1. What vulgar fraction is equivalent to .138 ?

OPERATION.

.138=16% +580 = £53 +080 = £88 = 3 Ans. As this is a mixed circulate, we divide it into its finite and circulating parts; thus .138.13, the finite part, and .008 the repetend or circulating part; but .13=13; and .008 would be equal to, if the circulate began immediately after the place of units; but, as it begins after the place of hundreds, it is & of 160 = 80. Therefore .138 = 16% +80=JJZ+ 980=136=1 Ans. Ans. Q. E. D.

RULE. To as many nines as there are figures in the repetend, annex as many ciphers as there are finite places for a denominator; multiply the nines in the denominator by the finite part, and add the repeating decimal to the product for the numerator. If the repetend begins in some integral place, the finite value of the circulating must be added to the finite part.

2. What is the least vulgar fraction equivalent to .53 ?

Ans.

3. What is the least vulgar fraction equivalent to .5925?

Ans. 19.

4. What is the least vulgar fraction equivalent to .008497133 ?

Ans..

5. What is the finite number equivalent to 31.62 ?

CASE III.

Ans. 313.

To make any number of dissimilar repetends similar and conterminous.

1. Dissimilar made similar and conterminous.

OPERATION.

Any given repetend whatever, 9.1679.61767676 whether single, compound, pure, or mixed, may be transformed into another repetend, that shall consist of an equal or greater number of figures at pleasure; thus 4 may be changed into .44 or .444; and 29 into .2929 or 2929. And as some of the circu

14.6 = 14.60000000 3.1653.16555555 12.432 = 12.43243243 8.181 8.18181818 1.307 1.30730730

=

lates in this question consist of one, some of two, and others of three places; and as the least common multiple of 1, 2, and 3

is 6, we know that the new repetend will consist of 6 places, and will begin just so far from unity as is the farthest among the dissimilar repetends, which, in the present example, is the third place.

RULE. Change the given repetends into other repetends, which shall consist of as many figures as the least common multiple of the several number of places found in all the repetends contains units.

2. Make 3.671, 1.0071, 8.52, and 7.616325 similar and conterminous.

3. Make 1.52, 8.7156, 3.567, and 1.378 similar and conter

minous.

4. Make .0007,.141414, and 887.i similar and conterminous.

CASE IV.

To find whether the decimal fraction equal to a given vulgar fraction be finite or infinite, and of how many places the repetend will consist.

RULE. - Reduce the given fraction to its least terms, and divide the denominator by 2, 5, or 10, as often as possible. If the whole denominator vanish in dividing by 2, 5, or 10, the decimal will be finite, and will consist of so many places as you perform divisions. If it do not vanish, divide 9999, &c., by the result till nothing remain, and the number of 9's used will show the number of places in the repetend; which will begin after so many places of figures as there are 10's, 2's, or 5's used in dividing.

NOTE. In dividing 1.0000, &c., by any prime number whatever, except 2 or 5, the quotient will begin to repeat as soon as the remainder is 1. And since 9999, &c., is less than 10000, &c., by 1, therefore 9999, &c., divided by any number whatever, will leave a 0 for a remainder, when the repeating figures are at their period. Now whatever number of repeating figures we have when the dividend is 1, there will be exactly the same number when the dividend is any other number whatever. For the product of any circulating number by any other given number will consist of the same number of repeating figures as before. Thus, let .378137813781, &c., be a circulate, whose repeating part is 3781. Now every repetend (3781), being equally multiplied, must produce the same product. For these products will consist of more places, yet the overplus in each, being alike, will be carried to the next, by which means each product will be equally increased, and consequently every four places will continue alike. And the same will hold for any other number whatever. Hence it appears, that the dividend may be altered at pleasure, and the number of places in the repetend will be still the same; thus, = .09, and 1.27, where the number of places in each are alike; and the same will be true in all cases.

P

EXAMPLES.

1. Required to find whether the decimal equal to 20% be finite or infinite; and if infinite, of how many places the repetend will consist.

210

3

(2) (2) (2)

2)168=4=2=1; therefore, because the denominator vanishes in dividing, the decimal is finite, and consists of four places; thus, 16)3. 3:08990.

2. Required to find whether the decimal equal to 475 be finite or infinite; and, if infinite, of how many places that repetend will consist.

7)999999

475 = 12, 2)112=36-28-14-7. Thus, ??????;

therefore, because the denominator, 112, did not vanish in dividing by 2, the decimal is infinite; and as six 9's were used, the circulate consists of six places, beginning at the fifth place, because four 2's were used in dividing.

3. Let 4. Let

be the fraction proposed.

be the fraction proposed.

SECTION XXXI.

ADDITION OF CIRCULATING DECIMALS.

EXAMPLE.

1. Let 3.5+7.651+1.765+6.173+51.7+3.7+27.631 and 1.003 be added together.

51.7

=

Having made all the numbers similar and conterminous by Sect. XXX., Case III., we add the first six columns, as in Simple Addition, and find the sum to

999999

51.7777777 be 3591224=35912243.591227. 3.7 = 3.7000000 The repeating decimals .591227 we 27.63127.6316316 write in their proper place, and carry 3 1.003 1.0030030 to the next column, and then proceed

103.2591227

as in whole numbers.

RULE. · Make the repetends similar and conterminous, and find their sum, as in common Addition. Divide this sum by as many 9's as there are places in the repetend, and the remainder is the repetend of the sum, which must be set under the figures added, with ciphers on the left when it has not so many places as the repetends. Carry the quotient of this division to the next column, and proceed with the rest as with finite decimals.

2. Add 27.56+5.632 + 6.7 + 16.356+.71 and 6.1234 together. Ans. 63.1690670868888. 3. Add 2.765+7.16674 +3.671+.7 and .1728 together.

4. Add 5.16345 +8.6381 +3.75 together.

Ans. 14.55436.

Ans. 17.55919120847374090302.

5. Reduce the following numbers to decimals, and find their

OPERATION.

87.1645 = 87.164545 19.47916719.479167

67.685377

Having made the numbers similar and conterminous, we subtract as in whole numbers, and find the remainder of the circulate to be 5378, from which we subtract 1, and write the remainder in its place, and proceed with the other part of the question as in whole numbers. The reason why 1 should be added to the repetend may be shown as follows. The minuend may be considered 163515, and the subtrahend 78187; we then proceed with these numbers as in Case II. of Subtraction of Vulgar Fractions; and the numerator 5377 will be the re88377 peating decimal. Q. E. D.

OPERATION.

78

4545
9999

9999

RULE. - Make the repetends similar and conterminous, and subtract as usual; observing, that if the repetend of the subtrahend be greater than the repetend of the minuend, then the remainder on the right must be less by unity than it would be if the expressions were finite.

2. From 7.1 take 5.02.

Ans. 2.08.

5. From 16.1347 take 11.0884.

3. From 315.87 take 78.0378. Ans. 237.838072095497.

4. Subtract from 3. ·

6. From 18.1678 take 3.27.

7. From 3.123 take 0.71.

Ans. 14.8951. Ans. 2.405951.

Ans. .079365.

Ans. 5.0462.

10. From

take 17.

Ans. .246753.

Ans. .158730.

Ans..176470588235294i.

11. From 5.12345 take 2.3523456.

Ans. 2.7711055821666927777988888599994.

2776 =

=

999

RULE. Turn both the terms into their equivalent vulgar fractions, and find the product of those fractions as usual. Then change the vulgar fraction expressing the product into an equivalent decimal, and it will be the product required. But, if the multiplicand ONLY has a repetend, multiply as in whole numbers, and add to the right-hand place of the product as many units as there are tens in the product of the lefthand place of the repetend. The product will then contain a repetend whose places are equal to those in the multiplicand.