| L. | Stock. gain. Stock. | |----|---------------------| | A's stock 12 | A. If 20 : 5 :: 12 | | B's stock 8 | 5 | | Total stock 20 | 2(0)60 |

Stock. gain. Stock. A's gain 3l.

B. If 20 : 5 :: 8

| L. | Stock. gain. Stock. | |----|---------------------| | 8 | A's gain 3 | | 2(0)40 | B's gain 3 |

B's gain 2l. Total gain 5 proof.

Note 1. When in any question there happen to be remainders, they must be reduced equally low, so as to be all of one name; and then their sum will be either equal to the divisor, or exactly double, triple, &c. of it: and accordingly 1, 2, 3, &c. carried from the sum of the remainders, and added to the particular gains, will make up the total gain; or the divisor will always divide the sum of the remainders exactly, and the quotient added to the particular gains will give the total gain.

Note 2. When the partners have equal shares of stock or capital, their shares of gain, loss, or neat proceeds, is found readily by dividing the total gain, loss, &c. by the number of partners.

II. Fellowship with time.

In fellowship with time, the gain or loss is divided among the partners, both in proportion to the stocks themselves, and also in proportion to the times of their continuance in company: For the same stock continued a double time, procures a double share of gain; and continued a triple time, procures a triple share of gain; that is, the shares of gain or loss are as the products of the several stocks multiplied into their respective times: and accordingly questions belonging to this rule are wrought by the following proportion.

As the sum of the products of the several stocks into their respective times To the total gain or loss, So the product of each man's stock into his time To his share of the gain or loss.

Quest. 1. A put into company 40l. for 3 months, B 75l. for 4 months; they gain 70l.: What share must each man have?

A 40 × 3 = 120, third term for A's share. B 75 × 4 = 300, third term for B's share.

420, first term:

| L. | L. | |----|----| | A. If 420 : 70 :: 120 | B. If 420 : 70 :: 300 | | 120 | 300 | | 42(0)840 | (20l.) | | 84 | 210 |

(0) (0)

A's gain 20 B's gain 50

Total gain 70 proof.

Quest. 2. A put into company 560l. for 8 months, B 279l. for 10 months, and C 735l. for 6 months; they gained 1000l.: What share of the gain must each have?

A 560 × 8 = 4480, third term for A's share. B 279 × 10 = 2790, third term for B's share. C 735 × 6 = 4410, third term for C's share.

11680, first term.

| L. | L. s. d. f. Rem. | |----|------------------| | A If 11680 : 1000 :: 4480 : 382—11—2—3—208 | | B If 11680 : 1000 :: 2790 : 238—17—4—3—80 | | C If 11680 : 1000 :: 4410 : 377—11—4—1—880 |

Proof 1000—00—0—0—1168.

CHAP. IX. VULGAR FRACTIONS.

A Fraction is a part or parts of an unit, or of any integer or whole; and is expressed by two numbers, one above and the other below a line drawn between them; as, $\frac{1}{2}$.

The number under the line shews into how many parts the unit or integer is divided; and is called the denominator, because it gives name to the fraction: The number above the line shews or tells how many of these parts the fraction contains; and is therefore called the numerator.

In the fraction $\frac{3}{4}$, a pound Sterling is the unit, integer, or whole; and the denominator 4 shews that the pound is broken or divided into four equal parts, viz. 4 crowns; and the numerator 3 shews that the fraction contains three of these parts, that is, three crowns; and so the value of this fraction is fifteen shillings.

Cor. 1. Hence it follows, 1. When the numerator of a fraction is less than the denominator, the value of such a fraction is less than unity, or the integer. 2. When the numerator is equal to the denominator, the value of the fraction is exactly an unit or integer. 3. When the numerator is greater than the denominator, the value of the fraction is more than an unit; and so often as the denominator is contained in the numerator, so many units or wholes are contained in the fraction. If, therefore the numerator of a fraction be divided by the denominator, the quot will be a number of units or integers, and the remainder so many parts.

The numerator of a fraction is to be considered as a dividend, and the denominator as a divisor; and the fraction itself may be taken to denote the quotient.

Cor. 2. From this view of a fraction, it is evident, that if the numerator and denominator of a fraction be either both multiplied or both divided by the same number, the products or quotients will retain the same proportion to one another; and consequently the new fraction thence arising will be of the same value with the given one. Thus the numerator and denominator of the fraction $\frac{3}{4}$ multiplied by 2 produces $\frac{6}{8}$, and divided by 2 quots $\frac{1}{2}$, both which fractions are of the same value with $\frac{3}{4}$.

Fractions having 10, 100, 1000, or 1, with any number of ciphers annexed to it, for a denominator, are call- ed decimal fractions; and fractions having any other denominator are called vulgar fractions.

1. A proper fraction is that whose numerator is less than its denominator, and consequently is in value less than unity; as \( \frac{3}{4} \).

2. An improper fraction is that whose numerator is equal to or greater than its denominator; and consequently is in value equal to or greater than an unit; as \( \frac{5}{4} \), \( \frac{7}{4} \).

3. A simple fraction is that which has but one numerator, and one denominator; and may be either proper or improper; as \( \frac{5}{9} \) or \( \frac{9}{5} \).

4. A compound fraction is made up of two or more simple fractions, coupled together with the particle of, and is a fraction of a fraction; as \( \frac{3}{4} \) of \( \frac{4}{5} \), or \( \frac{3}{4} \) of \( \frac{3}{4} \) of \( \frac{4}{5} \).

5. A mixt number consists of an integer, and a fraction joined with it; as \( 7\frac{1}{2} \).

Because in most cases fractions can neither be added nor subtracted, till they be reduced, we begin with reduction.

Reduction of Vulgar Fractions.

Problem I. To reduce an improper fraction to an integer, or mixt number.

Rule. Divide the numerator by the denominator, the quot gives integers; and the remainder, if there be any, placed over the divisor or denominator, gives the fraction to be annexed.

Examples.

1. \( \frac{85}{9} = 9 \) integers, there being no remainder. 2. \( \frac{43}{7} = 6 \) integers, the remainder being 5. 3. \( \frac{98}{12} = 8 \) integers, the remainder being 10. 4. \( \frac{173}{36} = 4 \) integers, the remainder being 48.

Problem II. To reduce a mixt number to an improper fraction.

Rule. Multiply the integer by the denominator; to the product add the numerator: The sum is the numerator of the improper fraction; and the denominator is the same as before.

Examples.

1. \( 54\frac{5}{8} = 437 \); for \( 54 \times 8 = 432 \) \[ \begin{array}{c} + 5 \\ \hline \text{Numerator } 437 \end{array} \]

2. \( 98\frac{1}{2} = 138\frac{1}{2} \); for \( 98 \times 14 = 1372 \) \[ \begin{array}{c} + 10 \\ \hline \text{Numerator } 1382 \end{array} \]

Problem III. To reduce a whole number to a fraction of a given denominator.

Rule. Multiply the whole number by the given denominator; and place the product by way of numerator over the given denominator.

Examples.

1. Reduce 9 to a fraction whose denomination is 5. \[ 9 \times 5 = 45; \text{ so the fraction is } \frac{45}{5}. \]

2. Reduce 36 to a fraction whose denominator is 4. \[ 36 \times 4 = 144; \text{ so the fraction is } \frac{144}{4}. \]

3. Reduce 8 to a fraction whose denominator is 1. \[ 8 \times 1 = 8; \text{ so the fraction is } \frac{8}{1}. \]

The reason of the rule appears by reversing the operation; for if the numerator be divided by the denominator, it will quot the integer, or whole number.

Problem IV. To reduce a compound fraction to a simple one.

Rule. Multiply the numerators continually for the numerator of the simple fraction; and multiply the denominators continually for its denominator.

Examples.

Ex. 1. \( \frac{3}{4} \) of \( \frac{4}{5} = \frac{8}{5} \). Ex. 2. \( \frac{1}{2} \) of \( \frac{3}{4} \) of \( \frac{1}{4} = \frac{6}{32} \).

Cor. From this problem may be deduced a method of reducing a fraction of a lesser denomination to a fraction of a greater denomination; namely,

Form a compound fraction, by comparing the given fraction with the superior denominations; and then reduce the compound fraction to a simple one.

Examples.

1. What fraction of a pound Sterling is \( \frac{3}{4} \) of a penny? \[ \frac{3}{4} \text{ d. is } \frac{1}{4} \text{ of } \frac{1}{12} \text{ of } \frac{1}{10} \text{ L. } = \frac{3}{480} \text{ L. } \]

2. What fraction of a C. is \( \frac{7}{8} \) of a pound? \[ \frac{7}{8} \text{ lb. is } \frac{7}{8} \text{ of } \frac{1}{2} \text{ of } \frac{1}{4} \text{ C. } = \frac{7}{32} \text{ C. } \]

Problem V. To reduce a fraction of a greater denomination to a fraction of a lesser denomination.

Rule. Multiply the numerator of the given fraction, as in reduction of integers descending; and the product is the numerator, to be placed over the denominator of the given fraction.

Examples.

1. What fraction of a shilling is \( \frac{3}{4} \) of a pound?

Here, as in reduction descending, multiply the numerator 3 by 20, because 20 shillings make a pound; as under.

\[ \frac{3 \times 20}{4} = \frac{60}{4} \text{ shilling. } \]

2. What fraction of a penny is \( \frac{4}{5} \) L.?

\[ \frac{4 \times 20 \times 12}{5} = \frac{960}{5} \text{ d. } \]

The reason of this rule will appear by observing, that every fraction may be considered in two views. Thus, \( \frac{1}{4} \) may either be considered as expressing three fourths of one unit, or as denoting the fourth part of three units.

Now, if the unit be a pound Sterling, the fraction, in the latter view, will denote the fourth part of three pounds; and by reducing the numerator L. 3 to shillings, we have \( \frac{60}{4} \) s.; and again reducing 60 shillings to pence, we have \( \frac{720}{4} \) d. equal to \( \frac{60}{4} \) s., or to \( \frac{3}{4} \) L.

Problem VI. To find the value of a fraction.

Rule. Reduce the numerator to the next inferior denomination; divide by the denominator; and the quot, if nothing remain, is the value complete.

If there be any remainder, it is the numerator of a fraction whose denominator is the divisor. This fraction may either be annexed to the quotient, or reduced to value, if there be any lower denomination.

Examples. **ARITHMETICK**

**Examples.**

What is the value of \( \frac{3}{20} \) L.?

Here consider \( \frac{3}{20} \) L. as expressing the fourth part of three pounds Sterling; so reduce L 3, the numerator, to shillings, and divide by the denominator 4; and as nothing remains, the quot, viz. 15 shillings, is the value complete.

The reason of this rule is the same with that in the preceding problem. It is by the practice of this problem that remainders in the rule of three are reduced to value.

**Prob. VII.** To reduce a fraction to its lowest terms.

**Rule.** Divide both numerator and denominator by their greatest common divisor; the two quots make the new fraction.

The greatest common divisor of the numerator and denominator of a fraction is found by the following

**Rule.** Divide the greater of these two numbers by the lesser; and again divide the divisor by the remainder; and so on, continually, till 0 remains. The last divisor is their greatest common divisor.

**Example.** Reduce \( \frac{784}{952} \) to its lowest terms.

First find the greatest common divisor of the numerator and denominator, as follows:

\[ \begin{array}{c} 784 \\ \hline 168 \quad (4) \\ \hline 672 \\ \hline 112 \quad (1) \\ \end{array} \]

Greatest common divisor = 112

Then proceed to reduce the given fraction to its lowest terms, by dividing both numerator and denominator by 56, the greatest common divisor.

\[ \begin{array}{c} 56 \quad (14 \text{ new num.}) \\ \hline 56 \\ \hline 224 \\ \hline 392 \\ \end{array} \]

So \( \frac{224}{392} = \frac{1}{2} \).

**Prob. VIII.** To reduce fractions of different denominators to a common denominator.

**Rule.** Multiply the denominators continually for the common denominator; and multiply each numerator into all the denominators, except its own, for the several numerators.

**Examples.**

Reduce \( \frac{1}{4} \) and \( \frac{3}{5} \) to a common denominator.

\( 4 \times 5 = 20 \), the common denominator.

\( 3 \times 5 = 15 \), the first numerator.

\( 4 \times 4 = 16 \), the second numerator.

So the new fractions are \( \frac{15}{20} \) and \( \frac{16}{20} \).

When the denominator of one fraction happens to be an aliquot part of the denominator of another fraction, the former may be reduced to the same denominator with the latter, by multiplying both its numerator and denominator by the number which denotes how often the lesser denominator is contained in the greater.

Thus, \( \frac{3}{4} + \frac{1}{8} = \frac{6}{8} + \frac{1}{8} = \frac{7}{8} \).

Here 3 is contained in 12 four times; so multiply both 2 and 3 by 4, and you have \( \frac{6}{8} + \frac{1}{8} = \frac{7}{8} \).

Again, \( \frac{1}{4} + \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} + \frac{1}{8} = \frac{4}{8} \).

Sometimes too, the fraction that has the greater denominator may, in like manner, be reduced to the same denominator with that which has the lesser, by division.

Thus, \( \frac{3}{4} + \frac{1}{8} = \frac{6}{8} + \frac{1}{8} = \frac{7}{8} \).

And \( \frac{4}{8} + \frac{1}{8} + \frac{1}{8} = \frac{4}{8} + \frac{1}{8} + \frac{1}{8} = \frac{6}{8} \).

The reason of the above rule for reducing fractions to a common denominator is evident from Corollary II.; for both numerator and denominator of every fraction are multiplied by the same number, or by the same numbers.

After fractions are reduced to a common denominator, they may frequently be reduced to lower terms, by dividing all the numerators, and also the common denominator, by any divisor that leaves no remainder, or by cutting off an equal number of ciphers from both.

**Addition of Vulgar Fractions.**

**Rule I.** If the given fractions have all the same denominator, add the numerators, and place the sum over the denominator.

**Ex. 1.** What is the sum of \( \frac{1}{2} + \frac{1}{4} \)? **Ans.** \( \frac{3}{4} \).

**2.** What is the sum of \( \frac{1}{3} + \frac{1}{3} \)? **Ans.** \( \frac{2}{3} \), by Prob. VII.

**Rule II.** If the given fractions have different denominators, reduce them to a common denominator, by Prob. VIII., then add the numerators, and place the sum over the common denominator.

**Ex.** What is the sum of \( \frac{1}{2} + \frac{1}{3} \)?

\( \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \), by Prob. VIII.

and \( \frac{1}{6} + \frac{1}{6} = \frac{2}{6} \).

**Rule III.** If mixt numbers be given, or if mixt numbers and fractions be given, reduce the mixt numbers to improper fractions, by Prob. II.; then reduce the fractions to a common denominator, by Prob. VIII., and add the numerators.

**Ex.** What is the sum of \( \frac{7}{4} + \frac{5}{3} \)?

\( \frac{7}{4} + \frac{5}{3} = \frac{21}{12} + \frac{20}{12} = \frac{41}{12} \), by Prob. II.

and \( \frac{1}{4} + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12} \), by Prob. VIII.

and \( \frac{3}{12} + \frac{4}{12} = \frac{7}{12} \), by Prob. I.

When mixt numbers, or mixt numbers and fractions, are given, you may, with greater expedition, work by the following rule, viz. reduce only the fractions to a common denominator, and add the sum of the fractions. to the integers. The above example wrought in this manner follows.

Ex. What is the sum of \( \frac{7}{4} + \frac{5}{2} \)?

\[ \frac{3}{4} + \frac{5}{2} = \frac{7}{4} + \frac{10}{4} = \frac{17}{4} \]

and \( 7 + 5 + 1 = 13 \).

Rule IV. If any, or all of the given fractions, be compound, first reduce the compound fractions to simple ones, by Prob. IV.; then reduce the simple fractions to a common denominator, by Prob. VIII., and add the numerators.

Ex. What is the sum of \( \frac{3}{4} \) of \( \frac{4}{7} + \frac{1}{4} \)?

\[ \frac{3}{4} \times \frac{4}{7} = \frac{3}{7}, \text{ by Prob. IV.} \]

and \( \frac{3}{7} + \frac{1}{4} = \frac{12}{28} + \frac{7}{28} = \frac{19}{28}, \text{ by Prob. VIII.} \]

and \( \frac{19}{28} + \frac{1}{4} = \frac{19}{28} + \frac{7}{28} = \frac{26}{28} = \frac{13}{14}, \text{ by Prob. I.} \)

Rule V. If the given fractions be of different denominations, first reduce them to the same denomination, by Cor. of Prob. IV. or by Prob. V.; then reduce the fractions, now of one denomination, to a common denominator, by Prob. VIII., and add the numerators; or reduce each of the given fractions separately to value, by Prob. VI., and then add their values.

Ex. What is the sum of \( \frac{3}{4} s. \) and \( \frac{1}{4} l. \)?

Method I.

\[ \frac{3}{4} s. = \frac{3}{4} \times \frac{1}{4} l. = \frac{3}{16} l., \text{ by Cor. Prob. IV.} \]

and \( \frac{3}{16} l. + \frac{1}{4} l. = \frac{3}{16} l. + \frac{4}{16} l. = \frac{7}{16} l., \text{ by Prob. VIII.} \]

and \( \frac{7}{16} l. + \frac{1}{4} l. = \frac{7}{16} l. + \frac{4}{16} l. = \frac{11}{16} l., \text{ by Prob. VI.} \)

Method II.

\[ \frac{3}{4} l. = \frac{7 \times 20}{8} s. = \frac{70}{8} s., \text{ by Prob. V.} \]

and \( \frac{70}{8} s. + \frac{1}{4} l. = \frac{70}{8} s. + \frac{2}{8} s. = \frac{72}{8} s., \text{ by Prob. VIII.} \]

and \( \frac{72}{8} s. + \frac{1}{4} l. = \frac{72}{8} s. + \frac{2}{8} s. = \frac{74}{8} s., \text{ by Prob. I. and VI.} \)

Method III.

\[ \frac{3}{4} s. = \frac{3 \times 12}{4} d. = \frac{36}{4} d. = 9 \]

\[ \frac{3}{4} l. = \frac{7 \times 20}{8} s. = \frac{140}{8} s. = 17 - 6 \]

Subtraction of Vulgar Fractions.

Rule I. If the given fractions have the same denominator, subtract the lesser numerator from the greater, and place the remainder over the denominator.

Ex. From \( \frac{7}{4} \) subtract \( \frac{3}{4} \).

\[ \frac{7}{4} - \frac{3}{4} = \frac{4}{4} = 1 \]

Rule II. If the given fractions have different denominators, reduce them to a common denominator, by Prob. VIII.; then subtract the lesser numerator from the greater, and place the remainder over the common denominator.

Ex. From \( \frac{3}{4} \) subtract \( \frac{1}{2} \).

\[ \frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4} \]

Rule III. If it be required to subtract one mixt number from another, or to subtract a fraction from a mixt number, reduce the mixt numbers to improper fractions, by Prob. II.; then reduce the fractions to a common denominator, by Prob. VIII., and subtract the one numerator from the other.

Ex. From \( \frac{7}{4} \) subtract \( \frac{5}{2} \).

\[ \frac{7}{4} - \frac{5}{2} = \frac{7}{4} - \frac{10}{4} = \frac{3}{4} \]

and \( \frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4} \), by Prob. II.

and \( \frac{5}{4} - \frac{1}{2} = \frac{5}{4} - \frac{2}{4} = \frac{3}{4} \), by Prob. VIII.

and \( \frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4} \), by Prob. I. and VII.

Rule IV. If it be required to subtract a mixt number, or a fraction, from an integer, first subtract the fraction from an unit borrowed; that is, subtract the numerator from the denominator, and place the remainder, as a numerator, over the denominator, for the fractional part of the answer: Then, for the unit borrowed, add 1 to the integral part of the mixt number; subtract the sum from the given integer; and prefix the remainder to the fractional part of the answer. But when a fraction is subtracted from an integer, for the unit borrowed, take 1 from the given integer, and prefix the remainder to the fractional part of the answer.

Ex. 1. From 14 subtract \( \frac{7}{4} \).

Here say, \( 5 - 3 = 2 \); so \( \frac{2}{4} \) is the fractional part of the answer: Then say, 1 borrowed and 7 make 8, and 8 subtracted from 14 leaves 6; which prefix to the fractional part: So the difference or answer is \( 6 \frac{2}{4} \).

Ex. 2. From 12 subtract \( \frac{4}{4} \).

Here say, \( 7 - 3 = 4 \); so \( \frac{4}{4} \) is the fractional part; then say, 1 borrowed from 12, and 11 remains: So \( 11 \frac{4}{4} \) is the difference, or answer.

Note. When an integer is given to be subtracted from a mixt number, you have only to subtract the given integer from the integral part of the mixt number; and to the remainder annex the fractional part. Thus,

\[ \frac{9}{4} - 5 = 4 \frac{1}{4} \]

Rule V. If one or both of the given fractions be compound, first reduce the compound fractions to simple ones, by Prob. IV.; then reduce the simple fractions to a common denominator, by Prob. VIII.; and subtract the one numerator from the other.

Ex. From \( \frac{7}{4} \) subtract \( \frac{3}{4} \) of \( \frac{2}{4} \).

\[ \frac{7}{4} \times \frac{3}{4} = \frac{21}{16}, \text{ by Prob. IV.} \]

and \( \frac{21}{16} - \frac{3}{4} = \frac{21}{16} - \frac{12}{16} = \frac{9}{16}, \text{ by Prob. VIII.} \]

and \( \frac{9}{16} - \frac{3}{4} = \frac{9}{16} - \frac{12}{16} = \frac{3}{16}, \text{ by Prob. VII.} \)

Rule VI. When the given fractions are of different denominations, first reduce them to the same denomination, by Cor. of Prob. IV. or by Prob. V., then reduce the fractions, now of one denomination, to a common denominator, by Prob. VIII.; and subtract the one numerator from the other. Or, reduce each of the given fractions, separately, to value, by Prob. VI.; and subtract the one value from the other.

Ex. From \( \frac{3}{4} l. \) subtract \( \frac{7}{4} s. \)

Method I.

\[ \frac{3}{4} s. = \frac{3}{4} \times \frac{1}{4} l. = \frac{3}{16} l., \text{ by Cor. Prob. IV.} \]

and \( \frac{3}{16} l. + \frac{1}{4} l. = \frac{3}{16} l. + \frac{4}{16} l. = \frac{7}{16} l., \text{ by Prob. VIII.} \]

and \( \frac{7}{16} l. + \frac{1}{4} l. = \frac{7}{16} l. + \frac{4}{16} l. = \frac{11}{16} l., \text{ by Prob. I. and VI.} \)

Method II.

\[ \frac{3}{4} l. = \frac{7 \times 20}{8} s. = \frac{70}{8} s., \text{ by Prob. V.} \]

and \( \frac{70}{8} s. + \frac{1}{4} l. = \frac{70}{8} s. + \frac{2}{8} s. = \frac{72}{8} s., \text{ by Prob. VIII.} \]

and \( \frac{72}{8} s. + \frac{1}{4} l. = \frac{72}{8} s. + \frac{2}{8} s. = \frac{74}{8} s., \text{ by Prob. I. and VI.} \)

Method Multiplication of Vulgar Fractions.

In multiplication of fractions there is no occasion to reduce the given fractions to a common denominator, as in addition and subtraction: only if a mixt number be given, reduce it to an improper fraction; if an integer be given, reduce it to an improper fraction, by putting an unit for its denominator; if a compound fraction be given, you may either reduce it to a simple one, or, instead of the particle of, infert the sign of multiplication: then work by the following

Rule. Multiply the numerators for the numerator of the product, and multiply the denominators for its denominator.

Example 1. \( \frac{3}{4} \times \frac{4}{5} = \frac{12}{20} \)

Note 1. If any number be multiplied by a proper fraction, the product will be less than the multiplicand; for multiplication is the taking of the multiplicand as often as the multiplier contains unity; and consequently, if the multiplier be greater than unity, the product will be greater than the multiplicand; if the multiplier be unity, the product will be equal to the multiplicand; and if the multiplier be less than unity, the product will, in the same proportion, be less than the multiplicand. Thus, supposing the multiplier to be \( \frac{1}{2} \) or \( \frac{1}{3} \), the product, in this case, will be equal to one half or to one third of the multiplicand.

2. Mixt numbers may be multiplied without reducing them to improper fractions, by working as in the margin; where, first multiply the integral parts, viz. \( \frac{54}{24} \) by \( \frac{24}{24} \); then multiply the integral parts cross-ways into their alternate fractions, viz. \( \frac{54}{24} \times \frac{24}{24} \) and the product set down; in like manner multiply \( \frac{24}{24} \) by \( \frac{24}{24} \) and the product likewise set down; then add; and to the sum annex \( \frac{1}{2} \), the product of the two fractions.

3. In multiplying a fraction by an integer, you have only to multiply the numerator by the integer, the putting one for the denominator being only matter of form. And to multiply a fraction by its denominator is to take away the denominator, the product being an integer, the same with, or equal to the numerator. Thus, \( \frac{7}{8} \times 8 = 7 \). For \( \frac{7}{8} \times 8 = \frac{56}{8} = 7 \).

4. If the numerators and denominators of two equal fractions be multiplied cross-ways, the products will be equal. Thus, if \( \frac{3}{4} = \frac{4}{3} \), then will \( 3 \times 12 = 9 \times 4 \); for multiplying both by 9, we have \( \frac{3}{4} \times \frac{9}{4} = \frac{9}{12} \); and multiplying these by 12, we have \( 3 \times 12 = 9 \times 4 \). Hence, if four numbers be proportional, the product of the extremes will be equal to the product of the means: for if \( 3 : 9 :: 4 : 12 \), then \( \frac{3}{4} = \frac{9}{12} \); and it has been proved, that \( 3 \times 12 = 9 \times 4 \). Therefore, if, of four proportional numbers, any three be given, the fourth may easily be found, viz. when one of the extremes is sought, divide the product of the means by the given extreme; and when one of the means is sought, divide the product of the extremes by the given mean.

5. In multiplying fractions, equal factors above and below may be divided or dropped. Thus, \( \frac{3}{4} \times \frac{4}{3} \times \frac{4}{3} \times \frac{3}{4} \); and dropping the factors 2, 3, 4, both above and below, the product is \( \frac{1}{2} \). In like manner, to facilitate an operation, a factor above and another below may be divided by the same number: Thus, \( \frac{6}{7} \times \frac{7}{6} = \frac{6}{7} \times \frac{7}{6} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \). Or we may exchange one numerator for another: Thus, \( \frac{6}{7} \times \frac{7}{6} = \frac{6}{7} \times \frac{7}{6} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).

6. To take any part of a given number, is to multiply the said number by the fraction. Thus, \( \frac{5}{6} \) of 320 is found thus, \( \frac{5}{6} \times \frac{1}{2} = \frac{5}{6} \times \frac{1}{2} = \frac{5}{6} \times \frac{1}{2} = \frac{5}{6} \times \frac{1}{2} = 200 \). In like manner, \( \frac{5}{6} \) of 45\( \frac{1}{2} \) is \( \frac{5}{6} \times \frac{1}{2} = \frac{5}{6} \times \frac{1}{2} = \frac{5}{6} \times \frac{1}{2} = \frac{5}{6} \times \frac{1}{2} = 30 \). Hence, to reduce a compound fraction to a simple one, is to multiply the parts of it into one another.

7. If a multiplicand of two or more denominations be given to be multiplied by a fraction, reduce the higher part or parts of the multiplicand to the lowest species, and then multiply. Thus, to multiply 8 l. 10 s. by \( \frac{3}{4} \), say, 8 l. = 8 x 20 s. = 160 s. and 160 + 10 s. = 170 s. \( \frac{3}{4} \), and \( \frac{3}{4} \times \frac{6}{4} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{2} \times \frac{1}{2} = 113 \frac{1}{2} \) s. L. 5 : 13 : 10. Or, without reducing, you may multiply the given multiplicand by the numerator of the fraction, and divide the product by the denominator.

Example 1. Multiply \( \frac{7}{8} \) by \( \frac{9}{7} \). Prod. \( \frac{63}{56} \)

2. Multiply \( \frac{7}{9} \) by \( \frac{4}{7} \). Prod. \( \frac{28}{63} \)

3. Multiply \( \frac{8}{3} \) by \( \frac{9}{4} \). Prod. \( \frac{72}{12} \)

The reason of the rule may also be shewn thus: Assume two fractions equal to two integers, such as, \( \frac{9}{2} \), and \( \frac{9}{2} \), equal to 2 and 3, and the product of the fractions will be equal to the product of the integers; for \( \frac{9}{2} \times \frac{9}{2} = \frac{81}{4} = 6 \), and \( 2 \times 3 = 6 \).

Division of Vulgar Fractions.

In division of fractions, if a mixt number be given, reduce it to an improper fraction; if an integer be given, put an unit for its denominator; if a compound fraction be given, reduce it to a simple one, and then work by the following

Rule. Multiply cross-ways; viz. the numerator of the divisor into the denominator of the dividend, for the denominator of the quot; and the denominator of the divisor into the numerator of the dividend, for the numerator of the quot.

Example 1. \( \frac{3}{4} \times \frac{4}{3} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{2} \)

2. \( \frac{4}{3} \times \frac{3}{4} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{2} \)

3. \( \frac{4}{3} \times \frac{3}{4} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{2} \)

Note. Note 1. Instead of working division of fractions as taught above, you may invert the divisor, and then multiply it into the dividend. Thus, in Example 1, instead \( \frac{3}{4} \times \frac{5}{6} = \frac{15}{24} = \frac{5}{8} \).

2. If any number be divided by a proper fraction, the quot will be greater than the dividend: for in division the quot shews how often the divisor is contained in the dividend; and consequently if the divisor be greater than unity, the quot will be less than the dividend; if the divisor be unity, the quot will be equal to the dividend; and if the divisor be less than unity, the quot will, in the same proportion, be greater than the dividend. Thus, supposing the divisor to be \( \frac{1}{2} \), or \( \frac{1}{3} \), the quot in this case will be double or triple of the dividend.

3. To divide a fraction by an integer, is only to multiply the integer into the denominator of the fraction, the numerator being continued. Thus, \( \frac{7}{9} \div \frac{3}{4} = \frac{28}{27} \).

4. A mixt number may sometimes be divided by an integer, with more ease, in the following manner. Divide the integral part of the mixt number by the given integer: and if there be no remainder, divide likewise the fraction of the mixt number by the given integer, and annex the quot to the integral quot formerly found. But if, in dividing the integral part, there happen to be a remainder, prefix this remainder to the fraction for a new mixt number; which reduce to an improper fraction: then divide the improper fraction by the given integer, and annex the quot to the integral quot formerly found. Thus, if it be required to divide \( \frac{15}{4} \) by 8, say, 8 \( \times \frac{1}{4} \), and 7 remains; which 7, prefixed to the fraction, gives \( \frac{7}{4} \) for a new mixt number; and this, reduced to an improper fraction, is \( \frac{15}{4} \); and 8 \( \times \frac{1}{4} \) \( \times \frac{15}{4} \): so the complete quot is \( \frac{15}{4} \).

5. If the factors of the numerator and denominator of the quot, instead of being actually multiplied, be only connected with the sign of multiplication, it will be easy to drop such factors, above and below, as happen to be the same, thus: \( \frac{4}{5} \times \frac{3}{4} \times \frac{5}{8} = \frac{3}{8} \). Or a factor above and below may be divided by the same number thus: \( \frac{6}{5} \times \frac{7}{5} \times \frac{5}{2} = \frac{7}{2} \). Or the factors of the numerator of the quot may be exchanged, thus: \( \frac{3}{5} \times \frac{5}{2} \times \frac{5}{2} = \frac{5}{2} \).

6. To divide an integer by a fraction, is to divide the product of the denominator and integer by the numerator, thus: \( \frac{4}{5} \times \frac{8}{4} = \frac{5}{2} \times 2 = 10 \).

7. If the divisor and dividend have the same denominator, you have only to divide the numerator of the dividend by the numerator of the divisor, thus: \( \frac{4}{5} \times \frac{5}{2} = \frac{1}{2} \); for \( \frac{8}{5} \times \frac{3}{5} = \frac{3}{5} \).

8. If a dividend of two or more denominations be given to be divided by a fraction, reduce the higher part or parts of the dividend to the lowest species, and then divide. Thus, to divide 61.9s. by \( \frac{2}{3} \), say, 61 \( \times \frac{2}{3} = 120 \); and 120 \( + \frac{9}{2} = 129 \frac{1}{2} s. = \frac{5}{4} \); and \( \frac{5}{4} \times \frac{7}{8} = \frac{35}{32} = 91.14s.7\frac{1}{2}d \).

Or, Divide the given multiplicand by the numerator of the fraction, and multiply the quot by the denominator.

Example. Divide L. 276 : 16 : 8 among four men, A, B, C, D, so that A, B, C, may have equal shares, and D only two thirds of one of their shares.

\[ \begin{array}{cccc} L. & s. & d. & L. & s. & d. & L. & s. & d. \\ 11 & 276 & 16 & 8 & (25 & 3 & 4 & 3 & 75 & 10 & A. \\ & & & & & & & & 3 & 75 & 10 & B. \\ & & & & & & & & 3 & 75 & 10 & C. \\ & & & & & & & & 2 & 50 & 6 & 8 & D. \end{array} \]

Proof 276 16 8

The reason of the rule will appear by considering, that the method here used is nothing else but the reducing the divisor and dividend to a common denominator, and then dividing the one numerator by the other. Thus, \( \frac{2}{3} \times \frac{3}{5} = \frac{2}{5} \).

The truth of the rule may also be proved by assuming two fractions equal to two integers, such as, \( \frac{5}{6} \) and \( \frac{1}{6} \), equal to 2 and 4, and the quot of the fractions will be equal to the quot of the integers. Thus, \( \frac{5}{6} \times \frac{1}{6} = \frac{2}{4} \), and 2 \( \times \frac{1}{4} = 2 \).

The Simple Rule of Three in Vulgar Fractions.

The question is stated as formerly taught in the rule of three. The extremes must be of one denomination. Reduce mixt numbers and integers to improper fractions, compound fractions to simple ones, and then work by the following rule, viz.

Multiply the second and third terms, and divide the product by the first term; that is, multiply the numerator of the first term into the denominators of the second and third, for the denominator of the answer; and multiply the denominator of the first term into the numerators of the second and third, for the numerator of the answer.

I. Direct.

Quest. If \( \frac{3}{4} \) yard cost \( \frac{1}{2} \) l., what will \( \frac{2}{3} \) yard cost?

\[ \begin{array}{ccc} Yd. & L. & Yd. \\ \frac{3}{4} & : & \frac{1}{2} \\ \frac{2}{3} & : & ? \end{array} \]

Ans. \( \frac{4}{3} \times \frac{5}{8} \times \frac{9}{10} = \frac{9}{10} \times \frac{3}{2} \times \frac{2}{2} = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4} \times \frac{20}{4} = \frac{6}{8} = 15s. \)

II. Inverse.

Quest. If \( \frac{3}{4} \) yard of cloth that is 2 yards wide, will make a garment, how much of any other cloth that is \( \frac{1}{2} \) yard wide will make the same garment?

Bread. len. Bread.

\[ \begin{array}{ccc} \frac{3}{4} & : & \frac{1}{2} \\ \frac{1}{2} & : & ? \end{array} \]

Ans. \( \frac{5}{3} \times \frac{3}{2} \times \frac{2}{5} = \frac{5}{2} \times \frac{2}{4} = 2\frac{1}{2} \text{ yards.} \) The Compound Rule of Three in Vulgar Fractions.

QUEST. If \( \frac{1}{4} \) acre of grass be cut down by 2 men in \( \frac{1}{4} \) day, how many acres shall be cut down by 6 men in \( 3 \frac{1}{3} \) days?

Men. acr. men. \( \frac{3}{4} : \frac{1}{4} :: \frac{6}{5} \) day \( \frac{3}{4} : \frac{1}{4} :: \frac{9}{5} \) days.

Ans. \( \frac{3 \times 3 \times 60}{4 \times 4 \times 3} = \frac{3 \times 60}{4 \times 4} = \frac{3 \times 15}{4} = 11 \frac{1}{4} \) acr.

Or thus:

Ans. \( \frac{3 \times 1 \times 3 \times 6 \times 10}{2 \times 2 \times 4 \times 1 \times 3} = \frac{3 \times 6 \times 10}{2 \times 2 \times 4} = \frac{3 \times 3 \times 5}{2 \times 2} = \frac{43}{4} = 11 \frac{1}{4} \) acres.

Chap. X. Rules of Practice.

When the first term of a question in the rule of three happens to be unity, the answer may frequently be found more speedily and easily than by a formal stating or working of the rule of three; and the directions to be observed in such operations are called Rules of Practice.

The rules of practice naturally follow the doctrine of vulgar fractions, the operation being nothing else but a multiplying the number whose price is required, by such a fraction of a pound, of a shilling, or of a penny, as denotes the rate or price of one.

Thus, if the price of 24 yards, at 6s. 8d. per yard, be demanded, the answer is found by multiplying 24 by \( \frac{1}{4} \), the fraction of a pound equivalent to 6s. 8d., viz. \( \frac{1}{4} \times \frac{7}{4} = \frac{24}{4} = 6 \).

Hence, it is obvious, that to multiply a number by a fraction whose numerator is unity, is to divide the said number by the denominator of the fraction. But if the numerator of the fraction be not unity, you must first multiply the given number by the numerator, and then divide the product by the denominator. Thus, if the rate be 13s. 4d., the price of 24 yards is found by saying, \( \frac{13}{4} \times \frac{3}{4} = \frac{39}{4} = 16 \frac{1}{4} \); or take \( \frac{3}{4} \) of the given number twice.

When the fraction denoting the rate happens to be compound, the product or answer is found by dividing the given number by one of the denominators of the compound fraction, the quot by another, and the next quot by the third, &c. Thus, if the rate be 2 farthings \( = \frac{1}{2} \) of \( \frac{1}{2} \) of \( \frac{1}{2} \) l., the price of 1440 yards is found by saying, \( \frac{1440}{2} = 720 \), and \( \frac{720}{2} = 360 \), and \( \frac{360}{2} = 180 \).

When the rate is expressed by two or more simple fractions, connected with the sign +, the product or answer is found by dividing the given number successively by the several denominators, and then adding the quot. Thus, if the rate be 3s. \( = \frac{1}{3} + \frac{1}{3} \), the price of 80 yards is found by saying, \( \frac{80}{3} = 26 \frac{2}{3} \), and \( \frac{26}{3} = 8 \), and \( \frac{8}{3} = 2 \frac{2}{3} \).

The fractions equivalent to any number of farthings under 4, to any number of pence under 12, and to any number of shillings under 20, are exhibited in the following tables.

| Table I. | Table II. | Table III. | |----------|-----------|------------| | Farthings | of a penny | of a shilling | of a pound | | 1 | \( \frac{1}{4} \) | \( \frac{1}{4} \) | \( \frac{1}{4} \) | | 2 | \( \frac{1}{2} \) | \( \frac{1}{2} \) | \( \frac{1}{2} \) | | 3 | \( \frac{3}{4} \) | \( \frac{3}{4} \) | \( \frac{3}{4} \) | | 4 | \( \frac{1}{4} \) | \( \frac{1}{4} \) | \( \frac{1}{4} \) | | 5 | \( \frac{1}{5} \) | \( \frac{1}{5} \) | \( \frac{1}{5} \) | | 6 | \( \frac{1}{6} \) | \( \frac{1}{6} \) | \( \frac{1}{6} \) | | 7 | \( \frac{1}{7} \) | \( \frac{1}{7} \) | \( \frac{1}{7} \) | | 8 | \( \frac{1}{8} \) | \( \frac{1}{8} \) | \( \frac{1}{8} \) | | 9 | \( \frac{1}{9} \) | \( \frac{1}{9} \) | \( \frac{1}{9} \) | | 10 | \( \frac{1}{10} \) | \( \frac{1}{10} \) | \( \frac{1}{10} \) | | 11 | \( \frac{1}{11} \) | \( \frac{1}{11} \) | \( \frac{1}{11} \) |

The fractions in Table II. become compound fractions of a pound, by annexing (of \( \frac{1}{2} \)) to each of them. Thus, 1d. is \( \frac{1}{2} \) of \( \frac{1}{2} \) l.; and 5d. is \( \frac{1}{2} \) of \( \frac{1}{2} \) of \( \frac{1}{2} \) l., &c.

The variety that occurs in the rules of practice arises chiefly from the different rates, or prices, of one thing, as a yard, a pound, an ounce, &c. and may be reduced to the eight cases following, viz.

1. Farthings under four. 2. Pence under twelve. 3. Pence and farthings. 4. Shillings under twenty. 5. Shillings, pence, and farthings. 6. Pounds. 7. Pounds, shillings, pence, and farthings. 8. The given number may consist of integers and parts.

Case I. When the rate is farthings, under four.

Rule. Divide the given number by the denominator of the fraction denoting the rate, as contained in Tab. I! viz. if the rate be 1 or 2 farthings, divide by 4 or 2, the quot will be pence; and the remainder, in dividing by 4, will be farthings, and in dividing by 2, it will be 1 halfpenny: then divide the pence by 12, the quot will be shillings, and the remainder pence. Lastly, divide the shillings by 20, the quot will be pounds, and the remainder shillings. But if the rate is 3 farthings, first multiply the given number by the numerator 3, and then divide as above directed.

| Ex. 1. | Ex. 2. | |--------|--------| | \( \frac{1}{4} \times 4859 \), at 1 f. | \( \frac{1}{4} \times 8347 \), at 2 f. | | \( \frac{1}{2} \times 1214 \), 3 f. | \( \frac{1}{2} \times 4173 \), 4 d. | | \( \frac{1}{2} \times 101 \), 2 d. | \( \frac{1}{2} \times 317 \), 9 d. | | L.5 1 2½ | L.17 7 9½ |

Case II. When the rate is pence, under twelve.

Rule. Divide the given number by the denominator of the fraction denoting the rate, as contained in Table II, and you have the answer in shillings; which reduce into pounds, by dividing by 20. Note, The remainders at the first division in the above examples are the same with the rate. Thus, in Ex. 1. every remainder is 1 d.

Case III. When the rate is pence and farthings.

Rule. The pence must be some aliquot part of a shilling; and, at the same time, the farthings some aliquot part of the pence; and if they be not so given, divide the pence into two or more such parts, so as the farthings may be some aliquot part of the lowest division of the pence. Then, beginning with the highest division of the pence, divide by the denominators of the fractions denoting the aliquot parts.

| Ex. 1. | Ex. 2. | |-------|-------| | 532, at 1½ d. | 1753, at 1½ d. | | 44—4 | 146—1 | | 11—1 | 73—0½ | | 5½—5 | 21½—1½ |

Explication.

In Ex. 1. work first for 1 d.; which being ¼ s., divide the given number by the denominator 12, and the quotient is shillings, and the remainder pence; then, because 1 farthing is ¼ d., divide the former quotient by 4, and the sum of the quotients is the price in shillings; which divide by 20.

In Ex. 2. the rate 1½ d. being an aliquot part of a shilling, the second method is shorter and better than the first.

Case IV. When the rate is shillings under twenty.

Rule. Multiply the given number by the numerator of the fractions contained in Tab. III. and divide the product by the denominators. Or, instead of this general rule, take the two particular ones following.

1. If the rate be an even number of shillings, multiply the given number by half the number of shillings in the rate, always doubling the right-hand figure of the product for shillings, and the rest are pounds.

2. If the rate be an odd number of shillings, work for the next lesser even number of shillings, as above; and for the odd shilling take ½ of the given number.

Examp. I. When the rate is an even number of shillings.

| Ex. 1. | Ex. 2. | |-------|-------| | 436, at 2 s. | 127, at 4 s. | | 1 | 2 | | L. 43, 12 s. | L. 25, 8 s. |

2. When the rate is an odd number of shillings.

| Ex. 1. | Ex. 2. | |-------|-------| | 635, at 1 s. | 422, at 3 s. | | 6 | 4 | | L. 31, 15 s. | L. 63, 6 s. |

Note 1. The reason of multiplying by half the number of shillings in the rate will appear by considering, that these are the numerators of the fractions denoting the rate. Thus, 2 s. is ¼ l. and 4 s. is ½ l. and 6 s. is ¾ l. and each unit in the product is two shillings. The division by the denominator 10 is performed by cutting off the right-hand figure of the product, and the figure so cut off is the remainder; and as each unit in the remainder is two shillings, the double of them is the remainder in shillings.

Note 2. From Ex. 1. we may learn, that when the rate is 2 s., the price is found by doubling the right-hand figure of the given number for shillings, and the other figure or figures are pounds.

Note 3. In Ex. 2. the price may also be had by taking ½ of the given number; and in this way every remainder will be 4 s.

Note 4. By reversing the operation, from the price and any even rate given, we may readily find the quantity of goods, viz. Multiply the price by 10, that is, to the price annex a cipher, and divide the product by half the rate.

Ex. 1. How many yards, at 14 s. may be bought for 49 l. 7)490(70 yards. Ans.

Ex. 2. How many gallons, at 8 s. may be bought for 500 l.? 4)5000(1250 gallons. Ans.

Case V. When the rate is shillings and pence, or shillings, pence, and farthings.

Rule I. If the rate be shillings and pence which make an aliquot part of a pound, divide the given number by the denominator of the fraction denoting the rate; the quotient is pounds, and each unit of the remainder is equal to the rate.

| Ex. 1. | Ex. 2. | |-------|-------| | 354, at 1 s. 8 d. | 443, at 2 s. 6 d. | | L. 29, 10. | L. 55 7 6 |

Rule II. If the rate be no aliquot part of a pound, but may be divided into such parts, divide it accordingly; work for the parts separately, and then add.

| Ex. 1. | Ex. 2. | |-------|-------| | 427, at 8 s. | 540, at 5 s. | | 6 d. | 4 d. | | 128 2 | 3 s. 4 d. | | 53 7 6 | 2 s. 10 | | L. 181 9 6 | L. 144 |

Rule III. **ARITHMETICK**

**Rule III.** If the rate be no aliquot part of a pound, and cannot readily be divided into such parts, divide it into parts whereof one at least may be an aliquot part of a pound, and the subsequent part, or parts, each an aliquot part of some prior part.

| Ex. 1. | Ex. 2. | |--------|--------| | 350s.6d., at 1s. 3d. | 9½s., at 1s. 10½d. | | 1s. 3d. | 4½s. 16½d. | | 175 6 | 4 15 | | 43 16 6 | 6d. 2 7 6 | | L. 219 2 6 | 3d. 1 3 9 | | | 1½d. 11 10½ | | | L. 8 18 1½ |

**Case VI.** When the rate is pounds.

**Rule.** Multiply the given number by the rate, and the product is the price in pounds.

| Ex. 1. | Ex. 2. | |--------|--------| | 42, at 2l. | 13, at 8l. | | L. 84 | L. 104 |

**Case VII.** When the rate is pounds and shillings, or pounds, shillings, pence, and farthings.

**Rule I.** If the rate be pounds and shillings, multiply the given number by the pounds, and work for the shillings as in Case IV.

| Ex. 1. | Ex. 2. | |--------|--------| | 1l. 46, at 1l. 4s. | 82, at 4l. 10s. | | 4s. 9 4 | 4l. 328 | | L. 55 4 | 10s. 41 | | | L. 369 |

**Note.** When the rate is more than 1l. and less than 2l. as in Ex. 1, we have no occasion to draw a line under the given number, it being esteemed so many pounds, and the parts for the shillings or pence are added up with it.

**Rule II.** If the rate be pounds, with shillings and pence that make some aliquot part of a pound, or are divisible into aliquot parts, or into shillings and some aliquot part or parts; then multiply the given number by the pounds, and work for the shillings and pence as in Case V. Rule I. or II.

| Ex. 1. | Ex. 2. | |--------|--------| | 54, at L. 3:2:6. | 43, at L. 5:3:4. | | 3l. 162 | 5l. 215 | | 23.6d. 6 15 | 3s. 4d. 7 3 4 | | L.168 15 | L.222 3 4 |

**Rule III.** If the rate be pounds, with shillings, pence, and farthings, that cannot readily be resolved into aliquot parts of a pound; multiply the given number by the pounds; and then work for the shillings, pence, and farthings, as in Case V. Rule III.

| Ex. 1. | Ex. 2. | |--------|--------| | 213, at 1l. 13s. 4½d. | 37, at 3l. 8s. 10½d. | | 10s. 106 10 | 3l. 111 | | 2s. 21 6 | 6s. 11 2 | | 1s. 10 13 | 2s. 6d. 4 12 6 | | 3d. 2 13 3 | 3d. 9 3 | | 1½d. 1 6 7½ | 1d. 3 1 | | L. 355 8 10½ | ¼d. 9½ | | | L. 127 7 7½ |

**Case VIII.** When the given number consists of integers and parts.

**Rule.** Work for the price of the integers as already taught; and for the part or parts, take a proportional part or parts of the rate.

| Ex. 1. | Ex. 2. | |--------|--------| | Yards. 720½, at 6s. 8d. per yd. | Yards. 116½, at 4s. 6d. per yd. | | 6s. 8d. 240 | 2s. 6d. 14 10 | | ¼yd. 0 1 8 | 2s. 11 12 | | L. 240 1 8 | ½yd. 2 3 | | | L. 26 4 3 |

An operation in the rules of practice may be proved by running over the several steps a second time, by working the same question a different way, or by the rule of three.

**Chap. XI. Of Decimals.**

I. **Notation.**

A Fraction having 10, 100, 1000, or unity with any number of ciphers annexed to it, for a denominator, is called a decimal fraction; such as, \( \frac{7}{10} \), \( \frac{25}{100} \), \( \frac{3}{1000} \).

In decimal fractions, as in vulgar, the denominator shews into how many parts the unit or integer is divided, and the numerator shews how many of these parts the fraction contains. Thus, if the fraction be \( \frac{9}{10} \), the unit is divided into ten equal parts, and the fraction contains nine of these parts; and consequently, if the unit or integer be a pound Sterling, the value of such a fraction is eighteen shillings.

We may conceive the denominator of a decimal fraction to be formed by dividing the unit into 10 equal parts, and each of these parts into 10 other equal parts, each of these again into 10 other equal parts, and so on, as far as necessary; and hence a decimal fraction will always be so many tenths, or so many tenths of \( \frac{1}{10} \), or so many tenths of \( \frac{1}{10} \) of \( \frac{1}{10} \), &c.; and by reducing the compound fraction to a simple one, we have the decimal. Thus, \( \frac{9}{10} \) of \( \frac{1}{10} \) of \( \frac{1}{10} = \frac{9}{1000} \). Or we may conceive the denominator of a decimal to be formed by the continual multiplication of unity into 10, as often as there are ciphers in it. Thus, \(1 \times 10 = 10\), and \(1 \times 10 \times 10 = 100\), and \(1 \times 10 \times 10 \times 10 = 1000\), &c. And because the fractions \(\frac{2}{10}\), \(\frac{3}{10}\), \(\frac{4}{10}\), &c., have the highest numerators possible, it is plain, that the number of figures or places in the numerator of a decimal can never exceed the number of ciphers in the denominator.

It is usual to write down only the numerator of a decimal fraction, omitting the denominator; and when the numerator has the same number of figures or places as the denominator has ciphers, it is done by writing down the figures of the numerator, and prefixing a point, to distinguish them from a whole number. So \(\frac{7}{10}\) is written thus, .7; and \(\frac{3}{10}\) is written thus, .25. The point thus prefixed is called the decimal point.

But when the numerator has not so many figures or places as there are ciphers in the denominator, the defect is supplied by prefixing a cipher for every figure wanting, and then placing the decimal point on the left. So \(\frac{1}{100}\) is written thus, .03; and \(\frac{1}{1000}\) thus, .0075; and \(\frac{1}{10000}\) thus, .0005.

From this manner of notation, it is easy to read a decimal, or to know its denominator, viz., imagine 1 to stand under the decimal point, and a cipher under every decimal place. Thus, .9 is \(\frac{9}{10}\), and .48 is \(\frac{48}{100}\), and .05 is \(\frac{5}{100}\), and .007 is \(\frac{7}{1000}\), and .00056 is \(\frac{56}{10000}\).

Hence it is plain, that decimals, like integers, decrease from the left to the right, and increase from the right to the left, in a decuple proportion. On the contrary, any decimal figure, by being removed one place toward the left, becomes ten times greater.

An integer, by annexing ciphers, is raised to higher places on the left, and may by this means have its value increased to infinity. On the other hand, a decimal, by prefixing ciphers, is depressed to lower places on the right, and may by this means have its value diminished to infinity.

Ciphers annexed to decimals do not change the value of the decimals. Thus, .50 = 5, and .500 = 5, for .50 = \(\frac{50}{100} = \frac{5}{10} = .5\); and \(\frac{50}{100} = \frac{5}{10} = .5\).

Decimals may be resolved into constituent parts, and the parts may be read, separately, thus, .847 = 8 + .04 + .007 = \(\frac{8}{10} + \frac{4}{100} + \frac{7}{1000}\).

In decimals the figure next the point, being the first decimal place, is sometimes called primes, and the second figure from the point is called seconds, the next thirds, &c. Thus, in .875 the figure 8 is primes, 7 is seconds, and 5 is thirds.

From this brief account of the nature of decimals, it follows, that the manner of operation in decimals will be the same as in whole numbers; and also, that the same number may be differently expressed, according as the integer is chosen. Thus, the time since our Saviour's birth may be written thus, 1769; or thus, 1769.9; or thus, 1769.99; or thus, 17699; according as one year, a decad, a century, a chilad, or myriad, is used as the integer. Hence arises the superior excellency of decimal arithmetic, above every other sort of numerical computation; as will appear in the sequel.

II. Reduction of Decimals.

Prob. I. To reduce a vulgar fraction to a decimal.

Rule. To the numerator of the vulgar fraction affix a point or comma, then annex a competent number of ciphers, and divide by the denominator; the quot is the numerator of the decimal, and the cyphers annexed show the number of decimal places.

Examp. I. Reduce \(\frac{1}{2}\) to a decimal?

Here to the numerator 1 annex one cipher, and dividing by the denominator 2, the quot is 5, and 0 remains; and because a single cipher only was annexed to the numerator, the decimal numerator will consist but of one figure, namely 5; to which, therefore, prefix the decimal point. So \(\frac{1}{2} = .5\).

Hence appears the reason of the rule; namely, \(2 : 1 :: 10 : 5\); that is, as the vulgar denominator to the vulgar numerator, so is the decimal denominator to the decimal numerator.

Examp. II. Reduce \(\frac{3}{4}\) to a decimal.

To the numerator 3, annex two ciphers; and, dividing by the denominator, the quot gives 75 for the numerator of the decimal, two ciphers having been annexed. So \(\frac{3}{4} = .75\).

Though ciphers may be annexed at pleasure, yet it is the ciphers used that determine the number of decimal places in the quot; and at first it is sufficient to annex so many as serve to complete the first dividend, leaving room to annex more as you proceed in the operation; or rather annex the other ciphers to the remainders, without giving them a place in the dividend.

The first dividend also shows whether ciphers ought to be prefixed to the quot, and how many. Thus, if the first dividend take in only one of the annexed ciphers, the figure put in the quot is primes, and no cipher to be prefixed. If the first dividend comprehend two of the annexed ciphers, the figure put in the quot is seconds, and one cipher must be prefixed. If the first dividend comprehend three of the annexed ciphers, the figure put in the quot is thirds, and two ciphers must be prefixed, &c. Hence, in reducing a vulgar fraction to a decimal, the natural and easy way is, to place first the decimal point in the quot, and after it a cipher or ciphers, or the quotient-figure, as the first dividend directs.

In reducing a vulgar fraction to a decimal, if 0 at last remains, as in all the above examples, the decimal is precisely equal to the vulgar fraction, and is called a finite or terminant decimal.

In finite decimals, the denominator is always some aliquot part of the numerator increased by annexing ciphers; and such decimals take their rise from vulgar fractions whose denominator is 2 or 5, or some power of 2 or 5, or the product of some of their powers. See Chap. XII. and Algebra, Chap. III.

The powers of numbers are sometimes expressed by indices or exponents placed at the corners of the numbers. Thus, $2^3$ signifies the second power of 2, and $5^3$ signifies the third power of 5; and $10^4$ signifies the fourth power of 10, &c. The index of the root or first power is seldom expressed.

Any power of 2 multiplied into the like power of 5 gives a product equal to the same power of 10; as appears from the following specimen of the powers of 2, 5, and 10.

\[ \begin{array}{cccc} 2^1 & = & 2 & \times & 5^1 & = & 5 \\ 2^2 & = & 4 & \times & 5^2 & = & 25 \\ 2^3 & = & 8 & \times & 5^3 & = & 125 \\ 2^4 & = & 16 & \times & 5^4 & = & 625 \\ 2^5 & = & 32 & \times & 5^5 & = & 3125 \\ 2^6 & = & 64 & \times & 5^6 & = & 15625 \\ 2^7 & = & 128 & \times & 5^7 & = & 78125 \\ \end{array} \]

The product of two different powers of 2 and 5, is equal to the product that will arise by raising 10 to the power denoted by the lesser given index, and then multiplying this power of 10 into that power of the other number which is denoted by the difference of the two given exponents. Thus,

\[ \begin{align*} 2^6 \times 5^3 &= 64 \times 25 = 10^3 \times 2^4 = 100 \times 16 = 1600 \\ 2^3 \times 5^6 &= 4 \times 15625 = 10^3 \times 5^4 = 100 \times 625 = 62500 \end{align*} \]

From these remarks it is easy to perceive, that 2 or 5, or any of their powers, or product of their powers, will measure 10 or its powers, viz. 100, 1000, &c., or their multiples, such as, 20, 200, 2000, &c.; and such every numerator becomes by having ciphers annexed; and therefore 2 or 5, or their powers, or product of their powers, used as a denominator, will divide any numerator with a competent number of ciphers annexed, and leave no remainder; and consequently the decimal thence resulting will be finite.

If the numerator of the vulgar fraction be unity, and the denominator any single power of 2 or 5, there will be as many decimal places in the quoit as there are units in the index of the given power. Thus, $16 = 2^4$ gives a decimal of four places, viz. $\frac{1}{16} = .0625$; and, $125 = 5^3$ gives a decimal of three places, viz. $\frac{1}{125} = .008$.

When the denominator is the product of like powers of 2 and 5; in this case, such a product being equal to the like power of 10, and any power of 10 being equal to 1, with as many ciphers annexed as there are units in the index, it follows, that there will still be as many decimal places in the quoit as there are units in the index, either of 2, of 5, or 10. Thus, $8 \times 125 = 2^3 \times 5^3 = 10^3 = 1000$, gives a decimal of three places, viz. $\frac{1}{1000} = .001$.

When the denominator is the product of different powers of 2 or 5, find what power of 10, and what power of 2 or 5, upon being multiplied, will give the same product, as is taught above; and the sum of the indices shews the number of decimal places; thus,

\[ 2^6 \times 5^1 = 10^3 \times 2^4; \text{ and the sum of the indices, } 2 + 4 = 6, \text{ gives the number of decimal places, viz. } \frac{1}{1000} = .000625. \]

And, in general, to find what number of decimal places any such vulgar fraction will give, divide the denominator by 2, 5, or 10, till the last quotient be 1, and the remainder 0; and the number of divisors shews the number of decimal places. Thus, $\frac{1}{10}$ gives a decimal of four places; for $2)16(8(4(2(1. And $\frac{1}{5}$ gives a decimal of three places; for $5)125(25(5(1. And $\frac{1}{100}$ gives a decimal of three places; for $10)1000(10(10)$

$\frac{1}{100}(10(1. And $\frac{1}{1000}$ gives a decimal of six places; for $10)1600(160(16(8(4(2(1.$

If the denominator of a vulgar fraction be neither 2 nor 5, nor any of their powers, nor product of their powers, such a denominator will not divide the numerator with annexed ciphers without a remainder; and the decimal thence resulting is called infinite, or interminate.

Of infinite or interminate decimals, there are two sorts. For some constantly repeat the same figure; and are called repeating decimals, repeaters, or single repends. Others repeat a circle of figures; and on that account are called circulating decimals, circulates, or compound repeends.

Examp. III. Reduce $\frac{1}{3}$ to a decimal.

Here the remainder being still the same, viz. 1, the same figure will constantly be repeated in the quoit.

Repeating decimals are of two kinds: viz. some consist only of the repeating figures, such as the examples above; and these are called pure repeaters; others have one or more digits or ciphers betwixt the decimal point and the repeating figure; and these are called mixt repeaters; and the digits or ciphers on the left of the repeating figures are called the finite part of such decimals.

Pure repeaters take their rise from vulgar fractions whose denominator is 3, or its multiple 9; and are but few in number.

Mixt repeaters derive their origin from vulgar fractions whose denominator is the product of 3 into 2 or 5, or into some of their powers, or product of their powers; and such denominators may be considered as the product of two component parts, whereof one is 2 or 5, or some of their powers, or product of their powers; and hence the finite part. The other component part is 3; and hence the repeating figure.

Examp. IV. Reduce $\frac{4}{5}$ to a decimal.

Here the repeater is mixt, the finite part being 2, and the repeating figure 6. We now resolve such denominators into their component parts, and divide the numerator by one of these parts, and then divide the quotient by the other. Thus,

\[ \frac{15}{5} = 3 \times 3. \]

The number of places in the finite part of a mixed repeater may be ascertained from the number of units in the index of the powers of 2 or 5.

And, universally, to find the number of places in the finite part of such fractions, divide the denominator first by 3, and then divide the quotient by 2, 5, or 10, till the last quotient be 1, and 0 remain; and the number of divisors, excluding 3, shows the number of places in the finite part.

Repeating decimals are usually marked by a dash through the right-hand figure, as in the examples above; but some choose to mark them by a point set over the repeating figure, thus, \( \frac{3}{7}, \frac{26}{9}. \) The remainder where the repetition begins is commonly marked with an asterisk.

Because any quotient multiplied by the divisor reproduces the dividend, it follows, that any decimal multiplied by the denominator of the vulgar fraction from which it resulted, will reproduce the numerator with the annexed ciphers. Thus, if \( \frac{75}{100} \), the decimal of \( \frac{3}{4} \), be multiplied by 4, it will reproduce the numerator 3 and the two annexed ciphers.

Now, suppose the given decimal to be a repeater; such as \( \frac{3}{7} \), resulting from the vulgar fraction \( \frac{3}{7} \); if the repeating decimal be multiplied by the denominator 3, it will, by carrying at 9 on the right hand, reproduce the numerator 1 with the annexed cipher. In like manner, if the repeater \( \frac{6}{7} \), be multiplied by 3, it will, by carrying at 9 on the right hand, reproduce the numerator 2 with the annexed cipher. Again, if the repeater \( \frac{1}{7} \), be multiplied by the denominator 9, it will, by carrying at 9 on the right hand, reproduce the numerator 1 with the annexed cipher. And, if the mixed repeater \( \frac{26}{9} \), be multiplied by the denominator 15, it will, by carrying at 9 on the right hand, reproduce the numerator 4 with the two annexed ciphers.

From these remarks we may conclude, that the right-hand figure of every repeating decimal is ninth-parts; and the same truth may be evinced by resolving the decimal into its constituent parts, in the following manner.

The vulgar fraction \( \frac{3}{7} \) reduced to a decimal gives \( \frac{4}{7}, \frac{6}{7}, \ldots \); and this repeater resolved into decimal constituent parts, becomes \( \frac{1}{7} + \frac{2}{7} + \frac{3}{7} + \frac{4}{7} + \frac{5}{7} + \frac{6}{7} + \frac{7}{7} \), etc., to infinity. But if we esteem the right-hand figure to be ninth-parts, we have \( \frac{1}{9} + \frac{2}{9} + \frac{3}{9} + \frac{4}{9} + \frac{5}{9} + \frac{6}{9} + \frac{7}{9} + \frac{8}{9} + \frac{9}{9} = 1 \), the given vulgar fraction. And as the vulgar fraction \( \frac{3}{7} \) gives \( \frac{4}{7} \), so \( \frac{6}{7} \) gives \( \frac{9}{9} \); that is, \( \frac{9}{9} = \frac{1}{9} = 1 \). And, universally, a series of nines infinitely continued is equal to unity in the place on the left hand; thus, \( \frac{9}{9} = 1 \); and \( \frac{9}{9} = 1 \); and \( \frac{9}{9} = 1 \); and \( \frac{9}{9} = 1 \).

Hence may be ascertained the value of an infinite series decreasing in a decuple proportion. Thus, \( \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \ldots \), etc. \( = \frac{1}{9} \). And \( \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \ldots \), etc. \( = \frac{1}{9} \).

If the denominator of a vulgar fraction be neither 2 nor 5, nor any power of 2 or 5, nor any product of their powers; nor 3, nor 9, nor any product of 3 into 2 or 5, or into some of their powers, or product of their powers, the decimal resulting from such a vulgar fraction will circulate.

Circulates, like repeaters, are of two sorts, viz. pure and mixt. A pure circulate consists of the figures of the circle only; as .09, 09, etc. or 18, 18, etc. A mixt circulate has a finite part betwixt the decimal point and the figure that begins the circle; as .0, 45, 45, etc.; or .32, 142857, 142857, etc. Some chuse to distinguish the finite part from the circle, and one circle from another, by a comma, as above. Others dash the first and last figure of the circle. It is likewise usual to mark the remainder where the new circle begins, by affixing an asterisk.

Examp. V. Reduce \( \frac{1}{7} \) to a decimal.

The denominator 11 gives a pure \( \frac{1}{11} \), 09, 09, circle of two figures.

\[ \begin{array}{c} \text{11} \\ \times 9 \\ \hline \text{99} \\ \end{array} \]

It is easy to perceive, that if any of the vulgar fractions in the above specimen have both its numerator and denominator multiplied by 9, there will arise a new vulgar fraction of the same value, whose numerator will be the figures of the circle, and its denominator the like number of 9's. Thus,

\[ \begin{array}{c} \text{3} \times \text{9} = \text{27}, \quad \text{and} \quad \text{7} \times \text{9} = \text{63}. \\ \text{11} \times \text{9} = \text{99}, \quad \text{and} \quad \text{11} \times \text{9} = \text{99}. \\ \end{array} \]

As the denominator 11, whereof 99 is a multiple, gives a pure circulate of two places, to any denominator, whereof 999, or 9999, or 99999, etc. are multiples, will give a pure circulate of three, four, five, etc. places; that is, of as many places as there are 9's in the multiple. And such denominators are all the prime numbers, except 2, 3, and 5, viz. 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, etc.; also their products into 3, viz. 21, 33, 39, 51, 57, 69, etc. Such too are all the powers of 3, except 3 and 9, viz. 27, 81, 243, 729, 2187, etc.

The reason is plain: for if any divisor, as 37, divide 999, without a remainder, it will also divide 1000, and leave a remainder of 1, to begin a new circle.

To find how many places the circle will consist of, divide a competent number of 9's by any of the above denominators, continuing the operation till 0 remain; and the number of 9's used will show the number of places. Thus, \(7\overline{99999}\) six places. Thus, \(27\overline{955}\) three places.

The number of figures in a circle, when some power of 3 is the denominator, may also be found thus: Divide the given denominator by 9, and the number of units in the quotient will be equal to the number of figures in the circle. Thus, \(9\overline{27}\) (3 places). Thus, \(9\overline{81}\) (9 places), &c.

If 3 divide a repeater whose repeating figure is not a multiple of 3, the quotient will be a pure circulate of three places. Thus, \(3\overline{11}(.037)\), and \(3\overline{855}(.185)\), and \(3\overline{77}(.259)\).

If 3 divide a pure circulate, the circle not being a multiple of 3, the quotient will be a pure circulate of thrice as many places as the circle of the dividend. Thus, \(3\overline{037}(.037)\overline{012345679}\).

Mixt circulates take their rise from fractions whose denominators are the prime numbers 7, 11, 13, 17, 19, 23, 29, &c., multiplied into 2, 5, or 10, or into some of their powers, or product of their powers.

**Example VI.** Reduce \(\frac{2}{28}\) to a decimal.

\[ \begin{array}{cccc} 28)9(0.3214285714 \\ 84 \\ 60 \\ 56 \\ \end{array} \]

The denominator \(28 = 7 \times 2 \times 2\), gives a mixt circulate, consisting of the finite part 32, and a circle of six figures or places, whose sum is equal to the product of 9 into half the number of figures; that is, \(9 \times 3 = 27\).

The number of places, both in the finite part and in the circle, may be ascertained thus: Divide the denominator of the vulgar fraction by 10, 5, or 2, as often as possible, and the number of divisors will show the number of places in the finite part; make the last quotient a divisor, and the dividend any competent number of 9's; continue the operation till 0 remain, and the number of 9's used will be equal to the number of places in the circle. Thus, \(10\overline{0500}(0500)(20514)\); and \(41\overline{9999}(2439)\), and e remains. So you may conclude, that the finite part will consist of three places, and the circle of five.

Universally, any vulgar fraction being given, we may determine whether the decimal thence resulting will be finite or infinite; and if infinite, whether pure or mixt; with the number of places, &c., in the following manner.

Reduce the given vulgar fraction to its lowest terms, then divide the denominator by 10, 5, or 2, as often as possible; and if the last quotient be unity, without any remainder, the decimal is finite, and the number of divisors shows the number of decimal places.

If the last quotient be 3, or any power of 3, the resulting decimal will be a mixt repeater, the number of whose finite places will be equal to the number of divisors.

If the last quotient cannot be divided by 2, 5, 10, or 3, the resulting decimal will be a mixt circulate; and the way of finding the number of places, both in the finite part and circle, is taught above.

If the denominator of the given vulgar fraction can be divided, neither by 2, 5, nor 10, the resulting decimal will be a pure repeater, or a pure circulate, according as the denominator is 3 or 9, or some of the prime numbers, 7, 11, 13, &c.; as has been already explained.

Every vulgar fraction may be reduced to a decimal, finite or infinite; that is, to a finite decimal, to a repeater, or a circulate. For if the denominator divide the numerator with ciphers annexed, so as to leave no remainder, the resulting decimal is finite. If the remaining figure be always the same, the resulting decimal will be a repeater. If neither of these be the case, yet, because the divisor is a finite number, the remainder at last must either be the same with the numerator of the vulgar fraction, or the same with some preceding remainder, and then a new circle begins; and consequently the resulting decimal will be a circulate.

Because in circulates the circle runs on sometimes to 16, 18, 22, 28, 81, 243, &c., places, and because, in decimals of every sort, the finite part runs sometimes on too many places, such circulates, or finite parts, may, without any sensible error, be limited at five or six places, and used as finites: for five decimal places, divide the integer into 100,000 equal parts, and all the loss that can be occasioned by such limitation, is less than one hundred thousandth part of the integer. And in most cases, the decimal may be limited at three places, which divide the integer into 1000 equal parts.

Circulates, or finite parts, thus limited, are called approximate decimals; and are sometimes marked with + or — annexed, according as the right-hand figure is taken less or greater than just; for in limiting the decimal, if you foresee that the succeeding figure of the quotient would be 6 or 7, or any figure above 5, you lessen the error by increasing the right-hand figure of the approximate by unity.

**Problem II.** To reduce the parts of coin, &c., to decimals.

**Rule.** Convert the given part or parts to a vulgar fraction. fraction of the integer, and then reduce the vulgar fraction to a decimal.

Ex. 1. Reduce 9 pence to the decimal of a shilling.

\[ \frac{9}{2} = \frac{9}{2} \times \frac{1}{2} = \frac{9}{4} \]

Here the fraction \( \frac{9}{4} = 2.25 \); and

the denominator \( 4 = 2 \times 2 \) gives

a finite decimal of two places.

\[ \frac{9}{4} = 2.25 \]

Ex. 2. Reduce 9 pence to the decimal of a pound.

\[ \frac{9}{2} = \frac{9}{2} \times \frac{1}{2} = \frac{9}{4} \]

The fraction \( \frac{9}{4} = 2.25 \);

and the denominator \( 80 = 10 \times 2 \times 2 \times 2 \) gives a finite decimal of four places.

\[ \frac{9}{4} = 2.25 \]

Ex. 3. Reduce 16 s. 6 d. to the decimal of a pound.

\[ \frac{16}{2} = \frac{16}{2} \times \frac{1}{2} = \frac{16}{4} \]

The fraction \( \frac{16}{4} = 4 \);

and the denominator \( 40 = 10 \times 2 \times 2 \) gives a finite decimal of three places.

\[ \frac{16}{4} = 4 \]

Prob. III. To reduce the remainder of a division to a decimal.

Rule. The remainder being the numerator, and the divisor the denominator of a vulgar fraction, after placing the decimal point on the right of the integral part of the quotient, annex ciphers to the remainder; then continue the division till it remain, or till the quotient repeat or circulate, or till you think proper to limit the decimal; and the number on the right of the point is a decimal of the integer expressed in the quotient.

Example 1.

Divide 513 l. among 36 men.

\[ \frac{513}{36} = 14.25 \]

Example 2.

Divide 176 s. among 24 boys.

\[ \frac{176}{24} = 7.33 \]

Rem. 80

Prob. IV. To reduce a decimal to value.

Rule. Multiply the given decimal by the number of parts of the next inferior denomination contained in an unit of the integer; and from the product point off so many figures to the right hand as there are places in the given decimal. On the left hand of the point are parts, and on the right a decimal of one of these parts; which decimal must be reduced in the same manner to the next inferior denomination, and from that to the next, and so on to the lowest; the several figures on the left of the points are parts; and if there be still some figure or figures on the right, they are a decimal of the lowest of the parts.

Example 1.

Reduce .875 l. to value.

\[ \frac{875}{1000} = 0.875 \]

Example 2.

Reduce .7691 l. to value.

\[ \frac{7691}{10000} = 0.7691 \]

The reason of pointing the product, as the rule directs, is plain. For, in Ex. 1, as 1000 : 875 :: 20 : 17; that is, as the decimal denominator to the decimal numerator, so the vulgar denominator to the vulgar numerator.

In Ex. 1, the full value of the decimal comes out in parts, the decimal being quite exhausted; but in Ex. 2, besides the parts, there is a decimal of a farthing, viz. .336 f.

The decimal of a pound Sterling may be reduced to value by inspection, in the following manner.

Double the figure in the place of primes for shillings; and if the figure in the place of seconds be 5, or exceed 5, reckon 1 shilling more; and rejecting 5 in the second place, the figures in the second and third places are so many farthings, abating 1 for every 25.

\[ \frac{718}{14} = 4.2 \]

\[ \frac{759}{15} = 2.1 \]

\[ \frac{894}{17} = 10.3 \]

In Example 1, the figure 7 doubled gives 14 s.; the two following figures 18 are farthings, equal to 4 d. 2 f.

In Example 2, the figure 7 doubled gives 14 s. and 5 in the place of seconds gives 1 shilling more, in all 15 s.; and the other figure 9 is farthings, viz. 2 d. 1 f.

In Example 3, the figure 8 in the place of primes, and 5 in the place of seconds, give 17 s.; the remaining figures 44, abating 1, are farthings, viz. 10 d. 3 f.

When the figures in the second and third place to be converted into farthings are 25, the answer, by inspection, comes out exact, viz. 24 f. or 6 d.; but in all other cases, the answer, by inspection, is too great, no allowance or correction being made till the convertible number. number amount to 25, and afford a deduction of 1 farthing complete. Hence, by inspection, we have frequently 1 farthing more than by the common method; but the two methods will agree, or give the same answer, if, from the figures to be turned into farthings, we subtract their 25th part, esteeming the remainder farthings and decimal parts of a farthing.

Thus, .7181 = 14s. 4d. 2f. by inspection; but by the common method, and by inspection corrected, the answer comes out 1 farthing less, as follows.

Common method. Inspection corrected.

| L. | If 25 : 1 :: 18 : .72. | |----|----------------------| | .718 | that is, 25)18.0(.72 |

| s. 14.360 | 50 and 18 | | 12 | 50 - .72 | | d. 4.32 | (0) 17.28 | | 4 | d. f. |

f. 1.28

And 17.28 = 4 1.28

To conclude, instead of dividing by 25, we may multiply by .04; and then the exact value of any decimal of a pound Sterling may be found as follows.

From the primes and seconds set off the shillings; multiply the remainder by 4, setting the product two places to the right; subtract the product from the first remainder; and from the second remainder point off so many places to the right as there are figures in the first remainder. The number on the left of the point is farthings, and the figures on the right are a decimal of a farthing.

Example 1.

| s. d. f. | |----------| | .7181 = 14 4 1.28 |

1 Rem. .18

72 = 18 × 4

2 Rem. 17.28

Example 2.

| s. d. f. | |----------| | .7691 = 15 4 2.336 |

1 Rem. 191

764 = 191 × 4

2 Rem. 18.336

Prob. V. To reduce a decimal to its primitive vulgar fraction.

Case I. When the given decimal is finite.

Rule. Divide both numerator and denominator of the given decimal by their greatest common measure; the quotient is the vulgar fraction required.

Thus, .875 = \(\frac{875}{1000} = \frac{7}{8}\). For \(875\) \(1000\)

Greatest common measure 125

\(875\) \(7\)

And \(125\) \(875\) \(8\).

Case II. When the given decimal is a pure repeater, or a pure circulate.

Rule. Make the repeating figure, or the figures of the circle, the numerator of the vulgar fraction; the denominator is 9 for the repeating figure, or 9 for every figure of the circle; and then, if occasion require, reduce this fraction to its lowest terms.

Thus, .3 = \(\frac{3}{10}\), and .6 = \(\frac{6}{10}\), and .9 = \(\frac{9}{10}\).

Again, .27 = \(\frac{27}{99}\), and .714285 = \(\frac{714285}{999999}\).

Case III. When the given decimal is a mixt repeater, or a mixt circulate.

Rule. From the mixt repeater, or mixt circulate, subtract the finite part, and the remainder is the numerator of the vulgar fraction; the denominator is 9 for the repeating figure, or 9 for every figure of the circle, with as many ciphers annexed as there are figures in the finite part.

Thus, .03 = \(\frac{3}{100}\), and .16 = \(\frac{16}{100}\), and .083 = \(\frac{83}{1000}\).

The reason of the rule may be shown thus: Esteem the finite part of the last example an integer, and then the mixt number \(\frac{571428}{999999}\) will be equal to the given circulate. Again, reduce this mixt number to an improper fraction, viz., multiply the integer 3 by the denominator 999999, and to the product add the numerator, as directed in reduction of vulgar fractions.

Multiply the integer 3 into 999999 by the method of multiplying any number by 9, 99, 999, etc., taught in multiplication of integers, and to the product add the numerator, and the sum shall be the numerator of the improper fraction, as in the margin.

Now it is evident that the same numerator will be found, if, in the upper line, instead of the six ciphers, you place the figures of the circle, and from them subtract 3, the finite part.

To the numerator thus found, the denominator is 999999; and so the vulgar fraction is \(\frac{571428}{999999}\). But we esteemed 3 an integer; whereas, in fact, it is \(\frac{3}{100}\); and so our vulgar fraction will be 100 times greater than it ought to be: to correct this error, we must multiply the denominator by 100, which is done by annexing two ciphers to it; and the true fraction comes out to be \(\frac{571428}{999999}\), as by the rule.

Because this rule is of great importance, and will often occur in practice, we shall here subjoin another example.

Reduce .0416 to a vulgar fraction.

\[ \begin{array}{c} .0416 \\ \hline \end{array} \]

Num. 375

Den. 9000

In this manner too may any mixt number, consisting of an integer with a repeater or circulate, be reduced to an improper vulgar fraction; but no ciphers are to be annexed to the denominator for the figures of the integer. Ex. Reduce $8.\overline{3}$ to an improper vulgar fraction.

$$\frac{8.\overline{3}}{8}$$

Num. $\frac{75}{9} = \frac{25}{3} = 8\frac{1}{3}$

Den. 9

Approximate decimals being imperfect, cannot be exactly reduced back to the vulgar fractions from which they resulted. But if the approximate be completed by annexing to it a vulgar fraction, whereof the remainder of the division is the numerator, and the divisor the denominator, you shall have a mixt number, which you may reduce to an improper vulgar fraction; then to the denominator annex as many ciphers as there are figures in the approximate; and this fraction reduced to its lowest terms, will be the primitive vulgar fraction required.

Prob. VI. To reduce unlike circles to others that are similar and conterminous.

Similar or like circles are such as consist of an equal number of places.

Thus, .27, and .09, are similar circles, as consisting of two places each. But .63, and .148, are unlike; the former consisting of two, and the latter of three places.

Conterminous circles are such as begin and end at the same distance from the decimal point.

Thus, .153846, and .384615, are conterminous; because they both begin at the place of primes, and have an equal number of places. And .0714285, and .7142857142, are conterminous, because they both begin at the place of seconds, and have the same number of places. But .81, and .136, are not conterminous, the former beginning at the place of primes, and the latter at the place of seconds. Again, .63, and .481, are not conterminous, because they have not the same number of places; for circles cannot be conterminous unless they be at the same time similar.

Unlike circles are reduced to similar ones by the following

Rule. Find the least multiple of the numbers denoting the number of places in the several given circles, and extend each of the given circles to as many places as there are units in the least multiple.

Thus, to reduce the unlike circles .63, = .636363, .63, and .148, to similar ones, extend both circles to six places, because 6 is the least multiple of 2 and 3, the number of places in the given circles.

In a circle any one of the circulating figures may be made the first of the circle. Thus, 7.592, may be expressed thus, 7.59292; or thus, 7.592929; and that without changing its value: consequently a pure circulate may put on the form of a mixt circulate, if one or more figures on the left be set aside for the finite part; thus, .72 = .727, where .72 is the finite part.

That the value is not changed may be thus demonstrated.

$$\frac{72}{99} = \frac{72}{99} = \frac{72}{99} = .72$$

Hence two or more given circles may be made conterminous, by the following

Rule. Set aside by a comma on the left, as many figures as there are places in the longest finite part, and then prolong the several circles to as many places as will make them similar.

Ex. To make .5463, and .5463, = .54636363, .9148, conterminous.

Here, because .54, the longest finite part, consists of two places, set aside .91, in the other circulate, for a finite part, and then prolong both circles to six places, which renders them similar.

III. Addition of Decimals.

Rule I. Place the given decimals so that the points may stand directly under each other, and consequently tenths under tenths, hundredths under hundredths &c.; then, if the given decimals be all finite or approximate, add them as integers, inserting the decimal point directly under the column of points. The figures on the left of the point are integers, and those on the right are a decimal of the integer, consisting of as many places as there are figures in the longest of the given decimals.

The operation is the same here as in addition of vulgar fractions; for a cipher on the right of a decimal does not change its value: If, therefore, ciphers be annexed, so as to give every decimal the same number of places, as is done in the margin, they will by this means be reduced to a common denominator, viz. 1000.

Note. If the decimals to be added are of different denominations, first reduce them to one denomination, and then add. The reason is, because like things only can be added or subtracted.

Ex. What is the sum of .7251, and .625?

Here you may either reduce the decimal of a shilling to that of a pound, or you may reduce the decimal of a pound to that of a shilling.

First reduce the decimal of a shilling to that of a pound, by reduction-ascending, viz. divide by 20, as follows.

| L. | 20 | .62500 | .03125 | |----|----|--------|--------| | | | | .725 |

Sum .75625 = 15 1\(\frac{1}{2}\)

Secondly, reduce the decimal of a pound to that of a shilling, by reduction-descending; that is, multiply by 20, as follows.

| L. | 20 | .7251 | |----|----|-------| | | | .14500 | | | | 625 | | | | .15125 sum. |

d. 1500

f. 2.0

APPROXI- APPROXIMATES.

If the decimals to be added run on to a great many places, it will be sufficient in most cases to use only four or five places, and observe to increase the figure at which you break off by an unit, if the rejected figure on the right exceed 5. And in adding such approximates, omit the right-hand figure of the sum, as uncertain, but take in the carriage. Follows an example at large, and the same contracted.

Ex. at large. | contracted. ---|--- 12.2352946 | 12.23529 + 8.15789325 | 8.15789 + 7.08698435 | 7.08696 + 6.32143482 | 6.32143 + 4.75 | 4.75

38.55159105 | 38.5515 certain.

RULE II. When all or any of the given decimals are repeaters, give every repeater the same number of places, and one place more than the longest finite; and for every nine in the right-hand column carry 1, or to its sum add 1 for every nine, and then carry at ten.

Examp. 748\(\frac{1}{3}\) = 748.83 653\(\frac{1}{3}\) = 653.96 84\(\frac{1}{3}\) = 84.11 25\(\frac{1}{3}\) = 25.83 37\(\frac{1}{3}\) = 37.95 8\(\frac{1}{3}\) = 8.16

1557\(\frac{1}{3}\) = 1557.86

In this example the sum of the right-hand column is 24, which contains 9 twice, and 6 over; so set down 6 and carry 2; Or to the sum 24 add 2, for the two nines, which makes 26; so set down 6 and carry 2. Proceed with the rest as in integers.

The sums, differences, and products, of intermediate decimals, are always intermediate, unless they end in a cipher.

A repeating digit is the numerator of a vulgar fraction, whose denominator is 9; and hence, in adding a column of repeating digits, every 9 of the sum is \(\frac{1}{9}\), or an unit, to be carried; and what is over a just number of nines is so many ninth-parts.

Or, if to the sum of a column of repeating digits, 1 for every 9 contained in it be added, we then carry 1 for every ten; but what is over a just number of tens will still continue to be ninth-parts.

If in any example the repeating figures happen all to be reiterated, the carriage from the right-hand column adjusts the column on the left, or makes every ten of them equal to an unit of the next superior column, &c. Thus, if we imagine a column of the repeating figures reiterated on the right of any example, the carriage from it would adjust the right-hand column of the example.

RULE III. When all or any of the given decimals are circulates, make all the circles conterminous, find the number of tens to be carried from the left-hand column of the circles, add this carriage to the right hand column, and proceed as in addition of integers.

If repeaters be mixed with the circulates, give the repeaters the form of circulates, by extending the repeating figures till they become conterminous with the other circles.

If finite decimals are joined with the circulates, extend the finite parts of all the circulates to as many places as there are figures in the longest finite.

Examp. \( \frac{3}{9} = .428571, = .428571, \) \( \frac{9}{9} = .857142, = .857142, \) \( \frac{5}{15} = .45, = .454545, \) \( \frac{12}{12} = .370, = .370370, \)

2.110630

In order to find the carriage from the left-hand column of the circles, add the columns next to it on the right, saying, 7 + 5 + 5 + 2 = 19; from which carry 1, and say, 1 + 3 + 4 + 8 + 4 = 20; from which carry 2, and go on to add the right-hand column of the circle, saying, the carriage 2 + 5 + 2 + 1 = 10; so set down 0, and carry 1, and proceed with the rest as in integers.

The adding the carriage from the left-hand column of the circles to the column on the right hand, arises from the flux of numbers; for as the circles repeat infinitely, if we suppose a new feet of the same circles to be repeated upon the right of our examples, it is plain, that in adding them the carriage from the left-hand column of the new feet would naturally fall into the right-hand column of our example.

The operation here is the same as in addition of vulgar fractions; for every circle is the numerator of a vulgar fraction, whose common denominator is 999999; and if the circles or numerators be added, without minding any carriage from the left-hand column, the sum will be 2110628.

And 999999(2110628(2110628

But, by pointing off from the sum of the circles six figures towards the right, we divide by 1000000, instead of dividing by 999999; which gives indeed the same quot, but makes the remainder too small.

Now, that the carriage-figure from the left-hand column of the circles, is the integral part of the quot, and at the same time the difference between the true and false remainder, is evident; for the quotient-figure 2, multiplied into the two divisors 1000000 and 999999, gives two products, whose difference is 2; and consequently, if the greatest product, viz. 2 X 1000000 = 2000000, be subtracted from the dividend, the result will want 2 of the true remainder. To prevent such errors, and to put the work on a sure footing, find the carriage from the left-hand column of the circles, add this carriage to the right-hand column, divide the sum by 1000000, and you will have a true quot, and a true remainder. The learner may look back to division of integers. integers, where the method of dividing by 9, 99, 999, &c., is explained.

Hence it follows, that if we add the circles as they stand, without minding any carriage from the left, and to the sum add the excrement figure on the left of the decimal point, we shall have the full sum of the circles, both as to the integral and fractional part, as in the margin.

Pure repeaters, being the numerators of vulgar fractions, whose denominator is 9 as often taken as the digit is repeated, may be added in the same manner as circles. But in examples clear of circulates, the method prescribed in Case II. is preferable.

In adding circles and pure repeaters by the method now explained, it will sometimes happen that the fractional part of the sum will be a series of nines, as in the margin: And in this case, the numerator of the fraction being the same with the denominator, its value will be unity; and accordingly it must be added to the integral part. But in adding pure repeaters by the method in Case II. this cannot happen.

By way of proof, we shall here add all the vulgar fractions in Examp. I. and reduce their sum to a mixt number, continuing the division to a decimal.

\[ \frac{1}{4} + \frac{6}{7} + \frac{5}{8} + \frac{1}{2} = \frac{6237}{14553} + \frac{12474}{14553} + \frac{6615}{14553} + \frac{5300}{14553} = \frac{30716}{14553} \]

Here the dividual being the same with the second, a new circle begins.

IV. Subtraction of Decimals.

Rule I. Place the minor under the major, so that that the points may be in one column; and then, if the given decimals be finite or approximate, work as in subtraction of integers.

If the major and minor have not the same number of places, imagine the void places to be filled up with ciphers.

Example I.

| L. | r. | d. | L. | |----|----|----|----| | From | 48 | 10 | 6 = 48.525 | | Sub. | 18 | 12 | 8\(\frac{1}{4}\) = 18.634375 | | Rem. | 29 | 17 | 9\(\frac{1}{4}\) = 29.800625 |

Example II.

| C. | Q. | lb. | C. | |----|----|----|----| | From | 54 | 2 | 21 = 54.6875 | | Sub. | 36 | 3 | 14 = 36.875 | | Rem. | 17 | 3 | 17 = 17.8125 |

Approximates.

In subtracting approximates, neglect the right-hand figure of the remainder, as uncertain; but an unit borrowed on the right must be repaid, as in the two examples following.

Ex. 1.

| From | 783.0625 | | Sub. | 495.28571 | | Rem. | 287.7767 |

Rem. 287.7767 certain.

Ex. 2.

| From | 549.4043 | | Sub. | 78.0875 | | Rem. | 471.3168 |

Rem. 471.3168 certain.

Rule II. If one of the given decimals is a repeater, and the other a finite decimal, give the repeater one place more than the finite decimal, and in subtracting borrow 9 on the right hand.

But if both major and minor repeat, give them an equal number of places, and then subtract as above.

Ex. 1.

| From | .7145833 | | Sub. | .634375 | | Rem. | .0802083 |

Ex. 2.

| From | .525 | | Sub. | .3333 | | Rem. | .1916 |

Ex. 3.

| From | .9989583 | | Sub. | .0291666 | | Rem. | .9697916 |

In Ex. 1. and 2. you give the repeater one place more than the finite decimal, and by this means you obtain the repeating figure of the remainder. But in Ex. 3. you give the two repeaters an equal number of places.

In Ex. 2. and 3. you borrow 9 on the right hand.

Rule III. If both the given decimals be circulates, make the circles conterminous, and work as in integers; only if, in the left-hand column of the circles, you foresee, that, in subtracting the figure of the minor from that of the major, one must be borrowed, in this case add 1 to the right-hand figure of the minor, and then subtract.

If one of the given decimals be a circulate, and the other a repeater, give the repeater the form of a conterminous circulate, and then subtract as above.

If one of the given decimals be a circulate, and the other a finite decimal, extend the finite part of the circulate to as many places as there are figures in the finite decimal, and then subtract.

Example I.

From \( \frac{1}{2} = .6428571 \), \( \frac{1}{2} = .6428571 \),

Sub. \( \frac{1}{2} = .1785714 \), \( \frac{1}{2} = .1785714 \),

Rem. \( .46428571 \)

In this example, because, in the left-hand column of the the circles; 8 cannot be subtracted from 2 without borrowing; therefore add 1 to the right-hand figure of the minor, and say, 1+2=3, and 3 from 4, and 1 remains. The reason is obvious: for, supposing the circles reiterated on the right of the example, it would be, 8 from 2 you cannot, but 8 from 12 and 4 remains; 1 borrowed, and 2, make 3, &c.

**Example II.**

From \( \frac{1}{3} = .9285714 \), \( \frac{2}{3} = .6666666 \),

Rem. \( .2619047 \).

In the above example the repeaters are given in the form of conterminous circulates.

**Example III.**

From \( \frac{5}{11} = .384615 \), \( \frac{6}{11} = .5454545 \),

Rem. \( .259615384 \).

In the last example the finite part of the circulate is extended to as many places as there are figures in the finite decimal, by which means like things come to be subtracted, and you obtain the exact circle of the remainder.

**V. Multiplication of Decimals.**

In multiplication and division there may happen nine varieties, arising from the different nature of the numbers that may occur in the operation; and these are of three sorts, viz. integers, mixt numbers, and pure decimals.

Now, since the multiplier or divisor may be of three kinds, and the multiplicand or dividend of as many, there must of consequence be nine varieties; which are these following.

An integer may multiply or divide

- an integer, - a mixt number, - a pure decimal.

A mixt number may multiply or divide

- an integer, - a mixt number, - a pure decimal.

A pure decimal may multiply or divide

- an integer, - a mixt number, - a pure decimal.

Of these varieties, the first belongs properly to vulgar arithmetic, the other eight occur in decimal operations.

But in multiplication and division of decimals, there will occur other nine varieties, arising likewise from the nature of the numbers; which may either be finite, repeating, or circulating.

And since the multiplier or divisor may be of three sorts, and the multiplicand or dividend of as many, there must of course be nine varieties; and these are so obvious, that it would be losing time here to enumerate them.

Before entering on multiplication, we shall lay down a rule for pointing the product, which is of a general nature, and extends to decimals of every sort, whether finite, repeating, or circulating; and is as follows.

**General Rule.**

Give so many decimal places to the product, on the right, as are in both factors; and if the product has not so many figures, supply that defect by prefixing ciphers.

We now proceed to multiplication.

**Rule I.** If both factors are finite or approximate, work exactly as in multiplication of integers.

| Ex. 1. | Ex. 2. | |-------|-------| | .785 | .125 | | .75 | .25 | | 3925 | 625 | | 5495 | 250 | | .58875| .03125|

In Ex. 2. the product not affording so many decimal places as are in the multiplicand and multiplier, the defect is supplied by prefixing ciphers.

The reason of giving as many decimal places to the product as are in both factors, appears by considering that the operation is the same here as in multiplication of vulgar fractions. Thus, \( .785 \times .75 = \frac{785}{1000} \times \frac{75}{100} = \frac{58875}{100000} = .58875 \).

To multiply by 10, 100, 1000, &c., move the decimal point so many places toward the right hand as there are ciphers in the multiplier.

Thus:

\[ \begin{align*} 4375 \times 10 &= 4375 \\ 4375 \times 100 &= 43750 \end{align*} \]

And thus:

\[ \begin{align*} 6875 \times 10 &= 68750 \\ 6875 \times 100 &= 687500 \end{align*} \]

**Approximates.**

In multiplying approximates, the certain places of the product may be determined by one or other of the two rules following, viz.

1. If both factors are approximates, the uncertain places of the product will be one more than the number of places in the longest factor.

2. If one of the factors be finite, and the other approximate, the uncertain places of the product will be one more than the number of places in the finite factor.

| Ex. 1. | Ex. 2. | |-------|-------| | 245.118 | .2105264 | | .3529+ | 2.875 | | 2206062 | 1052630 | | 490236 | 1473682 | | 1225590 | 1684208 | | 735354 | 421052 | | 86.5021422 | .605262250 |

In Ex. 1. the integral part of the product, viz. 86, is certain, and all the decimal places on the right are uncertain. In Ex. 2. only four places on the left, viz. .6052, are certain, and all the other places uncertain.

The reason of Rule 1. is plain. For if in Ex. 1. we make the longest factor the multiplier, and the total product will be the same either way, it is obvious, that in this case we shall have six particular products, in each of which the right-hand figure will be uncertain, and consequently... consequently we shall have six uncertain places in the total product toward the right, and also one uncertain place more on account of the uncertain carriage from the column in which the right-hand figure of the last particular product stands.

The reason of Rule 2. is also obvious. For in Ex. 2., by making the finite factor the multiplier, we have four particular products, in each of which the right-hand figure is uncertain; and so we have four uncertain places in the total product, and one uncertain place more arising from the uncertain carriage.

The carriage in some cases may affect several columns on the left, and thereby render so many more figures uncertain.

The surest way therefore to determine the certain places in the product of approximates, is by a second operation, giving the approximates contrary signs; for then, so far as the two products agree, the figures are certain. The second operations of the two former examples follow.

| Ex. 1. | Ex. 2. | |-------|-------| | 245.117+ | .210527- | | -353- | 2.875 | | 235351 | 1052635 | | 1225585 | 1473689 | | 735351 | 1684216 | | 86.526301 | 421054 |

In Ex. 1. 865 is certain, and all the other figures uncertain. In Ex. 2., .60526 are the only certain places.

Because the multiplication of decimals that consist of many places, proves, in the way hitherto practised, a tedious operation, we shall here explain a method whereby decimals of this sort, whether finite or approximate, may be multiplied expeditiously, and at the same time have the decimal places in the product limited to any number proposed. This may be effected by the following

**Rule.** Under the multiplicand place the multiplier inverted, so that its units place may stand under that place of the multiplicand to which you propose to limit the product; then multiply the right-hand figure of the multiplier into that figure of the multiplicand which stands directly over it, taking in the carriage from the right, and go on to multiply it into all the other figures on the left. Proceed in like manner with every other figure of the multiplier, placing the right-hand figures of all the particular products directly under other. The total product will be approximate, and the right-hand figure uncertain.

To make this rule more easily understood, the reader may look back to the multiplication of integers; where it was observed, that instead of beginning with the right-hand figure of the multiplier, we may begin with the left, and still have a just product, provided the right-hand figure of every particular product be placed directly under the multiplying figure. Now, the working an example, both in this manner, and also by the rule, and comparing the steps and results of the two opera-

tions, will throw a light upon the matter, and unfold the reason of the rule. Which take as follows.

Multiply 18.634375 into 9.875, and limit the product to four decimal places.

By the rule. 18.634375 578.9 1677093 149075 13044 931

184.0143+

By the other method. 18.634375 A 9.875 167709375 149075000 130440625 9317875

184.014453125

In working by this rule, you invert the multiplier, and place 9, the units figure, under 3, the fourth place of decimals, because the product is limited to four decimal places; then multiply, saying, \(9 \times 3 = 27\), and 6 carried from the right makes 33, &c. In multiplying by 8, say, \(8 \times 4 = 32\), and 3 of carriage makes 35; to set 5 under 3; and proceed in like manner to multiply the figures on the left. The right-hand figure of the product is defective, as wanting the carriage from the columns cut off on the right by the line A B. The figures expressing the sum of the columns so cut off, are so many uncertain places of the product, when the factors are approximate, and on that account to be rejected as useless. The figures, moreover, on the right of the line A B, show how far the operation is contracted, or how much labour is saved in working by the rule.

If there be no units in the multiplier, in this case set the right-hand figure of the inverted multiplier under that figure of the multiplicand, below which it would have stood had there been units.

**Ex.** Multiply .825 by .825, limiting the product to three decimal places.

By the rule. .825 528. 660 16 4

.680+

The common way at large. .825 .825 640 16 4

.680625

The decimal places of the factors may either be retained at full length, or turned into approximates before you begin to multiply.

**Ex.** Multiply 25.849013625 by 42.97235, limiting the product to two decimal places.

25.849013625 53279.24 103396 5169 2326 130

1110.76+ We shall next turn the decimals of the former example into approximates, and then the operation will be as follows.

By the rule. Common way at large.

\[ \begin{array}{c} 25.85 \\ +79.24 \\ \hline 105.09 \end{array} \]

\[ \begin{array}{c} 25.85 \\ +79.24 \\ \hline 105.09 \end{array} \]

Prod. 1110.76

It remains to be observed, that the want of carriage from the right hand may sometimes affect more columns on the left than one, and thereby occasion more uncertain figures in the product than that on the right hand. The best security on this head is, never to limit the product to fewer than four or five decimal places.

To conclude, when decimals to be multiplied are long, you may frequently perform the operation more easily in vulgar fractions, and then reduce the product to a decimal.

**Rule II.** If the multiplier be finite, and the multiplicand repeat, in multiplying carry at 9 on the right hand; and before you add, prolong the repetends of the particular products, till their right-hand figures stand directly under one another; and in adding, carry at 9 on the right hand.

The product repeats, as in Ex. 1, 2, &c.; or turns out finite, as in Ex. 6.

| Ex. 1 | Ex. 2 | Ex. 3 | Ex. 4 | |-------|-------|-------|-------| | .3 | .16 | 27.08 | 354.26 | | 4 | 7 | .5 | .08 | | 1.3 | 1.16 | 13.5416 | 28.3413 |

| Ex. 5 | Ex. 6 | |-------|-------| | 4.09 | 6.43 | | 5.2 | 123 | | 81.3 | 1930 | | 20333 | 12866 | | | 64333 | | 21.146 | 791.30 |

Note. If the multiplier has ciphers on the right, instead of annexing ciphers to the product, reiterate its right-hand figure so many times as there are ciphers.

| Ex. 1 | Ex. 2 | |-------|-------| | 79.8 | 874.3 | | 50 | 900 | | 3983.3 | 786900.0 |

**Rule III.** If the multiplier be finite, and the multi-

plicand circulate, to the product of the right-hand figure of the circle add the carriage from the left, then proceed as in multiplication of integers; but before you add the particular products, make them conterminous, and then add as in addition of circulates.

The product commonly circulates; and then its circle is similar to the circle of the multiplicand, as in Ex. 1, and 2.; but the product sometimes repeats, as in Ex. 5.; or it may turn out finite, as in Ex. 6.

Here it is obvious, that in multiplying .481 by 7, the carriage from the left would be 3; to say, \(7 \times 1 = 7\), and 3 of carriage, make 10, &c. The product circulates, and its circle .370, is similar to .481, the circle of the multiplicand.

Ex. 2. Ex. 3. Ex. 4. 7.518, 7.518, 7.518, -5 -05 -005 3.7592, 3.7592, 3.7592,

In the above three examples the products are mixt circulates, the three figures on the right being the circle, and the figures on the left the finite parts.

**Rule IV.** If the multiplier be interminate, reduce it to a vulgar fraction, as directed in reduction of decimals, Prob. V.; then multiply the given multiplicand by the numerator, (working as in integers, if the multiplicand be finite; or as directed in Rule 2. if it repeat; or as prescribed in Rule 3. if it circulate); and divide the product by the denominator.

Here there are six cases; for the multiplier may repeat or circulate, and may multiply a finite, a repeating, or circulating multiplicand.

**Case I.** When a repeating multiplier multiplies a finite multiplicand.

**Example.** Multiply 638.25 by .4 = 3

\[ \begin{array}{c} 638.25 \\ \times 0.4 \\ \hline 2553.00 \end{array} \]

**Case II.** When both factors repeat:

**Example.** **ARITHMETICK**

**Examp.** Multiply 6.83 by \( \frac{7}{9} = \frac{2}{3} \)

\[ \begin{array}{c} 6.83 \\ \times 7 \\ \hline 47.83 \\ \end{array} \]

9) 47.83 (5.3148, prod.

\[ \begin{array}{c} 45 \\ -23 \\ 27 \\ \times 13 \\ 9 \\ \end{array} \]

\[ \begin{array}{c} 43 \\ 36 \\ \end{array} \]

\[ \begin{array}{c} 73 \\ 72 \\ \end{array} \]

* 1

**Case III.** When a repeating multiplier multiplies a circulating multiplicand.

**Examp.** Multiply 24.36, by \( \frac{4}{9} = \frac{1}{3} \)

\[ \begin{array}{c} 24.36 \\ \times 4 \\ \hline 97.45 \\ \end{array} \]

9) 97.45 (10.82, prod.

\[ \begin{array}{c} 9 \\ \times 74 \\ 72 \\ \end{array} \]

\[ \begin{array}{c} 25 \\ 18 \\ \end{array} \]

* 74

**Case IV.** When a circulate multiplies a finite multiplicand.

**Examp.** Multiply 825 by \( .36 = \frac{16}{99} \)

\[ \begin{array}{c} 825 \\ \times 36 \\ \hline 4950 \\ 2475 \\ \end{array} \]

99) 29700 (300 product

297

**Case V.** When a circulate multiplies a repeating multiplicand.

**Examp.** Multiply 8.02083 by \( .72 = \frac{16}{99} \)

\[ \begin{array}{c} 8.02083 \\ \times 72 \\ \hline 1604166 \\ 56145833 \\ \end{array} \]

99) 577.50000 (5.83

495

\[ \begin{array}{c} 825 \\ 792 \\ \end{array} \]

\[ \begin{array}{c} 330 \\ 297 \\ \end{array} \]

* 33

**Case VI.** When both factors circulate.

**Examp.** Multiply \( .714285 \), by \( .36 = \frac{16}{99} \)

\[ \begin{array}{c} .714285 \\ \times 36 \\ \hline 4.285714 \\ 21.428571 \\ \end{array} \]

99) 25.7142857

2571428

25714

257

Prod. \( .25974025 = .25974025 \)

The circle of the first product is always similar to that of the multiplicand, and in the above example consists of six places; but to secure the carriage from right to left, and thereby complete the circle of the quot or total product, transfer 7, the left-hand figure of the circle of the first product, to the right, and fill up the places under it with the figures that come in course, and from the sum of these figures on the right carry 2, which completes the circle of the total product.

**VI. Division of Decimals.**

Before we enter on division, it will be proper to observe, that there are two rules for pointing the quot, both which are general in their nature, and extend to decimals of every sort, whether terminate or interminate; but unwilling to perplex the learner with too many things at once, we shall at present lay down only one of these rules; and afterwards, when the rule now to be assigned appears to be sufficiently exemplified, shall then bring the other rule upon the field.

**General Rule.**

The decimal places in the divisor and quot together must always be equal in number to those of the dividend. The five following practical directions will make the application of the general rule easy.

1. When the divisor and dividend have an equal number of decimal places, the quot comes out an integer; as in Ex. 2.

2. When the decimal places of the dividend are more than those of the divisor, the number of decimal places in the quot must be equal to the excess; as in Ex. 1.

3. When the decimal places of the divisor are more than those of the dividend, annex ciphers to the dividend, so as to make them equal, and the quot, by direction 1, will be integers; as in Ex. 3, 5, and 7.

4. When, after division is finished, the quot has not so many figures, as, by the general rule, it ought to have decimal places, supply that defect by prefixing ciphers; as in Ex. 6.

5. If, after the dividend is exhausted, there be a remainder, annex a cipher, or ciphers, to the remainder, and continue the division till o remain, or till the quot repeat or circulate, or till you think proper to limit it; as in Ex. 9, 10, 11, and 12.

We now proceed to division.

Rule I. If the divisor and dividend are both finite or approximate, work exactly as in division of integers.

Ex. 1. \[ \begin{array}{c} \frac{75}{58875} \\ \end{array} \]

Ex. 2. \[ \begin{array}{c} \frac{25}{182573} \\ \end{array} \]

In Ex. 1. A decimal divides a decimal; and because the dividend has five decimal places, and the divisor only two, give three decimal places to the quot, according to Direction 2.

In Ex. 2. A mixt number divides a mixt number, and the divisor and dividend having an equal number of decimal places, the quot comes out an integer, according to Direction 1.

The reason of the rule for pointing the quot is obvious; for multiplication gives as many decimal places to the product as are in both factors; but the dividend is the product of the divisor and quot, and so has as many decimal places as are in both; consequently the decimal places in the divisor and quot together must be equal in number to those of the dividend.

Ex. 3. \[ \begin{array}{c} \frac{85476}{8547600} \\ \end{array} \]

Ex. 4. \[ \begin{array}{c} \frac{7875}{787500} \\ \end{array} \]

In Ex. 3. A decimal divides an integer; and the dividend having no decimal place, annex two ciphers, be-

Vol. I. No. 18. In Ex. 11. An integer divides an integer; and the dividend being less than the divisor, annex a cipher to it; again, after the dividend is exhausted, annex a cipher to the remainder, and continue the division till you find the quot circulates.

In Ex. 12. A mixt number divides a mixt number: and after the dividend is exhausted, by annexing ciphers to the remainder, continue the division till the quot has three decimal places; and as there is still a remainder, it might be carried further; but three decimal places being in most cases sufficiently accurate, here you may limit it; so the quot is approximate.

In division of decimals, the place of the first figure of the quot may likewise be known from the first dividend, much after the same manner as in division of integers, by the following

II. General Rule.

The place of the first figure of the quot is the same with the place of that figure in the dividend which stands over the units of the first product.

Thus, in the example of integers in the margin, the figure o, that stands over 5, the units of the product of $9 \times 35$, is in the place of hundreds; and therefore o, the first figure of the quot, is likewise hundreds; and so the quot is 917 integers.

To illustrate the rule, we shall give decimal places to the dividend of the above example; and thereby exhibit the varieties that will occur in pointing the quot.

| Variety 1 | Var. 2 | |-----------|--------| | 35)32095(917 | 35)32095(917 | | 315 | 315 | | 59 | 59 | | 35 | 35 | | 245 | 245 | | 245 | 245 |

In Var. 1. say, $9 \times 3.5 = 31.5$; and the unit 1 standing under the place of thousands, the figure 9 is also thousands; and as o at last remains, annex a cipher for the decimal .5 in the divisor; then dividing, you get o to the quot; and because 9 stands in the place of thousands, the quot is wholly integers. In Var. 2, the unit 3 stands under the place of ten-thousands, and so 9 is ten-thousands; and to the remainder 0 annex . . . , for the two decimal places in the divisor; then dividing, you get 00 to the quot; and because 9 stands in the place of ten-thousands, the quot continues to be wholly integers. The process is the same in Var. 3, and 4.

Lastly, we shall allow decimal places to both dividend and divisor.

| Var. 1 | Var. 2 | |-------|-------| | 3.5)320.95(.917 | 3.5)32.095(.917 | | 31.5 | 31.5 | | 59 | 59 | | 35 | 35 | | 245 | 245 | | 245 | 245 |

Var. 3:

| Var. 4 | |-------| | 3.5)32005(.917 | 3.5)32095(.917 | | 31.5 | 31.5 | | 59 | 59 | | 35 | 35 | | 245 | 245 | | 245 | 245 |

In Var. 1, the units of the first product stand under tens of the dividend; and so 9, the first figure of the quot, is tens. In Var. 2, the units of the first product stand under units of the dividend, and so 9 is units. In Var. 3, the units of the first product stand under primes, and so 9 is primes, &c.

To divide by 10, 100, 1000, &c. is to move the decimal point one place toward the left for every cipher in the divisor.

Thus,

| 10)768(76.8 | 10)17.28(1.728 | | 100)768(76.8 | 100)17.28(1.728 | | 1000)768(768 | 1000)17.28(1.728 | | 10000)768(.0768 | 10000)17.28(.001728 |

APPROXIMATES.

In dividing approximates, the certain places of the quot may be determined by the following

RULE. Place the divisor under the first dividual, and the number of certain figures in the quot shall be one less than the number of places from the left of the divisor to the first + or —, whether in the divisor or in the dividend.

Ex. 1.

Dividend 1110.79286078— Dividend 1110.7929—

Divisor 42.9723+ Divisor 25.8490136+

Certain places five, where- Certain places six, where-

of three are decimals. of four are decimals.

But here it is to be observed, that the uncertain carriage may, in some cases, effect several columns on the left, and thereby render more figures of the quot uncertain than the rule prescribes. The surest way, therefore, is, to make two operations with contrary signs, and then the figures in which the two quots agree are certain.

In order to make the reason of the rule appear, it will be necessary to work an example.

Example.

A

42.9723+1110.79286078—(25.849.

859446

2513468

2148615

3648536

3437784

2107520

1718892

3886287

3867507

18780

B

Here we stop, no more places being certain. The reason is obvious; for the right-hand figure of the first product, viz., 6, is uncertain; and consequently all the figures under it, on the right of the line A B, will be so too; that is, the last remainder and new dividual are uncertain, and of course the figure that would go next to the quot.

From this example it appears, that all the figures on the right of the line A B are uncertain and useless; if therefore a way of working, without writing down these useless figures, can be found, we shall then have a method of dividing long decimals, whether finite or approximate, so as to contract the operation, and limit the product to any number of decimal places proposed. And this may be effected by observing the following

RULE. Write the product of the first quotient-figure under the dividend; and from the situation of the units place, consider how many figures of the dividend must be retained to give the quot the number of decimal places intended; cut off the other figures on the right, and also the figures corresponding to them on the right of the divisor; then subtract; esteem this and every following remainder a new dividual; and for each new dividual drop a figure on the right of the divisor; but in multiplying the quotient-figures into the divisor, take in the carriage from the right hand; as in the following examples.

Ex. I. Divide 95.432756463275 by 3.4637528; and limit the quot to four decimal places. In working the same example at large, the line A B shows how far the operation is contracted, and how much labour is saved.

But here observe, that by the rule for approximates the certain places of the quot are no more than five, viz. 27-551. And therefore, in all operations of this kind, care should be taken to limit the quot to so many places certain; as is done in the following example.

**Exampl. II.** Divide 87.0763264525 by 9.365407024; limiting the quot to four decimal places certain.

\[ \begin{array}{c} 9.365407024)87.0763264525(9.2976 \\ \hline 84.288663216 \\ \end{array} \]

\[ \begin{array}{c} 278766 \\ 187308 \\ \end{array} \]

\[ \begin{array}{c} 91458 \\ 84288 \\ \end{array} \]

\[ \begin{array}{c} 7170 \\ 6555 \\ \end{array} \]

\[ \begin{array}{c} 615 \\ 561 \\ \end{array} \]

Here we put a stop to the operation; because, by the rule for approximates, the next figure of the quot would be uncertain.

We shall conclude division of finite decimals with two very useful problems.

**Prob. I.** From a given multiplier to find a divisor that gives a quot equal to the product.

**Rule.** Divide an unit with ciphers annexed by the given multiplier, and the quot will be the divisor sought.

**Exampl.** What divisor will give a quot equal to the product of 125 into the dividend?

Given multiplier 125)1.000(.008 divisor sought.

Now, if any number be divided by .008, and the same number be multiplied by 125, the quot and product will be equal.

\[ \begin{array}{c} .008)7315.000(914375 quot. \\ \hline 72 \\ \end{array} \]

\[ \begin{array}{c} 11 & 7315 \\ 8 & 125 \\ \end{array} \]

\[ \begin{array}{c} 35 & 36575 \\ 32 & 14630 \\ \end{array} \]

\[ \begin{array}{c} 30 & 9315 \\ 24 & 914375 product. \\ \end{array} \]

The reason is plain: for an unit contains the quot .008 just 125 times; and consequently .008 dividing any number