figurines eleven, LXV sixty-five, MDCXXVIII one thousand six hundred and twenty-eight.

3d. When a numeral letter of lesser value is placed before one of greater, the value of the lesser is taken from that of the greater; thus IV signifies four, XL forty, XC ninety, CD four hundred.

Sometimes IO is used instead of D for 500, and the value is increased ten times by annexing O to the right hand.

Thus IO signifies 500 Afto IOO is used for 1000

IOO 5000 CCIOO for 10000

IOOO 50000 CCCIOO for 100000

Sometimes thousands are represented by drawing a line over the top of the numeral V being used for five thousand, L for fifty thousand, CC two hundred thousand.

About the year of Christ 200, a new kind of arithmetic, called sexagesimal, was invented, as is supposed, by Claudius Ptolemaeus. The design of it was to remedy the difficulties of the common method, especially with regard to fractions. In this kind of arithmetic, every unit was supposed to be divided into 60 parts, and each of these into 60 others, and so on: hence any number of such parts were called sexagesimal fractions; and to make the computation in whole numbers more easy, he made the progression in these also sexagesimal. Thus from one to 59 were marked in the common way: then 60 was called a sexagesima prima, or first sexagesimal integer, and had one single dash over it; so 60 was expressed thus I'; and so on to 59 times 60, or 3540, which was thus expressed LIX'. He now proceeded to 60 times 60, which he called a sexagesima secunda, and was thus expressed II'. In like manner, twice 60 times 60, or 7200, was expressed by II''; and so on till he came to 60 times 6000, which was a third sexagesimal, and expressed thus, I'''. If any number less than 60 was joined with these sexagesimals, it was added in its proper characters without any dash; thus IXV represented 60 and 15, or 75; IVXCV is four times 60 and 25, or 265; X'IVXV, is ten times 3600, twice 60 and 15, or 36135, &c. Sexagesimal fractions were marked by putting the dash at the foot, or on the left hand of the letter: thus I, or I' denoted $\frac{1}{60}$; I', or $I'\frac{1}{60}$, &c.

The most perfect method of notation, which we now use, came into Europe from the Arabs, by the way of Spain. The Arabs, however, do not pretend to be the inventors of them, but acknowledge that they received them from the Indians. Some there are, indeed, who contend that neither the Arabs nor the Indians were the inventors, but that they were found out by the Greeks. But this is by no means probable; as Maximus Planudes, who lived towards the close of the 13th century, is the first Greek who makes use of them: and he is plainly not the inventor; for Dr Wallis mentions an inscription on a chimney in the parsonage house of Helendon in Northamptonshire, where the date is expressed by M° 133, instead of 1133. Mr Luffkin furnishes a still earlier instance of their use, in the window of a house, part of which is a Roman wall, near the market place in Colchester; where between two carved lions stands an escutcheon with the figures 1090. Dr Wallis is of opinion that these characters must have been used in England at least as long ago as the year 1552, if not in ordinary affairs, at least in mathematical ones, and in astronomical tables. How these characters came to be originally invented by the Indians we are entirely ignorant.

The introduction of the Arabian characters in notation did not immediately put an end to the sexagesimal arithmetic. As this had been used in all the astronomical tables, it was for their sakes retained for a considerable time. The sexagesimal integers went first out, but the fractions continued till the invention of decimals.

The oldest treatises extant, upon the theory of arithmetic, are the seventh, eighth, and ninth books of Euclid's Elements, where he treats of proportion and of prime and composite numbers; both of which have received improvements since his time, especially the former. The next, of whom we know anything, is Nicomachus the Pythagorean, who wrote a treatise of the theory of arithmetic, consisting chiefly of the definitions and divisions of numbers into classes, as plain, solid, triangular, quadrangular, and the rest of the figurate numbers as they are called, numbers odd and even, &c., with some of the more general properties of the several kinds. This author is, by some, said to have lived before the time of Euclid; by others, not long after. His arithmetic was published at Paris in 1538. The next remarkable writer on this subject is Boethius, who lived at Rome in the time of Theodoric the Goth. He is supposed to have copied most of his works from Nicomachus.

From this time no remarkable writer on arithmetic appeared till about the year 1200, when Jordanus of Namur wrote a treatise on this subject, which was published and demonstrated by Johannes Faber Stapulensis in the 15th century, soon after the invention of printing. The same author also wrote upon the new art of computation by the Arabic figures, and called this book Algorithmus Demonstratus. Dr Wallis says, this manuscript is in the Savilian library at Oxford, but it hath never yet been printed. As learning advanced in Europe, so did the knowledge of numbers; and the writers on arithmetic soon became innumerable. About the year 1464, Regiomontanus, in his triangular tables, divided the radius into 10,000 parts, instead of 60,000; and thus tacitly expelled the sexagesimal arithmetic. Part of it, however, still remains in the division of time, as of an hour into 60 minutes, a minute into 60 seconds, &c. Ramus in his arithmetic, written about the year 1550, and published by Lazarus Schonerus in 1586, used decimal periods in carrying on the square and cube roots to fractions. The same had been done before by our countrymen Buckley and Record; but the first who published an express treatise on decimals was Simon Stevinus, about the year 1582. As to the circulating decimals, Dr Wallis is the first who took much notice of them. He is also the author of the arithmetic of infinities, which has been very usefully applied to geometry. The greatest improvement, however, which the art of computation ever received, is the invention of logarithms. The honour of this invention is unquestionably due to Baron Napier of Merchiston in Scotland, about the end of the 16th or beginning of the Chap. I. ARITHMETIC.

Notation in the 17th century. By these means arithmetic has advanced to a degree of perfection which the ancients could never have imagined possible, much less hoped to attain; and we believe it may now be reckoned one of those few sciences which have arrived at their utmost height, and which is in its nature capable of little further improvement.

Chap. I. NOTATION AND NUMERATION.

The first elements of arithmetic are acquired during our infancy. The idea of one, though the simplest of any, and suggested by every single object, is perhaps rather of the negative kind, and consists partly in the exclusion of plurality, and is not attended to till that of number be acquired. Two is formed by placing one object near another; three, four, and every higher number, by adding one continually to the former collection. As we thus advance from lower numbers to higher, we soon perceive that there is no limit to this increasing operation; and that, whatever number of objects be collected together, more may be added, at least, in imagination; so that we can never reach the highest possible number, nor approach near it. As we are led to understand and add numbers by collecting objects, so we learn to diminish them by removing the objects collected; and if we remove them one by one, the number decreases through all the steps by which it advanced, till only one remain, or none at all. When a child gathers as many stones together as suits his fancy, and then throws them away, he acquires the first elements of the two capital operations in arithmetic. The idea of numbers, which is first acquired by the observation of sensible objects, is afterwards extended to measures of space and time, affections of the mind, and other immaterial qualities.

Small numbers are most easily apprehended: a child soon knows what two and what three is; but has not any distinct notion of seventeen. Experience removes this difficulty in some degree; as we become accustomed to handle larger collections, we apprehend clearly the number of a dozen or a score; but perhaps could hardly advance to a hundred without the aid of classical arrangement, which is the art of forming so many units into a class, and so many of these classes into one of a higher kind, and thus advancing through as many ranks of classes as occasion requires. If a boy arrange an hundred stones in one row, he would be tired before he could reckon them; but if he placed them in ten rows of ten stones each, he will reckon an hundred with ease; and if he collect ten such parcels, he will reckon a thousand. In this case, ten is the lowest class, a hundred is a class of the second rank, and a thousand is a class of the third rank.

There does not seem to be any number naturally adapted for constituting a class of the lowest, or any higher rank, to the exclusion of others. However, as ten has been universally used for this purpose by the Hebrews, Greeks, Romans, and Arabians, and by all nations who have cultivated this science, it is probably the most convenient for general use. Other scales, however, may be assumed, perhaps on some occasions, with superior advantage; and the principles of notation in arithmetic will appear in their full extent, if the student can adapt them to any scale whatever: thus, if eight were the scale, 6 times 3 would be two classes, and two units, and the number 18 would then be represented by 22. If 12 were the scale, 5 times 9 would be three classes and nine units, and 45 would be represented by 39, &c.

It is proper, whatever number of units constitutes a class of the lower rank, that the same number of each class should make one of the next higher. This is observed in our arithmetic, ten being the universal scale: but it is not regarded in the various kinds of moneys, weights, and the like, which do not advance by any universal measure; and much of the difficulty in the practice of arithmetic arises from that irregularity.

As higher numbers are somewhat difficult to apprehend, we naturally fall on contrivances to fix them in our minds, and render them familiar: but notwithstanding all the expedients we can fall upon, our ideas of high numbers are still imperfect, and generally far short of the reality; and though we can perform any computation with exactness, the answer we obtain is often incompletely apprehended.

It may not be amiss to illustrate, by a few examples, the extent of numbers which are frequently named without being attended to. If a person employed in telling money reckon an hundred pieces in a minute, and continue at work ten hours each day, he will take seventeen days to reckon a million; a thousand men would take 45 years to reckon a billion. If we supposed the whole earth to be as well peopled as Britain, and to have been so from the creation, and that the whole race of mankind had constantly spent their time in telling from a heap consisting of a quadrillion of pieces, they would hardly have yet reckoned the thousandth part of that quantity.

All numbers are represented by the ten following characters.

1 2 3 4 5 6 7 8 9 0

One, two, three, four, five, six, seven, eight, nine, cypher.

The nine first are called significant figures, or digits; and sometimes represent units, sometimes tens, hundreds, or higher classes. When placed singly, they denote the simple numbers subjacent to the characters. When several are placed together, the first or right-hand figure only is to be taken for its simple value: the second signifies so many tens, the third so many hundreds, and the others so many higher classes, according to the order they stand in. And as it may sometimes be required to express a number consisting of tens, hundreds, or higher classes, without any units or classes of a lower rank annexed; and as this can only be done by figures standing in the second, third, or higher place, while there are none to fill up the lower ones; therefore an additional character or cypher (0) is necessary, which has no signification when placed by itself, but serves to supply the vacant places, and bring the figures to their proper station.

The following table shows the names and divisions of the classes. The first five figures from the right hand are called the unit period; the next five the million period, after which the trillion, quadrillion, quintillion, sextillion, septillion, octillion, and nonillion periods follow in their order.

It is proper to divide any number, before we reckon it, into periods and half periods, by different marks. We then begin at the left hand, and read the figures in their order, with the names of their places, from the table. In writing any number, we must be careful to mark the figures in their proper places, and supply the vacant places with cyphers.

As there are no possible ways of changing numbers, except by enlarging or diminishing them according to some given rule, it follows, that the whole art of arithmetic is comprehended in two operations, Addition and Subtraction. However, as it is frequently required to add several equal numbers together, or to subtract several equal ones from a greater, till it be exhausted, proper methods have been invented for facilitating the operation in these cases, and distinguished by the names of Multiplication and Division; and these four rules are the foundation of all arithmetical operations whatever.

As the idea of number is acquired by observing several objects collected, so is that of fractions by observing an object divided into several parts. As we sometimes meet with objects broken into two, three, or more parts, we may consider any or all of these divisions promiscuously, which is done in the doctrine of vulgar fractions, for which a chapter will be allotted. However, since the practice of collecting units into parcels of tens has prevailed universally, it has been found convenient to follow a like method in the consideration of fractions, by dividing each unit into ten equal parts, and each of these into ten smaller parts; and so on. Numbers divided in this manner are called Decimal Fractions.

**Chap. II. ADDITION.**

Addition is that operation by which we find the amount of two or more numbers. The method of doing this in simple cases is obvious, as soon as the meaning of number is known, and admits of no illustration. A young learner will begin at one of the numbers, and reckon up as many units separately as there are in the other, and practice will enable him to do it at once. It is impossible, strictly speaking, to add more than two numbers at a time. We must first find the sum of the first and second; then we add the third to that number; and so on. However, as the several sums obtained are easily retained in the memory, it is neither necessary nor usual to mark them down. When the numbers consist of more figures than one, we add the units together, the tens together, and so on. But if the sum of the units exceed ten, or contain ten several times, we add the number of tens it contains to the next column, and only set down the number of units that are over. In like manner we carry the tens of every column to the next higher. And the reason of this is obvious from the value of the places; since an unit, in any higher place, signifies the same thing as ten in the place immediately lower.

**Example.**

**Rule.** "Write the numbers distinctly, 346863 "units under units, tens under tens; and 876734 "so on. Then reckon the amount of the 123467 "right-hand column. If it be under ten, 3124213 "mark it down. If it exceed ten, mark 712316 "the units only, and carry the tens to the 438987 "next place. In like manner, carry the 279654 "tens of each column to the next, and mark "down the full sum of the left-hand col. 3092234 "lumn."

As it is of great consequence in business to perform addition readily and exactly, the learner ought to practice it till it become quite familiar. If the learner can readily add any two digits, he will soon add a digit to a higher number with equal ease. It is only to add the unit place of that number to the digit; and, if it exceed ten, it raises the amount accordingly. Thus, because 8 and 6 is 14, 48 and 6 is 54. It will be proper to mark down under the sums of each column, in a small hand, the figure that is carried to the next column. This prevents the trouble of going over the whole operation again, in case of interruption or mistake. If you want to keep the account clean, mark down the sum and figure you carry on a separate paper, and after revising them, transcribe the sum only. After some practice, we ought to acquire the habit of adding two or more figures at one glance. This is particularly useful when two figures which amount to 10, as 6 and 4, or 7 and 3, stand together in the column.

Every operation in arithmetic ought to be revised, to prevent mistakes; and as one is apt to fall into the same mistake, if he revise it in the same manner he performed it, it is proper either to alter the order, or else to trace back the steps by which the operation advanced, which will lead us at last to the number we began with. Every method of proving accounts may be referred to one or other of these heads.

1st, Addition may be proven by any of the following methods: Repeat the operation, beginning at the top of the column, if you began at the foot when you wrought it.

2d, Divide the account into several parts; add these separately, and then add the sums together. If their amount correspond with the sum of the account, when added at once, it may be presumed right. This method is particularly proper when you want to know the sums of the parts, as well as that of the whole.

3d, Subtract the number successively from the sum; if the account be right, you will exhaust it exactly, and find no remainder. When the given number consists of articles of different value, as pounds, shillings, and pence, or the like, which are called different denominations, the operations in arithmetic must be regulated by the value of the articles. We shall give here a few of the most useful tables for the learner's information.

I. Sterling Money. 4 Farthings = 1 penny, 12 Pence = 1 shilling, 20 Shillings = 1 pound, L. Also, 6s. 8d. = 1 noble 12s. = 1 angel 13s. 4d. or two-thirds of a pound = 1 merk.

Scots money is divided in the same manner as sterling, and has one twelfth of its value. A pound Scots is equal to 1s. 8d. sterling, a shilling Scots to a penny sterling, and a penny Scots to a twelfth part of a penny sterling; a merk Scots is two-thirds of a pound Scots, or 13½d. Sterling.

II. Avoirdupois Weight. 16 Drams = 1 ounce, oz. 16 Ounces = 1 pound, lb. 28 Pounds = 1 quarter, qr. 4 Quart = 1 hun. wght, C. 20 Hun. weight = 1 ton, T.

III. Troy Weight. 20 Mites = 1 grain, gr. 24 Grains = 1 pen.wt, dwt. 20 Penny wt = 1 ounce, oz. 12 Ounces = 1 pound, lib.

IV. Apothecaries Weight. 20 Grains = 1 scruple, 3 3 Scruples = 1 dram, 3 8 Drams = 1 ounce, 3

V. English Dry Measure. 2 Pints = 1 quart 4 Quarts = 1 gallon 2 Gallons = 1 peck 4 Pecks = 1 bushel 8 Bushels = 1 quarter

VI. Scots Dry Measure. 4 Lippies = 1 peck 4 Pecks = 1 firlot 4 Firlots = 1 boll 16 Bolls = 1 chaldar

VII. English Land Measure. 30½ Square yards = 1 pole or perch 40 Poles = 1 rood 4 Rods = 1 acre

VIII. Scots Land Measure. 36 Square ells = 1 fall 40 Falls = 1 rood 4 Roods = 1 acre

IX. Long Measure. 12 Inches = 1 foot 3 Feet = 1 yard 5½ Yards = 1 pole 40 Poles = 1 furlong 8 Furlongs = 1 mile 3 Miles = 1 league

X. Time. 60 Seconds = 1 minute 60 Minutes = 1 hour 24 Hours = 1 day 7 Days = 1 week 365 Days = 1 year 52 Weeks & 1 day = 1 year

Rule for Compound Addition. "Arrange like quantities under like, and carry according to the value of the higher place."

Note 1. When you add a denomination, which contains more columns than one, and from which you carry to the higher by 20, 30, or any even number of tens, first add the units of that column, and mark down their sum, carrying the tens to the next column; then add the tens, and carry to the higher denomination, by the number of tens that it contains of the lower. For example, in adding shillings, carry by 10 from the units to the tens, and by 2 from the tens to the pounds.

Note 2. If you do not carry by an even number of tens, first find the complete sum of the lower denomination, then inquire how many of the higher that sum contains, and carry accordingly, and mark the remainder, if any, under the column. For example, if the addition of a column of pence be 43, which is three shillings and sevenpence, mark 7 under the pence column, and carry 3 to that of the shillings.

Note 3. Some add the lower denominations after the following method: when they have reckoned as many as amounts to one of the higher denomination, or upwards, they mark a dot, and begin again with the excess of the number reckoned above the value of the denomination. The number of dots shows how many are carried, and the last reckoned number is placed under the column.

Examples in Sterling Money.

| L. | 145 | 6 | 8 | |----|-----|---|---| | | 215 | 3 | 9 | | | 172 | 18| 4 | | | 645 | 7 | 7 | | | 737 | 2 | 3 | | | 35 | 3 | 9 | | | 9 | - | 7 | | | 1764| 12| 3 | | | 780 | - | - | | | 99 | 9 | 9 | | | 150 | 10| - | | | 844 | 9 | 7 |

| L. | 16 | 9 | 11 ¼ | |----|---|---|------| | | 169| 16| 10 | | | 36 | 12| 9½ | | | 54 | 7 | 6 | | | 30 | - | 14 | | | 7 | 19| 6 | | | 707| 19| 11 | | | 14 | 14| 4 | | | 84 | 18| 8½ | | | 125| 3 | 7 | | | 16 | 16| 8½ | | | 62 | 5 | 3 |

In Avoirdupois Weight.

| T. C. | qr. | lb. | |-------|-----|-----| | 1 | 19 | 3 | | | 14 | 1 | | 2 | 18 | 1 | | | 1 | 2 | | 3 | 1 | 10 | | | 17 | 2 | | | 15 | 3 | | 4 | 6 | - | | | 6 | 3 | | 5 | 4 | - | | | 5 | 3 | | 6 | 4 | - | | | 6 | 4 | | 5 | 5 | - | | | 5 | 5 |

| T. C. | qr. | lb. | |-------|-----|-----| | 3 | 15 | 2 | | | 6 | 3 | | 5 | 7 | 3 | | | 3 | 2 | | 4 | 3 | 1 | | | 18 | 1 | | | 1 | 1 | | 5 | 3 | 7 | | | 6 | 4 | | 4 | 6 | - | | | 2 | 1 | | 3 | 4 | - |

When one page will not contain the whole account, we add the articles it contains, and write against their sum Carried forward; and we begin the next page with the sum of the foregoing, writing against it, Brought forward.

When the articles fill several pages, and their whole sum is known, which is the case in transcribing accounts, it is best to proceed in the following manner: Add the pages, placing the sums on a separate paper; then add the sums, and if the amount of the whole be right, it only remains to find what number should be placed at the foot and top of the pages. For this purpose, repeat the sum of the first page on the same line; add the sums of the first and second, placing the amount in a line with the second; to this add the sum of the third, placing the amount in a line with the third. Proceed in the like manner with the others; and if the last sum corresponds with the amount of the page, it is right. These sums are transcribed at the foot of the respective pages, and tops of the following ones. Examples.

| L. | 134 6 8 | |----|---------| | | 42 3 9 | | | 175 4 9 |

| L. | 170 5 4 | |----|---------| | | 66 9 8 | | | 73 8 6 |

| L. | 70 4 2 | |----|---------| | | 12 13 2 | | | 15 3 9 |

| L. | 15 3 9 | |----|---------| | | 12 2 6 | | | 7 5 4 |

| L. | 149 5 8 | |----|---------| | | 78 7 9 | | | 17 5 4 |

| L. | 149 5 8 | |----|---------| | | 18 6 8 | | | 5 10 |

L. 149 17

Then we transfer the 77s. 16s. at the foot of the first and top of the second pages, 1224l. 10s. 5d. at the foot of the second and top of the third; and so on.

Chap. III. SUBTRACTION.

Subtraction is the operation by which we take a lesser number from a greater, and find their difference. It is exactly opposite to addition, and is performed by learners in a like manner, beginning at the greater, and reckoning downwards the units of the lesser. The greater is called the minuend, and the lesser the subtrahend.

If any figure of the subtrahend be greater than the corresponding figure of the minuend, we add ten to that of the minuend, and having found and marked the difference, we add one to the next place of the subtrahend. This is called borrowing ten. The reason will appear, if we consider that, when two numbers are equally increased by adding the same to both, their difference will not be altered. When we proceed as directed above, we add ten to the minuend, and likewise add one to the higher place of the subtrahend, which is equal to ten of the lower place.

Rule. "Subtract units from units, tens from tens, and so on. If any figure of the subtrahend be greater than the corresponding one of the minuend, borrow ten."

Example. Minuend 173694 Subtrahend 21453

Remainder 152241

To prove subtraction, add the subtrahend and remainder together; if their sum be equal to the minuend, the account is right.

Or subtract the remainder from the minuend. If the difference be equal to the subtrahend, the account is right.

Rule for Compound Subtraction. "Place like denominations under like; and borrow, when necessary, according to the value of the higher place."

Example.

| C. | qr. | lib. | A. | R. | F. | E. | |----|-----|------|----|----|----|----| | L. | 146 | 3 3 | 12 | 3 19 | 15 | 2 24 | 18 | | | 58 | 7 6 | 4 | 3 24 | 12 | 2 36 | 7 |

L. 87 15 9 7 3 23 2 3 28 11

Note 1. The reason for borrowing is the same as in simple subtraction. Thus, in subtracting pence, we add 12 pence when necessary to the minuend, and at the next step, we add one shilling to the subtrahend.

Note 2. When there are two places in the same denomination, if the next higher contain exactly so many tens, it is best to subtract the units first, borrowing ten when necessary; and then subtract the tens, borrowing, if there is occasion, according to the number of tens in the higher denomination.

Note 3. If the value of the higher denomination be not an even number of tens, subtract the units and tens at once, borrowing according to the value of the higher denomination.

Note 4. Some choose to subtract the place in the subtrahend, when it exceeds that of the minuend, from the value of the higher denomination, and add the minuend to the difference. This is only a different order of proceeding, and gives the same answer.

Note 5. As custom has established the method of placing the subtrahend under the minuend, we follow it when there is no reason for doing otherwise; the minuend may be placed under the subtrahend with equal propriety; and the learner should be able to work it either way, with equal readiness, as this last is sometimes more convenient; of which instances will occur afterwards.

Note 6. The learner should also acquire the habit, when two numbers are marked down, of placing such a number under the lesser, that, when added together, the sum may be equal to the greater. The operation is the same as subtraction, though conceived in a different manner, and is useful in balancing accounts and on other occasions.

It is often necessary to place the sums in different columns, in order to exhibit a clear view of what is required. For instance, if the values of several parcels of goods are to be added, and each parcel consists of several articles, the particular articles should be placed in an inner column, and the sum of each parcel extended to the outer column, and the total added there.

If any person be owing an account, and has made some partial payments, the payments must be placed in an inner column, and their sum extended under that of the account in the outer column, and subtracted there.

An example or two will make this plain.

1st.] 30 yards linen at 2s. L. 3 45 ditto at 1s. 6d. 3 7 6

120 lb. thread at 4s. L. 24 40 ditto at 3s. 6 30 ditto at 2s. 6d. 3 15

33 15

L. 40 2 6

2d.] 1773. Jan. 15. Lent James Smith L. 50 22. Lent him further 70

Feb. 3. Received in part L. 62 5. Received further In gold L. 10 10 In silver 13

23 10

Balance due me L. 34 10 In multiplication, two numbers are given, and it is required to find how much the first amounts to, when reckoned as many times as there are units in the second. Thus, 8 multiplied by 5, or 5 times 8, is 40. The given numbers (8 and 5) are called factors; the first (8) the multiplicand; the second (5) the multiplier; and the amount (40) the product.

This operation is nothing else than addition of the same number several times repeated. If we mark 8 five times under each other, and add them, the sum is 40. But, as this kind of addition is of frequent and extensive use, in order to shorten the operation, we mark down the number only once, and conceive it to be repeated as often as there are units in the multiplier.

For this purpose, the learner must be thoroughly acquainted with the following multiplication table, which is composed by adding each digit twelve times.

| Twice | Thrice | Four times | Five times | Six times | Seven times | |-------|--------|------------|------------|-----------|-------------| | 1 is 2 | 1 is 3 | 1 is 4 | 1 is 5 | 1 is 6 | 1 is 7 | | 2 | 4 | 6 | 8 | 10 | 12 | | 3 | 6 | 9 | 12 | 15 | 18 | | 4 | 8 | 12 | 16 | 20 | 24 | | 5 | 10 | 15 | 20 | 25 | 30 | | 6 | 12 | 18 | 24 | 30 | 36 | | 7 | 14 | 21 | 28 | 35 | 42 | | 8 | 16 | 24 | 32 | 40 | 48 | | 9 | 18 | 27 | 36 | 45 | 54 | | 10 | 20 | 30 | 40 | 50 | 60 | | 11 | 22 | 33 | 44 | 55 | 66 | | 12 | 24 | 36 | 48 | 60 | 72 |

If both factors be under 12, the table exhibits the product at once. If the multiplier only be under 12, we begin at the unit place, and multiply the figures in their order, carrying the tens to the higher place, as in addition.

Ex. 76859 multiplied by 4, or 76859 added 4 times.

\[ \begin{array}{c} 4 \\ \times 76859 \\ \hline 307436 \end{array} \]

If the multiplier be 10, we annex a cypher to the multiplicand. If the multiplier be 100, we annex two cyphers; and so on. The reason is obvious, from the rule of cyphers in notation.

If the multiplier be any digit, with one or more cyphers on the right hand, we multiply by the figure, and annex an equal number of cyphers to the product. Thus, if it be required to multiply by 50, we first multiply by 5, and then annex a cypher. It is the same thing as to add the multiplicand 50 times; and this might be done by writing the account at large, dividing the column into 10 parts of 5 lines, finding the sum of each part, and adding these ten sums together.

If the multiplier consist of several significant figures, we multiply separately by each, and add the products. It is the same as if we divided a long account of addition into parts corresponding to the figures of the multiplier.

Example. To multiply 7329 by 365

\[ \begin{array}{cccc} 7329 & 7329 & 7329 & 36645 = 5 times \\ 5 & 60 & 300 & 439740 = 60 times \\ \hline 36645 & 439740 & 2198700 & 2675085 = 365 times \end{array} \]

It is obvious that 5 times the multiplicand added to 60 times, and to 300 times, the same must amount to the product required. In practice, we place the products at once under each other; and as the cyphers arising from the higher places of the multiplier are lost in the addition, we omit them. Hence may be inferred the following

Rule. "Place the multiplier under the multiplicand, and multiply the latter successively by the significant figures of the former; placing the right-hand figure of each product under the figure of the multiplier from which it arises; then add the product."

Ex. 7329 \times 42785

\[ \begin{array}{cccc} 7329 & 42785 & 37846 & 93956 \\ 365 & 91 & 235 & 8704 \\ \hline 36645 & 42785 & 189230 & 375824 \\ 43974 & 385965 & 113538 & 657692 \\ 21987 & 75692 & 751648 \\ \hline 2675085 & 8893810 & 817793024 \end{array} \]

A number which cannot be produced by the multiplication of two others is called a prime number; as 3, 5, 7, 11, and many others.

A number which may be produced by the multiplication of two or more smaller ones, is called a composite number. For example 27, which arises from the multiplication of 9 by 3; and these numbers (9 and 3) are called the component parts of 27.

Contractions and Varieties in Multiplication.

First, If the multiplier be a composite number, we may multiply successively by the component parts.

Ex. 7638 by 45 or 5 times 9

\[ \begin{array}{cccc} 7638 & 7638 & 7638 & 7638 \\ 45 & 9 & 2d, & 13759 by 56 \\ 38190 & 38190 & 38190 & 38190 \\ 68742 & 68742 & 68742 & 68742 \\ 30552 & 30552 & 30552 & 30552 \\ 342710 & 342710 & 342710 & 342710 \end{array} \]

Because the second product is equal to five times the first, and the first is equal to nine times the multiplicand, Multiplicand, it is obvious that the second product must be five times nine, or forty-five times as great as the multiplicand.

Secondly, If the multiplier be 5, which is the half of 10, we may annex a cypher, and divide by 2. If it be 25, which is the fourth part of 100, we may annex two cyphers, and divide by 4. Other contractions of the like kind will readily occur to the learner.

Thirdly, To multiply by 9, which is one less than 10, we may annex a cypher; and subtract the multiplicand from the number it composes. To multiply by 99,999, or any number of 9's, annex as many cyphers, and subtract the multiplicand. The reason is obvious; and a like rule may be found, though the unite place be different from 9.

Fourthly, Sometimes a line of the product is more easily obtained from a former line of the same than from the multiplicand.

Ex. 1st.] 1372 2d.] 1348

| 84 | 36 | |----|----| | 5488 | 8088 | | 10976 | 4044 | | 115248 | 48528 |

In the first example, instead of multiplying by 5, we may multiply 5480 by 2; and, in the second, instead of multiplying by 3, we may divide 8088 by 2.

Fifthly, Sometimes the product of two or more figures may be obtained at once, from the product of a figure already found.

Ex. 1st.] 14356 2d.] 3462321

| 648 | 90484 | |----|----| | 114848 | 13849284 | | 918784 | 166191408 | | 9392688 | 332382816 | | 334058579364 |

In the second example, we multiply first by 4; then because 12 times 4 is 48, we multiply the first line of the product by 12, instead of multiplying separately by 8 and 4; lastly, because twice 48 is 96, we multiply the second line of the product by 2, instead of multiplying separately by 6 and 9.

When we follow this method, we must be careful to place the right-hand figure of each product under the right-hand figure of that part of the multiplier which it is derived from.

It would answer equally well in all cases, to begin the work at the highest place of the multiplier; and contractions are sometimes obtained by following that order.

Ex. 1st.] 3125 or 3125 2d.] 32452

| 642 | 52575 | |----|----| | 18750 | 18750 | | 12500 | 131250 | | 6250 | 811300 | | 2006250 | 2433900 | | 2006250 | 1706163900 |

It is a matter of indifference which of the factors Multiplication be used as the multiplier; for 4 multiplied by 3 gives the same product as 3 multiplied by 4; and the like holds universally true. To illustrate this, we may make three rows of points, four in each row, placing the rows under each other; and we shall have also four rows, containing three points each, if we reckon the rows downwards.

Multiplication is proven by repeating the operation, using the multiplier for the multiplicand, and the multiplicand for the multiplier. It may also be proven by division, or by casting out the 9's; of which afterwards; and an account, wrought by any contraction, may be proven by performing the operation at large, or by a different contraction.

**Compound Multiplication.**

**Rule I.** "If the multiplier do not exceed 12, the operation is performed at once, beginning at the lowest place, and carrying according to the value of the place."

[Examples.]

| Cwt. qr. lb. | A. R. P. Lb. ozs. dwt. | |--------------|------------------------| | L. 13 6 7 12 2 8 | 13 3 18 7 5 9 | | 9 | 5 | | 6 | 12 |

L.119 19 3 62 3 12 83 — 28 89 5 8

**Rule II.** "If the multiplier be a composite number, whose component parts do not exceed 12, multiply first by one of these parts, then multiply the product by the other. Proceed in the same manner if there be more than two."

Ex. 1st.] L. 15 3 8 by 32=8×4

| 8 | | L.121 9 4 = 8 times. | | 4 |

L.485 17 4 = 32 times.

2d.] L. 17 3 8 by 75=5×5×5

| 3 | | L. 51 11 = 3 times. | | 5 |

L.257 15 = 15 times.

| 5 | | L.1288 15 = 75 times. |

Note 1. Although the component parts will answer in any order, it is best, when it can be done, to take them in such order as may clear off some of the lower places at the first multiplication, as is done in Ex. 2d.

Note 2. The operation may be proved, by taking the component parts in a different order, or dividing the multiplier in a different manner.

**Rule III.** "If the multiplier be a prime number, multiply first by the composite number next lower, then by the difference, and add the products." Here because 8 times 8 is 64, we multiply twice by 8, which gives 2296l. 16s. equal to 64 times the multiplicand; then we find the amount of 3 times the multiplicand, which is 107l. 13s. 2d.; and it is evident that these added, amount to 67, the multiplicand.

**Rule IV.** "If there be a composite number a little above the multiplier, we may multiply by that number, and by the difference, and subtract the second product from the first."

L. 17 4 5 by 106 = 108 - 2 Here we multiply 12 108 = 9 × 12 by 12 and 9, the component parts of 108, and obtain a product of 1860l. 6s. equal to 108 times the multiplicand; and, as this is twice oftener than was required, we subtract the multiplicand doubled, and the remainder is the number sought.

**Example.** 34l. 8s. 2½d. by 3405.

**Rule V.** "If the multiplier be large, multiply by 10, and multiply the product again by 10; by which means you obtain an hundred times the given number. If the multiplier exceed 1000, multiply by 10 again; and continue it farther, if the multiplier require it; then multiply the given number by the unit place of the multiplier; the first product by the ten-place, the second product by the hundred place; and so on. Add the products thus obtained together."

L. 34 8 2½ by 5 = L. 172 1 0½ = 5 times

10 times L. 344 2 1 by 6 = 2064 12 6 = 60 times

100 times L. 3441 10 by 4 = 13764 3 4 = 400 times

1000 times L. 34410 8 4 by 3 = 103231 5 = 3000 times

L. 103231 9 10½ = 13,65 times

The use of multiplication is to compute the amount of any number of equal articles, either in respect of measure, weight, value, or any other consideration. The multiplicand expresses how much is to be reckoned for each article; and the multiplier expresses how many times that is to be reckoned. As the multiplier points out the number of articles to be added, it is always an abstract number, and has no reference to any value or measure whatever. It is therefore quite improper to attempt the multiplication of shillings by shillings, or to consider the multiplier as expressive of any denomination. The most common instances in which the practice of this operation is required, are, to find the amount of any number of parcels, to find the value of any number of articles, to find the weight or measure of a number of articles, &c.

This computation, for changing any sum of money, weight, or measure, into a different kind, is called Reduction. When the quantity given is expressed in different denominations, we reduce the highest to the next lower, and add thereto the given number of that denomination; and proceed in like manner till we have reduced it to the lowest denomination.

**Example.** To reduce 46l. 13s. 8½d. to farthings.

L. 46 20 Or thus:

920 shillings in L. 46 13 4½

13

933 shillings in L. 46 13

12

11196 pence in L. 46 13

8

11204

4

44186 farthings in L. 46 13 8

3

44819 farthings in L. 46 13 8½

It is easy to take in or add the higher denomination at the same time we multiply the lower.

**Chap. V. Division.**

In division, two numbers are given; and it is required to find how often the former contains the latter. Thus, it may be asked how often 21 contains 7, and the answer is exactly 3 times. The former given number (21) is called the Dividend; the latter (7) the Divisor; and the number required (3) the Quotient. It frequently happens that the division cannot be completed exactly without fractions. Thus it may be asked, how often 8 is contained in 19? the answer is twice, and the remainder of 3.

This operation consists in subtracting the divisor from the dividend, and again from the remainder, as often as it can be done, and reckoning the number of subtractions; as,

| 21 | 19 | |----|----| | 7 first subtraction | 8 first subtraction | | 14 | 11 | | 7 second subtraction | 8 second subtraction | | 7 third subtraction | 3 remainder |

As this operation, performed at large, would be very tedious, when the quotient is a high number, it is proper to shorten it by every convenient method; and, for this purpose, we may multiply the divisor by any number whose product is not greater than the dividend, and so subtract it twice or thrice, or oftener at the same time. The best way is to multiply it by the greatest number, that does not raise the product too high, and that number is also the quotient. For example, to divide 45 by 7, we inquire what is the greatest multiplier for 7, that does not Division. not give a product above 45; and we shall find that it is 6; and 6 times 7 is 42, which, subtracted from 45, leaves a remainder of 3. Therefore 7 may be subtracted 6 times from 45; or, which is the same thing, 45, divided by 7, gives a quotient of 6, and a remainder of three.

If the divisor do not exceed 12, we readily find the highest multiplier that can be used from the multiplication table. If it exceed 12, we may try any multiplier that we think will answer. If the product be greater than the dividend, the multiplier is too great; and if the remainder, after the product is subtracted from the dividend, be greater than the divisor, the multiplier is too small. In either of these cases, we must try another. But the attentive learner, after some practice, will generally hit on the right multiplier at first.

If the divisor be contained oftener than ten times in the dividend, the operation requires as many steps as there are figures in the quotient. For instance, if the quotient be greater than 100, but less than 1000, it requires 3 steps. We first inquire how many hundred times the divisor is contained in the dividend, and subtract the amount of these hundreds. Then we inquire how often it is contained ten times in the remainder, and subtract the amount of these tens. Lastly, we inquire how many single times it is contained in the remainder. The method of proceeding will appear from the following example:

To divide 5936 by 8.

From 5936 Take 5600 = 700 times 8

Rem. 336 From which take 320 = 40 times 8

Rem. 16 From which take 16 = 2 times 8

0 742 times 8 in all.

It is obvious, that as often as 8 is contained in 59, so many hundred times it will be contained in 5900, or in 5936; and, as often as it is contained in 33, so many ten times it will be contained in 330, or in 336; and thus the higher places of the quotient will be obtained with equal ease as the lower. The operation might be performed by subtracting 8 continually from the dividend, which will lead to the same conclusion by a very tedious process. After 700 subtractions, the remainder would be 336; after 40 more, it would be 16; and after 2 more, the dividend would be entirely exhausted. In practice, we omit the cyphers, and proceed by the following rules:

Rules. 1st, "Assume as many figures on the left hand of the multiplier as contain the divisor once or oftener: find how many times they contain it, and place the answer as the highest figure of the quotient."

2d, "Multiply the divisor by the figure you have found, and place the product under the part of the dividend from which it is obtained."

3d, "Subtract the product from the figures above it."

4th, "Bring down the next figure of the dividend to the remainder, and divide the number it makes up, as before."

Examples. 1st. 8)5936(742 2d. 63)30114(478 Division.

| 56 | 252 | |----|-----| | 33 | 491 | | 32 | 441 |

| 16 | 504 | |----|-----| | 16 | 504 |

| 3d. 365)974932(2671 77 | |------------------------| | 730 | ... | | 2449 | | 2190 |

| 2593 | |------| | 2555 |

| 382 | |-----| | 365 |

Remainder 17

The numbers which we divide, as 59, 33, and 16, in the first example, are called individuals.

It is usual to mark a point under the figures of the dividend, as they are brought down, to prevent mistakes.

If there be a remainder, the division is completed by a vulgar fraction, whose numerator is the remainder, and its denominator the divisor. Thus, in Ex. 3, the quotient is 2671, and the remainder 17; and the quotient completed is 2671\(\frac{17}{33}\).

A number which divides another without a remainder is said to measure it; and the several numbers which measure another, are called its aliquot parts. Thus, 2, 4, 6, 8, and 12, are aliquot parts of 24. As it is often useful to discover numbers which measure others, we may observe,

1st, Every number ending with an even figure, that is, with 2, 4, 6, 8, or 0, is measured by 2.

2d, Every number ending with 5 or 0, is measured by 5.

3d, Every number, whose figures, when added, amount to an even number of 3's or 9's, is measured by 3 or 9, respectively.

Contractions and Varieties in Division.

First, When the divisor does not exceed 12, the whole computation may be performed without setting down any figures except the quotient.

Ex. 7)35868(5124 or 7)35868 5124

Secondly, when the divisor is a composite number, and one of the component parts also measures the dividend, we may divide successively by the component parts.

Ex. 1st.] 30114 by 63 2d.) 975 by 105=5×7×3 | 9)30114 | 5)975 | |---------|-------| | 7)3346 | 3)195 |

Quotient 478 Quotient 97

This method might be also used, although the component parts of the divisor do not measure the dividend; but the learner will not understand how to manage Division. manage the remainder till he be acquainted with the doctrine of vulgar fractions.

Thirdly, When there are cyphers annexed to the divisor, cut them off; and cut off an equal number of figures from the dividend; annex those figures to the remainder.

Ex. To divide 378643 by 5200.

\[ \begin{array}{c} 52|00|378643 \\ \hline 364 \\ 146 \\ 104 \\ \hline 4243 \end{array} \]

The reason will appear by performing the operation at large, and comparing the steps.

To divide by 10, 100, 1000, or the like. Cut off as many figures on the right hand of the dividend as there are cyphers in the divisor. The figures which remain on the left hand compose the quotient, and the figures cut off compose the remainder.

Fourthly, When the divisor consists of several figures we may try them separately, by inquiring how often the first figure of the divisor is contained in the first figure of the dividend, and then considering whether the second and following figures of the divisor be contained as often in the corresponding ones of the dividend with the remainder (if any) prefixed. If not, we must begin again, and make trial of a lower number. When the remainder is nine, or upwards, we may be sure the division will hold through the lower places; and it is unnecessary to continue the trial farther.

Fifthly, We may make a table of the products of the divisor, multiplied by the nine digits, in order to discover more readily how often it is contained in each individual. This is convenient when the dividend is very long; or when it is required to divide frequently by the same divisor.

\[ \begin{array}{c} 73 \times 2 = 146 \\ 3 = 216 \\ 4 = 292 \\ 5 = 365 \\ 6 = 438 \\ 7 = 511 \\ 8 = 584 \\ 9 = 657 \\ \end{array} \]

Rem. 8

Sixthly, To divide by 9, 99, 999, or any number of 9's, transcribe under the dividend part of the same, shifting the highest figure as many places to the right hand as there are 9's in the divisor. Transcribe it again, with the like change of place, as often as the length of the dividend admits; add these together, and cut off as many figures from the right hand of the sum as there are 9's in the divisor. The figures which remain on the left hand compose the Division quotient, and those cut off the remainder.

If there be any carriage to the unit place of the quotient, add the number carried likewise to the remainder, as in Ex. 2.; and if the figures cut off be all 9's, add 1 to the quotient, and there is no remainder.

Examples. 1st.] 99|324123 2d.] 99|547825

\[ \begin{array}{c} 3241 \\ 32 \\ \hline 327396 \\ \end{array} \]

Quotient 3273 and rem. 96.

Quotient 5533|58 rem.

3d.] 999|476523

\[ \begin{array}{c} 476 \\ \hline 476999 \\ \end{array} \]

Quotient 477

To explain the reason of this, we must recollect, that whatever number of hundreds any dividend contains, it contains an equal number of 99's, together with an equal number of units. In Ex. 1. the dividend contains 3241 hundreds, and a remainder of 23. It therefore contains 3241 times 99, and also 3241, besides the remainder already mentioned.—Again, 3241 contains 32 hundreds, and a remainder of 51: it therefore contains 32 99's, and also 32, besides the remainder of 41. Consequently the dividend contains 99, altogether, 3241 times, and 32 times, that is, 3273 times, and the remainder consists of 23, 41, and 32, added, which makes 96.

As multiplication supplies the place of frequent additions, and division of frequent subtractions, they are only repetitions and contractions of the simple rules, and when compared together, their tendency is exactly opposite. As numbers, increased by addition, are diminished and brought back to their original quantity by subtraction; in like manner, numbers compounded by multiplication are reduced by division to the parts from which they were compounded. The multiplier shows how many additions are necessary to produce the number; and the quotient shows how many subtractions are necessary to exhaust it. It follows, that the product, divided by the multiplicand, will quote the multiplier; and because either factor may be assumed for the multiplicand, therefore the product divided by either factor, quotes the other. It follows, also, that the dividend is equal to the product of the divisor and quotient multiplied together; and hence these operations mutually prove each other.

To prove multiplication. Divide the product by either factor. If the operation be right, the quotient is the other factor, and there is no remainder.

To prove division. Multiply the divisor and quotient together; to the product add the remainder, if any; and, if the operation be right, it makes up the dividend. Otherwise divide the dividend (after subtracting the remainder, if any) by the quotient. If the operation be right, it will quote the divisor. The reason of all these rules may be collected from the last paragraph. **Division.**

**Compound Division.**

**Rule I.** "When the dividend only consists of different denominations, divide the higher denomination, and reduce the remainder to the next lower, taking in (p. 629. Rule V.) the given number of that denomination, and continue the division."

**Examples.**

Divide L. 465 : 12 : 8 by 72.

| L. s. d. | L. s. d. | |----------|----------| | 72)465 | 12 8 | | | 6 9 4 |

Divide 345 cwt. 1 q. 8 lb. by 22.

| Cwt. q. lb. | Cwt. q. lb. | |-------------|-------------| | 22)345 | 1 8 | | | (15 2 21) |

Or we might divide by the component parts of 72, (as explained under Thirdly, p. 631).

| 432 | 125 | |------------|-----------| | 33 | 110 | | 20 | |

| 72)672 | 15 | |------------|-----------| | 648 | 4 | | 24 | 44 | | 14 | |

| 72)296 | 17 | |------------|-----------| | 288 | 28 | | | |

| 8 Rem. | | |------------|-----------| | | |

**Rule II.** "When the divisor is in different denominations, reduce both divisor and dividend to the lowest denomination, and proceed as in simple division. The quotient is an abstract number."

To divide 38l. 13s. by 3l. 4s. 5d.

| L. 3 4 5 | L. 3 8 13 | |----------|-----------| | 20 | 20 |

To divide 96 cwt. 1 q. 20 lb. by 3cwt. 2q. 8lb.

| Cwt. q. lb. | Cwt. q. lb. | |-------------|-------------| | 3 2 8 | 96 1 20 |

It is best not to reduce the terms lower than is necessary to render them equal. For instance, if each of them consists of an even number of sixpences, fourpences, or the like, we reduce them to sixpences, or fourpences, but not to pence.

The use of division is to find either of the factors by whose multiplication a given number is produced, when the other factor is given; and therefore is of two kinds, since either the multiplier or the multiplicand may be given. If the former be given, it discovers what that number is which is contained so many times in another. If the latter be given, it discovers how many times one number is contained in another. Thus, it answers the questions of an opposite kind to those mentioned under Rule IV. p. 629, as, To find the quantity of a single parcel or share; to find the value, weight, or measure, of a single article; to find how much work is done, provisions consumed, interest incurred, or the like, in a single day, &c.

The last use of division is a kind of reduction exactly opposite to that described under Rule V. p. 629. The manner of conducting and arranging it, when there are several denominations in the question, will appear from the following examples.

1. To reduce 15783 pence to pounds, sh. and pence. 2. To reduce 174865 grs. to lb. oz. and dwt. Troy.

| 12)15783(1315(65 | |------------------| | 12 | 120 |

| 24)174865(7286(364(30 | |----------------------| | 168 | 360 |

Answer, 65l. 15s. 3d. Ans. 38lb. 4oz. 6dwt. 1gr.

In the first example, we reduce 15783 pence to shillings, by dividing by 12, and obtain 1315 shillings, and a remainder of 3 pence. Then we reduce 1315 shillings to pounds, by dividing by 20, and obtain 65 pounds and a remainder of 15 shillings. The divisions might have been contracted.

In the practice of arithmetic, questions often occur which require both multiplication and division to resolve. This happens in reduction, when the higher denomination does not contain an exact number of the lower.

**Rule for mixed reduction.** "Reduce the given denomination by multiplication to some lower one, which is an aliquot part of both; then reduce that by division to the denomination required."

Ex. Reduce 31742l. to guineas.

Here we multiply by 20, which reduces the pounds to shillings; and divide the product by 21, which reduces the shillings to guineas.

| 31742 | |-------| | 20 |

| 634840 | |--------| | 21 |

Answer, 30230 guineas and 10 shillings. As Portuguese money frequently passes here in payments, we shall give a table of the pieces, and their value.

| Piece | Value | |----------------|-------| | A moidore | L.1 | | A half moidore | -13 | | A quarter moidore | -6 | | A double Joannes | 3 | | A Joannes | 1 | | A half ditto | -18 | | A quarter ditto | -9 | | An eighth ditto | -4 |

Note 1. Guineas may be reduced to pounds, by adding one-twentieth part of the number.

2. Pounds may be reduced to merks by adding one half.

3. Merks may be reduced to pounds by subtracting one-third.

4. Four moidores are equal to three Joannes; therefore moidores may be reduced to Joannes, by subtracting one-fourth; and Joannes to moidores, by adding one-third.

5. Five Joannes are equal to 9l. Hence it is easy to reduce Portuguese money to sterling.

Another case, which requires both multiplication and division, is, when the value, weight, measure, or duration of any quantity is given, and the value, &c., of a different quantity required, we first find the value, &c., of a single article by division, and then the value, &c., of the quantity required, by multiplication.

Ex. If 3 yards cost 15s. 9d. what will 7 yards cost, at the same rate?

| s. | d. | |----|----| | 15 | 9 | | 5 | 3 |

Price of 1 yard, by Rule IV. p. 629.

L. 1 16 9 Price of 7 yards (by par. ult. p. 632. col. 1.).

Many other instances might be adduced, where the operation, and the reason of it, are equally obvious. These are generally, though unnecessarily, referred to the rule of Proportion.

We shall now offer a general observation on all the operations in arithmetic. When a computation requires several steps, we obtain a just answer, whatever order we follow. Some arrangements may be preferable to others in point of ease, but all of them lead to the same conclusion. In addition, or subtraction, we may take the articles in any order, as is evident from the idea of number; or, we may collect them into several sums, and add or subtract these, either separately or together. When both the simple operations are required to be repeated, we may either complete one of them first, or may introduce them promiscuously, and the compound operations admit of the same variety. When several numbers are to be multiplied together, we may take the factors in any order, or we may arrange them into several clauses, find the product of each clause, and then multiply the products together. When a number is to be divided by several others, we may take the divisors in any order, or we may multiply them into each other, and divide by the product; or we may multiply them into several parcels, and divide by the products successively. Lastly, When multiplication and division are both required, we may begin with either; and when both are repeatedly necessary, we may collect the multipliers into one product, and the divisors into another; or, we may collect them into parcels, or use them singly, and that in any order. Still we shall obtain the proper answer, if none of the terms be neglected.

When both multiplication and division are necessary to obtain the answer of a question, it is generally best to begin with the multiplication, as this order keeps the account as clear as possible from fractions. The example last given may be wrought accordingly as follows:

| s. | d. | |----|----| | 15 | 9 | | 7 | |

3) 5 10 3

Some accountants prove the operations of arithmetic by a method which they call casting out the 9's, depending on the following principles:

Firstly, If several numbers be divided by any divisor (the remainders being always added to the next number), the sum of the quotients, and the last remainder, will be the same as those obtained when the sum of the number is divided by the same divisor. Thus, 19, 15, and 23, contain, together, as many 5's, as many 7's, &c., as their sum 57 does, and the remainders are the same; and, in this way, addition may be proven by division. It is from the correspondence of the remainders, that the proof by casting out the 9's is deduced.

Secondly, If any figure with cyphers annexed, be divided by 9, the quotient consists entirely of that figure; and the remainder is also the same. Thus, 40, divided by 9, quotes 4, remainder 4; and 400 divided by 9, quotes 44, remainder 4. The same holds with all the digits; and the reason will be easily understood: every digit, with a cypher annexed, contains exactly so many tens; it must therefore contain an equal number of 9's, besides a remainder of an equal number of units.

Thirdly, If any number be divided by 9, the remainder is equal to the sum of the figures of the number, or to the remainder obtained, when that sum is divided by 9. For instance, 3765, divided by 9, leaves a remainder of 3; and the sum of 3, 7, 6, and 5, is 21, which divided by 9, leaves a remainder of 3. The reason of this will appear from the following illustration:

\[ \begin{array}{cccc} 3000 & \text{divided by } 9 & \text{quotes} & 333; \\ 700 & \text{quotes} & 77; \\ 60 & \text{quotes} & 6; \\ 5 & \text{quotes} & 0; \\ \hline 3765 & \text{Sum of rem.} & 21 \\ \end{array} \]

Again: 21 divided by 9 quotes 2; remainder 3

wherefore, 3765 divided by 9 quotes 418; remainder 3; for the reason given. Hence we may collect the following rules for practice.

To cast the 9's out of any number, or to find what remainder will be left when any number is divided by 9: Add the figures; and when the sum exceeds 9, add the figures which would express it. Pass by the 9's; and when the sum comes exactly to 9, neglect it, and begin anew. For example, if it be required to cast the 9's out of 3573294, we reckon thus; 3 and 4 L. Division. 5 is 8, and 7 is 15; 1 and 5 is 6, and 3 is 9, which we neglect; 2 and (passing by 9) 4 is 6; which is the remainder or RESULT. If the article out of which the 9's are to be cast contains more denominations than one, we cast the 9's out of the higher, and multiply the result by the value of the lower, and carry on the product (casting out the 9's, if necessary), to the lower.

To prove addition, cast the 9's out of the several articles, carrying the results to the following articles; cast them also out of the sum. If the operation be right, the results will agree.

To prove subtraction, cast the 9's out of the minuend; cast them also out of the subtrahend and remainder together; and if you obtain the same result, the operation is presumed right.

To prove multiplication, cast the 9's out of the multiplicand, and also out of the multiplier, if above 9. Multiply the results together, and cast the 9's, if necessary, out of their product. Then cast the 9's out of the product, and observe if this result correspond with the former.

Ex. 1st.] 9276 ref. 6 × 8 = 48 ref. 3.

\[ \begin{array}{c} 8 \\ 74208 \text{ ref. } 3. \end{array} \]

2nd.] 7898 ref. 5 × 3 = 15 ref. 6.

\[ \begin{array}{c} 48 \text{ ref. } 3 \\ 63184 \\ 31592 \\ 379104 \text{ ref. } 6. \end{array} \]

The reason of this will be evident, if we consider multiplication under the view of repeated addition. In the first example it is obviously the same. In the second, we may suppose the multiplicand repeated 48 times. If this be done, and the 9's cast out, the result, at the end of the 9th line, will be 0; for any number, repeated 9 times, and divided by 9, leaves no remainder. The same must happen at the end of the 18th, 27th, 36th, and 45th lines; and the last result will be the same as if the multiplicand had only been repeated 3 times. This is the reason for casting out the 9's from the multiplier as well as the multiplicand.

To prove division, cast the 9's out of the divisor, and also out of the quotient; multiply the results, and cast the 9's out of the product. If there be any remainder, add to it the result, casting out the 9's, if necessary. If the account be right, the last result will agree with that obtained from the dividend.

Ex.] 42) 2490 (59 ref. 5 × 6 = 30 ref. 3.

\[ \begin{array}{c} \text{ref. } 6 \\ 210 \\ 390 \\ 378 \\ \end{array} \]

Rem. 12 - - ref. 3.

And the result of the dividend is 6

This depends on the same reason as the last; for the dividend is equal to the product of the divisor and quotient added to the remainder.

We cannot recommend this method, as it lies under proportion. The following disadvantages.

Firstly, If an error of 9, or any of its multiples, be committed, the results will nevertheless agree; and so the error will remain undiscovered. And this will always be the case, when a figure is placed or reckoned in a wrong column; which is one of the most frequent causes of error.

Secondly, When it appears by the disagreement of the results, that an error has been committed, the particular figure or figures in which the error lies are not pointed out; and, consequently, it is not easily corrected.

Chap. VI. RULE OF PROPORTION.

Sect. I. Simple Proportion.

Quantities are reckoned proportional to each other, when they are connected in such a manner, that if one of them be increased or diminished, the other increases or diminishes at the same time; and the degree of the alteration on each is a like part of its original measure; thus four numbers are in the same proportion, the first to the second as the third to the fourth, when the first contains the second, or any part of it, as often as the third contains the fourth, or the like part of it. In either of these cases, the quotient of the first divided by the second, is equal to that of the third divided by the fourth; and this quotient may be called the measure of the proportion.

Proportions are marked down in the following manner:

\[ \begin{array}{c} 6 : 3 :: 8 : 4 \\ 12 : 36 :: 9 : 27 \\ 9 : 6 :: 24 : 18 \\ 16 : 24 :: 10 : 15 \end{array} \]

The rule of Proportion directs us, when three numbers are given, how to find a fourth, to which the third may have the same proportion that the first has to the second. It is sometimes called the Rule of Three, from the three numbers given; and sometimes the Golden Rule, from its various and extensive utility.

Rule. "Multiply the second and third terms together, and divide the product by the first."

Ex. To find a fourth proportional to 18, 27, and 34.

\[ \begin{array}{c} 18 : 27 :: 34 : 51 \\ 34 \\ 108 \\ 81 \\ \end{array} \]

\[ \begin{array}{c} 18918(51) \\ 90 \\ 18 \\ 18 \\ 0 \end{array} \]

To explain the reason of this, we must observe that if two or more numbers be multiplied or divided alike, the products or quotients will have the same proportion.

\[ \begin{array}{c} 18 : 27 \\ \text{Multiplied by } 34, 612 : 918 \\ \text{Divided by } 18, 34 : 51 \end{array} \] The products 612, 918, and the quotients 34, 51, have therefore the same proportion to each other that 18 has to 27. In the course of this operation, the products of the first and third term are divided by the first; therefore the quotient is equal to the third.

The first and second terms must always be of the same kind; that is, either both moneys, weights, measures, both abstract numbers, or the like. The fourth, or number sought, is of the same kind as the third.

When any of the terms is in more denominations than one, we may reduce them all to the lowest. But this is not always necessary. The first and second should not be reduced lower than directed p. 632, col. 1, par. penult.; and, when either the second or third is a simple number, the other, though in different denominations, may be multiplied without reduction.

| L. | s. | d. | |----|----|----| | Ex. 5 : 7 :: 25 11 3 |

| L. | s. | d. | |----|----|----| | 7 |

| 5) 178 18 9 (35 19 9 |

The accountant must consider the nature of every question, and observe the circumstance which the proportion depends on; and common sense will direct him to this if the terms of the question be understood. It is evident that the value, weight, and measure of any commodity is proportioned to its quantity; that the amount of work or consumption is proportioned to the time; that gain, loss, or interest, when the rate and time are fixed, is proportioned to the capital sum from which it arises; and that the effect produced by any cause is proportioned to the extent of the cause. In these, and many other cases, the proportion is direct, and the number sought increases or diminishes along with the term from which it is derived.

In some questions, the number sought becomes less, when the circumstances from which it is derived become greater. Thus, when the price of goods increases, the quantity which may be bought for a given sum is smaller. When the number of men employed at work is increased, the time in which they may complete it becomes shorter; and, when the activity of any cause is increased, the quantity necessary to produce a given effect is diminished. In these, and the like, the proportion is said to be inverse.

**General Rule for stating all questions, whether direct or inverse.** "Place that number for the third term which signifies the same kind of thing with what is sought, and consider whether the number sought will be greater or less. If greater, place the least of the other terms for the first; but, if less, place the greatest for the first."

Ex. 1st.] If 30 horses plough 12 acres, how many will 42 plough in the same time?

H. H. A.

30 :: 42 :: 12

Here, because the thing sought is a number of acres, we place 12, the given number of acres, for the third term; and, because 42 horses will plough more than 12, we make the lesser number 30, the first term, and the greater number, 42, the second term.

Ex. 2d.] If 40 horses be maintained for a certain sum on hay, at 5d. per stone, how many will be maintained on the same sum when the price of hay rises to 8d.

d. d. H.

8 : 5 :: 40

Here, because a number of horses is sought, we make the given number of horses, 40, the third term; and, because fewer will be maintained for the same money, when the price of hay is dearer, we make the greater price, 8l., the first term; and the lesser price 5d., the second term.

The first of these examples is direct, the second inverse. Every question consists of a supposition and demand. In the first, the supposition is, that 30 horses plough 12 acres, and the demand how many 42 will plough? and the first term of the proportion, 30, is found in the supposition, in this and every other direct question. In the second, the supposition is, that 40 horses are maintained on hay at 5d., and the demand, how many will be maintained on hay at 8l.? and the first term of the proportion, 8, is found in the demand, in this and every other inverse question.

When an account is stated, if the first and second term, or first and third, be measured by the same number, we may divide them by that measure, and use the quotients in their stead.

Ex. If 36 yards cost 42 shillings, what will 27 cost?

Y. Y. th.

Here 36 and 27 are both measured by 9, and we work

| 36 : 27 :: 42 | | 4 : 3 :: 42 |

s. d.

4)126(31 6

**Sect. II. Compound Proportion.**

Sometimes the proportion depends upon several circumstances. Thus, it may be asked, if 18 men consume 6 bolls corn in 28 days, how much will 24 men consume in 56 days? Here the quantity required depends partly on the number of men, partly on the time; and the question may be resolved into the two following ones:

1st. If 18 men consume 6 bolls, in a certain time, how many will 24 men consume in the same time?

M. M. B. B.

18 : 24 : 6 : 8

Answer, 24 men will consume

6

8 bolls in the same time.

18)144(8

2d. If a certain number of men consume 8 bolls in 28 days, how many will they consume in 56 days?

D. D. B. B.

28 : 56 :: 8 : 16

Answer The same number of men

will consume 16 bolls in 56 days.

28)448(16

In the course of this operation, the original number of bolls, 6, is first multiplied into 24, then divided by 18, then multiplied into 8, then divided by 28. It would answer the same purpose to collect the multipliers into one product, and the divisors into another; and then to multiply the given number of bolls by the former, and divide the product by the latter, p. 633, col. 1, par. ult.

The above question may therefore be stated and wrought as follows:

4 L. 2 Men Here we multiply 18 into 28 for a divisor, and 6 into the product of 24 by 56, for a dividend.

\[ \begin{align*} 144 & \quad 144 \\ 36 & \quad 120 \\ \hline 504 & \quad 1344 \\ 6 & \\ \hline 504)8064(16 \end{align*} \]

"In general, state the several particulars on which the question depends, as so many simple proportions, attending to the sense of the question to discover whether the proportions should be stated directly or inversely; then multiply all the terms in the first rank together, and all those in the second rank together; and work with the products as directed in the simple rule (Sect. i. p. 634.)"

**Example.** If 100 men make 3 miles of road in 27 days, in how many days will 150 men make 5 miles?

Men \(150 : 100 :: 27\) days.

Here the first stating is inverse, because more men will do it in fewer days; but the second is direct, because more miles will require more days.

The following contraction is often useful. After stating the proportion, if the same number occurs in both ranks, dash it out from both; or, if any term in the first rank, and another in the second rank are measured by the same numbers, dash out the original terms, and use the quotients in their stead.

**Ex.** If 18 men consume 30l. value of corn in 9 months, when the price is 16s. per boll, how many will consume 54l. value in 6 months, when the price is 12s. per boll? In this question, the proportion depends upon three particulars, the value of corn, the time and the price. The first of which is direct, because the more the value of provisions is, the more time is required to consume them; but the second and third are inverse, for the greater the time and price is, fewer men will consume an equal value.

Value \(36 : 24 :: 18\) men.

Months \(6 : 9\)

Price \(12 : 18\)

\[ \begin{align*} 10 & \quad 9 \\ 3 & \quad 3 \\ 4 & \\ \hline 36 & \quad 18 \\ 288 & \quad 36 \\ \hline 10)648(64\frac{8}{9} \end{align*} \]

The moneys, weights, and measures of different countries, may be reduced from the proportion which they bear to each other.

**Ex.** If 112 lb. avoirdupois make 104 lb. of Holland, and 100 lb. of Holland make 89 of Geneva, and 110 of Geneva make 117 of Seville, how many lbs. of Seville will make 100 lb. avoirdupois?

\[ \begin{align*} 112 & : 104 :: 100 \\ 100 & : 89 \\ 110 & : 117 \end{align*} \]

If it be required, how many lb. avoirdupois will make 100 of Seville, the terms would have been placed in the different columns thus:

\[ \begin{align*} 104 & : 112 :: 100 \\ 89 & : 100 \\ 117 & : 110 \end{align*} \]

**Sect. III. Distributive Proportion.**

If it be required to divide a number into parts, which have the same proportion to each other, that several other given numbers have, we add these numbers together, and state the following proposition: As the sum is to the particular numbers, so is the number required to be divided to the several parts sought.

**Ex. [a.]** Four partners engage to trade in company; A's stock is 150l. B's 320l. C's 350l. D's 500l.; and they gain 730l.: Required how much belongs to each, if the gain be divided among them in proportion to their stocks?

A's stock L. 150 1320 : 150 :: 730 : L. 82 19 1 = 120

B's 320 1320 : 320 :: 730 : 176 19 4 = 960

C's 350 1320 : 350 :: 730 : 193 11 2 = 720

D's 500 1320 : 500 :: 730 : 276 10 3 = 840

Whole stock 1320 Proof L. 730

This account is proved by adding the gains of the partners; the sum of which will be equal to the whole gain, if the operation be right; but, if there be remainders, they must be added, their sum divided by the common divisor, and the quotient carried to the lowest place.

**Ex. [b.]** A bankrupt owes A 146l., B 170l., C. 45l., D 480l. and E 72l.; his whole effects are only 342l. 75s. 6d. How much should each have?

A's debt L.146 913 : 146 :: L.342 7 6 : L.54 15 A's share.

B's 170 913 : 170 :: 342 7 6 : 63 15 B's

C's 45 913 : 45 :: 342 7 6 : 16 17 6 C's

D's 480 913 : 480 :: 342 7 6 : 180 D's

E's 72 913 : 72 :: 342 7 6 : 27 E's

L.913 L.342 7 6

This might also be calculated, by finding what composition the bankrupt was able to pay per pound; which is obtained by dividing the amount of his effects by the amount of his debts, and comes to 75s. 6d. and then finding by the rules of practice, how much each debt came to at that rate.

**Chap. VII. Rules for Practice.**

The operations explained in the foregoing chapters comprehend the whole system of arithmetic, and are sufficient for every computation. In many cases, however, the work may be contracted, by adverting to the particular circumstances of the question. We shall explain, in this chapter, the most useful methods which practice has suggested for rendering mercantile computations easy; in which the four elementary rules of arithmetic are sometimes jointly, sometimes separately employed.

**Sect. I. Computation of Prices.**

The value of any number of articles, at a pound, a shilling, Chap. VII.

Practice. shilling, or a penny, is an equal number of pounds, shillings, or pence; and these two last are easily reduced to pounds. The value, at any other rate, may be calculated by easy methods, depending on some contraction already explained, or on one or more of the following principles.

1st. If the rate be an aliquot part of a pound, a shilling, or a penny, then an exact number of articles may be bought for a pound, a shilling, or a penny; and the value is found by dividing the given number accordingly. Thus, to find the price of so many yards at 2s. 6d. which is the eighth part of a pound, we divide the quantity by eight, because every eight yards cost L.1.

2nd. If the rate be equal to the sum of two other rates which are easily calculated, the value may be found by computing these separately, and adding the sums obtained. Thus, the price of so many yards, at 9d. is found, by adding their prices, at 6d. and 3d. together.

3rd. If the rate be equal to the difference of two easy rates, they may be calculated separately, and the lesser subtracted from the greater. Thus, the value of so many articles at 11d. is found, by subtracting their value at a penny from their value at a shilling. We may suppose that a shilling was paid for each article, and then a penny returned on each.

4th. If the rate be a composite number, the value may be found by calculating what it comes to at one of the component parts, and multiplying the same by the other.

CASE I. "When the rate is an aliquot part of a pound, divide the quantity by the number which may be bought for a pound."

Table of the aliquot parts of L.1.

| 10 shillings = | 1/2 of L.1. | |----------------|-------------| | 6s. 8d. | 1/3 | | 5s. | 1/4 | | 4s. | 1/5 | | 3s. 4d. | 1/6 | | 2s. 6d. | 1/7 | | 1s. 8d. | 1/8 |

Ex. 1st.] What is the value of 7463 yards, at 4s.? 5)7463

L. 1492 12s.

In the first example we divide by 5 because 4s. is 1/5 of a pound; the quotient 1492 shows how many pounds they amount to; besides which there remain three yards at 4s., and these come to 12s. In the second example, we divide by 8s., as directed, and the quotient gives L.22, and the remainder 13 yards, which at 3d. come to 3s. 3d.

This method can only be used in calculating for the particular prices specified in the table. The following 6 cases comprehend all possible rates, and will therefore exhibit different methods of solving the foregoing questions.

CASE II. "When the rate consists of shillings only, multiply the quantity by the number of shillings, and divide the product by 20: Or, if the number of shillings be even, multiply by half the number, and divide the product by 10."

Ex. 1st.] 4573 at 13s. 13

13719

4573

20)59449

L. 2972 9s.

The learner will easily perceive, that the method in which the second example is wrought, must give the same answer as if the quantity had been multiplied by 14, and divided by 20; and as the division by 10 doubles the last figure for shillings, and continues all the rest unchanged for pounds, we may obtain the answer at once, by doubling the right-hand figure of the product before we set it down.

If the rate be the sum of two or more aliquot parts of a pound, we may calculate these as directed in Case I, and add them. If it be any odd number of shillings, we may calculate for the even number next lower, and add thereto the value of a shilling. If it be 19s., we may subtract the value at a shilling, from the value at a pound.

CASE III. "When the rate consists of pence only."

Method 1. If the rate be an aliquot part of a shilling, divide the quantity accordingly, which gives the answer in shillings; if not, it may be divided into two or more aliquot parts: calculate these separately, and add the values; reduce the answer to pounds.

1 penny is 1/12 of a shilling

2d. 1/6 of ditto.

3d. 1/4 of ditto.

4d. 1/3 of ditto.

6d. 1/2 of ditto.

5d. is the sum of 4d. and 1d. or of 2d. and 3d.

7d. is the sum of 4d. and 3d. or of 6d. and 1d.

8d. is the sum of 6d. and 2d. or the double of 4d.

9d. is the sum of 6d. and 3d.

10d. is the sum of 6d. and 4d.

11d. is the sum of 6d. 3d. and 2d.

Ex. 1st.] 7423 at 4d.

3)7423

20)2474 4

L. 123 14 4

2d.)9786 at 9d.

At 6d. = 1/6 of 1s. 4893

At 3d. = 1/3 of 6d. 2446

At 9d. 7339 6

L. 366 19

3d.)4856 at 11d.

At 6d. = 1/6 of 1s. 2428

At 3d. = 1/3 of 6d. 1214

At 2d. = 1/2 of 6d. 809

11d. 4451 4

L. 222 11 4

It is sometimes easier to calculate at two rates, whose difference is the rate required, and subtract the lesser value from the greater. Thus, the last example may be wrought by subtracting the value at a penny from the value at a shilling. The remainder must be the value

Chap. VII.

Practice. value at 1½d.

At 1s. 4856s. At 1½d. = 1s. 404 8 At 1½d. 4451 4 L.222 11 4

Meth. 2. Multiply the quantity by the number of pence, the product is the answer in pence. Reduce it to pounds.

Meth. 3. Find the value at a penny by division, and multiply the same by the number of pence.

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

Ex. 11th. J 378 13 at 1 farthing. 2d. J 23754 at ¼d. 4)378 13 farthing. 2)23754 halfpence 12)9 6¾ pence 12)11877 pence 788 8 4¾ 989 9 L. 39 8 4¾ L. 49 9 9 3d. J 72564 at ¼d. Or, 72564

At ¼d. 3682 d. 4)217692 farthing. 1)54423 pence 4535 3 12)54423 d. L. 226 15 3 4535 3 L. 226 15 3

We may also find the amount in twopences, threepences, fourpences, or sixpences, by one division, and reduce these as directed in Case I.

Case V. "When the rate consists of pence and farthings, find the value of the pence, as directed in Case III. and that of the farthings from the proportion which they bear to the pounds. Add these together, and reduce."

Ex. 11th. J 3287 at 5½d.

At 4d. = ¼ of 1s. 1095 8 At 1½d. = ¼ of 4d. 273 11 At 1½d. = ¼ of 1d. 68 5¾

At 5½ 1438 ¾ L. 71 18 ¾ 2d. J 4573 at 2¾

At 2d. = ¼ of 1s. 762 2 At ¼d. = ¼ of 2d. 190 6½ At ¼d. = ¼ of ¼d. 85 3¾

At 2½ 1037 11¾ L. 51 17 11¾ 3d. J 2842 at 3¾d.

At 3d. = ¼ of 1s. 710 6 At 3½d. = ¼ of 3d. 176 7¾

At 3½ 887 1¾ L. 44 8 1¾ 4½h. J 3572 at 7½d.

At 6½d. = ¼ of 1s. 1386 At 1½d. = ¼ of 6d. 346 6

At 7½ 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 6d. and at 3 halfpence, which makes 7½d.

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

Case VI. "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. And add reduce.

Ex. 11th. J 4258 at 17s. 3d.

17 29806 4258

17s. 72386

3d. = ¼ of 1s. 1064 6 17s. 3d. 73450 6 L. 3672 10 6 2d. J 5482 at 12s. 4½d.

12

12s. 65784 1370 6 685 3

12s. 4½d. 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. 11th. J 3894 at 17 6 2d. J 1765 at 9 2

10s. = ¼ L. 1947 5s. = ¼ 973 10 2s. 6d. = ¼ 486 15

17s. 6d. 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.

Ex. 11th. J 63790 at 5 4 4th. J 3664 at 14 9

4s. = ¼ of L. 1278 1s. 4d. = ¼ of 4s. 4252 13 4 5s. 4d. L. 17010 13 4

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 2s. 11d. or 35d.

12)5628 20)469 L. 23 9 5 L. 117 5 7 L. 820 15

Case 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. 1ft.] 3592 at L.3 : 12 : 8.

L.3

12s. = \(\frac{1}{3}\) of L.3

8l. = \(\frac{1}{16}\) of 12s.

L.3 12 8

L.13050 18 8

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

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

72 20

7184 72

25144

At 4s. 165735

At 1d. = \(\frac{1}{12}\) s. 307 11

44s. 11d. 165427 - 1

L.13050 18 8

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 halfpenny 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 by several denominations, and the rate given for the higher; calculate the higher, consider the lower one 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.

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.

2d, 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.

3d, 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 groat 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.

Cloff, 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. 1ft. What is the nett weight of 17 C. 2 q. 14lb. tare 18 lb. per cwt.

| C. q. lb. | C. q. lb. | |-----------|-----------| | 17 2 14 groats, or 17 2 14 | 6 | | 16lb. = \(\frac{1}{2}\) C. | 2 2 2 | | 2lb. = \(\frac{1}{8}\) of 16lb. | 1 7 | | 18 lb. | 2 3 9\(\frac{1}{4}\) tare |

In the first method, we add the tare at 16lb. which is \(\frac{1}{2}\) of the groat weight, to the tare, at 2 lb. which is \(\frac{1}{8}\) of the former. In the second, we multiply the groat weight by 18; the tare is 1 lb. for each cwt. of the product, and is reduced by division to higher denominations.

2d.] What is tret of 158 C. 2 q. 24 lb.

| C. q. lb. | C. q. lb. | |-----------|-----------| | 158 2 26 | (6 - 11 Tret. |

Because tret is always 4 lb. in 104, or 1 lb. in 26, it is obtained by dividing by 26.

3d.] 3d.] 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 C's and multiplying it by 2. It is better to begin with multiplication, for the reason given, p. 633. 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 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 Stock, 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 728l. at 2\(\frac{1}{4}\) per cent.?

\[ \begin{array}{c} 728 \\ \times 2\frac{1}{4} \\ \hline 1456 \\ 364 \\ \hline 182 \\ \hline 20 \\ \hline 40 \\ 12 \\ \hline 480 \\ \hline 4 Answ. L. 20 — 4\(\frac{1}{4}\) \end{array} \]

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 10l. at \(\frac{1}{4}\) per cent. is a shilling; and at other rates in proportion. Thus, 170l. at \(\frac{1}{4}\) per cent. is 17s.; and, at \(\frac{1}{4}\) per cent. 8s. 6d.

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 34675l. 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 34675l. 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 34675l. for 17 days, is 17 times 95 shillings or 1615 shillings; and this divided by 20, in order to reduce it, comes to 80l. 15s.

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 inserted p. 631. 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 34675l. for 17 days, at 5 per cent.?

\[ \begin{array}{c} 34675 \\ \times 17 \\ \hline 242725 \\ 34675 \\ \hline L. s. 73|00)5894|75(80 15 384 \\ 5475 \\ 20 \\ 1095|00 \\ 73 \\ 365 \\ 365 \\ 0 \end{array} \]

Ex. 2d. Interest on 304l. 3s. 4d. for 8 days, at 4 per cent.

\[ \begin{array}{c} L. 304 3 4 \\ \times 8 \\ \hline s. d. 73|00)2433 6 8(6 8 20 \\ 486|66 \\ 438 \\ 4866 \\ 12 \\ 584|00 \\ 584 \\ 0 \end{array} \] When partial payments are made, we proceed in the following manner: Let us suppose a bill of £170l. was due 12th August, that £4l. was paid on the 18th September, £6l. on the 17th October, and the balance on the 14th November; and let it be required to find how much interest is due.

| Days | |------| | Aug. 12. L.170 | 37 | | Sept. 18. pd. | 54 | | Oct. 17. pd. | 56 | | Nov. 14. pd. | 60 |

Here we subtract the several payments from the original sum in their order, placing the dates in the margin; and from this it appears that there is interest due on £170l. from 12th August to 18th September, on £16l. from 18th September to 17th October, and on £6l. from 17th October to 14th November. We next compute the number of days in each of these periods, and mark it against the respective sum. Then we multiply each sum by the number of days; reserving a column, when necessary, for the products of the several figures in the multiplier. Lastly, we add these products, and divide their sum by 7300.

Interest on current accounts is calculated nearly in the same manner. For example, let the interest due on the following account be required to 31st July, at 4 per cent:

Dr. Mr A. Baird, his account current with W. Neil, Cr.

| Date | Dr. | L. | |------|-----|----| | Jan. 15. | 16c. | 50 | | Mar. 12. | 36 | 960 | | June 23. | 13 | 6 | | July 19. | 26 | 13 |

| Date | Cr. | L. | |------|-----|----| | Jan. 15. | 16c. | 50 | | Mar. 12. | 36 | 960 | | June 23. | 13 | 6 | | July 19. | 26 | 13 |

| Date | Dr. | L. | |------|-----|----| | Jan. 15. | 16c. | 50 | | Mar. 12. | 36 | 960 | | June 23. | 13 | 6 | | July 19. | 26 | 13 |

| Date | Cr. | L. | |------|-----|----| | Jan. 15. | 16c. | 50 | | Mar. 12. | 36 | 960 | | June 23. | 13 | 6 | | July 19. | 26 | 13 |

Interest at 4 per cent. L.2 16 7

Here the sums on either side of the account are introduced according to the order of their dates. Those on the Dr. side are added to the former balance, and those on the Cr. side subtracted. Before we calculate the days, we try if the last sum 91l. 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 1286. As the computation on the odd threepences 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 threepences exceed 10, and neglect them when below it.

2d.] 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.

Dr. Mr J. T., his account current with N. W. Cr.

| Date | Dr. | L. | |------|-----|----| | Dec. 31. | 150 | 71 | | Mar. 12. | 120 | 120 | | April 9. | 70 | 70 | | May 12. | 300 | 300 | | June 3. | 240 | 240 | | June 17. | 165 | 165 | | Sept. 24. | 242 | 242 | | Oct. 9. | 178 | 178 |

| Date | Cr. | L. | |------|-----|----| | Dec. 31. | 150 | 71 | | Mar. 12. | 120 | 120 | | April 9. | 70 | 70 | | May 12. | 300 | 300 | | June 3. | 240 | 240 | | June 17. | 165 | 165 | | Sept. 24. | 242 | 242 | | Oct. 9. | 178 | 178 |

Interest due to N. W. at 5 per cent. L.6 8 9 Deduce 3/5 part 1 5 9

Due to N. W. at 4 per cent. L.5 3 0 Due to J. T. at 5 per cent. 3 7 11 5

Balance due to N. W. L.1 15 0 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 200l. due him. On the 12th of May, J. T. pays him 300l. which discharges what he owed, and leaves a balance of 100l. due him. The balance continues in J. T.'s favour till the 24th of September, when N. W. pays 242l. These changes are distinguished by the marks Dr. and Cr. 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 540l. was due the 18th Aug. 1772; and there was paid 19th March 1773, 50l.; and 19th December 1773, 25l.; and 23d September 1774, 25l.; and 18th August 1775, 110l. Required the interest and balance due on the 11th November 1775?

| Date | Amount | |-----------------------|--------| | 18th August 1772 | L 540 | | Interest to 19th March 1773, 218 days | L 16 2 6 | | Paid 19th March 1773 | L 566 2 6 | | Balance due 19th March 1773 | L 506 2 6 | | Interest to 19th December 1773, 272 days | L 19 1 2 | | Paid 19th December 1773 | L 525 3 8 | | Balance due 19th December 1773 | L 25 0 0 | | Interest to 23d September 1774, 278 days | L 9 19 0 9 | | Paid 23d September 1774 | L 519 4 5 | | Balance due 23d September 1774 | L 494 4 5 | | Interest to 18th August 1775, 329 days | L 5 3 22 5 3 | | Paid 18th August 1775 | L 516 9 8 | | Balance due 18th August 1775 | L 406 9 8 | | Interest to 11th November 1775, 85 days | L 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-parts, or $5\frac{2}{3}$. A fraction whose numerator is greater than its denominator is called an improper fraction. For example, seventeen third-parts, or $7\frac{1}{3}$. 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. $5\frac{2}{3} = \frac{17}{3}$

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 an assigned denominator: The numerator is obtained by multiplying the integer into the denominator. Hence the reason of the foregoing rule is evident: 5, 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. $\frac{17}{3} = 5\frac{2}{3}$

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. $\frac{135}{360} = \frac{3}{8}$

Here we observe that 135 and 360 are both measured by 5, and the quotient form $\frac{27}{72}$, which is a fraction of the same value as $\frac{135}{360}$ 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. 629, col. 2. But, if it be required to reduce a fraction to the lowest possible terms, we must divide... 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. 1st.] \( \frac{1}{3} + \frac{2}{5} = \frac{7}{15} \) by Problem IV.

2nd.] \( \frac{1}{3} + \frac{2}{5} + \frac{1}{6} = \frac{19}{30} \)

By Problem II. \( \frac{19}{30} \)

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 \( \frac{3}{7} \) from \( \frac{1}{2} \) remainder \( \frac{1}{14} \).

\( \frac{1}{2} - \frac{3}{7} = \frac{1}{14} \) by Prob. IV.

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 \( \frac{3}{7} \) from 12. remainder \( \frac{11}{7} \).

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. Here, because \( \frac{27}{3} \), the numerator of the fraction in the minuend, is less than \( \frac{35}{5} \), 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. \( \frac{1}{3} \times \frac{5}{7} = \frac{5}{21} \)

\( 2d. \) \( \frac{18}{5} \times \frac{7}{4} = \frac{126}{20} \)

\( 2 \times 5 = 10 \) num.

\( 3 \times 7 = 21 \) den.

\( 42 \times 31 = 1302 \)

\( 5 \times 4 = 20 \)

To multiply \( \frac{5}{7} \) by \( \frac{5}{7} \), is the same as to find what two third parts of \( \frac{5}{7} \) 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 \( \frac{5}{21} \); and, because two-thirds were required, we must double that fraction, which is done by multiplying the numerator by 2, and comes to \( \frac{10}{21} \). Hence we infer, that fractions of fractions, or compound fractions, such as \( \frac{5}{7} \) of \( \frac{5}{7} \), 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. 2d.; 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. \( 375 \times \frac{3}{5} = 225 \)

\( 5 \times 126 = 630 \)

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

3d, 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. \( \frac{15}{2} \times \frac{5}{6} = \frac{3}{2} \times \frac{5}{6} \)

\( \frac{3}{2} \times \frac{5}{6} = \frac{15}{12} = \frac{5}{4} \)

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. \( \frac{8}{3} \times \frac{7}{4} = \frac{56}{12} \)

\( \frac{8}{3} \times \frac{1}{2} = \frac{4}{3} \)

\( \frac{8}{3} \times \frac{7}{4} = \frac{14}{3} \)

\( \frac{8}{3} \times \frac{1}{4} = \frac{2}{3} \)

product \( 6 \frac{5}{12} \) 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 \( \frac{3}{5} \) by \( \frac{7}{9} \).

Quotient \( \frac{27}{35} \)

\( 2 \times 9 = 18 \)

\( 5 \times 7 = 35 \)

To explain the reason of this operation, let us suppose it required to divide \( \frac{5}{7} \) by 7, 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 \( \frac{5}{49} \). Again, because \( \frac{5}{7} \) is nine times less than seven, the quotient of any number divided by \( \frac{5}{7} \) will be nine times greater than the quotient of the same number divided by 7. Therefore we multiply \( \frac{5}{7} \) by 9, and obtain \( \frac{45}{7} \).

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

Ex. \( \frac{12}{7} \) divided by \( \frac{4}{7} \) 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 In the first example, we multiply the numerator by 20, the number of shillings in a pound, and divide the product 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 8l. 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 an 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 \( \frac{3}{8} \) of a pound is equal to \( \frac{19}{8} \) of a shilling, or \( \frac{340}{8} \) of a penny, \( \frac{660}{8} \) 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 signification, 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 signification; 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 .047 thousandth parts. - **.0047** Four tenth parts, and seven thousandth parts, or .0047 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.**

---

**Ex. Divide 36 by \( \frac{4}{7} \)**

\( 36 \div \frac{4}{7} = 63 \)

---

**Ex. Divide \( \frac{3}{5} \) by 5**

\( \frac{3}{5} \times 5 = \frac{3}{1} \)

---

**Ex. To divide 576 by 7**

\( 576 \div 7 = 82 \text{ remainder } 2 \)

Here we divide 576 by 7, the quotient is 82, and the remainder 2, to which we annex the fraction \( \frac{2}{7} \); and reduce \( \frac{2}{7} \) to an improper fraction \( \frac{16}{7} \), and multiply its denominator by 7, which gives \( \frac{16}{7} \).

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 wrought 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. \( \frac{7}{12} \). 3. Reduce 1s. 7d. to a fraction of a pound. Ans. \( \frac{17}{12} \).

**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. [Required the value of \( \frac{1}{8} \) of 1l.]

\( 17 \div 8 = 2 \text{ remainder } 1 \)

\( 20 \div 9 = 2 \text{ remainder } 2 \)

\( 300 \div 27 = 11 \text{ remainder } 3 \)

\( 40 \div 5 = 8 \)

\( 12 \div 28 = 0 \text{ remainder } 12 \)

\( 60 \div 480 = 0 \text{ remainder } 60 \)

\( 480 \div 9 = 53 \text{ remainder } 3 \)

\( 0 \div 50 = 0 \text{ remainder } 0 \)

\( 50 \div 45 = 1 \text{ remainder } 5 \) 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 2d, 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. 3d 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. 3d 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?

\[ \frac{28}{210} = 0.75 \] \[ \frac{4}{275} = 0.6875 \] \[ \frac{20}{56875} = 0.284375 \]

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?

\[ \frac{425}{20} = 21.25 \] \[ \text{fh. } 8,500 \] \[ \text{d. } 3,000 \]

Sect. II. Arithmetic of Terminate Decimals.

The value of decimal places decreases 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."

\[ \begin{align*} 32.35 & \quad \text{from} \quad 13.348 \quad \text{and} \quad 12.248 \\ 116.374 & \quad \text{take} \quad 9.2993 \quad 10.6752 \\ 160.63 & \\ 12.3645 & \quad 4.0487 \quad 1.5728 \\ \hline 331.4035 \end{align*} \]

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 2d.] 43.75 3d.] .1572

\[ \begin{align*} 1.37 & \quad .48 \\ 1096 & \quad 35000 \\ 137 & \quad 17500 \\ 2.466 & \quad 21.0000 \\ \end{align*} \]

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 dividend as in the divisor and quotient together."

Therefore, if there be the same 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 \( \frac{1}{5} \), we obtain the same answer as when we divide by 2, and every integer has a correspondent decimal, which may be called its reciprocal. Multiplication by that decimal supplies the place of division by the integer, and division supplies the place of multiplication.

To find the reciprocal of any number, divide 1 with cyphers annexed by that number.

Ex. Required the reciprocal of 625.

\[ \begin{align*} 625)1.000(.0016 \\ 625 \\ \hline 3750 \\ 3750 \\ \end{align*} \]

The product of any number multiplied by .0016 is the same as the quotient divided by 625.

Example.

\[ \begin{align*} 625)9375(15 \\ 625 \\ \hline 3125 \\ 3125 \\ \end{align*} \]

Because .0016 is \( \frac{1}{625} \) of unity, any number multiplied by that fraction will be diminished 625 times. For a like reason, the quotient of any number divided by .0016, will be equal to the product of the same multiplied by 625.

Example.

\[ \begin{align*} .0016)516.0000(322500 \\ 48 \\ \hline 36 \\ 32 \\ \end{align*} \]

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 of 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 one, 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- Decimal 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 \( \frac{1}{2} \), 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 deducting 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 little deduction of \( \frac{1}{2} \) 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 = 3.01964 | 2 - 11 = 2.09821 |

14.324 = 14.06427

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 qrs. 22lb. is .946428571, and therefore .943 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. 1st | Meth. 2nd | Meth. 3rd | |----------|----------|----------| | 8278 | 8278 | 8278 | | 2153 | 2153 | 3512 | | 24834 | 16556 | 16556 | | 41390 | 8278 | 827 | | 8278 | 41390 | 413 | | 16556 | 24834 | 24 | | 17822534 | 17822534 | 1782 |

Here the last four 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 them, 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. 629, col. i, fifth variety) at the highest place of the multiplier. We may perceive that all the figures on the right hand of the line on Meth. 2 serve no purpose, and may be left out, if we only multiply the figures on the multiplicand, whose products are placed on the right hand of the line. This is readily done by inverting the multiplier in Meth. 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.**

-374—79864327+(2144 or 3724)79864327(2144

7448

5384

3724

16602

14896

17063

14892

2171

4

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 dividends 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.

CHAP. X. INTERMINATE DECIMALS.

Sect. I. Reduction of Interminate Decimals.

As the arithmetic of interminate decimals, otherwise called the arithmetic of infiniter, 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 \( \frac{1}{7} \) is .142857

which multiplied by

produces the decimal equal to \( \frac{3}{7} = .428571 \)

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,

\( \frac{1}{7} = .142857 \), \( \frac{3}{7} = .428571 \), \( \frac{1}{2} = .5 \), \( \frac{1}{3} = .333333 \), \( \frac{1}{4} = .25 \), \( \frac{1}{5} = .2 \), \( \frac{1}{6} = .166666 \), \( \frac{1}{7} = .142857 \), \( \frac{1}{8} = .125 \), \( \frac{1}{9} = .111111 \)

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:

\[ \begin{array}{ccc} 2 & \times & 5 \\ 2^2 & \times & 5^2 \\ 2^3 & \times & 5^3 \\ 2^4 & \times & 5^4 \\ 2^5 & \times & 5^5 \\ \end{array} \]

And the reason is easily pointed out: for \( 2^3 \times 5^3 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 \); or, because the factors may be taken in any order, \( 2 \times 5 \times 2 \times 5 \times 2 \times 5 \); 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.

Ex. \( \frac{1}{7} = .03125 \), a decimal of 5 places, and \( \frac{3}{7} = .42857 \).

Again, No denominators except 2, 5, or their powers, produce terminate decimals. It is obvious from p. 631, 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. 630, 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 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 \( \frac{3}{9} \), or 1.

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 \( \frac{1}{10} \), the next \( \frac{1}{100} \), the next \( \frac{1}{1000} \). The sum of the two first places is \( \frac{1}{100} \) of the three places \( \frac{1}{1000} \); 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, informing the decimal, we divide the numerator successively by the component parts of the denominator, as directed p. 630, col. 1, par. ult. The first divisor is 2, 5, or so of their powers, and consequently gives a finite quotient by p. 649, 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. 649, col. 2, par. 2.

Ex. \( \frac{144}{14} = 16 \times 9 \)

\[ \begin{array}{c|c} 144) & 1000(00694 \\ 864 & 96.00694 \\ \hline 1360 & 40 \\ 1296 & 32 \\ \hline *640 & 80 \\ 576 & 80 \\ \hline 640 & 0 \\ \end{array} \]

In order to illustrate this subject further, we shall explain the operation of casting out the threes, which resembles that for casting out the nines, formerly laid down, p. 633, col. 2, par. 4.—p. 634, 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 allotted 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 composing 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 cast 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 casting out the 3's may be derived from those obtained by casting 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 cast out the 3's from any number, add the figures, neglecting 3, 6, or 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 casting out the 3's we compute thus; 2 and 8 is 10, which is three times 3, and 1 over; 1 and (passing 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 cast out from 2 or 4, the result is 1; from 2 or 8, the result is 2; from 2 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 casting 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 cast out, or) 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 cast 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. 649, 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 casting out the 9's from the corresponding power, by 5; and if it be multiplied by 5, or any of its powers, by casting 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. 649, 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 \( \frac{1}{7} \) 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 remainder is one less than the divisor: But frequently the circle is much shorter. Thus \( \frac{1}{7} = .09 \), a circle of 2 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. 631, 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, is equal to one whose numerator is the circulating 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 \( \frac{1}{9} = .142857 \), \( \frac{2}{9} = .285714 \), \( \frac{3}{9} = .375 \), \( \frac{4}{9} = .428571 \), \( \frac{5}{9} = .542857 \), \( \frac{6}{9} = .628571 \), \( \frac{7}{9} = .714285 \), \( \frac{8}{9} = .857142 \), \( \frac{9}{9} = .928571 \).

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.

\[ \begin{array}{c} 999)375000(.375, \\ 2997. \\ \hline 7530 \\ 6993 \\ \hline 5370 \\ 4995 \\ \hline *375 \end{array} \]

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. 649, 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.

\( \frac{1}{81} = .012345679 \), 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 individual 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 cast out, the result is 2), and, for a like reason, the third individual clears off without a remainder. If the first remainder be 2, the second is (twice 2 or 4, with the 3's cast out, or) 1, and the third 2; 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 casting 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.

\[ \begin{array}{c} 27 = 9 \times 3, \text{and } 9)1.0(.111x, \text{and } 3)111x(.037, \\ 81 = 27 \times 3, \text{and } 3)237.037.037(.012345679. \end{array} \]

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. 650, 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.

\[ \begin{array}{c} Ex. \frac{1}{18} = 216 = 27 \times 8, \\ 216)1,000(.004,629, \\ 864 \\ \hline 1360 \\ 1296 \\ \hline *170 \\ 162 \\ \hline 640 \\ 432 \\ \hline 80 \\ 54 \\ \hline 260 \\ 1944 \\ \hline 243 \\ *1360 \\ \hline 17 \end{array} \]

All mixed circulates are derived from vulgar frac- tions 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 1 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 $\frac{1}{7}$ multiplied by 3, would be a whole number, and $5 \times 3$ 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. 642, col. 2. par. ult. &c. But it was proven, p. 649, 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. 649, 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. 649, col. 2. par. 2.) The products of 2, 5, and their powers, by 3 or 9, produce mixed repeaters by p. 649, col. 2. par. ult.; their products by other multipliers, produce mixed circulates by p. 649, 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. 651, 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. 650, col. 2. par. 5.

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

\[ \begin{align*} \frac{1}{2} &= .5 \\ \frac{1}{3} &= .333 \\ \frac{1}{4} &= .25 \\ \frac{1}{5} &= .2 \\ \frac{1}{6} &= .1666 \\ \frac{1}{7} &= .142857 \\ \frac{1}{8} &= .125 \\ \frac{1}{9} &= .111 \\ \frac{1}{10} &= .1 \\ \frac{1}{11} &= .0909 \\ \frac{1}{12} &= .08333 \\ \frac{1}{13} &= .076923 \\ \frac{1}{14} &= .0714285 \\ \frac{1}{15} &= .0666 \\ \frac{1}{16} &= .0625 \\ \frac{1}{17} &= .0588235294117047 \\ \frac{1}{18} &= .05555 \\ \frac{1}{19} &= .052631578947368421 \\ \frac{1}{20} &= .05 \\ \frac{1}{21} &= .047619 \\ \frac{1}{22} &= .0454545 \\ \frac{1}{23} &= .0434782608695652173913 \\ \frac{1}{24} &= .041666 \\ \frac{1}{25} &= .04 \\ \frac{1}{26} &= .0384615 \\ \frac{1}{27} &= .037 \\ \frac{1}{28} &= .03571428 \\ \frac{1}{29} &= .0344827586206896551724137931 \\ \frac{1}{30} &= .0333 \end{align*} \]

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, $235,62 = \frac{23562}{99999}$

From the whole decimal $23562$

We subtract the finite part $235$

and the remainder $23527$ 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.

$235,62$ may be divided into $235,111$ by Rule I. and $.000,62 = \frac{62}{99999}$ 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 $\frac{33335}{99999}$; then we add the numerators:

| Numerator | Sum of numerators | |-----------|-------------------| | 2115 | 23500 | | 2115 | 235 |

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 $.272$, which is reduced by the last of the foregoing rules to $\frac{27}{99}$, or $\frac{3}{11}$, 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, $2727 = \frac{2727}{999}$, which is reduced by dividing the terms by 101 to $\frac{27}{99}$. 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, .34725725725725, 1.4562, 14.562456245624,

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 .23 .296 .8 88888 .3.8 .42 .643 643 .409 .7548 .73 73333 .36 .31

264246

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

.0356 .646 .7358 .7382 .469 .84728 .53428 .62363 .68 .38

.0877 .11172

The reason of these rules will be obvious, if we recollect that repeating figures signify nine 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. 652, 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. [ii.] Extended. Ex. 2d.] Extended. .574, .574-574, .874, .874.874.874, .2698, .269.869, .1463 .146.333333, .428, .428, .158, .158.858.858, .37983, .379,839, .32, .323.232323,

1.652,284, 1.503,026390,

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.

.3686, .0842, .368 .003094, .4375, .0842, .57, .765, .853492, .0842, .895, .76, .62, .0842, .742, .765

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

.572, .974, or .974974, .8135, or .8135135, .486, .86, .868686, .452907 or .4529074, .085, .106288, .3606060, or .360

Sect. III. Multiplication of 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 .376 9446x 809668 4048333

.0495246x

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, X .235 1. .674, X .78 235, 2. .37, X .86 3. .025, X .42 187,32, 4. .4793, X .48 1123,93, 5. .375, X .124 7492,92, 6. .2963, X .36

.08804,19,

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

Ex. [iit.] .958+.8 2d.] .784+.36 8 7664 4704 7664 2352 7664 7664 28224 7664 28224 .8513 .284,09, 3d.] The multiplication of interminate decimals may be often facilitated, by reducing the multiplier to a vulgar fraction, and proceeding as directed p. 643, col. 1, par. 6.

Thus,

\[ \begin{align*} \text{4th.] } & \quad .3824 \times \frac{7}{9} = \frac{3}{5} \\ & \quad 7 \\ & \quad 9) 2.6768 \\ & \quad 9742 \\ & \quad 768 \\ \text{5th.] } & \quad .384 \times .23 = \frac{11}{23} \\ & \quad 23 \\ & \quad 90) 8.832 \\ & \quad 9813 \\ \end{align*} \]

Therefore, in order to multiply by \( \frac{3}{5} \), we take one-third part of the multiplier; and, to multiply by \( \frac{2}{3} \), we take two-thirds of the same. Thus,

\[ \begin{align*} \text{6th.] } & \quad .784 \times \frac{3}{7} = \frac{2}{7} \\ & \quad 3) 784 \\ & \quad 2613 \\ \text{7th.] } & \quad .8761 \times \frac{6}{7} = \frac{2}{7} \\ & \quad 3) 17522 \\ & \quad 38403 \\ \end{align*} \]

As the denominator of the vulgar fractions always consists of 8's, or of 9's with cyphers annexed, we may use the contraction explained p. 631, col. 1, par. ult. &c.; and this will lead us exactly to the same operation which was explained p. 653, col. 2, par. ult. &c., on the principles of decimal arithmetic.

\[ \begin{align*} \text{8th.] } & \quad .735 \times 3.26 = \frac{311}{300} \\ & \quad 323 \\ & \quad 2205 \\ & \quad 1470 \\ & \quad 2205 \\ & \quad 999(237425 \\ & \quad 2374.05 \\ & \quad 2374 \\ & \quad ,23 \\ & \quad 239803 \\ \text{9th.] } & \quad .278 \times 365 = \frac{999}{999} \\ & \quad 365 \\ & \quad 1390 \\ & \quad 1668 \\ & \quad 834 \\ & \quad 999(101470 \\ & \quad 101 \\ & \quad 101,571 \\ & \quad ,23 \\ & \quad 239803 \\ \end{align*} \]

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, \( .384 \times .23 \) or by \( .2 = .0768 \) or thus, \( .384 \)

and by \( .05 = .02133 \)

\[ \begin{align*} & \quad .09813 \\ & \quad 9) 1920 \\ & \quad 2133 \\ & \quad 768 \\ & \quad 9813 \\ \end{align*} \]

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. 653, col. 2, par. 2.

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 2d. As the learner may not apprehend it so readily in Ex. 3d. 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 38,961037, and there is a carriage of 1 from the right-hand column, which completes 38,961038. 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 line 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. 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, 369 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 are 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 .10, 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 dividends, 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 intermediate decimals may be proven, by altering the order of the factors, (p. 628, col. 2, par. 2.) or by reducing them both to vulgar fractions in their lowest terms, multiplying these as directed p. 643, col. 2, par. 3, and reducing the product to a decimal.

Sect IV. Division of Intermediate Decimals.

Case I. "When the dividend only is intermediate, 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."

Ex.

Chap. X.

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 $\frac{9}{5}$.

$\begin{array}{c} \phantom{5}3\phantom{0}6\phantom{0}0\phantom{5}(68121) \\ \end{array}$

2d.] Divide .37845 by $\frac{6}{5}$.

$\begin{array}{c} \phantom{2}1\phantom{3}5\phantom{3}6(56768). \end{array}$

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. 628, col. 1, par. 1. It may be wrought in the same way, when the divisor repeats, and the denominator, of 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 divisors as explained at large, p. 649, col. 1, par. 1.—p. 651, 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. 655, col. 2, par. u/t., &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, No 66. 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. 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 dividior or resolvend."

III. "Double the quot for the first part of the dividior; 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 dividior complete."

IV. "Multiply the dividior 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 1st. If the first part of the dividior, 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 dividior; to the resolvend bring down another period; and proceed to divide as before.

Note 2. If the product of the quotient figure into the dividior 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.

Vol. II. Part II.

2 div. 725)3625 resolvend.

3625 product.

2d.] Required the square root of 72, to eight decimal places.

72.0000000

64

164(800

656

1688)14400

13504

16965)89600

84825

169702)477500

339404

169702)138096

135763

2333

1697

636

599

127

118

(9)

3d.] Required the square root of .2916.

.2916(54 root.

25

104 416

416

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 \( \frac{4}{9} \) is \( \frac{2}{3} \); and the root of \( \frac{4}{9} \) is \( \frac{2}{3} \); and thus the root of \( \frac{2}{3} \) (=\( \frac{2}{3} \)) is \( \frac{2}{3} \).

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. 3d. 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 shows 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 dividior or resolvend."

The dividior consists of three parts, which may be found as follows:

40 III. 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 0 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 1380 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 \( \frac{8}{27} \) is \( \frac{2}{3} \), and the cube root of \( \frac{4}{27} \) is \( \frac{2}{3} \); and thus the cube root of \( \frac{13}{8} \) (\( = \frac{15}{8} \)) is \( \frac{2}{3} \).

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.