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Aryabhata

Āryabhaṭa (आर्यभट), author of the Āryabhaṭīya

A course in two strands: what can be known of the young astronomer who wrote at Kusumapura in the Gupta age, and the 121 verses in which he packed a place-value arithmetic, a sine table, a method for indeterminate equations, a rotating Earth and a physical theory of eclipses.

Born 476 AD (from his own statement); place uncertain Died unknown; c. 550 is a conventional estimate Signature result π ≈ 62832/20000 = 3.1416 (Gaṇitapāda 10) Level advanced undergraduate
Āryabhaṭa’s half-chords (jyā). A quadrant of radius 3438′ is cut into 24 arcs of 225′; the bars show his tabulated differences, from 225 down to 7, whose running sums give the sines.
Lesson 00

How to use this course

Āryabhaṭa is the first Indian mathematician and astronomer whose own words survive in a work under his name. That work, the Āryabhaṭīya, is short: 121 verses, written to be memorized, stating rules without proofs. Almost everything else about him comes from one autobiographical verse and from what later astronomers, admirers and critics, said about him. The course therefore keeps two kinds of evidence apart: the text itself, and later tradition.

Prerequisites for Part II
  • High-school trigonometry: sine, cosine, the addition formulas.
  • Elementary number theory: divisibility, greatest common divisors, the Euclidean algorithm.
  • Sums of arithmetic progressions and proof by induction.
  • For Lessons 08–09: the daily motion of the sky, the ecliptic, and the geometry of similar triangles.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Verse references follow the critical edition of K. S. Shukla and K. V. Sarma (1976); W. E. Clark’s translation (1930) numbers the first section slightly differently. Gītikā, Gaṇita, Kālakriyā and Gola denote the four sections. Angles are often measured in minutes of arc, written ′, as Āryabhaṭa did.

Reference

Timeline

All dates AD unless marked. Only the year 499 and the birth year 476 rest on Āryabhaṭa’s own words.

3102 BCTraditional start of the Kali Yuga (18 February), the epoch from which Āryabhaṭa counts.
c. 320–550Gupta dynasty in northern India, with its capital at Pāṭaliputra.
476Born, by his statement that he was 23 years old when 3600 years of the Kali Yuga had elapsed.
499Kali year 3600: the reference year of the Āryabhaṭīya, written at or for Kusumapura.
after 499A second work, the Āryabhaṭa-siddhānta, counting the day from midnight; now lost.
6th c.Varāhamihira’s Pañcasiddhāntikā summarizes five earlier astronomical systems.
628Brahmagupta’s Brāhmasphuṭasiddhānta attacks Āryabhaṭa on many points.
629Bhāskara I writes his commentary on the Āryabhaṭīya, the oldest that survives.
665Brahmagupta’s Khaṇḍakhādyaka, a handbook based on the midnight system of Āryabhaṭa.
8th c.Indian astronomy reaches Baghdad; Āryabhaṭa’s parameters appear in Arabic as the Zīj al-Arjabhar.
c. 1030Al-Bīrūnī’s India discusses Āryabhaṭa and his followers’ view that the Earth rotates.
12th c.Gerard of Cremona’s Latin translations fix the word sinus, from Sanskrit jyā through Arabic.
c. 1500Nīlakaṇṭha Somayājī of the Kerala school writes a major commentary.
1874Hendrik Kern publishes the first printed edition, at Leiden.
1930Walter Eugene Clark’s English translation.
197519 April: India’s first satellite, Aryabhata, is launched.
1976Critical edition and translation by K. S. Shukla and K. V. Sarma for the Indian National Science Academy, marking the 1500th anniversary of his birth.
Part I · Formation
Lesson 01

Kusumapura and the sources

Everything we know from Āryabhaṭa himself fits in two verses. In the time-reckoning section (Kālakriyā 10) he says that when sixty times sixty years of the Kali Yuga had passed, he was twenty-three years old. Kali year 3600 corresponds to AD 499, so he was born in 476. The verse fixes his age, not necessarily the year in which he wrote: 499 may be the epoch chosen for the calculations. At the start of the mathematics section (Gaṇita 1) he says he sets out the knowledge honoured at Kusumapura.

Kusumapura

Kusumapura, “city of flowers”, is usually identified with Pāṭaliputra, modern Patna on the Ganges, capital of the Gupta empire. Bhāskara I, his commentator in 629, confirms the link with Kusumapura. The phrase does not say he was born there, only that he wrote in its tradition.

Aśmaka and Kerala

Bhāskara I also calls him Āśmaka, “of the Aśmaka country”, and names his book the Āśmakatantra. Aśmaka was a region in the Deccan between the Godāvarī and Narmadā rivers. Nīlakaṇṭha, writing in Kerala around 1500, also calls him a man of Aśmaka. Since most surviving manuscripts and commentaries of the Āryabhaṭīya come from South India, some scholars have argued for a southern, even Keralan, origin. None of this is settled. A modern claim that he headed the university of Nālandā rests on a single word in a later verse calling him a kulapa, head of an institution, and is speculation.

Exercise 1.1The Kali Yuga began at midnight between 17 and 18 February 3102 BC (Julian). Show that 3600 full years later falls in AD 499. Why is there no year 0 in this calculation?

Show solution
In astronomical year numbering, 3102 BC is year \(-3101\), since 1 BC is year 0. Then \(-3101+3600=499\), so 3600 years after the start of the Kali Yuga the year is AD 499. Counting with historians’ BC/AD years, where there is no year 0, one must subtract 1 when crossing from BC to AD: \(3600-3102=498\), plus 1 gives 499.
Lesson 02

Indian astronomy before Āryabhaṭa

The oldest Indian astronomical text, the Vedāṅga Jyotiṣa (late first millennium BC), regulated the calendar for Vedic ritual with a five-year cycle and simple arithmetic. In the early centuries AD, Indian astronomers absorbed Greek methods transmitted through texts in Sanskrit: the zodiac of twelve signs, the division of the circle into 360 degrees, epicycles and eccentrics, and the use of chord tables. The Yavanajātaka (“Greek horoscopy”), a Sanskrit versification made in western India in the second or third century, is one witness to this contact.

By Āryabhaṭa’s time several systems called siddhāntas (“established conclusions”) were in use. Varāhamihira of Ujjain, a younger contemporary, summarized five of them in his Pañcasiddhāntikā (“Five Siddhāntas”), including the Romaka, the Pauliśa and an early version of the Sūryasiddhānta. These gave rules for computing planetary positions from large cycles of revolutions.

What was new

Āryabhaṭa’s book was different in three ways. It bears an individual author’s name. It puts mathematics in its own section, as a discipline serving astronomy but stated in general form. And it changed the parameters: new revolution numbers for the planets, a mahāyuga divided into four equal quarters instead of the traditional proportions 4:3:2:1, and a rotating Earth. These innovations created a school, the Āryapakṣa, which competed for centuries with the Brāhmapakṣa that followed Brahmagupta.

Exercise 2.1The traditional mahāyuga of 4,320,000 years is divided into four ages in the ratio 4:3:2:1. Āryabhaṭa divides it into four equal ages. Compute the length of the Kali Yuga under each scheme, and explain why an astronomer whose epoch is the start of the Kali Yuga might prefer the equal division.

Show solution
Traditional: Kali \(=\tfrac{1}{10}\times4{,}320{,}000=432{,}000\) years. Equal division: \(1{,}080{,}000\) years each. The equal quarters make the start of every quarter a point at which the integer revolution numbers divide evenly by 4 (Lesson 08), so mean positions at the epoch are simple. In Āryabhaṭa’s system all the mean planets are at the start of Aries at the start of the Kali Yuga, which requires the elapsed revolutions to be whole numbers at that moment. His revolution numbers are indeed all divisible by 4: for example \(57{,}753{,}336=4\times14{,}438{,}334\) for the Moon.
Lesson 03

The book of 121 verses

The Āryabhaṭīya has four sections called pādas (“quarters”):

  • Gītikāpāda, 13 verses in the gīti metre: the alphabetic numeral system, the revolution numbers of the planets in a mahāyuga, the sizes of the Earth and planets, and the sine table in a single verse.
  • Gaṇitapāda, 33 verses on mathematics: place value, squares and cubes and their roots, areas and volumes, π, the sine rule, shadows of the gnomon, series, interest, the rule of three, linear equations and the kuṭṭaka.
  • Kālakriyāpāda, 25 verses on time: units, the conjunctions of planets, intercalary months, and the motions of planets on eccentrics and epicycles.
  • Golapāda, 50 verses on the sphere: the celestial sphere and its circles, the spherical Earth and its rotation, day and night at different latitudes, and eclipses.

Bhāskara I calls the first section the “ten gīti” verses, though it has thirteen; which three are framing or later additions is debated. Another early name, Āryāṣṭaśata, “the 108 āryā verses”, refers to the last three sections.

A text for memory

The verses are compressed to the limit. Numbers are encoded in syllables (Lesson 04), rules are stated without proofs or examples, and a single line can hold a full algorithm. The book was meant to be learned by heart and explained by a teacher. This is why the commentaries matter so much: Bhāskara I in 629, Someśvara, Sūryadeva Yajvan in the 12th–13th centuries and Nīlakaṇṭha around 1500 supply the examples and arguments that the verses leave out. Several key rules are ambiguous, and modern interpretations sometimes depend on which commentator one follows.

The lost second work

Later astronomers quote a second treatise, usually called the Āryabhaṭa-siddhānta, in which the day begins at midnight rather than at sunrise. It described instruments including the gnomon, a water clock and shadow devices. Brahmagupta based his Khaṇḍakhādyaka on it, and Arabic sources knew its parameters.

Exercise 3.1Count the verses: \(13+33+25+50\). Then explain how the names “ten gīti” and “108 āryā” relate to the total.

Show solution
The total is 121. The last three sections have \(33+25+50=108\) verses in the āryā metre, which explains Āryāṣṭaśata. The first section has 13 verses in place of the ten its name promises, so three of its verses must be framing (such as the invocation) or additions. The names describe a core of \(10+108=118\) verses inside a 121-verse text, which is why some accounts, MacTutor among them, speak of 118 verses.
Part II · Mathematics and astronomy
Lesson 04

Numbers in syllables: place value

For the general reader. Astronomy needs huge numbers: the Moon goes round the Earth more than 57 million times in Āryabhaṭa’s great cycle. To fit such numbers into memorizable verse, he invented a code in which every syllable is a number. The code only works because he thought in powers of ten, the decimal place value that India later gave the world.

Goals
  • State the rule of Gītikā 2 and decode syllables.
  • Understand the decimal places named in Gaṇita 2.
  • Weigh the evidence that Āryabhaṭa computed with a place-value system including zero.

The places (Gaṇita 2)

Āryabhaṭa names the decimal places: eka (1), daśa (10), śata (100), sahasra (1000), and so on up to vṛnda (\(10^9\)), and says each place is ten times the one before. This is the principle of decimal place value, stated as a rule.

The syllabic code (Gītikā 2)

The Sanskrit consonants are arranged in five groups of five (varga, “classed” letters, from k to m) followed by eight others (avarga: y r l v ś ṣ s h).

Rule (Gītikā 2, in modern form)
  1. The classed consonants \(k, kh, g, gh, \dots, m\) have the values \(1,2,3,\dots,25\).
  2. The unclassed consonants \(y,r,l,v,ś,ṣ,s,h\) have the values \(30,40,50,\dots,100\).
  3. The vowels \(a,i,u,ṛ,ḷ,e,ai,o,au\) multiply the consonant they follow by \(10^0,10^2,10^4,\dots,10^{16}\).
  4. The value of a word is the sum of the values of its syllables; a conjunct consonant takes the vowel that follows it.

Each vowel thus governs two decimal places: a classed consonant fills the lower one (values 1–9 there, or 10–25 spilling into the upper), and an unclassed consonant fills the upper one (30–90, with h = 100 reaching into the next vowel’s range). For example, khyughṛ is \(kh\cdot u + y\cdot u + gh\cdot ṛ = 2\cdot10^4+30\cdot10^4+4\cdot10^6=4{,}320{,}000\), the number of solar revolutions in a mahāyuga.

Zero

The code has no symbol for zero: an empty place is simply omitted. But arithmetic in the Gaṇitapāda, particularly the square-root and cube-root algorithms (Lesson 05), proceeds digit by digit in a way that presupposes a written place-value notation, and Bhāskara I in 629 uses written numerals with zero in his commentary. Historians such as Georges Ifrah argue that Āryabhaṭa must have used such a notation; it is not attested in his own text.

Exercise 4.1Decode makhi and bhakhi, the first two entries of the sine table (Lesson 06). (Values: m = 25, bh = 24, kh = 2.)

Show solution
ma \(=25\times10^0=25\) and khi \(=2\times10^2=200\), so makhi \(=225\). Likewise bhakhi \(=24+200=224\). These are the first two sine differences, in minutes of arc.

Exercise 4.2Explain why the code is ambiguous neither for 4,320,000 nor for any number below \(10^{18}\), and why the order of syllables within a word does not matter.

Show solution
Write the number in base 100: \(N=\sum_{k=0}^{8} c_k\,100^k\) with \(0\le c_k\le 99\). Each \(c_k=10t+u\) is represented by an unclassed consonant for the tens \(t\ge3\) (or by a classed consonant for 10–25) and a classed consonant for the units \(u\), both carrying vowel number \(k\). Since the value is a sum, the order of syllables is irrelevant, and the base-100 digits are unique, so decoding is unambiguous. With nine vowels the largest place is \(100^8=10^{16}\), times at most 99, so all numbers below \(10^{18}\) are covered. (Some choices, such as writing 20 as one classed consonant or as tens and units, give several spellings of the same number, which is harmless.)
Lesson 05

Roots, series and the circle

For the general reader. The Gaṇitapāda is a compact handbook of the mathematics an astronomer needs: how to extract square and cube roots, find areas and volumes, sum series, and relate a circle’s circumference to its diameter. Most rules are right; a few volume rules are wrong. His value of π, correct to four decimal places, became famous.

Goals
  • Run the digit-by-digit square root of Gaṇita 4.
  • Prove the sum formulas of Gaṇita 19–22.
  • Read the π rule of Gaṇita 10 and the word “approximate”.

Square roots (Gaṇita 4)

Split the number into pairs of digits from the right. Find the largest square not exceeding the leading group; its root is the first digit. Then repeatedly bring down the next digit, divide by twice the root found so far, subtract the square of the new quotient from what remains, and continue. The method is the one taught in schools until recently, and it rests on \((10a+b)^2=100a^2+2\cdot10a\cdot b+b^2\).

Series (Gaṇita 19–22)

Rules (Gaṇita 21–22)

For a positive integer \(n\):

\[ \sum_{k=1}^{n}\frac{k(k+1)}{2}=\frac{n(n+1)(n+2)}{6},\qquad \sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6},\qquad \sum_{k=1}^{n}k^3=\Big(\frac{n(n+1)}{2}\Big)^2. \]

The first sum counts the balls in a triangular pile (citi). Gaṇita 19–20 give the sum of an arithmetic progression and solve for the number of terms, which requires a quadratic equation.

Areas and volumes (Gaṇita 6–7)

The area of a triangle is half the base times the height, and the area of a circle is half the circumference times half the diameter: both correct. The volume of a “six-edged solid” (a tetrahedron) is given as half the base area times the height, and the volume of a sphere as the area of a great circle times its own square root. Both are wrong: the true factors are \(1/3\) and \(\tfrac43\pi r^3\). Some modern readers have proposed reinterpretations, but the commentators read the rules this way.

The circle (Gaṇita 10)

Āryabhaṭa says: add four to a hundred, multiply by eight, and add sixty-two thousand; this is the approximate circumference of a circle of diameter twenty thousand. That is \(\pi\approx (8\times104+62000)/20000=62832/20000=3.1416\). The word he uses, āsanna, means “near” or “approaching”. Some historians read it as recognizing that no exact ratio exists; that π is irrational was proved by Johann Heinrich Lambert in 1761. How he obtained the value is unknown. A regular polygon of 384 sides inscribed in a circle of diameter 20,000 gives a perimeter close to 62,832.

Exercise 5.1Compute \(\sqrt{55225}\) by Āryabhaṭa’s method, showing each step.

Show solution
Pairs: 5 | 52 | 25. Largest square in 5 is 4, so the first digit is 2; remainder 1. Bring down 5: 15. Divide by \(2\times2=4\): quotient 3 (with \(3\times4=12\)), remainder 3; bring down 2 to get 32 and subtract \(3^2=9\): 23. The root so far is 23. Bring down 2: 232. Divide by \(2\times23=46\): quotient 5 (\(5\times46=230\)), remainder 2; bring down 5 to get 25 and subtract \(5^2=25\): 0. So \(\sqrt{55225}=235\).

Exercise 5.2Prove the sum-of-cubes rule by induction, and check the citi rule for \(n=4\).

Show solution
For \(n=1\) both sides are 1. If \(\sum_{k\le n}k^3=T_n^2\) with \(T_n=n(n+1)/2\), then \(T_{n+1}^2-T_n^2=(T_{n+1}-T_n)(T_{n+1}+T_n)=(n+1)\cdot(n+1)^2=(n+1)^3\), since \(T_{n+1}+T_n=(n+1)^2\). For \(n=4\): \(1+3+6+10=20=\tfrac{4\cdot5\cdot6}{6}\).

Exercise 5.3Let \(p_n\) be the perimeter of a regular \(n\)-gon inscribed in a circle of diameter 20,000. Compute \(p_{384}\) and compare with 62,832. How many doublings from the hexagon is this?

Show solution
\(p_n=20000\,n\sin(\pi/n)\). For \(n=384\): \(\sin(\pi/384)\approx0.00818114\), so \(p_{384}\approx20000\times3.1415576\approx62831.15\), which rounds to 62,831, just below 62,832. (The circumscribed 384-gon gives about 62,833.) Since \(384=6\times2^6\), this is six doublings from the hexagon. The agreement does not prove that Āryabhaṭa used this route, but it shows the value was within reach of such a calculation.
Lesson 06

The sine table

For the general reader. Greek astronomers used tables of chords: the length of the straight line across a circle for each angle. Indian astronomers halved the chord and the angle, which gives the modern sine. Āryabhaṭa’s table of 24 sines in a quarter circle is the earliest that survives, and his rule for computing it uses what we would call a second difference, a first step toward the calculus of the sine.

Goals
  • Understand the half-chord and the radius 3438.
  • Rebuild the table from its differences.
  • Derive the exact second-difference relation and compare it with the rule of Gaṇita 12.

Half-chords and the radius

If a circle has radius \(R\), the chord subtending an arc \(2\theta\) is \(2R\sin\theta\). The Sanskrit jyā or ardhajyā (“bow-string”, “half bow-string”) is the half-chord \(R\sin\theta\). Āryabhaṭa measures the radius in the same unit as the arc. With the circle divided into \(21600′\), the radius is \(R=21600/2\pi\approx3437.75\), which he rounds to 3438.

The table (Gītikā, one verse)

The quadrant is divided into 24 arcs of \(225′=3^\circ45′\). The verse gives not the sines but their differences \(\Delta_n=J_n-J_{n-1}\), where \(J_n=R\sin(n\cdot225′)\):

\[ 225,\ 224,\ 222,\ 219,\ 215,\ 210,\ 205,\ 199,\ 191,\ 183,\ 174,\ 164,\ 154,\ 143,\ 131,\ 119,\ 106,\ 93,\ 79,\ 65,\ 51,\ 37,\ 22,\ 7. \]

Their sum is 3438, the radius, as it must be since \(J_{24}=R\sin90^\circ\). The running sums agree with \(3438\sin\theta\) to within one minute; the largest error is at \(67^\circ30′\), where the table gives 3177 against 3176.3.

Proposition (second differences of the sine)

Let \(J_n=R\sin(n\alpha)\). Then for \(n\ge1\)

\[ \Delta_{n+1}-\Delta_n = -\,2(1-\cos\alpha)\,J_n. \]

For \(\alpha=225′\), \(2(1-\cos\alpha)\approx 1/233.5\).

Proof. \(\Delta_{n+1}-\Delta_n=J_{n+1}-2J_n+J_{n-1}=R[\sin(n\alpha+\alpha)+\sin(n\alpha-\alpha)]-2R\sin n\alpha=2R\sin n\alpha\cos\alpha-2R\sin n\alpha\).

Āryabhaṭa’s rule (Gaṇita 12)

The verse is terse and its reading is disputed. On the reading of Bhāskara I and most modern scholars, it says that each difference is obtained from the previous one by subtracting the accumulated sum of the sines so far divided by the first sine, 225:

\[ \Delta_{n+1}=\Delta_n-\frac{J_n}{225}. \]

This is the proposition with the factor \(2(1-\cos\alpha)\approx\alpha^2=(225/3438)^2\approx1/233.5\) replaced by \(1/225\), a slightly larger factor that is easy to remember because 225 is the first sine. Applied literally the rule drifts below the table after the first few entries, so the tabulated values were probably computed or corrected by other means, for example from the half-angle formula. The idea that the change in the sine’s increments is proportional to the sine itself is the discrete form of \(\frac{d^2}{d\theta^2}\sin\theta=-\sin\theta\).

Exercise 6.1Use the table to find \(R\sin30^\circ\) and \(R\sin45^\circ\), and compare with \(R/2\) and \(R/\sqrt2\).

Show solution
\(30^\circ=8\times225′\): \(225+224+222+219+215+210+205+199=1719=3438/2\) exactly. \(45^\circ=12\times225′\): adding \(191+183+174+164\) gives \(2431\), and \(3438/\sqrt2\approx2431.03\).

Exercise 6.2Starting from \(\Delta_1=225\), apply the exact recurrence with \(2(1-\cos\alpha)\approx1/233.5\) to compute \(\Delta_2,\Delta_3,\Delta_4\), rounding at the end. Do the same with the divisor 225. Which matches the table better?

Show solution
Divisor 233.5: \(\Delta_2=225-225/233.5\approx224.04\); \(J_2\approx449.04\), \(\Delta_3\approx224.04-1.923\approx222.11\); \(J_3\approx671.15\), \(\Delta_4\approx222.11-2.874\approx219.24\). Rounded: 224, 222, 219, as in the table. Divisor 225 gives \(224.00,\ 222.00,\ 219.02\), also matching here. The two diverge later: by \(\Delta_{10}\) the divisor-225 rule gives 181 against the table’s 183, and its last difference is about 1 instead of 7, while the divisor-233.5 rule reproduces the whole table to within a unit.

Exercise 6.3Show that \(R=21600/(2\times3.1416)\) rounds to 3438, and explain why measuring the radius in arc-minutes makes \(R\sin\theta\approx R\theta\) for small \(\theta\) (in minutes), in agreement with \(\Delta_1=225\).

Show solution
\(21600/6.2832\approx3437.74\), which rounds to 3438. When the radius and arc are measured in the same unit, an arc of length \(s\) subtends the angle \(s/R\) radians, so the half-chord is \(R\sin(s/R)\approx s\) for small \(s\). Hence the first sine equals the first arc, 225′, to the nearest minute: \(3438\sin(225/3438)\approx224.84\).
Lesson 07

The pulverizer: indeterminate equations

For the general reader. Suppose you know that a planet returns to the same place every so many days, and another every so many days, and you want to know when both will be in given positions together. Questions like this lead to equations whose solutions must be whole numbers. Āryabhaṭa gave a general method, the kuṭṭaka or “pulverizer”, which grinds large numbers down to small ones by repeated division.

Goals
  • State the problem of Gaṇita 32–33 precisely.
  • Run the pulverizer and see why it works.
  • Connect it with the Euclidean algorithm, continued fractions and the Chinese remainder theorem.

The problem

Find a number \(N\) that leaves remainder \(r_1\) on division by \(a\) and remainder \(r_2\) on division by \(b\). Writing \(N=ax+r_1=by+r_2\), this asks for non-negative integers \(x,y\) with

\[ ax-by=c,\qquad c=r_2-r_1. \]
Theorem (solvability)

Let \(a,b\) be positive integers and \(d=\gcd(a,b)\). The equation \(ax-by=c\) has integer solutions if and only if \(d\mid c\). If \((x_0,y_0)\) is one solution, all solutions are \(x=x_0+(b/d)t\), \(y=y_0+(a/d)t\), \(t\in\mathbb Z\).

The method

Divide the larger divisor by the smaller, the smaller by the remainder, and so on, as in the Euclidean algorithm, recording the quotients \(q_1,q_2,\dots,q_k\). Āryabhaṭa’s verse then tells how to build the solution by working back up through the quotients, multiplying and adding at each stage; Bhāskara I explains it with worked examples. In modern terms the quotients are the partial quotients of the continued fraction of \(a/b\), and the back-substitution computes a convergent \(p/q\) with \(aq-bp=\pm1\). Multiplying by \(c\) and reducing gives the smallest solution.

Why it works. If \(a=q_1b+r\), then \(ax-by=c\) becomes \(b(q_1x-y)+rx=c\); setting \(y'=q_1x-y\) gives an equation in the smaller pair \((b,r)\). Each step reduces the coefficients, and when the remainder reaches 0 or 1 the equation is solved by inspection. Reversing the substitutions recovers \(x,y\). This is the “pulverizing”.

Astronomical use

Indian astronomers used the kuṭṭaka to recover the number of days elapsed since the epoch from observed fractional positions of planets, and to choose revolution numbers that make all planets align at the epoch. Brahmagupta and Bhāskara I extended it to systems of congruences, which is the content of the Chinese remainder theorem.

Exercise 7.1Find the smallest positive \(N\) with \(N\equiv1\pmod{7}\) and \(N\equiv5\pmod{8}\) by the pulverizer, then add the condition \(N\equiv4\pmod9\).

Show solution
\(N=7x+1=8y+5\) gives \(7x-8y=4\). Euclid on \((8,7)\): \(8=1\cdot7+1\). So \(8-7=1\), that is \(7\cdot(-1)-8\cdot(-1)=1\); multiply by 4: \(x=-4,y=-4\). General \(x=-4+8t\); \(t=1\) gives \(x=4\), \(N=29\). All solutions are \(N\equiv29\pmod{56}\). Now \(29+56s\equiv4\pmod9\): \(29\equiv2\), \(56\equiv2\), so \(2+2s\equiv4\), \(s\equiv1\pmod 9\). Thus \(N=85\), and the solutions are \(N\equiv85\pmod{504}\). Check: \(85=12\cdot7+1=10\cdot8+5=9\cdot9+4\).

Exercise 7.2Solve \(137x-60y=3\) in positive integers using the continued fraction of \(137/60\).

Show solution
\(137=2\cdot60+17\), \(60=3\cdot17+9\), \(17=1\cdot9+8\), \(9=1\cdot8+1\). So \(137/60=[2;3,1,1,8]\), with convergents \(2,\ 7/3,\ 9/4,\ 16/7,\ 137/60\). The convergent before last is \(16/7\): \(137\cdot7-60\cdot16=959-960=-1\). Hence \(137\cdot(-7)-60\cdot(-16)=1\), and multiplying by 3, \(x=-21\), \(y=-48\). Add multiples of \((60,137)\): \(x=39\), \(y=89\). Check: \(137\cdot39=5343\), \(60\cdot89=5340\), difference 3.
Lesson 08

The yuga and the rotating Earth

For the general reader. Every day the stars rise in the east and set in the west. Almost every ancient astronomer took this at face value: the sky turns. Āryabhaṭa said instead that the Earth turns, and the stars only seem to move, as the shore seems to move to someone on a boat. His later followers mostly abandoned the idea, but his numbers for the length of the day and the year were among the most accurate of antiquity.

Goals
  • Derive the civil days, sidereal day and year from the revolution numbers.
  • Read the boat verse and the tension with the following verse.
  • Understand mean motions and the structure of the planetary model.

The revolution numbers (Gītikā 3–4)

In a mahāyuga of 4,320,000 years the Sun makes 4,320,000 revolutions, the Moon 57,753,336, and the Earth 1,582,237,500 rotations eastward. For the outer planets he gives Mars 2,296,824, Jupiter 364,224 and Saturn 146,564. Mercury and Venus, like the Sun, complete 4,320,000 mean revolutions; their faster cycles appear as the revolutions of their śīghra epicycles. All these are mean motions, which epicycles then correct.

Derived constants

The number of sunrises in a mahāyuga is the number of rotations minus the number of solar revolutions, since each revolution of the Sun eastward costs one sunrise:

\[ D = 1{,}582{,}237{,}500-4{,}320{,}000 = 1{,}577{,}917{,}500 \ \text{civil days}. \]

Hence the sidereal year is \(D/4{,}320{,}000\approx365.25868\) days, and one rotation of the Earth relative to the stars lasts \(D/1{,}582{,}237{,}500\) days \(\approx 23^{\text h}\,56^{\text m}\,4.1^{\text s}\).

The modern sidereal year is 365.25636 days, so Āryabhaṭa’s is about 3 minutes 20 seconds too long. His sidereal day agrees with the modern value, \(23^{\text h}56^{\text m}4.09^{\text s}\), to a tenth of a second.

The boat (Gola 9–10)

In Gola 9 Āryabhaṭa compares the observer to a man in a moving boat who sees fixed objects on the shore moving backward; in the same way, he says, the fixed stars are seen at Laṅkā moving westward. Laṅkā here is a reference point on the equator at the prime meridian of Ujjain, not the island. The next verse says that the rising and setting are caused by the sphere of stars, with the planets, being driven westward by the cosmic wind. The two verses are hard to reconcile. Some commentators took the second as describing appearances; some later copyists and commentators altered the first to remove the rotating Earth. Brahmagupta and later critics rejected the rotation on physical grounds; al-Bīrūnī later reported that Āryabhaṭa’s followers held the Earth to rotate and judged that the question did not affect the astronomical calculations.

Planets

Āryabhaṭa’s model is geocentric. Each planet’s mean position is corrected by two epicycles: the manda, which accounts for the unequal speed around the zodiac (like Ptolemy’s eccentric), and the śīghra, which accounts for retrograde motion (equivalent to the Earth’s motion around the Sun in a modern description). Unusually, he lets the epicycles change size around the circle. Because the śīghra cycle of an outer planet equals the solar year, some historians, notably Bartel van der Waerden, argued for a hidden heliocentric source; most historians reject this. He also says the Moon and planets shine by reflected sunlight.

Exercise 8.1Compute the sidereal and synodic months from the revolution numbers.

Show solution
Sidereal month: \(D/57{,}753{,}336\approx27.32167\) days. The synodic month is the time for the Moon to gain one revolution on the Sun: \(D/(57{,}753{,}336-4{,}320{,}000)=1{,}577{,}917{,}500/53{,}433{,}336\approx29.530582\) days. The modern mean values are 27.32166 and 29.530589 days.

Exercise 8.2Find the sidereal periods of Mars, Jupiter and Saturn in years implied by the revolution numbers, and compare with the modern values 1.881, 11.862 and 29.457 years.

Show solution
Period \(=4{,}320{,}000/n\) years. Mars: \(4320000/2296824\approx1.8809\). Jupiter: \(4320000/364224\approx11.861\). Saturn: \(4320000/146564\approx29.475\). The first two agree to about 0.01%, Saturn to about 0.06%.

Exercise 8.3Show that the number of rotations of the Earth relative to the Sun and relative to the stars differ by exactly the number of years, and deduce the general relation between a solar day \(d_\odot\) and a sidereal day \(d_*\): \(1/d_*=1/d_\odot+1/Y\), with \(Y\) the year in solar days.

Show solution
In time \(T\) the Earth turns \(T/d_*\) times relative to the stars. The Sun moves eastward one revolution per year, so relative to the Sun the Earth turns one time fewer each year: \(T/d_\odot=T/d_*-T/Y\). Dividing by \(T\) gives \(1/d_*=1/d_\odot+1/Y\). With \(d_\odot=1\) and \(Y=365.25868\): \(d_*=Y/(Y+1)\approx0.9972696\) days \(\approx23^{\text h}56^{\text m}4.1^{\text s}\), matching the derived constant.
Lesson 09

Eclipses and shadows

For the general reader. Indian tradition explained eclipses by a demon, Rāhu, swallowing the Sun or Moon. Āryabhaṭa gave the physical explanation: a solar eclipse occurs when the Moon covers the Sun, a lunar eclipse when the Moon enters the Earth’s shadow. He then gave rules for computing the size of the shadow and the duration of an eclipse, using nothing but similar triangles.

Goals
  • State Āryabhaṭa’s explanation of eclipses.
  • Derive the length of the Earth’s shadow and its width at the Moon.
  • Compute the half-duration of a central lunar eclipse.

The explanation (Golapāda)

The Moon covers the Sun, and the great shadow of the Earth covers the Moon. Eclipses happen only near the nodes, where the Moon’s path crosses the ecliptic, at new Moon (solar) or full Moon (lunar). The verses then give rules for the length of the shadow, its diameter at the Moon’s distance, the half-duration of the eclipse and the part of the disc obscured. Āryabhaṭa’s diameters, given in the Gītikāpāda in yojanas, are: Earth 1050, Sun 4410, Moon 315. The length of a yojana in his system is uncertain, but the ratios are what matter.

Proposition (the Earth’s shadow)

Let the Sun and Earth have diameters \(S>E\) and be at distance \(d\). The Earth’s shadow is a cone of length

\[ L=\frac{E\,d}{S-E}. \]

At distance \(m\) from the Earth, the diameter of the shadow is \(E\,(L-m)/L\).

Proof. The shadow cone is tangent to both spheres. By similar triangles, the diameters of Sun and Earth are proportional to their distances from the cone’s apex: \(S/(L+d)=E/L\), so \(SL=EL+Ed\) and \(L=Ed/(S-E)\). Along the cone the diameter falls linearly from \(E\) at the Earth to 0 at the apex.

Duration

For a central lunar eclipse, the Moon moves relative to the shadow at the difference of the daily motions of the Moon and the Sun (the shadow’s centre is opposite the Sun). From first contact to mid-eclipse the Moon’s centre travels the sum of the radii of shadow and Moon. Āryabhaṭa’s rule divides this distance by the relative daily motion to obtain the half-duration.

Exercise 9.1Take Āryabhaṭa’s diameters and, as illustrative values chosen for this exercise (not his own), a Sun distance of 459,600 yojanas and a Moon distance of 34,380 yojanas. Compute the length of the Earth’s shadow and its diameter at the Moon. How many Moon diameters wide is the shadow there?

Show solution
\(L=1050\times459600/(4410-1050)=1050\times459600/3360\approx143{,}600\) yojanas. At the Moon: \(1050\times(143600-34380)/143600\approx1050\times0.7606\approx798.6\) yojanas. Ratio \(798.6/315\approx2.54\). The modern ratio of the umbra to the Moon’s diameter at mean distance is about 2.6–2.7, so the geometry is sound.

Exercise 9.2With the shadow diameter 798.6 and Moon diameter 315 (both in yojanas at the Moon’s distance), and taking the Moon’s motion relative to the shadow as about \(12.2^\circ\) per day on an orbit of radius 34,380 yojanas, estimate the half-duration of a central lunar eclipse. Compare with the observed maximum of a little under two hours.

Show solution
Distance to travel from first contact to centre: \((798.6+315)/2\approx556.8\) yojanas. Relative speed: \(34380\times12.2\times\pi/180\approx7320\) yojanas per day. Half-duration \(\approx556.8/7320\approx0.076\) days \(\approx1.8\) hours. The total duration of about 3.6 hours from first to last contact matches the longest real lunar eclipses (about 3.8 hours for the partial phases).

Exercise 9.3Āryabhaṭa’s Sun and Moon diameters are 4410 and 315 yojanas. For a total solar eclipse the Moon must appear at least as large as the Sun. Show that this requires the ratio of the Sun’s distance to the Moon’s to be at least 14, and compare with the ratio of the distances in Exercise 9.1.

Show solution
Apparent sizes are proportional to diameter over distance: the Moon covers the Sun if \(315/m\ge4410/d\), i.e. \(d/m\ge4410/315=14\). With the illustrative distances of Exercise 9.1, \(459600/34380\approx13.4\), slightly less than 14: the Moon would appear slightly smaller than the Sun, and a central solar eclipse would be annular. In reality both annular and total eclipses occur, because the distances vary.
Lesson 10

Gnomons, lamps and latitude

For the general reader. The simplest astronomical instrument is a vertical stick, the gnomon, and the length of its shadow tells the time and the latitude. Āryabhaṭa gives a group of rules about shadows cast by a lamp and by the Sun. They rest on two ideas, similar triangles and the theorem of the right triangle, and he states both in general form.

Goals
  • State the right-triangle rule and the chord rule of Gaṇita 17.
  • Derive the lamp-and-gnomon rules of Gaṇita 15–16.
  • Find latitude from the noon equinoctial shadow.
Rules (Gaṇita 17)

(i) In a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides. (ii) If a chord of a circle crosses a diameter at right angles, the square of the half-chord equals the product of the two segments of the diameter: \(c^2=a\,b\).

Rule (ii) follows from (i): if the circle has radius \(r\) and the chord meets the diameter at distance \(x\) from the centre, the segments are \(r+x\) and \(r-x\), and \(c^2=r^2-x^2=(r+x)(r-x)\). It is the tool for the versine, the “reversed sine” \(r-r\cos\theta\) that Āryabhaṭa also used.

The lamp and the gnomon (Gaṇita 15–16)

A gnomon of height \(g\) stands at horizontal distance \(d\) from the foot of a lamp of height \(H>g\). Its shadow has length

\[ s=\frac{g\,d}{H-g}. \]

This is Gaṇita 15. The next rule solves the inverse problem: from the lengths of two shadows of the same gnomon at two positions along a line from the lamp, find the distance and height of the lamp. If the shadows are \(s_1,s_2\) and the tips of the shadows are a distance \(t\) apart, then the distance from the tip of the first shadow to the foot of the lamp is \(s_1t/(s_2-s_1)\), and the lamp’s height is that distance times \(g/s_1\).

Derivation. The lamp’s top, the gnomon’s top and the shadow’s tip are collinear, so by similar triangles \(H/(d+s)=g/s\), which gives the first rule. For the second, let \(D_i\) be the distance from the lamp’s foot to the tip of shadow \(i\). Then \(H/D_i=g/s_i\), so \(D_1/s_1=D_2/s_2\). With \(D_2=D_1+t\), \(D_1(s_2-s_1)=s_1t\).

Latitude

At noon on an equinox the Sun stands on the celestial equator, and its altitude is \(90^\circ-\varphi\), where \(\varphi\) is the latitude. A gnomon of height \(g\) then casts a shadow \(s\) with \(s/g=\tan\varphi\). In the Indian tradition the gnomon was 12 units (aṅgulas) tall, and the “equinoctial shadow” of a place was a standard datum. The Golapāda uses the corresponding sine form: \(R\sin\varphi:R\cos\varphi=s:g\).

Exercise 10.1A lamp 9 units high stands 12 units from a gnomon of height 3. Find the shadow. Then the gnomon is moved 4 units farther away; find the new shadow and check that the two-shadow rule recovers the lamp.

Show solution
\(s_1=3\cdot12/(9-3)=6\). At distance 16: \(s_2=3\cdot16/6=8\). The shadow tips are at 18 and 24 from the lamp’s foot, so \(t=6\). Rule: \(D_1=s_1t/(s_2-s_1)=6\cdot6/2=18\), correct; height \(=D_1\cdot g/s_1=18\cdot3/6=9\), correct.

Exercise 10.2At Pāṭaliputra (latitude about \(25.6^\circ\)), what is the equinoctial noon shadow of a 12-aṅgula gnomon? If a town’s equinoctial shadow is 5 aṅgulas, what is its latitude?

Show solution
\(s=12\tan25.6^\circ\approx12\times0.479\approx5.75\) aṅgulas. Then \(\tan\varphi=5/12\), so \(\varphi\approx22.6^\circ\). The hypotenuse of the shadow triangle is then exactly 13, a 5–12–13 right triangle, which made the value convenient.

Exercise 10.3Use the chord rule to find the versine \(v\) of an arc whose half-chord is \(c\) in a circle of radius \(r\), and check it with Āryabhaṭa’s \(r=3438\), \(c=1719\).

Show solution
The segments of the diameter are \(v\) and \(2r-v\), so \(c^2=v(2r-v)\), giving \(v=r-\sqrt{r^2-c^2}\) (the smaller root). With \(c=1719=r/2\): \(v=3438-3438\cdot\tfrac{\sqrt3}{2}\approx3438-2977.4=460.6\), the versine of \(30^\circ\). Indeed \(3438(1-\cos30^\circ)\approx460.6\), and \(1719\cdot(6876-460.6)\approx1719^2\).
Part III · Life and legacy
Lesson 11

Pupils, critics and commentators

Nothing is known of Āryabhaṭa’s later life or death. The date 550 often given for his death is a convention without a source. His influence, however, can be traced in detail.

Pupils

Bhāskara I names Āryabhaṭa’s pupils, including Pāṇḍuraṅgasvāmin, Lāṭadeva, Prabhākara and Niḥśaṅku. Lāṭadeva is usually identified with the Lāṭācārya whom Varāhamihira cites on the older systems. Bhāskara I himself, a century later, called Āryabhaṭa the master who, having reached the far shore of the ocean of mathematics, astronomy and spherics, handed down their jewels. His commentary of 629 is our best guide to how the text was taught, and his own Mahābhāskarīya gave a celebrated rational approximation to the sine.

Brahmagupta

Brahmagupta, writing at Bhillamāla in 628, devoted much of his Brāhmasphuṭasiddhānta to attacking Āryabhaṭa: his equal yugas, his rotating Earth, his eclipse theory in places, and his parameters. Yet in 665 Brahmagupta based his practical handbook, the Khaṇḍakhādyaka, on Āryabhaṭa’s midnight system, which suggests that the attacks were partly about school allegiance. The Āryapakṣa and Brāhmapakṣa remained rival traditions.

The South

The Āryabhaṭīya was taught and copied most intensively in South India. Lalla in the 8th century followed it closely. In Kerala, the school founded by Mādhava of Saṅgamagrāma in the 14th century worked within the Āryabhaṭīya tradition, and Nīlakaṇṭha Somayājī’s commentary around 1500 is one of its major works. The Kerala school went on to find infinite series for the sine, cosine and arctangent, growing directly from the sine-difference ideas of Lesson 06.

Exercise 11.1Bhāskara I’s rational approximation is \(\sin x\approx\dfrac{16x(\pi-x)}{5\pi^2-4x(\pi-x)}\) for \(0\le x\le\pi\). Check it at \(x=\pi/6\) and \(x=\pi/2\).

Show solution
At \(x=\pi/2\): \(x(\pi-x)=\pi^2/4\), so the value is \(4\pi^2/(5\pi^2-\pi^2)=1\), exact. At \(x=\pi/6\): \(x(\pi-x)=5\pi^2/36\), giving \(\dfrac{80\pi^2/36}{5\pi^2-20\pi^2/36}=\dfrac{80/36}{160/36}=\tfrac12\), exact. Its maximum error over the range is under 0.002.

DiscussionBrahmagupta attacked Āryabhaṭa, then built a handbook on his methods. What does this suggest about how scientific disagreement worked in a culture of rival schools and memorized texts? Compare with a modern priority dispute.

Lesson 12

From jyā to sine: the long legacy

Baghdad

In the 8th century Indian astronomical works reached the Abbasid court. Āryabhaṭa’s parameters were known in Arabic as the system of the Zīj al-Arjabhar, and Muḥammad ibn Mūsā al-Khwārizmī’s astronomical tables drew on Indian methods. Al-Bīrūnī, around 1030, discussed Āryabhaṭa in his book on India, including the question of the Earth’s rotation.

The word “sine”

Sanskrit jyā was transliterated into Arabic as jība, a word without other meaning. Written without vowels as jyb, it was read as the ordinary word jayb, “pocket” or “fold”. In the 12th century Gerard of Cremona and others translated jayb into Latin as sinus, which also means a fold or bay. The cosine and versine have parallel Indian ancestors (koṭijyā, utkramajyā).

India today

Traditional Indian calendars (pañcāṅgas) still use rules descended from the Āryabhaṭa and Brahmagupta schools. India’s first satellite, launched on 19 April 1975, was named Aryabhata; Aryabhatta Knowledge University in Patna and the Aryabhatta Research Institute of Observational Sciences at Nainital also carry his name, as does a lunar crater.

Assessment

  • Mathematics: a systematic decimal arithmetic, the first surviving sine table and a second-difference rule for it, the general solution of linear indeterminate equations, and sum formulas for powers.
  • Astronomy: highly accurate mean motions, a rotating Earth, and a physical theory of eclipses with quantitative rules.
  • Limits: wrong volume formulas, no proofs in the text, and a geocentric model.

Exercise 12.1Trace the history of the word “sine” as given in this lesson, and explain how a transliteration error can turn a technical term into a metaphor.

Show solution
Jyā (bow-string, half-chord) → Arabic jība (a loan-word) → written jyb without vowels → read as jayb (“pocket, fold”) → Latin sinus (“fold, bay”) → “sine”. The Arabic consonantal script allowed two readings; a Latin translator who knew only the common word translated its meaning, so a bow-string became a fold. The original geometric image of the half-chord was lost.

Final projectChoose one verse-rule of the Āryabhaṭīya (π, the sine differences, the pulverizer or the eclipse shadow). Write two pages for a strong undergraduate that give the rule in paraphrase, show how Bhāskara I or a modern scholar interprets it, reconstruct the mathematics in modern notation, and say where the interpretation is uncertain.

Reference

Connections

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

Students

  • Latadeva
  • Nihsanku
  • Panduranga-svamin
  • Prabhakara

Influenced by

Influenced

Colleagues and collaborators

  • Varahamihira

Rivals

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Texts and translations

  1. W. E. Clark, The Āryabhaṭīya of Āryabhaṭa: An Ancient Indian Work on Mathematics and Astronomy, University of Chicago Press, 1930.
  2. K. S. Shukla and K. V. Sarma, Āryabhaṭīya of Āryabhaṭa, critical edition with translation and notes, Indian National Science Academy, New Delhi, 1976.
  3. K. S. Shukla, Āryabhaṭīya of Āryabhaṭa with the Commentary of Bhāskara I and Someśvara, INSA, 1976.
  4. H. Kern, The Aryabhatiya, with the commentary Bhatadipika of Paramadiçvara, Leiden, 1874.

Biography and history

  1. Aryabhata I, MacTutor History of Mathematics.
  2. Aryabhata and Aryabhatiya, Wikipedia.
  3. K. Plofker, Āryabhaṭa I, Biographical Encyclopedia of Astronomers.
  4. K. Plofker, Mathematics in India, Princeton University Press, 2009, ch. 4.
  5. D. Pingree, Āryabhaṭa I, Dictionary of Scientific Biography, vol. 1, 1970.

Mathematics and astronomy

  1. Āryabhaṭa’s sine table, Wikipedia (with the 24 differences and a comparison with modern values).
  2. J. Q. Jacobs, The Āryabhaṭīya of Āryabhaṭa: the oldest precise astronomical constant? (revolution numbers and derived periods).
  3. G. G. Joseph, The Crest of the Peacock: Non-European Roots of Mathematics, 3rd ed., Princeton, 2011.
  4. B. L. van der Waerden, The heliocentric system in Greek, Persian and Hindu astronomy, Annals of the New York Academy of Sciences 500 (1987).
  5. G. Ifrah, The Universal History of Numbers, Wiley, 2000.