Case IV. "When the rate consists of farthings only, find the value in pence, and reduce it by division to pounds."

Ex. 1st.] 37843 at 1 farthing.

At 4 d. = 1 of 1 s. 1095 8 At 1 d. = 1 of 4 d. 273 11 At 1 f. = 1 of 1 d. 68 5

At 5 l. 1438 1 L. 71 18 1 2d.] 4573 at 2½ d.

At 2 d. = 1 of 1 s. 762 2 At ½ d. = 1 of 2 d. 190 6 At ¼ d. = 1 of ½ d. 85 3

At 2½ l. 1037 11 L. 51 17 11 3d.] 2842 at 3½ d.

At 3 d. = 1 of 1 s. 710 6 At 3f. = 1 of 3 d. 177 7½

At 3½ d. 887 1½ L. 44 8 1½ 4th.] 3572 at 7½ d.

At 6 d. = 1 of 1 s. 1386 At 1½ d. = 1 of 6 d. 346 6

At 7½ d. 1732 6 L. 87 12 6

It is sometimes best to join some of the pence with the farthings in the calculation. Thus, in Ex. 4, we reckon the value at 6 d. and at 3 halfpence which makes 7½ d.

If the rate be 1½ d. which is an eighth part of a shilling, the value is found in shillings, by dividing the quantity by 8.

Case V. "When the rate consists of shillings and lower denominations."

Method 1. Multiply the quantity by the shillings, and find the value of the pence and farthings, if any, from the proportion which they bear to the shillings. Add and reduce.

Ex. 1st.] 4258 at 17 s. 3 d.

17 29806 4258 72386 17 s. 72386 3d. = 1 of 1 s. 1064 6 17 s. 3 d. 73450 6 L. 3672 10 6 2½] 5482 at 12 s. 4½ d.

12 65784 3 d. = 1 of 1 s. 1370 6 1½ d. = 1 of 3 d. 685 3 12 s. 4½ d. 67839 9 L. 3391 19 9

Method 2. Divide the rate into aliquot parts of a pound; calculate the values corresponding to these, as directed in Case I., and add them.

Ex. 1st.] 3894 at 17 6 2½] 1765 at 9 2

10s. = 1 L. 1947 5 = 1 973 10 2 6d. = 1 486 15 17½ d. L. 3407 5

Sometimes part of the value is more readily obtained from a part already found; and sometimes it is easiest to calculate at a higher rate, and subtract the value at the difference.

3d.] 63790 at 5 4 4th.] 3664 at 14 9

4s. = 1 L. 12758 1s. 4d. = 1 of 4s. 4252 13 4 5s. = 1 of 10s. 916 5½ 4d. L. 17010 13 4 15s. 2748 3d. = 1 of 5s. 45 16

Method 3. If the price contain a composite number of pence, we may multiply the value at a penny by the component parts.

Ex. 5628 at 2 s. 11 d. or 35 d.

12] 5628 20) 469 L. 23 9 5 L. 117 5 7 L. 820 15 Case VII. "When the rate consists of pounds and lower denominations."

Method 1. Multiply by the pounds, and find the value of the other denominations from the proportion which they bear to the pounds.

Ex. 1st.] 3592 at L. 3 : 12 : 8.

\[ \begin{align*} L. & \quad 3 \\ & \quad 12 \\ & \quad 8d = \frac{1}{4} \text{ of } L. 3 \\ & \quad 119 \quad 14 \quad 8 \end{align*} \]

L. 3 12 8

\[ \begin{align*} L. & \quad 13050 \quad 18 \quad 8 \\ & \quad 543 \text{ at } L. 2 : 5 : 10. \end{align*} \]

L. 2

\[ \begin{align*} & \quad 1086 \\ & \quad 5s = \frac{1}{4} \text{ of } L. 1 \\ & \quad 135 \quad 15 \\ & \quad 10d = \frac{1}{8} \text{ of } 5s \\ & \quad 22 \quad 12 \quad 6 \\ & \quad \frac{1}{4} d = \frac{1}{16} \text{ of } 10d \\ & \quad 1 \quad 2 \quad 7 \frac{1}{2} \end{align*} \]

L. 1245 10 1\(\frac{1}{2}\)

Method 2. Reduce the pounds to shillings, and proceed as in Case VI.

Ex. 1st.] 3592 at L. 3 : 12 : 8

\[ \begin{align*} & \quad 72 \quad 20 \quad 45 \\ & \quad 7184 \quad 72 \quad 18415 \\ & \quad 25144 \quad 14732 \end{align*} \]

At 45s. 165735

\[ \begin{align*} & \quad 4d = \frac{1}{8} s. \quad 1197 \quad 4 \\ & \quad 4d = \frac{1}{8} s. \quad 1197 \quad 4 \end{align*} \]

44s. 11d. 165427 1

8d. 261018 8

L. 13050 18 8

L. 8271 7 1

The learner should at first try every calculation more ways than one; which will not only serve the purpose of proving the operation, but will render him expert at discovering the best method for solving each question, and will lead him to invent other methods; for we have not exhausted the subject.

Thus, if the number of articles be 20, each shilling of the rate makes a pound of the amount. If it be 12, each penny of the rate makes a shilling of the amount. If 240, each penny of the rate makes a pound of the amount. If 480, each half-penny makes a pound. If 960, each farthing makes a pound. If the number of articles be a multiple, or an aliquot part of any of these, the amount is easily calculated. And if it be near to any such number, we may calculate for that number, and add or subtract for the difference.

We have hitherto explained the various methods of computation, when the quantity is a whole number, and in one denomination. It remains to give the proper directions when the quantity contains a fraction, or is expressed in several denominations.

When the quantity contains a fraction, work for the integers by the preceding rules, and for the fraction take proportional parts.

When the quantity is expressed in several denominations, and the rate given for the higher; calculate the higher, consider the lower ones as fractions, and work by the last rule.

When the rate is given for the lower denomination, reduce the higher denomination to the lower, and calculate accordingly.

Vol. I.

Note 1st. 7 lb. 14 lb. and 21 lb. are aliquot parts of 1 qr.; and 16 lb. is \(\frac{1}{4}\) of 1 cwt.; and are therefore easily calculated.

2nd. If the price of a dozen be so many shillings, that of an article is as many pence; and if the price of a groat be so many shillings, that of a dozen is as many pence.

3rd. If the price of a ton or score be so many pounds, that of 1 cwt. or a single article, is as many shillings.

4th. Though a fraction less than a farthing is of no consequence, and may be rejected, the learner must be careful lest he lose more than a farthing, by rejecting several remainders in the same calculation.

Sect. ii. Deductions on Weights, &c.

The full weight of any merchandise, together with that of the cask, box, or other package, in which it is contained, is called the gross weight. From this we must make proper deductions, in order to discover the quantity, for which price or duty should be charged, which is called the nett weight.

Tare is the allowance for the weight of the package; and this should be ascertained by weighing it before the goods are packed. Sometimes, however, particularly in payment of duty, it is customary to allow so much per C. or so much per 100 lb. in place of tare.

Tret is an allowance of 4 lb. on 104 granted on currants, and other goods on which there is waste, in order that the weight may answer when the goods are retailed.

Cluff, or Draught, is a further allowance granted on some goods in London, of 2 lb. on every 3 C. to turn the scale in favour of the purchaser. The method of calculating these and the like will appear from the following examples.

Ex. 1st. What is the nett weight of 17 C. 2 q. 14 lb. tare 18 lb. per cwt.

C. q. lb. C. q. lb.

\[ \begin{align*} & \quad 17 \quad 2 \quad 14 \text{ grofs. or } 17 \quad 2 \quad 14 \\ & \quad 16lb. = \frac{1}{2} C. \quad 2 \quad 2 \\ & \quad 2lb. = \frac{1}{4} \text{ of } 16lb. \quad 1 \quad 7 \\ & \quad 18lb. \quad 2 \quad 3 \quad 9 \frac{1}{2} \text{ tare. } \end{align*} \]

\[ \begin{align*} & \quad 317 \quad 1 \\ & \quad 14 \quad 3 \quad 4 \frac{1}{4} \text{ nett. } 28 \quad 317 \frac{1}{4} \text{ lb. C. q. lb. } \\ & \quad 4) \quad 11 \quad 9 \frac{1}{2} (2 \quad 3 \quad 9 \frac{1}{2} \text{ tare. } \end{align*} \]

In the first method, we add the tare at 16 lb., which is \(\frac{1}{4}\) of the gross weight to the tare, at 2 lb., which is \(\frac{1}{4}\) of the former. In the second, we multiply the gross weight by 18; the tare is 1 lb. for each cwt. of the product, and is reduced by division to higher denominations.

2nd.] What is Tret of 158 C. 2 q. 24 lb.?

C. q. lb. C. q. lb.

\[ \begin{align*} & \quad 158 \quad 2 \quad 26 \\ & \quad 26) \quad 158 \quad 2 \quad 26 \quad (6 - 11 \text{ Tret. } \end{align*} \]

Because tret is always 4 lb. in 104, or 1 lb. in 26, it is obtained by dividing by 26. 3rd.] What is the cloff on 28 C. 2 q.?

C. q. 28 2

This allowance being 2 lb. on every 3 C. might be found by taking \( \frac{1}{3} \) of the number of Cs and multiplying it by 2. It is better to begin with multiplication, for the reason given p. 663, col. 2 par. 1.

Sect iii. Commission, &c.

It is frequently required to calculate allowances on sums of money, at the rate of so many per L. 100. Of this kind is Commission, or the allowance due to a factor for buying or selling goods, or transacting any other business; Premium of Insurance, or allowance given for engaging to repay one's losses at sea, or otherwise; Exchange, or the allowance necessary to be added or subtracted for reducing the money of one place to that of another; Premiums on Stocks, or the allowance given for any share of a public stock above the original value. All these and others of a like kind are calculated by the following

Rule. "Multiply the sum by the rate, and divide the product by 100. If the rate contain a fraction, take proportional parts."

Ex. What is the commission on L. 728, at \( \frac{1}{4} \) per cent.

\[ \begin{array}{c} 728 \\ \times \frac{1}{4} \\ \hline 182 \\ \times 100 \\ \hline 18200 \\ \div 20 \\ \hline 910 \\ \end{array} \]

Answ. L. 910

When the rate is given in guineas, which is common in cases of insurance, you may add a twentieth part to the sum before you calculate. Or you may calculate at an equal number of pounds, and add a twentieth part to the answer.

When the given sum is an exact number of 10 pounds, the calculation may be done without setting down any figures. Every L. 10, at \( \frac{1}{4} \) per cent. is a shilling; and at other rates in proportion. Thus, L. 170, at \( \frac{1}{4} \) per cent. is 17 s.; and, at \( \frac{1}{4} \) per cent., 8 s. 6 d.

Sect iv. Interest.

Interest is the allowance given for the use of money by the borrower to the lender. This is computed at so many pounds for each hundred lent for a year, and a like proportion for a greater or a less time. The highest rate is limited by our laws to 5 per cent. which is called the legal interest; and is due on all debts constituted by bond or bill, which are not paid at the proper term, and is always understood when no other rate is mentioned.

The interest of any sum for a year, at any rate, is found by the method explained in the last section.

The interest of any number of pounds for a year, at 5 per cent. is one twentieth part, or an equal number of shillings. Thus, the interest of L. 34675 for a year is 34675 shillings.

The interest for a day is obtained by dividing the interest for a year, by the number of days in a year. Thus, the interest of L. 34675 for a day is found by dividing 34675 shillings by 365, and comes to 95 shillings.

The interest for any number of days is obtained by multiplying the daily interest by the number of days. Thus, the interest of L. 34675 for 17 days, is 17 times 95 shillings, or 1615 shillings; and this divided by 20, in order to reduce it, comes to L. 80 : 15 s.

It would have served the same purpose, and been easier to multiply at first by 17, the number of days; and, instead of dividing separately by 365, and by 20, to divide at once by 7300, the product of 365 multiplied by 20; and this division may be facilitated by the table inferred p. 661, col. 1.

The following practical rules may be inferred from the foregoing observations.

I. To calculate interest at 5 per cent. "Multiply the principal by the number of days, and divide the product by 7300."

II. To calculate interest at any other rate. "Find what it comes to at 5 per cent. and take a proper proportion of the same for the rate required."

Ex. 1st. Interest on L. 34675 for 17 days, at 5 per cent.

\[ \begin{array}{c} 34675 \\ \times 17 \\ \hline 242725 \\ \end{array} \]

L. s.

7300(589475)(80 15)

\[ \begin{array}{c} 5475 \\ \times 20 \\ \hline 109500 \\ \end{array} \]

Ex. 2nd. Interest on 304 : 3 : 4 for 8 days, at 4 percent.

L. 304 3 4

\[ \begin{array}{c} 7300(2433)68(68) \\ \end{array} \] Here the sums on either side of the account are introduced according to the order of the dates. Those on the D' side are added to the former balance, and those on the C' side subtracted. Before we calculate the days, we try if the last sum L. 91, be equal to the balance of the account, which proves the additions and subtractions; and, before multiplying, we try if the sum of the column of days be equal to the number of days, from 1st January to 31st July.

In the 5th and 6th multiplications, we begin at the pence-column, and take in the carriage. In the 7th, instead of multiplying the 6s. 8d. by 21, we add the third part 21 to the product, because 6s. 8d. is the third of a pound. This is done by marking down the second line 1287, instead of 1280. As the computation on the odd shillings and pence is troublesome, and makes a very small increase of the interest, some neglect them altogether; others add one to the pound, when the shillings exceed 10; and neglect them when below it.

Required interest on the following account to 31st December, allowing 5 per cent, when the balance is due to J. T. and 4 per cent, when due to N. W. D'. Mr J. T., his account current with N. W. C'.

Dec. 31. To balance L. 150 April 9. By cash. L. 70 Mar. 12. To cash 120 May 12. By cash 300 June 17. To cash 165 June 3. By cash 240 Sept. 24. To cash 242 Aug. 2. By cash 10 Oct. 9. To cash 178

Interest due to N. W. at 5 per cent. L. 6 8 9 Deduct 1/4 part 1 5 5

Due to N. W. at 4 per cent. L. 5 3 4 Due to J. T. at 5 per cent. 3 7 11 1/2

Balance due to N. W. L. 1 15 4 1/2 In this account, the balance is sometimes due to the one party, sometimes to the other. At the beginning, there is a balance due to N. W.; and, on the 9th of April, there is L. 200 due him. On the 12th of May, J. T. pays him L. 300, which discharges what he owed, and leaves a balance of L. 100 due him. The balance continues in J. T.'s favour till the 24th of September, when N. W. pays L. 242. These changes are distinguished by the marks D' and C'. The products are extended in different columns, and divided separately.

When payments are made on constituted debts, at considerable distances of time, it is usual to calculate the interest to the date of each payment, and add it to the principal, and then subtract the payment from the amount.

Ex. A bond for L. 540 was due the 18th Aug. 1772; and there was paid 19th March 1773 L. 50; and 19th December 1773 L. 25; and 23rd September 1774 L. 25; and 18th August 1775 L. 110. Required the interest and balance due on the 11th November 1775?

A bond due 18th August 1772 L. 540 Interest to 19th March 1773, 218 days L. 6 16 16 16 Paid 19th March 1773 L. 50

Balance due 19th March 1773 L. 506 16 Interest to 19th December 1773, 275 days 19 1 19 1 2 Paid 19th December 1773 L. 25

Balance due 19th December 1773 L. 500 3 8 Interest to 23rd September 1774, 278 days 19 9 19 9 Paid 23rd September 1774 L. 25

Balance due 23rd September 1774 L. 474 4 5 Interest to 18th August 1775, 39 days 23 5 3 23 5 3 Paid 18th August 1775 L. 110

Balance due 18th August 1775 L. 406 9 8 Interest to 11th November 1775, 85 days 4 14 6 4 14 6 Balance due 11th November 1775 L. 411 4 2 Amount of the interest L. 81 4 2

CHAP. VIII. VULGAR FRACTIONS.

In order to understand the nature of vulgar fractions, we must suppose unity (or the number 1) divided into several equal parts. One or more of these parts is called a fraction, and is represented by placing one number in a small character above a line, and another under it: For example, two fifth parts is written thus, \( \frac{2}{5} \). The number under the line (5) shows how many parts unity is divided into, and is called the denominator. The number above the line (2) shows how many of these parts are represented, and is called the numerator.

It follows from the manner of representing fractions, that, when the numerator is increased, the value of the fraction becomes greater; but, when the denominator is increased, the value becomes less. Hence we may infer, that, if the numerator and denominator be both increased, or both diminished, in the same proportion, the value is not altered; and therefore, if we multiply both by any number whatever, or divide them by any number which measures both, we shall obtain other fractions of equal value. Thus, every fraction may be expressed in a variety of forms, which have all the same signification.

A fraction annexed to an integer, or whole number, makes a mixed number. For example, five and two thirds, or \( \frac{5}{3} \). A fraction whose numerator is greater than its denominator is called an improper fraction. For example, seventeen third-parts, or \( \frac{17}{3} \). Fractions of this kind are greater than unity. Mixed numbers may be represented in the form of improper fractions, and improper fractions may be reduced to mixed numbers, and sometimes to integers. As fractions, whether proper or improper, may be represented in different forms, we must explain the method of reducing them from one form to another, before we consider the other operations.

Problem I. "To reduce mixed numbers to improper fractions; Multiply the integer by the denominator of the fraction, and to the product add the numerator. The sum is the numerator of the improper fraction sought, and is placed above the given denominator."

Ex. \( \frac{5}{3} = \frac{17}{3} \) 5 integer. 3 denominator. 15 product. 2 numerator given.

17 numerator sought.

Because one is equal to two halves, or 3 third parts, or 4 quarters, and every integer is equal to twice as many halves, or four times as many quarters, and so on; therefore, every integer may be expressed in the form of an improper fraction, having any assigned denominator: The numerator is obtained by multiplying the integer into the denominator. Hence the reason of the foregoing rule is evident. \( \frac{5}{3} \), reduced to an improper fraction, whose denominator is 3, makes \( \frac{15}{3} \), and this added to \( \frac{2}{3} \), amounts to \( \frac{17}{3} \).

Problem II. "To reduce improper fractions to whole or mixed numbers: Divide the numerator by the denominator."

Ex. \( \frac{11}{17} = \frac{6}{17} \) \( \frac{17}{11} \times \frac{12}{17} = \frac{1}{17} \) \( \frac{102}{17} \times \frac{1}{17} = \frac{6}{17} \)

This problem is the converse of the former, and the reason may be illustrated in the same manner.

Problem III. "To reduce fractions to lower terms. Divide both numerator and denominator by any number which measures both, and place the quotients in the form of a fraction."

Example. \( \frac{11}{18} = \frac{2}{3} = \frac{1}{3} \)

Here we observe that 135 and 360 are both measured by 5, and the quotients form \( \frac{27}{72} \), which is a fraction of the same value as \( \frac{1}{3} \) in lower terms. Again, 27 and 72 are both measured by 9, and the quotients form \( \frac{3}{8} \), which is still of equal value, and in lower terms.

It is generally sufficient, in practice, to divide by such measures as are found to answer on inspection, or by the rules given p. 659, col. 2. But, if it be required to reduce a fraction to the lowest possible terms, we must divide Chap. VIII.

Vulgar Fractions.

vide the nominator and denominator by the greatest number which measures both. What number this is may not be obvious, but will always be found by the following rule.

To find the greatest common measure of two numbers, divide the greater by the lesser, and the divisor by the remainder continually, till nothing remain; the last divisor is the greatest common measure.

Example. Required the greatest number which measures 475 and 589?

Here we divide 589 by 475, and the remainder is 114; then we divide 475 by 114, and the remainder is 19; then we divide 114 by 19, and there is no remainder: from which we infer, that 19, the last divisor, is the greatest common measure.

To explain the reason of this, we must observe, that any number which measures two others, will also measure their sum, and their difference, and will measure any multiple of either. In the foregoing example, any number which measures 589, and 475, will measure their difference 114, and will measure 456, which is a multiple of 114; and any number which measures 475, and 456, will also measure their difference 19. Consequently, no number greater than 19 can measure 589 and 475. Again, 19 will measure them both, for it measures 114, and therefore measures 456, which is a multiple of 114, and 475, which is just 19 more than 456; and, because it measures 475 and 114, it will measure their sum 589. To reduce \(\frac{4}{3}\) to the lowest possible terms, we divide both by numbers 19, and it comes to \(\frac{2}{1}\).

If there be no common measure greater than 1, the fraction is already in the lowest terms.

If the greatest common measure of 3 numbers be required, we find the greatest measure of the two first, and then the greatest measure of that number, and the third. If there be more numbers, we proceed in the same manner.

Problem IV. "To reduce fractions to others of equal value that have the same denominator."

Multiply the numerator of each fraction by all the denominators except its own. The products are numerators to the respective fractions sought."

Multiply all the denominators into each other; the product is the common denominator.

Ex. \(\frac{4}{9}\) and \(\frac{5}{9}\) and \(\frac{1}{9} = \frac{4}{9} + \frac{5}{9} + \frac{1}{9} = \frac{10}{9}\)

\(4 \times 9 \times 8 = 288\) first numerator.

\(7 \times 5 \times 8 = 280\) second numerator.

\(3 \times 5 \times 9 = 135\) third numerator.

\(5 \times 9 \times 8 = 360\) common denominator.

Here we multiply 4, the numerator of the first fraction, by 9 and 3 the denominators of the two others; and the product 288 is the numerator of the fraction sought, equivalent to the first. The other numerators are found in like manner, and the common denominator 360, is obtained by multiplying the given denominators 5, 9, 8, into each other. In the course of the whole operation, the numerators and denominators of each fraction are multiplied by the same number, and therefore their value is not altered.

The fractions thus obtained may be reduced to lower terms, if the several numerators and denominators have a common measure greater than unity. Or, after arranging the number for multiplication, as is done above, if the same number occur in each rank, we may dash them out and neglect them; and, if numbers which have a common measure occur in each, we may dash them out and use the quotients in their stead; or any number, which is a multiple of all the given denominators, may be used as a common denominator. Sometimes a number of this kind will occur on inspection, and the new numerators are found by multiplying the given ones by the common denominator, and dividing the products by the respective given denominators.

If the articles given for any operation be mixed numbers, they are reduced to improper fractions by problem I. If the answer obtained be an improper fraction, it is reduced to a mixed number by problem II. And, it is convenient to reduce fractions to lower terms, when it can be done, by problem III. which makes their value better apprehended, and facilitates any following operation. The reduction of fractions to the same denominator by problem IV. is necessary to prepare them for addition or subtraction, but not for multiplication or division.

1. Addition of Vulgar Fractions.

Rule. "Reduce them, if necessary, to a common denominator; add the numerators, and place the sum above the denominator."

Ex. \(1\). \(\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}\) by problem IV.

\(2\). \(\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1\) by problem II.

The numerators of fractions that have the same denominator signify like parts; and the reason for adding them is equally obvious, as that for adding shillings or any other inferior denomination.

Mixed numbers may be added, by annexing the sum of the fractions to the sum of the integers. If the former be a mixed number, its integer is added to the other integers.

2. Subtraction of Vulgar Fractions.

Rule. "Reduce the fractions to a common denominator; subtract the numerator of the subtrahend from the numerator of the minuend, and place the remainder above the denominator."

Ex. Subtract \(\frac{1}{2}\) from \(\frac{1}{2}\) remainder \(\frac{1}{2}\)

\(\frac{1}{2} - \frac{1}{2} = \frac{1}{2}\) by Prob. IV. take 24

Rem. 11

To subtract a fraction from an integer: subtract the numerator from the denominator, and place the remainder above the denominator; prefix to this the integer diminished by unity.

Ex. Subtract \(\frac{1}{2}\) from 12 remainder 11\(\frac{1}{2}\).

To subtract mixed numbers, proceed with the fractions by the foregoing rule, and with the integers in the common method. If the numerator of the fraction in the subtrahend exceed that in the minuend, borrow the value of the denominator, and repay it by adding 1 to the unit-place of the subtrahend.

Ex. Ex. Subtract $145\frac{1}{2}$ from $248\frac{1}{2}$

$\frac{1}{2} = \frac{3}{4}$

by Prob. IV.

$145\frac{1}{2} - 102\frac{1}{2} = 43$

Here, because $27$ the numerator of the fraction in the minuend is less than $35$, the numerator of the subtrahend, we borrow $45$ the denominator; $27$ and $45$ make $72$, from which we subtract $35$, and obtain $37$ for the numerator of the fraction in the remainder, and we repay what was borrowed, by adding $1$ to $5$ in the unit-place of the subtrahend.

The reason of the operations in adding or subtracting fractions will be fully understood, if we place the numerators of the fractions in a column like a lower denomination, and add or subtract them as integers, carrying or borrowing according to the value of the higher denomination.

3. Multiplication of Vulgar Fractions.

Rule. "Multiply the numerators of the factors together for the numerator of the product, and the denominators together for the denominator of the product."

Ex. $1\frac{1}{2} \times \frac{3}{4} = \frac{9}{8} = 1\frac{1}{8}$

$2 \times 5 = 10$ num.

$8\frac{1}{2} \times 7\frac{1}{2} = 65\frac{1}{2}$

$3 \times 7 = 21$ den.

$7\frac{1}{2} = \frac{15}{2}$ by prob. I.

$42 \times 31 = 1302$

$5 \times 4 = 20$

To multiply $\frac{1}{2}$ by $\frac{1}{2}$ is the same, as to find what two third parts of $\frac{1}{2}$ comes to; if one third part only had been required, it would have been obtained by multiplying the denominator $7$ by $3$, because the value of fractions is lessened when their denominators are increased; and this comes to $\frac{1}{4}$; and, because two thirds were required, we must double that fraction, which is done by multiplying the numerator by $2$, and comes to $\frac{1}{2}$. Hence we infer, that fractions of fractions, or compound fractions, such as $\frac{1}{2}$ of $\frac{1}{2}$ are reduced to simple ones by multiplication. The same method is followed when the compound fraction is expressed in three parts or more.

If a number be multiplied by any integer, its value is increased: if it be multiplied by $1$, or taken one time, it undergoes no alteration. If it be multiplied by a proper fraction, or taken for one half, two thirds, or the like, its value is diminished, and the product is less than the number multiplied.

The foregoing rule extends to every case, when there are fractions in either factor. For mixed numbers may be reduced to improper fractions, as is done in Ex. 2nd; and integers may be written, or understood to be written, in the form of fractions whose numerator is $1$. It will be convenient, however, to give some further directions for proceeding, when one of the factors is an integer, or when one or both are mixed numbers.

1st. To multiply an integer by a fraction, multiply it by the numerator, and divide the product by the denominator.

Ex. $3756 \times \frac{1}{2} = 2253\frac{1}{2}$

$5)11268(2253\frac{1}{2}$

2nd. To multiply an integer by a mixed number, we multiply it first by the integer, and then by the fraction, and add the products.

Ex. $138 \times 5\frac{1}{2} = 793\frac{1}{2}$

$138 \times \frac{1}{2} = 690$

$3$

$4)414(103\frac{1}{2}$

$793\frac{1}{2}$

3rd. To multiply a mixed number by a fraction, we may multiply the integer by the fraction, and the two fractions together, and add the products.

Ex. $15\frac{1}{2} \times \frac{3}{4} = 3\frac{3}{4}$

$15 \times \frac{3}{4} = 3\frac{3}{4}$

$\frac{1}{2} \times \frac{3}{4} = \frac{3}{8}$

$3\frac{3}{4}$

4th. When both factors are mixed numbers, we may multiply each part of the multiplicand first by the integer of the multiplier, and then by the fraction, and add the four products.

Ex. $8\frac{1}{2} \times 7\frac{1}{2}$

$8 \times 7 = 56$

$8 \times \frac{1}{2} = \frac{4}{2} = 2$ by prob. II.

$\frac{1}{2} \times 7 = \frac{7}{2} = 3\frac{1}{2}$

$\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

product $65\frac{1}{2}$ as before.

4. Division of Vulgar Fractions.

Rule I. "Multiply the numerator of the dividend by the denominator of the divisor. The product is the numerator of the quotient."

II. "Multiply the denominator of the dividend by the numerator of the divisor. The product is the denominator of the quotient."

Ex. Divide $\frac{1}{2}$ by $\frac{1}{2}$ Quotient $\frac{1}{2}$

$2 \times 9 = 18$

$5 \times 7 = 35$

To explain the reason of this operation, let us suppose it required to divide $\frac{1}{2}$ by $\frac{1}{2}$, or to take one seventh part of that fraction. This is obtained by multiplying the denominator by $7$; for the value of fractions is diminished by increasing their denominators, and comes to $\frac{1}{7}$. Again, because $\frac{1}{2}$ is nine times less than $7$, the quotient of any number divided by $\frac{1}{2}$ will be nine times greater than the quotient of the same number divided by $7$. Therefore we multiply $\frac{1}{7}$ by $9$, and obtain $\frac{1}{2}$.

If the divisor and dividend have the same denominator, it is sufficient to divide the numerators.

Ex. $\frac{1}{2}$ divided by $\frac{1}{2}$ quotes $4$.

The quotient of any number divided by a proper fraction is greater than the dividend. It is obvious, that any integer contains more halves, more third parts, and the like, than it contains units; and, if an integer and fraction be divided alike, the quotients will have the same proportion to the numbers divided; but the value of an integer is increased when the divisor is a proper fraction; therefore, the value of a fraction in the like case is increased also.

The foregoing rule may be extended to every case, by reducing integers and mixed numbers to the form of improper fractions. We shall add some directions for shortening the operation when integers and mixed numbers are concerned.

1st. When the dividend is an integer, multiply it by by the denominator of the divisor, and divide the product by the numerator.

Ex. Divide 368 by \( \frac{1}{4} \)

\[ \begin{array}{c} 7 \\ 5)2576 (515\text{ quotient}. \end{array} \]

2d. When the divisor is an integer, and the dividend a fraction, multiply the denominator by the divisor, and place the product under the numerator.

Ex. Divide \( \frac{1}{5} \) by 5 quotient \( \frac{1}{25} \)

\[ \begin{array}{c} 8 \times 5 = 40 \end{array} \]

3d. When the divisor is an integer, and the dividend a mixed number, divide the integer, and annex the fraction to the remainder; then reduce the mixed number, thus formed, to an improper fraction, and multiply its denominator by the divisor.

Ex. To divide \( 576\frac{4}{7} \) by 7 quotient \( 82\frac{4}{7} \)

\[ \begin{array}{c} 7)576 (82 \\ 56 \\ 16 \\ 14 \\ \end{array} \]

Here we divide 576 by 7, the quotient is 82, and the remainder 2, to which we annex the fraction \( \frac{4}{7} \); and reduce \( 2\frac{4}{7} \) to an improper fraction \( \frac{18}{7} \), and multiply its denominator by 7, which gives \( \frac{126}{7} \).

Hitherto we have considered the fractions as abstract numbers, and laid down the necessary rules accordingly. We now proceed to apply these to practice. Shillings and pence may be considered as fractions of pounds, and lower denominations of any kind as fractions of higher; and any operation, where different denominations occur, may be brought by expressing the lower ones in the form of vulgar fractions, and proceeding by the foregoing rules. For this purpose, the two following problems are necessary.

**Problem V.** "To reduce lower denominations to fractions of higher, place the given number for the numerator, and the value of the higher for the denominator."

Examples:

1. Reduce 7d. to the fraction of a shilling. Ans. \( \frac{7}{12} \) 2. Reduce 7d. to a fraction of a pound. Ans. \( \frac{7}{20} \) 3. Reduce 15s. 7d. to a fraction of a pound. Ans. \( \frac{157}{20} \)

**Problem VI.** "To value fractions of higher denominations, multiply the numerator by the value of the given denomination, and divide the product by the denominator; if there be a remainder, multiply it by the value of the next denomination, and continue the division."

Ex. 1st.] Required the value of \( \frac{1}{20} \) of L. 1.

\[ \begin{array}{c} 17 \\ 20 \\ \end{array} \]

Ex. 2nd.] Required the value of \( \frac{1}{20} \) of 1 Cwt.

\[ \begin{array}{c} 8 \\ 9 \\ \end{array} \]

\[ \begin{array}{c} s. d. \\ 4 \\ \end{array} \]

\[ \begin{array}{c} qrs. lb. \\ 3 \\ \end{array} \]

\[ \begin{array}{c} 60)340(5 \\ 300 \\ \end{array} \]

\[ \begin{array}{c} 27 \\ \end{array} \]

\[ \begin{array}{c} 40 \\ 12 \\ \end{array} \]

\[ \begin{array}{c} 28 \\ \end{array} \]

\[ \begin{array}{c} 60)480(9 \\ 480 \\ \end{array} \]

\[ \begin{array}{c} 9 \\ \end{array} \]

\[ \begin{array}{c} 50 \\ 45 \\ \end{array} \]

In the first example, we multiply the numerator 17 by 20, the number of shillings in a pound, and divide the product 340 by 60, the denominator of the fraction, and obtain a quotient of 5 shillings; then we multiply the remainder 40 by 12, the number of pence in a shilling, which produces 480, which divided by 60 quotes 8d. without a remainder. In the second example we proceed in the same manner; but as there is a remainder, the quotient is completed by a fraction.

Sometimes the value of the fraction does not amount to a unit of the lowest denomination; but it may be reduced to a fraction of that or any other denomination, by multiplying the numerator according to the value of the places. Thus \( \frac{1}{12} \) of a pound is equal to \( \frac{1}{12} \) of a shilling, or \( \frac{1}{12} \) of a penny, \( \frac{1}{12} \) of a farthing.

**Chap. IX. Decimal Fractions.**

**Sect. i. Notation and Reduction.**

The arithmetic of vulgar fractions is tedious, and even intricate to beginners. The difficulty arises chiefly from the variety of denominators; for when numbers are divided into different kinds of parts, they cannot be easily compared. This consideration gave rise to the invention of decimal fractions, where the units are divided into like parts, and the divisions and subdivisions are regulated by the same scale which is used in the Arithmetic of Integers. The first figure of a decimal fraction signifies tenth parts, the next hundredth parts, the next thousandth parts, and so on; and the columns may be titled accordingly. Decimals are distinguished by a point, which separates them from integers, if any be prefixed.

The use of cyphers in decimals, as well as in integers, is to bring the significant figures to their proper places, on which their value depends. As cyphers, when placed on the left hand of an integer, have no significance, but when placed on the right hand, increase the value ten times each; so cyphers, when placed on the right hand of a decimal, have no significance; but when placed on the left hand, diminish the value ten times each.

The notation and numeration of decimals will be obvious from the following examples:

- 4.7 signifies Four, and seven tenth-parts. - .47 Four tenth-parts, and seven hundredth-parts, or 47 hundredth-parts. - .047 Four hundredth-parts, and seven thousandth-parts, or 47 thousandth-parts. - .407 Four tenth-parts, and seven thousandth-parts, or 407 thousandth-parts. - 4.07 Four, and seven hundredth-parts. - 4.007 Four, and seven thousandth-parts.

The column next the decimal point is sometimes called decimal primes, the next decimal seconds; and so on.

To reduce vulgar fractions to decimal ones: "Annex a cypher to the numerator, and divide it by the denominator, annexing a cypher continually to the remainder."

Ex. The reason of this operation will be evident, if we consider that the numerator of a vulgar fraction is understood to be divided by the denominator; and this division is actually performed when it is reduced to a decimal.

In like manner, when there is a remainder left in division, we may extend the quotient to a decimal, instead of completing it by a vulgar fraction, as in the following example.

\[ \frac{25}{64} = 0.390625 \]

From the foregoing examples, we may distinguish the several kinds of decimals. Some vulgar fractions may be reduced exactly to decimals, as Ex. 1st. and 2nd., and are called **terminate** or **finite decimals**. Others cannot be exactly reduced, because the division always leaves a remainder; but, by continuing the division, we will perceive how the decimal may be extended to any length whatever. These are called **infinite decimals**. If the same figure continually returns, as in Ex. 3rd. and 4th., they are called **repeaters**. If two or more figures return in their order, they are called **circulates**. If this regular succession go on from the beginning, they are called **pure repeaters**, or **circulates**, as Ex. 3rd. and 5th. If otherwise, as Ex. 4th. and 6th., they are mixed repeaters or circulates, and the figures prefixed to those in regular succession are called the **finite part**. Repeating figures are generally distinguished by a dash, and circulates by a comma, or other mark, at the beginning and end of the circle; and the beginning of a repeater or circulate is pointed out in the division by an asterisk.

Lower denominations may be considered as fractions of higher ones, and reduced to decimals accordingly. We may proceed by the following rule, which is the same, in effect, as the former.

To reduce lower denominations to decimals of higher:

"Annex a cypher to the lower denomination, and divide it by the value of the higher. When there are several denominations, begin at the lowest, and reduce them in their order."

Ex. To reduce 5 cwt. 2 qr. 21 lb. to a decimal of a ton?

\[ \begin{array}{cccc} 28 & 210 & (75) & 4 \\ & 196 & 24 & 40 \\ & 140 & 35 & 168 \\ & 140 & 32 & 160 \\ & 0 & 30 & 87 \\ & 28 & 80 & \\ & 20 & 75 & 60 \\ & 0 & 150 & \\ & 140 & & \\ & 100 & & \\ & 100 & & \\ \end{array} \]

Here, in order to reduce 21 lb. to a decimal of 1 qr. we annex a cypher, and divide by 28, the value of 1 qr. This gives .75. Then we reduce 2.75 qrs. to a decimal of 1 cwt. by dividing by 4, the value of 1 cwt., and it comes to .6875. Lastly, 5.6875 cwt. is reduced to a decimal of a ton by dividing by 20, and comes to .284375.

To value a decimal fraction: "Multiply it by the value of the denomination, and cut off as many decimal places from the product as there are in the multiplicand. The rest are integers of the lower denomination."

Example. What is the value of .425 of L.1.

\[ \begin{array}{c} .425 \\ \times 20 \\ \hline 8.500 \\ \end{array} \]

d. 3,000

Sect. ii. Arithmetic of Terminate Decimals.

The value of decimal places decrease like that of integers, ten of the lower place in either being equal to one of the next higher; and the same holds in passing from decimals to integers. Therefore, all the operations are performed in the same way with decimals, whether... whether placed by themselves, or annexed to integers, as with pure integers. The only peculiarity lies in the arrangement and pointing of the decimals.

In addition and subtraction, "Arrange units under units, tenth-parts under tenth-parts, and proceed as in integers."

\[ \begin{array}{cccc} 321.4035 & \text{from} & 13.348 & \text{and} & 12.248 \\ 116.374 & \text{take} & 9.2993 & \text{and} & 10.6752 \\ 160.63 & & & & \\ 12.3045 & & 4.0487 & & 1.5728 \\ \end{array} \]

In multiplication, "Allow as many decimal places in the product as there are in both factors. If the product has not so many places, supply them by prefixing cyphers on the left hand."

Ex. 1st. \(1.37 \times 1.8 = 2.466\)

The reason of this rule may be explained, by observing, that the value of the product depends on the value of the factors; and since each decimal place in either factor diminishes its value ten times, it must equally diminish the value of the product.

To multiply decimals by 10, move the decimal point one place to the right; to multiply by 100, 1000, or the like, move it as many places to the right as there are cyphers in the multiplier.

In division, "Point the quotient so, that there may be an equal number of decimal places in the divisor as in the divisor and quotient together."

Therefore, if there be the same number of decimal places in the divisor and dividend, there will be as many in the quotient.

If there be more in the dividend, the quotient will have as many as the dividend has more than the divisor.

If there be more in the divisor, we must annex (or suppose annexed) as many cyphers to the dividend, as may complete the number in the divisor, and all the figures of the quotient are integers.

If the division leave a remainder, the quotient may be extended to more decimal places; but these are not regarded in fixing the decimal point.

The reason for fixing the decimal point, as directed, may be inferred from the rule followed in multiplication. The quotient multiplied by the divisor produces the dividend; and therefore the number of decimal places in the dividend is equal to those in the divisor and quotient together.

The first figure of the quotient is always at the same distance from the decimal point, and on the same side as the figure of the dividend, which stands above the unit place of the first product. This also takes place in integers; and the reason is the same in both.

It was formerly observed, that numbers were diminished when multiplied by proper fractions, and increased when divided by the same. Thus, multiplication by fractions corresponds with division by integers; and division by fractions with multiplication by integers; when we multiply by \(\frac{1}{2}\) or \(\frac{1}{3}\), we obtain the same an-

Sect. iii. Approximate Decimals.

It has been shown, that some decimals, though extended to any length, are never complete; and others, which terminate at last, sometimes consist of so many places, that it would be difficult in practice to extend them fully. In these cases, we may extend the decimal to three, four, or more places, according to the nature of the articles, and the degree of accuracy required, and reject the rest of it as inconsiderable. In this manner we may perform any operation with ease by the common rules, and the answers we obtain are sufficiently exact for any purpose in business. Decimals thus restricted are called approximate.

Shillings, pence, and farthings, may be easily reduced to decimals of three places, by the following rule. Take half the shillings for the first decimal place, and the number of farthings increased by 1, if it amount to 24, or upwards; by two, if it amount to 48 or upwards; and by three, if it amount to 72, or upwards, for the two next places.

The reason of this is, that 20 shillings make a pound, two shillings is the tenth part of a pound; and there- fore half the number of shillings makes the first decimal place. If there were 50 farthings in a shilling, or 1000 in a pound, the units of the farthings in the remainder would be thousandth-parts, and the tens would be hundredth-parts, and so would give the two next decimal places; but because there are only 48 farthings in a shilling, or 960 in a pound, every farthing is a little more than the thousandth-part of a pound; and since 24 farthings make 25 thousandth-parts, allowance is made for that excess by adding 1 for every 24 farthings, as directed.

If the number of farthings be 24, 48, or 72, and consequently the second and third decimal places 25, 50, and 75, they are exactly right; otherwise they are not quite complete, since there should be an allowance of \(\frac{1}{4}\) not only for 24, 48, and 72 farthings, but for every other single farthing. They may be completed by the following rule: Multiply the second and third decimal places, or their excess above 25, 50, 75, by 4. If the product amount to 24 or upwards, add 1; if 48, add 2; if 72, add 3. By this operation we obtain two decimal places more; and by continuing the same operation, we may extend the decimal till it terminate in 25, 50, 75, or in a repeater.

Decimals of sterling money of three places may easily be reduced to shillings, pence, and farthings, by the following rule. Double the first decimal place, and if the second be 5 or upwards, add 1 thereto for shillings. Then divide the second and third decimal places, or their excess above 50, by 4, first deducing 1, if it amount to 25, or upwards; the quotient is pence, and the remainder farthings.

As this rule is the converse of the former one, the reason of the one may be inferred from that of the other. The value obtained by it, unless the decimal terminate in 25, 50, or 75, is a little more than the true value; for there should be a deduction not only of 1 for 25, but a like deduction of \(\frac{1}{4}\) on the remaining figures of these places.

We proceed to give some examples of the arithmetic of approximates, and subjoin any necessary observations.

**Addition:**

| Cwt. qrs. lb. | Cwt. qrs. lb. | |---------------|---------------| | 3 2 14 = 3.625 | 3 2 2 = 3.51785 | | 2 3 22 = 2.94642 | 1 1 19 = 1.41964 | | 3 3 19 = 3.91964 | | | 4 1 25 = 4.47321 | 2 — 9 = 2.09821 |

14. 3 24 14.96427

If we value the sum of the approximates, it will fall a little short of the sum of the articles, because the decimals are not complete.

Some add 1 to the last decimal place of the approximate, when the following figure would have been 5, or upwards. Thus the full decimal of 3 cwt. 22 lb. is .946428571, and therefore .94643 is nearer to it than .94642. Approximates, thus regulated, will in general give exact answers, and sometimes above the true one, sometimes below it.

The mark + signifies that the approximate is less than the exact decimal, or requires something to be added. The mark — signifies that it is greater, or requires something to be subtracted.

**Multiplication:**

| Meth. 2d. | Meth. 3d. | |----------|----------| | 8278 | 8278 | | 2153 | 2153 | | 24834 | 16556 | | 41390 | 8278 | | 8278 | 41390 | | 16556 | 24834 | | 17822534 | 17822534 |

Here the four last places are quite uncertain. The right-hand figure of each particular product is obtained by multiplying 8 into the figures of the multiplier; but if the multiplicand had been extended, the carriage from the right-hand place would have been taken in; consequently the right-hand place of each particular product, and the four places of the total product, which depend on these, are quite uncertain. Since part of the operation, therefore, is useless, we may omit it; and, for this purpose, it will be convenient to begin (as in p. 638, col. 1, fifth variety) at the highest place of the multiplier.

We may perceive that all the figures on the right hand of the line in Method 2, serve no purpose, and may be left out, if we only multiply the figures of the multiplicand, whose products are placed on the right-hand of the line. This is readily done by inverting the multiplier in Method 3, and beginning each product with the multiplication of that figure which stands above the figure of the multiplier that produces it, and including the carriage from the right-hand place.

If both factors be approximates, there are as many uncertain places, at least in the product, as in the longest factor. If only one be an approximate, there are as many uncertain places as there are figures in that factor, and sometimes a place or two more, which might be affected by the carriage. Hence we may infer, how far it is necessary to extend the approximates, in order to obtain the requisite number of certain places in the product.

**Division:**

-3724—79864327+(2144 or 3#24)79864327(2144

| 7448 | 7448 | |------|------| | 5384 | 538 | | 3724 | 372 | | 16602 | 166 | | 14896 | 148 | | 17063 | 18 | | 14892 | 14 |

Here all the figures on the right hand of the line are uncertain; for the right-hand figure of the first product 7448 might be altered by the carriage, if the divisor were extended; and all the remainders and dividuals that follow are thereby rendered uncertain. We may omit these useless figures; for which purpose, we dash a figure on the right hand of the divisor at each step, and neglect it when we multiply by the figure of the quotient next obtained; but we include the carriage. The operation, and the reason of it, will appear clear, by comparing the operation at large, and contracted, in the above example. Chap. X. INTERMINATE DECIMALS.

Sect. i. REDUCTION OF INTERMINATE DECIMALS.

As the arithmetic of interminate decimals, otherwise called the arithmetic of infinities, is facilitated by comparing them with vulgar fractions, it will be proper to inquire what vulgar fractions produce the several kinds of decimals, terminate or interminate, repeaters or circulates, pure or mixed. And, first, we may observe, that vulgar fractions, which have the same denominator, produce decimals of the same kind. If the decimals corresponding to the numerator 1 be known, all others are obtained by multiplying these into any given numerator, and always retain the same form, providing the vulgar fraction be in its lowest terms.

Thus, the decimal equal to $\frac{1}{4}$ is .142857, which multiplied by

$$\frac{3}{4}$$

produces the decimal equal to $\frac{3}{4}$.142857.

Secondly, If there be cyphers annexed to the significant figures of the denominator, there will be an equal number of additional cyphers prefixed to the decimal. The reason of this will be evident, if we reduce these vulgar fractions to decimals, or if we consider that each cypher annexed to the denominator diminishes the value of the vulgar fraction ten times, and each cypher prefixed has a like effect on the value of the decimal.

Thus, $\frac{1}{10} = .142857$, $\frac{1}{100} = .0142857$, $\frac{1}{1000} = .00142857$, $\frac{1}{10000} = .000142857$.

We may therefore confine our attention to vulgar fractions, whose numerator is 1, and which have no cyphers annexed to the significant figures of the denominator.

Thirdly, Vulgar fractions, whose denominators are 2 or 5, or any of their powers, produce terminate decimals; for, if any power of 2 be multiplied by the same power of 5, the product is an equal power of 10, as appears from the following table:

| $2^x \times 5^y$ | $= 10^{x+y}$ | |-----------------|--------------| | $2^2 \times 5^1$ | $= 100$ | | $2^3 \times 5^2$ | $= 1000$ | | $2^4 \times 5^3$ | $= 10000$ |

And the reason is easily pointed out; for $2^x \times 5^y = 2^{x+y}$; or, because the factors may be taken in any order, $2^x \times 5^y = 2^{x+y}$; and this, if we multiply the factors by pairs, becomes $10 \times 10 \times 10$, or $10^3$.

The like may be shown of any other power. And we may infer, that, if any power of 10 be divided by a like power of 2 or 5, the quotient will be an equal power of 5 or 2 respectively, and will come out exact, without a remainder; and, since the vulgar fractions above mentioned are reduced to decimals by some such division, it follows that the equivalent decimals are terminate.

The number of places in the decimal is pointed out by the exponent of the power; for the dividend must be a like power of 10, or must have an equal number of cyphers annexed to 1, and each cypher of the dividend gives a place of the quotient.

Again, no denominators except 2, 5, or their powers, produce terminate decimals. It is obvious from p. 661, col. 2, par. 4, that, if any denominator which produces a terminate decimal be multiplied thereby, the product will consist of 1, with cyphers annexed; and consequently the lowest places of the factors, multiplied into each other, must amount to 10, 20, or the like, in order to supply a cypher for the lowest place of the product; but none of the digits give a product of this kind, except 5 multiplied by the even numbers; therefore one of the factors must terminate in 5, and the other in an even number. The former is measured by 5, and the latter by 2, as was observed p. 660, col. 2, par. 7. Let them be divided accordingly, and let the quotients be multiplied. This last product will be exactly one tenth-part of the former; and therefore will consist of 1, with cyphers annexed, and the factors which produce it are measured by 5 and 2, as was shewn before. This operation may be repeated; and one of the factors may be divided by 5, and the other by 2, till they be exhausted; consequently they are powers of 5 and 2.

Fourthly, Vulgar fractions, whose denominators are 3 or 9, produce pure repeating decimals.

Thus, $\frac{1}{3} = .333333$, $\frac{2}{3} = .666666$, $\frac{1}{9} = .111111$, $\frac{2}{9} = .222222$, $\frac{3}{9} = .333333$, $\frac{4}{9} = .444444$, $\frac{5}{9} = .555555$, $\frac{6}{9} = .666666$, $\frac{7}{9} = .777777$, $\frac{8}{9} = .888888$.

The repeating figure is always the same as the numerator. Hence we infer, that repeating figures signify ninth-parts; a repeating 3 signifies $\frac{1}{3}$; a repeating 6 signifies $\frac{2}{3}$; and a repeating 9 signifies $\frac{1}{9}$.

The value of repeating decimals may also be illustrated by collecting the values of the different places; for example, let the value of $\frac{1}{11}$ be required; the first decimal place signifies $\frac{1}{11}$, the next $\frac{1}{110}$, the next $\frac{1}{1100}$. The sum of the two first places is $\frac{1}{11}$ of the three places $\frac{1}{1100}$; and so on. If we subtract these values successively from $\frac{1}{11}$, the first remainder is $\frac{1}{11}$, the second $\frac{1}{110}$, the third $\frac{1}{1100}$. Thus, when the value of the successive figures is reckoned, the amount of them approaches nearer and nearer to $\frac{1}{11}$, and the difference becomes 10 times less for each figure assumed; and, since the decimal may be extended to any length, the difference will at last become so small, that it need not be regarded. This may give a notion of a decreasing series, whose sum may be exactly ascertained, though the number of terms be unlimited.

Fifthly, Vulgar fractions, whose denominators are a product of 3 or 9 multiplied by 2, 5, or any of their powers, produce mixed repeaters. The reason of this will be evident, if, in forming the decimal, we divide the numerator successively by the component parts of the denominator, as directed p. 660, col. 1, par. ult. The first divisor is 2, 5, or some of their powers, and consequently gives a finite quotient by p. 679, col. 1, par. 3, &c. The second divisor is 3 or 9; and therefore, when the figures of the dividend are exhausted, and figures annexed to the remainder, the quotient will repeat, by p. 679, col. 2, par. 2.

Ex. \( \frac{1}{14} = \frac{1}{14} \times 9 \)

\[ \begin{array}{cccc} 144)1.000(0.0693 & \text{or} & 16)1.000(0.0625 \\ 864 & & 96.00693 \\ 1360 & & 40 \\ 1296 & & 32 \\ 640 & & 80 \\ 576 & & 80 \\ 640 & & 0 \\ \end{array} \]

In order to illustrate this subject further, we shall explain the operation of calling out the threes, which resembles that for calling out the nines, formerly laid down, p. 663, col. 2, par. 4—p. 664, col. 2, par. 3, and depends on the same principles, being a method of finding the remainder of a number divided by 3. If the same number be divided by 3 and by 9, the remainders will either agree, or the second remainder will exceed the first by 3 or by 6. The reason of this will be obvious, if we suppose a collection of articles assorted into parcels of 3, and afterwards into parcels of 9, by joining three of the former together. If the lesser parcels be all taken up in compounding the greater ones, the remainder will be the same at the end of the second assortment as before; but, if one of these lesser parcels be left over, the remainder will be more, and if two of them be left over, the remainder will be 6 more. Therefore, when the nines are called out from any number, and the result divided by 3, the remainder is the same as when the number is divided by 3: Thus, the results on calling out the 3's may be derived from those obtained by calling out the 9's; and the same correspondence which was pointed out with respect to the latter, for proving the operations of arithmetic, applies also to the former.

To call out the 3's from any number, add the figures, neglecting 3, 6, and 9; and when the sum amounts to 3, 6, or 9, reject them, and carry on the computation with the excess only. For example, take 286754: in calling out the 3's, we compute thus, 2 and 8 is 10, which is three times 3, and 1 over; 1 and (falling by 6) 7 is 8, which is twice 3, and 2 over; 2 and 5 is 7, which is twice 3, and 1 over; Lastly, 1 and 4 is 5, which contains 3 once, and 2 over, so the result is 2.

If the 3's be called out from 2^2 or 4, the result is 1; from 2^3 or 8, the result is 2; from 2^4 or 16, the result is 1; and universally the odd powers of 2 give a result of 2, and the even powers give a result of 1. As every higher power is produced by multiplying the next lower by 2, the result of the product may be found by multiplying the result of the lower power by 2, and calling out the 3's, if necessary. Therefore, if the result of any power be 1, that of the next higher is 2, and that of the next higher (4 with the 3's called out) is 1. Thus the results of the powers of 2 are 1 and 2 by turns; also, because the result of 5, when the 3's are called out, is 2, its powers will have the same results as the corresponding powers of 2.

If the denominator be a product of an even power of 2 or 5, multiplied by 3, the repeating figure of the corresponding decimal is 3; but, if it be the product of an odd power, the repeating figure is 6. For, in forming the decimal, we may divide by the component parts of the denominator, and the first divisor is a power of 2 or 5; therefore the first quotient is a like power of 4 or 2, (p. 679, col. 1, par. 3, &c.), and this power is again divided by 3. If it be an even power, the remainder or result is 1, as was demonstrated above; and if cyphers be annexed to the remainder, and the division continued, it quotes a repeating 3; but if it be an odd power, the remainder is 2, and the quotient continued by annexing cyphers is a repeating 6.

If the denominator be 9, multiplied by 2, or any of its powers, the repeating figure may be found by calling out the 9's from the corresponding power of 5; and, if it be multiplied by 5 or any of its powers, by calling out the 9's from the corresponding power of 2. For if the decimal be formed by two divisions, the first quotes the corresponding power; and the second, because the divisor is 9, repeats the resulting figure after the dividend is exhausted.

If any mixed repeater be multiplied by 9, the product is a terminate decimal, and may be reduced (p. 679, col. 1, par. 3, &c.) to a vulgar fraction, whose denominator is 2, 5, or some of their powers; therefore all mixed repeaters are derived from vulgar fractions, whose denominators are products of 2, 5, or their powers, multiplied by 3 or 9.

Sixthly, All denominators, except 2, 5, 3, 9, the powers of 2 and 5, and the products of these powers, multiplied by 3 or 9, produce circulating decimals. We have already shown, that all terminate decimals are derived from 2, 5, or their powers; all pure repeaters, from 3 or 9; and all mixed repeaters, from the products of the former multiplied by the latter. The number of places in the circle is never greater than the denominator diminished by unity. Thus \( \frac{1}{7} \) produces .142857, a decimal of 6 places; and \( \frac{1}{17} \) produces .0588235294117647, a decimal of 16 places. The reason of this limit may be inferred from the division; for whenever a remainder which has recurred before returns again, the decimal must circulate, and the greatest number of possible remainders is one less than the divisor: But frequently the circle is much shorter. Thus \( \frac{1}{7} = .142857 \), a circle of 6 places.

When a vulgar fraction, whose numerator is 1, produces a pure circulate, the product of the circle multiplied by the denominator will consist of as many 9's as there are places in the circle. Thus \( \frac{1}{7} = .142857 \), which multiplied by 7 produces 999999. The like holds in every decimal of the same kind; for they are formed by dividing 10, or 100, or 1000, or some like number, by the denominator, and the remainder is 1, when the decimal begins to circulate; for the division must be then exactly in the same state as at the beginning: Therefore if the dividend had been less by 1, or had consisted entirely of 9's, the division would have come out without a remainder; and, since the quotient, multiplied by the divisor, produces the dividend, as was shown p. 661, col. 2, par. 3, it follows, that the circulating figures, multiplied by the denominator, produce an equal number of 9's.

Every vulgar fraction, which produces a pure circulate, Interminiate late, is equal to one whose numerator is the circulating decimals, figures, and its denominator a like number of 9's. If the numerator be 1, the vulgar fraction is reduced to that form by multiplying both terms into the circle of the decimal; and, if the numerator be more than 1, the equivalent decimal is found by multiplying that which corresponds to the numerator 1 into any other numerator.

Thus $\frac{1}{4} = .285714$, $\frac{3}{4} = .714285$, $\frac{5}{7} = .714285$, $\frac{6}{7} = .857142$, $\frac{7}{9} = .216$

Hence we may infer, that pure circulates are equal in value to vulgar fractions whose numerators consist of the circulating figures, and denominators of as many 9's as there are places in the circle. To place this in another point of view, we shall reduce a vulgar fraction, whose numerator consists entirely of 9's, to a decimal.

\[ \begin{array}{c} 999)375000(-375 \\ 2997 \\ \hline 753 \\ 6993 \\ \hline 5370 \\ 4995 \\ \hline 375 \end{array} \]

The remainder is now the same as the dividend, and therefore the quotient must circulate; and, in general, since any number with 3 cyphers annexed, may be divided by 1000, without a remainder, and quotes the significant figures; therefore, when divided by 999, it must quote the same figures, and leave an equal remainder. This also applies to every divisor which consists entirely of 9's. Circles of two places, therefore, signify ninety ninth-parts; circles of 3 places signify nine hundred and ninety ninth-parts; and so on.

The value of circulating decimals may also be illustrated by adding the values of the places. Thus, if two figures circulate, the first circle signifies hundredth-parts, and every following circle signifies one hundred times less than the preceding; and their values added, as in p. 679, col. 2, par. 3, will approach nearer to ninety ninth-parts than any assigned difference, but will never exactly complete it.

All denominators which are powers of 3, except 9, produce pure circulates; and the number of places in the circle is equal to the quotient of the denominator divided by 9.

Thus, $\frac{1}{9} = .037$, a circle of 3 places, and 27 divided by 9 = 3.

$\frac{1}{9} = .012345679$, a circle of 9 places, and 81 divided by 9 = 9.

These decimals may be formed, by dividing the numerator by the component parts of the denominator. In the first example, the component parts of the numerator are 9 and 3. The division by 9 quotes a pure circulate, and the circulating figure is not 3, 6, or 9, if the vulgar fraction be in its lowest terms. And any other repeating figure divided by 3, quotes a pure circulate of 3 places; for the first dividend must leave a remainder of 1 or 2. If the first remainder be 1, the second remainder is 2, (because, if 1 be prefixed to the repeating figure, and the 3's be called out, the result is 2); and, for a like reason, the third dividend clears off without a remainder. If the first remainder be 2, the second is twice 2 or 4, with the 3's called out, or 1, and the third 0; so the circle is always complete at 3 places, and the division begins anew.

The sum of such a circle cannot be a multiple of 3; for, since the repeating figure is not 3, nor any of its multiples, the sum of 3 places is not a multiple of 9, and therefore cannot be divided by 9, nor twice by 3, without a remainder.

Again, if the decimal equal to $\frac{1}{9}$ be divided by 3, we shall obtain the decimal equal to $\frac{1}{27}$. The dividend, as we have shown already, is a pure circulate of 3 places, whose sum is not a multiple of 3. Therefore, when divided by 3, the first circle leaves a remainder of 1 or 2, which being prefixed to the second, and the division continued, the remainder, at the end of the second circle, is 2 or 1, and, at the end of the third circle, there is no remainder; all which may be illustrated by calling out the 3's. The division being completed at 9 places, finishes the circle; and it may be shown, as before, that the sum of these places is not a multiple of 3. The learner will apprehend all this if he reduce these, or the like vulgar fractions, to decimals, by successive divisions.

\[ \begin{array}{c} 27 = 9 \times 3, \text{and } 610(111x), \text{and } 31111(037, \\ 81 = 27 \times 3, \text{and } 3037037037(012345679). \end{array} \]

For the same reason, if any circulating decimal, not a multiple of 3, be divided by 3, the quotient will circulate thrice as many places as the dividend; and, if any circulate obtained by such division be multiplied by 3, the circle of the product will be restricted to one third of the places in the multiplicand.

All vulgar fractions, whose denominators are multiples of 2, 5, or their powers, except those already considered, produce mixed circulates; for they may be reduced by dividing by the component parts of the denominator. The first divisor is 2, 5, or some of their powers, and therefore gives a finite quotient. The second divisor is none of the numbers enumerated p. 680, col. 2, par. 2, and therefore gives a circulating quotient when the significant figures of the dividend are exhausted, and cyphers annexed to the remainder.

\[ \begin{array}{c} Ex. \frac{1}{27} = 216 = 27 \times 8, \\ 216)1.000(004,629, \text{or } 8)1.000 \\ 864 \\ \hline 1360 \\ 1296 \\ \hline 640 \\ 432 \\ \hline 2080 \\ 1944 \\ \hline 1360 \end{array} \]

All mixed circulates are derived from vulgar fractions. Intermediate decimals of this kind, whose denominators are multiples of 2, 5, or their powers; and therefore all other denominators, except 3 and 9, produce pure circulates.

The reader will easily perceive, that, when a decimal is formed from a vulgar fraction, whose numerator is 1, when the remainder occurs in the division, the decimal is a pure circulate; but, if any other remainder occurs twice, the decimal is a mixed circulate. We are to show that this last will never happen, unless the divisor be a multiple of 2, 5, or their powers. If two numbers be prime to each other, their product will be prime to both; and, if two numbers be proposed, whereof the first does not measure the second, it will not measure any product of the second, if the multiplier be prime to the first. Thus, because 7 does not measure 12 it will not measure any product of 12 by a multiplier prime to 7. For instance, it will not measure \(12 \times 3\), or 36. Otherwise, the quotient of 12 divided by 7, or 14 multiplied by 3, would be a whole number, and 5 × 3 would be measured by 7, which it cannot be, since 5 and 3 are both prime to 7.

Now, if we inspect the foregoing operation, we shall perceive that the product of 136, the remainder, where the decimal begins to circulate, multiplied by 999, is measured by the denominator 216. But 999 is not measured by the denominator, otherwise the decimal would have been a pure circulate; therefore 126, and 136, are not prime to each other, but have a common measure, and that measure must apply to 864, a multiple of 126, and to 1000, the sum of 136 and 864; see p. 672, col. 2, par. ult. &c. But it was proven, p. 679, col. 1, par. 1, that no numbers, except the powers of 5 and 2, measure a number consisting of 1 with cyphers annexed; consequently the denominator must be measured by a power of 2 or 5. The reader will perceive, that the exponent of the power must be the same as the number of cyphers annexed to 1, or as the number of figures in the finite part of the decimal.

We shall now recapitulate the substance of what has been said with respect to the formation of decimals. 2, 5, and their powers, produce finite decimals, by p. 679, col. 1, par. 3, &c. and the number of places is measured by the exponent of the power. 3 and 9 produce pure repeaters (p. 679, col. 2, par. 2.) The products of 2, 5, and their powers, by 3 or 9, produce mixed repeaters by p. 679, col. 2, par. ult.; their products by other multipliers, produce mixed circulates by p. 679, col. 2, par. ult.; and all numbers of which 2 and 5 are not aliquot parts, except 3 and 9, produce pure circulates.

To find the form of a decimal corresponding to any denominator, divide by 2, 5, and 10, as often as can be done without a remainder; the number of divisions shows how many finite places there are in the decimal, by p. 681, col. 2, par. 3. If the dividend be not exhausted by these divisions, divide a competent number of 9's by the last quotient, till the division be completed without a remainder; the number of 9's required shows how many places there are in the circle, and the reason may be inferred from p. 680, col. 2, par. 5.

We shall conclude this subject by marking down the decimals produced by vulgar fractions, whose numerator is 1, and denominators 30; and under that the reader may observe their connection with the denominators.

| Denominator | Decimal | |-------------|---------| | 2 | 0.5 | | 3 | 0.333 | | 4 | 0.25 | | 5 | 0.2 | | 6 | 0.1666 | | 7 | 0.142857 | | 8 | 0.125 | | 9 | 0.1111 | | 10 | 0.1 | | 11 | 0.0909 | | 12 | 0.08333 | | 13 | 0.076923 | | 14 | 0.0714285 | | 15 | 0.066667 | | 16 | 0.0625 | | 17 | 0.0588235294117647 | | 18 | 0.055555 | | 19 | 0.052631578947368421 | | 20 | 0.05 | | 21 | 0.047619 | | 22 | 0.0454545 | | 23 | 0.0434782608695652173913 | | 24 | 0.0416666 | | 25 | 0.04 | | 26 | 0.0384615 | | 27 | 0.037 | | 28 | 0.03571428 | | 29 | 0.034827586206896552724137931 | | 30 | 0.0333 |

Rules for reducing intermediate decimals to vulgar fractions:

I. "If the decimal be a pure repeater, place the repeating figure for the numerator, and 9 for the denominator."

II. "If the decimal be a pure circulate, place the circulating figures for the numerator, and as many 9's as there are places in the circle for the denominator."

III. "If there be cyphers prefixed to the repeating or circulating figures, annex a like number to the 9's in the denominator."

IV. "If the decimal be mixed, subtract the finite part from the whole decimal. The remainder is the numerator; and the denominator consists of as many 9's as there are places in the circle, together with as many cyphers as there are finite places before the circle."

Thus, \(235_{\text{finite}} = 235_{\text{finite}}\)

From the whole decimal \(235_{\text{finite}}\) we subtract the finite part \(235\)

and the remainder \(23327\) is the numerator.

The reason may be illustrated by dividing the decimal into two parts, whereof one is finite, and the other a pure repeater or circulate, with cyphers prefixed. The sum of the vulgar fractions corresponding to these will be the value of the decimal sought.

\(235_{\text{finite}}\) may be divided into \(235_{\text{finite}}\) by rule I, and \(000_{\text{finite}} = 000_{\text{finite}}\) by rules II, III

In order to add these vulgar fractions, we reduce them to a common denominator; and, for that purpose, we multiply both terms of the former by 99, which gives \(23327\); then we add the numerators.

\[ \begin{array}{cccc} 2115 & 23500 & 23265 & 23562 \\ 2115 & 235 & 62 & 235 \\ \hline 23265 & 23265 & 23327 & 23327 \end{array} \]

Sum of numerators.

The value of circulating decimals is not altered, though one or more places be separated from the circle, and considered as a finite part, providing the circle be completed. For example, .27 may be written \(27_{\text{finite}}\), which is reduced by the half of the foregoing rules to \(27_{\text{finite}}\) or \(27_{\text{finite}}\), which is also the value of .27. And, if two or more circles be joined, the value of the decimal is still the same. Thus, \(27_{\text{finite}} = 27_{\text{finite}}\), which is reduced by dividing the terms by 101 to \(27_{\text{finite}}\). All circulating decimals may be reduced to a similar form, having a like number both of finite and circulating places. For this purpose, we extend the finite part of each as far as the longest, and then extend all the circles to so many places as may be a multiple of the number of places in each.

Ex. .34725, extended .34725725725725725,

Here the finite part of both is extended to two places, and the circle to 12 places, which is the least multiple for circles of 3 and 4 places.

Sect. ii. Addition and Subtraction of Intermediate Decimals.

To add repeating decimals, "Extend the repeating figures one place beyond the longest finite ones, and when you add the right-hand column, carry to the next by 9."

Ex. .37524 or .37524 .28 .296

To subtract repeating decimals, "Extend them as directed for addition, and borrow at the right-hand place, if necessary, by 9."

The reason of these rules will be obvious, if we recollect that repeating figures signify ninth-parts. If the right-hand figure of the sum or remainder be 0, the decimal obtained is finite; otherwise it is a repeater.

To add circulating decimals, "Extend them till they become similar (p. 682, col. 1, par. ult. &c.) and, when you add the right-hand column, include the figure which would have been carried if the circle had been extended further.

Ex. 1st. Extended. Ex. 2nd. Extended.

Note 1. Repeaters mixed with circulates are extended and added as circulates.

Note 2. Sometimes it is necessary to inspect two or more columns for ascertaining the carriage; because the carriage from a lower column will sometimes raise the sum of the higher, so as to alter the carriage from it to a new circle. This occurs in Ex. 2.

Note 3. The sum of the circles must be considered as a similar circle. If it consist entirely of cyphers, the amount is terminate. If all the figures be the same, the amount is a repeater. If they can be divided into parts exactly alike, the amount is a circle of fewer places; but, for the most part, the circle of the sum is similar to the extended circles.

To subtract circulating decimals, "Extend them till intermediate they become similar; and, when you subtract the circulating right-hand figure, consider whether I would have been borrowed if the circles had been extended further, and make allowance accordingly.

To multiply intermediate decimals:

Case I. "When the multiplier is finite, and the multiplicand repeats, carry by 9 when you multiply the repeating figure: The right-hand figure of each line of the product is a repeater; and they must be extended and added accordingly."

Ex. .13494

If the sum of the right-hand column be an even number of 9's, the product is finite; otherwise, it is a repeater.

Case II. "When the multiplier is finite, and the multiplicand circulates, add to each product of the right-hand figure the carriage which would have been brought to it if the circle had been extended. Each line of the product is a circle similar to the multiplicand, and therefore they must be extended and added accordingly."

The product is commonly a circulate similar to the multiplicand; sometimes it circulates fewer places, repeats, or becomes finite; it never circulates more places.

Ex. .3746 × 235

Case III. "When the multiplier repeats or circulates, find the product as in finite multipliers, and place under it the products which would have arisen from the repeating or circulating figures, if extended."

Ex. 1st. .958 × 8

3rd. 784 × 36 It is evident, that, if a repeating multiplier be extended to any length, the product arising from each figure will be the same as the first, and each will stand one place to the right hand of the former. In like manner, if a circulating multiplier be extended, the product arising from each circle will be alike, and will stand as many places to the right hand of the former as there are figures in the circle. In the foregoing examples, there are as many of these products repeated as is necessary for finding the total product. If we place down more, or extend them further, it will only give a continuation of the repeaters or circulates.

This is obvious in Ex. 1st and 2nd. As the learner may not apprehend it so readily in Ex. 3rd, when the multiplicand is a circulate, and consequently each line of the product is also a circulate, we have divided it into columns, whose sums exhibit the successive circles. The sum of the first column is 28,961037, and there is a carriage of 1 from the right-hand column, which completes 28,961038. This one is supplied from the three first lines of the second column, the sum of which is 999999, and being increased by 1, in consequence of the carriage from the third column, amounts to 1,000000, and therefore carries 1 to the first column, and does not affect the sum of the remaining lines, which are the same as those of the first column. The third column contains two sets of these lines, which amount to 999999, besides the lines which compose the circle. Each of these sets would be completed into 1,000000 by the carriage from the 4th column, if extended, and each would carry 1 to the second column. One of these would complete the sum of the three first lines, and the other would complete the sum of the circle. In like manner, if the circles be extended ever so far, the increasing carriages will exactly answer for the increasing deficiencies, and the sum will be always a continuation of the circle; but the product could not circulate, unless the sum of the lines marked off in the second column had consisted entirely of 9's, or had been some multiple of a number of 9's; and the circles must be extended till this take place, in order to find the complete product.

The multiplication of intermediate decimals may be often facilitated, by reducing the multiplier to a vulgar fraction, and proceeding as directed p. 674, col. 1, par. 6.

Thus,

\[ \begin{array}{c} \text{4th.] } \frac{3824}{7} \times \frac{5}{6} = \frac{384}{23} \\ \text{9)2.6768} \\ \text{.9742} \\ \text{90)8.832} \\ \text{.09813} \end{array} \]

Therefore, in order to multiply by 3, we take one third-part of the multiplier; and, to multiply by 9, we take two thirds of the same. Thus,

\[ \begin{array}{c} \text{6th.] } \frac{784}{3} \times \frac{3}{7} = \frac{876}{2} \\ \text{3)784} \\ \text{.2613} \\ \text{3)1.7522} \\ \text{.58406} \end{array} \]

As the denominator of the vulgar fractions always consists of 9's, or of 9's with cyphers annexed, we may use the contraction explained p. 661, col. 1, par. ult., &c.; and this will lead us exactly to the same operation which was explained p. 683, col. 2, par. ult., &c. on the principles of decimal arithmetic.

\[ \begin{array}{c} \text{8th.] } \frac{735}{323} \times \frac{326}{3} = \frac{278}{365} \\ \text{323} \\ \text{323} \\ \text{1470} \\ \text{2205} \\ \text{99(0)237405} \\ \text{2374,05} \\ \text{2374} \\ \text{2374} \\ \text{.239803} \end{array} \]

When the multiplier is a mixed repeater or circulate, we may proceed as in Ex. 5th and 8th; or we may divide the multiplier into two parts, of which the first is finite, and the second a pure repeater or circulate, with cyphers prefixed, and multiply separately by these, and add the products.

Thus, \( .384 \times .28 \) or by .2 = .0768 or thus, .384

and by .05 = .02133

\[ \begin{array}{c} \text{.09813} \\ \text{9)1920} \\ \text{2133} \\ \text{768} \\ \text{.09813} \end{array} \]

In the following examples, the multiplicand is a repeater; and therefore the multiplication by the numerator of the vulgar fraction is performed as directed p. 683, col. 2, par. 2. In Ex. 13th, we have omitted the products of the divisor, and only marked down the remainders. These are found, by adding the left-hand figure of the dividend to the remaining figures of the same. Thus, 363 is the first dividend; and 3, the left-hand figure, added to 63, the remaining figures, gives 66 for the first remainder; and the second dividend, 666, is completed by annexing the circulating figure 6. The reason of which may be explained as follows. The highest place of each dividend shows, in this example, how many hundreds it contains; and, as it must contain an equal number of ninety-nines, and also an equal number of units, it follows, that these units, added to the lower places, must show how far the dividend exceeds that number of ninety-nines. The figure of the quotient is generally the same as the first place of the dividend, sometimes one more. This happens in the last step of the foregoing example, and is discovered when the remainder found, as here directed, would amount to 99, or upwards; and the excess, above 99 only, must in that case be taken to complete the next dividend.

The number of places in the circle of the product is sometimes very great, though there be few places in the factors: but it never exceeds the product of the denominator of the multiplier, multiplied by the number of places in the circle of the multiplicand. Therefore, if the multiplier be 3 or 6, the product may circulate three times as many places as the multiplicand; if the multiplier be any other repeater, nine times as many; if the multiplier be a circulate of two places, ninety-nine times as many: thus, in the last example, .01, a circulate of two places, multiplied by .01, a circulate of two places, produces a circulate of twice 99, or 198 places. And the reason of this limit may be inferred from the nature of the operation; for the greatest possible number of remainders, including 0, is equal to the divisor 99; and each remainder may afford two dividuals, if both the circulating figures, 3 and 6, occur to be annexed to it. If the multiplier circulate three places, the circle of the product, for a like reason, may extend nine hundred and ninety-nine times as far as that of the multiplicand. But the number of places is often much less.

The multiplication of interminate decimals may be proven, by altering the order of the factors, (p. 658, col. 2, par. 2.) or by reducing them both to vulgar fractions in their lowest terms, multiplying these as directed p. 673, col. 2, par. 3, and reducing the product to a decimal.

Sect. iv. Division of Interminate Decimals.

Case I. "When the dividend only is interminate, proceed as in common arithmetic; but, when the figures of the dividend are exhausted, annex the repeating figure, or the circulating figures in their order, instead of cyphers, to the remainder."

Chap. XI.

The foregoing method is the only one which properly depends on the principles of decimal arithmetic; but it is generally shorter to proceed by the following rule.

"Reduce the divisor to a vulgar fraction, multiply the dividend by the denominator, and divide the product by the numerator."

Ex. 1st.] Divide .37845 by $\frac{6}{9}$

$5)3.40605(\cdot68121$

2nd.] Divide .37845 by $\frac{6}{9}$

$2)1.1536(\cdot567683$

Note 1. Division by 3 triples the dividend, and division by 6 increases the dividend one half.

Note 2. When the divisor circulates, the denominator of the vulgar fraction consists of 9's, and the multiplication is sooner performed by the contraction explained p. 658, col. 1, par. 3. It may be wrought in the same way, when the divisor repeats, and the denominator, or consequence, is 9.

Note 3. If a repeating dividend be divided by a repeating or circulating divisor; or, if a circulating dividend be divided by a similar circulating dividend; or, if the number of places in the circle of the divisor be a multiple of the number in the dividend; then the product of the dividend multiplied by the denominator of the divisor will be terminate, since like figures are subtracted from like in the contracted multiplication, and consequently no remainder left. The form of the quotient depends on the divisor, as explained at large, p. 679, col. 1, par. 1.—p. 681, col. 2, par. 3.

Note 4. In other cases, the original and multiplied dividend are similar, and the form of the quotient is the same as in the case of a finite divisor. See p. 685, col. 2, par. ult., &c.

Note 5. If the terms be similar, or extended till they become so, the quotient is the same as if they were finite, and the operation may be conducted accordingly; for the quotient of vulgar fractions that have the same denominator is equal to the quotient of their numerators.

CHAP. XI. OF THE EXTRACTION OF ROOTS.

The origin of powers by involution has already been explained under the article ALGEBRA, p. 8 and 9. There now remains therefore only to give the most expeditious methods of extracting the square and cube roots; the reasons of which will readily appear from what is said under that article. As for all powers above the cube, unless such as are multiples of either the square and cube, the extraction of their roots admits of no deviation from the algebraic canon which must be always constructed on purpose for them.

If the root of any power not exceeding the seventh power, be a single digit, it may be obtained by inspection, from the following table of powers. **Chap. XI. ARITHMETIC**

### Extraction of Roots

#### Sect. i. Extraction of the Square Root.

**Rule I.** "Divide the given number into periods of two figures, beginning at the right hand in integers, and pointing toward the left. But in decimals, begin at the place of hundreds, and point toward the right. Every period will give one figure in the root."

**II.** "Find by the table of powers, or by trial, the nearest lesser root of the left-hand period, place the figure so found in the quot; subtract its square from the said period, and to the remainder bring down the next period for a dividend or resolvend."

**III.** "Double the quot; for the first part of the divisor; inquire how often this first part is contained in the whole resolvend, excluding the units place; and place the figure denoting the answer both in the quot; and on the right of the first part; and you have the divisor complete."

**IV.** "Multiply the divisor thus completed by the figure put in the quot; subtract the product from the resolvend, and to the remainder bring down the following period for a new resolvend, and then proceed as before."

**Note 1.** If the first part of the divisor, with unity supposed to be annexed to it, happen to be greater than the resolvend, in this case place 0 in the quot; and also on the right of the partial divisor; to the resolvend bring down another period; and proceed to divide as before.

**Note 2.** If the product of the quotient-figure into the divisor happen to be greater than the resolvend, you must go back, and give a lesser figure to the quot;.

**Note 3.** If, after every period of the given number is brought down, there happen at last to be a remainder, you may continue the operation, by annexing periods or pairs of cyphers, till there be no remainder, or till the decimal part of the quot; repeat or circulate, or till you think proper to limit it.

**Ex. 1st.** Required the square root of 133225.

| Square number | 133225 | (365 root) | |---------------|--------|------------| | | 9 | 365 |

1 div. 66) 432 resolvend.

396 product.

1095

If the square root of a vulgar fraction be required, find the root of the given numerator for a new numerator, and find the root of the given denominator for a new denominator. Thus, the square root of \( \frac{3}{7} \) is \( \frac{3}{7} \), and the root of \( \frac{3}{7} \) is \( \frac{3}{7} \); and thus the root of \( \frac{3}{7} \left(=6\frac{1}{2}\right) \) is \( \frac{3}{7} = 2\frac{1}{2} \).

But if the root of either the numerator or denominator cannot be extracted without a remainder, reduce the vulgar fraction to a decimal, and then extract the root, as in Ex. 3rd. above.

#### Sect. ii. Extraction of the Cube Root.

**Rule I.** "Divide the given number into periods of three figures, beginning at the right hand in integers, and pointing toward the left. But in decimals, begin at the place of thousands, and point toward the right. The number of periods shews the number of figures in the root."

**II.** "Find by the table of powers, or by trial, the nearest lesser root of the left-hand period; place the figure so found in the quot; subtract its cube from the said period; and to the remainder bring down the next period for a dividend or resolvend."

The divisor consists of three parts which may be found as follows. III. "The first part of the divisor is found thus: Multiply the square of the quot by 3, and to the product annex two cyphers; then inquire how often this first part of the divisor is contained in the resolvend, and place the figure denoting the answer in the quot."

IV. "Multiply the former quot by 3, and the product by the figure now put in the quot; to this last product annex a cypher; and you have the second part of the divisor. Again, square the figure now put in the quot for the third part of the divisor; place these three parts under one another, as in addition; and their sum will be the divisor complete."

V. "Multiply the divisor, thus completed, by the figure last put in the quot, subtract the product from the resolvend, and to the remainder bring down the following period for a new resolvend, and then proceed as before."

Note 1. If the first part of the divisor happen to be equal to or greater than the resolvend, in this case, place o in the quot, annex two cyphers to the said first part of the divisor, to the resolvend bring down another period, and proceed to divide as before.

Note 2. If the product of the quotient-figure into the divisor happen to be greater than the resolvend, you must go back, and give a lesser figure to the quot.

Note 3. If, after every period of the given number is brought down, there happen at last to be a remainder, you may continue the operation by annexing periods of three cyphers till there be no remainder, or till you have as many decimal places in the root as you judge necessary.

Ex. 1st. Required the cube root of 12812904.

Cube number 12812904 (234 root)

1st part 1200 2nd part 180 3rd part 9

1 divisor 1389 x 3 = 4167 product

1st part 158700 2nd part 2760 3rd part 16

2 divisor 161476 x 4 = 645904 product.

If the cube root of a vulgar fraction be required, find the cube root of the given numerator for a new numerator, and the cube root of the given denominator for a new denominator. Thus, the cube root of \( \frac{3}{4} \) is \( \frac{3}{4} \); and the cube root of \( \frac{5}{6} \) is \( \frac{5}{6} \); and thus the cube root of \( \frac{3}{4} \times \frac{5}{6} \) is \( \frac{3}{4} \times \frac{5}{6} \).

But if the root of either the numerator or denominator cannot be extracted without a remainder, reduce the vulgar fraction to a decimal, and then extract the root.