ARITHMETICK

A R I T H M E T I C K.

ARITHMETICK is a science which explains the properties of numbers, and shews the method or art of computing them.

VOL. I. No. 16.

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We have very little intelligence about the origin and invention of arithmetick; but probably it must have taken its rise from the introduction of commerce, and conse-

4 Z

quently

quently be of Tyrian invention. From Asia it passed into Egypt, where it was greatly cultivated. From thence it was transmitted to the Greeks, who conveyed it to the Romans with additional improvements. But, from some treatises of the ancients remaining on this subject, it appears that their arithmetick was much inferior to that of the moderns.

NUMBER, which is the object of arithmetick, is that which answers directly to the question, How many? and is either an unit, or some part or parts of an unit, or a multitude of units.

To a person having the idea of number in his mind, the following questions naturally occur, viz. 1. How is such a number to be expressed or written? Hence we have Notation. 2. What is the sum of two or more numbers? Hence Addition. 3. What is the difference of two given numbers? Hence Subtraction. 4. What will be the result or product of a given number repeated or taken a certain number of times? Hence Multiplication. 5. How often is one given number contained in another? Hence Division.

These five, viz. Notation, Addition, Subtraction, Multiplication, and Division, are the chief parts, or rather the whole of arithmetic; as every arithmetical operation requires the use of some of them, and nothing but a proper mixture of them is necessary in any operation whatever; and, by an Arabic term, these are called the algorithm.

CHAP. I. NOTATION.

NOTATION is that part of arithmetic which explains the method of writing down, by characters or symbols, any number expressed in words; as also the way of reading or expressing, in words, any number given in characters or symbols. But the first of these is properly notation, and the last is more usually called numeration.

The things then proper to be comprised in this chapter are, 1. The figural notation. 2. Numeration, or the way of reading numbers. 3. Descriptions of the kinds or species of numbers.

I. Figural Notation.

AN unit, or unity, is that number by which any thing is called one of its kind. It is the first number; and if to it be added another unit, we shall have another number called two; and if to this last another unit be added, we shall have another number called three; and thus, by the continual addition of an unit, there will arise an infinite increase of numbers. On the other hand, if from unity any part be subtracted, and again from that part another part be taken away, and this be done continually, we shall have an infinite decrease of numbers. But though number, with respect to increase and decrease, be infinite, and knows no limits; yet ten figures, variously combined or repeated, are found sufficient to express any number whatsoever. These, with

the method of notation by them, were originally invented by some of the eastern nations, probably the Indians; afterwards improved by the Arabians; and at last brought over to Europe, particularly into Britain, betwixt the tenth and twelfth century. From the ten fingers of the hands, on which it hath been usual to compute numbers, figures were called digits. Their form, order, and value, are as follows:

1 One, an unit, or unity, 2 two, 3 three, 4 four, 5 five, 6 six, 7 seven, 8 eight, 9 nine, 0 cipher, nought, null, or nothing. Of these, the first nine, in contradistinction to the cipher, are called significant figures.

The value of the figures now assigned is called their simple value, as being that which they have in themselves, or when they stand alone. But when two or more figures are joined as in a line, the figures then receive also a local value from the place in which they stand, reckoning the order of places from the right-hand towards the left, thus,

A figure standing in the first place has only its simple value; but a figure in the second place has ten times the value it would have in the first place; and a figure in the third place has ten times the value it would have in the second place; and universally a figure in any superior place has ten times the value it would have in the next inferior place.

Hence it is plain, that a figure in the first place simply signifies so many units as the figure expresses; but the same figure advanced to the second place will signify so many tens; in the third place, it will signify so many hundreds; in the fourth place, so many thousands; in the fifth place, so many ten thousands; in the sixth place, so many hundred thousands; and in the seventh place, so many millions, &c. Thus, 7 in the first place, will denote seven units; in the second place, seven tens, or seventy; in the third place, seven hundred; in the fourth place, seven thousand, &c.

Every three places, reckoning from the right-hand, make a half period; and the right-hand figures of these half-periods are termed units and thousands by turns; the middle figure is always tens, and the left-hand figure always hundreds.

Two half-periods, or six places, make a full period; and the periods, reckoning from the right-hand towards the left, are titled as follows, viz. the first is the period of units; the second, that of millions; the third is titled bimillions, or billions; the fourth, trimillions, or trillions; the fifth, quadrillions; the sixth, quintillions; the seventh sextillions; the eighth, septillions; the ninth, octillions; the tenth, nonillions, &c.

Half-periods are usually distinguished from one another by a comma, and full periods by a point or colon; as in the following

T A B L E.

The table may be expressed in a more concise form thus,

From the table it is obvious, that though a cipher signifies nothing of itself, yet it serves to supply vacant places, and raises the value of significant figures on its left hand, by throwing them into higher places. Thus, in the first period, by a cipher's filling the place of units, the figure 4 is thrown into the place of tens, and signifies forty. But a cipher does not change the value of a significant figure on its right-hand. Thus, 07, or 007, is the same as 7.

II. Numeration.

NOTATION and numeration are so nearly allied, that he who understands the one cannot fail soon to acquire the other. The method of reading numbers, expressed by figures, may be easily learned from the table of the figural notation; in which observe the following

RULE. Beginning at the left hand, and reading toward the right; to the simple value of every figure join the name of its place, and conclude each period by expressing its title, every where omitting the ciphers.

III. Descriptions of the kinds or species of numbers.

1. An integer, or whole number, is an unit; or any multitude of units; as 1, 7, 48, 100, 125.

2. A fraction, or broken number, is any part or parts of an unit; and is expressed by two numbers, which are separated from one another by a line drawn betwixt them; the under number being called the denominator, and the upper one the numerator, of the fraction; as $\backslash frac\{1\}\{2\}$, $\backslash frac\{1\}\{3\}$, $\backslash frac\{2\}\{3\}$.

3. A mixed number is an integer with a fraction joined to it; as $4\backslash frac\{1\}\{2\}$, $7\backslash frac\{1\}\{3\}$, $48\backslash frac\{2\}\{3\}$.

4. A number is said to measure another number, when it is contained in that other number a certain number of times, or when it divides that other number without any remainder. Thus, 3 measures 6, 9, or 12.

5. An even number is that which is measured by 2, or which 2 divides without any remainder; as 2, 4, 6, 8, 10, 12.

6. An odd number is that which 2 does not measure, or which cannot be divided by 2, without a remainder; as 1, 3, 5, 7, 9, 11, 13.

7. A prime number is that which unity, or itself, only measures; as 3, 5, 7, 11, 13, 17, 19.

8. A composite number is that which is measured by some other number than itself, or unity; as 12, which is measured by 2, 3, 4, or 6.

9. Numbers are called prime to one another, when unity only measures them. Thus 13 and 36 are prime to one another; for no number, except unity, measures both.

10. Numbers are called composite to one another, when some number, besides unity, measures them. Thus 12 and 18 are composite to one another; for 3 or 6 measures both of them.

11. A number which measures another is called an aliquot part of that other. Thus 6 is an aliquot part of 18, and 3 of 12, and 5 of 20.

12. The number measured, or which contains the aliquot part a certain number of times, is called a multiple of that aliquot part. Thus 18 is a multiple of 6, and 12 of 3.

13. A number is called an aliquant part of another, when it does not divide that other without a remainder. Thus 7 is an aliquant part of 24.

14. Two, three, or more numbers, which, multiplied together, produce another number, are called the component parts of the number produced. Thus 3 and 4, 2 and 6, are the component parts of 12; and 2, 3, and 4, are the component parts of 24.

15. The product of a number multiplied into itself is called the square, or second power, of that number; and the number itself is in this case called the root. And if the square be multiplied into the root, the product is called the cube, or third power, of that number. And if the cube be multiplied into the root, the product thence arising is called the biquadrate, or fourth power, &c.

CHAP. II. ADDITION.

ADDITION is the collecting of two or more numbers into one sum or total.

I. Addition of Integers:

RULE I. Set figures of like places under other, viz. units under units, tens under tens, &c.

II. Beginning at the lowest place, set down the right-hand figures of the sum of every column, and carry the rest as so many units to the next superior place.

EXAMP. I. Because similar or like things only can be added, place the numbers as directed in Rule I. viz. units under units, tens under tens, &c. as in the margin. Then beginning at the lowest place, viz. that of units; say, 4 units and 3 units make 7 units, which set below in the place of units; then 3 tens and 5 tens make 8 tens, which set below in the place of tens; then 2 hundreds and 4 hundreds make 6 hundreds, which set below in the place of

of hundreds, and you will find the sum or total to be 687.

EXAMP. II. Having placed the numbers, units under units, &c. as in the margin, say 2 and 1 make 3, and 3 make 6, and 4 make 10; which being just 1 ten, and nothing over, set the right-hand figure 0 in the place of units; and because ten in any lower place makes but one in the next superior place, carry 1 ten, as directed in Rule II. — saying, 1 ten, collected out of the units, and 6 24180 tens, make 7 tens, and 4 make 11, and 0 makes but still 11, and 7 make 18; here again set down the right-hand figure 8, in the place of tens, and carry the remaining figure 1, being 1 hundred, to the next place, viz. that of hundreds; and having in like manner added up this column, the amount is 31; set down the right-hand figure 1 in the place of hundreds, and carry the remaining figure 3 to the next place or column; which being also added, amounts to 24; set the right-hand figure 4 below, in its proper place, and the remaining figure 2, which belongs to the next place, set on the left hand, there being no figure in the next place to which it can be carried. So the sum or total is 24180.

II. Addition of the parts of integers, such as shillings, pence, farthings, ounces, &c.

RULE I. Place like parts under other; viz. farthings under farthings, pence under pence, &c.

II. Begin at the lowest of the parts, and carry according to the value of an unit of the next superior denomination; viz. for every four in the sum of farthings carry 1 to the pence, and for every twelve in the pence carry 1 to the shillings, &c.

III. If you carry at 20, 30, 40, 60, or any just number of tens, as in adding shillings, degrees, poles, minutes, seconds, &c. proceed with the column of units as in addition of integers, and from the sum of the column of tens carry 1 for every two, or 1 for every three, &c. according as 20 or two tens, thirty or three tens, &c. make an unit of the next superior denomination. The reason appears plain in the following operations.

I. M O N E Y.

T A B L E.

Marked thus.

Note, The above mark signifies equal to.

l. is put for libra, a pound; d. for denarius, a penny; and q. for quadrans, a fourth-part; but f is now the more usual mark for farthings.

That the learner may proceed in addition of money with the greater ease, it will be proper he get the following table by heart.

MONEY-TABLE.

EXAMP. Having, according to Rule I. placed like parts under other, viz. farthings under farthings, pence under pence, &c. and in each of these denominations, units under units, tens under tens, as in the margin, begin with the lowest of the parts, viz. the farthings; and say, 2 farthings and 1 farthing make 3 farthings, and 2 make 5, and 3 make 8; which, by the money-table, is 2 fours, or 2 pence, and nothing over; wherefore place 0 below in the place of farthings, or rather leave that place blank, and carry 2 pence to the place of pence, as directed in Rule II, saying, 2 pence, collected out of the farthings, and 9 make 11, and 8 make 19, and 1 (passing the 0) make 20; to this sum of units add the tens. Thus, 20 and 1 ten make 30, and 1 ten more make 40 pence; which, by the money-table, is 3 twelves, or 3 shillings, and 4 pence over; these 4 pence set below in the place of pence, and carry 3 shillings to the place of shillings. Thus, 3 shillings, collected out of the pence, and 1 shilling make 4, and 7 make 11, and 9 make 20, and 8 make 28; and because in shillings we carry at a just number of tens, viz. at 20, set the right-hand figure 8 below in the place of units, as directed in Rule III. and carry the 2 tens to the place of tens. Thus 2 tens collected out of the units, and 1 ten make 3 tens, and 1 make 4, and 1 make 5 tens, or 2 twenties, and 1 ten over; and because 2 tens, or 1 twenty, make an unit in the next place, viz. that of pounds, set the 1 ten below in the place of tens, and carry the 2 twenty shillings, or 2 pounds, to the place of pounds; which, being integers, are added as taught in addition of integers.

It is usual to subjoin the farthings to the pence by way of fraction, as in the margin, where the former example is transcribed in this form for the learner's instruction; in which $\backslash frac\{1\}\{4\}$ denotes one farthing, $\backslash frac\{1\}\{2\}$ two farthings, and $\backslash frac\{3\}\{4\}$ three farthings.

In adding up large accounts, some dot at 60 in the pence, and for every dot carry 5 to the shillings; and in adding the shillings they dot likewise at 60, and for every dot carry 3 to the pounds. Others chuse to divide them

them into parcels, then cast up each parcel separately, and afterwards add the sums of the several parcels into one total.

Marked thus.

By Avoirdupois weight are weighed butter, cheese, rosin, wax, pitch, tar, tallow, soap, salt, hemp, flax, beef, brass, iron, steel, tin, copper, lead, allum, and all grocery wares.

Note, 19½ C. of lead make a fodder.

In adding the following example, begin with the ounces, and say, 15 and 10 make 25; which being above 16, dot, and carry away the excess 9, saying, 9 of excess and 6 make 15, and 8 make 23; where again dot, and carry away the excess 7, saying, 7 and 2 is 9, and 1 ten on the left is 19; where dot, and proceed with the excess 3, saying, 3 and 4 is 7, and 1 ten on the left is 17; where dot, and carry the excess 1, saying, 1 and 5 is 6, and 1 ten on the left is 16; where again dot, and there being no excess, you have nothing to let down.

Proceed now to add the pounds; saying 5 carried from the ounces, viz. one for every dot, and 3 make 8, and 6 make 14, and 1 ten on the left is 24, and 8 make 32; which being above 28, dot, and go on, saying, 4 of excess and 1 ten on the left is 14, and 9 is 23, and 1 ten on the left is 33; where again dot, and go on, saying, 5 of excess and 20 is 25, and 4 is 29; where dot, and proceed, saying, 1 of excess and 2 tens on the left make 21, and 7 make 28; where dot, and the 2 tens, or 20, on the left, set below.

We should now proceed to add the quarters; saying, 4 carried from the pounds and 1 make 5, &c.; but as you carry here 1 for every four, the quarters are added exactly as the farthings in addition of money. In the hundreds you carry at 20; which, therefore, are added as shillings. The tuns are integers; and added accordingly.

ADDITION may be proved several ways.

1. Merchants and men of business usually add each column first upwards, and then downwards, and, upon finding the sum to be the same both ways, they conclude the work to be right: and this is all the proof that their time, or the hurry of business, will admit of.

2. It is a common practice in schools, to prove the work by a second summing without the top-line; and if thus sum added to the top-line makes the first total, the work is supposed to be right; as in the following example.

Note, This mark + signifies added to.

3. Addition is also proved by casting out the 9's; for if the excess above the 9's in the total be the same as the excess in the items, the work may be presumed right. Thus, to prove the example in the margin, begin with the items, and say, 3+4=7, and 7+7=14=1+4=5; with this 5 pass to the next item, and say, 5+6=11=1+1=2, and 2+8=10=1, and 1+4=5; which 5 being the excess of the items, place at the top of the cross, and proceed to cast the 9's out of the total, saying, 1+3=4, and 4+1=5; which 5, being the excess of the total, place at the foot of the cross; and because it is the same with the figure at the top, you conclude the work to be right.

If the items are of different denominations; as pounds, shillings, pence, &c.; you must begin with the highest denomination; and, after casting out the 9's, reduce the excess to the next inferior denomination; and then casting out the 9's, reduce the excess to the next inferior denomination; proceed in like manner with this, and all the other lower denominations, placing the last excess at the top of the cross; then, in the same manner, cast the 9's out of the total, placing the excess at the foot of the cross; and if the figure at the foot and top be the same, the work may be presumed right.

If any operation, whether in addition, subtraction, multiplication, or division, be right, this kind of proof will always show it to be so; but if an operation be wrong, by a figure or figures being misplaced, or by miscounting 9, or any just number of 9's, this kind of proof will not discover the mistake.

SUBTRACTION is the taking a lesser number from a greater, in order to discover their difference, or the remainder.

RULE I. Set figures of like place under other, viz. units under units, tens under tens, &c. and the greater of the given numbers uppermost.

II. Beginning at the place of units, take the lower figures from those above, borrowing and paying ten, as need requires, and write the remainders below.

EXAMP. I. Because similar or like things only can be subtracted, place the numbers as directed in Rule I. viz. units under units, tens under tens, &c. and the greatest uppermost, as in the margin.

Then, beginning at the place of units, say, 2 units from 7 units, and 5 units remain; which set below in the place of units; then 6 tens from 6 tens, and nothing remains; wherefore set 0 below, in the place of tens; then 5 hundred from 8 hundred, and 3 hundred remain; which set below, in the place of hundreds; and you will find the total difference or remainder to be 305.

II. Having placed the numbers, units under units, &c. as in the margin, say, 5 units from 2 units, you cannot, but, because an unit in the next superior place makes ten in this place, you must borrow 1, viz. 1 ten, from the said next place, as directed in Rule II.; which 1 ten being added to 2 makes 12; then say, 5 from 12, and 7 remains; which 7 set below in the place of units; then proceed, and pay the unit borrowed, either by esteeming 3, the next figure in the major, to be only 2, or, which is more usual, and the same in effect, by adding 1 to the next figure in the minor, thus, 1 that you borrowed and 8 make 9, from 3 you cannot, but, borrowing as before, you say, 9 from 13 and 4 remains; which 4 set below: proceed, and say, 1 that you borrowed and 7 make 8, from 4 you cannot, but from 14, and 6 remains; which 6 set below: go on, and say, 1 borrowed and 2 make 3, from 7, and 4 remains; which 4 set below. So the difference or remainder is 4647.

RULE I. Place like parts under other, viz. farthings under farthings, pence under pence, &c. and the greater of the given numbers uppermost.

II. Begin at the lowest of the parts, and borrow according to the value of an unit of the next superior denomination; viz. in farthings borrow 4, in pence borrow 12, &c. as the tables of money and weights direct.

III. If you borrow 20, 30, 40, 60, or any just number of tens, as in subtracting shillings, degrees, poles, minutes, seconds, &c. proceed with the right-hand column, as in subtraction of integers; and then subtract

your tens, borrowing, if need be, the number of tens contained in an unit of the next superior denomination. The reason appears plain in the following operations.

Having, according to (10) (20) (12) (4)
Rule I. placed like parts L. s. d. f.
under other, viz. farthings under farthings, 73 15 10 2 major.
pence under pence, 48 12 6 2 minor.
and in each of these denominations, units under 25 3 4 remainder.
units, tens under tens, and the greater of the given numbers uppermost, as in the margin, begin with the farthings, and say, 2 from 2, and 0 remains; and proceed to the pence, saying, 6 from 10 and 4 remains; which 4 set down, and go on to the shillings, saying 2 from 5 and 3 remains, and 1 from 1, and 0 remains; or you may say at once, 12 from 15, and 3 remains; which 3 being set down, proceed to the pounds, which are integers, and subtracted as such.

In this example say, 3 farthings from 1 farthing you cannot, but as directed in Rule II. you say, 3 from 4, the number of farthings in 1 penny borrowed, and 1 remains; which 1 added to 1 in the major gives 2 farthings for a remainder; which set down, and proceed to the pence, saying, 1 penny borrowed and 10 make 11, which from 6 you cannot, but from 12, the number of pence in 1 shilling, and 1 remains; which 1 added to 6 in the major gives a remainder of 7; which set down, and go on to the shillings; and because in subtracting shillings we borrow a just number of tens, viz. 2 tens, or 20, work as directed in Rule III.; and in the right-hand column say, 1 borrowed and 7 make 8, which from 4 you cannot, but from 14, and 6 remains; which being set down, go on to the left-hand column, and say, 1 borrowed and 1 make 2, which from 1 you cannot, but from 2, the number of tens in 1 pound, and nothing remains, which 0 added to 1 in the major gives 1 for a remainder; which set down, and proceed to the pounds, saying, 1 borrowed and 8 make 9, which from 8 you cannot, but from 18, &c.

Note. Some add the number borrowed to the figure or number in the major, and then subtract from their sum. Thus, in the farthings they add the 4 borrowed to 1 in the major, and then from the sum 5 they subtract the 3 in the minor; and in the pence they add the 12 borrowed to 6 in the major, and subtract from the sum 18, &c.; but the method taught above is the easiest and most usual.

BEGIN with the pounds, and (10) (4) (28)
say, 24 from 22 you cannot, but C. Q. lb.
from 28, the number of pounds in 84 1 22 major.
1 quarter, and 4 remains, which 49 3 24 minor.
added to 22 in the major, gives 34 1 26 rem.
26 for

26 for a remainder; which set below, and proceed to the quarters, saying, 1 quarter borrowed and 3 make 4, which from 1 you cannot, but from 4, the number of quarters in 1 C. and 0 remains, which 0 added to 1 in the major gives 1 for a remainder; which set down, and go on to the C. which are integers, saying, 1 C. borrowed and 9 make 10, which from 4 you cannot, but from 14, &c.

III. The Proof of Subtraction.

MERCHANTS and men of business use no other proof besides a revival of the work, or running over it a second time; but it is usual in schools to put the learner upon proving the operation, by some of the three methods following, viz.

1. The work may be proved by addition; for if you add the remainder to the minor, the sum will be equal to the major, as in the following example.

2. By subtraction; for if you subtract the remainder from the major, the difference will be equal to the minor, as follows.

3. By casting out the 9's; for the major being equal to the sum of the minor and remainder, if you cast the 9's out of the major, and place the excess at the top of the cross, and then cast the 9's out of the minor and remainder, as if they were items in addition, and place the excess at the foot of the cross, it is plain the figure at the top and foot, if the work be right, will be the same. Only, in proving subtraction of money, Avoidupois weight, &c. care must be taken to begin with the highest denomination, reducing always the excess to the next inferior denomination, as taught in the proof of addition. See the following example.

CHAP. IV. MULTIPLICATION.

In multiplication there are two numbers given, viz. one to be multiplied, called the multiplicand; and another that multiplies it, called the multiplier; these two go

under the common name of factors; and the number arising from the multiplication of the one by the other is called the product, and sometimes the fact, or the rectangle. If a multiplier consists of two or more figures, the numbers arising from the multiplication of these several figures into the multiplicand, are called particular, or partial products; and their sum is called the total product.

Multiplication then is the taking or repeating of the multiplicand, as often as the multiplier contains unity. Or,

Multiplication, from a multiplicand and a multiplier given, finds a third number, called the product, which contains the multiplicand as often as the multiplier contains unity.

Hence multiplication supplies the place of many additions; for if the multiplicand be repeated or set down as often as there are units in the multiplier, the sum of these, taken by addition, will be equal to the product by multiplication. Thus, $5\; \backslash times\; 3\; =\; 15\; =\; 5\; +\; 5\; +\; 5$.

The first and lowest step in multiplication is, to multiply one digit by another; and the fact or number thence arising is called a single product. This elementary step may be learned from the following table, commonly called Pythagoras's table of multiplication: which is consulted thus; seek one of the digits or numbers on the head, and the other on the left side, and in the angle of meeting you have their product. The learner, before he proceed further, ought to get the table by heart.

To Pythagoras's table are here added, on account of their usefulness, the products of the numbers 10, 11, 12.

T A B L E.

I. Multiplication of Integers.

RULE I. Set the multiplier below the multiplicand, so as like places may stand under other, viz. units under units.

units, tens under tens, &c. : but if either or both of the factors have ciphers on the right hand, set their first significant figures under other.

The order prescribed in this rule is not absolutely necessary, but very convenient as will appear in the examples.

II. Beginning at the right hand, multiply each figure of the multiplier into the whole multiplicand, carrying as in addition, and placing the right-hand figure of each particular product directly under the multiplying figure.

III. Add the particular products, and their sum will be the total product.

EXAMP. I. Having placed the multiplier under the multiplicand, as directed in Rule I. proceed to the operation, and say, 7 times 4 make 28; set the 8 below in the place of units, and carry the 2 tens to the next place, as directed in Rule II. saying 7 times

9 make 63, and 2 that I carried make 65; set 5 below in the place of tens, and the 6, which belongs to the next place, set on its left hand, there being no further place to which it can be carried; so the product is 658.

II. Here first multiply the right-hand figure 8 into the whole multiplicand, as in the former example; then proceed, and multiply likewise the 6 tens into the whole multiplicand, saying 6 times 2 make 12; set the 2 below under the multiplying figure, viz. in the place of tens, and carry the 1 to the next place, as directed in Rule II. The reason why the 2 is set under the multiplying figure, or in the place of tens, is, because the multiplying figure 6 is really 6 tens or 60, and 60 times 2 make 120; so that by carrying the 1 to the next place, and setting down 20, the 0 would fall into the place of units, and throw the 2 into the place of tens; but as 0 can make no alteration in the addition of the partial products, the setting of it down is safely and justly omitted.

III. When the multiplier has ciphers on the right hand, as it would be evidently lost labour to multiply by the ciphers, their only use being to throw figures on their left hand into higher places, set the first significant figures of the factors under other; and, after the operation is finished, annex the ciphers of the multiplier to the right hand of the product.

IV. When the multiplier has ciphers intermixed with significant figures, omit the ciphers, because the multiplying by them would only produce so many lines of ciphers and so be labour in vain; wherefore multiply by the significant figures only; but take care to place the right-hand figure of each particular product directly under the multiplying figure.

Contractions, and simple ways of working multiplication of integers.

1. To multiply any number by 10, by 100, by 1000, &c. to the given number annex one, two, three ciphers, &c. Thus, $23\; \backslash times\; 10\; =\; 230$; and $384\; \backslash times\; 100\; =\; 38400$; and $745\; \backslash times\; 1000\; =\; 745000$.

2. To multiply any number by 9, by 99, by 999, &c. multiply the given number first by 10, by 100, by 1000, &c. that is, annex one, two, three, &c. ciphers to it; from this subtract the given number, and the remainder is the product; as in the following examples.

From Ex. 3. we may learn, in general, that to multiply any number consisting entirely of 9's by itself, is to set 1 in the place of units, then as many ciphers, save one, as there are 9's in the given number; then 8, and on the left hand of 8 as many 9's as there are ciphers on its right.

3. To multiply any number by 5; first multiply it by 10, that is, annex a cipher to it, and then halve it: and to multiply any number by 15, use the same method; and add both numbers together, as in the following examples.

4. To multiply any number by 11, 12, 13, 14, 15, 16, &c. multiply by the unit's figure, and add the back-figure of the multiplicand to the product; and to multiply by 21, 22, 23, 24, 25, 26, 27, &c. add the double of the back-figure; and to multiply by 31, 32, 33, 34, &c. add the triple of it; and to multiply by 112, 113, 114, &c. add the two back-figures; and to multiply by 101, 102, 103, 104, &c. add the next back-figure save one: as in the following examples.

In multiplying by 12, as in Ex. 8. it is more usual, and equally easy, to proceed by saying, twelve times 8 make 96, and, setting down the 6, say, twelve times 4 is 48, and 9 carried is 57; which set down, and the product is 576.

5. If the multiplier consist of the same figure repeated, as 111, 222, 333, 777, &c. multiply by the unit's figure, and out of that product make up the total product, thus. Begin at the right hand, and first take one figure, then the sum of two, then the sum of three, &c. repeating the operation still from the right hand, as often as there are figures in the multiplier; then neglecting the right-hand figure, or figure in the first place, take the sum of as many figures toward the left hand as the multiplier has places; and if there be not so many, take the sum of all the figures there are; then, neglecting the figures in the first and second place, begin at the figure in the third place, proceed as before; and thus go on till the last or left-hand figure is taken in alone; as in the following examples.

6. The operation may frequently be rendered shorter or easier, either by addition, subtraction, or a more simple multiplication; and the cases of this kind are so numerous and various, that they admit of no limitation. Consult the following examples and directions.

Work the above examples as follows.
Ex. 1. Multiply by 7, and add that product to the multiplicand, instead of multiplying by 8.

Ex. 2. Multiply by 6, and out of that product subtract the multiplicand, instead of multiplying by 5.

Ex. 3. Multiply by 4, and double that product for 8.

II. Multiplication of the parts of Integers.

Here there are three cases.
1. If your multiplier is a single digit, set it under the units figure of the lowest denomination, multiply it into all the parts of the multiplicand, beginning at the lowest, and carrying always as in addition, or according to the value of the next superior place.

EXAMP. What is the price of 7 packs of cloth at L. 64, 8 s. 10½ d. per pack?

L. s. d. Here say, 7 times 2 is 14, which is 64 8 10½ 3 pence and 2 farthings over; set down the 2 farthings, and carry 3 to the place of pence, saying, 7 times 10 is 70, and 451 2 1½ 3 that I carried makes 73, which is 6 shillings and 1 penny; set down the

1 penny, and carry 6 to the place of shillings, saying, 7 times 8 is 56, and 6 that I carried is 62, which makes 3 pounds and 2 shillings; set down the 2 shillings, and carry 3 to the place of pounds which are integers.

2. If your multiplier consists of two or more figures, multiply continually by its component parts, or by the component parts of the composite number that comes nearest to it, and then multiply the given multiplicand by the difference of the multiplier, and the nearest composite number: the sum or difference of these two products is the answer.

EXAMP. I. What is the price of 56 C. tobacco, at L. 2 : 14 : 9½ per C.

Here the component parts are 8 and 7; for $8\; \backslash times\; 7\; =\; 56$: therefore,

Multiply first by 8, and that product by 7; or, which will give the same answer, multiply first by 7, and then that product by 8.

EXAMP. II. What is the price of 126 yards of velvet, at L. 3 : 8 : 4 per yard?

Here multiply first by 6, that product by 7, and that product again by 3: but as the component parts are various, and may be chosen at pleasure, you would have had the same answer, had you multiplied by $9\; \backslash times\; 7\; \backslash times\; 2$; or by $7\; \backslash times\; 3\; \backslash times\; 3\; \backslash times\; 2$.

From the above example may be deduced a general and easy rule for working all questions of this kind; and is of excellent use when the multiplier happens to be a number; viz.

Multiply continually so many times by 10 as there are figures in the multiplier, save one; then multiply the given price by the right-hand figure of the multiplier; and again, the first product of 10 by the following figure of the multiplier; and so on, till you have multiplied by all the figures in the multiplier. The sum of these products is the answer.

EXAMP. III. What is the price of 8604 yards of cloth, at 19 s. 6½ d. per yard?

Price of 8604 yards, 8406 16 6

5 B 3. If

3. If your multiplier consists of integers and parts, the operation is performed by a cross multiplication of the several parts of the multiplier into all the parts of the multiplicand.

The contents of mason and joiners work are frequently cast up by this kind of multiplication; for understanding of which observe, that

The superficial content of any rectangle is found by multiplying the length into the breadth; and the content of a right-angled triangle is found by multiplying the base into half the perpendicular or height.

The dimensions are usually taken in lineal feet, inches, and lines; and the operation is performed by the following rules.

I. Any lineal measure multiplied into the same lineal measure produces squares of that name. Thus, lineal feet multiplied into lineal feet produce square feet; lineal inches into lineal inches produce square inches, &c.

II. Lineal feet into lineal inches produce rectangles 1 foot long and 1 inch broad, which divided by 12 quote square feet; and the remainder multiplied by 12, produces square inches.

III. Lineal feet into lineal lines produce rectangles 1 foot long and 1 line broad, which divided by 144 quote square feet; and the remainders are rectangles equal to square inches.

IV. Lineal inches into lineal lines produce small rectangles 1 inch long and 1 line broad, which divided by 12 quote square inches; and the remainder, multiplied by 12, produces square lines.

EXAMP. I. In an area, pavement, or piece of plaster-work, in length 24 feet 7 inches, and in breadth 18 feet 5 inches, how many square feet?

Here multiply 18 lineal feet into 24 lineal feet, and the product is 432 square feet; then multiply 5 lineal inches into 7 lineal inches, and the product is 35 square inches, by Rule I.; then multiply 18 lineal feet into 7 lineal inches, and the product is 126; and again multiply 24 lineal feet into 5 lineal inches, and the product is 120; which added to the former product, gives 246 rectangles, each being 1 foot in length and one inch in breadth; these divided by 12 quote 20 square feet; and the remainder 6 multiplied by 12, produces 72 square inches, according to Rule II.; these add to the former square feet and inches, and you'll find the answer or total product to be 452 square feet, and 107 square inches.

EXAMP. II. In an area or floor, in length 38 feet 9 inches 6 lines, and in breadth 23 feet 8 inches 6 lines, how many square feet?

Because the sum of the inches exceeds 144, carry 1 from them to the column of feet, and set down the overplus, viz. 98.

The operation may be rendered easier and shorter by previously reducing the factors to two denominations, viz. inches and lines. Thus the former example may be proposed and wrought as follows.

In an area or floor, in length 465 inches 6 lines, and in breadth 284 inches 6 lines, how many square inches and feet?

The answer here is 132434 square inches, and 108 square lines; and if the inches be divided by 144, you will have 919 square feet and a remainder of 98 square inches, as before.

Or the factors may be reduced to the lowest denomination, viz. lines, and then the product will be square lines, which, divided by 144, will quote square inches, and the remainder will be square lines; and the square inches, divided by 144, will quote square feet, and the remainder will be square inches. Again, the square feet, divided by 9, will quote square yards, and the remainder will be square feet; and the square yards, divided by 36, will quote square roods, and the remainder will be square yards.

If this cross multiplication be extended to the mensuration of solids, the content of which is found by multiplying the superficial content of the base into the height, depth, length, or thickness, the operation must be conducted by the following rules.

V. Any superficial measure multiplied into the same lineal measure produces a solid of the same name. Thus superficial feet multiplied into lineal feet produce solid feet; superficial inches multiplied into lineal inches produce solid inches, &c.

VI. Superficial feet into lineal inches produce parallelopipeds, whose base is 1 square foot, and their height 1 inch; which divided by 12 quote solid feet; and the remainder, multiplied by 144, produces solid inches.

VII. Superficial feet into lineal lines produce parallelopipeds, whose base is 1 square foot, and their height 1 line; which divided by 144 quote solid feet; and the remainder multiplied by 12 produces solid inches.

VIII. Superficial inches into lineal lines produce parallelopipeds, whose base is 1 square inch, and their height 1 line; which divided by 12 quote solid inches; and the remainder multiplied by 12 produces solid inches.

IX. Lineal feet into superficial inches produce parallelopipeds, whose base is one square inch, and their height 1 foot; which divided by 144 quote solid feet; and the remainder multiplied by 12 produces solid lines.

X. Lineal feet into superficial lines produce parallelopipeds, whose base is 1 square line, and their height 1 foot; which divided by 12 quote solid inches; and the remainder multiplied by 144 produces solid lines.

XI. Lineal inches into superficial lines produce parallelopipeds, whose base is 1 square line, and their height 1 inch; which divided by 144 quote solid inches; and the remainder multiplied by 12 produces solid lines.

EXAMP. III. In a piece of timber, whose length is 18 feet 16 inches, breadth 2 feet 4 inches, and thickness 2 feet 3 inches, how many solid feet?

EXAMP. IV. How many solid feet in a polished stone, that is 8 feet 9 inches 5 lines long, 7 feet 3 inches broad, and 3 feet 5 lines thick?

The operation may be facilitated by previously reducing the three factors to two denominations, viz. inches and lines, as was done in Example II. on superficial measure.

Or the three factors may be reduced to the lowest denomination, viz. lines, which being multiplied continually, will produce solid lines, which divided by 1728, will quote solid inches, the remainder being solid lines; and the solid inches divided by 1728 will quote solid feet, the remainder being solid inches; and the solid feet divided by 27 will quote solid yards, the remainder being solid feet; and the solid yards divided by 216 will quote solid roods, the remainder being solid yards.

We shall only further observe, that as the rules for working questions by cross multiplication are numerous, and the operation tedious, it is easier to convert the parts into a decimal fraction of their integer, and then work as taught in the multiplication of decimals.

III. The Proof of Multiplication.

MULTIPLICATION may be proved several ways; viz. by multiplication, by division, and by casting out the 9's.

1. By multiplication: Change the places of the factors, and make that the multiplier which before was the multiplicand; and if the work be right, you will have the same product as before; but this method is tedious.

2. By division: When the work is right, the product divided by the multiplier quotes the multiplicand; or, divided by the multiplicand, quotes the multiplier. But this supposes the learner acquainted with division.

3. The most usual method therefore of proving multiplication is by casting out the 9's; which is done thus: Cast the 9's out of the multiplicand and multiplier, and place the excesses on the right and left sides of a cross; multiply these two figures into one another, casting the 9's out of their product, if need be, and place the excess at the top of the cross; then casting the 9's also out of the product of your multiplication, place its excess at the bottom.

Here first multiply 18 feet 6 inches into 2 feet 4 inches, as formerly, and the product is 43 feet 24 inches superficial; which next multiply into 2 feet 3 inches lineal, thus, 43 superficial feet into 2 lineal feet produce 86 solid feet, and 24 superficial inches into 3 lineal inches produce 72 solid inches, by Rule V.; then 43 superficial feet into 3 lineal inches produce 129 parallelopipeds, whose base is 1 square foot, and their height 1 inch; which divided by 12 quotes 10 solid feet; and the remainder 9 multiplied into 144 produces 1296 solid inches, by Rule VI. Again, 2 lineal feet into 24 superficial inches produce 48; which, being less than 144, you esteem a remainder, and multiplying it into 12 you have a product of 576 solid inches, by Rule IX.

Because the sum of the inches exceeds 1728, carry 1 from thence to the feet, and the overplus 216 set down.

bottom; and if the work be right, the figures at top and bottom will agree, or be the same.

EXAMP. I. Here cast the 9's out of the multiplicand, and place the excess 7 on the right side of the cross; then cast the 9's out of the multiplier, and place the excess 2 on the left side of the cross; next multiply these excesses 2 and 7 into one another, cast the 9's out of their product, and place the excess five at the top of the cross; lastly, cast the 9's out of the product, and place the excess 5 at the foot of the cross; which being the same with the figure at the top, you may conclude the work to be right.

that of the pence to farthings. The multiplier being an abstract number, needs no reduction; but if a multiplier be a mixt number, or consist of integers and parts, as feet and inches, &c. the excess of the higher denomination must always be reduced to the lower.

CHAP. V. DIVISION.

DIVISION discovers how often one number is contained in another: or,

Division, from two numbers given, finds a third, which contains unity as often as the one given number contains the other.

The number to be divided, or which contains the other, is called the dividend; the number by which we divide, or which is contained in the dividend, is called the divisor; and the number found by division, or which expresses how often the dividend contains the divisor, is called the quotient or quot.

As multiplication supplies the place of many additions, so division, which is the reverse of multiplication, serves instead of many subtractions; as will thus appear: Suppose it were required to divide 18 by 6, that is, to find how often 6 is contained in 18, the work by subtraction will stand as in the margin: by which it appears, that 6 is contained 3 times in the number 18. But this, by division, may be found at one trial: thus,

Set the divisor on the left of the dividend, leaving room on the right hand for the quotient, as in the margin; and then say, How often 6 in 18? Ans. 3 times: this 3 set in the quotient; then multiply the quotient figure 3 into the divisor 6, saying, 3 times 6 make 18; which set down below the dividend, and subtract it from the dividend, and 0 remains.

I. Division of Integers.

RULE I. From the left-hand part of the dividend point off the first dividual, viz. so many figures as will contain the divisor.

II. Ask how often the divisor is contained in the dividual, and put the answer in the quotient.

III. Multiply the divisor by the figure set in the quotient, and subtract the product from the dividual.

IV. To the right of the remainder bring down the next figure of the dividend for a new dividual; and then proceed as before.

EXAMP. I. Here, because the divisor 7 is contained in 8, the left-hand figure of the dividend, point it off as the first dividual, according to Rule I; and then say, How often 7 in 8? Ans. 1 time; which 1 set in the quotient, as directed in Rule II.; then multiply the divisor 7 by this quotient figure 1, and subtract the product 7 from the dividual 8, as directed in Rule III.; to the remainder 1 bring down the following figure of the dividend, for the second dividual, as directed in Rule IV.; then proceed as before, and say, How often 7 in 17? Ans. 2 times; wherefore, setting two in the quotient, multiply and subtract and find the next remainder to be 3; to which bring down the following figure of the dividend, and you have 35 for the third dividual; then say, How often 7 in 35? Ans. 5 times; which 5 being placed in the quotient, multiply and subtract, and 0 remains; so the quotient is 125.

By reviewing the steps of the preceding operation, and reducing the dividuals and quotient-figures to their separate values, the reason of the rules will be obvious; for,

is 35, in which the divisor 7 is contained 5 times. Now it is evident, that the sum of the partial quots, 125, is the total quot, or a number expressing how often the dividend 875 contains the divisor 7.

From the above example we may learn, that there are always just so many figures in the quotient as there are dividuals; or the first dividual, with the number of subsequent figures in the dividend, is equal to the number of places or figures in the quotient.

Hence likewise may be inferred, that no divisor is contained in any dividual oftener than 9 times; for the dividual, excluding the right-hand figure, is always less than the divisor by 1 at least; and if both be multiplied by 10, or have a cipher annexed to each of them, the product of the dividual will be less than the product of the divisor by 10 at least; but no right-hand figure can supply this defect of 10; therefore the divisor is not contained 10 times in any dividual, and consequently not oftener than 9 times.

Here too observe, that the right-hand figure of the first dividual, and all the subsequent figures of the dividend, have a point or dot set below them, as they are brought down; which is done to prevent mistakes, by distinguishing them, in this manner, from the figures not yet brought down.

8)56032897(7004112 $\backslash frac\{1\}\{8\}$ numer.
.....
56

.....
032

32

.....
8

8

.....
9

8

.....
17

16

.....
(1)

and, as multiplying and subtracting is in this case needless, you bring down the next figure of the dividend 3; and as you cannot have 8 in 3, put another 0 in the quotient, and bring down the next figure of the dividend 2: Then say, How often 8 in 32? Ans. 4; which put in the quotient: Then multiply and subtract; and as nothing remains, bring down the next figure of the dividend 8, and say, How often 8 in 8? Ans. 1; which put in the quotient: then multiply and subtract; and as nothing remains, bring down the next figure of the dividend 9, and say, How often 8 in 9? Ans. 1; which put in the quotient: then multiply and subtract; and to the remainder 1 bring down the next and last figure of the dividend 7, and say, How often 8 in 17? Ans. 2; which put in the quotient: then multiply and subtract, and 1 remains.

To complete the quotient, draw a line on the right hand, and for the remainder above the line, and the di-

visor 8 below it, signifying that 1 remains to be divided by 8; or this part of the quotient may be considered as a fraction, whose numerator is 1, and its denominator 8; and the quotient thus completed shews, that the dividend contains the divisor 7004112 times, and one eighth part of a time.

Here observe, that not only the last remainder, but every other remainder, must be less than the divisor; for if it be either greater or equal, the divisor might have been oftener got, and the quotient-figure is too little. And should any one in this case attempt to continue the operation, the quotient-figures would be all 9's, the dividuals would prove inexhaustible, and the remainders would constantly increase.

Hence also learn, that if any dividual happen to be less than the divisor, you must put 0 in the quotient, and bring down the next figure of the dividend; and if it be still less than the divisor, you must put another 0 in the quotient, and bring down the following figure of the dividend, &c.

III. Here the divisor consists of two figures; and because it

36)789426(21928 $\backslash frac\{1\}\{6\}$
.....

72

.....
69

36

.....
334

324

.....
102

72

.....
306

288

.....
(18)

is contained in the two left-hand figures of the dividend 78, point them off as the first dividual; and say, How often 3 in 7? Ans. 2, and 1 remains; which 1 placed, or conceived as placed, on the left hand of the following figure 8, makes 18: then say, Can I have the following figure of the divisor 6 also 2 times in 18? Ans. Yes; consequently I get 36 the divisor 2 times in 78 the dividual; wherefore put 2 in the quotient, and multiply that 2 into the divisor 36, and subtract the product 72 from the dividual 78; and to the remainder 6 bring down the following figure of the dividend 9, for a new dividual: then say, How often 3 in 6? Ans. 2, and 0 remains; again you say, Can I have 6 also 2 times in 9? Ans. No; therefore you can have 36 in 69 only 1 time, which 1 you put in the quotient: then multiply and subtract as before; and to the remainder 33 bring down the next figure 4 for a new dividual: Then, because the dividual consists of a figure more than the divisor, say, How often the first figure of the divisor 3 in the first two figures of the dividual 33? Ans. 9, and 6 remains; which 6 placed on the left hand of the following figure 4 makes 64: Again, say, Can I have 6 also 9 times in 64? Ans. Yes; consequently 36 can be had 9 times in 334; wherefore you put 9 in the quotient: Then multiply and subtract; and to the remainder 10 bring down the next figure 2 for a new dividual: Here likewise, because the dividual has a figure more than the divisor, say, How often 3 in 10? Ans. 3, and 1 remains; which 1 placed on the left hand of the following figure 2 makes 12: Again say, Can I have 6 also 3 times in 12? Ans. No; consequently 36 cannot be had 3 times in 102; wherefore try if

you can have it 2 times; saying, 2 times 3 is 6 from 10, and 4 remains; which 4 placed on the left hand of the next figure 2 makes 42. And again say, Can I have 6 also 2 times in 42? Ans. Yes; consequently 36 can be had 2 times in 102; accordingly put 2 in the quotient, multiply and subtract; and to the remainder 30 bring down the next and last figure of the dividend 6, for a new dividend: Then, because the dividend has a figure more than the divisor, say, How often 3 in 30? Ans. 9, and 3 remains; which 3 placed on the left hand of the following figure 6 make 36: And again say, Can I have 6 also 9 times in 36? Ans. No; consequently 36 cannot be had 9 times in 306; therefore try if it can be had 8 times, saying, 8 times 3 is 24 from 30, and 6 remains; which 6 placed on the left hand of the following figure 6 makes 66: Again say, Can I have 6 also 8 times in 66? Ans. Yes; consequently 36 can be had 8 times in 306; therefore put 8 in the quotient, and multiply and subtract as before: The last remainder 18 is the numerator of a fraction, and the divisor its denominator, to be annexed to the integral part of the quotient; as was taught in the former example.

The preceding operation points out the manner of procedure when the divisor consists of more figures than one, viz. you must take the first figure of the divisor out of the first figure of the dividend, or out of the first two figures of the dividend in case the dividend have a figure more than the divisor: Then imagine the remainder to be prefixed to the next figure of the dividend, and try if you can have the second figure of the divisor as often out of this number; if you can, imagine again the remainder to be prefixed to the following figure of the dividend, and try if you can have the third figure of the divisor as often out of this number, &c.; but if you find you cannot have some subsequent figure of the divisor so often as you took the first, you must go back, and take the first figure of the divisor 1 time less, or some number of times less out of the first, or out of the first two figures of the dividend: Then proceed as before, repeating the trial till you find you have the second and all the subsequent figures of the divisor as often as you took the first.

But here observe, that if, in trying how often the divisor can be had in the dividend, either 9, or a number greater than 9, any where remain, you may conclude, without further trial, that all the subsequent figures of the divisor can be had as often as you took the first; as may be thus demonstrated.

Suppose the subsequent figures of the divisor to be the highest possible, that is, all 9's, and the following figures of the dividend the lowest possible, that is, all 0's; again, imagine the remainder 9 prefixed to the following figure of the dividend 0, that it will make 90; now it is plain, that the subsequent figure of the divisor 9 can be had in 90, the highest number of times possible, viz. 9 times, and 9 will remain; which prefixed to the next figure of the dividend 0, makes 90, in which the subsequent figure of the divisor 9 can again be had 9 times, and 9 will remain as before; therefore all the subsequent figures of the divisor can be had as often as you took the first; and if they can be had in this case, much more can they be had when a number greater than 9 remains.

IV If, as in the margin, a cipher or ciphers, possess the right hand of the divisor, cut them off, and cut off as many figures, viz. in this example, the figure 2 from the right hand of the dividend; then divide the remaining figures of the dividend, viz. 89678, by the remaining figures of the divisor, viz. 648, and you have the integral part of the quotient; but to the remainder 254 annex the figure cut off from the dividend, and you have 2542 for the numerator of your fraction, and the whole divisor 6480 is the denominator.

The reason will appear obvious by working a question in this manner, and also at full length, without cutting off the cipher or ciphers, and then comparing the two operations.

V. If, as in the margin, the figures cut off from the right hand of the dividend, happen to be all ciphers; in this case, the last remainder, without regarding the ciphers cut off, is the numerator of your fraction, and the significant figures of the divisor the denominator. The reason is assigned in the doctrine of fractions.

In like manner, if there be cut off from the dividend any number of significant figures, with a cipher or ciphers on their right hand; in this case the last remainder, with the significant figures cut off, make the numerator of your fraction; and the significant figures of the divisor, with as many ciphers as the number of significant figures cut off from the dividend, make the denominator. Thus, if, in the above example, the figures cut off from the dividend had been 50, the numerator of your fraction would have been 365, and the denominator 480.

Contractions in working Division of Integers.

1. To divide any number by 10, 100, 1000, &c. you have only to point off for a remainder as many figures on the right hand of the dividend as the divisor has ciphers, and the other figures on the left of the point or separatrix are the quotient. Thus, 7489634 divided by 10, 100, 1000, &c. stands as follows.

2. If the figures of the divisor are all 9's, or all except the units figure, as 9, 99, 999, 98, 997, 9996, &c. work as follows:

Find a new divisor, by annexing to unity as many ciphers as there are figures in the given divisor, subtract the given from the new divisor, and the remainder or difference is the complement. Divide the given dividend by the new divisor, viz. point off so many figures on the

the right hand as there are ciphers in the said divisor; the figures thus pointed off are to be esteemed a remainder, and the other figures on the left hand are to be accounted a quotient; then multiply this quotient by the complement, placing the units of the product under the units of the former remainder; again, divide this product by the new divisor, by pointing off from the right hand the same number of figures as in the former remainder, and the figures to the left are to be esteemed another quotient; which quotient you are again to multiply by the complement, and divide as before. And in this manner proceed till the last quotient is nothing; then add as in addition of integers, observing the carriage from the left hand column of the remainders; to the remainders add the product of the said carriage and complement, and the sum is the total remainder; and the sum of the several quotients is the total quotient required.

First, to unity annex two ciphers, because the given divisor consists of two figures, and so the new divisor is 100; from which subtract the given divisor 98, and there remains 2 for the complement.

Next divide the given dividend by the new divisor, viz. point off 98, the two figures next the right hand, for the first remainder; and the figures on the left, namely, 746, is the quotient.

Then multiply the said first quotient 746 by the complement 2; and by the new divisor divide the product 1492, viz. point off 92 for the second remainder, and 14 is the second quotient.

Again, multiply the second quotient 14 by the complement 2, and the product 28, divided by 100, gives 28 for the third remainder, but nothing to the quotient.

Then add the several remainders and quotients, and find the total quotient amounts to 762, and the remainders to 18.

Lastly, multiply 2, the carriage from the left-hand column of the remainders, by the complement 2; and the product 4 add to the remainders 18, and the sum 22 is the total remainder.

Here there are three cases.

1. If the divisor be a digit, by it divide the integers of the dividend, reduce the remainder to the parts of the next inferior denomination, and add it, when thus reduced, to the parts; then divide the sum, reducing and adding the remainder to the parts of the following denomination, &c.

Note. If the integral part of the dividend be less than the divisor, you must, in the first place, reduce it to the parts of the next denomination.

EXAMP. I. If L. 274:13:8:3 be equally divided among 8 men, what will each man's share be?

Here first divide the integers L. 274 by 8, and the quotient is L. 34, and L. 2 8)274 13 8 3 dividend. remains; which reduced to the next denomination makes 40 shillings; and these added to 13 shillings make 53 shillings; which divided by 8 gives 6 shillings to the quotient, and 5 shillings remains; which 5 shillings reduced make 60d. and 60d. added to 8d. make 68d.; which divided by 8 gives 8d. to the quotient, and 4d. remains, &c.

The operation may, if you please, be drawn out at large; as in the following

EXAMP. II. If C. 42:2:8 of tobacco be made up into 5 equal hhd's, what will be the neat weight of each hhd?

Here divide the C. 43 by 5, and the quotient is C. 8, and C. 3 remains; which reduced, and added to the 2 Q. makes 14 Q. which divide by 5, &c.

2. If the divisor consists of two or more figures, and be a composite number, resolve it into its component parts, and divide the given dividend by one of these parts, the quotient by another, &c. and the last quotient is the answer.

3. If the divisor consists of integers and parts, reduce both divisor and dividend to the same denomination, and then proceed as in division of integers.

DIVISION may be proved several ways, viz. by multiplication, by division, and by casting out the 9's.

1. By multiplication: Multiply the quotient by the divisor, or the divisor by the quotient; and the product with the remainder added to it, will be equal to the dividend: Or, take the products of the quotient-figures into the divisor, add them in the order they stand under the individuals; and their sum, with the remainder, will be equal to the dividend.

2. By division: Divide the difference of the dividend and remainder by the quotient, and your next quotient will be equal to your first divisor, without any remainder. But this method is tedious.

3. By casting out the 9's: Cast the 9's out of the divisor.

visor and quotient, place the excesses on the right and left sides of a cross; then multiply these two figures into one another, and cast the 9's out of their product; add the excess to the remainder; and, casting out the 9's if need be, place the sum or excess at the top of the cross; then cast the 9's out of the dividend, and set the excess at the bottom. If the work be right, the figures at the top and bottom of the cross will agree, or be the same.

These methods of proof are a proper exercise to the learner in schools; but, in business, the only proof used is a careful revival of the operation.

CHAP. VI. REDUCTION.

REDUCTION teacheth how to bring a number of one name or denomination to another of the same value; and is either descending, ascending, or mixt.

I. Reduction descending brings a number of a higher denomination to a lower, when the lower is some aliquot part of the higher; as pounds to shillings, pence, or farthings; and is performed by multiplication.

II. Reduction ascending brings a number of a lower denomination to a higher, when the lower is some aliquot part of the higher; as shillings, pence, or farthings, to pounds; and is performed by division.

III. Mixt reduction brings a number of one denomination to another, when the one is no aliquot part of the other; as pounds to guineas, and requires the use of both multiplication and division.

In treating of reduction we shall conjoin the descending and ascending, the one serving as a proof of the other; and shall afterwards treat of mixt reduction by itself.

In working reduction, of whatever kind, the following rule is to be observed, viz.

Multiply or divide as the tables of money and weights direct.

Reduction descending and ascending.

I. MONEY.

QUEST. 1. In L. 472 how many shillings, pence, and farthings?

Proof by Reduction ascending.

In 453120 farthings how many pence, shillings, and pounds?

Note 1. To reduce pounds to pence at one operation, multiply by 240, the number of pence in 1 pound.

Note 2. To reduce pounds to farthings at one operation, multiply by 960, the number of farthings in 1 pound.

Note 3. To reduce shillings to farthings at one operation, multiply by 48, the number of farthings in 1 shilling.

Note 4. To reduce pence to pounds at one operation, divide by 240, the pence in 1 pound.

Note 5. To reduce farthings to pounds at one operation, divide by 960, the farthings in 1 pound.

Note 6. To reduce farthings to shillings at one operation, divide by 48, the farthings in 1 shilling.

Here follows the farthings of Quest. 1. reduced back to pounds by these notes.

By note 6.

In working mixt reduction observe the following RULE. By reduction descending bring the given name to some such third name as is an aliquot part both of the name given and of the name sought, and then by reduction ascending bring the third name to the name sought.

Mixt reduction, as well as reduction descending and ascending, extends to money, as follows.

Here the given name is 764 pounds.
pounds, the name sought is 20
guineas, and the third name,
to which the pounds are re- 21) 15280 (727 guineas.
duced, is shillings; for a 147...
shilling is an aliquot part
both of a pound and of a 58
guinea. 42
— — — — —
160
147
— — — — —
(13) shillings.

THE Rule of Three, called also, on account of its excellence, the Golden Rule, from certain numbers given finds another; and is divided into simple and compound, or into single and double.

THE simple rule of three, from three numbers given, finds a fourth, to which the third bears the same proportion as the first does to the second.

The nature and properties of proportional numbers may be understood sufficiently for our purpose from the following observations.

In comparing any two numbers, with respect to the proportion which the one bears to the other, the first number, or that which bears proportion, is called the antecedent; and the other, to which it bears proportion, is called the consequent; and the quantity of the proportion or ratio is estimated from the quot arising from dividing the antecedent by the consequent. Thus the ratio or proportion betwixt 6 and 3 is the quot arising from dividing the antecedent 6 by the consequent 3; namely, 2; and the ratio or proportion betwixt 1 and 2 is the quot arising from the division of the antecedent 1 by the consequent 2; namely $\backslash frac\{1\}\{2\}$, or one half.

Four numbers are said to be proportional when the ratio of the first to the second is the same as that of the third to the fourth; and the proportional numbers are usually distinguished from one another as in the following examples.

Proportional numbers, or numbers in proportion, are usually denominated terms; of which the first and last are called extremes, and the intermediate ones get the name of means, or middle terms.

If four numbers are proportional, they will also be inversely proportional; that is, the first consequent will be to its own antecedent as the second consequent is to its antecedent; or the fourth term will be to the third as the second is to the first. Thus, if $6\; :\; 3\; ::\; 10\; :\; 5$, then by inversion, $3\; :\; 6\; ::\; 5\; :\; 10$, or $5\; :\; 10\; ::\; 3\; :\; 6$. Euclid v. 4. cor. By either of these kinds of inversion may any question in the rule of three be proved.

If four numbers are proportional, they will also be alternately proportional; that is, the first antecedent will

be to the second antecedent as the first consequent is to the second consequent; or the first term will be to the third term as the second term is to the fourth. Thus, if $8\; :\; 4\; ::\; 24\; :\; 12$, then, by alternation, $8\; :\; 24\; ::\; 4\; :\; 12$. Euclid v. 16.

But the celebrated property of four proportional numbers is, that the product of the extremes is equal to the product of the means. Thus, if $2\; :\; 3\; ::\; 6\; :\; 9$, then $2\; \backslash times\; 9\; =\; 3\; \backslash times\; 6\; =\; 18$. Euclid vi. 16.

Hence we have an easy method of finding a fourth proportional to three numbers given, viz.

Multiply the middle number by the last, and divide the product by the first, the quot gives the fourth proportional.

EXAMP. Given 6, 5, and 36, to find a fourth proportional; put $x$ equal to the fourth proportional, then $6\; :\; 5\; ::\; 36\; :\; x$, and $5\; \backslash times\; 36\; =\; 180\; =\; 6\; \backslash times\; x$; wherefore, dividing the product 180 by the factor 6, the quot gives the other factor $x$, namely 30, the fourth proportional sought.

Every question in the rule of three may be divided into two parts, viz. a supposition and a demand; and of the three given numbers, two are always found in the supposition, and only one in the demand.

EXAMP. If 4 yards cost 12 shillings, what will 6 yards cost at that rate?

In this question the supposition is, If 4 yards cost 12 shillings; and the two terms contained in it are 4 yards and 12 shillings: The demand lies in these words, What will 6 yards cost? and the only term found in it is 6 yards.

The supposition and demand being thus distinguished, proceed to state the question, or to put the terms in due order for operation, as the following rules direct.

RULE I. Place that term of the supposition, which is of the same kind with the number sought, in the middle. The two remaining terms are extremes, and always of the same kind.

II. Consider, from the nature of the question, whether the answer must be greater or less than the middle term; and if the answer must be greater, the least extreme is the divisor; but if the answer must be less than the middle term, the greatest extreme is the divisor.

III. Place the divisor on the left hand, and the other extreme on the right; then multiply the second and third terms, and divide their product by the first; and the quot gives the answer; which is always of the same name with the middle term.

When the divisor happens to be the extreme found in the supposition, the proportion is called direct; but when the divisor happens to be the extreme in the demand, the proportion is inverse.

The three rules delivered above are indeed so framed, as to preclude the distinction of direct and inverse, or render it needless; the left-hand term being always the divisor; but yet the direct questions being plainer in their own nature, and more easily comprehended by a learner, we shall, in the first place, exemplify the rules by a set of questions of the direct kind, and shall afterwards adduce an example or two of such as are inverse.

I. The Simple Rule of Three Direct.

QUEST. 1. If 4 yards cost 12 shillings, what will 6 yards cost at that rate?

The supposition and demand of this question have already been distinguished, and the two terms in the former are 4 yards 12 shillings, and the only term in the latter is 6 yards.

The number sought is the price of six yards, and the term in the supposition of the same kind is the price of 4 yards, viz. 12 shillings, which place in the middle, as directed in Rule I. and the two remaining terms are extremes, and of the same kind, viz. both lengths.

Wherefore place the divisor 4 yards on the left hand, and the other extreme 6 yards on the right; and multiplying the second and third terms, divide their product by the first term, and the quot 18 is the answer, and of the same name with the middle term, viz. shillings, according to Rule III.

And because the divisor is the extreme found in the supposition, the proportion is direct.

QUEST. 2. If 7 C. of pepper cost 21 l. how much will 5 C. cost at that rate?

The supposition in this question is, that 7 C. of pepper costs 21 l. and the two terms in it are 7 C. and 21 l.; the demand is, How much will 5 C. cost? and the term in it is 5 C.

The number sought is the price of 5 C. and the term in the supposition of the same kind is the price of 7 C. viz. 21 l. which place in the middle. The two remaining terms are extremes, and of the same kind, viz. quantities of pepper.

Accordingly place the divisor 7 C. on the left hand, and the other extreme 5 C. on the right; and having multiplied the second and third terms, divide their product by

by the first term, and the quot 15 is the answer, of the same name with the middle term, viz. L. Sterling.

And because the divisor happens to be the extreme in the supposition, the proportion is direct.

QUEST. 3. If 13 yards of velvet cost L. 21, what will 27 yards cost at that rate?

* Rem. 4 s.

Such remainders are always of the same name with the preceding part of the quot. Thus, the first remainder 8, and the first part of the quot 43, are both pounds; and the second remainder 4, and the second part of the quot 12, are both shillings; and the third remainder 9, and the third part of the quot 3, are both pence; and the fourth remainder 10, and the fourth part of the quot 2, are both farthings.

As we have no money under farthings, the last remainder cannot be reduced any lower; so there remains 10 farthings to be divided by 13; that is, there is wanting to complete the quot, the thirteenth part of 10 farthings, or the thirteenth part of every remaining farthing; that is, ten thirteenth parts of one farthing; so you set the remainder 10 above, and the divisor 13 below a line drawn between them, in the form of a fraction, of which the remainder is the numerator, and the divisor the denominator.

II. The Simple Rule of Three Inverse.

QUEST. 1. If 8 men can do a piece of work in 12 days, in how many days will 16 men do the same?

In this question the supposition is, If 8 men do a piece of work in 12 days, and the two terms contained in it are 8 men and 12 days: The demand lies in these words, In how many days will 16 men do the same? and the only term contained in it is 16 men.

The number sought here is the days in which 16 men will do the work, and the term in the supposition of the same kind is 12 days; wherefore I place 12 days as the middle term, according to Rule I. the two remaining

terms are extremes, and of the same kind, viz. both of them men.

It is obvious that the answer must be less than the middle term; for 16 men will do the work in fewer days than 8 men; and therefore, by Rule II. the greatest extreme, viz. 16, is the divisor; which place on the left hand, and the other extreme on the right, as directed in Rule III. Then multiplying the second and third, and dividing their product by the first, the quot comes out in days; that is, of the same name with the middle term.

And because the extreme found in the demand happens to be the divisor, the proportion is inverse.

QUEST. 2. How much plush of 3 quarters wide will line a cloak that hath in it 4 yards of 7 quarters wide?

Here the answer must be greater than the middle term; for the plush being narrower than the cloth of which the cloak is made, will require more length.

QUEST. 3. If 36 yards be a rood of mason-work, at 3 feet high, how many yards will make a rood at 9 feet high?

SECT. II. The Compound Rule of Three.

THE Compound Rule of Three, from five given numbers finds a sixth, or from seven given numbers finds an eighth, or from eleven finds a twelfth, &c.

This rule easily and naturally admits of subdivisions, which, from the number of the terms given, may be denominated the rule of Five, the rule of Seven, the rule of Nine, the rule of Eleven, &c.

Questions in the Compound rule of three are also resolved into two parts, viz. a supposition and a demand.

If five terms be given, three of these are always found in the supposition, and two in the demand; if seven terms be given, four of these are in the supposition, and three in the demand; if nine terms are given, five of these are in the supposition, and four in the demand; if eleven terms be given, six of these are in the supposition, and five in the demand, &c.

The supposition and demand being distinguished, proceed to state the question; that is, to put the terms in due order for operation, as the following rules direct.

RULE I. Place that term of the supposition which is of the same kind with the number sought, in the middle.

The

The remaining terms are extremes, which must be classed into similar pairs, by making each pair consist of one term taken from the supposition, and another of the same kind taken from the demand.

II. Out of each similar pair, joined with the middle term, form a simple question; and in each simple question, so formed, find the divisor; viz. consider from the nature of the question, whether the answer must be greater or less than the middle term; and if the answer must be greater, the least extreme is the divisor; but if the answer must be less than the middle term, the greatest extreme is the divisor.

III. Place all the divisors on the left hand, and the other extremes on the right; then multiply the divisors, or extremes on the left, continually, for a divisor, and multiply the extremes on the right hand and the middle term, continually, for a dividend; and, lastly, divide the dividend by the divisor; and the quot is the answer, of the same name with the middle term.

The answer to questions in the compound rule of three may also be had by working the simple questions separately, or by themselves, in the following manner, viz.

The middle term, with any one pair of similar extremes, make the first simple question, and the answer to this question must be made the middle term to the next similar pair of extremes; and the answer to this second question, must in like manner be made the middle term to the following similar pair of extremes, &c.; and the answer to the last simple question is the number sought.

But the joint operation prescribed in Rule III. is the shorter as well as the easier method; for in working some of the simple questions, there may happen to be a remainder, and consequently the middle term of the next simple question will have some fractional part; which inconvenience is avoided by working jointly.

In every simple question, when the divisor is an extreme found in the supposition, the proportion is direct; but when the divisor is an extreme found in the demand, the proportion is inverse.

The three rules delivered above are indeed so calculated, as to make no difference between direct and inverse, or so as to render that distinction needless, the left-hand extremes being all divisors; but yet, as questions consisting entirely of direct proportions are the plainest and easiest, it will be proper, in the first place, to exemplify the rules by questions of the direct kind, and afterwards introduce such as are inverse.

And as questions in the rule of five are by far more numerous, and occur much oftener, than questions in the rule of seven, nine, or eleven; we shall, first of all, give questions in the rule of five, wherein both proportions are direct; then those wherein one or both proportions are inverse; and, lastly, give a few examples of the rules of seven, nine, and eleven.

I. The Rule of Five Direct.

QUEST. 1. If 14 horses eat 56 bushels of corn in

16 days, how many bushels will 20 horses eat in 24 days?

The supposition in this question is, If 14 horses eat 56 bushels in 16 days; and the three terms contained in it are, 14 horses, 56 bushels, and 16 days: The demand is, How many bushels will 20 horses eat in 24 days? and the two terms contained in it are 20 horses, and 24 days.

The number sought is bushels, and the term in the supposition of the same kind is 56 bushels; wherefore, according to Rule I. place 56 bushels in the middle. The remaining four terms are extremes, which you class into similar pairs, by making each pair consist of one term taken from the supposition, and another of the same kind taken from the demand. Thus, 14 horses, and 20 horses make one pair; again, 16 days, and 24 days make another pair.

Out of the several similar pairs, joined with the middle term, you form so many simple questions, according to Rule II. viz. by saying,

1. If 14 horses eat 56 bushels in a certain number of days, how many bushels will 20 horses eat in the same time?

2. If 16 days eat up, or consume, 56, or any other number of bushels, how many bushels will 24 days consume?

In the first simple question it is obvious, that the answer will be greater than the middle term; for 20 horses will eat more bushels than 14 horses will do in the same time; and so the least extreme, viz. 14, is the divisor; and because 14 is an extreme found in the supposition, the proportion is direct.

In the second simple question it is also plain, that the answer will be greater than the middle term; for 24 days will consume more bushels than 16 days; and consequently the least extreme, viz. 16, is the divisor; and because 16 is an extreme found in the supposition, the proportion is direct.

According to Rule III. place the divisors on the left hand, and the other extremes on the right, and both of them under one another, so that the two upper ones make a pair, or be of one kind, and the two lower ones make another pair, or be of one kind; and no matter which of the pairs be uppermost: then multiply the divisors, or the extremes on the left hand, for a divisor; and again multiply the extremes on the right, and the middle term, continually, for a dividend; and dividing the dividend by the divisor, the quot or answer comes out of the same name with the middle term, viz. 120 bushels.

Joint operation.

Ans. 120 bushels.

The two simple questions into which the compound question is resolved, are stated, and wrought separately, as follows.

Ans. 120 bushels, as before.

II. The Rule of Five Inverse.

THE questions that fall under this rule have commonly one of the proportions inverse, and the other direct, and sometimes the upper, and sometimes the lower, is the inverse proportion; and in some few questions both proportions are inverse. Now, though the three rules delivered above make no difference betwixt direct and inverse; yet, to bring the learner to some measure of acquaintance with this useful distinction, we shall, in stating the following questions, expose the same to view, by affixing an asterisk to the extremes of every inverse proportion.

Quest. If 14 horses eat 56 bushels of corn in 16 days, in how many days will 20 horses eat 120 bushels at that rate?

In this question the supposition is, that 14 horses eat 56 bushels in 16 days; and the demand is, In how many days 20 horses will eat 120 bushels.

The number sought is days, and the term in the supposition of the same kind is 16 days; and accordingly place 16 days in the middle. The remaining four terms are extremes; which class into similar pairs, by making each pair consist of one term taken from the supposition, and another of the same kind taken from the demand. Thus, 14 horses and 20 horses make one pair; again, 56 bushels and 120 bushels make another pair.

Out of the similar pairs, joined with the middle term, form so many simple questions; namely,

1. If 14 horses eat a certain number of bushels in 16 days, in how many days will 20 horses eat the same quantity?

2. If 56 bushels are eat up in 16 days, in how many days will 120 bushels be eat up by the same eaters?

In the first simple question it is plain, that the answer must be less than the middle term; for 20 horses will eat the same number of bushels in fewer days than 14 horses; and so the greatest extreme, viz. 20, is the divisor; and because 20 is an extreme found in the demand, the proportion is inverse.

In the second simple question it is also obvious, that the answer must be greater than the middle term; for 120 bushels will require more days to be eat up in than 56 bushels; and therefore the least extreme, viz. 56, is the divisor; and because 56 is an extreme found in the supposition, the proportion is direct.

We now proceed to state the question, by placing the divisors on the left hand, and the other extremes on the right; then multiply and divide, as directed in Rule III. and the answer comes out of the same name with the middle term, viz. 24 days.

The two simple questions into which the compound question is resolved, are stated and wrought separately, as follows.

III. The Rule of Seven, Nine, &c.

Quest. If 15 men eat 156 d. worth of bread in 6 days, when wheat is sold at 12 s. per bushel, in how many days will 30 men eat 520 d. worth of bread when wheat is at 10 s. per bushel?

This question belongs to the rule of seven, the number sought is days, and the term of the same kind in the supposition is 6 days, which place in the middle. The remaining six terms are extremes, which class into similar pairs, by taking one term of each pair out of the supposition, and another of the same kind out of the demand.

Out of the similar pairs, joined with the middle term, form so many simple questions, in each of which you find the divisor by Rule II.; then place the divisors on the left hand, and the other extremes on the right, as directed in Rule III. and multiply and divide, as follows.

This compound question is resolved into three simple ones, as follows.

EXAMP. If 100 lb. of Venice weigh 70 lb. of Lyons, and 120 lb. of Lyons weigh 100 lb. of Roan, and 80 lb. of Roan weigh 100 lb. of Toulouse, and 100 lb. of Toulouse weigh 74 lb. of Geneva, how many pounds of Geneva will 100 lb. of Venice weigh?

This question belongs to the rule of nine; and because pounds of Geneva is the number sought, the given pounds of Geneva, viz. 74, must be the middle term: the remaining terms are extremes; which may be classed into similar pairs, and stated as follows.

But the question becomes more simple, and is wrought with greater ease and advantage, by being stated in the fractional form, as follows.

We shall conclude by observing, that every compound question, whether in the rule of five, seven, nine, or eleven, &c. properly speaking, consists but of three given terms. For the first term, or divisor, is to be considered as one compound term made up, or produced, by the continual multiplication of the extremes on the left hand, as so many component parts. In like manner, the third term is to be considered as one compound term, made up by the continual multiplication of the extremes on the right, as component parts. Suppose the question to be,

If L. 100 in 12 months gain L. 5 interest, what will L. 75 gain in 19 months?

Here it is obvious, that it is neither the L. 100 principal, nor the 12 months of time, taken separately, that gains the L. 5 interest, but both contribute their share; that is, they conspire, as joint causes, to produce one effect; and therefore their product, viz. the first term, is to be considered as the cause producing the effect; that is, the first term, viz. $100\; \backslash times\; 12$, causeth, produceth, or gains L. 5 of interest. And in like manner, the product of the extremes on the right hand, or the third term, viz. $75\; \backslash times\; 9$, is to be esteemed the cause that produceth a similar effect; that is, gains a like sum of interest, namely, the fourth term, or answer. In reference to this way of considering the first and third terms, the question might be stated as under.

CHAP. VIII. FELLOWSHIP.

FELLOWSHIP, called also Company, or Partnership, is when two or more persons join their stocks, and trade together, dividing the gain or loss proportionally among the partners.

Fellowship is either without or with time, called also Single or Double.

I. Fellowship without time.

Questions in fellowship without time are wrought by the following proportion.

As the total stock

To the total gain or loss,

So each man's particular stock

To his share of the gain or loss.

Quest. A and B make a joint stock: A puts in 12 l. and B 8 l.; they gain 5 l.: What is each man's share?

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

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

II. Fellowship with time.

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

As the sum of the products of the several stocks into their respective times.

To the total gain or loss.

So the product of each man's stock into his time.

To his share of the gain or loss.

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

A $40\; \backslash times\; 3\; =\; 120$, third term for A's share.

B $75\; \backslash times\; 4\; =\; 300$, third term for B's share.

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

A $560\; \backslash times\; 8\; =\; 4480$, third term for A's share.

B $279\; \backslash times\; 10\; =\; 2790$, third term for B's share.

C $735\; \backslash times\; 6\; =\; 4410$, third term for C's share.

11680, first term.

Proof 1000—00—0—0—1162.

CHAP. IX. VULGAR FRACTIONS.

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

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

In the fraction $\backslash frac\{1\}\{4\}$ l. a pound Sterling is the unit, integer, or whole; and the denominator 4 shews that the pound is broken or divided into four equal parts, viz. 4 crowns; and the numerator 3 shews that the fraction contains three of these parts, that is, three crowns; and so the value of this fraction is fifteen shillings.

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

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

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

Fractions having 10, 100, 1000, or 1, with any number of ciphers annexed to it, for a denominator, are called

ed decimal fractions; and fractions having any other denominator are called vulgar fractions.

1. A proper fraction is that whose numerator is less than its denominator, and consequently is in value less than unity; as $\backslash frac\{1\}\{2\}$.

2. An improper fraction is that whose numerator is equal to or greater than its denominator; and consequently is in value equal to or greater than an unit; as $\backslash frac\{3\}\{2\}$, $\backslash frac\{7\}\{4\}$.

3. A simple fraction is that which has but one numerator, and one denominator; and may be either proper or improper; as $\backslash frac\{1\}\{2\}$ or $\backslash frac\{3\}\{4\}$.

4. A compound fraction is made up of two or more simple fractions, coupled together with the particle of, and is a fraction of a fraction; as $\backslash frac\{1\}\{2\}$ of $\backslash frac\{1\}\{2\}$, or $\backslash frac\{1\}\{4\}$ of $\backslash frac\{1\}\{2\}$.

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

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

Reduction of Vulgar Fractions.

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

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

EXAMPLES.

1. $5\backslash frac\{2\}\{3\}\; =\; 85$ integers, there being no remainder.

2. $4\backslash frac\{1\}\{2\}\; =\; 54\backslash frac\{1\}\{2\}$, the remainder being 5.

3. $2\backslash frac\{1\}\{2\}\; =\; 98\backslash frac\{1\}\{2\}$, the remainder being 10.

4. $3\backslash frac\{1\}\{2\}\; =\; 173\backslash frac\{1\}\{2\}$, the remainder being 48.

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

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

EXAMPLES.

1. $54\backslash frac\{1\}\{2\}\; =\; 4\backslash frac\{1\}\{2\}\; \backslash times\; 8\; =\; 432$
$+\; 5$
Numerator 437

2. $98\backslash frac\{1\}\{2\}\; =\; 1\backslash frac\{1\}\{2\}\; \backslash times\; 14\; =\; 1372$
$+\; 10$
Numerator 1382

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

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

EXAMPLES.

1. Reduce 9 to a fraction whose denominator is 5.
$9\; \backslash times\; 5\; =\; 45$; so the fraction is $\backslash frac\{45\}\{5\}$.

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

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

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

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

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

EXAMPLES.

Ex. 1. $\backslash frac\{1\}\{2\}$ of $\backslash frac\{4\}\{5\}\; =\; \backslash frac\{2\}\{5\}$. Ex. 2. $\backslash frac\{1\}\{2\}$ of $\backslash frac\{1\}\{3\}$ of $\backslash frac\{1\}\{4\}\; =\; \backslash frac\{1\}\{12\}$.

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

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

EXAMPLES.

1. What fraction of a pound Sterling is $\backslash frac\{1\}\{4\}$ of a penny?
$\backslash frac\{1\}\{4\}$ d. is $\backslash frac\{1\}\{4\}$ of $\backslash frac\{1\}\{20\}$ of $\backslash frac\{1\}\{20\}$ L. = $\backslash frac\{1\}\{400\}$ L.

2. What fraction of a C. is $\backslash frac\{1\}\{4\}$ of a pound?
$\backslash frac\{1\}\{4\}$ lb. is $\backslash frac\{1\}\{4\}$ of $\backslash frac\{1\}\{20\}$ of $\backslash frac\{1\}\{20\}$ C. = $\backslash frac\{1\}\{400\}$ C.

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

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

EXAMPLES.

1. What fraction of a shilling is $\backslash frac\{1\}\{4\}$ of a pound?

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

2. What fraction of a penny is $\backslash frac\{1\}\{4\}$ L.?

The reason of this rule will appear by observing, that every fraction may be considered in two views. Thus, $\backslash frac\{1\}\{4\}$ may either be considered as expressing three fourths of one unit, or as denoting the fourth part of three units. Now, if the unit be a pound Sterling, the fraction, in the latter view, will denote the fourth part of three pounds; and by reducing the numerator L. 3 to shillings, we have $\backslash frac\{60\}\{4\}$ s; and again reducing 60 shillings to pence, we have $\backslash frac\{720\}\{4\}$ d. equal to $\backslash frac\{60\}\{4\}$ s. or to $\backslash frac\{1\}\{4\}$ L.

PROB. VI. To find the value of a fraction.

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

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

EXAMP.

EXAMP. What is the value of $\backslash frac\{1\}\{4\}$ L.?

3
20
—

4) 60 (15 s. Here consider $\backslash frac\{1\}\{4\}$ L. as expressing the fourth part of three pounds Sterling; so reduce L 3, the numerator, to shillings, and divide by the denominator 4; and as nothing remains, the quot, viz. 15 shillings, is the value complete.

(o)

L. s. d.
$\backslash frac\{1\}\{4\}\; =\; 15$

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

PROB. VII. To reduce a fraction to its lowest terms.

RULE. Divide both numerator and denominator by their greatest common divisor; the two quot make the new fraction.

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

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

EXAMP. Reduce $\backslash frac\{784\}\{952\}$ to its lowest terms.

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

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

So $\backslash frac\{784\}\{952\}\; =\; \backslash frac\{14\}\{17\}$.

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

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

VOL. I. No. 17.

EXAMPLES.

Reduce $\backslash frac\{1\}\{2\}$ and $\backslash frac\{4\}\{5\}$ to a common denominator.

$4\; \backslash times\; 5\; =\; 20$, the common denominator.

$3\; \backslash times\; 5\; =\; 15$, the first numerator.

$4\; \backslash times\; 4\; =\; 16$, the second numerator.

So the new fractions are $\backslash frac\{15\}\{20\}$ and $\backslash frac\{16\}\{20\}$.

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

Thus, $\backslash frac\{1\}\{2\}\; +\; \backslash frac\{1\}\{3\}\; =\; \backslash frac\{5\}\{6\}\; +\; \backslash frac\{2\}\{6\}$.

Here 3 is contained in 12 four times; so multiply both 2 and 3 by 4, and you have $\backslash frac\{8\}\{12\}\; =\; \backslash frac\{2\}\{3\}$.

Again, $\backslash frac\{1\}\{2\}\; +\; \backslash frac\{1\}\{4\}\; +\; \backslash frac\{1\}\{8\}\; =\; \backslash frac\{4\}\{8\}\; +\; \backslash frac\{2\}\{8\}\; +\; \backslash frac\{1\}\{8\}$.

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

Thus, $\backslash frac\{1\}\{2\}\; +\; \backslash frac\{1\}\{3\}\; =\; \backslash frac\{1\}\{4\}\; +\; \backslash frac\{1\}\{3\}$.

And $\backslash frac\{1\}\{2\}\; +\; \backslash frac\{1\}\{3\}\; +\; \backslash frac\{1\}\{4\}\; =\; \backslash frac\{1\}\{4\}\; +\; \backslash frac\{1\}\{3\}\; +\; \backslash frac\{1\}\{4\}$.

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

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

Addition of Vulgar Fractions.

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

Ex. 1. What is the sum of $\backslash frac\{1\}\{2\}\; +\; \backslash frac\{1\}\{2\}$? Ans. $\backslash frac\{1\}\{1\}$.

2. What is the sum of $\backslash frac\{1\}\{3\}\; +\; \backslash frac\{1\}\{3\}$? Ans. $\backslash frac\{2\}\{3\}\; =\; \backslash frac\{1\}\{2\}$, by Prob. VII.

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

Ex. What is the sum of $\backslash frac\{1\}\{2\}\; +\; \backslash frac\{1\}\{3\}$?

$\backslash frac\{1\}\{2\}\; +\; \backslash frac\{1\}\{3\}\; =\; \backslash frac\{3\}\{6\}\; +\; \backslash frac\{2\}\{6\}$, by Prob. VIII.

and $\backslash frac\{3\}\{6\}\; +\; \backslash frac\{2\}\{6\}\; =\; \backslash frac\{5\}\{6\}$.

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

Ex. What is the sum of $7\backslash frac\{1\}\{2\}\; +\; 5\backslash frac\{2\}\{3\}$?

$7\backslash frac\{1\}\{2\}\; +\; 5\backslash frac\{2\}\{3\}\; =\; \backslash frac\{15\}\{2\}\; +\; \backslash frac\{17\}\{3\}$, by Prob. II.

and $\backslash frac\{15\}\{2\}\; +\; \backslash frac\{17\}\{3\}\; =\; \backslash frac\{45\}\{6\}\; +\; \backslash frac\{34\}\{6\}$, by Prob. VIII.

and $\backslash frac\{45\}\{6\}\; +\; \backslash frac\{34\}\{6\}\; =\; \backslash frac\{79\}\{6\}\; =\; 13\backslash frac\{1\}\{6\}$, by Prob. I.

When mixt numbers, or mixt numbers and fractions, are given, you may, with greater expedition, work by the following rule, viz. reduce only the fractions to a common denominator, and add the sum of the fractions

to the integers. The above example wrought in this manner follows.

Ex. What is the sum of $7\backslash frac\{1\}\{2\}\; +\; 5\backslash frac\{3\}\{4\}$?

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

Ex. What is the sum of $\backslash frac\{2\}\{3\}$ of $\backslash frac\{4\}\{5\}$ + $\backslash frac\{1\}\{2\}$?

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

Ex. What is the sum of $\backslash frac\{1\}\{2\}$s. and $\backslash frac\{1\}\{3\}$l.?

METHOD I.

METHOD II.

METHOD III.

Subtraction of Vulgar Fractions.

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

Ex. From $\backslash frac\{1\}\{2\}$ subtract $\backslash frac\{1\}\{3\}$.

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

Ex. From $\backslash frac\{1\}\{2\}$ subtract $\backslash frac\{1\}\{3\}$.

RULE III. If it be required to subtract one mixt number from another, or to subtract a fraction from a mixt number, reduce the mixt numbers to improper

fractions, by Prob. II; then reduce the fractions to a common denominator, by Prob. VIII. and subtract the one numerator from the other.

Ex. From $7\backslash frac\{1\}\{2\}$ subtract $5\backslash frac\{3\}\{4\}$.

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

Ex. 1. From 14 subtract $7\backslash frac\{1\}\{2\}$.

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

Ex. 2. From 12 subtract $\backslash frac\{1\}\{2\}$.

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

Note. When an integer is given to be subtracted from a mixt number, you have only to subtract the given integer from the integral part of the mixt number; and to the remainder annex the fractional part. Thus, $9\backslash frac\{3\}\{5\}\; -\; 5\; =\; 4\backslash frac\{3\}\{5\}$.

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

Ex. From $\backslash frac\{1\}\{2\}$ subtract $\backslash frac\{1\}\{3\}$ of $\backslash frac\{1\}\{2\}$.

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

Ex. From $\backslash frac\{1\}\{2\}$l. subtract $\backslash frac\{1\}\{3\}$s.

METHOD I.

METHOD II.

METHOD.

METHOD III.

Multiplication of Vulgar Fractions.

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

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

EXAMP. 1. $\backslash frac\{2\}\{3\}\; \backslash times\; \backslash frac\{4\}\{5\}\; =\; \backslash frac\{8\}\{15\}$

2. $\backslash frac\{1\}\{2\}\; \backslash times\; 5\backslash frac\{1\}\{2\}\; =\; \backslash frac\{1\}\{2\}\; \backslash times\; \backslash frac\{11\}\{2\}\; =\; \backslash frac\{11\}\{4\}\; =\; 4\backslash frac\{3\}\{4\}$

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

2. Mixt numbers may be multiplied without reducing them to improper fractions, by working as in the margin; where first multiply the integral parts, viz. 54 by 24; then multiply the integral parts cross-ways into their altern fractions, viz. 54 by $\backslash frac\{1\}\{2\}$, and the product 27 set down; in like manner multiply 24 by $\backslash frac\{1\}\{3\}$, and the product 6 likewise set down; then add; and to the sum annex $\backslash frac\{1\}\{3\}$, the product of the two fractions.

3. In multiplying a fraction by an integer, you have only to multiply the numerator by the integer, the putting one for the denominator being only matter of form. And to multiply a fraction by its denominator is to take away the denominator, the product being an integer, the same with, or equal to the numerator. Thus, $\backslash frac\{7\}\{8\}\; \backslash times\; 8\; =\; 7$. For $\backslash frac\{7\}\{8\}\; \backslash times\; \backslash frac\{8\}\{1\}\; =\; \backslash frac\{56\}\{8\}\; =\; 7$.

4. If the numerators and denominators of two equal fractions be multiplied cross-ways, the products will be equal. Thus, if $\backslash frac\{1\}\{3\}\; =\; \backslash frac\{4\}\{12\}$, then will $3\; \backslash times\; 12\; =\; 9\; \backslash times\; 4$; for multiplying both by 9, we have $3\; =\; \backslash frac\{9\; \backslash times\; 4\}\{12\}$; and multiplying these by 12, we have $3\; \backslash times\; 12\; =\; 9\; \backslash times\; 4$. Hence, if four numbers be proportional, the product of the extremes will be equal to the product of the means: for if

$3\; :\; 9\; ::\; 4\; :\; 12$, then $\backslash frac\{1\}\{3\}\; =\; \backslash frac\{4\}\{12\}$; and it has been proved, that $3\; \backslash times\; 12\; =\; 9\; \backslash times\; 4$. Therefore if, of four proportional numbers, any three be given, the fourth may easily be found, viz. when one of the extremes is sought, divide the product of the means by the given extreme; and when one of the means is sought, divide the product of the extremes by the given mean.

5. In multiplying fractions, equal factors above and below may be dashed or dropt. Thus, $\backslash frac\{1\}\{2\}$ of $\backslash frac\{2\}\{3\}\; \backslash times\; \backslash frac\{1\}\{4\}$ of $\backslash frac\{4\}\{5\}\; =\; \backslash frac\{1\}\{2\}\; \backslash times\; \backslash frac\{2\}\{1\}\; \backslash times\; \backslash frac\{1\}\{4\}\; \backslash times\; \backslash frac\{4\}\{1\}$; and dropping the factors 2, 3, 4, both above and below, the product is $\backslash frac\{1\}\{5\}$. In like manner, to facilitate an operation, a factor above and another below may be divided by the same number: Thus,

Or we may exchange one numerator for another: Thus, $\backslash frac\{6\}\{7\}\; \backslash times\; \backslash frac\{1\}\{2\}\; =\; \backslash frac\{6\}\{2\}\; \backslash times\; \backslash frac\{1\}\{7\}\; =\; \backslash frac\{3\}\{7\}\; \backslash times\; \backslash frac\{1\}\{1\}\; =\; \backslash frac\{3\}\{7\}$.

6. To take any part of a given number, is to multiply the said number by the fraction. Thus, $\backslash frac\{1\}\{2\}$ of 320 is found thus, $\backslash frac\{1\}\{2\}\; \backslash times\; \backslash frac\{320\}\{1\}\; =\; \backslash frac\{320\}\{2\}\; =\; \backslash frac\{160\}\{1\}\; =\; 160$. In like manner, $\backslash frac\{2\}\{3\}$ of $45\backslash frac\{1\}\{2\}$, is $\backslash frac\{2\}\{3\}\; \backslash times\; 45\backslash frac\{1\}\{2\}\; =\; \backslash frac\{2\}\{3\}\; \backslash times\; \backslash frac\{91\}\{2\}\; =\; \backslash frac\{182\}\{6\}\; =\; \backslash frac\{91\}\{3\}\; =\; 30\backslash frac\{2\}\{3\}$. Hence, to reduce a compound fraction to a simple one, is to multiply the parts of it into one another.

7. If a multiplicand of two or more denominations be given to be multiplied by a fraction, reduce the higher part or parts of the multiplicand to the lowest species, and then multiply. Thus, to multiply 8 l. 10 $\backslash frac\{1\}\{2\}$ s. by $\backslash frac\{1\}\{2\}$, say, $8\; \backslash text\{\; l.\}\; =\; 8\; \backslash times\; 20\; \backslash text\{\; s.\}\; =\; 160\; \backslash text\{\; s.\}$ and $10\; \backslash frac\{1\}\{2\}\; \backslash text\{\; s.\}\; =\; 10\; \backslash frac\{1\}\{2\}\; \backslash times\; 12\; \backslash text\{\; d.\}\; =\; 126\; \backslash text\{\; d.\}\; =\; 10\; \backslash frac\{1\}\{2\}\; \backslash text\{\; s.\}$, and $\backslash frac\{1\}\{2\}\; \backslash times\; \backslash frac\{160\}\{2\}\; =\; \backslash frac\{160\}\{4\}\; =\; 40\; \backslash text\{\; s.\}$ and $\backslash frac\{1\}\{2\}\; \backslash times\; \backslash frac\{126\}\{2\}\; =\; \backslash frac\{126\}\{4\}\; =\; 31\; \backslash frac\{1\}\{2\}\; \backslash text\{\; d.\}\; =\; 2\; \backslash frac\{1\}\{2\}\; \backslash text\{\; s.\}$. Or, without reducing, you may multiply the given multiplicand by the numerator of the fraction, and divide the product by the denominator.

EXAMP. 1. Multiply $\backslash frac\{1\}\{2\}$ by $\backslash frac\{1\}\{2\}$. Prod. $\backslash frac\{1\}\{4\}$.
2. Multiply $7\backslash frac\{1\}\{2\}$ by $\backslash frac\{1\}\{2\}$. Prod. $6\backslash frac\{1\}\{4\}$.
3. Multiply $8\backslash frac\{1\}\{2\}$ by $9\backslash frac\{1\}\{2\}$. Prod. $84\backslash frac\{1\}\{4\}$.

The reason of the rule may be shewn thus: $\backslash frac\{1\}\{2\}\; \backslash times\; \backslash frac\{1\}\{2\}\; =\; \backslash frac\{1\}\{4\}$; for $\backslash frac\{1\}\{2\}\; =\; \backslash frac\{1\}\{2\}$, and $\backslash frac\{1\}\{2\}$ of $\backslash frac\{1\}\{2\}$ is $\backslash frac\{1\}\{4\}$; and consequently $\backslash frac\{1\}\{2\}$ of $\backslash frac\{1\}\{2\}$ is $\backslash frac\{1\}\{4\}$.

The truth of the rule may also be proved thus: Assume two fractions equal to two integers, such as, $\backslash frac\{2\}\{3\}$, and $\backslash frac\{3\}\{4\}$, equal to 2 and 3, and the product of the fractions will be equal to the product of the integers; for $\backslash frac\{2\}\{3\}\; \backslash times\; \backslash frac\{3\}\{4\}\; =\; \backslash frac\{6\}\{12\}\; =\; 6$, and $2\; \backslash times\; 3\; =\; 6$.

Division of Vulgar Fractions.

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

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

EXAMP. 1. $\backslash frac\{2\}\{3\}\; \backslash div\; \backslash frac\{1\}\{2\}\; =\; \backslash frac\{2\}\{3\}\; \backslash times\; \backslash frac\{2\}\{1\}\; =\; \backslash frac\{4\}\{3\}\; =\; 1\backslash frac\{1\}\{3\}$.
2. $\backslash frac\{1\}\{2\}\; \backslash div\; 4\backslash frac\{1\}\{2\}\; =\; \backslash frac\{1\}\{2\}\; \backslash times\; \backslash frac\{9\}\{2\}\; =\; \backslash frac\{9\}\{4\}\; =\; 2\backslash frac\{1\}\{4\}$.
3. $4\backslash frac\{1\}\{2\}\; \backslash div\; \backslash frac\{2\}\{3\}\; =\; \backslash frac\{9\}\{2\}\; \backslash times\; \backslash frac\{3\}\{2\}\; =\; \backslash frac\{27\}\{4\}\; =\; 6\backslash frac\{3\}\{4\}$.

Note.

Note 1. Instead of working division of fractions as taught above, you may invert the divisor, and then multiply it into the dividend. Thus, in Example 1. instead $\backslash frac\{1\}\{2\}\; \backslash div\; \backslash frac\{1\}\{3\}$, you may say, $\backslash frac\{1\}\{2\}\; \backslash times\; \backslash frac\{3\}\{1\}\; =\; \backslash frac\{3\}\{2\}\; =\; 1\backslash frac\{1\}\{2\}$.

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

3. To divide a fraction by an integer, is only to multiply the integer into the denominator of the fraction, the numerator being continued. Thus, $7\; \backslash frac\{1\}\{2\}\; \backslash div\; 8$.

4. A mixt number may sometimes be divided by an integer, with more ease, in the following manner. Divide the integral part of the mixt number by the given integer: and if there be no remainder, divide likewise the fraction of the mixt number by the given integer, and annex the quot to the integral quot formerly found. But if, in dividing the integral part, there happen to be a remainder, prefix this remainder to the fraction for a new mixt number; which reduce to an improper fraction: then divide the improper fraction by the given integer, and annex the quot to the integral quot formerly found. Thus, if it be required to divide $15\backslash frac\{1\}\{2\}$ by 8, say, $8)15$ (1, and 7 remains; which 7, prefixed to the fraction, gives $7\backslash frac\{1\}\{2\}$ for a new mixt number; and this, reduced to an improper fraction, is $\backslash frac\{15\}\{2\}$, and $8)\; \backslash frac\{15\}\{2\}\; (1\backslash frac\{1\}\{2\}$: so the complete quot is $1\backslash frac\{1\}\{2\}$.

5. If the factors of the numerator and denominator of the quot, instead of being actually multiplied, be only connected with the sign of multiplication, it will be easy to drop such factors, above and below, as happen to be the same, thus: $\backslash frac\{1\}\{2\}\; \backslash div\; \backslash frac\{1\}\{3\}$ of $\backslash frac\{4\; \backslash times\; 4\; \backslash times\; 3\}\{3\; \backslash times\; 5\; \backslash times\; 8\}\; =\; \backslash frac\{4\; \backslash times\; 4\}\{5\; \backslash times\; 8\}\; =\; \backslash frac\{2\}\{5\}$. Or a factor above and below may be divided by the same number thus: $\backslash frac\{6\; \backslash times\; 7\}\{5\; \backslash times\; 12\}\; =\; \backslash frac\{7\}\{5\; \backslash times\; 2\}\; =\; \backslash frac\{7\}\{10\}$. Or the factors of the numerator of the quot may be exchanged, thus: $\backslash frac\{3\; \backslash times\; 5\}\{2\; \backslash times\; 9\}\; =\; \backslash frac\{5\; \backslash times\; 3\}\{2\; \backslash times\; 9\}\; =\; \backslash frac\{5\}\{2\; \backslash times\; 3\}\; =\; \backslash frac\{5\}\{6\}$.

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

7. If the divisor and dividend have the same denominator, you have only to divide the numerator of the dividend by the numerator of the divisor, thus: $\backslash frac\{1\}\{8\}\; \backslash div\; \backslash frac\{1\}\{5\}\; =\; \backslash frac\{5\}\{8\}$; for $\backslash frac\{1\}\{8\}\; \backslash div\; \backslash frac\{1\}\{5\}\; =\; \backslash frac\{5\; \backslash times\; 3\}\{8\; \backslash times\; 3\}\; =\; \backslash frac\{15\}\{24\}\; =\; \backslash frac\{5\}\{8\}$.

8. If a dividend of two or more denominations be given to be divided by a fraction, reduce the higher part or parts of the dividend to the lowest species, and then divide. Thus, to divide 6 l. 9 s. by $\backslash frac\{1\}\{4\}$, say, $6\; \backslash times\; 20s.\; =\; 120$; and $120\; +\; 9s.\; =\; 129s.\; =\; \backslash frac\{519\}\{4\}$; and $\backslash frac\{519\}\{4\}\; \backslash div\; \backslash frac\{1\}\{4\}\; =\; \backslash frac\{519\}\{1\}\; =\; 519$ d. = 9 l. 14 s. 7 d.

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

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

Proof 276 16 8

The reason of the rule will appear by considering, that the method here used is nothing else but the reducing the divisor and dividend to a common denominator, and then dividing the one numerator by the other. Thus, $\backslash frac\{1\}\{2\}\; \backslash div\; \backslash frac\{1\}\{3\}$, for reducing the divisor and dividend to a common denominator, we have $\backslash frac\{3\}\{2\}\; \backslash div\; \backslash frac\{3\}\{3\}\; =\; \backslash frac\{3\}\{2\}\; \backslash times\; \backslash frac\{3\}\{3\}\; =\; \backslash frac\{9\}\{6\}\; =\; \backslash frac\{3\}\{2\}$.

The truth of the rule may also be proved by assuming two fractions equal to two integers, such as, $\backslash frac\{2\}\{1\}$ and $\backslash frac\{4\}\{1\}$, equal to 2 and 4, and the quot of the fractions will be equal to the quot of the integers. Thus, $\backslash frac\{2\}\{1\}\; \backslash div\; \backslash frac\{4\}\{1\}\; =\; \backslash frac\{2\}\{4\}\; =\; \backslash frac\{1\}\{2\}$, and $2\; \backslash div\; 4\; =\; \backslash frac\{1\}\{2\}$.

The Simple Rule of Three in Vulgar Fractions.

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

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

I. Direct.

QUEST. If $\backslash frac\{1\}\{2\}$ yard cost $\backslash frac\{1\}\{4\}$ l. what will $\backslash frac\{3\}\{4\}$ yard cost?

Yd. L. Yd.
If $\backslash frac\{1\}\{2\}\; :\; \backslash frac\{1\}\{4\}\; ::\; \backslash frac\{3\}\{4\}\; :\; ?$

Ans. $\backslash frac\{4\; \backslash times\; 5\; \backslash times\; 9\}\{3\; \backslash times\; 8\; \backslash times\; 10\}\; =\; \backslash frac\{5\; \backslash times\; 9\}\{3\; \backslash times\; 2\; \backslash times\; 10\}\; =\; \backslash frac\{9\}\{3\; \backslash times\; 2\; \backslash times\; 2\}\; =\; \backslash frac\{3\}\{2\; \backslash times\; 2\}\; =\; \backslash frac\{3\}\{4\}$ l. = $\backslash frac\{3\; \backslash times\; 20s.\}\{4\}\; =\; \backslash frac\{60s.\}\{4\}\; =\; 15s.$

II. Inverse.

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

Bread. len. Bread.
$\backslash frac\{1\}\{2\}\; :\; \backslash frac\{1\}\{4\}\; ::\; \backslash frac\{2\}\{1\}\; :\; ?$

Ans. $\backslash frac\{5\; \backslash times\; 3\; \backslash times\; 2\}\{3\; \backslash times\; 4\; \backslash times\; 1\}\; =\; \backslash frac\{5\; \backslash times\; 2\}\{4\}\; =\; \backslash frac\{10\}\{4\}\; =\; 2\backslash frac\{1\}\{2\}$ yards.

The Compound Rule of Three in Vulgar Fractions.

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

Ans. $\backslash frac\{3\; \backslash times\; 3\; \backslash times\; 60\}\{4\; \backslash times\; 4\; \backslash times\; 3\}\; =\; \backslash frac\{3\; \backslash times\; 60\}\{4\; \backslash times\; 4\}\; =\; \backslash frac\{3\; \backslash times\; 15\}\{4\}\; =\; \backslash frac\{45\}\{4\}\; =\; 11\backslash frac\{1\}\{4\}\; \backslash text\{\; acr.\}$

Or thus:

Ans. $\backslash frac\{3\; \backslash times\; 1\; \backslash times\; 3\; \backslash times\; 6\; \backslash times\; 10\}\{2\; \backslash times\; 2\; \backslash times\; 4\; \backslash times\; 1\; \backslash times\; 3\}\; =\; \backslash frac\{3\; \backslash times\; 6\; \backslash times\; 10\}\{2\; \backslash times\; 2\; \backslash times\; 4\}\; =\; \backslash frac\{3\; \backslash times\; 3\; \backslash times\; 5\}\{2\; \backslash times\; 2\}\; =\; \backslash frac\{45\}\{4\}\; =\; 11\backslash frac\{1\}\{4\}\; \backslash text\{\; acres.\}$

CHAP. X. RULES OF PRACTICE.

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

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

Thus, if the price of 24 yards, at 6s. 8d. per yard, be demanded, the answer is found by multiplying 24 by $\backslash frac\{1\}\{4\}$, the fraction of a pound equivalent to 6s. 8d. viz. $\backslash frac\{1\}\{4\}\; \backslash times\; \backslash frac\{1\}\{2\}\; =\; \backslash frac\{1\}\{8\}\; =\; 6s.\; 8d.$

Hence, it is obvious, that to multiply a number by a fraction whose numerator is unity, is to divide the said number by the denominator of the fraction. But if the numerator of the fraction be not unity, you must first multiply the given number by the numerator, and then divide the product by the denominator. Thus, if the rate be 13s. 4d. = $\backslash frac\{27\}\{20\}$ l. the price of 24 yards is found by saying, $\backslash frac\{24\}\{1\}\; \backslash times\; \backslash frac\{27\}\{20\}\; =\; \backslash frac\{648\}\{20\}\; =\; 32\backslash frac\{1\}\{2\}$ l.; or take $\backslash frac\{1\}\{2\}$ of the given number twice.

When the fraction denoting the rate happens to be compound, the product or answer is found by dividing the given number by one of the denominators of the compound fraction, the quot by another, and the next quot by the third, &c. Thus, if the rate be 2 farthings = $\backslash frac\{1\}\{4\}$ of $\backslash frac\{1\}\{4\}$ of $\backslash frac\{1\}\{2\}$ l. the price of 1440 yards is found by saying, $\backslash frac\{1440\}\{\backslash frac\{1\}\{4\}\}\; =\; 5760$, and $\backslash frac\{5760\}\{\backslash frac\{1\}\{4\}\}\; =\; 14400$, and $\backslash frac\{14400\}\{\backslash frac\{1\}\{2\}\}\; =\; 28800$.

When the rate is expressed by two or more simple fractions, connected with the sign $\backslash frac\{1\}\{2\}$, the product or answer is found by dividing the given number successively by the several denominators, and then adding the quot. Thus, if the rate be 3s. = $\backslash frac\{1\}\{2\}$ of $\backslash frac\{1\}\{2\}$ l. the price of 80 yards is found by saying, $\backslash frac\{80\}\{\backslash frac\{1\}\{2\}\}\; =\; 160$, and $\backslash frac\{160\}\{\backslash frac\{1\}\{2\}\}\; =\; 320$, and $320\; +\; \backslash frac\{1\}\{2\}\; =\; 320\backslash frac\{1\}\{2\}$.

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

TABLE I.

TABLE II.

Pen. of a shill.

TABLE III.

s. d. of a pound.

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

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

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

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

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

Ex. 1.
$\backslash frac\{1\}\{4\}\; 4859$, at 1 f.
$\backslash frac\{1\}\{2\}\; 1214\; -\; 3\; f.$
$\backslash frac\{1\}\{2\}\; 10\; |\; 1\; -\; 2\; d.$
L. 5 1 2 $\backslash frac\{1\}\{2\}$

Ex. 2.
$\backslash frac\{1\}\{2\}\; 8347$, at 2 f.
$\backslash frac\{1\}\{2\}\; 4173\; -\; \backslash frac\{1\}\{2\}\; d.$
$\backslash frac\{1\}\{2\}\; 3\; |\; 7\; -\; 9\; d.$
L. 17 7 9 $\backslash frac\{1\}\{2\}$

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

RULE. Divide the given number by the denominator of the fraction denoting the rate, as contained in Table II. and you have the answer in shillings; which reduce into pounds, by dividing by 20.

Note. The remainders at the first division in the above examples are the same with the rate. Thus, in Ex. 1. every remainder is 1 d.

CASE III. When the rate is pence and farthings.

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

EXPLICATION.

In Ex. 1. work first for 1 d.; which being $\backslash frac\{1\}\{12\}$ s. divide the given number by the denominator 12, and the quot is shillings, and the remainder pence; then, because 1 farthing is $\backslash frac\{1\}\{4\}$ d. divide the former quot by 4, and the sum of the quots is the price in shillings; which divide by 20.

In Ex. 2. the rate $1\backslash frac\{1\}\{2\}$ d. being an aliquot part of a shilling, the second method is shorter and better than the first.

CASE IV. When the rate is shillings under twenty.

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

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

2. If the rate be an odd number of shillings, work for the next lesser even number of shillings, as above; and for the odd shilling take $\backslash frac\{1\}\{20\}$ of the given number.

EXAMP. 1. When the rate is an even number of shillings.

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

Note 1. The reason of multiplying by half the number of shillings in the rate will appear by considering, that these are the numerators of the fractions denoting the rate. Thus, 2 s. is $\backslash frac\{1\}\{10\}$ l. and 4 s. is $\backslash frac\{2\}\{10\}$ l. and 6 s. is $\backslash frac\{3\}\{10\}$ l. and each unit in the product is two shillings. The division by the denominator 10 is performed by cutting off the right-hand figure of the product, and the figure so cut off is the remainder; and as each unit in the remainder is two shillings, the double of them is the remainder in shillings.

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

Note 3. In Ex. 2. the price may also be had by taking $\backslash frac\{1\}\{3\}$ of the given number; and in this way every remainder will be 4 s.

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

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

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

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

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

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

RULE III.

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

CASE VI. When the rate is pounds.

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

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

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

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

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

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

CASE VIII. When the given number consists of integers and parts.

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

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

CHAP. XI. Of DECIMALS.

I. Notation.

A FRACTION having 10, 100, 1000, or unity with any number of ciphers annexed to it, for a denominator, is called a decimal fraction; such as, $\backslash frac\{1\}\{10\}$, $\backslash frac\{2\}\{10\}$, $\backslash frac\{3\}\{10\}$, $\backslash frac\{14\}\{100\}$, $\backslash frac\{86\}\{1000\}$.

In decimal fractions, as in vulgar, the denominator shews into how many parts the unit or integer is divided, and the numerator shews how many of these parts the fraction contains. Thus, if the fraction be $\backslash frac\{1\}\{10\}$, the unit is divided into ten equal parts, and the fraction contains nine of these parts; and consequently, if the unit or integer be a pound Sterling, the value of such a fraction is eighteen shillings.

We may conceive the denominator of a decimal fraction to be formed by dividing the unit into 10 equal parts, and each of these parts into 10 other equal parts, each of these again into 10 other equal parts, and so on, as far as necessary; and hence a decimal fraction will always be so many tenths, or so many tenths of $\backslash frac\{1\}\{10\}$, or so many tenths of $\backslash frac\{1\}\{10\}$ of $\backslash frac\{1\}\{10\}$, &c.; and by reducing the compound fraction to a simple one, we have the decimal. Thus, $\backslash frac\{8\}\{10\}$ of $\backslash frac\{1\}\{10\}$ of $\backslash frac\{1\}\{10\}$ = $\backslash frac\{8\}\{1000\}$.

Or

Or we may conceive the denominator of a decimal to be formed by the continual multiplication of unity into 10, as often as there are ciphers in it. Thus, $1\; \backslash times\; 10\; =\; 10$, and $1\; \backslash times\; 10\; \backslash times\; 10\; =\; 100$, and $1\; \backslash times\; 10\; \backslash times\; 10\; \backslash times\; 10\; =\; 1000$, &c. And because the fractions $\backslash frac\{1\}\{10\}$, $\backslash frac\{1\}\{100\}$, $\backslash frac\{1\}\{1000\}$, &c. have the highest numerators possible, it is plain, that the number of figures or places in the numerator of a decimal can never exceed the number of ciphers in the denominator.

It is usual to write down only the numerator of a decimal fraction, omitting the denominator; and when the numerator has the same number of figures or places as the denominator has ciphers, it is done by writing down the figures of the numerator, and prefixing a point, to distinguish them from a whole number. So $\backslash frac\{7\}\{10\}$ is written thus, .7; and $\backslash frac\{25\}\{100\}$ is written thus, .25. The point thus prefixed is called the decimal point.

But when the numerator has not so many figures or places as there are ciphers in the denominator, the defect is supplied by prefixing a cipher for every figure wanting, and then placing the decimal point on the left. So $\backslash frac\{3\}\{100\}$ is written thus, .03; and $\backslash frac\{75\}\{1000\}$ thus, .0075; and $\backslash frac\{36\}\{10000\}$ thus, .00036.

From this manner of notation, it is easy to read a decimal, or to know its denominator, viz. imagine 1 to stand under the decimal point, and a cipher under every decimal place. Thus, .9 is $\backslash frac\{9\}\{10\}$, and .48 is $\backslash frac\{48\}\{100\}$, and .05 is $\backslash frac\{5\}\{100\}$, and .007 is $\backslash frac\{7\}\{1000\}$, and .00036 is $\backslash frac\{36\}\{10000\}$.

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

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

Ciphers annexed to decimals do not change the value of the decimals. Thus, .50 = 5, and .500 = 5, for $.50\; =\; \backslash frac\{50\}\{100\}\; =\; \backslash frac\{5\}\{10\}\; =\; .5$; and $\backslash frac\{500\}\{1000\}\; =\; \backslash frac\{5\}\{10\}\; =\; .5$.

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

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

From this brief account of the nature of decimals, it follows, that the manner of operation in decimals will be the same as in whole numbers; and also, that the same number may be differently expressed, according as the integer is chosen. Thus, the time since our Saviour's birth may be written thus, 1769; or thus, 176.9; or thus, 17.69; or thus, 1.769; or thus, .1769, according as one year, a decad, a century, a chiliad, or myriad, is used as the integer. Hence arises the superior excellency of decimal arithmetic, above every

other sort of numerical computation; as will appear in the sequel.

II. Reduction of Decimals.

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

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

EXAMP. I. Reduce $\backslash frac\{1\}\{2\}$ to a decimal?

Here to the numerator 1 annex one cipher, 2) 1.0(.5 and dividing by the denominator 2, the quot is 5, and 0 remains; and because a single cipher only was annexed to the numerator, the decimal numerator will consist but of one figure, namely 5; to which, therefore, prefix the decimal point. So $\backslash frac\{1\}\{2\}\; =\; .5$.

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

EXAMP. II. Reduce $\backslash frac\{1\}\{3\}$ to a decimal.

To the numerator 3, annex two ciphers; 4) 3.00(.75 and, dividing by the denominator, the quot gives 75 for the numerator of the decimal, two ciphers having been annexed. So $\backslash frac\{1\}\{3\}\; =\; .75$.

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

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

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

In finite decimals, the denominator is always some aliquot part of the numerator increased by annexing ciphers; and such decimals take their rise from vulgar fractions whose denominator is 2 or 5, or some power of

2 or 5, or the product of some of their powers. See Chap. XII. and ALGEBRA, Chap. III.

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

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

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

From these remarks it is easy to perceive, that 2 or 5, or any of their powers, or product of their powers, will measure 10 or its powers, viz. 100, 1000, &c. or their multiples, such as, 20, 200, 2000, &c. 30, 300, 3000, &c.; and such every numerator becomes by having ciphers annexed; and therefore 2 or 5, or their powers, or product of their powers, used as a denominator, will divide any numerator with a competent number of ciphers annexed, and leave no remainder; and consequently the decimal thence resulting will be finite.

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

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

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

$2^6\; \backslash times\; 5^4\; =\; 10^4\; \backslash times\; 2^4$; and the sum of the indices, $2\; +\; 4\; =\; 6$, gives the number of decimal places, viz. $\backslash frac\{1\}\{1000000\}\; =\; .0000625$.

And, in general, to find what number of decimal places any such vulgar fraction will give, divide the denominator by 2, 5, or 10, till the last quotient be 1, and the remainder 0; and the number of divisors shews the number of decimal places. Thus, $\backslash frac\{1\}\{125\}$ gives a decimal of four places; for $125\; =\; 5^3$.

And $\backslash frac\{1\}\{1000\}$ gives a decimal of three places; for $1000\; =\; 10^3$. And $\backslash frac\{1\}\{1000000\}$ gives a decimal of six places; for $1000000\; =\; 10^6$.

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

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

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

Here the remainder being still the same, 3)1.0(.3 viz. 1, the same figure will constantly be repeated in the quot.

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

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

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

EXAMP. IV. Reduce $\backslash frac\{1\}\{15\}$ to a decimal.

Here the repeater is mixed, the finite part 15)4.0(.26 being 2, and the repeating figure 6.

We now resolve such denominators into their component parts, and divide the numerator by one of these parts, and then divide the quot by the other. Thus, $15\; =\; 5\; \backslash times\; 3$.

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

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

Repeating decimals are usually marked by a dash through the right-hand figure, as in the examples above: But some chuse to mark them by a point set over the repeating figure, thus, $.3,\; .26$. The remainder where the repetition begins is commonly marked with an asterisk.

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

Now, suppose the given decimal to be a repeater; such as $.3$, resulting from the vulgar fraction $\backslash frac\{3\}\{10\}$, if the repeating decimal be multiplied by the denominator 3, it will, by carrying at 9 on the right hand, reproduce the numerator 1 with the annexed cipher. In like manner, if the repeater $.6\; =\; \backslash frac\{3\}\{5\}$, be multiplied by 3, it will, by carrying at 9 on the right hand, reproduce the numerator 2 with the annexed cipher. Again, if the repeater $.4\; =\; \backslash frac\{2\}\{5\}$, be multiplied by the denominator 9, it will, by carrying at 9 on the right hand, reproduce the numerator 1 with the annexed cipher. And, if the mixt repeater $.26\; =\; \backslash frac\{26\}\{15\}$, be multiplied by the denominator 15, it will, by carrying at 9 on the right hand, reproduce the numerator 4 with the two annexed ciphers.

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

The vulgar fraction $\backslash frac\{3\}\{5\}$ reduced to a decimal gives ; and this repeater resolved into decimal constituent parts, becomes to infinity. But if we esteem the right-hand figure to be ninth-parts, we have $\backslash frac\{3\}\{5\}\; +\; \backslash frac\{3\}\{50\}\; +\; \backslash frac\{3\}\{500\}\; =\; \backslash frac\{3\}\{5\}\; +\; \backslash frac\{3\}\{50\}\; =\; \backslash frac\{3\}\{5\}\; +\; \backslash frac\{3\}\{50\}\; =\; \backslash frac\{3\}\{5\}$, the given vulgar fraction. And as the vulgar fraction $\backslash frac\{3\}\{5\}$ gives so $\backslash frac\{3\}\{5\}$ gives $.69$; that is, $.69\; =\; \backslash frac\{3\}\{5\}\; =\; 1$. And, universally, a series of nines infinitely continued is equal to unity in the place on the left hand; thus, $.999\; =$

1; and $.0999\; =\; .1$; and $.0099\; =\; .01$, and $.4444\; =\; .45$.

Hence may be ascertained the value of an infinite series decreasing in a decuple proportion. Thus, . And .

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

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

EXAMP. V. Reduce $\backslash frac\{1\}\{11\}$ to a decimal.

The denominator 11 gives a pure circle of two figures.

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

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

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

To find how many places the circle will consist of, divide a competent number of 9's by any of the above denominators, continuing the operation till 0 remain; and the number of 9's used will show the number of places.

Thus,

Thus, 7)999999 six places. Thus, 27)999 three places.

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

If 3 divide a repeater whose repeating figure is not a multiple of 3, the quot will be a pure circulate of three places. Thus, 3).11(.037, and 3).55(.185, and 3).77(.259.

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

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

EXAMP. VI. Reduce $\backslash frac\{28\}\{9\}$ to a decimal.

The number of places, both in the finite part and in the circle, may be ascertained thus: Divide the denominator of the vulgar fraction by 10, 5, or 2, as often as possible, and the number of divisors will show the number of places in the finite part; make the last quot a divisor, and the dividend any competent number of 9's; continue the operation till 0 remains, and the number of 9's used will be equal to the number of places in the

circle. Thus, 10)0300(2050(205(41; and 41)99999(2439, and 0 remains. So you may conclude, that the finite part will consist of three places, and the circle of five.

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

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

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

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

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

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

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

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

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

RULE. Convert the given part or parts to a vulgar fraction.

fraction of the integer, and then reduce the vulgar fraction to a decimal.

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

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

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

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

RULE. The remainder being the numerator, and the divisor the denominator of a vulgar fraction, after placing the decimal point on the right of the integral part of the quot., annex ciphers to the remainder; then continue the division till 0 remain, or till the quot. repeat or circulate, or till you think proper to limit the decimal; and the number on the right of the point is a decimal of the integer expressed in the quot.

PROB. IV. To reduce a decimal to value.

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

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

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

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

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

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

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

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

When the figures in the second and third place to be converted into farthings are 25, the answer, by inspection, comes out exact, viz. 24 f. or 6 d.; but in all other cases, the answer, by inspection, is too great, no allowance or correction being made till the convertible number

number amount to 25, and afford a deduction of 1 farthing complete. Hence, by inspection, we have frequently 1 farthing more than by the common method; but the two methods will agree, or give the same answer, if, from the figures to be turned into farthings, we subtract their 25th part, esteeming the remainder farthings and decimal parts of a farthing.

Thus, .718 l. = 14s. 4d. 2f. by inspection; but by the common method, and by inspection corrected, the answer comes out 1 farthing less, as follows.

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

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

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

CASE I. When the given decimal is finite.

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

Thus, $.875\; =\; \backslash frac\{875\}\{1000\}\; =\; \backslash frac\{7\}\{8\}$. For $875)1000(1$
875

Greatest common measure 125)875(7
875

And $125)875(\backslash frac\{7\}\{8\}$. (0)

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

RULE. Make the repeating figure, or the figures of the circle, the numerator of the vulgar fraction; the de-

nominator is 9 for the repeating figure, or 9 for every figure of the circle; and then, if occasion require, reduce this fraction to its lowest terms.

Thus, $.3\; =\; \backslash frac\{3\}\{9\}\; =\; \backslash frac\{1\}\{3\}$, and $.6\; =\; \backslash frac\{6\}\{9\}\; =\; \backslash frac\{2\}\{3\}$, and $.7\; =\; \backslash frac\{7\}\{9\}$.

Again, $.27\; =\; \backslash frac\{27\}\{99\}\; =\; \backslash frac\{3\}\{11\}$, and $.714285\; =\; \backslash frac\{714285\}\{999999\}\; =\; \backslash frac\{5\}\{7\}$.

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

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

Thus, $.03\; =\; \backslash frac\{3\}\{99\}\; =\; \backslash frac\{1\}\{33\}$, and $.16\; =\; \backslash frac\{16\}\{99\}\; =\; \backslash frac\{16\}\{99\}$, and $.083\; =\; \backslash frac\{83\}\{999\}\; =\; \backslash frac\{83\}\{999\}$.

The reason of the rule may be shewn thus: Esteem the finite part of the last example an integer, and then the mixt number $3\backslash frac\{1\}\{33\}$ will be equal to the given circulate. Again, reduce this mixt number to an improper fraction, viz. multiply the integer 3 by the denominator 99999, and to the product add the numerator, as directed in reduction of vulgar fractions.

Multiply the integer 3 into 99999 by the method of multiplying any number by 9, 99, 999, &c. taught in multiplication of integers, and to the product add the numerator, and the sum shall be the numerator of the improper fraction, as in the margin.

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

To the numerator thus found, the denominator is 99999; and so the vulgar fraction is $\backslash frac\{3571425\}\{99999\}$. But we esteemed 3 an integer; whereas, in fact, it is $\backslash frac\{1\}\{3\}$; and so our vulgar fraction will be 100 times greater than it ought to be: to correct this error, we must multiply the denominator by 100, which is done by annexing two ciphers to it; and the true fraction comes out to be $\backslash frac\{3571425\}\{9999900\}$, as by the rule.

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

Reduce .0416 to a vulgar fraction.

In this manner too may any mixt number, consisting of an integer with a repeater or circulate, be reduced to an improper vulgar fraction; but no ciphers are to be annexed to the denominator for the figures of the integer.

Ex. Reduce 8.3 to an improper vulgar fraction.

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

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

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

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

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

Thus, .153846, and .384615, are conterminous; because they both begin at the place of primes, and have an equal number of places. And .0714285, and .7857142, are conterminous, because they both begin at the place of seconds, and have the same number of places. But .81, and .136, are not conterminous, the former beginning at the place of primes, and the latter at the place of seconds. Again, .63, and .481, are not conterminous, because they have not the same number of places; for circles cannot be conterminous unless they be at the same time similar.

Unlike circles are reduced to similar ones by the following

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

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

In a circle any one of the circulating figures may be made the first of the circle. Thus, 7.592, may be expressed thus, 7.5925,; or thus, 7.59259,; and that without changing its value: consequently a pure circulate may put on the form of a mixt circulate, if one or more figures on the left be set aside for the finite part; thus, .72, = .727, where .7, is the finite part.

That the value is not changed may be thus demonstrated.

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

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

Ex. To make .54,63, and .9,148, conterminous.

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

III. Addition of Decimals.

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

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

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

Ex. What is the sum of .7251. and .625s.? Here you may either reduce the decimal of a shilling to that of a pound, or you may reduce the decimal of a pound to that of a shilling.

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

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

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

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

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

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

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

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

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

RULE III. When all or any of the given decimals are circulates, make all the circles conterminous, find the number of tens to be carried from the left-hand column

of the circles, add this carriage to the right hand column, and proceed as in addition of integers.

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

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

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

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

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

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

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

integers, where the method of dividing by 9, 99, 999, &c. is explained.

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

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

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

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

IV. Subtraction of Decimals.

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

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

EXAMPLE I.

EXAMPLE II.

APPROXIMATES.

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

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

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

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

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

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

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

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

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

EXAMPLE II.

Rem. .2,619047,

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

EXAMPLE III.

Rem. .259,615384,

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

V. Multiplication of Decimals.

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

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

• An integer may multiply or divide { an integer,
• { a mixt number,
• { a pure decimal.
• A mixt number may multiply or divide { an integer,
• { a mixt number,
• { a pure decimal,
• A pure decimal may multiply or divide { an integer,
• { a mixt number,
• { a pure decimal.

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

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

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

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

GENERAL RULE.

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

We now proceed to multiplication.

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

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

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

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

Thus:

And thus:

APPROXIMATES.

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

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

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

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

The reason of Rule 1. is plain. For if in Ex. 1. we make the longest factor the multiplier, and the total product will be the same either way, it is obvious, that in this case we shall have six particular products, in each of which the right-hand figure will be uncertain, and consequently

consequently we shall have six uncertain places in the total product toward the right, and also one uncertain place more on account of the uncertain carriage from the column in which the right-hand figure of the last particular product stands.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

RULE III. If the multiplier be finite, and the multi-

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

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

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

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

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

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

CASE I. When a repeating multiplier multiplies a finite multiplicand.

EXAMP. Multiply 638.25 by $.4\; =\; \backslash frac\{2\}\{5\}$

CASE II. When both factors repeat:

EXAMP. Multiply 6.83 by $\backslash frac\{1\}\{7\}\; =\; \backslash frac\{1\}\{7\}$

CASE III. When a repeating multiplier multiplies a circulating multiplicand.

EXAMP. Multiply 24.36, by $\backslash frac\{4\}\{9\}\; =\; \backslash frac\{4\}\{9\}$

CASE IV. When a circulate multiplies a finite multiplicand.

EXAMP. Multiply 825 by $.36\; =\; \backslash frac\{36\}\{100\}$

CASE V. When a circulate multiplies a repeating multiplicand.

EXAMP. Multiply 8.02083 by $.72\; =\; \backslash frac\{72\}\{100\}$

CASE VI. When both factors circulate.

EXAMP. Multiply .714285, by $.36\; =\; \backslash frac\{36\}\{100\}$

Prod. .25,974025, = .259740.25

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

VI. Division of Decimals.

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

GENERAL RULE.

The decimal places in the divisor and quot together must always be equal in number to those of the dividend.

The

The five following practical directions will make the application of the general rule easy.

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

2. When the decimal places of the dividend are more than those of the divisor, the number of decimal places in the quot must be equal to the excess; as in Ex. 1. 4. and 8.

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

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

5. If, after the dividend is exhausted, there be a remainder, annex a cipher, or ciphers, to the remainder, and continue the division till 0 remain, or till the quot repeat or circulate, or till you think proper to limit it; as in Ex. 9. 10. 11. and 12.

We now proceed to division.

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

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

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

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

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

cause there are two decimal places in the divisor, and the quot comes out an integer, according to Direction 3.

In Ex. 4. An integer divides a decimal; and because the dividend has three decimal places, and the divisor none, give the quot three, by Direction 2.

In Ex. 5. A decimal divides a mixt number; and the divisor having three decimal places, and the dividend but two, supply that defect by annexing a cipher, and the quot comes out an integer, by Direction 3.

In Ex. 6. A mixt number divides a decimal; and because the dividend has four decimal places more than the divisor, and the quot, after the division is finished, has only three figures, supply this defect by prefixing a cipher to it, according to Direction 4.

In Ex. 7. A mixt number divides an integer; and the dividend having no decimal places, supply that defect by annexing two ciphers, the number of decimal places in the divisor, and the quot is an integer, by Direction 3.

In Ex. 8. An integer divides a mixt number; and the divisor having no decimal place, and the dividend only one, give one to the quot, according to Direction 2.

In Ex. 9. A decimal divides an integer; and after the dividend is exhausted, annex a cipher to the remainder, and continue the division till 0 remain, according to Direction 5.

In Ex. 10. A decimal divides a decimal; and after the dividend is exhausted, annex a cipher to the remainder, and continue the division till you find the quot repeats.

In Ex. 11. An integer divides an integer; and the dividend being less than the divisor, annex a cipher to it; again, after the dividend is exhausted, annex a cipher to the remainder, and continue the division till you find the quot circulates.

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

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

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

Thus, in the example of integers in $35)32095(917$ the margin, the figure 0, that stands over 5, the units of the product of $9\; \backslash times\; 35$, is in the place of hundreds; and therefore 9, the first figure of the quot, is likewise hundreds; and so the quot is 917 integers.

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

In all the above varieties, the figure 0 in the dividend stands over the units of the first product: and in Var. 1. the figure 0 is in the place of tens, and accordingly 9, the first figure of the quot, is tens; in Var. 2. the figure 0 is in the place of units, and so 9 is units; in Var. 3. the figure 0 is in the place of primes, and so 9 is primes, &c.

Here observe that 9, the first significant figure of the quot, in all the above varieties, as well as in the varieties that follow, must always be considered, in multiplying the divisor, as an integer; and, in pointing the first product, no decimal place is to be allowed for it.

We shall now keep the dividend an integer, and give decimal places to the divisor.

In Var. 1. say, $9\; \backslash times\; 3.5\; =\; 31.5$; and the unit 1 standing under the place of thousands, the figure 9 is also thousands; and as 0 at last remains, annex a cipher for the decimal .5 in the divisor; then dividing, you get 0 to the quot; and because 9 stands in the place of thousands, the quot is wholly integers.

In Var. 2. the unit 3 stands under the place of ten-thousands, and so 9 is ten-thousands; and to the remainder 0 annex .00, for the two decimal places in the divisor; then dividing, you get 00 to the quot; and because 9 stands in the place of ten-thousands, the quot continues to be wholly integers. The process is the same in Var. 3. and 4.

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

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

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

APPROXIMATES.

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

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

But here it is to be observed, that the uncertain car-

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

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

Example.

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

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

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

Ex. 1. Divide 95.432756463275 by 3.4637528; and limit the quot to four decimal places.

Contracted

In the above example the units of the first product standing under the place of tens, the first figure of the quot is tens; and hence it is easy to foresee, that six figures of the dividend retained will give four decimal places to the quot; and accordingly cut off all the other figures on the right of the dividend; cut off likewise from the divisor two figures that correspond to them.

At every new dividend, drop or omit a figure on the right of the divisor, and mark the figure so dropped by setting a point under it; and in multiplying the quotient-figure 7 into the divisor, say, 7 times 7 is 49, and 3 of carriage from the right, (arising from $7\; \backslash times\; 5\; =\; 35$), makes 52; so set down 2, and carry 5. The same method is observed in multiplying every other quotient-figure into the divisor.

In working the same example at large, the line AB shows how far the operation is contracted, and how much labour is saved.

But here observe, that by the rule for approximates the certain places of the quot are no more than five, viz. 27.551. And therefore, in all operations of this kind, care should be taken to limit the quot to so many places certain; as is done in the following example.

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

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

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

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

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

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

Given multiplier 125) 1.000(.008 divisor sought.

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

.008) 7315.000(914375 quot.

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

her will give a quot. 125 times greater than the dividend; that is, the quot will be equal to the product of the dividend multiplied by 125.

PROB. II. From a given divisor to find a multiplier that gives a product equal to the quot.

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

EXAMP. What multiplier will give a product equal to the quot arising from the same number divided by .008?

Given divisor .008)1.000(125 multiplier sought.

Now, if any number be multiplied by 125, and the same number be divided by .008, the product and quot will be equal; as appears in the example following.

RULE II. If a finite divisor divide a repeating dividend, work as in integers; but in continuing the division, instead of annexing ciphers to the remainder, annex the repeating figure of the dividend.

RULE III. If a finite divisor divide a circulating dividend, work as in integers; but in continuing the division, instead of annexing ciphers to the remainder, annex the circulating figures of the dividend.

RULE IV. If the divisor be interminate, reduce it to a vulgar fraction, as taught in reduction of decimals, Prob. V.; then multiply the given dividend by the denominator, and divide the product by the numerator.

Here there are six cases; for the divisor may either repeat or circulate, and may divide a finite, a repeating, or circulating dividend.

CASE I. When a repeating divisor divides a finite dividend.

EXAMP. Divide 23.5 by $.4\; =\; \backslash frac\{4\}\{10\}$

CASE II. When a repeating divisor divides a repeating dividend.

EXAMP. Divide 43.28 by $.3\; =\; \backslash frac\{3\}\{10\}$

CASE. III. When a repeating divisor divides a circulate.

EXAMP. Divide 92.518, by $.4\; =\; \backslash frac\{4\}\{10\}$

CASE IV. When a circulate divides a finite dividend.

EXAMP. Divide 9 by $.45\; =\; \backslash frac\{45\}\{100\}$

CASE V. When a circulate divides a repeating dividend.

EXAMP. Divide 5.83 by $.72\; =\; \backslash frac\{72\}\{100\}$

CASE VI. When a circulate divides a circulate.

EXAMP. Divide .962, by $.18\; =\; \backslash frac\{18\}\{100\}$

When the circle of the quot is likely to run on to many places, you may stop the operation, and complete the quot by a vulgar fraction; as in the following example.

EXAMP. Divide 34.56097, by $3.592\; =\; \backslash frac\{3592\}\{1000\}$

The quot would run on to 49 figures of a finite part, and then a circle of 65 places; but limit it at seven places of decimals, and then complete it by a vulgar fraction; as follows, viz.

Complete the partial remainder 2724 by annexing to it the circle of the dividend, and placing both, by way of numerator, over the divisor 3589.

The numerator of this complex fraction being a mixt number, reduce it to an improper fraction, by multiplying 2724 by the denominator 99999, and adding the numerator 46341 to the product; as in the margin: and then, instead of the mixt number, the numerator of the complex fraction will be $\backslash frac\{272446341\}\{99999\}$. Or rather work thus: Esteem 2724.46341, a circulate; and then you find the numerator of the vulgar fraction by subtracting the finite part.

Next divide this fractional numerator by the denominator; which is done by multiplying 3589 by 99999, as in the margin; and now the simple vulgar fraction to be annexed to the partial quot is

If the quot thus completed be multiplied by the divisor, it will produce the dividend.

VII. Decimal Practice.

THE price of goods or merchandise may be cast up decimally by any of the methods following.

METHOD I. Find the decimal of the rate, viz. the value of one yard, one pound, one piece, &c.; and this decimal of the rate multiplied into the number or quantity of the goods gives the price.

Ex. 1. At 3 s. 4 d. what cost 346?

TABLE of RATES and DIVISORS.

In the above table the pence stand in the left-hand column, and the farthings on the head, and the divisors in the angle of meeting; which are to be understood and read as follows.

3, 8, 40. Divide the given number of goods by 3, divide the quot by 8, and again divide this last quot by 40.

Ex. 2. At 6 s. 8 d. what cost 439?

The decimal of the rate is $\backslash frac\{2\}\{3\}$ L. s. d.
$3)439(146.\backslash frac\{2\}\{3\}\; =\; 146\; \backslash quad\; 6\; \backslash quad\; 8$

METHOD II. When the rate consists of pence and farthings, find how often it is contained in one pound Sterling, divide the given number of goods by this number, or by its component parts, or work by aliquot parts, and the result will be the price sought.

We confine this method to such rates as consist of pence and farthings, because when the rate consists of shillings, pence, and farthings, or of pounds, shillings, pence, &c. it is shorter and easier to work by Method I.

To make the practice ready and easy, it will be proper to have at hand a table of rates and divisors, such as the following one.

4, 30, -8. Divide the number of goods by 4, divide the quot by 30, and from this last quot subtract one 8th of itself.

80, +12. To an 80th add a 12th of that 80th.

4, 30, +4. To a 30th of a 4th add a 4th of that 30th.

60, + 4 of 4, Divide by 60, divide again the quot by 4, and to the first quot add a 4th of the second quot.

80 x 3, — 12 of 80. Divide by 80, multiply the quot by 3, and from the product subtract a 12th of the first quot.

Ex. 1. At 1 s. per yard, what cost 432 yards?

Ex. 2. At 3 s. what cost 728.5?

METH. III. The third method is by decimal tables of rates suited to the nine digits; such as those composed and published by the Rev. Mr George Brown in 1718, under the title of Arithmetica Infinita, and recommended by Dr John Keill professor of astronomy in the university of Oxford.

These tables are still extant, and extend from 1 farthing to 20s; a short specimen of which, with their construction, and the manner of using them, we shall here subjoin.

Decimal Table of Rates, 1 l. the integer.

In the left-hand column stand the nine digits; and on the right of 1 are the decimals of the respective rates on the head. Thus, .57083 is the decimal of 11s. 5d. one pound being the integer; and .571875 is the decimal of 11s. 5$\backslash frac\{1\}\{4\}$d. &c. Those decimals opposite to 1 being multiplied through the nine digits, make up or compose the rest of the table.

The superior excellency of tables thus constructed is, that we multiply or divide by 10, 100, 1000, &c. by moving the decimal point so many places to the right or left as there are ciphers in the multiplier or divisor.

Hence the price or value of any number of yards, or other things, denoted by a single digit, or by any of its decuples, may be readily found. Thus, the price of 7, 70, 700, 7000, 70000, 700000 yards, at 11s. 5d. per yard, is found as follows.

Now, every number may be resolved into decuples of the several digits of which it is composed; find therefore the price of each decuple by itself, as already taught, and their sum will be the price of the whole.

EXAMP. 1. Required the price of 7956 yards, at 11s. 5$\backslash frac\{1\}\{4\}$d. per yard.

EXAMP. 2. How much money will one spend in a year, or 365 days, at the rate of 11s. 5$\backslash frac\{1\}\{4\}$d. per day?

Tables of this sort may be framed for a great variety of useful purposes, and are easily constructed.

Thus, suppose a table wanted for showing the daily income of any annuity, or yearly pension; in this case, divide 1 by 365, and the quot is the income of 1 l. annuity for one day; and by multiplying this quot through the nine digits, the table is constructed as follows.

TABLE.

The use of the table will best appear by examples; which take as follows.

Example 1.

If one has a yearly pension of 375 l. what is his daily income?

Example 2.

The yearly rent of a gentleman's estate is 963 l. 10s. what can he afford to spend per day?

L. 900

If the income for any number of days be required, find the income for one day as above; and multiply the decimal answer by the given number of days. Or, multiply the yearly pension by the given number of days, and use the product as the yearly pension. Thus, in Ex. 2. if the gentleman's income for 64 days be demanded, you may either multiply 2.6532 by 64; or multiply 968.5 by 64; and then work for the product as follows.

The decimals in the table being circles of eight figures, we have used them as approximates, by confining the operations to four decimal places; which, in affairs of this kind, is sufficiently accurate.

If the annual interest of any principal sum be considered as the yearly pension, the interest of the same principal for any number of days may be found by the table as taught above.

The interest of any principal sum for a year is easily found, as being always the hundredth part of the product of the principal multiplied by the rate per cent.

EXAMPLE. Required the interest for 26 days of 685 l. principal, at 5 per cent.

DUODECIMALS.

Decimal practice may be used with great advantage in the multiplication and division of duodecimals, where the integer is divided into twelve equal parts, called primes, and each prime into twelve seconds, each second into twelve thirds, &c.

For the ready conversion of primes, seconds, thirds, &c. into decimals of the integer, the following table is constructed.

Decimal table of primes, seconds, &c.

In the column of fourths, the decimals run on to eight places of a finite part, and nine figures of a circle; but the finite part by itself, which alone is inserted in the table, will be found sufficient; and in the column of thirds too, the circle of three figures may in most cases be neglected.

I. Multiplication.

Example 1.

What is the product of 247 by 18 5/7.

In working by the inverted method, for the repeating $\backslash beta$ in the multiplier, take $\backslash frac\{1\}\{3\}$ of the multiplicand. The result wants very little of the true answer.

Examp. 2.

Multiply 18 6 by 2 4, and 2 3 continually.

II. Division.

Examp. 1.

Divide 452 8 11 = 452.74308
by 18 5 = 18.41$\backslash beta$ = $18.\backslash overline\{41\}$

Examp. 2.

Divide 97 1 6 = 97.125
by 2 3 = 2.25 and the quot
by 18 6 = 18.5

SEXAGESIMALS.

Decimal practice might likewise be used to good purpose in the arithmetic of sexagesimals, as it would shorten and facilitate the operations.

Sexagesimals, strictly speaking, are degrees, minutes, seconds, thirds, &c. where each degree is divided into 60 minutes, and each minute into 60 seconds, &c.; but under this title is also usually comprehended the division of a sign into 30 degrees. They are commonly marked as under.

Signs. deg. min. sec. thirds.

7 24 36 54 48 &c.

Sexagesimals properly belong to astronomy, being used in computations of motion and time, where the degree of motion, and hour of time, are equally divided into 60 minutes. The preference of the decimal method to that of the sexagesimal will appear from the following example of addition done both ways.