an obsolete name of a species of clupea. See CLUPEA.
ARITHMETICK.
Arithmetick is a science which explains the properties of numbers, and shews the method or art of computing them.
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 consequently 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 contradiction to the cipher, are called significant figures.
The value of the figures now aligned 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
Table. TABLE.
3d Period. 2d Period. 1st Period.
Billions. Millions. Units.
Hundred thousands. Ten thousands. Hundreds. Tens. Units. Hundred thousands. Ten thousands. Hundreds. Tens. Units. Hundred thousands. Ten thousands. Hundreds. Tens. Units.
8 1 3 , 7 0 0 : 2 3 7 , 8 9 4 : 6 7 8 , 0 4 0 .
The table may be expressed in a more concise form thus,
3. 2. 1. Per.
Billions. Millions. Units.
8 1 3 , 7 0 0 : 2 3 7 , 8 9 4 : 6 7 8 , 0 4 0 .
From the table it is obvious, that though a cipher signify 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 $\frac{1}{2}$, $\frac{3}{4}$, $\frac{9}{10}$.
3. A mixt number is an integer with a fraction joined to it; as $4\frac{1}{2}$, $7\frac{3}{4}$, $48\frac{9}{10}$.
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 composites 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.
Example 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 let below in the place of units; then 3 tens and 5 tens make 8 tens, which let below in the place of tens; then 2 hundreds and 4 hundreds make 6 hundreds, which let below in the place... 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 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 37; 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.
### Money
#### Table
| 4 farthings | 1 penny | |-------------|---------| | 12 pence | 1 shilling | | 20 shillings | 1 pound |
Marked thus.
1. s. d. f. or q.
1 = 20 = 240 = 960
Note. The above mark signifies equal to.
1. is put for libra, a pound; d. for denarii, 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
| f. | d. | s. | l. | |----|----|----|---| | 4 | 1 | 12 | 20 | | 8 | 2 | 24 | 40 | | 12 | 3 | 36 | 60 | | 16 | 4 | 48 | 80 | | 20 | 5 | 60 | 100 | | 24 | 6 | 72 | 120 | | 28 | 7 | 84 | 140 | | 32 | 8 | 96 | 160 | | 36 | 9 | 108| 180 | | 40 | 10 | 120| 200 |
**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 \( \frac{1}{4} \) denotes one farthing, \( \frac{3}{4} \) two farthings, and \( \frac{3}{4} \) 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 choose to divide them... them into parcels; then cast up each parcel separately, and afterwards add the sums of the several parcels into one total.
2. AVOIRDUPOIS WEIGHT.
TABLE.
| 16 drams | 1 ounce. | |----------|---------| | 16 ounces | 1 pound. | | 28 pounds | make 1 quarter. | | 4 quarters | 1 hundred. | | 20 hundreds | 1 tun. |
Marked thus.
| T. C. | lb. | oz. | dr. | |-------|-----|----|----| | 1 = 20 | 80 = 2240 = 35840 = 573490 | | 1 = 4 | 112 = 1792 = 28672 |
By Avordupois weight are weighed butter, cheese, rosin, wax, pitch, tar, tallow, soap, salt, hemp, flax, beef, brafs, iron, steel, tin, copper, lead, alum, and all grocery wares.
Note, 10½ 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 excesses 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 set down.
| (10) (20) (4) (28) (16) | |------------------------| | T. C. | lb. | oz. | |-------|-----|----| | 74 | 19 | 3 | | 85 | 17 | 2 | | 68 | 13 | 1 | | 52 | 18 | 3 | | 48 | 9 | 3 | | 97 | 5 | 1 |
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 excesses 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 excesses and 20 is 25, and 4 is 29; where dot, and proceed, saying, 1 of excesses 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.
III. Proof of Addition.
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.
| Top-line | L. s. d. | |----------|---------| | 748 | 15 | 10½ | | 674 | 13 | 11½ | | 835 | 17 | 9½ | | 90 | 18 | 8 |
Total 2350 6 3½
Total without the top-line 1601 10 4½
Proof 2350 6 3½
Note, This mark + signifies added to.
3. Addition is also proved by calling 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's, or any just number of 9's, this kind of proof will not discover the mistake. CHAP. III. SUBTRACTION.
Subtraction is the taking a lesser number from a greater, in order to discover their difference, or the remainder.
I. Subtraction of Integers.
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.
Example 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 effecting 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.
II. Subtraction 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., 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.
1. MONEY.
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, &c., 48 12 6 2 minor. and in each of these denominations, units under 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 (10) (20) (12) 4, the number of farthings L. s. d. in 1 penny borrowed, and 708 14 6½ major. 1 remains; which 1 added 170 17 10¼ minor. to 1 in the major gives 2 farthings for a remainder. 429 16 7½ 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.
2. AVOIRDUPOIS WEIGHT.
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. 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 quar- ters 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. borrow- ed 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 revisal of the work, or running over it, a se- cond time; but it is usual in schools to put the learner upon proving the operation, by some of the three me- thods 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.
| Examp. | L. | s. | d. | |--------|----|----|----| | major | 73 | 15 | 10 | | minor | 48 | 12 | 6 | | rem. | 25 | 3 | 4 | | proof | 73 | 15 | 10 |
2. By subtraction; for if you subtract the remainder from the major, the difference will be equal to the minor, as follows.
| | L. | s. | d. | |--------|----|----|----| | 5847 | major | 73 | 15 | 10 | | 2569 | minor | 48 | 12 | 6 | | 3278 | rem. | 25 | 3 | 4 | | 2569 | proof | 48 | 12 | 6 |
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 cros, and then cast the 9's out of the minor and re- mainder, as if they were items in addition, and place the excess at the foot of the cros, 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.
| | L. | s. | d. | |--------|----|----|----| | major | 73 | 15 | 10 | | minor | 48 | 12 | 6 | | rem. | 25 | 3 | 4 |
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 fi- gures, the numbers arising from the multiplication of these several figures into the multiplicand, are called par- ticular, or partial product; 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 con- tains unity.
Hence multiplication supplies the place of many addi- tions; 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, $5 \times 3 = 15 = 5 + 5 + 5$.
The first and lowest step in multiplication is, to mul- tiply 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 con- sulted 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.
| TABLE | |-------| | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 | 22 | 24 | | 3 | 6 | 9 | 12 | 15 | 18 | 21 | 24 | 27 | 30 | 33 | 36 | | 4 | 8 | 12 | 16 | 20 | 24 | 28 | 32 | 36 | 40 | 44 | 48 | | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 | 55 | 60 | | 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | 54 | 60 | 66 | 72 | | 7 | 14 | 21 | 28 | 35 | 42 | 49 | 56 | 63 | 70 | 77 | 84 | | 8 | 16 | 24 | 32 | 40 | 48 | 56 | 64 | 72 | 80 | 88 | 96 | | 9 | 18 | 27 | 36 | 45 | 54 | 63 | 72 | 81 | 90 | 99 | 108 | | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 | 110 | 120 | | 11 | 22 | 33 | 44 | 55 | 66 | 77 | 88 | 99 | 110 | 121 | 132 | | 12 | 24 | 36 | 48 | 60 | 72 | 84 | 96 | 108 | 120 | 132 | 144 |
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 tens, 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, \(23 \times 10 = 230\); and \(384 \times 100 = 38400\); and \(745 \times 1000 = 745000\).
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.
| Ex. 1 | Ex. 2 | Ex. 3 | |-------|-------|-------| | Mult. | 47 | Mult. | 627 | Mult. | 999 | | by 9 | 470 | by 99 | 62700 | by 999| 999000| | Sub. | 47 | Sub. | 627 | Sub. | 999 | | Prod. | 423 | Prod. | 62073 | Prod. | 998001|
From Ex. 3, we may learn, in general, that to multiply any number consisting entirely of 9's by itself, is to let 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:
Multiply \(7439\) by 5 — \(74390\) Product \(37195\)
Multiply \(9856\) by 15 — \(98560\) add \(49280\) Product \(147840\)
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.
Ex. 1. \(876\) or multiply by \(876\) or thus, \(876\) \(11\) \(11\) thus \(876\) \(9636\) Ex. 3. \(435\) \(27\) \(11745\) Ex. 5. \(7234\) or thus, \(7234\) \(112\) \(810208\) Ex. 2. \(876\) \(876\) \(9636\) Ex. 4. \(241\) \(34\) \(8194\) Ex. 6. \(263\) \(119\) \(31297\) Ex. 7. \(745\) \(103\) \(76735\) 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 —— 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.
Ex. 1. Ex. 2. Ex. 3. 7645 4983 38 33 666 4444 —— ——— ——— 22935 pr. by 3. 29898 pr. by 6. 152 pr. by 4. 25228 total. 3318679 total. 168872 total.
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.
Ex. 1. Ex. 2. Ex. 3. 438 374 746 87 56 84 —— ——— ——— 3066 2244 2984 3504 1870 5968 —— ——— ——— 38106 20944 62664
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 7 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; let 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 × 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.
| L. | s. | d. | |----|----|----| | 2 | 14 | 9½ |
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 × 7 × 2; or by 7 × 3 × 3 × 2.
| L. | s. | d. | |----|----|----| | 3 | 8 | 4 |
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?
| L. | s. | d. | Price of | |----|----|----|----------| | 19 | 6½| 1 yd. | × 4 = 3 18 2 4 yds. | | | | | | | 9 | 15| 5 | 10 yds. | × 0 = | | | | | | | 97 | 14| 2 | 100 yds.| × 6 = 586 5 600 yds. | | | | | | | 977| 1 | 8 | 1000 yds.| × 8 = 7816 13 4 8000 yds. |
Price of 8604 yards, 8406 16 6
Vol. I. No. 16. 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.
Example 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?
| F. | in. | li. | |----|-----|----| | 24 | 7 | | | 18 | 5 | | | 432| 35 | | | 20 | 72 | | | 452| 107 | |
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.
Example 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?
| F. | in. | li. | |----|-----|----| | 38 | 9 | | | 23 | 8 | | | 874| 72 | | | 42 | 84 | | | 278| 8 | | | 919| 98 | |
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?
| Inch. | lin. | |-------|------| | 465 | 6 | | 284 | 6 | | 132060| 36 | | 374 | 72 | | 132434| 108 |
The answer here is 132434 square inches, and 108 square lines; and if the inches be divided by 144, you will have 910 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 1 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.
Example III. In a piece of timber, whose length is 18 feet 6 inches, breadth 2 feet 4 inches, and thickness 2 feet 3 inches, how many solid feet?
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.
Example 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?
| F. | in. | l. | |----|-----|---| | 7 | 9 | 5 | | 56 | 27 | 1287(7F.) | | 7 | 36 | 12x3=36 in. | | 35 | | 7x5=35 in. by Rule III. | | | 36 | 3x5=15 | | | | 12)15(1 in. by Rule IV. | | 63 | 95 | 36 sup. 12x3=36 li. | | | | 5 lin. | | 189| 180 | 63x5=315, and 144)315(2F.) by R.VII. | | | | 12x27=324 in. | | 2124| 2108| 3x99=297, and 144)297(2F.) by R. IX. | | | | 12x9=108 in. | | 41432| 9 | 3x36=108, and 12)108(9 in. by Rule X. | | | | 5x99=495, and 12)495(41 in. by R.VIII. | | 193482| 612| 144x3=432 lines. |
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 cros 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 cros; 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 cros; then casting the 9's also out of the product of your multiplication, place its excess at the bottom; 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 excesses 7 on the right side of the cross; then cast the 9's out of the multiplier, and place the excesses 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.
**Examp. II.** Here, in casting the 9's out of the multiplicand, and out of the product, begin with the pounds, and reduce the excess to shillings, and in like manner the excess of the shillings is reduced to pence, and 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.
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**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:
\[ \begin{array}{c} 18 \\ - 6 \\ \hline 12 \\ - 6 \\ \hline 6 \\ \end{array} \]
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?
\[ \begin{array}{c} 18 \\ \end{array} \]
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.
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**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,
\[ \begin{array}{c} 7)875(125 \\ \text{Dividend} \\ \text{Quotient} \\ \end{array} \]
\[ \begin{array}{c} 7)875(100 \\ \text{1st dividual} 800 \\ \text{Partial quoits} \\ \end{array} \]
\[ \begin{array}{c} 7)875(100 \\ \text{Rem. 100 125 total quoit.} \\ \end{array} \]
\[ \begin{array}{c} 7)875(100 \\ \text{Add 70} \\ \end{array} \]
\[ \begin{array}{c} 7)875(100 \\ \text{Rem. 30 add 5} \\ \end{array} \]
\[ \begin{array}{c} 7)875(100 \\ \text{3rd dividual 35} \\ \end{array} \] is 35, in which the divisor 7 is contained 5 times. Now it is evident, that the sum of the partial quotients, 125, is the total quotient, 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.
Examp. II. Here,
\[ \begin{array}{c} 8)56032897(7004112 \\ \hline 56 \\ \hline '032 \\ 32 \\ \hline '8 \\ 8 \\ \hline '9 \\ 8 \\ \hline 17 \\ 16 \\ \hline (1) \end{array} \]
num. denom.
because 8 is not contained in 5, point off 56 as the first dividual, and say, How often 8 in 56? Ans. 7; which put in the quotient; then multiply 7 into the divisor 8, and subtract the product 56 from the dividual; and as nothing remains, bring down the next figure of the dividend, which happens to be a cipher; and as you cannot have 8 in 0, put 0 in the quotient; 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 divisor 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 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 I 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 I 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 I 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; wherefore 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:
Quot rem. 10)748963.4 100)74896.34 1000)7489.634 10000)748.9634
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.
**Example.**
Divide 74678 by 98.
New divisor 100
Given divisor 98
Complement 2
Carriage 2×2 complement = 4
**Explanation.**
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.
**II. Division of the parts of Integers.**
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.
**Example. 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
L. s. d. f. 34 6 8 2½ quotient.
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
**Example. II.** If C. 42 : 2 : 8 of tobacco be made up into 5 equal hds., what will be the neat weight of each hd?
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.
C. Q. lb. C. Q. lb.
5)43 2 8 (8 2 24)
— 40
— 3 rem.
— 4
— 14
— 10
— 28
— 120
— 10
— 20
— 20
— (o)
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.
**III. The Proof of Division.**
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 crofs; 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 crofs; 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 crofs 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 revisal 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.**
1. **Money.**
**Quest. 1.** In L. 472 how many shillings, pence, and farthings?
This reduction is descending, therefore multiply the pounds by 20, because 20 shillings make 1 pound, and the product is shillings: then multiply the shillings by 12, because 12 pence make 1 shilling, and the product is pence: lastly, multiply the pence by 4, because 4 farthings make 1 penny, and the product is farthings.
| | | |--------|--------| | 472 | | | 9440 | | | 113280 | | | 453120 | |
**Proof by Reduction ascending.**
In 453120 farthings how many pence, shillings, and pounds?
Here divide the farthings by 4, because 4 farthings make 1 penny, and the quotient is pence: then divide the pence by 12, because 12 pence make 1 shilling, and the quotient is shillings: lastly, divide the shillings by 20, because 20 shillings make 1 pound, and the quotient is 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 4.
| | | |--------|--------| | 4)453120 | | | 113280 | | | 9440 | | | 472 | |
By note 5.
| | | |--------|--------| | 96)453120 | | | 384 | | | 691 | | | 672 | | | 192 | | | 192 | |
By note 6.
| | | |--------|--------| | 48)453120 | | | 9440 | | | 432 | | | 472 | | | 192 | | | 192 | |
2. **Avoir-** 2. AVOIRDUPOIS WEIGHT.
Quest. 1. In C. 47 : 1 : 20 how many ounces?
| C. | Q. | lb. | |----|----|-----| | 47 | 1 | 20 |
\[ \frac{4}{189} \]
\[ \frac{28}{1512} \]
\[ \frac{380}{5312} \]
\[ \frac{16}{31872} \]
\[ \frac{5312}{84992} \]
PROOF.
In 84992 ounces how many lb. Q. and C.
\[ \frac{28}{16(84992)} \]
\[ \frac{4}{5312} \]
\[ \frac{189}{1512} \]
\[ \frac{47}{380} \]
\[ \frac{1}{20} \]
\[ \frac{80}{28} \]
\[ \frac{28}{16} \]
\[ \frac{49}{251} \]
\[ \frac{48}{224} \]
\[ \frac{28}{252} \]
\[ \frac{19}{272} \]
\[ \frac{16}{252} \]
\[ \frac{32}{(20) \text{ lb.}} \]
\[ (0) \]
Mixt Reduction.
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.
Quest. In 764l. how many guineas?
Here the given name is pounds, the name sought is guineas, and the third name, to which the pounds are reduced, is shillings; for a shilling is an aliquot part both of a pound and of a guinea.
\[ \frac{21}{15280} \]
\[ \frac{727}{764} \]
\[ \frac{147}{58} \]
\[ \frac{42}{160} \]
\[ \frac{147}{(13) \text{ shillings.}} \]
PROOF.
In 727 guineas 13 shillings, how many pounds?
Guineas. Shill.
\[ \frac{727}{21} \]
\[ \frac{730}{1455} \]
\[ \frac{20}{15280} \]
\[ 764 \text{ pounds.} \]
CHAP. VII. The Rule of Three.
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.
Sect. I. The Simple or Single Rule of Three.
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 \( \frac{1}{2} \), 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.
\[ 4 : 2 :: 16 : 8 \]
\[ 6 : 9 :: 12 : 18. \]
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 $2 \times 9 = 3 \times 6 = 18$. 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 $5 \times 36 = 180 = 6 \times x$; therefore, 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.
It is easy to perceive that the answer must be greater than the middle term; for 6 yards will cost more than 4 yards; therefore the least extreme, viz. 4 yards, is the divisor, according to Rule II.
| Yds. | s. | yds. | |------|----|-----| | 4 | 12 | 6 |
$\frac{4}{72}(18$ shillings Anf.
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.
It is obvious, that the answer must be less than the middle term; for 5 C. will cost less than 7 C.; and therefore the greatest extreme, viz. 7 C. is the divisor.
| C. | L. | C. | |----|----|----| | 7 | 21 | 5 |
$\frac{7}{105}(15$ l. Anf.
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?
\[ \text{If } 13 : 21 :: 27 \]
*Rem. 4 s.*
\[ \begin{array}{c} 27 \\ 147 \\ 42 \\ \hline 13)569(43L. \end{array} \]
Rem. 9 d.
\[ \begin{array}{c} 52 \\ 47 \\ 39 \\ \hline 13)36(2f. \end{array} \]
Rem. 8 L.
\[ \begin{array}{c} 20 \\ 13)160(12s. \end{array} \]
Rem. 10 f.
\[ \begin{array}{c} 30 \\ Anf. 43. 12 3 2\frac{1}{3} \end{array} \]
*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?
\[ \begin{array}{c} Feet. yds. feet. 9 : 36 :: 3. \end{array} \]
\[ \begin{array}{c} 3 \\ 9)108. \end{array} \]
Anf. 12 yards.
**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 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 quotient 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. I. 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; therefore, according to Rule I. place 56 bushels in the middle.
The remaining four terms are extremes, which you clas 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.
Joint operation.
| Horses | bushels | horses | |--------|---------|--------| | 14 | 56 | 20 | | da. | 16 | 24 da. |
84 480
14 56
224 288
240
224)26880(120
224
448
448
(c)
Ans. 120 bushels.
The quotient or answer comes out of the same name with the middle term, viz. 120 bushels. The two simple questions into which the compound question is resolved, are stated, and wrought separately, as follows.
| H. B. | H. Days. B. Days. | |-------|------------------| | If 14 : 56 :: 20 | If 16 : 80 :: 24 |
\[ \begin{align*} & 20 \\ & \frac{14}{112} \text{(80 B.)} \\ & \frac{16}{112} \text{(120 B.)} \end{align*} \]
Anf. 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 clasps 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 eaten up in 16 days, in how many days will 120 bushels be eaten 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 eaten 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.
\[ \begin{align*} & \text{Joint operation.} \\ & \text{Horf. days. horf.} \\ & * 20 : 16 :: 14 * \\ & \text{bush. 56} \\ & \text{1120} \\ & \text{16} \\ & \text{1008} \\ & \text{168} \\ & \text{days.} \\ & \text{112 (o) 2683 (o) (24 An.)} \\ & \text{224} \\ & \text{448} \\ & \text{448} \end{align*} \]
III. The Rule of Seven, Nine, &c.
Quest. If 15 men eat 150 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 five terms are extremes, which clasps 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.
**Joint operation.**
| Men. days. men. | |-----------------| | *30 : 6 :: 15* | | *10 : d. 156* | | 520 d. : 12* |
\[ \begin{align*} 4680 & \quad 30 \\ 10 & \quad 75 \\ \hline 46800 & \quad 7800 \\ 12 & \quad 93600 \\ 6 & \quad 936 \\ 936 & \quad 936 \\ (0) & \end{align*} \]
This compound question is resolved into three simple ones, as follows.
\[ \begin{align*} 30 : 6 :: 15 : 3 \\ 156 : 3 :: 520 : 10 \\ 10 : 10 :: 12 : 12 \text{ days. Ans.} \end{align*} \]
**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 Tolouse, and 100 lb. of Tolouse 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 clasped into similar pairs, and stated as follows.
\[ \begin{align*} \text{Ven. Ly.} & \quad 100 : 74 :: 100 \\ \text{Roan} & \quad 80 : 100 \\ \text{Roan} & \quad 70 : 100 \\ 8000 & \quad 10000 \\ 120 & \quad 70 \\ 960000 & \quad 700000 \\ 100 & \quad 100 \\ 9600000 & \quad 7000000 \\ 74 & \quad 74 \\ 96000000 & \quad 518000000 (53\frac{2}{3} \text{ lb. of Geneva. Ans.}) \end{align*} \]
But the question becomes more simple, and is wrought with greater ease and advantage, by being stated in the fractional form, as follows.
\[ \begin{align*} 100 \times 70 \times 100 \times 100 \times 74 = \frac{70 \times 100 \times 74}{100 \times 120 \times 80 \times 100 \times 120 \times 80} \\ 7 \times 10 \times 74 = \frac{5180}{53\frac{2}{3}} \text{ lb. of Geneva. Ans.} \end{align*} \]
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. \(100 \times 12\), 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. \(75 \times 9\), 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.
If \(100 \times 12 : 5 :: 75 \times 9\)
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**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.
**Ques.** A and B make a joint stock: A puts in 12l. and B 8l.; they gain 5l.: What is each man's share?