Book Title: Acharya Shantisagar Janma Shatabdi Mahotsav Smruti Granth
Author(s): Jinwani Jirnoddharak Sanstha Faltan
Publisher: Jinwani Jirnoddharak Sanstha Faltan

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Page 517
________________ ३१२ आ. शांतिसागरजी जन्मशताब्दि स्मृतिग्रंथ used, names of operations and numerals. Rules governing the use of negative numbers are correctly stated those regarding the use of zero may be stated in modern notation thus : a + o= a; axo = 0; 2:0 = a. The last part is obviously wrong. As regards the square root of a negative number, the author observes that since squares of positive and negative numbers are positive, square root of a negative number cannot exist. Considering the limitations of his time, Mahaviracharya could not have reached a more sensible conclusion. We may note, in this context, the necessary extension of the concept of number which assimilates square roots of negative numbers into the number system, was achieved as late as in 1797 by c. Wessel a Norwegian Surveyor (Bell's “The Development of Mathematics" Page 177). Chapter II deals, in respect of integers, with operations of multiplication, division, squaring and its inverse, cubing and its inverse, arithmetic and geometric series. Problem II. 17. In this problem, put down in order (from the unit's place upwards) 1,1,0, 1, 1, 0, 1 and 1, which (figures so placed ) give the measure of a number and (then) if this number is multiplied by 91, there results that necklace which is worthy of a prince. The Necklace' referred to, may be displayed thus : 11011011 x 91 = 1002002001. Two more 'garlands worthy of a prince' are : (II. 11, 15): 333333666667 x 33 = 11000011000011; and 752207 x 73 = 11,111,111. Chapters III and IV are devoted to elementary operations with fractions, Mahaviracharya has paid considerable attension to the problem of expression of a unit fraction as the sum of unit fraction. This problem has interested mathematicians from remote antiquity (Ahmes Papyrus, 1650 B. C.). Here are three relevant problems (II 75, 77, 78 ) set in modern notation. (1) 1 = 1 + 1 + 32 + .***** +31+3n-3.3 (2) 1 = 2.3.1+3.4.+****+ (2n – 1)2n.4 +2n. (3) a n (na,(n +a) (na, +2, +..... а ag n n (n+ai) (n +a ) (n+a, +a, 2.-1 (n+a? + a2 + ...... + ar-2) (n + a1 + a2 + ar-1) ac a: (n + a1 + a2 + ......ar) Jain Education International For Private & Personal Use Only www.jainelibrary.org

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