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Indian and Vedic Mathematics Explained: 16 Sutras to the Kerala School

The 16 Vedic sutras are a 1965 system of mental shortcuts. Real Indian mathematics gave the world zero, sine tables and infinite series. Both, worked.

Safeguard Security Team
55 min read

"Vedic Mathematics" is a set of 16 sutras and 13 sub-sutras for fast mental arithmetic, published in 1965 by Bharati Krishna Tirtha. Historians find no trace of them in the Vedas. Classical Indian mathematics is a separate and older history: altar geometry, place value with zero, sine tables, Pell equations and infinite series for π.

This post works both, step by step, on the same fixed problem set our mathematics traditions series uses for Chinese and Japanese methods, so you can compare like with like.

Key takeaways

  • The 16 sutras are short Sanskrit phrases attached to calculation tricks. They come from one book, Vedic Mathematics by Bharati Krishna Tirtha, published in 1965 after his death in 1960. Mathematicians and historians, including S. G. Dani and Kim Plofker, note that the sutras are not found in the Vedic texts.
  • The tricks still work, and some are very good. Nikhilam multiplies numbers near 100 or 1000 in two small steps. Ekādhikena Pūrveṇa squares any number ending in 5 with one multiplication.
  • The shortcuts are special purpose. Nikhilam on 386 × 524 creates a harder problem than the original; Ūrdhva-Tiryagbhyām (vertically and crosswise) is the general method that scales.
  • Real classical Indian mathematics is the bigger story: the Śulba Sūtras' √2 ≈ 577/408, Piṅgala's binary patterns of metre, the decimal place value system with zero, Āryabhaṭa's π of 3.1416 and his sine table, Brahmagupta's rules for zero and negative numbers, Bhāskara II's chakravala solution of 61y² + 1 = x², and Mādhava's infinite series for π more than two hundred years before they appeared in Europe.

The problem set used in this series

Every post in the series works these problems; here are the answers.

ProblemAnswerBest Indian method in this post
98 × 979,506Nikhilam, base 100
47 × 532,491Nikhilam with working base 50, or Ūrdhva-Tiryagbhyām
386 × 524202,264Ūrdhva-Tiryagbhyām, 3 by 3 digits
85²7,225Ekādhikena Pūrveṇa
√7,74488Duplex square root, or observation (Vilokanam)
7,429 ÷ 37200 remainder 29Flag (dhvajāṅka) division
1/190.052631578947368421 repeating (18 digits)Ekādhikena Pūrveṇa
x + y + z = 6, 2y + 5z = −4, 2x + 5y − z = 27x = 5, y = 3, z = −2Elimination with cross multiplication

Part 1: Vedic Mathematics, the 1965 system

What is Vedic Mathematics?

Vedic Mathematics (वैदिक गणित, Vaidika Gaṇita) is a book by Jagadguru Swami Bharati Krishna Tirtha, the Shankaracharya of Govardhan Math in Puri from 1925 until his death in 1960. The book was published posthumously in 1965 by Motilal Banarsidass. It presents 16 sutras (सूत्र, short aphorisms) and 13 sub-sutras, each two to four Sanskrit words, attached to methods for multiplication, division, squares, roots, recurring decimals and simple equations.

The methods are clever and mostly correct. They work because they exploit place value and algebraic identities, for example (100 − a)(100 − b) = 100(100 − a − b) + ab. What the book does not do is show where in the Vedas the sutras appear; we cover that after the worked examples.

The 16 sutras and 13 sub-sutras

Names as in Tirtha's book; spellings vary between editions.

#SutraUsual translation
1Ekādhikena Pūrveṇa (एकाधिकेन पूर्वेण)By one more than the previous one
2Nikhilam Navataścaramaṁ Daśataḥ (निखिलं नवतश्चरमं दशतः)All from 9 and the last from 10
3Ūrdhva-Tiryagbhyām (ऊर्ध्वतिर्यग्भ्याम्)Vertically and crosswise
4Parāvartya YojayetTranspose and apply
5Śūnyaṁ SāmyasamuccayeWhen the sum is the same, that sum is zero
6(Ānurūpye) ŚūnyamanyatIf one is in ratio, the other is zero
7Saṅkalana-vyavakalanābhyāmBy addition and by subtraction
8PūraṇāpūraṇābhyāmBy completion or non-completion
9Calana-KalanābhyāmDifferences and similarities
10YāvadūnamWhatever the extent of its deficiency
11VyaṣṭisamaṣṭiḥPart and whole
12Śeṣāṇyaṅkena CaramenaThe remainders by the last digit
13SopāntyadvayamantyamThe ultimate and twice the penultimate
14Ekanyūnena PūrveṇaBy one less than the previous one
15GuṇitasamuccayaḥThe product of the sum equals the sum of the product
16GuṇakasamuccayaḥThe factors of the sum equal the sum of the factors

The 13 sub-sutras (upa-sūtra) are:

#Sub-sutraUsual translation
1ĀnurūpyeṇaProportionately
2Śiṣyate ŚeṣasaṁjñaḥThe remainder remains constant
3ĀdyamādyenantyamantyenaThe first by the first and the last by the last
4Kevalaiḥ Saptakaṁ GuṇyātFor 7 the multiplicand is 143
5VeṣṭanamBy osculation
6Yāvadūnaṁ TāvadūnamLessen by the deficiency
7Yāvadūnaṁ Tāvadūnīkṛtya Vargaṁ Ca YojayetLessen by the deficiency and set up the square of the deficiency
8Antyayordaśake'piLast digits summing to 10
9AntyayorevaOnly the last terms
10SamuccayaguṇitaḥThe sum of the coefficients in the product
11LopanasthāpanābhyāmBy alternate elimination and retention
12VilokanamBy mere observation
13Guṇitasamuccayaḥ SamuccayaguṇitaḥThe product of the sums of coefficients in the factors equals the sum of coefficients in the product

Each worked example shows the paper layout, then the algebra behind it.

Nikhilam: "all from 9 and the last from 10"

Nikhilam is the most famous sutra. Its name describes how to find the complement of a number from a power of 10: take every digit from 9 and the last digit from 10. The complement of 386 from 1000 is 9−3, 9−8, 10−6, which is 614. Once you have complements (the book calls them deviations), multiplication near a base becomes a subtraction and one small product.

98 × 97 with base 100

StepLeft partRight partWhat you do
198−298 is 2 below 100
297−397 is 3 below 100
398 − 3 = 95Cross subtract: either number minus the other's deviation (97 − 2 gives 95 too)
4(−2) × (−3) = 6, written 06Multiply the deviations; base 100 needs two digits
59506Read off 9,506
   98   -2
 x 97   -3
 ---------
   95 | 06      ->  9506

Why it works: (100 − 2)(100 − 3) = 100 × (100 − 2 − 3) + 2 × 3 = 100 × 95 + 6. The left part counts hundreds; the right part needs as many digits as the base has zeros, so 6 becomes 06.

998 × 997 with base 1000

StepValueNote
Deviations−2 and −3from 1000
Left part998 − 3 = 995thousands
Right part2 × 3 = 6, written 006base 1000 needs three digits
Answer995,006

The work did not grow when the numbers grew: Nikhilam's effort depends on the deviations, not the size of the numbers.

47 × 53 with a working base of 50

The base does not have to be a power of 10. The sub-sutra Ānurūpyeṇa (proportionately) lets you work from 50, which is 100 ÷ 2.

StepLeft partRight partWhat you do
147−33 below 50
253+33 above 50
347 + 3 = 50Cross: 47 minus (−3), or 53 plus (−3)
450 ÷ 2 = 25Working base is 100 ÷ 2, so halve the left part; it now counts hundreds
5(−3) × (+3) = −9Multiply the deviations
625 hundreds − 92,500 − 9
72,491Answer
 base 50 = 100 / 2
   47   -3
 x 53   +3
 ---------
   50 | -9
   50 / 2 = 25  ->  2500 - 9  =  2491

The identity (50 − 3)(50 + 3) = 50² − 3² gives the same 2,500 − 9.

Where Nikhilam breaks: 386 × 524

Far from any base, Nikhilam gets worse, not better.

StepValue
Deviations from 1000386 is −614; 524 is −476
Left part386 − 476 = −90, so −90 thousands = −90,000
Right part614 × 476 = 292,264
Combine−90,000 + 292,264 = 202,264

The answer is right, but you had to multiply 614 × 476, which is harder than the original 386 × 524. A practitioner would switch to Ūrdhva-Tiryagbhyām, the general method in the next section. The shortcut is only a shortcut when the numbers suit it.

Ūrdhva-Tiryagbhyām: vertically and crosswise

Ūrdhva-Tiryagbhyām multiplies any two numbers column by column, right to left. Each column adds every digit product whose places sum to that column; write one digit, carry the rest.

47 × 53, two digits by two digits

Digits: 4 and 7 on top, 5 and 3 below.

ColumnProductsSumPlus carryWriteCarry
Units (vertical)7 × 3212112
Tens (crosswise)4 × 3 + 7 × 5 = 12 + 35474994
Hundreds (vertical)4 × 5202424

Reading the written digits from left: 24, 9, 1, which is 2,491.

   4   7
   | X |        units: 7x3          = 21  -> write 1, carry 2
   5   3        tens:  4x3 + 7x5    = 47  -> 47+2 = 49, write 9, carry 4
                hundreds: 4x5       = 20  -> 20+4 = 24
   answer 2491

For 98 × 97 the columns are 8 × 7 = 56, then 9 × 7 + 8 × 9 = 63 + 72 = 135, then 9 × 9 = 81. Write 6 carry 5; 135 + 5 = 140, write 0 carry 14; 81 + 14 = 95. The answer reads 95, 0, 6, which is 9,506. It works, but Nikhilam needed one small product where this needed four.

386 × 524, three digits by three digits

Top digits a = 3, 8, 6. Bottom digits b = 5, 2, 4. There are five columns.

ColumnPatternProductsSumPlus carryWriteCarry
1 (units)right × right6 × 4242442
2 (tens)cross of the right pair8 × 4 + 6 × 2 = 32 + 12444664
3 (hundreds)outer cross plus middle3 × 4 + 8 × 2 + 6 × 5 = 12 + 16 + 30586226
4 (thousands)cross of the left pair3 × 2 + 8 × 5 = 6 + 40465225
5 (ten thousands)left × left3 × 5152020

Reading from the top of the last column down: 20, 2, 2, 6, 4, which is 202,264.

     3   8   6
     5   2   4
 col 5: 3x5                    = 15
 col 4: 3x2 + 8x5              = 46
 col 3: 3x4 + 8x2 + 6x5        = 58
 col 2: 8x4 + 6x2              = 44
 col 1: 6x4                    = 24
 carry right to left: 24 -> 4 c2 | 46 -> 6 c4 | 62 -> 2 c6 | 52 -> 2 c5 | 20
 answer 202264

Why it works: write 386 as 3x² + 8x + 6 and 524 as 5x² + 2x + 4 with x = 10. Multiplying the two polynomials gives 15x⁴ + 46x³ + 58x² + 44x + 24. The column sums are exactly those coefficients, and the carries turn them back into base 10. Long multiplication computes the same nine products but stores them as three partial rows instead of five column sums.

The practical difference is memory: done mentally, Ūrdhva-Tiryagbhyām holds one column sum and one carry at a time.

Ekādhikena Pūrveṇa: "by one more than the previous one"

85² in one multiplication

For a number ending in 5, take the digits before the 5, multiply them by one more than themselves, and write 25 after the result.

StepValue
Digits before the 58
"One more than the previous"9
8 × 972
Append 257,225

Why it works: (10a + 5)² = 100a² + 100a + 25 = 100 × a(a + 1) + 25. With a = 8 that is 100 × 72 + 25. The trick also works for 115² (11 × 12 = 132, so 13,225) and for any number ending in 5.

1/19 as a decimal, two ways

The denominator 19 ends in 9. The digit before the 9 is 1, and "one more than the previous" makes it 2. The book gives two procedures that both use this 2.

Method A, multiply from the right. Start with the last digit, 1, and keep doubling, carrying as you go. Write each new digit to the left of the previous one.

StepCalculationWriteCarryDigits so far (right to left)
1start11
21 × 2221
32 × 24421
44 × 288421
58 × 2 = 166168421
66 × 2 + 1 = 1331368421
73 × 2 + 1 = 777368421
87 × 2 = 144147368421
94 × 2 + 1 = 99947368421
109 × 2 = 18818947368421
118 × 2 + 1 = 177178947368421
127 × 2 + 1 = 1551578947368421
135 × 2 + 1 = 11111578947368421
141 × 2 + 1 = 3331578947368421
153 × 26631578947368421
166 × 2 = 12212631578947368421
172 × 2 + 1 = 5552631578947368421
185 × 2 = 1001052631578947368421

The next step would be 0 × 2 + 1 = 1, which is where we started, so the 18 digits repeat: 1/19 = 0.052631578947368421 052631578947368421 and so on.

Method B, divide from the left. Divide by 2, and prefix each remainder to the next digit you produce.

 1 / 2 = 0 r1   -> digit 0, next dividend 10
10 / 2 = 5 r0   -> digit 5, next dividend 5
 5 / 2 = 2 r1   -> digit 2, next dividend 12
12 / 2 = 6 r0   -> digit 6, next dividend 6
 6 / 2 = 3 r0   -> digit 3, next dividend 3
 3 / 2 = 1 r1   -> digit 1, next dividend 11
11 / 2 = 5 r1   -> digit 5, next dividend 15
15 / 2 = 7 r1   -> digit 7 ...
 digits: 0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1 (then repeat)

Why it works: 1/19 = 0.05 / 0.95, and repeated halving with the remainder carried in front is the same as dividing by 20 while feeding back the digits already produced. A built-in check: the second half of the period is the nines complement of the first half. 052631578 plus 947368421 gives 999999999.

Long division by 19 also needs 18 steps, but each needs a trial multiple of 19. Same length, easier steps: a fair, measurable win.

Yāvadūnam: squares near a base

Yāvadūnam ("whatever the deficiency") squares a number near a base. Take the deviation away from the number once more, then append the square of the deviation.

NumberBaseDeviationLeft partRight partAnswer
98100−298 − 2 = 962² = 049,604
103100+3103 + 3 = 1063² = 0910,609
9961000−4996 − 4 = 9924² = 016992,016
85100−1585 − 15 = 7015² = 225, carry 270 + 2 = 72, then 25: 7,225

The last row carries: base 100 allows two right digits, so the 2 in 225 moves left. It matches Ekādhikena Pūrveṇa's 7,225 with more work, and the agreement is a useful check.

Division: Parāvartya Yojayet and the flag method

Parāvartya Yojayet for divisors just above a base

"Transpose and apply" divides by a number like 12 or 112. You replace each digit after the leading 1 with its negative, then multiply and add down the columns. The last columns (as many as the divisor has digits after the 1) form the remainder.

1,458 ÷ 12. The divisor's digits after the 1 are "2", transposed to "−2". One remainder column.

ColumnDividend digitAddedTotalRole
111quotient digit
24−2 × 1 = −22quotient digit
35−2 × 2 = −41quotient digit
48−2 × 1 = −26remainder

Quotient 121, remainder 6. Check: 12 × 121 + 6 = 1,458.

Why it works: dividing by 12 is dividing by 10 + 2. Each quotient digit you find means "subtract that many 2s from the column to its right", and the transposed −2 does exactly that subtraction in place. The same layout handles 112 with transposed digits −1 and −2: 13,456 ÷ 112 gives quotient 120, remainder 16.

7,429 ÷ 37 with the flag (dhvajāṅka) method

Parāvartya suits divisors like 12 or 112, not 37. For a general divisor the book uses straight division, which it ties to Ūrdhva-Tiryagbhyām and which teachers call the flag method. You divide only by the first digit of the divisor (3) and treat the rest (7) as a "flag" that you subtract along the way. Because the flag has one digit, the last one digit of the dividend belongs to the remainder.

 divisor 37  ->  operator 3, flag 7
 dividend  7 4 2 | 9
StepGross dividendMinus flag × previous quotient digitNet÷ 3Quotient digitRemainder carried
17none72 r121
214 (carry 1 with digit 4)7 × 2 = 1400 r000
302 (carry 0 with digit 2)7 × 0 = 020 r202
Remainder29 (carry 2 with digit 9)7 × 0 = 029

Quotient 200, remainder 29. Check: 37 × 200 + 29 = 7,429.

On this problem plain long division is just as short: 74 ÷ 37 = 2 exactly, then two zeros, then 29 is left over. The flag method's real advantage shows on long divisors, where it lets you divide by one digit instead of trial-multiplying a multi-digit divisor.

Square roots: observation and the duplex method on √7,744

Vilokanam: the observation route for perfect squares

ClueReasoningResult
Last digit is 4Only numbers ending in 2 or 8 have squares ending in 4Root ends in 2 or 8
First pair is 778² = 64 and 9² = 81, so 77 lies between themRoot is in the 80s
Compare with 85² = 7,2257,744 is largerRoot is above 85, so 88
Check88² = 7,744√7,744 = 88

It only works for known perfect squares.

The duplex method

The duplex method works for any number, one digit at a time. The duplex D of a string of digits is: for one digit a, a²; for two digits a b, 2ab; for three digits a b c, 2ac + b²; and so on, pairing from the outside in.

Pair the digits from the right: 77 | 44.

StepWhat you doWorkingResult
1Largest square not above 778² = 64, remainder 13First root digit 8
2Divisor is twice the first root digit2 × 816
3Gross dividend: remainder 13 joined to next digit 4134; no duplex to subtract yet134 ÷ 16 = 8 r6, second root digit 8
4Gross dividend: remainder 6 joined to next digit 464; subtract D(8) = 640 ÷ 16 = 0, next digit 0
5Nothing left, and D(8, 0) = 00Root is exactly 88.0
        7 7 | 4 4
 root   8   | 8 . 0
 div 16     13 -> 134 - 0      = 134 ; 134/16 = 8 r 6
            6  -> 64  - D(8)   = 0   ; 0/16   = 0 r 0
 sqrt(7744) = 88

The duplex method is the Ūrdhva-Tiryagbhyām pattern run backwards: squaring 88 vertically and crosswise gives columns 64, 128, 64, and the duplex steps peel those columns off one at a time.

Digit sums and casting out nines

The digit sum (remainder after division by 9) is a check, not a method. Add digits until one is left; 9 counts as 0. Since 10 leaves remainder 1 on division by 9, a number and its digit sum share the same remainder mod 9.

CheckLeft sideRight sideMatch
386 × 524 = 202,264386 gives 17, then 8. 524 gives 11, then 2. 8 × 2 = 16, then 72 + 0 + 2 + 2 + 6 + 4 = 16, then 7Yes
7,429 = 37 × 200 + 297 + 4 + 2 + 9 = 22, then 437 gives 1; 200 gives 2; 29 gives 2. 1 × 2 + 2 = 4Yes
98 × 97 = 9,50698 gives 8; 97 gives 7. 8 × 7 = 56, then 29 + 5 + 0 + 6 = 20, then 2Yes

The limit matters. Casting out nines catches any error that changes the remainder mod 9, which is most slips. It misses errors that are multiples of 9, including swapped digits: 202,246 also reduces to 7, yet it is wrong. If errors were random, about one in nine would slip through.

Equations: the 3 by 3 system with cross multiplication

The book treats simultaneous equations with Parāvartya Yojayet, which for two unknowns is the cross multiplication formula, and Saṅkalana-vyavakalanābhyām (by addition and subtraction) for elimination. The fixed system is:

 x +  y +  z =  6     (1)
      2y + 5z = -4    (2)
 2x + 5y -  z = 27    (3)
StepOperationResult
1(3) − 2 × (1)3y − 3z = 15, so y − z = 5 (4)
2Pair (4) and (2): 1y − 1z = 5 and 2y + 5z = −4Two equations in y, z
3Denominator by cross multiplication: 1 × 5 − 2 × (−1)7
4y = (5 × 5 − (−4) × (−1)) ÷ 7 = (25 − 4) ÷ 7y = 3
5z = (1 × (−4) − 2 × 5) ÷ 7 = −14 ÷ 7z = −2
6From (1): x = 6 − 3 − (−2)x = 5
7Check in (3): 2 × 5 + 5 × 3 − (−2) = 10 + 15 + 227

Cross multiplication for two unknowns is the 2 by 2 case of what Europe later called Cramer's rule. The 3 by 3 determinant of this system is −21, and x = −105 ÷ −21, y = −63 ÷ −21 and z = 42 ÷ −21 give the same answer.

Where do the sutras come from?

This is the part of the story that needs care, because many people learn these methods as ancient.

Tirtha's preface says that the sutras are contained in a Pariśiṣṭa, an appendix, of the Atharva Veda. S. G. Dani, a mathematician at the Tata Institute of Fundamental Research, examined that claim in a two-part article in Frontline in 1993, later updated as "Myths and reality: On 'Vedic mathematics'". He points out that the passing reference to the Atharva Veda is the only source the book ever gives, and that the known Pariśiṣṭas of the Atharva Veda do not contain the sutras. Dani also reports that the historian of mathematics K. S. Shukla met Tirtha in Lucknow in 1950, carrying the standard edition of the Atharva Veda Pariśiṣṭas, and asked him to point out the sutras. According to Shukla's account, Tirtha said they were not in those texts and occurred in "his own" Pariśiṣṭa.

Kim Plofker, author of Mathematics in India (Princeton University Press, 2009), the standard modern history of the subject, is among the historians who note that the sutras are not found in Vedic texts, and her history of Vedic period mathematics is built from the Śulba Sūtras and related texts instead. Other scholars note that the Sanskrit of the sutras reads as modern, not Vedic, and that the book's content has little in common with the mathematics that survives from the Vedic period or the classical period that followed.

None of this makes the methods wrong, or criticises the teachers and students who enjoy them. The fair summary: they are a 20th century compilation by a learned author, with real teaching value, but not evidence of what Vedic India knew. Calling them ancient also hides the documented achievements of Indian mathematicians, which are far more significant. That is the second half of this post.

Is Vedic Mathematics "more powerful"? Same problems, measured

"More powerful" can mean faster, more general, less memory, or better at catching mistakes. Here is each problem measured three ways: the number of single-digit multiplications (or equivalent steps), what you must hold in your head at the worst point, and whether the method has a natural check. Counts are for a person working mentally or with minimal writing.

ProblemMethodCore stepsHeld in memory at worstVerdict
98 × 97Long multiplication4 digit products, 2 rows, 1 additionTwo partial rows (686 and 8,820)General, slower here
98 × 97Nikhilam, base 1001 subtraction, 1 tiny product (2 × 3)95 and 06Clear win
47 × 53Long multiplication4 digit products, 2 rows (141 and 2,350)Two rowsGeneral
47 × 53Nikhilam, base 501 addition, 1 halving, 1 product, 1 subtraction25 and −9Win if you spot the base
386 × 524Long multiplication9 digit products, 3 rows, 1 additionThree rows (1,544, 7,720, 193,000)General
386 × 524Ūrdhva-Tiryagbhyām9 digit products, 5 column sumsOne column sum and one carrySame work, less memory
386 × 524Nikhilam, base 1000Requires 614 × 476LargeLoses badly
85²Long multiplication4 digit products, 2 rowsTwo rows (425 and 6,800)General
85²Ekādhikena Pūrveṇa1 product (8 × 9), append 2572Clear win
√7,744Long square root2 root digits with trial divisorsRemainder and trial divisorGeneral
√7,744Observation3 comparisonsNothingWin, perfect squares only
7,429 ÷ 37Long division3 quotient digitsRemainderEqual here
7,429 ÷ 37Flag method3 steps of ÷3 with flag subtractionRemainderEqual here; better on long divisors
1/19Long division18 steps with multiples of 19Remainder below 19General
1/19Ekādhikena Pūrveṇa18 doublings or halvingsOne digit and carryEasier steps

So the honest answer is that some sutras are faster for special numbers (near a base, ending in 5, denominators ending in 9), one is a genuinely good general method (vertically and crosswise), and the rest are standard algebra in memorable clothing. They are not more general than the school algorithms, which handle every case with one procedure. We found no large, independent, peer reviewed trials that would support claims about intelligence, brain development or exam results from Vedic maths, so we make no such claims.

Part 2: classical Indian mathematics, worked

Dates below follow MacTutor and the scholarly sources at the end; early dates are approximate.

PeriodWho or whatMain contribution
About 800 to 200 BCE (approximate)Śulba Sūtras of Baudhāyana, Āpastamba, Kātyāyana, MānavaAltar geometry, the diagonal rule, √2 ≈ 577/408
About 3rd to 2nd century BCE (approximate)Piṅgala, ChandaḥśāstraEnumerating metres with two syllable types: binary patterns
Disputed: between the 3rd and 12th centuries CEBakhshali manuscriptDecimal place value with a dot for zero; arithmetic and algebra
499 CEĀryabhaṭa, Āryabhaṭīyaπ ≈ 3.1416, a sine table, place value computation
628 CEBrahmagupta, BrāhmasphuṭasiddhāntaRules for zero and negative numbers; his composition identity
876 CEGwalior inscriptionEarliest dated, universally accepted zero written as a numeral
About 1150 CEBhāskara II, Līlāvatī and BījagaṇitaChakravala method for Pell equations
About 1400 CEMādhava of Saṅgamagrāma, KeralaInfinite series for sine, cosine, arctan and π
1501 and the 16th centuryNīlakaṇṭha, Tantrasaṅgraha; Jyeṣṭhadeva, YuktibhāṣāRecords and proofs of the Kerala series

The Śulba Sūtras: geometry with cords

The Śulba Sūtras (शुल्बसूत्र, "rules of the cord") are manuals for building fire altars of exact shapes and areas, using stakes and measuring cords. MacTutor dates Baudhāyana, author of one of the earliest, to around 800 BCE; that date is uncertain, and other scholars place the texts across the middle of the first millennium BCE.

The diagonal rule. Baudhāyana states that the cord stretched across the diagonal of a rectangle produces both areas that the length and breadth produce separately: the square on the diagonal equals the sum of the squares on the sides. The texts use many whole number examples that MacTutor lists, including (5, 12, 13), (8, 15, 17), (12, 16, 20), (12, 35, 37) and (15, 36, 39). For the last: 15² + 36² = 225 + 1,296 = 1,521 = 39².

Doubling a square. An altar sometimes had to be twice the area of another. The cord rule gives the answer directly: the square built on the diagonal of a square has twice its area, so the side you need is the diagonal, √2 times the original side.

The value of √2. The Āpastamba and Kātyāyana Śulba Sūtras give a rule MacTutor translates as: increase a unit length by its third, and this third by its own fourth, less the thirty-fourth part of that fourth.

StepTermRunning valueDecimal
1111.000000
2+ 1/34/31.333333
3+ 1/(3 × 4) = 1/1217/121.416667
4− 1/(3 × 4 × 34) = 1/408577/4081.414216
True value√21.414214

577/408 = 1.4142156..., which is off from √2 = 1.4142136... by about 0.0000021, roughly two parts in a million. Two facts make it striking:

  • 577² = 332,929 and 2 × 408² = 332,928. So 577² − 2 × 408² = 1, which makes (577, 408) a solution of the Pell equation x² − 2y² = 1, the same kind of equation Bhāskara II solved for 61 two thousand years later.
  • 577/408 is exactly what one step of the "average a guess with 2 divided by the guess" rule gives from 17/12: (17/12 + 24/17) ÷ 2 = 577/408. Nobody knows how the Śulba authors found it, so treat this as a modern way to see why it is so good, not as their method.

Piṅgala's metres: binary patterns before binary numbers

Piṅgala's Chandaḥśāstra (छन्दःशास्त्र), usually dated to roughly the 3rd or 2nd century BCE, is a treatise on Sanskrit prosody. Each syllable of a verse is light (laghu, L) or heavy (guru, G). Piṅgala gives procedures for listing every possible pattern of a given length (prastāra), counting them (saṅkhyā), finding the position of a pattern in the list, and recovering a pattern from its position. Those procedures treat L and G as two symbols, which is why modern writers connect them with binary numbers.

Worked example: all metres of 3 syllables, in the prastāra order where the first syllable alternates fastest.

PositionPatternRead as binary with G = 0, L = 1, first syllable lowestValue + 1
1G G G0001
2L G G0012
3G L G0103
4L L G0114
5G G L1005
6L G L1016
7G L L1107
8L L L1118

There are 2³ = 8 patterns, and the position of each pattern is its binary value plus one. A related count, the number of ways to fill a line of n beats when L takes one beat and G takes two, gives the sequence 1, 2, 3, 5, 8, 13, which later Indian writers on metre, including Virahāṅka, Gopāla and Hemacandra, described centuries before Leonardo of Pisa. The triangle of binomial coefficients appears as the Meru Prastāra in Halāyudha's 10th century commentary on Piṅgala. Exact attributions in this area are debated, so we give the later writers credit only for what their own texts show.

Place value and zero

The decimal place value system with a zero, the one this page is written in, took shape in India. Three pieces of evidence matter most.

The Gwalior inscription (876 CE). MacTutor describes it as the first record of the Indian use of zero that is dated and agreed by all to be genuine. A stone tablet records a garden 187 by 270 hastas that would supply 50 garlands a day to the local temple. Both 270 and 50 are written almost as they appear today, except that the 0 is smaller and slightly raised.

Brahmagupta (628 CE). Brahmagupta treated zero as a number with its own arithmetic rules, covered below. That is a conceptual step, not just a notation.

The Bakhshali manuscript and its dating dispute. The Bakhshali manuscript was dug up in 1881 near the village of Bakhshālī, about 80 km northeast of Peshawar in what is now Pakistan. It is 70 badly damaged birch-bark leaves in Śāradā script, held at the Bodleian Library in Oxford as MS. Sansk. d. 14, and it uses a dot for zero hundreds of times. Its editor G. R. Kaye placed it in the 12th century; others argued for the 3rd or 4th.

DateEvent
14 Sep 2017The Bodleian announces radiocarbon results dating it "from as early as the 3rd or 4th century"
6 Oct 2017Kim Plofker, Agathe Keller, Takao Hayashi, Clemency Montelle and Dominik Wujastyk publish a response in History of Science in South Asia, criticising how the results were announced and read
2018Oxford dates more folios and resamples folio 16; the retest shows the first folio 16 result was inaccurate
Later revisionA paper by Chivall, Lladó-Buisán, Howell and Evison (Oxford Research Archive) models all the dates: production started 799 to 892 CE and ended 900 to 1102 CE (95.4% ranges)

So the headline from 2017, that the manuscript held the world's oldest zero symbol from the 3rd or 4th century, is not supported by Oxford's own later analysis. The manuscript is still a very important early witness to Indian place value arithmetic. The text it copies may be older than the bark, a question radiocarbon cannot answer.

A worked Bakhshali style square root. The manuscript gives a rule for square roots of non-squares that, in modern notation, refines a first guess A for √(A² + b): take A + b/(2A), then subtract the square of b/(2A) divided by twice that first result. Applied to √2 with A = 1 and b = 1:

StepCalculationValue
1A + b/(2A) = 1 + 1/23/2
2(b/(2A))² = (1/2)²1/4
3Divide by 2 × 3/2 = 31/12
43/2 − 1/1217/12 = 1.41667

Run it again from A = 17/12, where b = 2 − 289/144 = −1/144. The first part, A + b/(2A) = 17/12 − 1/408, is exactly 577/408, the Śulba value. The full correction then goes further, to 665,857/470,832, which is within about 0.0000000000016 of √2. Two traditions, centuries apart, pass through the same fraction.

Āryabhaṭa: π ≈ 3.1416 and the first sine table

Āryabhaṭa (476 to 550 CE according to MacTutor) finished the Āryabhaṭīya in 499 CE. He tells us he was 23 when he wrote it, which is how his birth year is known. Its mathematics chapter covers arithmetic, algebra, plane and spherical trigonometry, and a table of sines.

His value of π. MacTutor translates the verse: add four to one hundred, multiply by eight, and then add sixty-two thousand; the result is approximately the circumference of a circle of diameter twenty thousand.

StepCalculationValue
1100 + 4104
2104 × 8832
3832 + 62,00062,832
4Circumference ÷ diameter = 62,832 ÷ 20,0003.1416
Error3.1416 − 3.14159265...about 0.0000073

Āryabhaṭa presents the value as approximate, not exact, and he does not explain how he found it. MacTutor notes that historians have proposed several reconstructions, including half the perimeter of a regular polygon with 256 sides, and that this is disputed.

His sine table. Indian astronomers used the jyā, a half chord, which is R × sin θ for a circle of radius R. Āryabhaṭa used R = 3,438 because a circle of 21,600 arc minutes (360 × 60) has a radius of 21,600 ÷ 2π ≈ 3,437.75 arc minutes. He tabulated 24 values at steps of 3°45′ (90° ÷ 24), listing the differences between successive values rather than the values themselves.

His difference rule, in modern terms: each new difference is the previous difference minus the running total divided by 225. The table below runs that rule and compares it with modern values of 3438 × sin θ.

kAngleRule: previous difference − (running total ÷ 225)Āryabhaṭa's differenceRunning total (his jyā)Modern 3438 sin θ, rounded
13°45′start225225225
27°30′225 − 225/225 = 224224449449
311°15′224 − 449/225 ≈ 222.0222671671
415°00′222 − 671/225 ≈ 219.0219890890
518°45′219 − 890/225 ≈ 215.02151,1051,105
622°30′215 − 1105/225 ≈ 210.12101,3151,316

All 24 of his differences are 225, 224, 222, 219, 215, 210, 205, 199, 191, 183, 174, 164, 154, 143, 131, 119, 106, 93, 79, 65, 51, 37, 22 and 7. They add to exactly 3,438, the radius, which is the sine of 90°. Most of his running totals match modern rounded values; the few that differ are off by 1, because his difference rule is itself an approximation. MacTutor describes it as a formula for sin(n + 1)x − sin nx in terms of sin nx and sin(n − 1)x, which is a second difference rule: the change in the differences is proportional to the sine itself.

Brahmagupta: zero, negative numbers and the composition identity

Brahmagupta (598 to 670 CE per MacTutor) wrote the Brāhmasphuṭasiddhānta in 628 CE. He gave arithmetic rules for "fortunes" (positive numbers), "debts" (negative numbers) and zero.

Brahmagupta's rule (MacTutor's translation, paraphrased)Modern formStatus today
A debt minus zero is a debt; a fortune minus zero is a fortunea − 0 = aCorrect
Zero minus zero is zero0 − 0 = 0Correct
The product of zero and a debt or fortune is zero0 × a = 0Correct
The product or quotient of two debts is a fortune(−a)(−b) = abCorrect
The product or quotient of a debt and a fortune is a debt(−a)(b) = −abCorrect
A number divided by zero is a fraction with zero as denominatora ÷ 0 = a/0Left undefined; not a real number
Zero divided by zero is zero0 ÷ 0 = 0Incorrect by modern standards; undefined

Six of seven right in 628 CE, including sign rules students still find confusing, is a major achievement; the 0 ÷ 0 rule is wrong, and saying so is part of giving credit.

The composition identity (bhāvanā). Brahmagupta found that if two pairs solve equations of the form x² − N y² = k, you can combine them:

 (a^2 - N b^2)(c^2 - N d^2) = (ac + N bd)^2 - N (ad + bc)^2

So from x² − Ny² = k₁ and x² − Ny² = k₂ you get a new solution with k₁k₂. Composing a pair with itself makes k a perfect square, which you can divide away. MacTutor's Pell's equation page works his method for N = 23: since 23 × 1² + 2 = 5², the pair (5, 1) gives k = 2; composing and halving gives 24² − 23 × 5² = 576 − 575 = 1, and composing (24, 5) with itself gives 1,151² − 23 × 240² = 1. Here is the same method on N = 92, which is 4 × 23, so expect the number 1,151 to appear again.

StepPair (x, y)x² − 92y²How
1(10, 1)100 − 92 = 8Observation
2Compose (10, 1) with itself: (10 × 10 + 92 × 1 × 1, 10 × 1 + 1 × 10) = (192, 20)36,864 − 36,800 = 64Bhāvanā
3Divide both by 8: (24, 5/2)576 − 575 = 1Since 64 = 8²
4Compose (24, 5/2) with itself: (24² + 92 × 25/4, 2 × 24 × 5/2) = (1151, 120)1,324,801 − 1,324,800 = 1Bhāvanā again

So x = 1,151 and y = 120 solve x² − 92y² = 1. We checked: 1151² = 1,324,801 and 92 × 120² = 1,324,800.

Bhāskara II and the chakravala: solving 61y² + 1 = x²

Bhāskara II (1114 to 1185 CE per MacTutor) completed the cyclic method, chakravala (चक्रवाल), in his Bījagaṇita around 1150. It solves x² − N y² = 1 for any non-square N. MacTutor records that he worked the cases N = 8, 11, 32, 61 and 67. The case N = 61 is famous because its smallest solution is enormous.

The method keeps a triple (a, b, k) with a² − N b² = k. At each step you choose a whole number m so that (a + bm) is divisible by k and m² − N is as small as possible in size, then move to:

 a' = (a m + N b) / |k|      b' = (a + b m) / |k|      k' = (m^2 - N) / k

For N = 61, start with a = 8, b = 1, since 8² − 61 = 3.

Stepm chosenabkCheck a² − 61b²
081364 − 61 = 3
17 (m² − 61 = −12)395−41,521 − 1,525 = −4
29 (m² − 61 = 20)16421−526,896 − 26,901 = −5
36453585205,209 − 205,204 = 5
491,52319544
575,639722−3−3
6829,7183,805−1−1

At step 6, k = −1. Brahmagupta's composition of (29,718, 3,805) with itself turns −1 into (−1)² = 1:

QuantityFormulaValue
x29,718² + 61 × 3,805²1,766,319,049
y2 × 29,718 × 3,805226,153,980
Checkx² − 61y²1

So 61 × 226,153,980² + 1 = 1,766,319,049². The equation is still called Pell's equation, a name MacTutor says is generally believed to come from Euler confusing John Pell with William Brouncker; whether Pell contributed anything to it is a question MacTutor examines in detail. (MacTutor's Bhāskara II page prints the larger number as 1,776,319,049, apparently a typo; its Pell's equation page gives the correct 1,766,319,049.)

Mādhava and the Kerala school: infinite series for π and arctan

Mādhava of Saṅgamagrāma (about 1350 to 1425 CE, per MacTutor) founded a school of astronomer mathematicians in Kerala. His own writings on series are mostly lost; we know his results through later members of the school, especially Nīlakaṇṭha Somayāji's Tantrasaṅgraha (1501) and Jyeṣṭhadeva's Yuktibhāṣā (16th century), a survey of Kerala mathematics written in Malayalam that gives derivations. MacTutor states that Mādhava found the series now known as the Maclaurin expansions of sine, cosine and arctan around 1400, more than two hundred years before they were found again in Europe.

The arctan series (known in Europe as Gregory's series):

 arctan x = x - x^3/3 + x^5/5 - x^7/7 + ...

Putting x = 1 gives π/4 = 1 − 1/3 + 1/5 − 1/7 + ..., usually called the Leibniz series in Europe. It converges very slowly. After 10 terms, 4 × (1 − 1/3 + ... − 1/19) = 3.0418, which is wrong in the first decimal.

Mādhava's correction terms. The Kerala texts add a correction after stopping. With p the even number after the last odd denominator (here p = 20), the corrections are:

CorrectionFormulaFor p = 20π estimate after 10 termsError
None3.0418396−0.0997530
First order1/(2p)1/403.1418396+0.0002470
Second order(p/2) / (p² + 1)10/4013.1415902−0.0000024
Third order((p/2)² + 1) / ((p² + 5) × p/2)101/40503.14159271+0.00000005

Ten terms plus the third order correction give π to about seven decimal places; the plain series would need millions of terms.

The faster series. Putting x = 1/√3, where arctan is π/6, gives:

 pi = sqrt(12) * (1 - 1/(3*3) + 1/(5*3^2) - 1/(7*3^3) + ...)
Termsπ estimateError
13.4641+0.3225
23.0792−0.0624
33.1562+0.0146
113.1415933+0.00000065
213.1415926535956about 0.000000000006

MacTutor credits Mādhava with π = 3.14159265359, correct to 11 decimal places, and notes that this series reaches it with 21 terms.

How these ideas travelled

  • To Baghdad. MacTutor quotes the 12th century chronicler al-Qifti on a scholar from India who came before the Caliph al-Mansur in 776 CE, versed in siddhānta astronomy and calculation with the half chord, essentially the sine. Al-Khwārizmī (about 780 to 850) then wrote on Hindu numerals; MacTutor notes that the word "algorithm" comes from his name.
  • To Europe. Leonardo of Pisa (Fibonacci) learned the system in North Africa, at Bugia, and set it out in Liber Abaci (1202), which MacTutor credits with introducing the Hindu-Arabic place value system into Europe.
  • Pell's equation. MacTutor notes that the work of Brahmagupta and Bhāskara II on x² − Ny² = 1 was unknown in 17th century Europe. Fermat's challenge of 1657 restarted the problem there, and Lagrange gave a polished continued fraction solution in 1766.
  • The Kerala series. Whether the Kerala results reached Europe through 16th century Jesuit contacts is an open question; some historians argue for it, but no document settles it. Kerala had these series first; a transmission route is unproven.

What to practise, and in what order

A four week plan at 15 minutes a day:

WeekPractiseTarget
1Nikhilam with base 100 and 1000; Yāvadūnam squares; Ekādhikena squares ending in 5Any product of two numbers from 88 to 112 in under 5 seconds
2Ūrdhva-Tiryagbhyām for 2 by 2, then 3 by 3; check every answer by casting out nines386 × 524 mentally, writing only the answer digits
3Flag division and the duplex square root on 4 to 6 digit numbers; 1/19, 1/29, 1/39 by Ekādhika√7,744 and 7,429 ÷ 37 without paper
4Rederive 577/408; run chakravala for N = 61; sum Mādhava's series with corrections in a spreadsheetKnow why each method works

Where Vedic shortcuts genuinely beat the standard methods

In short: Nikhilam for numbers near a base, Ekādhikena Pūrveṇa for squares ending in 5 and for denominators ending in 9, Ūrdhva-Tiryagbhyām for general mental multiplication, and digit sums for checking. For everything else, the standard algorithms are just as fast and work every time. The best mental calculators mix both: they recognise the special case when it appears and fall back to a general method when it does not.

Where this meets software security

Safeguard does not do mathematics. Our models are single purpose security models, and Safeguard's own benchmarks page says that general purpose models outperform Safeguard's on math problems. The overlap is a habit of mind. Casting out nines is a checksum: a cheap test that catches most errors and misses a known class of them, exactly like a simple checksum on a downloaded file. Knowing exactly which components went into a build, and being able to check them against known vulnerabilities, is the software equivalent of showing your working.

The mathematics here has no vulnerabilities, but the software people use to practise, compute and publish it does: learning apps, notebooks and numerical libraries are built from open source dependencies that receive security fixes like any other code. The section below is our standard guidance for closing that gap in the code you build and ship; it applies to those dependencies, not to mathematics itself.

Fix it without waiting for upstream

Whatever the vulnerability, most of the risk sits in two gaps. The first is the time between a fix being published and that fix actually running in your production builds. The second, and worse, is the time before any fix exists at all, while a maintainer works out a patch and a release schedule. Waiting through both gaps is a habit, not a requirement. Safeguard gives teams three ways to close them.

1. Autonomous remediation: pull requests, not tickets

When a fixed version exists, Safeguard's Griffin models open a tested, policy gated pull request that moves the affected dependency to the safe version. You can run it on one repository or in bulk across every project that carries the vulnerable version. Remediation categories include dependencies, licenses, security posture and findings, and each one is set separately. Reachability analysis ranks the vulnerable code paths your application actually calls first, so the first pull requests you merge are the ones that matter most. Autonomous remediation ships switched off, with every category set to Manual, so your team decides how much runs on its own and what still needs a human approval.

2. Gold Open Source: the safe version in seconds, free

gold.safeguard.sh is free and needs no login. Search any CVE or package to see affected and fixed versions, CVSS, EPSS, CISA KEV status and a recommended safe version, or use the "Am I affected?" checker. The same data is available through a free read API, README status badges and RSS threat feeds.

3. Gold Registry: start from components that are already fixed

Gold Registry provides hardened, drop in replacements for upstream packages and container images. Every Gold artifact is held continuously to zero exploitable critical or high CVEs and a completed license check, with signed build provenance (signed with Safeguard's key and verifiable with cosign, SLSA Build Level 3 as the design target) and a public evidence page you can hand to an auditor. An artifact loses its Gold badge if it regresses. Teams can point their own package directory at the registry so hardened equivalents resolve without changing how developers install anything.

When upstream has no fix yet, or will not ship one, Safeguard first works to get the fix accepted upstream. If the issue is severe or upstream will not move, Safeguard builds a fork, fixes the vulnerability there and publishes a hardened Gold build, so you are not stuck on a maintainer's release schedule. You can also request a hardened, compatibility tested build of a specific package. Ask which packages it covers before you plan around it.

Your situationWhat to use
A fixed version existsAutonomous remediation opens the upgrade pull request; Gold Open Source confirms the safe version
You need to know today whether you are affectedGold Open Source lookup and "Am I affected?" checker, free and without login
You want vulnerable versions to stop shippingSafeguard policies and gates in CI
Upstream has no fix yet, or will not ship oneA custom Gold request; fork and patch
You want to start clean on new projectsPull hardened components and images from Gold Registry

One honest limit: when the vulnerable code lives in a vendor appliance, a hosted service or an operating system runtime you do not build yourself, the fix still has to come from that vendor or distribution. The steps above apply to the open source dependencies and container images you build and ship, which is where most teams carry the largest and least tracked share of their exposure.

Common mistakes

  • Calling the 16 sutras ancient. They come from a 1965 book.
  • Using Nikhilam far from a base. 386 × 524 by base 1000 needs 614 × 476. Switch to vertically and crosswise.
  • Dropping the digit count on the right part. In base 100, 2 × 3 is written 06, not 6.
  • Trusting casting out nines completely. It cannot see swapped digits or errors that are multiples of 9.
  • Repeating the 2017 Bakhshali headline. Oxford's own later analysis dates the manuscript to between about 800 and 1100 CE.
  • Saying Kerala calculus "was copied" by Europe. The Kerala priority is documented; the transmission route is not.

FAQ

Are the Vedic Maths sutras really from the Vedas?

No surviving Vedic text contains them. They appear in Bharati Krishna Tirtha's Vedic Mathematics (1965). According to K. S. Shukla, Tirtha told him in 1950 that the sutras were in his own Pariśiṣṭa, not the known ones.

Is Vedic Maths faster than normal methods?

For special cases, yes: numbers near 100 or 1000, squares ending in 5, multipliers of nines and denominators ending in 9. For general problems, vertically and crosswise does the same digit products as long multiplication with less to remember.

Did India invent zero?

The decimal place value system with zero as a written digit and as a number with arithmetic rules developed in India. The earliest dated, universally accepted Indian zero is the Gwalior inscription of 876 CE, and Brahmagupta gave rules for zero in 628 CE.

How old is the Bakhshali manuscript?

The 2017 announcement suggested parts dated to the 3rd or 4th century. After more tests in 2018 found the first result for folio 16 inaccurate, Oxford's revised model places production between about 799 and 1102 CE. The mathematics it copies could be older.

What did Āryabhaṭa calculate?

In 499 CE he gave π ≈ 62,832/20,000 = 3.1416, a table of 24 sine differences at 3°45′ steps for a radius of 3,438.

What is the chakravala method?

A cyclic algorithm, completed by Bhāskara II around 1150, for solving x² − Ny² = 1. For N = 61 it gives x = 1,766,319,049 and y = 226,153,980.

Did Indians discover calculus before Newton?

Mādhava and the Kerala school found infinite series for sine, cosine, arctan and π around 1400, and Jyeṣṭhadeva's Yuktibhāṣā gives derivations. That is a central part of calculus, discovered first in Kerala. The general theory of derivatives and integrals that Newton and Leibniz built is a different achievement, so Kerala is best described as finding key results first.

Is Vedic Maths good for children?

The near-base and squaring tricks are fun and can build number sense, and checking with digit sums is a good habit. We found no large, independent, peer reviewed trials showing effects on intelligence or exam scores, so treat claims like that with caution. Teach it alongside the standard methods.

See how Safeguard would help

None of these demos do mathematics; they show how Safeguard handles open source dependencies.

The first demo is a real Safeguard CLI session. Asked to fix a lodash CVE, it reads the finding and advisory, asks before any command that changes something, edits package.json to lodash 4.17.21, runs npm install and npm test with approval, and pushes nothing. The same flow applies to a learning app you maintain.

Click through it at your own pace, or open it full screen.

The second demo sets up Safeguard's local runner on your own machine, approves it with its device code, and runs a scan from the terminal. On a laptop full of notebooks, that shows which library versions your work depends on.

Click through it at your own pace, or open it full screen.

The third demo connects Claude to the Safeguard MCP server with OAuth and lets you choose which tools Claude can use. You can then ask which of your projects carry a particular library version, while keeping control over what the assistant can touch.

Click through it at your own pace, or open it full screen.

To try it on your own projects, download Safeguard or read the CLI documentation.

Quick check

Use Nikhilam to multiply 96 × 93.

8,928. The deviations from 100 are −4 and −7. Cross subtract: 96 − 7 = 89. Multiply the deviations: 4 × 7 = 28. Read off 89 | 28.

Why is 577/408 such a good value for √2?

Because 577² − 2 × 408² = 332,929 − 332,928 = 1. It solves the Pell equation x² − 2y² = 1, so its square is only 1/166,464 away from 2.

Sources

Last reviewed: 25 October 2026.

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