as Malfatti’s problem, and develops geometry with a compass of fixed opening.
Dividing the stakes
For the general reader. Two players agree to play until one has won a set number of points. The game is interrupted. How should they split the pot? Pacioli proposed a rule in 1494. Tartaglia, in the General trattato, showed it was absurd and proposed another, while doubting that any division would satisfy both players. The correct answer, based on each player’s chance of winning, came only with Pascal and Fermat in 1654.
- State Pacioli’s and Tartaglia’s rules and compute with them.
- Derive the Pascal–Fermat division.
- See exactly where Tartaglia’s rule fails.
Tartaglia’s example
A game is played to 60 points, scored 10 at a time. Each side stakes 22 ducats. Play stops at 50 to 30. Pacioli divides the pot in proportion to points won, 5 to 3. Tartaglia objects that if play stopped at 10 to 0 this rule would give everything to one side after a single goal. His own rule gives the leader his own stake plus a share of the other’s stake proportional to the lead as a fraction of the game.
In a game to \(c\) points with stakes \(S\) from each side, interrupted at \(a\) to \(b\) with \(a\ge b\), the leader receives \(S+\dfrac{a-b}{c}\,S\) and the other side \(S-\dfrac{a-b}{c}\,S\).
For 50 to 30: the lead 20 is a third of 60, so the leader gets \(22+\tfrac{22}{3}=29\tfrac13\) ducats and the other side \(14\tfrac23\). Tartaglia added that the question is one of judgment more than reason, and that any division would give cause for litigation.
Suppose each round is won by either side with probability \(\tfrac12\), independently, and the leader needs \(r\) more rounds and the other side \(s\). The fair share of the leader is the probability that he would have won,
\[ P=\sum_{k=r}^{r+s-1}\binom{r+s-1}{k}2^{-(r+s-1)}. \]Proof. Imagine that exactly \(r+s-1\) more rounds are played, whatever happens. The leader wins the real game if and only if he wins at least \(r\) of them, since then the other side wins at most \(s-1\). All \(2^{r+s-1}\) sequences are equally likely.
Exercise 10.1Compute the Pascal–Fermat division of the 44 ducats in Tartaglia’s example, and compare it with the rules of Pacioli and Tartaglia.
Show solution
Exercise 10.2In a game to 100 single points, compare Tartaglia’s rule and the fair division at 65 to 55 and at 99 to 89.
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Brescia again and the last years, 1548–1557
In 1548 Tartaglia was offered a lectureship in his native Brescia. MacTutor reports that he was expected to defend his reputation against Ferrari before taking it up, which is one reason he went to Milan in August. After the defeat he lectured in Brescia for about eighteen months. His salary was not paid. He sued, lost, and returned to Venice in 1549 in debt.
He kept writing. Travagliata inventione (1551), dedicated to the problem of raising sunken ships, also discussed diving apparatus, weather signs and specific weights, and contained his Italian Archimedes. Critics noted that his ship-raising procedure had been proposed by others. His most ambitious work, the General trattato di numeri et misure, ran to about 1,500 pages in six parts, running from commercial arithmetic through general arithmetic, with the triangle and root extraction, to geometry and measurement. He saw Parts I–III through the press in 1556.
Tartaglia died poor and alone in his house near the Rialto bridge in Venice on 13 December 1557, three days after dictating a will that mentions a brother named Fontana. Curtio Troiano published the remaining parts of the General trattato in 1560, and in 1565 brought out Archimedes’ On Floating Bodies and Jordanus de Nemore’s treatise on weights from Tartaglia’s papers. The book in which he had planned to publish the cubic rule himself never appeared.
Students
Tartaglia taught privately for most of his life. His pupils included the Englishman Richard Wentworth, Giovanni Antonio Rusconi, Maffeo Poveiano and, most important, Giovanni Battista Benedetti, whose criticisms of Aristotelian mechanics anticipated Galileo.
Legacy
Tartaglia lost the cubic dispute in the court of history for three centuries; the formula was called Cardano’s. Historians now usually call it the Cardano–Tartaglia formula, and many give del Ferro his place too. Wikipedia notes that later stories of Tartaglia devoting his life to ruining Cardano seem to be fabricated.
- Algebra. The rule of 1535 broke the barrier Pacioli had described and led directly to Ars Magna, to Ferrari’s quartic, and to Bombelli’s complex numbers.
- Mechanics. Nova scientia began the mathematical study of projectiles. Its correct 45° result and its claim that trajectories are curved throughout passed, through Benedetti and Galileo’s annotated copies, into the new mechanics of the seventeenth century.
- Texts. His Italian Euclid, reprinted many times, brought the Elements to readers without Latin, with a correct Book V. His Archimedes editions, whatever their debts, spread the texts of statics and hydrostatics.
- Practical mathematics. The General trattato contains the arithmetic triangle, root extraction, the tetrahedron volume, Malfatti’s problem, surveying with the surveyor’s cross, and the critique of Pacioli on dividing stakes.
Final projectWrite a two-page account of the cubic dispute from the point of view of a neutral observer in 1550 who has read Ars Magna, the Quesiti and Ferrari’s cartelli. Include one fully worked cubic of each type in the poem, and explain which of Tartaglia’s complaints you think were justified.
Connections
People linked to Niccolò Tartaglia across this collection. A name without a link has no page here yet. See Niccolò Tartaglia in the network of all 337 lives →
Students
- Giovanni Battista Benedetti
- Richard Wentworth
Influenced by
- Archimedes
- Euclid
- Luca Pacioli
Influenced
Colleagues and collaborators
- Scipione del Ferro
- Zuanne da Coi
Rivals
- Gerolamo Cardano 1539-1557
- Antonio Maria Fior 1535
- Lodovico Ferrari 1547-1548
Also mentioned in this life
Fields
On CoursesHub
Sources and further reading
Biography
- J. J. O’Connor and E. F. Robertson, Nicolo Tartaglia, MacTutor History of Mathematics.
- A. Masotti, Tartaglia, Niccolò, Dictionary of Scientific Biography (via MacTutor).
- J. J. O’Connor and E. F. Robertson, Tartaglia versus Cardan, MacTutor, with translated extracts from the Quesiti.
- Niccolò Fontana Tartaglia, Wikipedia.
- J. J. O’Connor and E. F. Robertson, Girolamo Cardano and Lodovico Ferrari, MacTutor.
Mathematics
- N. Tartaglia, Quesiti et inventioni diverse, Venice, 1546; Book IX contains the cubic correspondence and the poem.
- N. Tartaglia, Nova scientia, Venice, 1537.
- N. Tartaglia, General trattato di numeri et misure, Venice, 1556–1560. R. J. Pulskamp, translation of the passage on dividing stakes.
- G. Cardano, Ars Magna or the Rules of Algebra, trans. T. R. Witmer, MIT Press, 1968; Dover, 1993.
- Problem of points, Wikipedia.
- Cubic equation, Wikipedia, history section.
- J. Stillwell, Mathematics and Its History, Springer, 3rd ed. 2010.
- Galileo Galilei, Discorsi e dimostrazioni matematiche intorno a due nuove scienze, Leiden, 1638, Fourth Day.