Square root in Līlāvatī: Difference between revisions
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Here we will know how to find the Square root of a number as mentioned in Līlāvatī. | |||
Here we will know how to find the Square root of a number. | |||
==Verse No. 22 :== | ==Verse No. 22 :== | ||
Line 241: | Line 237: | ||
|}'''Answer: Square root of 88209 = 297''' | |}'''Answer: Square root of 88209 = 297''' | ||
==See Also== | ==See Also== | ||
[ | [[लीलावती में 'वर्गमूल']] | ||
==References== | ==References== | ||
<references /> | <references /> | ||
[[Category:Mathematics in Līlāvatī]] | |||
[[Category:General]] |
Latest revision as of 21:01, 30 August 2023
Here we will know how to find the Square root of a number as mentioned in Līlāvatī.
Verse No. 22 :
त्यक्त्वान्त्याद् विषमात् कृतिं द्विगुणयेन्मूलं समे तद्धृते
त्यक्त्वा लब्धकृतिं तदाद्यविषमाल्लब्धं द्विनिघ्नं न्यसेत् ।
पङ्क्त्यां पंक्तिहृते समेऽन्त्यविषमात्त्यक्त्वाप्तवर्गं फलम्
पङ्क्त्यां तद् द्विगुणं न्यसेदिति मुहुः पङ्क्तेर्दलं स्यात् पदम् ॥ २२ ॥
Translation :
Starting from the unit's place, mark alternatively vertical and horizontal bars above the digits so that the given number is divided into groups of two digits each[1]. The extreme left group may have one or two digits and will have a vertical bar on its top or on the right digit respectively.
From the group on the extreme left subtract the highest possible square (a1) and write 2a1 in the column called paṅkti (पङ्क्ति) meaning row.
To the right of the number obtained from the above subtraction, write the digit from the next group with a horizontal bar. Divide this number by 2a1. The quotient a2 should not be more than 9. Now write 2a2 below 2a1 after shifting it to one place to the right and add. This is the second paṅkti (पङ्क्ति).
Write the next digit to the right of the remainder so obtained and from that subtract the square of the second quotient a2 .Now to the right of the remainder so obtained, write the next digit and divide this by the second paṅkti (पङ्क्ति).This will give the third digit of the required square root. Now twice the third digit of the square root should be added on to the second paṅkti (पङ्क्ति) after shifting it by one place to the right. The result is the third paṅkti (पङ्क्ति). Then write the next digit of the given number to the right of the remainder and subtract from it the square of the third digit of the square root. Repeat this process. The result is the required square root.
Example: Square root of 196
Step 1: The given number to be marked as Odd (विषम) by the symbol " | " and Even (सम) by the symbol " - " . This marking should start from unit's place.
विषम (Odd) | सम (Even) | विषम (Odd) |
---|---|---|
| | - | | |
1 | 9 | 6 |
Here 6 is विषम ,9 is सम ,1 is विषम.
| | - | | | Steps:
From the last group, viz. 1 subtract the highest possible square (12) which is 1. We get the first remainder 0 = 1 - 1. Now write 9 (from the given number) to the right of the remainder 0 to get 09. Write 1 in the root column. 1 x 2 = 2 is the first pańkti. Divide 9 by 2 such that the highest one digit quotient does not exceed 9. Here, the quotient is 4. Write this 4 below 1 in the root column. In the same horizontal line write 2 × 4 = 8 with 0 below 2. Add the two to get 28 which is the second pańkti. Then subtract 8 = 2 × 4 from 9 to get 1. To its right write the next digit 6 and we get 16. From this subtract the square of 4 to get the remainder 0.The required square root is the number obtained by writing the digits from the root column in the order in which we derived them. Hence it is 14. We can get the same number as half of the second pańkti.(28 ÷ 2 = 14) | |||||
Divisor
भाजक |
To be divided
भाज्य |
मूलम् (Root) | पंक्ति (Paṅkti) | |||||
1 | 9 | 6 | ||||||
12 = 1 | 1 | 1 | (1 X 2 = 2)
2 |
1st | ||||
2) | 0 | 9 | (4 | 4 | (2 X 4 = 8)
08 |
2nd | ||
8 | 28 | |||||||
1 | 6 | 14 | 28 ÷ 2 = 14 | |||||
42 = 16 | 1 | 6 | ||||||
0 |
Answer: Square root of 196 = 14
Example: Square root of 88209
Step 1: The given number to be marked as Odd (विषम) by the symbol " | " and Even (सम) by the symbol " - ". This marking should start from unit's place.
विषम (Odd) | सम (Even) | विषम (Odd) | सम (Even) | विषम (Odd) |
---|---|---|---|---|
| | - | | | - | | |
8 | 8 | 2 | 0 | 9 |
Last group | 2nd group | 1st group |
| | - | | | - | | | Steps:
From the last group, viz. 8 subtract the highest possible square (22) which is 4. We get the first remainder 4 = 8 - 4. Now write 8 (from the given number) to the right of the remainder 4 to get 48. 2 x 2 = 4 is the first pańkti. Divide 48 by 4 such that the highest one digit quotient does not exceed 9. Here, the quotient is 9. Write this 9 below 2 in the root column. In the same horizontal line write 2× 9 = 18 with 1 below 4. Add the two to get 58 which is the second pańkti. Then subtract 36 = 9 × 4 from 48 to get 12. To its right write the next digit 2 and we get 122. From this subtract the square of 9 to get 41. To the right of 41 write 0, the next digit from the given number. Divide 410 by the second pańkti viz. 58 and get 7 as the quotient and 4 as remainder. Next we write this number 7 in the root column and 7 × 2 = 14 to its right with 1 below 8. Add the two to get 594 which is the third pańkti. Write the last digit 9 of the given number to the right of 4 to get 49. From this subtract 72 = 49 to get the remainder 0. The required square root is the number obtained by writing the digits from the root column in the order in which we derived them. Hence it is 297. We can get the same number as half of the third pańkti. | ||||||
Divisor
भाजक |
To be divided
भाज्य |
मूलम् (Root) | पंक्ति (Paṅkti) | ||||||||
8 | 8 | 2 | 0 | 9 | |||||||
22 = 4 | 4 | 2 | (2 X 2 = 4)
4 |
1st | |||||||
4) | 4 | 8 | (9 | 9 | (2 X 9 = 18)
18 | ||||||
3 | 6 | 58 | 2nd | ||||||||
1 | 2 | 2 | 7 | (2 X 7 = 14)
014 | |||||||
92=81 | 8 | 1 | 594 | 3rd | |||||||
58) | 4 | 1 | 0 | (7 | 594 ÷ 2 | ||||||
4 | 0 | 6 | 297 | 297 | |||||||
4 | 9 | ||||||||||
72=49 | 4 | 9 | |||||||||
0 | 0 |
Answer: Square root of 88209 = 297
See Also
References
- ↑ Līlāvatī Of Bhāskarācārya - A Treatise of Mathematics of Vedic Tradition. New Delhi: Motilal Banarsidass Publishers. 2001. pp. 23–25. ISBN 81-208-1420-7.