courseshub.world · Lives in Mathematics and Physics

Omar Khayyam

Ghiyath al-Din Abu’l-Fath Umar ibn Ibrahim al-Khayyami al-Nisaburi · عمر خیام

A course in two strands: the life of the Nishapur scholar who served Seljuq sultans as astronomer, and the mathematics he left behind: cubic equations solved by intersecting conics, a reformed solar calendar, and a searching critique of Euclid’s parallel postulate.

Born 18 May 1048 (reconstructed), Nishapur Died c. 1131, Nishapur Jalali calendar inaugurated 1079 Level undergraduate → advanced undergraduate
Khayyam’s construction for “a cube and sides equal a number”, y³ + 4y = 16. The parabola y² = 2x meets the circle on the diameter BH = 4 at D, whose height is the root y = 2.
Lesson 00

How to use this course

Most readers meet Omar Khayyam as the name on a Victorian book of quatrains. His contemporaries knew him as a philosopher, astronomer and geometer. His algebra gave the first systematic theory of cubic equations, his calendar was one of the most accurate ever designed, and his commentary on Euclid examined the parallel postulate with a quadrilateral that Europe rediscovered six centuries later.

Prerequisites for Part II
  • High-school algebra and Euclidean geometry: similar triangles, the power of a point, the circle on a diameter.
  • The equations of the parabola, circle and rectangular hyperbola in coordinates.
  • For Lesson 09, some acquaintance with the idea of non-Euclidean geometry is helpful.
  • For Lesson 10, the Euclidean algorithm and continued fractions.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proofs and exercises with folded solutions. Khayyam wrote in words, with no symbols. We translate his constructions into coordinates, and we say so each time. The biography is thin and partly legendary. Each biographical claim below names its source, and later stories are marked as such.

Reference

Timeline

1048Born in Nishapur, Khorasan. The date 18 May 1048 is a modern reconstruction from a horoscope recorded by his younger contemporary al-Bayhaqi.
c. 1068Goes to Bukhara; received at the court of the Qarakhanid ruler Shams al-Mulk Nasr (al-Bayhaqi).
c. 1070In Samarkand under the patronage of the chief judge Abu Tahir. Begins the treatise on algebra.
1074–75Summoned by Sultan Malik-Shah, through the vizier Nizam al-Mulk, to the new observatory at Isfahan (Ibn al-Athir, under the year 467 AH).
1077December: completes the commentary on the difficult postulates of Euclid at Isfahan.
1079March: the reformed solar calendar, the Jalali era, is inaugurated. Persian translation of a sermon of Ibn Sina.
1092Nizam al-Mulk is assassinated in October; Malik-Shah dies in November. Royal support for the observatory ends.
after 1092A pilgrimage to Mecca and a stay in Baghdad, reported by Ibn al-Qifti in the 13th century.
1112–13In Balkh with Abu Hatim al-Isfizari (Nizami Aruzi).
1114–15Winter in Merv, as a guest of the vizier (Nizami Aruzi).
1121Al-Khazini’s Balance of Wisdom preserves his work on the hydrostatic balance.
c. 1131Dies in Nishapur. The date 4 December 1131 is the one most often given; the sources leave it uncertain.
1250Nasir al-Din al-Tusi criticizes his parallel argument in al-Risala al-shafiya.
1851Franz Woepcke publishes the Arabic text and a French translation of the algebra.
1859Edward FitzGerald’s English version of quatrains attributed to Khayyam.
1925Iran adopts the Solar Hijri calendar, a simplified descendant of the Jalali reform.
Part I · Formation
Lesson 01

Nishapur, 1048–c. 1068

Omar Khayyam was born in Nishapur, in the province of Khorasan in north-eastern Iran, in 1048. The Seljuq Turks had taken the city less than a decade earlier, and it was one of the great centres of learning in the eastern Islamic world. His name al-Khayyami means “of the tent-maker”, which may record his father Ibrahim’s trade or an earlier family one.

What the sources say

Almost everything we know comes from a handful of medieval writers, and they disagree. The most valuable is Zahir al-Din al-Bayhaqi (d. 1169–70), who met Khayyam in his youth and wrote a biographical dictionary of philosophers. Al-Bayhaqi recorded Khayyam’s horoscope: born with the Sun and Mercury in Gemini. In the twentieth century the scholar Swami Govinda Tirtha used this horoscope to compute the date 18 May 1048. The date is a reconstruction, though most modern accounts adopt it.

The second key witness is Nizami Aruzi of Samarkand, whose Chahar Maqala (Four Discourses) reports meetings with Khayyam in Balkh and Merv and a visit to his grave. The third is Ibn al-Qifti, whose thirteenth-century history of learned men is increasingly hostile in tone.

Education

Later biographical traditions name teachers in Nishapur for Quran, Arabic and the religious sciences, and a teacher who took him through Ptolemy’s Almagest. These names cannot be checked against contemporary sources. What is certain from his own works is the shape of his learning: he wrote in Arabic for scholarly subjects and in Persian for some others, and he knew Euclid, Apollonius and Archimedes, the algebra of al-Khwarizmi and his successors, and the philosophy of Ibn Sina (Avicenna), whom he regarded as his master.

A legend to set aside

A famous story makes Khayyam a schoolfellow of Nizam al-Mulk, the future vizier, and of Hasan-i Sabbah, the future leader of the Assassins. The three supposedly swore that whoever rose first would help the others. The tale appears in Persian histories written two centuries and more after Khayyam’s death, and Edward FitzGerald repeated it in 1859. Nizam al-Mulk was born about 1018, some thirty years before Khayyam, so historians treat the story as fiction.

Lesson 02

Bukhara and Samarkand, c. 1068–1074

A scholar without inherited wealth needed a patron. Al-Bayhaqi places Khayyam at the court of the Qarakhanid ruler Shams al-Mulk Nasr in Bukhara, and around 1070 he was in Samarkand, supported by the chief judge Abu Tahir Abd al-Rahman ibn Alaq.

In the preface to his algebra Khayyam complains of the times. Scholars, he says, have been scattered or silenced, and few remain who care for real knowledge. He credits Abu Tahir with giving him the calm to write. The treatise he began in Samarkand, Risala fi’l-barahin ala masa’il al-jabr wa’l-muqabala (Treatise on the demonstration of problems of algebra), was most likely completed later, in Isfahan. It is the subject of Lessons 04 to 07.

The quadrant problem

Before the algebra Khayyam wrote a short treatise on the division of a quadrant of a circle. A geometric problem about choosing a point on a quarter circle led him to a cubic equation, which in modern form reads \(x^3+200x=20x^2+2000\). He solved it by intersecting a circle with a rectangular hyperbola and then found an approximate numerical value using trigonometric tables. In the same treatise he announced a plan: to classify all equations up to degree three and to solve each kind. The algebra carried out that plan.

Lesson 03

Isfahan and the observatory, 1074–1092

The Seljuq sultan Malik-Shah, acting through his powerful vizier Nizam al-Mulk, summoned astronomers to build an observatory at Isfahan. The historian Ibn al-Athir, writing in the thirteenth century, lists “Umar ibn Ibrahim al-Khayyami” among them under the year 467 AH (1074–75). Khayyam remained in Isfahan for about eighteen years, the most productive period of his life.

The observatory’s tasks were practical. It produced astronomical tables, later known as the Zij-i Malikshahi, of which only a fragment survives, and it reformed the Persian solar calendar (Lesson 08). Astronomers at court were also expected to cast horoscopes. Khayyam did so, though Nizami Aruzi reports him as sceptical of predictions.

Writing in Isfahan

  • 1077. The commentary on the difficult postulates of Euclid, Sharh ma ashkala min musadarat kitab Uqlidis, completed in December 1077 (Lessons 09 and 10).
  • 1079. A Persian translation of a sermon by Ibn Sina, made for friends in Isfahan.
  • c. 1080. Short philosophical treatises on existence and on obligation, one written for the same Abu Tahir, now a judge in Fars.

The balance

Khayyam also worked on the classical problem attributed to Archimedes: given a lump of gold alloyed with silver, find how much of each it contains without melting it. Weigh it in air and in water; the loss of weight in water is the weight of the displaced water. Al-Khazini’s Book of the Balance of Wisdom (1121) includes Khayyam’s treatment of this problem and a description of a balance he designed for it.

Exercise 3.1An alloy of gold and silver weighs 1000 g in air and 940 g in water. Take the densities of gold and silver as 19.3 and 10.5 g/cm³. How much gold does it contain?

Show solution
The loss of 60 g means the alloy displaces 60 cm³. With \(g+s=1000\) and \(g/19.3+s/10.5=60\), eliminate \(s\): \(s\left(\tfrac1{10.5}-\tfrac1{19.3}\right)=60-\tfrac{1000}{19.3}\), so \(s\approx188.5\) g and \(g\approx811.5\) g. The method needs only linear equations, which is why Khayyam could call it a purely arithmetical procedure.

In 1092 this world collapsed. Nizam al-Mulk was assassinated in October, and Malik-Shah died in November. The observatory lost its funding, and Khayyam lost his protectors. Lesson 11 takes up the story.

Part II · The mathematics
Lesson 04

Algebra before Khayyam and his classification

For the general reader. Algebra began in Baghdad around 820 as a method for solving quadratic equations, stated in words and justified by pictures of squares and rectangles. Khayyam asked the next question: what about cubes? He listed every possible equation of degree at most three, sorted them into types, and showed how to solve each type with geometry.

Goals
  • Understand why medieval algebra split equations into “types”.
  • Reconstruct Khayyam’s list of 25 types and his division into 11 and 14.
  • Prove a quadratic case the way al-Khwarizmi did.

Positive coefficients, positive roots

Muhammad ibn Musa al-Khwarizmi’s algebra (c. 820) had no negative numbers and no zero coefficient. An equation such as \(x^2+21=10x\) was a different species from \(x^2=10x+21\), because each side of an equation was a sum of positive quantities. Al-Khwarizmi listed six types for degree two and solved each by completing a square drawn in the plane. The words survive in Khayyam: mal (square), jidhr or dil (root, side), ka‘b (cube), adad (number).

Later algebraists, notably al-Karaji, extended the arithmetic of polynomials. Some cubic problems had appeared too. Al-Mahani in the ninth century reduced a problem of Archimedes on cutting a sphere to a cubic but could not solve it; according to Khayyam, Abu Ja‘far al-Khazin later solved it with conic sections. Khayyam was the first to treat all cubics as one subject.

Khayyam’s classification

Equations with positive coefficients, in which each of the terms number, root, square and cube appears at most once, fall into 25 types:

  1. 6 simple (binomial) equations, such as \(x^3=d\) or \(x^2=bx\);
  2. 12 trinomial equations, such as \(x^2+bx=d\) or \(x^3+cx=d\);
  3. 7 quadrinomial equations, such as \(x^3+cx=bx^2+d\).

Eleven types reduce to degree at most two and can be solved with ruler and compass, as in Euclid. The other 14 are genuine cubics. For these Khayyam gives constructions by conic sections.

Khayyam also states that the 14 cubic types cannot, in general, be constructed with ruler and compass alone. He gave no proof, and none was possible with the tools of his time. The impossibility of duplicating the cube with ruler and compass was proved by Pierre Wantzel in 1837.

The quadratic model

Proposition (al-Khwarizmi, type “squares and roots equal number”)

If \(b,d>0\), the equation \(x^2+bx=d\) has exactly one positive root, \(x=\sqrt{(b/2)^2+d}-b/2\).

Proof by picture. Draw a square of side \(x\). Attach to two adjacent sides rectangles of size \(x\times b/2\). The figure has area \(x^2+bx=d\). It lacks only a corner square of area \((b/2)^2\) to become a square of side \(x+b/2\). So \((x+b/2)^2=d+(b/2)^2\), and the result follows.

Exercise 4.1List all 12 trinomial types. Which three reduce to quadratics on division by \(x\)?

Show solution
Quadratic: \(x^2+bx=d\), \(x^2+d=bx\), \(bx+d=x^2\). Cubic with squares and roots: \(x^3+bx^2=cx\), \(x^3+cx=bx^2\), \(bx^2+cx=x^3\); dividing by \(x\) gives quadratics. Genuine cubics: \(x^3+cx=d\), \(x^3+d=cx\), \(x^3=cx+d\), \(x^3+bx^2=d\), \(x^3+d=bx^2\), \(x^3=bx^2+d\). The six genuine trinomial cubics, the seven quadrinomials and \(x^3=d\) make Khayyam’s 14.

Exercise 4.2Solve al-Khwarizmi’s own example, “a square and ten roots equal thirty-nine”, by completing the square. Why did he report only one answer?

Show solution
\(x^2+10x=39\) gives \((x+5)^2=64\), so \(x+5=8\) and \(x=3\). The other solution of the modern equation is \(x=-13\), which has no meaning as the side of a square.

Exercise 4.3Count the 7 quadrinomial types directly. (Hint: a quadrinomial has all four terms, and you must decide which side each falls on, with the cube’s side fixed.)

Show solution
Put \(x^3\) on the left. Each of the other three terms goes left or right, giving \(2^3=8\) cases. The case with all three on the left, \(x^3+bx^2+cx+d=0\), has no positive root and is discarded. That leaves 7.
Lesson 05

The conic toolkit

For the general reader. Greek geometers had found that some problems, like doubling the volume of a cube, could not be done with ruler and compass but could be done with curves cut from a cone. Khayyam took over these curves, mainly from Apollonius, and used them as a general machine for solving cubics.

Goals
  • Write the defining properties of the parabola and rectangular hyperbola as Apollonius stated them.
  • Understand Khayyam’s demand for homogeneity and the lemma that meets it.
  • Solve \(x^3=d\) by two parabolas.

Apollonius’ properties

Khayyam cites the Conics of Apollonius of Perga and assumes its first books. In coordinates, the two properties he uses most are:

The symptomata

Parabola with vertex \(B\), axis along the \(x\)-axis and parameter (latus rectum) \(p\): the square on the ordinate equals the rectangle on the abscissa and the parameter, \(y^2=px\).

Rectangular hyperbola referred to its asymptotes: the rectangle on the coordinates is constant, \(xy=k^2\).

Circle on a diameter \(AC\): for a point \(D\) on the circle with foot \(E\) on \(AC\), \(DE^2=AE\cdot EC\).

Homogeneity

For Khayyam a number in a cubic equation is a solid, the roots are lines, and the squares are areas. An equation must compare like with like. So before solving \(x^3+cx=d\) he writes \(c\) as a square \(b^2\) and turns \(d\) into a solid on base \(b^2\): a box with base \(b^2\) and some height \(h\), so that \(d=b^2h\). This needs a lemma.

Lemma (changing the base of a solid)

Given a box with square base \(a^2\) and height \(c\), and a square \(b^2\), one can construct with ruler and compass a height \(h\) such that \(b^2h=a^2c\).

Proof. Construct the fourth proportional \(k\) with \(b:a=c:k\), so \(k=ac/b\). Then construct the fourth proportional \(h\) with \(b:a=k:h\), so \(h=ak/b=a^2c/b^2\). Each step is Euclid VI.12, done with similar triangles, and \(b^2h=a^2c\).

The cube equal to a number

Proposition (Menaechmus’ method, as used by Khayyam)

Let \(d=a^2b\). The parabolas \(x^2=ay\) and \(y^2=bx\) meet at the origin and at exactly one other point, whose abscissa \(x\) satisfies \(x^3=d\).

Proof. Squaring the first, \(x^4=a^2y^2=a^2bx\). At the second point \(x\neq0\), so \(x^3=a^2b\). Equivalently, \(x\) and \(y\) are the two mean proportionals between \(a\) and \(b\): \(a:x=x:y=y:b\). Hippocrates of Chios had reduced doubling the cube to finding two mean proportionals, and Menaechmus found them with conics in the fourth century BC.

Exercise 5.1Use the proposition with \(a=1\), \(b=2\) to construct \(\sqrt[3]{2}\). Find the coordinates of the intersection point.

Show solution
\(x^2=y\) and \(y^2=2x\) give \(x^3=2\), so \(x=2^{1/3}\approx1.260\) and \(y=x^2=2^{2/3}\approx1.587\). The chain \(1:x=x:y=y:2\) holds since each ratio equals \(2^{-1/3}\).

Exercise 5.2Show that the circle on diameter \(AC\) satisfies \(DE^2=AE\cdot EC\), and write it in coordinates when \(A=(0,0)\), \(C=(h,0)\).

Show solution
Angle \(ADC\) is right (Thales), and \(DE\) is the altitude to the hypotenuse, so triangles \(AED\) and \(DEC\) are similar and \(AE:DE=DE:EC\). With \(D=(x,y)\), \(E=(x,0)\): \(y^2=x(h-x)\), i.e. \(x^2+y^2=hx\), the circle with centre \((h/2,0)\) and radius \(h/2\).
Lesson 06

A cube and sides equal a number

For the general reader. This is the type \(x^3+cx=d\), the one Gerolamo Cardano would solve by a formula four and a half centuries later. Khayyam found its root as the height of the point where a parabola crosses a circle. The picture at the top of this page is his construction.

Goals
  • Carry out Khayyam’s construction and prove it correct.
  • Show that the positive root exists and is unique.
  • Run the construction on numerical examples.
Theorem (Khayyam)

Let \(c=b^2\) and \(d=b^2h\) with \(b,h>0\). Place the vertex \(B\) of a parabola with parameter \(b\) at the origin, with axis along the positive \(x\)-axis: \(y^2=bx\). Lay off \(BH=h\) along the \(y\)-axis and draw the circle on diameter \(BH\): \(x^2=y(h-y)\). The circle meets the upper branch of the parabola at exactly one point \(D\) other than \(B\), and the height \(y\) of \(D\) is the unique positive root of \[ y^3+cy=d. \]

Proof

At an intersection point, the parabola gives \(x=y^2/b\). Substitute into the circle:

\[ \frac{y^4}{b^2}=y(h-y)\quad\Longrightarrow\quad y^3=b^2(h-y)\quad(y\neq0)\quad\Longrightarrow\quad y^3+b^2y=b^2h. \]

Khayyam argues with proportions instead. From the parabola, \(b:y=y:x\). From the circle, \(y:x=x:(h-y)\). So \(b,\ y,\ x,\ h-y\) are in continued proportion, and therefore \(b^2:y^2=y:(h-y)\). Cross-multiplying gives the cube \(y^3\) equal to the solid \(b^2(h-y)\), and adding \(b^2y\) to both sides gives the result.

Existence and uniqueness. The function \(f(y)=y^3+cy\) is strictly increasing on \(y>0\), with \(f(0)=0\) and \(f(y)\to\infty\), so it takes the value \(d\) exactly once. Geometrically, the circle lies in the strip \(0\le y\le h\), is tangent to the \(x\)-axis at \(B\), and the parabola leaves \(B\) tangent to the \(y\)-axis, so the parabola starts inside the circle and must cross it once before \(y\) reaches \(h\).

Exercise 6.1Carry out the construction for \(y^3+4y=16\). What are \(b\), \(h\), and the intersection point?

Show solution
\(b^2=4\) gives \(b=2\), and \(b^2h=16\) gives \(h=4\). The parabola \(y^2=2x\) and the circle \(x^2=y(4-y)\) meet at \(D=(2,2)\): \(2^2=2\cdot2\) and \(2^2=2(4-2)\). So \(y=2\), and indeed \(8+8=16\).

Exercise 6.2The equation \(x^3+6x=20\) is the first worked example in Cardano’s Ars Magna (1545). Set up Khayyam’s construction and verify the root \(x=2\).

Show solution
\(b=\sqrt6\), \(h=20/6=10/3\). At \(y=2\): \(x=y^2/b=4/\sqrt6\), so \(x^2=16/6=8/3\), and \(y(h-y)=2\cdot(10/3-2)=8/3\). The point lies on both curves. Cardano’s formula expresses the same root as \(\sqrt[3]{\sqrt{108}+10}-\sqrt[3]{\sqrt{108}-10}\), which also equals 2.

Exercise 6.3Show that the construction always gives \(0<y<h\), and explain this from the equation.

Show solution
From \(y^3=b^2(h-y)\) and \(y>0\) we get \(h-y>0\). In words: the cube \(y^3\) is positive, so the “sides” term \(b^2y\) alone is smaller than the number \(b^2h\).
Lesson 07

Circles, hyperbolas and the number of roots

For the general reader. Some cubic types can have no positive solution, or two. Khayyam saw this: in his constructions the two curves may miss each other, touch, or cross twice. He did not notice that some types can have three positive solutions. This lesson works through the quadrant equation of Lesson 02 and the question of how many roots there are.

Goals
  • Construct a root of a quadrinomial cubic with a circle and a rectangular hyperbola.
  • Relate the number of intersection points to the number of positive roots.
  • Find the case Khayyam missed.

A cube and sides equal squares and a number

Khayyam handles the type \(x^3+cx=ax^2+d\), to which the quadrant equation belongs, by a circle and a rectangular hyperbola. The following coordinate version follows his idea. Write \(c=k^2\) and \(m=d/k^2\). The equation becomes

\[ x^2(x-a)=k^2(m-x). \]
Proposition

Let \(\Gamma\) be the circle on the diameter joining \((a,0)\) and \((m,0)\), so that \(y^2=(x-a)(m-x)\), and let \(\mathcal H\) be the rectangular hyperbola \(xy=k(m-x)\). Every intersection point with \(x>0\) and \(x\neq m\) has abscissa a positive root of \(x^3+k^2x=ax^2+d\), and every positive root other than \(m\) arises this way.

Proof. On \(\mathcal H\), \(x^2y^2=k^2(m-x)^2\). On \(\Gamma\), \(y^2=(x-a)(m-x)\). Combining, \(x^2(x-a)(m-x)=k^2(m-x)^2\). Divide by \(m-x\). Conversely, a root \(x\neq m\) gives \(y=k(m-x)/x\), and the equation shows that this \(y\) satisfies \(y^2=(x-a)(m-x)\). The hyperbola is rectangular: it is \(x(y+k)=km\), with asymptotes \(x=0\) and \(y=-k\).

Any root must lie between \(a\) and \(m\), since \(x^2(x-a)\) and \(k^2(m-x)\) must have the same sign. Khayyam discusses such conditions in words: if the number is too large or too small relative to the other terms, the curves do not meet and the problem is impossible.

How many roots?

For several types Khayyam notes that the curves may meet in two points, giving two solutions, or touch, giving one, or miss, giving none. An example is \(x^3+d=cx\). The type \(x^3+cx=ax^2+d\), however, can have three positive roots, and Khayyam does not say so. Recognizing all real roots of cubics, and the role of negative and complex ones, waited for the sixteenth-century Italians and for René Descartes’ rule of signs (1637).

Exercise 7.1For the quadrant equation \(x^3+200x=20x^2+2000\), find \(k\), \(m\), the circle and the hyperbola. Show that the root lies between 10 and 20, and locate it to one decimal place.

Show solution
\(k^2=200\), \(m=2000/200=10\), \(a=20\). The circle has diameter from \((10,0)\) to \((20,0)\); the hyperbola is \(xy=\sqrt{200}\,(10-x)\). Roots lie between \(m=10\) and \(a=20\). Let \(g(x)=x^3-20x^2+200x-2000\). Then \(g(15)=-125\) and \(g(16)=176\), and \(g(15.4)\approx-10.9\), \(g(15.5)\approx18.9\). The root is about \(15.44\). There is only one real root, since \(g'(x)=3x^2-40x+200>0\) for all \(x\) (its discriminant is negative).

Exercise 7.2Expand \((x-1)(x-2)(x-3)\) and write the result as one of Khayyam’s types. Check that the circle and hyperbola above meet three times.

Show solution
\(x^3-6x^2+11x-6=0\), i.e. \(x^3+11x=6x^2+6\): a cube and sides equal squares and a number. Here \(k^2=11\), \(m=6/11\), \(a=6\). Each of \(x=1,2,3\) lies in \((m,a)\), and \(y=k(m-x)/x\) is defined; the identity \(x^2(x-a)=k^2(m-x)\) holds at each (e.g. \(x=2\): \(4\cdot(-4)=-16\) and \(11(6/11-2)=6-22=-16\)). Three intersection points, three positive roots.

Exercise 7.3Show that \(x^3+3=4x\) has two positive roots, and find them.

Show solution
\(x=1\) is a root: \(1+3=4\). Dividing, \(x^3-4x+3=(x-1)(x^2+x-3)\), with positive root \(x=(-1+\sqrt{13})/2\approx1.303\). The third root is negative. In Khayyam’s terms, the two conics cross twice in the positive quadrant.
Lesson 08

The Jalali calendar

For the general reader. A calendar must keep the seasons fixed, so the length of its average year must match the time the Sun takes to return to the same equinox. That interval is not a whole number of days. The Isfahan astronomers designed a calendar for Malik-Shah that tracked the spring equinox more closely than the Gregorian calendar adopted in Europe five centuries later.

Goals
  • Know what the reform did and what is uncertain about it.
  • Compare leap-year rules through continued fractions.
  • Compute the drift of several calendars.

What is documented

The old Persian solar year had 365 days with no leap day, so New Year (Nowruz) wandered through the seasons. Malik-Shah ordered a reform, carried out by a group of astronomers at Isfahan of whom Khayyam was one. The new era, named Jalali after the sultan’s title Jalal al-Dawla, was inaugurated in March 1079 (10 Ramadan 471 AH; modern conversions give 15 or 16 March). The year began at the spring equinox, and months were tied to the Sun’s passage through the signs of the zodiac. Khayyam is credited with a value of the tropical year of 365.24219858156 days, an extraordinary number of digits for an observed quantity; the modern mean tropical year is about 365.24219 days.

What is uncertain

As designed, the calendar placed Nowruz by astronomical computation of the equinox, so it needed no fixed leap-year rule. Later writers attributed to Khayyam an arithmetic rule of 8 leap years in every 33, and other cycles have been proposed. The surviving sources do not settle which rule, if any, the astronomers fixed. The calendar was used in Iran in various forms for centuries; in 1925 it was simplified into the present Solar Hijri calendar.

Continued fractions and leap rules

Let the fractional part of the year be \(\theta\approx0.24219\). Its continued fraction is \(\theta=[0;4,7,1,3,\dots]\) with convergents \[ \frac14,\quad\frac{7}{29},\quad\frac{8}{33},\quad\frac{31}{128},\ \dots \] A cycle of \(q\) years with \(p\) leap years gives mean year \(365+p/q\). Each convergent is the best approximation to \(\theta\) with denominator at most its own.

The Julian calendar uses \(1/4\). The 33-year rule uses \(8/33\). The Gregorian calendar uses \(97/400=0.2425\), which is not a convergent and is less accurate than \(8/33\). One subtlety: a calendar that keeps the spring equinox fixed should match the vernal equinox year, currently about 365.2424 days, which is slightly longer than the mean tropical year. The 33-year cycle fits that value even better.

Exercise 8.1Compute the error, in days per year and in years per day of drift, of the Julian rule, the Gregorian rule and the 33-year rule, against 365.24219 days.

Show solution
Julian \(365.25\): error \(0.00781\) d/yr, one day in about 128 years. Gregorian \(365.2425\): error \(0.00031\), one day in about 3200 years. 33-year \(365+8/33=365.242424\): error \(0.000234\), one day in about 4300 years.

Exercise 8.2Compute the first four partial quotients of \(\theta=0.24219\) and verify the convergents.

Show solution
\(1/0.24219=4.1290\to4\); \(1/0.1290=7.75\to7\); \(1/0.75\approx1.33\to1\); \(1/0.33\approx3\). Convergents: \([0;4]=1/4\), \([0;4,7]=7/29\), \([0;4,7,1]=8/33\), \([0;4,7,1,3]=31/128\).

Exercise 8.3A 33-year cycle with 8 leap years can be laid out as seven 4-year groups and one 5-year group. Show the count works, and explain why spreading the leap years evenly matters.

Show solution
\(7\times4+5=33\) years, with one leap year per group: 8 leap years. Spreading them evenly keeps the calendar date of the equinox within about a day at all times, not just on average over the cycle.
Lesson 09

The parallel postulate

For the general reader. Euclid’s fifth postulate, about when two lines meet, always seemed less obvious than the others. For two thousand years geometers tried to prove it from the rest. Khayyam’s attempt, written in Isfahan in 1077, introduced a four-sided figure that became the standard tool for the problem and examined the three possibilities for its angles. Those three possibilities correspond to the three geometries known today.

Goals
  • State Euclid’s fifth postulate and Khayyam’s replacement principle.
  • Prove the basic theorems about the Khayyam–Saccheri quadrilateral in absolute geometry.
  • See where the argument assumes what it set out to prove.
Euclid, Elements I, Postulate 5

If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two lines, if produced indefinitely, meet on that side.

Khayyam’s critique

Khayyam’s commentary opens by criticizing earlier attempts, including that of Ibn al-Haytham, for using motion in geometry: Ibn al-Haytham had a line segment slide along another, which Khayyam judged foreign to geometric proof. Khayyam then replaces the postulate with a principle he attributes to “the philosopher”, meaning Aristotle:

Khayyam’s principle

Two straight lines that converge must intersect, and two straight lines cannot converge in the direction in which they diverge.

The quadrilateral

Take a segment \(AB\). Erect equal perpendiculars \(AC\) and \(BD\) on the same side, and join \(CD\). This is the figure now called the Khayyam–Saccheri quadrilateral, after Giovanni Girolamo Saccheri, who used it in 1733. Khayyam proves, without the fifth postulate:

Theorem (absolute geometry)

In the quadrilateral \(ABDC\) with right angles at \(A\) and \(B\) and \(AC=BD\): (i) the summit angles at \(C\) and \(D\) are equal; (ii) the perpendicular bisector of \(AB\) is also the perpendicular bisector of \(CD\).

The summit angles are therefore both right, both obtuse or both acute. Khayyam rejects the obtuse and acute cases with his principle. In the obtuse case, he argues, the perpendiculars \(AC\) and \(BD\) would converge as one moves up from the base. Repeating the figure along the other side of the base would make them converge in the other direction too, so the lines would be diverging in the direction of convergence, which the principle forbids. The acute case is excluded symmetrically. From right summit angles he goes on to derive Euclid’s postulate.

Assessment

The principle is equivalent to the fifth postulate given the other axioms, so the argument replaces one assumption with another. The value of the work lies elsewhere. Khayyam isolated the three hypotheses on the summit angles and proved real theorems in absolute geometry. Nasir al-Din al-Tusi discussed Khayyam’s argument in 1250. A later work in al-Tusi’s school, printed in Rome in 1594, was read in Europe, and Saccheri’s book opens with a criticism of John Wallis, who had studied that text. Saccheri’s acute-angle hypothesis turned out to describe hyperbolic geometry, found by János Bolyai and Nikolai Lobachevsky around 1830. Whether Saccheri knew Khayyam’s work is not established.

Exercise 9.1Prove part (i) of the theorem without the parallel postulate.

Show solution
Triangles \(CAB\) and \(DBA\) are congruent (SAS: \(AC=BD\), right angles at \(A\) and \(B\), common side \(AB\)), so \(CB=DA\). Then triangles \(ACD\) and \(BDC\) are congruent (SSS: \(AC=BD\), \(AD=BC\), common \(CD\)), so \(\angle ACD=\angle BDC\). Only congruence is used, which holds in absolute geometry.

Exercise 9.2In the hyperbolic plane of curvature \(-1\), the area of a geodesic polygon with \(n\) sides is \((n-2)\pi\) minus its angle sum. Show that the summit angles of a Khayyam–Saccheri quadrilateral of area \(S\) are each \(\pi/2-S/2\).

Show solution
For \(n=4\): \(S=2\pi-(\pi/2+\pi/2+2\alpha)=\pi-2\alpha\), so \(\alpha=\pi/2-S/2<\pi/2\). The acute hypothesis holds, and larger quadrilaterals have smaller summit angles.

Exercise 9.3On a unit sphere, take the base \(AB\) on the equator and the sides along meridians up to latitude \(\varphi\). Are the summit angles acute or obtuse? What fails in Khayyam’s setting?

Show solution
The summit \(CD\) is a great-circle arc, which bulges toward the pole relative to the latitude circle, and the meridians converge toward the pole; the area is positive, and on the sphere area equals angle sum minus \(2\pi\), so the summit angles are obtuse. The sphere violates other assumptions Khayyam used: two great circles meet twice and lines are not infinitely extendable without returning. Saccheri likewise refuted the obtuse case using the infinitude of lines.
Lesson 10

Ratio, number and roots

For the general reader. Greek mathematics distinguished numbers from magnitudes such as lengths. A ratio of two lengths, like the diagonal of a square to its side, need not be a ratio of whole numbers. The second and third books of Khayyam’s Euclid commentary examine what it means for two ratios to be equal, and ask whether a ratio might itself be a kind of number.

Goals
  • State Eudoxus’ definition of equal ratios and the definition by repeated subtraction.
  • Understand Khayyam’s proof that they agree, and his question about ratios as numbers.
  • Know what is and is not documented about his work on roots and binomial coefficients.
Two definitions of equal ratios

Eudoxus (Euclid V, Def. 5). \(a:b=c:d\) when, for all positive integers \(m,n\), \(ma>nb\) exactly when \(mc>nd\), \(ma=nb\) exactly when \(mc=nd\), and \(ma<nb\) exactly when \(mc<nd\).

Anthyphairesis. Apply the Euclidean algorithm to \(a,b\): subtract the smaller from the larger as often as possible, then repeat with the remainder. \(a:b=c:d\) when the two sequences of quotients are identical, finite or infinite.

Euclid used the first in Book V; Arabic scholars, including al-Mahani, preferred the second, which they found closer to the meaning of ratio. Khayyam proved that the two definitions agree. In modern terms, the quotient sequence is the continued fraction of \(a/b\), and two positive reals have the same continued fraction exactly when they are equal, which is also what the Eudoxus test detects.

In Book III Khayyam treats the composition of ratios, \((a:b)\cdot(b:c)=a:c\). To handle it, he chooses a unit and considers, for any ratio, a magnitude standing to the unit in that ratio. He then asks whether such a magnitude should be counted among the numbers. He leaves the question open, but the step points toward treating any ratio, rational or not, as a number.

Roots and binomial coefficients

In the algebra Khayyam mentions a separate work, now lost, in which he extended Indian methods of extracting square and cube roots to roots of any order. Historians infer that such a method rests on the expansion of \((a+b)^n\) and its coefficients, the triangle known in Europe as Pascal’s. Al-Karaji had described that triangle earlier. Because the work is lost, Khayyam’s exact method is unknown.

Exercise 10.1Run anthyphairesis on the diagonal \(d\) and side \(s\) of a square, and show the quotient sequence is \(1,2,2,2,\dots\). Conclude that \(d:s\) is not a ratio of numbers.

Show solution
\(d=\sqrt2\,s\). \(d\) contains \(s\) once with remainder \(r=(\sqrt2-1)s\). Then \(s/r=\sqrt2+1=2+(\sqrt2-1)\), so quotient 2 with remainder \((\sqrt2-1)r\), and the pattern repeats forever. A ratio of whole numbers gives a terminating sequence, so \(d:s\) is not one.

Exercise 10.2Check by both definitions that \(14:10=21:15\).

Show solution
Anthyphairesis: \(14=1\cdot10+4\), \(10=2\cdot4+2\), \(4=2\cdot2\), quotients \(1,2,2\); \(21=1\cdot15+6\), \(15=2\cdot6+3\), \(6=2\cdot3\), quotients \(1,2,2\). Eudoxus: \(14m>10n\iff 7m>5n\iff 21m>15n\), and similarly for \(=\) and \(<\).

Exercise 10.3Use the binomial expansion to find the fifth root of 33 to two decimals, in the spirit of a digit-by-digit method: start from \(2^5=32\).

Show solution
Write \(33=(2+e)^5=32+5\cdot2^4e+10\cdot2^3e^2+\dots\). To first order \(80e\approx1\), \(e\approx0.0125\). The next term \(80e^2\approx0.0125\) changes \(e\) only slightly: \(e\approx(1-80e^2)/80\approx0.0123\). So \(\sqrt[5]{33}\approx2.012\).
Part III · Life and legacy
Lesson 11

After 1092: Merv and Nishapur

The deaths of Nizam al-Mulk and Malik-Shah in late 1092 opened years of struggle among the sultan’s heirs. Khayyam lost his position at court. The sources for the next forty years are scattered and sometimes contradictory.

  • Pilgrimage. Ibn al-Qifti, writing more than a century later, says Khayyam made the pilgrimage to Mecca and stayed in Baghdad. He presents it as a show of piety to silence critics, a reading coloured by his own hostility. The pilgrimage itself is commonly accepted; its motive is not known.
  • Sanjar. Al-Bayhaqi tells that Khayyam treated the young prince Sanjar during smallpox and gave a gloomy prognosis, and that Sanjar held this against him. Sanjar governed Khorasan from 1097 and became supreme sultan in 1118, with his capital at Merv.
  • Balkh and Merv. Nizami Aruzi met Khayyam in Balkh in 1112–13, in the company of Abu Hatim al-Isfizari, and again in Merv in the winter of 1114–15, where both were guests of the vizier. Al-Isfizari and Khayyam had worked together; al-Isfizari’s own studies of the balance appear in al-Khazini’s book.
  • Debate. The grammarian al-Zamakhshari describes a debate with Khayyam in Khwarazm and praises him while claiming victory.

Death

Khayyam died in Nishapur. The date 4 December 1131 is widely given; older literature offers c. 1123, and the medieval sources do not fix it. Nizami Aruzi wrote that he visited the grave in Nishapur in 1135–36 and found it beside a garden wall, under fruit trees whose blossom covered the stone, just as Khayyam had once predicted to him. The story is charming and impossible to check.

DiscussionAl-Bayhaqi, Nizami Aruzi and Ibn al-Qifti each had reasons to shape their portraits. For one claim in this lesson, say which source makes it, what that writer’s interests were, and how confident a historian can be.

Lesson 12

Afterlives

In mathematics

Khayyam’s geometric theory of cubics was taken further by Sharaf al-Din al-Tusi (d. c. 1213), who studied the existence and number of positive roots through the maximum of a cubic expression. In Europe the algebraic solution came from Scipione del Ferro, Niccolò Tartaglia and Gerolamo Cardano in the 1500s, without knowledge of Khayyam. René Descartes, in La Géométrie (1637), again solved cubics and quartics by intersecting a parabola with a circle, an approach close to Khayyam’s. The algebra became known in Europe only when Franz Woepcke edited and translated it in 1851.

The quadrilateral of the Euclid commentary passed to Nasir al-Din al-Tusi and his school. Giovanni Girolamo Saccheri’s Euclides ab omni naevo vindicatus (1733) used the same figure and the same three hypotheses, and Johann Heinrich Lambert pushed the acute case further. The acute hypothesis is consistent: it is hyperbolic geometry.

The calendar

The Jalali reform shaped Persian timekeeping for nine centuries. Iran adopted a solar calendar officially in 1911 and simplified it in 1925 into the Solar Hijri calendar still used in Iran and Afghanistan, with the year beginning at the spring equinox.

The quatrains

Persian quatrains (ruba‘iyat) were attributed to Khayyam from the late twelfth century, and manuscripts grew larger with time; a Bodleian manuscript copied in Shiraz in 1460 contains 158. Which, if any, are his is disputed, and some scholars doubt the attribution altogether. Edward FitzGerald’s free English version of 1859 sold slowly at first and then became one of the most popular poems in the language, making Khayyam’s name famous in the West for reasons that have little to do with his science.

Honours

  • A mausoleum designed by Houshang Seyhoun stands over his tomb in Nishapur.
  • The lunar crater Omar Khayyam (named 1970) and the asteroid 3095 Omarkhayyam (discovered 1980).
  • Iran donated a pavilion in his honour to the United Nations Office in Vienna in 2009.

Final projectChoose one of Khayyam’s 14 cubic types not covered in this course. Find a pair of conics whose intersection gives its positive roots, prove it, and discuss for which coefficients the curves meet zero, one, two or three times.

Reference

Connections

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

Influenced by

Influenced

Colleagues and collaborators

  • Abu Hatim al-Isfizari 1074-1115

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Omar Khayyam, MacTutor History of Mathematics.
  2. Khayyam, Omar: Life, Encyclopaedia Iranica (the medieval sources discussed one by one).
  3. Omar Khayyam, Wikipedia.
  4. Umar al-Khayyam, Muslim Heritage (works and dates).
  5. Nizami Aruzi, Chahar Maqala, trans. E. G. Browne, 1921.

Mathematics

  1. F. Woepcke, L’algèbre d’Omar Alkhayyâmî, Paris, 1851.
  2. D. S. Kasir, The Algebra of Omar Khayyam, Columbia University, 1931.
  3. R. Rashed and B. Vahabzadeh, Omar Khayyam the Mathematician, Bibliotheca Persica, 2000.
  4. J. L. Berggren, Episodes in the Mathematics of Medieval Islam, Springer, 2nd ed. 2016, chapter on algebra.
  5. Saccheri quadrilateral and Parallel postulate, Wikipedia (history sections).
  6. Jalali calendar, Wikipedia.
  7. B. A. Rosenfeld, A History of Non-Euclidean Geometry, Springer, 1988.
  8. Note sur le choix des courbes fait par al-Khayyam, arXiv:0912.1418 (Khayyam’s curves compared with Descartes’).