2. Make ·531, 7348, 07 and 0503 similar and conterminous.

CASE IV.

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

RULE.*

1. Reduce the given fraction to its least terms, and divide the denominator by 2, 5 or 10, as often as possible.

2. Divide 9999, &c. by the former result, till nothing remain, and the number of 9s used will show the number of places in the repetend; which will begin after so many places of figures as there were 10s, 23, or 5s, divided by.

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.

* In dividing 1000, &c. by any prime number whatever, except 2 or 5, the figures in the quotient will begin to repeat over again as soon as the remainder is 1: and since 999, &c. is less than 1000, &c. by 1, therefore 999, &c. divided by any number whatever, will, when the repeating figures are at their period, leave 0 for a remainder.

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.

Thus, let 390539053905, &c. be a circulate, whose repeating part is 3905. Now, every repetend (3905,) being equally multiplied, must give the same product: For although 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.

Now from hence it appears that the dividend may be altered at pleasure, and

the number of places in the repetend will still be the same;

thus,

1' or '¡ X4-26, whence the number of places in each are alike.

U a

09; and

EXAMPLES.

1. Required to find whether the decimal equal to

be finite

or infinite, and if infinite, how many places that repetend will consist of.

TheD,

2800
7)999999

; therefore, because the denominator 112 did 142857

not vanish in dividing by 2, the decimal is infinite, and, as six 9s were used, the circulate consists of 6 places, beginning at the fifth place, because four 2s were used in dividing.

2. Let be the fraction proposed.

3. Let be the fraction proposed.

ADDITION OF CIRCULATING DECIMALS.

RULE.

1. Make the repetends similar and conterminous, and find their sum as in common addition.

2. Divide this sum (of the repetends only) by so many nines as there are places in the repetend, and the remainder is the repetend of their sum; which must be set under the figures added, with cyphers on the left hand, when it has not so many places as the repetends.

3. Carry the quotient of this division to the next column, and proceed with the rest as infinite decimals.

EXAMPLES.

1. Let 5-3+59-4356+3976+519+39+217·5 be added together.

1199-3851305 the sum.

In this question, the sum of the repetends is 2851303, which divided by 999999, gives 2 to carry to the next column 5,3,0, &c. and the remainder is 851305.

2. Let 3275-319+36-45+123-19+5·3173+112-3513+11-131

1252910053 be added together.

Ans. 3593-00042.

SUBTRACTION OF CIRCULATING DECIMALS.

RULE.

Make the repetends similar and conterminous, and subtract as usual, observing, that if the repetend of the number to be subtracted be greater than the repetend of the number it is to be taken from, then the right hand of the remainder must be less by unity than it would be if the expressions were finite.

1. Turn both the terms into their equivalent vulgar fractions, and find the product of those fractions as usual.

2. Turn the vulgar fraction expressing the product, into an equivalent decimal one, and it will be the product required.

EXAMPLES.

1. Multiply 54 by 15. 54 and 15=y=3

1. Change both the divisor and dividend into their equivalent vulgar fractions, and find their quotient as usual.

2. Turn the vulgar fraction expressing the quotient, into its equivalent decimal, and it will be the quotient required.

1. Divide 54 by 15.

EXAMPLES.

•54=§}=}; and •15=}=}

fr÷3=&×¥=270=342=3·506493 the quotient.

IS the method of mixing two or more simples of different qualties, so that the composition may be of a mean or middle quality; It consists of two kinds, viz. Alligation Medial, and Alligation Alternate.

ALLIGATION MEDIAL

Is, when the quantities and prices of several things are given, to find the mean price of the mixture compounded of those things.

RULE.

As the sum of the quantities, or the whole composition, is to their total value; so is any part of the composition to its mean price or value.

EXAMPLES.

1. A Tobacconist would mix 60% of tobacco, at 6d. per

with

50 at 1s. 40 at 1s. 6d. and 30 at 2s. per : What is 1 of this mixture worth?

2. A farmer would mix 20 bushels of wheat at $1 per bushel, 16 bushels of rye at 75c. per bushel, 12 bushels of barley at 50c.

per bushel, and 8 bushels of oats at 40c. per bushel: What is the value of one bushel of this mixture? Ans. 73c. 5 m.

3. A wine merchant mixes 12 gallons of wine, at 75c. pér gallon, with 24 gallons at 90c. and 16 gallons at $1 10c.: What is a gallon of this composition worth?

Ans. 92c. 6m.

4. A goldsmith melted together 8oz. of gold of 22 carats fine, 1 8oz. of 21 carats fine, and 10oz. of 18 carats fine: Pray what is the quality, or fineness of the composition?

8×22+20×21+10×18,

8+20+10

=20, carats fine, Ans.

19

5. A refiner melts 5 of gold of 20 carats fine with 8 of 18 carats fine: How much alloy must be put to it, to make it 22 carats fine?

22-5×20+8×18÷÷÷5+8=3,3

Answer. It is not fine enough by 3, carats, so that no alloy must be added, but more gold.

ALLIGATION ALTERNATE*

Is the method of finding what quantity of each of the ingredients, whose rates are given, will compose a mixture of a given rate': So that it is the reverse of Alligation Medial, and may be proved by it.

CASE I.

The whole work of this case consists in linking the extremes truly together and taking the differences between them and the nean price, which differences are the quantities sought.

RULE.

1. Place the several prices of the simples, being reduced to one denomination, in a column under each other, the least uppermost, and so gradually downward, as they increase with a line of

Demon. By connecting the less rate with the greater, and placing the difference between them and the mean rate alternately, or one after the other in turn, the quantities resulting are such, that there is precisely as much gained by one quantity as is lost by the other, and therefore the gain and loss, upon the whole, are equal, and are exactly the proposed rate.

In like manner, let the number of simples be what it may, and with how many soever, each one is linked, since it is always a less with a greater than the mean price, there will be an equal balance of loss and gain between every two, and consequently an equal balance on the whole.

It is obvious from the rule, that questions of this sort admit of a great variety of answers; for having found one answer, we may find as many more as we please, by only multiplying or dividing each of the quantities found, by 2, 3, 4, &c, the reason of which is evident; for if two quantities of two simples make a balance of loss and gain with respect to the mean price, so must also the double or triple, the half or third part, or any other ratio of these quantities, and so on ad infinitum.

If any one of the simples be of little or no value with respect to the rest, its rate is supposed to be nothing, as water mixed with wine, and alloy with go and silver.