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51
CHAPTER III--FRACTIONS.

by dividing the denominations by means of a common factor thereof, (the quotient derived from the denominator of either of the fractions being used in the multiplication of the numerator and the denominator of the other fraction), those (fractions) become so reduced as to have equal denominators. (Then) removing one of these (equal) denominators, the numerators are to be added (to one another) or to be subtracted (from one another, so that the result may be the numerator in relation to the other equal denominator).

Another rule for arriving at the common denominator in another manner:--

6. The niruddha (or the least common multiple) is obtained by means of the continued multiplication of (all) the (possible) common factors of the denominators and (all) their (ultimate) quotients. In the case of (all) such multiples of the denominators and the numerators (of the given fractions), are obtained by multiplying those (denominators and numerators) by means of the quotients derived from the division of the niruddha by the (respective) denominators, the denominators become equal (in value).

Examples in illustration thereof.

57 and 58. A śrāvaka purchased, for the worship of Jina, jambu fruits, limes, oranges, cocoanuts, plantains, mangoes and pomegranates for ${\displaystyle {\tfrac {1}{2}},{\tfrac {1}{6}},{\tfrac {1}{12}},{\tfrac {1}{20}},{\tfrac {1}{30}},{\tfrac {1}{24}}}$ and ${\displaystyle {\tfrac {1}{8}}}$ of the golden coin in order ; tell (me) what the result is when these (fractions) are added together.

59. Add together ${\displaystyle {\tfrac {8}{15}},{\tfrac {1}{20}},{\tfrac {7}{36}},{\tfrac {1}{6}}{\tfrac {1}{3}}}$ and ${\displaystyle {\tfrac {1}{21}}}$

60. (There are 3 sets of fractions), tho denominators whereof begin with 1, 2 and 3, (respectively) and go on increasing gradually by one till the last (of such denominators) becomes 9, 10 and

60. The resulting problems are to find the values of--

(i)${\displaystyle {\tfrac {1}{1\times 2}}+{\tfrac {1}{2\times 3}}+{\tfrac {1}{3\times 4}}+....+{\tfrac {1}{8\times 9}}+{\tfrac {1}{9}}}$
(ii)${\displaystyle {\tfrac {2}{2\times 3}}+{\tfrac {2}{3\times 4}}+{\tfrac {2}{4\times 5}}+....+{\tfrac {2}{9\times 10}}+{\tfrac {2}{10}}}$
(iii)${\displaystyle {\tfrac {1}{3\times 4}}+{\tfrac {3}{4\times 5}}+{\tfrac {3}{5\times 6}}+....+{\tfrac {3}{15\times 16}}+{\tfrac {3}{16}}}$