RULE. Decompose the denominator, after the fraction is reduced to its lowest terms, into its prime factors, (by rule under Art. 7,) which factors cannot differ from 2 and 5, (by rule under Art. 43.) The highest exponent of 2, or 5, will be the number of decimal places sought. EXAMPLES. 1. How many places of decimals will be required to express? In this example, we find 40=23 × 5, where the highest exponent is 3; therefore, the number of decimal places is 3. 2. How many places of decimals will be required to express ? Ans. 3. Ans. 5. 3. How many places of decimals will be required to express! express 3125 4. How many places of decimals will be required to express? 5. How many places of decimals will be required to express? Ans. 5. Ans. 4. 6. How many places of decimals will be required to express ? Ans. 6. 45. When many figures in the decimal are required, we may proceed as follows: Required the decimal value of? 29 Following the rule under Art. 42, we get this Operation. 29)100(0.03448 87. 130 116 140 116 240 232 We have continued this process until we have found a remainder consisting of but one figure; placing this remainder, when divided by 29, at the right of the quotient, agreeably to the usual rules of division, we get, I. 003448. Multiplying this by 8, we get =0.27586. Substituting this value of in I., we get, 29 II. =0.0344827586; this, multiplied by 6, gives 2=0·20689655177; which, substituted in II., gives, 003448275862068965517. Again multi0.24137931034482758620. III. plying by 7, we get Substituting this in III., we get, IV. =0·0344827586206896551724137931034482758620. In the expression=0·03448, the numerator 8, of the vulgar fraction, is the fifth remainder; and in the expression=0.0344827586, the numerator 6, of the fraction, is the tenth remainder; but this remainder 6, was obtained by multiplying the 5th remainder, which is 8, into itself, and dividing the product, 64, by 29, we thus found the remainder, 6. Again, the 20th remainder, which is 7, was found by multiplying the 10th remainder into itself, and dividing the product by 29. For a similar reason, the remainder of the product of any two remainders will give the remainder corresponding with the sum of the numbers denoting their order; thus, the 5th multiplied by the 7th, will give the 12th; the 6th multiplied by the 9th will give the 15th, and so on for other combinations. 46. There is another way of decimating, which is as follows: According to rule under Art. 42, we find, 97)100(0.01 3 To continue this process, we must add ciphers to this remainder in the same way as we did to the numerator, 1. Now, the remainder being 3 times as large as the first numerator, it follows that the next two decimal figures must be 3 times the two just obtained, that is, 3×01 03; and, for a similar reason, we must multiply 03 by 3, to obtain the next two figures, and so on. Proceeding in this way, we find, 700103092781 243 729 2187 6561 &c. 0010309278350515, &c. Decimating by the above plan, we get =0·248 16 32 64 128 256 &c. =0·249999, &c., which will con stantly approximate towards 0.25; hence, Decimating by this method, we get, }=0·1248 0'25. 16 32 64 128 256 &c. 01249999, &c., which will con stantly approximate towards 0.125. 47. When the decimal figures, obtained by converting a vulgar fraction into decimals, do not terminate, they must recur in periods, whose number of terms cannot exceed the number of units in the denominator, less one. For, all the different remainders which occur must be less than the denominator; and, therefore, their number cannot exceed the denominator, less one; and, whenever we obtain a remainder like one that has previously occurred, then the decimal figures will begin to repeat. Decimals which recur in this way are called repetends. When the period begins with the first decimal figure, it is called a simple repetend. But when other decimal figures occur before the period commences, it is called a compound repetend A repetend is distinguished from ordinary decimals. by a period, or dot, placed over the first and last figure of the circulating period. 48. The following vulgar fractions give simple repetends : 1 = =0.3. T'T=0·09. r's=0·076923. =0·0588235294117647. =0·052631578947368421. =0·047619. =0·0434782608695652173913. 49. The following ones give compound repetends. † =0·16. =0.0714285 r's=0.06. T=0.05. =0.045. =0.0416. |