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Rafael Bombelli

Rafael Bombelli (Raffaele Bombelli), born Mazzoli

A course in two strands: the life of a Bolognese hydraulic engineer who wrote algebra while the marshes waited, and the book that came of it, L’Algebra (1572), which gave the first rules for calculating with complex numbers, resolved the paradox of the cubic, and brought Diophantus back to Western readers.

Born January 1526, Bologna (baptized 20 January) Died 1572, probably Rome L’Algebra Books I–III, 1572 Level undergraduate → advanced undergraduate
The three cube roots of 2 + √−121 (filled) and of 2 − √−121 (open) lie on a circle of radius √5. Each conjugate pair adds to a real number: 4 and −2 ± √3, the three roots of x³ = 15x + 4.
Lesson 00

How to use this course

Rafael Bombelli wrote one book. He was an engineer by trade, and he wrote L’Algebra to make the new algebra of Gerolamo Cardano’s Ars Magna clear and complete for readers without a university training. In doing so he had to face the strangest feature of Cardano’s cubic rule: for equations with three ordinary real roots, it passes through square roots of negative numbers. Bombelli decided to calculate with these quantities as if they were numbers, wrote down the rules, and showed that they give the right answers. That decision is the beginning of complex numbers.

Prerequisites for Part II
  • Algebra of polynomials, and the solution of the cubic (see the Cardano and Tartaglia courses, or Lesson 06 here).
  • Complex numbers in Cartesian and polar form.
  • Basic sequences and limits, for Lesson 08.
  • Euclidean geometry of similar triangles, for Lesson 10.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proofs and exercises with folded solutions. Bombelli stated rules in words, with a compact notation of his own. We use modern notation and write \(i\) for his più di meno, while explaining what he actually wrote. The documentary record of his life is thin: a baptism, contracts and letters about drainage works, and the prefaces to his book. Where sources disagree or speculate, we say so.

Reference

Timeline

1508A failed plot to restore the Bentivoglio to power in Bologna; a Mazzoli ancestor of Rafael is executed.
1526Baptized 20 January in Bologna, eldest of six children of Antonio Mazzoli, a wool merchant, and Diamante Scudieri.
1545Cardano’s Ars Magna appears in Nuremberg.
1549His patron Alessandro Rufini acquires rights to reclaim marshland in the Val di Chiana.
1551Bombelli is surveying the boundaries of the Val di Chiana lands.
1555Work in the Val di Chiana is suspended.
1557–60Writes the manuscript of L’Algebra, in five books.
1560Returns to the Val di Chiana and completes the reclamation.
1561Takes part in an unsuccessful attempt to repair the Ponte Santa Maria over the Tiber in Rome.
1560sIn Rome, Antonio Maria Pazzi shows him a codex of Diophantus’s Arithmetica in the Vatican Library. They begin a translation.
1560sConsultant on a proposed drainage of the Pontine Marshes under Pope Pius IV; the project is not carried out.
1572Books I–III of L’Algebra are published. Bombelli dies the same year, probably in Rome.
1579L’Algebra reissued in Bologna.
1923Ettore Bortolotti finds the manuscript, including the unpublished geometric Books IV and V, in a library in Bologna.
1929Bortolotti publishes Books IV and V.
Part I · Formation
Lesson 01

Bologna, 1526–1548

Rafael Bombelli was baptized in Bologna on 20 January 1526, the eldest of six children. His father, Antonio Mazzoli, was a wool merchant; his mother, Diamante Scudieri, was a tailor’s daughter. Bologna belonged to the Papal States. The Bentivoglio family had ruled it from 1443 until Pope Julius II took the city in 1506 and drove them out. The Mazzoli had supported the Bentivoglio, and in 1508, after a failed attempt to bring them back, one of Rafael’s forebears was executed and the family property was confiscated. Sources differ on whether it was his grandfather or his great-grandfather. The property was later restored to Antonio, who changed the family name to Bombelli, apparently to put distance between his household and the old affair.

Bombelli did not attend a university. His only known teacher was Pier Francesco Clementi of Corinaldo, an engineer and architect, from whom he learned surveying, hydraulics and the practical mathematics of the trade. Bologna was a natural place to absorb algebra too. Scipione del Ferro had found the solution of the cubic there in the 1510s, and Lodovico Ferrari, the solver of the quartic, was a Bolognese. Some historians suggest that Bombelli may have witnessed the disputation between Ferrari and Niccolò Tartaglia in Milan in 1548, but there is no direct evidence.

Lesson 02

Engineer of the Val di Chiana, 1549–1561

Bombelli’s patron was Alessandro Rufini, a Roman nobleman who later became Bishop of Melfi. In 1549 Rufini acquired the rights to reclaim marshland in the Val di Chiana, a broad valley between Arezzo and Lake Trasimeno whose drainage had been poor since antiquity. Bombelli was employed as engineer. By 1551 he was surveying and recording the boundaries of the land.

The work was suspended in 1555, for reasons that are not documented in the sources used here. Bombelli used the interruption to write. He resumed in 1560 and completed the reclamation that year. The project made his reputation as a hydraulic engineer. In 1561 he took part in an attempt to repair the Ponte Santa Maria, the ancient bridge over the Tiber in Rome, which failed. He was later consulted on a scheme to drain the Pontine Marshes south of Rome under Pope Pius IV, which was never carried out.

The engineer’s background shows in the book. Bombelli wanted algebra to be usable by people like himself, and he wrote it in Italian, not Latin.

Lesson 03

Writing the Algebra; finding Diophantus, 1557–1572

Bombelli’s aim, as his preface explains, was a complete and clear treatment. Cardano’s Ars Magna was the deepest book on the subject, but it was hard going for readers without a university training. Bombelli set out to write a self-contained treatment from first principles, so that a reader could learn algebra from it without another master. He wrote the manuscript between 1557 and 1560, in five books: three on arithmetic and algebra, two on geometry.

Diophantus

The decisive event came in Rome. Antonio Maria Pazzi, a lecturer in mathematics at the University of Rome, showed Bombelli a manuscript of Diophantus’s Arithmetica in the Vatican Library. The work had been almost unknown in the West. Bombelli and Pazzi began to translate it and completed a large part, five books by MacTutor’s account, but never finished or published the translation.

Bombelli saw that Diophantus was a better source of problems than the commercial exercises of the abacus schools. He rewrote Book III of his own algebra, replacing its practical problems with abstract ones. Of the 272 problems in the printed Book III, 143 come from Diophantus. Bombelli acknowledged his debt to Diophantus in general but did not mark which problems were taken from him.

DiscussionBombelli wrote in Italian for practitioners; Diophantus wrote abstract number problems in Greek. Why might an engineer have preferred Diophantus’s problems to the merchants’ problems of the abacus tradition for teaching algebra?

Part II · The mathematics
Lesson 04

Rules of signs

For the general reader. Why is minus times minus plus? Renaissance algebraists used the rule in calculations long before they accepted negative numbers as answers. Bombelli stated the rules for signed quantities plainly in Italian, and justified the hardest one with a geometric argument. Today we see the rule as forced by the ordinary laws of arithmetic.

Goals
  • State the rules of signs as Bombelli gave them.
  • Prove them from the axioms of a ring.
  • Use them in expanding binomials, which is where he needed them.

Bombelli states the four rules in words: plus times plus makes plus, minus times minus makes plus, plus times minus makes minus, minus times plus makes minus. In his calculations, “minus” first appears as a subtracted part of a binomial such as \(7-\sqrt3\), and the rules tell how such parts combine when two binomials are multiplied.

Proposition (rules of signs)

In any ring \(R\) (associative, with addition forming an abelian group and multiplication distributing over addition), for all \(a,b\in R\):

\[ a\cdot0=0,\qquad (-a)\,b=a\,(-b)=-(ab),\qquad (-a)(-b)=ab. \]

Proof. \(a\cdot0=a(0+0)=a\cdot0+a\cdot0\); subtract \(a\cdot0\). Next, \(ab+(-a)b=(a+(-a))b=0\cdot b=0\), so \((-a)b=-(ab)\), and similarly \(a(-b)=-(ab)\). Finally \((-a)(-b)=-(a(-b))=-(-(ab))=ab\).

The proof says that anyone who wants the distributive law to hold for subtracted quantities must accept minus times minus is plus. Bombelli’s justification was geometric: subtracting rectangles from a square and adding back the overlap that had been removed twice. Exercise 4.1 gives a version.

Exercise 4.1Let \(a>b>0\) and \(c>d>0\). By cutting a rectangle with sides \(a\) and \(c\), show geometrically that \((a-b)(c-d)=ac-ad-bc+bd\), and identify which term plays the role of “minus times minus”.

Show solution
Inside the \(a\times c\) rectangle, the rectangle \((a-b)\times(c-d)\) is what remains after removing a strip of width \(b\) (area \(bc\)) and a strip of height \(d\) (area \(ad\)). The two strips overlap in a \(b\times d\) corner, which has been removed twice, so it must be added back once: \(ac-ad-bc+bd\). The term \(+bd\) is \((-b)(-d)\).

Exercise 4.2Expand \((5-\sqrt2)(3-\sqrt8)\) and simplify, as Bombelli would with binomials.

Show solution
\(15-5\sqrt8-3\sqrt2+\sqrt{16}=15+4-10\sqrt2-3\sqrt2=19-13\sqrt2\). The \(+4\) is minus times minus.
Lesson 05

Plus of minus: complex arithmetic

For the general reader. Bombelli invented names for the square roots of negative numbers. A quantity like \(\sqrt{-121}\) he called “plus of minus 11”, and its negative “minus of minus 11”. Then he did something no one had done before: he wrote down exactly how these quantities add and multiply, and showed that with these rules they obey the same algebra as ordinary numbers. What he described is the field of complex numbers.

Goals
  • Translate Bombelli’s eight multiplication rules into modern form.
  • Prove that his arithmetic defines a field.
  • Use conjugates and the norm to divide.

The rules

Writing \(i\) for più di meno (abbreviated p. di m.) and \(-i\) for meno di meno (m. di m.), Bombelli’s rules of multiplication are:

\[ \begin{aligned} (+1)(+i)&=+i, & (-1)(+i)&=-i, & (+1)(-i)&=-i, & (-1)(-i)&=+i,\\ (+i)(+i)&=-1, & (+i)(-i)&=+1, & (-i)(+i)&=+1, & (-i)(-i)&=-1. \end{aligned} \]

For addition he kept the real and imaginary parts apart, never combining “plus” with “plus of minus”. He insisted that a più di meno quantity always comes paired with its meno di meno partner in the problems he met, which is the pairing of a complex number with its conjugate.

Theorem (the complex numbers form a field)

On \(\mathbb C=\{a+bi: a,b\in\mathbb R\}\) define \((a+bi)+(c+di)=(a+c)+(b+d)i\) and \((a+bi)(c+di)=(ac-bd)+(ad+bc)i\). Then \(\mathbb C\) is a field containing \(\mathbb R\), and \(i^2=-1\). Conjugation \(\overline{a+bi}=a-bi\) respects sums and products, and the norm \(N(z)=z\bar z=a^2+b^2\) is multiplicative: \(N(zw)=N(z)N(w)\).

Proof sketch. The product rule is what Bombelli’s rules give when expanded by distributivity. Commutativity is visible from the formula; associativity and distributivity are checked directly, or by identifying \(a+bi\) with the real matrix \(\begin{pmatrix}a&-b\\ b&a\end{pmatrix}\), for which they are inherited from matrix algebra. If \(z\ne0\) then \(N(z)>0\) and \(z^{-1}=\bar z/N(z)\). Multiplicativity of the norm follows from \(\overline{zw}=\bar z\,\bar w\), or from \(\det(AB)=\det A\det B\).

Multiplicativity of the norm is the identity \((a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2\), which Diophantus used to write numbers as sums of two squares in more than one way.

Exercise 5.1In Bombelli’s language, multiply “2 plus of minus 1” by itself, then by itself again. Translate to modern notation and check.

Show solution
\((2+i)^2=4+4i+i^2=3+4i\) (“3 plus of minus 4”). Then \((3+4i)(2+i)=6+3i+8i+4i^2=2+11i\) (“2 plus of minus 11”, that is \(2+\sqrt{-121}\)). This is the computation at the heart of Lesson 06.

Exercise 5.2Divide \(2+11i\) by \(3+4i\) using the conjugate. Verify the answer with the norm.

Show solution
\(\dfrac{(2+11i)(3-4i)}{9+16}=\dfrac{6-8i+33i+44}{25}=\dfrac{50+25i}{25}=2+i\). Norms: \(N(2+11i)=125\), \(N(3+4i)=25\), \(N(2+i)=5\), and \(125=25\cdot5\).

Exercise 5.3Use \(N(zw)=N(z)N(w)\) with \(z=2+i\), \(w=3+2i\) to write \(65\) as a sum of two squares in two different ways.

Show solution
\(N(z)=5\), \(N(w)=13\). \(zw=6+4i+3i+2i^2=4+7i\), so \(65=4^2+7^2\). \(z\bar w=(2+i)(3-2i)=6-4i+3i+2=8-i\), so \(65=8^2+1^2\).
Lesson 06

The irreducible cubic resolved

For the general reader. The equation \(x^3=15x+4\) has the obvious solution \(x=4\). Yet the cubic rule of del Ferro, Tartaglia and Cardano expresses its solution through \(\sqrt{-121}\). Cardano could not explain this. Bombelli guessed that the cube roots in the formula are themselves complex numbers, found them, and watched the imaginary parts cancel to leave 4. The same idea works for every cubic of this kind.

Goals
  • Recognize the irreducible case and why it has three real roots.
  • Follow Bombelli’s calculation of \(\sqrt[3]{2+\sqrt{-121}}\).
  • See why his method cannot avoid the cubic it started from.

The rule and the paradox

For \(x^3=px+q\) with \(p,q>0\), the substitution \(x=u+v\) with \(3uv=p\), \(u^3+v^3=q\) gives

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

For \(p=15\), \(q=4\): \(\tfrac{q^2}{4}-\tfrac{p^3}{27}=4-125=-121\), and the rule reads \(x=\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}\).

Theorem (casus irreducibilis)

Let \(p,q>0\) with \(\tfrac{q^2}{4}<\tfrac{p^3}{27}\). Then \(x^3=px+q\) has three distinct real roots. If \(w\) is any cube root of \(\tfrac q2+i\sqrt{\tfrac{p^3}{27}-\tfrac{q^2}{4}}\), then the three roots are \(w+\bar w\), \(\omega w+\overline{\omega w}\), \(\omega^2w+\overline{\omega^2w}\), with \(\omega=e^{2\pi i/3}\). If the cubic is irreducible over \(\mathbb Q\), no expression of a root by real radicals exists.

Proof of the formula. Let \(u=\omega^k w\) and \(v=\bar u\). Then \(u^3+v^3=2\,\mathrm{Re}(w^3)=q\) and \(uv=|w|^2=\big(\tfrac{p^3}{27}\big)^{1/3}=\tfrac p3\), so \(3uv=p\). The identity \((u+v)^3=u^3+v^3+3uv(u+v)\) shows \(x=u+v\) is a root, and it is real because \(v=\bar u\). The three values of \(k\) give three distinct real numbers. The final sentence was proved in the nineteenth century with Galois theory.

Bombelli’s calculation

Bombelli supposed that \(\sqrt[3]{2+\sqrt{-121}}=a+\sqrt{-b}\) and \(\sqrt[3]{2-\sqrt{-121}}=a-\sqrt{-b}\), with \(a,b\) real and positive. Taking norms, \(a^2+b=\sqrt[3]{4+121}=5\). Comparing real parts of the cube, \(a^3-3ab=2\). Trying small integers, \(a=2\), \(b=1\) satisfies both. So the cube roots are \(2\pm\sqrt{-1}\), and the rule gives

\[ x=(2+\sqrt{-1})+(2-\sqrt{-1})=4. \]

He checked the cube directly, exactly as in Exercise 5.1. The imaginary quantities were a route to a real answer, used and then discarded. This is the first time anyone showed that calculating through complex numbers yields correct real results.

Exercise 6.1Show that Bombelli’s two conditions \(a^2+b=5\), \(a^3-3ab=2\) reduce to the cubic \(4a^3-15a-2=0\), and that \(x=2a\) turns it into the original equation \(x^3=15x+4\). What does this say about his method?

Show solution
Substitute \(b=5-a^2\): \(a^3-3a(5-a^2)=4a^3-15a=2\). With \(a=x/2\): \(\tfrac{x^3}{2}-\tfrac{15x}{2}=2\), i.e. \(x^3=15x+4\). Finding the cube root of \(2+\sqrt{-121}\) algebraically is equivalent to solving the cubic itself. Bombelli’s method works when a root can be guessed, and then it explains the formula; it does not give a new way to find roots in real radicals, in line with the theorem.

Exercise 6.2Find the other two cube roots of \(2+11i\) and use them to get the other two roots of \(x^3=15x+4\).

Show solution
Multiply \(2+i\) by \(\omega=-\tfrac12+\tfrac{\sqrt3}{2}i\): \((2+i)\omega=-1-\tfrac{\sqrt3}{2}+\big(\sqrt3-\tfrac12\big)i\). Twice its real part is \(-2-\sqrt3\). With \(\omega^2\) one gets real part \(-1+\tfrac{\sqrt3}2\), giving \(-2+\sqrt3\). Check: \(x^3-15x-4=(x-4)(x^2+4x+1)\).

Exercise 6.3Apply Bombelli’s method to \(x^3=6x+4\). Find the cube roots of the complex numbers in the formula, all three real roots, and confirm by factoring.

Show solution
\(\tfrac{q^2}{4}-\tfrac{p^3}{27}=4-8=-4\), so \(x=\sqrt[3]{2+2i}+\sqrt[3]{2-2i}\). Try \(a+bi\) with \(a^2+b^2=\sqrt[3]{8}=2\) and \(a^3-3ab^2=2\): \(a=-1,\ b=1\) works, since \((-1+i)^3=2+2i\). So one root is \(2\,\mathrm{Re}(-1+i)=-2\). Multiplying by \(\omega,\omega^2\) gives real parts \(\tfrac{1-\sqrt3}{2}\) and \(\tfrac{1+\sqrt3}{2}\), hence roots \(1\mp\sqrt3\). Indeed \(x^3-6x-4=(x+2)(x^2-2x-2)\).
Lesson 07

Notation and the equations of Book II

For the general reader. Before symbols like \(x^3\) existed, algebra was written in abbreviated words. Bombelli took a large step toward modern notation: he wrote the power of the unknown as a small number on an arc, so that \(x^3=6x+40\) became something like “1 with 3 on an arc, equal to 6 with 1 on an arc, plus 40”. With powers written as numbers, the rules for multiplying them became arithmetic on exponents.

Goals
  • Read Bombelli’s notation and translate it.
  • Understand why he lists so many “types” of equation.
  • Solve representative equations from Book II.

Writing powers

In the manuscript, a power of the unknown was written as its exponent inside a small semicircle; the printed book reduced the semicircle to an arc for typographical reasons. The manuscript also used an exponent zero for the constant term, which the printed book dropped. “Plus” and “minus” are p. and m., square and cube roots are R.q. and R.c., and brackets around an aggregate under a root were marked by special L-shaped signs. In this notation the equation \(x^3=6x+40\) is written in a form that Wikipedia transcribes as “1U3 a. 6U1 p. 40”, where U stands for the arc and a. for “equal to”.

Bombelli’s exponent rule (in modern form)

For positive integers \(m,n\): \(x^m\cdot x^n=x^{m+n}\) and, for \(m>n\), \(x^m/x^n=x^{m-n}\). Hence in his notation, the product of “\(a\) on arc \(m\)” and “\(b\) on arc \(n\)” is “\(ab\) on arc \(m+n\)”.

Earlier Italian algebra had a separate name for each power: cosa, censo, cubo, censo di censo, and compounds beyond. Multiplying them meant consulting a table. Bombelli’s numerical exponents turned the table into addition. Simon Stevin adopted a modified form of the notation, with the exponent in a circle, and praised Bombelli as the great arithmetician of their time.

Many types

Book II solves equations of degree one to four, with all coefficients positive. Since negative coefficients were not allowed, \(x^2+10x=39\) and \(x^2+21=10x\) count as different types, each with its own rule. The Dictionary of Scientific Biography counts five types of quadratic, seven of cubic and forty-two of quartic in Book II, the quartics handled by Ferrari’s method of completing squares.

Exercise 7.1Translate “1U3 a. 6U1 p. 40” and solve the equation by the cubic rule.

Show solution
\(x^3=6x+40\). Then \(\tfrac{q^2}{4}-\tfrac{p^3}{27}=400-8=392\), and \(x=\sqrt[3]{20+\sqrt{392}}+\sqrt[3]{20-\sqrt{392}}\). Since \((2\pm\sqrt2)^3=20\pm14\sqrt2=20\pm\sqrt{392}\), \(x=4\). Check: \(64=24+40\). The other two roots are complex: \(x^2+4x+10=0\).

Exercise 7.2Count the types of complete quadratic, with all three terms \(x^2\), \(x\) and a number present, all coefficients positive, and terms placed on both sides of the equals sign. Which types have a positive root, and how many?

Show solution
One term must stand alone on one side: \(x^2=bx+c\), \(x^2+bx=c\), \(x^2+c=bx\). (Writing all three on one side gives no positive root.) The first two always have exactly one positive root, since the product of the roots of \(x^2-bx-c\) or \(x^2+bx-c\) is \(-c<0\). The third has two positive roots when \(b^2>4c\), one when \(b^2=4c\), and none otherwise. Bombelli’s count of five also includes incomplete forms. The case distinctions disappear once negative coefficients are allowed.
Lesson 08

Square roots by continued division

For the general reader. Book I of L’Algebra teaches how to approximate square roots. Bombelli’s method for \(\sqrt{13}\) gives the sequence of fractions \(3\tfrac23,\ 3\tfrac35,\ 3\tfrac{20}{33},\dots\) that close in on the root from alternate sides. In modern terms it is a continued fraction, a concept that Pietro Cataldi made explicit in 1613.

Goals
  • Derive Bombelli’s iteration for \(\sqrt{a^2+r}\).
  • Prove convergence and the alternating property.
  • Relate the method to continued fractions and Pell’s equation.

The method

To find \(\sqrt{13}\), take the nearest square below, \(9=3^2\), and write \(\sqrt{13}=3+x\). Squaring, \(9+6x+x^2=13\), so \(x(6+x)=4\) and

\[ x=\frac{4}{6+x}. \]

Ignoring \(x\) on the right gives \(x\approx\tfrac46=\tfrac23\). Putting \(\tfrac23\) back in gives \(\tfrac{4}{6+2/3}=\tfrac35\), then \(\tfrac{20}{33}\), \(\tfrac{66}{109}\), \(\tfrac{109}{180}\), and so on. Unwinding the substitution produces

\[ \sqrt{13}=3+\cfrac{4}{6+\cfrac{4}{6+\cfrac{4}{6+\cdots}}}. \]
Theorem (convergence of Bombelli’s iteration)

Let \(a\ge1\) and \(0<r\le2a\), and let \(x^*=\sqrt{a^2+r}-a\). Define \(x_0=0\) and \(x_{k+1}=\dfrac{r}{2a+x_k}\). Then \(x_k\to x^*\), the iterates lie alternately above and below \(x^*\) for \(k\ge1\), and \(|x_{k+1}-x^*|\le\dfrac{1}{2a}\,|x_k-x^*|\).

Proof. The fixed point of \(f(x)=r/(2a+x)\) on \([0,\infty)\) solves \(x^2+2ax-r=0\), so it is \(x^*\). For \(x,y\ge0\), \(f(x)-f(y)=\dfrac{-r\,(x-y)}{(2a+x)(2a+y)}\). Since \(f\) is decreasing, iterates alternate around \(x^*\). Since \((2a+x)(2a+y)\ge4a^2\) and \(r\le2a\), the factor is at most \(\tfrac{2a}{4a^2}=\tfrac1{2a}\le\tfrac12\), so the error shrinks geometrically.

For \(\sqrt{13}\), MacTutor lists the first ten iterates; the tenth, \(\tfrac{85674}{141481}\), agrees with \(\sqrt{13}-3\) to ten decimal places. The general-purpose continued fraction of \(\sqrt{13}\), with numerators 1, is \([3;\overline{1,1,1,1,6}]\), and its convergents give the solution \(649^2-13\cdot180^2=1\) of Pell’s equation.

Exercise 8.1Apply Bombelli’s method to \(\sqrt2\) with \(a=1\), \(r=1\). Write the first five approximations and show that each \(p/q\) satisfies \(p^2-2q^2=\pm1\).

Show solution
\(x_{k+1}=1/(2+x_k)\): \(x=\tfrac12,\tfrac25,\tfrac5{12},\tfrac{12}{29},\tfrac{29}{70}\), so \(\sqrt2\approx\tfrac32,\tfrac75,\tfrac{17}{12},\tfrac{41}{29},\tfrac{99}{70}\). Then \(9-8=1\), \(49-50=-1\), \(289-288=1\), \(1681-1682=-1\), \(9801-9800=1\). The signs alternate, matching the alternation around \(\sqrt2\).

Exercise 8.2Show that if \(x_k=p_k/q_k\) in lowest terms for \(\sqrt{13}\), then \(x_{k+1}=4q_k/(6q_k+p_k)\). Compute \(x_3\) and \(x_4\) from \(x_2=\tfrac{20}{33}\), reducing where possible.

Show solution
\(\dfrac{4}{6+p_k/q_k}=\dfrac{4q_k}{6q_k+p_k}\). From \(\tfrac{20}{33}\): \(\tfrac{132}{218}=\tfrac{66}{109}\). From \(\tfrac{66}{109}\): \(\tfrac{436}{720}=\tfrac{109}{180}\). So \(\sqrt{13}\approx3\tfrac{109}{180}=3.60556\), against \(3.60555\ldots\). The reductions by 2 and 4 explain why MacTutor’s list has irregular-looking fractions.
Lesson 09

Diophantus in Book III

For the general reader. Diophantus of Alexandria, probably in the third century, wrote the Arithmetica: problems asking for rational numbers with given properties, such as two squares that add to a given square. Bombelli met a manuscript in the Vatican and filled his Book III with such problems. It was the first time most Western readers saw them, nearly fifty years before Claude-Gaspard Bachet’s Greek and Latin edition of 1621, the copy in which Pierre de Fermat later wrote his famous marginal notes.

Goals
  • Solve typical Diophantine problems by Diophantus’s method of a single unknown.
  • Prove the rational parametrization of the circle.
  • Understand the kind of problem Bombelli took from Diophantus.

The method of one unknown

Diophantus works with one unknown (Bombelli’s tanto) and expresses every other quantity in terms of it by a clever choice. A classic example is Arithmetica I.28: find two numbers whose sum is 20 and the sum of whose squares is 208. Diophantus sets the numbers to \(10+x\) and \(10-x\), so that the sum condition holds automatically, and the second condition becomes \(200+2x^2=208\), giving \(x=2\) and the numbers 12 and 8.

Theorem (rational points on a circle)

Let \(a\) be a non-zero rational. Every rational solution of \(u^2+v^2=a^2\) other than \((u,v)=(0,-a)\) has the form

\[ u=\frac{2ta}{1+t^2},\qquad v=\frac{(1-t^2)\,a}{1+t^2}\qquad(t\in\mathbb Q), \]

and every rational \(t\) gives such a solution.

Proof. Draw the line through \((0,-a)\) with slope \(1/t\), i.e. \(v=u/t-a\). Substituting gives \(u^2+(u/t-a)^2=a^2\), so \(u\big(u(1+1/t^2)-2a/t\big)=0\). The second root is \(u=2ta/(1+t^2)\), rational when \(t\) is. Conversely, the line from \((0,-a)\) to a rational point other than itself has rational slope. This is Diophantus’s procedure in Arithmetica II.8, the problem of dividing a given square into two squares, in geometric dress.

Exercise 9.1Divide 16 into two rational squares, as in Arithmetica II.8. Diophantus takes one square to be \(x^2\) and the other to be \((2x-4)^2\). Find his answer and the value of \(t\) in the theorem.

Show solution
\(x^2+(2x-4)^2=16\) gives \(5x^2-16x=0\), so \(x=\tfrac{16}5\). The squares are \(\tfrac{256}{25}\) and \(\big(\tfrac{12}{5}\big)^2=\tfrac{144}{25}\), summing to \(\tfrac{400}{25}=16\). The line \(v=2u-4\) through \((0,-4)\) has slope 2, so \(t=\tfrac12\): \(u=\tfrac{2\cdot\frac12\cdot4}{5/4}=\tfrac{16}{5}\), \(v=\tfrac{(3/4)\cdot4}{5/4}=\tfrac{12}{5}\).

Exercise 9.2Find two numbers whose sum is 10 and whose product is 21 by Diophantus’s substitution. Then explain what goes wrong if the product is required to be 40, and how this connects with Cardano’s \(5\pm\sqrt{-15}\).

Show solution
Set the numbers \(5+x\), \(5-x\). Then \(25-x^2=21\), \(x=2\): the numbers are 7 and 3. With product 40, \(x^2=-15\). Diophantus would reject the problem; Cardano wrote \(5\pm\sqrt{-15}\) and called it useless; Bombelli’s rules make \((5+\sqrt{-15})(5-\sqrt{-15})=25+15=40\) a legitimate calculation, though the problem still has no real answer.
Lesson 10

Algebra with segments: Books IV and V

For the general reader. Bombelli planned two geometric books that were not printed in his lifetime. Book IV uses algebra to solve geometric problems; Book V uses geometry to solve algebraic ones. He represented numbers by line segments relative to a fixed unit, and showed that trisecting an angle leads to exactly the cubic that had baffled Cardano. The manuscript lay unread in Bologna until 1923.

Goals
  • Multiply and divide segments with a unit length.
  • Prove that angle trisection leads to the irreducible case.
  • Connect Bombelli’s geometry to Descartes and Viète.

Algebra linearia

Greek geometry treated the product of two segments as a rectangle, and the product of three as a solid. Bombelli, in what he called algebra linearia, fixed a unit segment and represented powers and products as segments, using similar triangles. The Dictionary of Scientific Biography also credits Book IV with the use of orthogonal coordinates for points. René Descartes made the unit segment the foundation of his Géométrie (1637), without access to Bombelli’s unpublished books.

Construction (product and quotient of segments)

Given segments of length \(1\), \(a\), \(b\): on one ray from a point \(O\) mark \(OU=1\) and \(OA=a\); on another ray mark \(OB=b\). The parallel to \(UB\) through \(A\) meets the second ray at \(P\) with \(OP=ab\). Interchanging the roles of \(1\) and \(b\) gives \(a/b\).

The proof is the similarity of triangles \(OUB\) and \(OAP\): \(OP/OB=OA/OU\), so \(OP=ab\).

Trisection

Theorem (trisection and the irreducible case)

Let \(0<\varphi<\pi/2\). Then \(x=2\cos(\varphi/3)\) is a root of \(x^3=3x+2\cos\varphi\), and this cubic is always in the irreducible case. Conversely, for \(p,q>0\) in the irreducible case, the substitution \(x=2\sqrt{p/3}\,y\) turns \(x^3=px+q\) into the trisection equation \(4y^3-3y=\cos\varphi\), where \(\cos\varphi=\tfrac q2\big(\tfrac3p\big)^{3/2}<1\).

Proof. From \(\cos3\theta=4\cos^3\theta-3\cos\theta\) with \(\theta=\varphi/3\) and \(x=2\cos\theta\): \(x^3-3x=8\cos^3\theta-6\cos\theta=2\cos\varphi\). Here \(p=3\), \(q=2\cos\varphi\), and \(\tfrac{q^2}{4}=\cos^2\varphi<1=\tfrac{p^3}{27}\). For the converse, substitute and divide by \(2(p/3)^{3/2}\) to get \(4y^3=3y+\tfrac q2\big(\tfrac3p\big)^{3/2}\); the irreducible-case inequality \(\tfrac{q^2}{4}<\tfrac{p^3}{27}\) is exactly \(\cos\varphi<1\). This is the link Book V establishes: the cubics Cardano could not solve with real radicals are the cubics of angle trisection. François Viète later turned the same identity into a practical method of solution.

Exercise 10.1Write the trisection equation for \(\varphi=60^\circ\) and find its three roots in trigonometric form.

Show solution
\(x^3=3x+1\). The roots are \(2\cos20^\circ\approx1.8794\), \(2\cos140^\circ\approx-1.5321\) and \(2\cos260^\circ\approx-0.3473\), since \(3\theta\) must be \(60^\circ\) modulo \(360^\circ\). Their sum is 0, as the missing \(x^2\) term requires.

Exercise 10.2Show that \(x^3=15x+4\) corresponds to trisecting the angle \(\varphi\) with \(\cos\varphi=\tfrac{2}{5\sqrt5}\), and recover the root 4.

Show solution
\(\cos\varphi=\tfrac q2\big(\tfrac3p\big)^{3/2}=2\cdot\big(\tfrac15\big)^{3/2}=\tfrac{2}{5\sqrt5}\approx0.1789\), so \(\varphi\approx79.70^\circ\) and \(\varphi/3\approx26.57^\circ\). Then \(x=2\sqrt5\cos(26.57^\circ)\). Since \(\tan26.57^\circ=\tfrac12\), \(\cos=\tfrac{2}{\sqrt5}\) and \(x=4\) exactly. The angle of \(2+i\) is \(\arctan\tfrac12\), which is the same \(26.57^\circ\): Bombelli’s cube root and the trisection are one computation.
Part III · Life and legacy
Lesson 11

Rome, publication and death, 1561–1572

After the Val di Chiana, Bombelli’s engineering career was less fortunate. The repair of the Ponte Santa Maria in 1561 failed, and the Pontine Marshes scheme was not carried out. He remained in the service of Rufini and spent much of the 1560s in Rome, where the encounter with Diophantus took place and where he revised his book.

The revision was substantial. Book III lost its commercial problems and gained Diophantus’s. The printed notation was simplified from the manuscript to suit the press. Only the first three books were printed, in 1572, under the title L’Algebra, with a promise that the geometric books would follow. Bombelli died the same year, probably in Rome; the exact date is not known. The geometric books did not follow. A reissue of the printed books appeared in Bologna in 1579.

In 1923 the historian Ettore Bortolotti found Bombelli’s manuscript in a library in Bologna. It contained an earlier version of Books I–III and the unpublished Books IV and V, which Bortolotti edited and published in 1929. The manuscript shows the original notation and the abacus-style problems that Diophantus displaced.

Lesson 12

Legacy

Opinions of Bombelli’s influence vary. Jean Dieudonné argued that imaginary quantities were used before him and that the book had little effect. Most historians, including the authors of the MacTutor biography, disagree. The evidence of readers is strong. Stevin used his notation. Gottfried Wilhelm Leibniz taught himself the theory of cubic equations from L’Algebra and called Bombelli an “outstanding master of the analytical art”. Historians such as John Crossley regard him as the first to give a real treatment of complex numbers.

  • Complex numbers. His rules for più di meno are, in substance, the arithmetic of \(\mathbb C\). It took two more centuries, through John Wallis, Abraham de Moivre and Leonhard Euler, to the geometric picture of Carl Friedrich Gauss and others, before complex numbers were fully accepted.
  • The cubic. He showed that the del Ferro–Tartaglia–Cardano formula is valid in every case once complex numbers are allowed, and connected the irreducible case to angle trisection.
  • Diophantus. L’Algebra brought Diophantine problems to Western readers before Bachet’s edition, and so stands at the start of the line that leads to Fermat and modern number theory.
  • Exposition and notation. A clear, systematic textbook in Italian, with numerical exponents that Stevin carried to the Low Countries.

Final projectChoose an irreducible cubic with integer coefficients and an integer root. Solve it three ways: by Bombelli’s guessed cube roots, by Viète’s trigonometric method, and by factoring. Then write a two-page account, for a reader of 1572, defending the use of più di meno quantities, and explain why Exercise 6.1 shows that the defence cannot rest on computational convenience alone.

Reference

Connections

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

Teachers and advisors

  • Pier Francesco Clementi

Influenced by

Influenced

Colleagues and collaborators

  • Alessandro Rufini
  • Antonio Maria Pazzi

Also mentioned in this life

Fields

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Rafael Bombelli, MacTutor History of Mathematics.
  2. S. A. Jayawardene, Bombelli, Rafael, Dictionary of Scientific Biography (via MacTutor).
  3. Rafael Bombelli, Wikipedia.
  4. H. Strick, Rafael Bombelli, MacTutor.

Mathematics

  1. R. Bombelli, L’Algebra, Books I–III, 1572; Books IV–V ed. E. Bortolotti, Bologna, 1929. Feltrinelli edition of the complete work, 1966.
  2. MacTutor, Bombelli: Algebra, extracts including the approximation of \(\sqrt{13}\).
  3. MAA Convergence, Mathematical Treasures: Bombelli’s Algebra.
  4. G. Cardano, Ars Magna or the Rules of Algebra, trans. T. R. Witmer, MIT Press, 1968; Dover, 1993.
  5. J. N. Crossley, The Emergence of Number, World Scientific, 1987.
  6. J. Stillwell, Mathematics and Its History, Springer, 3rd ed. 2010, chapters on polynomial equations and complex numbers.
  7. P. J. Nahin, An Imaginary Tale: The Story of √−1, Princeton University Press, 1998.