courseshub.world · Lives in Mathematics and Physics

Gerolamo Cardano

Gerolamo Cardano · Hieronymus Cardanus · Jerome Cardan

A course in two strands: the turbulent life of a Renaissance physician, astrologer and gambler, and the mathematics he put into print: the solution of the cubic and quartic, the first calculations with square roots of negative numbers, and the first book on the mathematics of chance.

Born 24 September 1501, Pavia Died 21 September 1576, Rome Ars Magna Nuremberg, 1545 Level undergraduate → advanced undergraduate
x³ = 15x + 4 has three real roots, 4 and −2 ± √3, yet Cardano’s rule expresses them through √−121. Rafael Bombelli resolved the paradox in 1572.
Lesson 00

How to use this course

Gerolamo Cardano was the most famous physician in Europe in the 1550s, a prolific writer on everything from astrology to the education of the deaf, and the author of one of the most candid autobiographies of the Renaissance. Mathematicians remember him for Ars Magna (1545), the book that published the solutions of the cubic and quartic equations and opened modern algebra. Probabilists remember him for a gambler’s manual written a century before Blaise Pascal and Pierre de Fermat.

Prerequisites for Part II
  • Algebra of polynomials, including factoring and completing the square.
  • Complex numbers in Cartesian and polar form, for Lesson 06.
  • Elementary counting and probability, for Lessons 08 and 09.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proofs and exercises with folded solutions. Cardano wrote in rhetorical Latin with only light abbreviation; we use modern notation throughout and say where his own formulation differs. Much of the biography comes from his autobiography, De propria vita, written in his last year. It is vivid and self-justifying, and we flag claims that rest on it alone.

Reference

Timeline

1501Born 24 September in Pavia, the illegitimate son of the Milanese jurist Fazio Cardano and Chiara Micheri.
1520Enters the University of Pavia to study medicine.
1524–25War closes Pavia; moves to Padua. Doctorate in medicine, 1525 (some sources give 1526).
1526–32Practises medicine at Sacco, near Padua, after the Milan College of Physicians refuses him.
1531Marries Lucia Bandarini.
1534Lecturer in mathematics at the Piatti foundation in Milan, his father’s former post.
153513 February: Niccolò Tartaglia defeats Antonio Maria Fior in a cubic-solving contest in Venice.
1539Admitted to the Milan College of Physicians. Publishes Practica arithmetice. March: Tartaglia reveals his rule to Cardano in Milan under an oath of secrecy.
1540His pupil Lodovico Ferrari solves the quartic.
1543In Bologna, Cardano and Ferrari see Scipione del Ferro’s earlier solution. Takes the chair of medicine at Pavia.
1545Artis magnae, sive de regulis algebraicis published by Petreius in Nuremberg.
1546His wife Lucia dies. Tartaglia’s Quesiti et inventioni diverse accuses Cardano of perjury.
154810 August: public disputation between Ferrari and Tartaglia in Milan.
1550De subtilitate, an encyclopedia of natural philosophy.
1552Travels to Scotland to treat John Hamilton, Archbishop of St Andrews.
1557De rerum varietate.
156013 April: his elder son Giovanni Battista is executed for poisoning his wife.
1562Professor of medicine at Bologna.
1570Imprisoned by the Inquisition for several months; loses his chair. Second edition of Ars Magna; Opus novum de proportionibus (Basel).
1571Moves to Rome; receives a papal pension.
1576Dies in Rome on 21 September.
1643De propria vita published in Paris by Gabriel Naudé.
1663Liber de ludo aleae published in his collected works, Lyon.
Part I · Formation
Lesson 01

Pavia and Padua, 1501–1525

Gerolamo Cardano was born in Pavia on 24 September 1501. His mother, Chiara Micheri, had left Milan to escape the plague, and she was not married to his father. Fazio Cardano was a Milanese lawyer with strong mathematical interests: he lectured on geometry at Pavia and at the Piatti foundation in Milan, edited a medieval treatise on optics, and was a friend of Leonardo da Vinci. He taught his son arithmetic, geometry and astrology.

Fazio hoped Gerolamo would study law. Gerolamo chose medicine and entered the University of Pavia in 1520. When the wars between France and the Habsburgs closed the university, he moved to Padua. There, by his own account, he was elected rector of the student body by a single vote, an expensive honour for a young man with little money. He received his doctorate in medicine in 1525.

The autobiography describes a sickly childhood, a hot temper and a passion for gambling at dice, cards and chess, which he says he practised for decades as a source of income. These habits supplied the material for his later book on chance.

Lesson 02

An outsider in Milan, 1525–1539

A physician in Milan needed membership in the College of Physicians. The college refused Cardano, officially because of his illegitimate birth and, by most accounts, also because of his abrasive manner. He practised for several years in the village of Sacco (Piove di Sacco) near Padua. In 1531 he married Lucia Bandarini. Their children were Giovanni Battista (born 1534), Chiara (1537) and Aldo (1543).

The family moved to Gallarate near Milan in 1532 and then to Milan, where they were for a time so poor that they lived in the poorhouse. A turning point came in 1534, when Cardano obtained his late father’s lectureship in mathematics at the Piatti foundation. He also practised medicine without a licence and won a reputation through some striking cures.

His first publication, a short polemic on the errors of contemporary physicians, appeared in Venice in 1536. In 1539 the College of Physicians changed its rule on legitimacy and admitted him. The same year he published Practica arithmetice et mensurandi singularis, a large arithmetic textbook. Its publisher would soon print a far more important book.

In 1536 a fourteen-year-old boy from Bologna, Lodovico Ferrari, entered Cardano’s household as a servant. Cardano saw his talent and taught him mathematics. Ferrari became his secretary, his collaborator and his fiercest defender.

Lesson 03

The secret of the cubic, 1539–1548

Luca Pacioli’s Summa (1494) had suggested that the cubic could not be solved by a general rule. Yet around the 1510s Scipione del Ferro, a professor at Bologna, found how to solve the case \(x^3+px=q\). He kept it secret, as was common for a weapon useful in public mathematical contests, and before he died in 1526 passed it to his student Antonio Maria Fior. In 1535 Fior challenged Niccolò Fontana, called Tartaglia (“the stammerer”), a self-taught teacher in Venice. Tartaglia found the rule for himself shortly before the deadline and in the early hours of 13 February 1535 solved all thirty of Fior’s problems.

Cardano was preparing a comprehensive algebra and wanted the rule. After persistent letters, Tartaglia visited him in Milan in March 1539 and gave him the method in the form of a poem, after Cardano swore not to publish it. Tartaglia’s own account of the meeting, published in 1546, is our main source for the oath.

With the rule in hand, Cardano and Ferrari extended it in every direction. Cardano proved it geometrically and adapted it to all thirteen types of cubic; Ferrari, in 1540, reduced the quartic to a cubic. They could publish none of it without breaking the oath. In 1543 they went to Bologna, where del Ferro’s son-in-law and successor, Annibale della Nave, showed them del Ferro’s notebook containing the rule. Cardano decided that he was not bound to keep secret a discovery that was not Tartaglia’s alone.

Ars Magna appeared in 1545. In it Cardano credits del Ferro with the first discovery and Tartaglia with an independent one made in competition, and he credits Ferrari with the quartic. Tartaglia was enraged. He printed the story of the oath in 1546, and Ferrari, not Cardano, answered with a series of printed challenges (cartelli). The quarrel ended with a public disputation in Milan on 10 August 1548, which went badly for Tartaglia; he left the city after the first day.

DiscussionCardano published with full attribution but broke an oath. Tartaglia kept a discovery secret that del Ferro had made first. Using the sixteenth-century practice of secret methods and public contests, argue each side.

Part II · The mathematics
Lesson 04

The rule for a cube and things

For the general reader. The quadratic formula was known in some form for millennia. A formula for the cubic eluded everyone until the early sixteenth century. The trick is to write the unknown as a difference of two new quantities chosen so that the equation splits into a quadratic. Cardano gave the first published proof, cutting a cube into pieces.

Goals
  • Derive the rule for \(x^3+px=q\).
  • Follow Cardano’s proof by dissection of a cube.
  • Apply the rule to his worked example.
Theorem (del Ferro, Tartaglia; Cardano’s proof, Ars Magna ch. XI)

Let \(p,q>0\). The equation \(x^3+px=q\) (“a cube and things equal to a number”) has exactly one positive root,

\[ x=\sqrt[3]{\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}+\frac q2}\;-\;\sqrt[3]{\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}-\frac q2}. \]

Proof

Look for \(x=u-v\) with \(u>v>0\). Expanding,

\[ (u-v)^3+3uv\,(u-v)=u^3-v^3. \]

So \(x=u-v\) solves \(x^3+px=q\) provided \(3uv=p\) and \(u^3-v^3=q\). Put \(U=u^3\), \(V=v^3\). Then \(U-V=q\) and \(UV=p^3/27\): two numbers with known difference and product, a quadratic problem. Solving, \(U=\sqrt{q^2/4+p^3/27}+q/2\) and \(V=\sqrt{q^2/4+p^3/27}-q/2\), both positive. Uniqueness holds because \(x^3+px\) is strictly increasing.

Cardano’s geometry

Cardano argued with solids. Take a cube of side \(u\) and mark off a smaller cube of side \(v\) in one corner. The large cube splits into the cube on \(u-v\), the small cube on \(v\), and three slabs, each a box with sides \(u\), \(v\) and \(u-v\), plus three more boxes. Collecting the pieces gives the identity \(u^3=(u-v)^3+v^3+3uv(u-v)\), which is the identity above. The same dissection stands behind the Babylonian and Islamic proofs of the quadratic rule, one dimension down.

Exercise 4.1Cardano’s example in chapter XI is “a cube and six things equal twenty”. Apply the rule and show the result equals 2.

Show solution
\(p=6\), \(q=20\): \(q^2/4+p^3/27=100+8=108\). So \(x=\sqrt[3]{\sqrt{108}+10}-\sqrt[3]{\sqrt{108}-10}\). Note \((1+\sqrt3)^3=1+3\sqrt3+9+3\sqrt3=10+6\sqrt3=10+\sqrt{108}\), and \((\sqrt3-1)^3=\sqrt{108}-10\) similarly. So \(x=(1+\sqrt3)-(\sqrt3-1)=2\). Check: \(8+12=20\).

Exercise 4.2Solve \(x^3+3x=4\) exactly with the rule, and compare with the obvious root.

Show solution
\(q^2/4+p^3/27=4+1=5\), so \(x=\sqrt[3]{\sqrt5+2}-\sqrt[3]{\sqrt5-2}\). The obvious root is \(x=1\). Indeed \(\left(\tfrac{1+\sqrt5}{2}\right)^3=2+\sqrt5\) and \(\left(\tfrac{\sqrt5-1}{2}\right)^3=\sqrt5-2\), so \(x=\tfrac{1+\sqrt5}2-\tfrac{\sqrt5-1}2=1\). The rule often hides simple roots inside nested radicals.
Lesson 05

Ars Magna: all the cubics

For the general reader. Without negative numbers, “a cube equal to things and a number” and “a cube and a number equal to things” were different problems. Ars Magna works through every such case. Along the way Cardano found tools that every algebra student now uses: shifting the unknown to remove a term, and reading off facts about the roots from the coefficients.

Goals
  • Remove the square term from a general cubic.
  • Reduce the 13 cubic types to the three depressed ones.
  • See Cardano’s observations on multiple roots and on roots and coefficients.

The plan of the book

Ars Magna has forty chapters. Chapters XI to XXIII treat thirteen types of cubic, each with a rule and a geometric demonstration. The first chapters set out general properties of equations; later ones treat transformations, approximation and the quartic.

Proposition (depressing the cubic)

The substitution \(x=y-a/3\) transforms \(x^3+ax^2+bx+c=0\) into \(y^3+py+q=0\), with

\[ p=b-\frac{a^2}{3},\qquad q=\frac{2a^3}{27}-\frac{ab}{3}+c. \]

Cardano carries out such substitutions type by type, in words, since he would not write a negative coefficient. With the square term gone, every cubic falls under one of three types with positive \(p,q\): \(x^3+px=q\), \(x^3=px+q\), \(x^3+q=px\). The first was Lesson 04. For the second, the substitution \(x=u+v\), \(3uv=p\), \(u^3+v^3=q\) gives

\[ x=\sqrt[3]{\frac q2+\sqrt{\frac{q^2}{4}-\frac{p^3}{27}}}+\sqrt[3]{\frac q2-\sqrt{\frac{q^2}{4}-\frac{p^3}{27}}}, \]

valid as a real formula only when \(q^2/4\ge p^3/27\). The third is reduced to the second: Cardano shows that if \(x_0\) solves \(x^3=px+q\), then the roots of \(x^3+q=px\) are found from \(x_0\) by solving a quadratic.

Several roots

Chapter I observes that an equation can have more than one solution, and Cardano allows negative roots, which he calls “fictitious”, to illustrate the point. His example \(x^3=12x+16\) has the true root 4 and the fictitious root \(-2\), counted twice in modern terms. He also notes that the true and fictitious solutions of \(x^3+q=px\) and \(x^3=px+q\) correspond with signs changed, and that in some cases the roots sum to the coefficient of the square term: early forms of the relations between roots and coefficients.

Exercise 5.1Depress \(x^3+6x^2+9x+4=0\) and solve.

Show solution
\(x=y-2\): \(p=9-12=-3\), \(q=\tfrac{2\cdot216}{27}-18+4=16-18+4=2\). So \(y^3-3y+2=0=(y-1)^2(y+2)\), giving \(y=1,1,-2\) and \(x=-1,-1,-4\). All roots are negative, matching the all-positive coefficients.

Exercise 5.2Use the second formula on Cardano’s \(x^3=12x+16\). What happens, and what are all the roots?

Show solution
\(q^2/4-p^3/27=64-64=0\), so \(x=2\sqrt[3]{8}=4\). Dividing, \(x^3-12x-16=(x-4)(x+2)^2\). A zero discriminant signals a repeated root, here \(-2\).

Exercise 5.3Show that if \(x_0\) is a root of \(x^3=px+q\), then \(-x_0\) is a root of \(x^3+q=px\), and find the other two roots of the latter by a quadratic.

Show solution
Substitute \(-x_0\): \(-x_0^3+q=-px_0\) is the first equation multiplied by \(-1\). Dividing \(x^3-px+q\) by \(x+x_0\) gives \(x^2-x_0x+(x_0^2-p)\), whose roots \(\tfrac{x_0}{2}\pm\sqrt{p-\tfrac{3x_0^2}{4}}\) are the positive roots when real. This is Cardano’s reduction in modern dress.
Lesson 06

Square roots of negatives

For the general reader. Cardano was the first to print a calculation with the square root of a negative number. He called it useless. Yet his own cubic rule forced such numbers on him: for some cubics with three perfectly ordinary roots, the formula passes through square roots of negatives. Making sense of this took another generation and led to complex numbers.

Goals
  • Work through Cardano’s problem of dividing 10 into two parts with product 40.
  • Understand the irreducible case of the cubic.
  • Follow Bombelli’s resolution and the trigonometric solution.

Chapter XXXVII

Cardano asks for two numbers whose sum is 10 and whose product is 40. The usual method gives \(5+\sqrt{-15}\) and \(5-\sqrt{-15}\). He calls the manipulation sophistic, then multiplies the two expressions and gets \(25-(-15)=40\), as required. He comments that the result is as subtle as it is useless. It was nonetheless the first time such quantities were treated as objects to calculate with.

The irreducible case

Theorem (casus irreducibilis)

Let \(x^3=px+q\) with \(p,q>0\) and \(q^2/4<p^3/27\). Then the equation has three distinct real roots, and the formula of Lesson 05 expresses them through cube roots of complex numbers. No expression in real radicals exists for them when the cubic is irreducible over \(\mathbb Q\).

The first sentence is elementary: \(f(x)=x^3-px-q\) has a local maximum at \(x=-\sqrt{p/3}\) of value \(\tfrac{2p}{3}\sqrt{p/3}-q\), which is positive exactly when \(q^2/4<p^3/27\), so \(f\) crosses the axis three times. The last sentence, that real radicals cannot help, was proved only in the nineteenth century using Galois theory. Cardano met the case and could not resolve it. His 1570 edition added a treatise, De regula aliza, that tried to work around it without success.

Bombelli’s resolution

Rafael Bombelli’s Algebra (1572) took Cardano’s formula at face value. For \(x^3=15x+4\) it gives

\[ x=\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}. \]

Bombelli guessed that \(\sqrt[3]{2+\sqrt{-121}}=2+\sqrt{-1}\), checked it by cubing with consistent rules for \(\sqrt{-1}\), and obtained \(x=(2+\sqrt{-1})+(2-\sqrt{-1})=4\), which is indeed a root. This was the first time imaginary numbers were used to reach a real answer that could be found no other way by formula.

Exercise 6.1Verify \((2+i)^3=2+11i\), and find the other two roots of \(x^3=15x+4\) both by factoring and by using the other cube roots of \(2+11i\).

Show solution
\((2+i)^2=3+4i\), and \((3+4i)(2+i)=6+3i+8i-4=2+11i\). Factoring, \(x^3-15x-4=(x-4)(x^2+4x+1)\), so \(x=-2\pm\sqrt3\). Alternatively, with \(\omega=e^{2\pi i/3}\), the pairs \((2+i)\omega+(2-i)\bar\omega\) give \(2\,\mathrm{Re}((2+i)\omega)=2(-1-\tfrac{\sqrt3}{2})=-2-\sqrt3\), and similarly \(-2+\sqrt3\).

Exercise 6.2François Viète later solved the irreducible case with trigonometry. Using \(\cos3\theta=4\cos^3\theta-3\cos\theta\), show that \(x=2\sqrt{p/3}\cos\theta\) solves \(x^3=px+q\) when \(\cos3\theta=\tfrac{q}{2}\left(\tfrac{3}{p}\right)^{3/2}\). Apply it to \(x^3=15x+4\).

Show solution
Substituting, \(x^3-px=2(p/3)^{3/2}(4\cos^3\theta-3\cos\theta)=2(p/3)^{3/2}\cos3\theta\), which equals \(q\) under the stated condition. For \(p=15,q=4\): \(x=2\sqrt5\cos\theta\) with \(\cos3\theta=2/5^{3/2}=0.1789\). Then \(3\theta\approx1.3909\), \(\theta\approx0.4636\), \(x\approx2\sqrt5\cdot0.8944=4.000\). The other roots come from \(\theta+2\pi/3\) and \(\theta+4\pi/3\).
Lesson 07

Ferrari’s quartic

For the general reader. Once cubics could be solved, the next step was the fourth degree. Lodovico Ferrari, still a teenager, found the method in 1540: rearrange the equation so that both sides become perfect squares, using a cubic to choose the right adjustment. Cardano published it in Ars Magna with full credit to Ferrari.

Goals
  • Derive Ferrari’s resolvent cubic.
  • Solve quartics by reducing to two quadratics.
  • See why the method stops at degree four.
Theorem (Ferrari, 1540)

Let \(x^4+px^2+qx+r=0\) with \(q\neq0\). Choose \(y\) to be any root of the resolvent cubic

\[ q^2=4\,(2y-p)\,(y^2-r). \]

Then \(2y-p\neq0\), and the quartic is equivalent to \((x^2+y)^2=\big(\sqrt{2y-p}\,x-\tfrac{q}{2\sqrt{2y-p}}\big)^2\). It therefore splits into two quadratic equations.

Proof. Rewrite the equation as \(x^4=-px^2-qx-r\) and add \(2yx^2+y^2\) to both sides:

\[ (x^2+y)^2=(2y-p)\,x^2-q\,x+(y^2-r). \]

The right side is a quadratic in \(x\). It is a perfect square exactly when its discriminant vanishes, \(q^2-4(2y-p)(y^2-r)=0\), which is the resolvent cubic. If \(2y-p\) were 0 the condition would force \(q=0\). Taking square roots of both sides gives two quadratic equations, \(x^2+y=\pm\big(\sqrt{2y-p}\,x-\tfrac{q}{2\sqrt{2y-p}}\big)\).

Any quartic can first be depressed by \(x\mapsto x-a/4\). In 1824 Niels Henrik Abel proved that no such general formula in radicals exists for degree five, and Évariste Galois explained why: the symmetric group \(S_4\) is solvable, and the resolvent cubic is a shadow of its normal subgroup chain, while \(S_5\) is not solvable.

Exercise 7.1Solve \(x^4=6x^2+8x+3\) by Ferrari’s method.

Show solution
Here \(p=-6,\ q=-8,\ r=-3\). Resolvent: \(64=4(2y+6)(y^2+3)\), i.e. \(16=(2y+6)(y^2+3)\); \(y=-1\) works. Then \((x^2-1)^2=4x^2+8x+4=(2x+2)^2\). So \(x^2-1=2x+2\), giving \(x=3,-1\), or \(x^2-1=-2x-2\), giving \(x=-1\) twice. Check: \(81=54+24+3\).

Exercise 7.2A problem treated in Ars Magna leads to \(x^4+6x^2+36=60x\). Find the resolvent cubic in the form used above.

Show solution
Here \(p=6,\ q=-60,\ r=36\). The condition \(3600=4(2y-6)(y^2-36)\) gives \(y^3-3y^2-36y-342=0\). Cardano himself completes the square differently, writing \((x^2+6)^2=6x^2+60x\) and adding \(2z(x^2+6)+z^2\); with \(y=z+6\) his resolvent \(z^3+15z^2+36z=450\) is the same equation. Its positive root is \(z\approx4.010\), after which two quadratics give the real roots \(x\approx0.644\) and \(x\approx3.100\).
Lesson 08

The book on games of chance

For the general reader. Cardano gambled for most of his life and wrote a manual on it, Liber de ludo aleae, probably finished by the 1560s and printed only in 1663. It mixes advice on cheating and on the morals of play with the first systematic attempt to compute chances by counting equally likely cases, a century before Pascal and Fermat.

Goals
  • Understand Cardano’s “circuit” and “equality”.
  • Count outcomes with two and three dice as he did.
  • Recognize the limits of his “reasoning on the mean”.
Cardano’s principle (in modern form)

If a game has a set \(\Omega\) of equally likely outcomes, the circuit, and \(A\subseteq\Omega\) is the set of favourable outcomes, the chance of \(A\) is \(|A|/|\Omega|\). Stakes are fair (“equality”) when they are in the ratio of favourable to unfavourable outcomes.

Cardano insists that the reasoning applies only if the die is honest. For one die the circuit has 6 outcomes; for two dice he counts 36, treating the dice as distinguishable; for three dice, 216. He tabulates how many of the 36 outcomes contain at least one given face, finding 11, and builds his analysis of the gambling games of his day on such counts.

Alongside correct counting, the book uses a heuristic now called reasoning on the mean: if a face comes up once in six throws on average, then three throws should give an even chance of seeing it. This is wrong, and the book contains other errors, some later corrected in the text itself. It was written over many years and never revised for print.

Exercise 8.1Verify Cardano’s count: of the 36 outcomes of two dice, how many contain at least one six? What is the chance with three dice?

Show solution
Outcomes with no six: \(5\cdot5=25\), so \(36-25=11\). With three dice: \(216-125=91\) of 216, about 0.421.

Exercise 8.2Show that reasoning on the mean is wrong for one die thrown three times, and find the least number of throws that gives at least an even chance of a six.

Show solution
\(P(\text{at least one six in }n)=1-(5/6)^n\). For \(n=3\) it is \(91/216\approx0.42<\tfrac12\). For \(n=4\) it is \(1-625/1296=671/1296\approx0.518\). So four throws are needed.

Exercise 8.3In Cardano’s counting, what are the fair stakes for a bet that the sum of two dice is 7?

Show solution
Six outcomes give 7, thirty do not. Equality requires stakes in the ratio 6 to 30, i.e. 1 to 5: the bettor on 7 should stake 1 against 5.
Lesson 09

Repeated trials and fair stakes

For the general reader. If an event has some chance in one play, what is the chance it happens every time in several plays? Cardano tried three answers in his manuscript before reaching the right one: multiply. That rule underlies everything from insurance to the famous problem posed to Pascal in 1654.

Goals
  • State and prove the multiplication rule for independent repetitions.
  • Compute odds for repeated events as Cardano did.
  • Connect his work to the problem of the Chevalier de Méré.
Multiplication rule

If a trial has \(s\) favourable outcomes in a circuit of \(t\) equally likely ones, and the trial is repeated \(n\) times independently, then the chance that every trial is favourable is \((s/t)^n\), and the fair odds against it are \(t^n-s^n\) to \(s^n\).

Proof. The circuit of \(n\) trials consists of the \(t^n\) sequences of outcomes, all equally likely, and exactly \(s^n\) of them are favourable at every step.

Cardano first proposed that the odds for repeated success could be found by multiplying the odds, an error, and worked through other attempts in the text before settling on the rule above. Historians such as Oystein Ore, who translated the book in 1953, regard this as a genuine discovery.

Exercise 9.1The chance of at least one six in a throw of two dice is \(11/36\). Find the chance of achieving this three times in a row, and the fair odds.

Show solution
\((11/36)^3=1331/46656\approx0.0285\). Odds against: \(46656-1331=45325\) to \(1331\), about 34 to 1.

Exercise 9.2The Chevalier de Méré’s question (1654): is it favourable to bet on at least one double six in 24 throws of two dice? Reasoning on the mean suggests 18 throws suffice. Compute the truth.

Show solution
\(1-(35/36)^{24}\approx0.4914<\tfrac12\): unfavourable. With 25 throws, \(1-(35/36)^{25}\approx0.5055\). Reasoning on the mean would claim \(36/2=18\) throws give equality, far from correct.
Lesson 10

Mechanisms: suspension, rolling circles, grilles

For the general reader. Several devices still carry Cardano’s name. Some he invented, and some he only described in his popular encyclopedias. The mathematics behind them is pleasant and short: a circle rolling inside another of twice its size makes each point move on a straight line.

Goals
  • Separate what Cardano invented from what he described.
  • Prove the theorem of Cardano’s circles.
  • Count the keys of a turning grille.

What bears his name

  • Cardan suspension (gimbal). Nested rings on perpendicular pivots that keep a compass or lamp level. Cardano described it in De subtilitate (1550); the device is older.
  • Cardan shaft. A shaft with universal joints, named after the suspension. Cardano did not describe the universal joint itself.
  • Cardan grille. A sheet with holes laid over a letter so that only the hidden message shows, described in De subtilitate.
  • Cardan’s movement. A gear that rolls inside a ring of twice its diameter, from Opus novum de proportionibus (1570).
Theorem (Cardano’s circles)

Let a circle of radius \(r\) roll without slipping inside a fixed circle of radius \(2r\). Then every point on the rolling circle moves back and forth along a diameter of the fixed circle.

Proof. Put the fixed circle at the origin and let the centre of the rolling circle be at \(r(\cos t,\sin t)\). Rolling without slipping inside a circle of radius \(R=2r\) means the small circle turns by \(-(R-r)t/r=-t\). A point that starts at \((2r,0)\) is then at

\[ r(\cos t,\sin t)+r(\cos(-t),\sin(-t))=(2r\cos t,\,0), \]

which moves on the \(x\)-axis. A point starting elsewhere on the rim moves on a different diameter. The device converts rotation to straight-line motion and was later used in printing presses.

Exercise 10.1Show that a point inside the rolling circle, at distance \(d<r\) from its centre, traces an ellipse. Find its semi-axes.

Show solution
Starting on the positive \(x\)-axis direction from the centre, the point is at \(r(\cos t,\sin t)+d(\cos t,-\sin t)=((r+d)\cos t,\,(r-d)\sin t)\): an ellipse with semi-axes \(r+d\) and \(r-d\). This is the principle of the trammel of Archimedes and of elliptic gear trains.

Exercise 10.2A later turning grille uses a \(4\times4\) card with 4 holes such that rotating it through 0°, 90°, 180°, 270° exposes each of the 16 cells exactly once. How many such grilles are there?

Show solution
The 16 cells fall into 4 orbits of size 4 under rotation. A valid grille chooses exactly one cell from each orbit: \(4^4=256\) grilles.
Part III · Life and legacy
Lesson 11

Fame and catastrophe, 1545–1570

The 1540s and 1550s made Cardano famous. In 1543 he accepted the chair of medicine at Pavia, which he held, with interruptions when salaries went unpaid, until 1560. His wife Lucia died in 1546. De subtilitate (1550) and De rerum varietate (1557) were encyclopedias of natural philosophy, technology and curiosities, widely read and widely attacked; the French scholar Julius Caesar Scaliger published a long rebuttal of the first. His medical books included early clinical descriptions, and his writings argued that the deaf could be taught to read and write.

Scotland

In 1552 John Hamilton, Archbishop of St Andrews, summoned him to treat a long-standing respiratory illness, usually described as asthma. Cardano left Milan on 23 February, reached Edinburgh on 29 June and left on 13 September. The archbishop recovered, Cardano was paid handsomely, and he declined offers to stay. On the way home he cast a horoscope for the young Edward VI of England, predicting a long life; Edward died the next year.

The disaster of 1560

His elder son Giovanni Battista, also a physician, married a woman the family considered unworthy. In 1560 he poisoned her. He was tried in Milan and executed on 13 April 1560. Cardano’s reputation in Milan and Pavia never recovered. His younger son Aldo gambled and stole from him, and Cardano had him banished from Bologna in 1569.

In 1562 Cardano moved to Bologna as professor of medicine. He continued to write prolifically. His collected works, published in 1663, fill ten folio volumes.

Lesson 12

Prison, Rome and legacy, 1570–1576

In 1570 the Inquisition arrested Cardano on a charge of heresy. The specific reasons are not fully documented; his astrological writings, including a horoscope of Christ published in the 1550s, are usually cited. He was held for several months, abjured, and was released. He lost his chair and was forbidden to teach or publish, and his non-medical works were placed on the Index of prohibited books.

In 1571 he moved to Rome. He was admitted to the College of Physicians there and received a pension from the papacy. In his last year he wrote De propria vita, an unsparing account of his illnesses, quarrels, dreams, gambling and misfortunes. He died in Rome on 21 September 1576. A story that he predicted the date of his death and starved himself to fulfil it circulated later; there is no good evidence for it.

Legacy

  • Algebra. Ars Magna is often taken as the start of modern algebra. It drove Bombelli’s complex arithmetic, Viète’s notation and trigonometric method, and ultimately the theory of equations of Lagrange, Abel and Galois.
  • Probability. Liber de ludo aleae was printed nine years after the Pascal–Fermat correspondence and had little influence, but it is the first known systematic treatment of chance by counting cases.
  • Mechanics and technology. The Cardan suspension and Cardan shaft keep his name in engineering vocabulary.
  • Autobiography. De propria vita remains a primary document of Renaissance mentality and was translated into English in 1930.

Final projectTake a cubic with three real roots and integer coefficients. Solve it three ways: by Cardano’s formula with complex cube roots, by Viète’s trigonometric method, and numerically. Write a two-page account explaining why Cardano could not finish the first method and what had to change in mathematics before anyone could.

Reference

Connections

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

Students

  • Lodovico Ferrari 1536-1548

Influenced by

  • Luca Pacioli
  • Scipione del Ferro

Influenced

Rivals

Family

  • Fazio Cardano

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Girolamo Cardano, MacTutor History of Mathematics.
  2. Cardano, Girolamo, Dictionary of Scientific Biography (via MacTutor).
  3. Gerolamo Cardano, Wikipedia.
  4. G. Cardano, The Book of My Life (De propria vita), trans. J. Stoner, 1930; New York Review Books, 2002.
  5. A. Grafton, Cardano’s Cosmos: The Worlds and Works of a Renaissance Astrologer, Harvard University Press, 1999.
  6. J. J. O’Connor and E. F. Robertson, Tartaglia and Ferrari, MacTutor.

Mathematics

  1. G. Cardano, Ars Magna or the Rules of Algebra, trans. T. R. Witmer, MIT Press, 1968; Dover, 1993.
  2. Ars Magna (Cardano book), Wikipedia.
  3. O. Ore, Cardano, the Gambling Scholar, Princeton University Press, 1953 (with a translation of Liber de ludo aleae by S. H. Gould).
  4. K. Bowman, Cardano’s Liber de Ludo Aleae, MAA.
  5. R. Bombelli, L’Algebra, Bologna, 1572.
  6. J. Stillwell, Mathematics and Its History, Springer, 3rd ed. 2010, chapters on polynomial equations and complex numbers.