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

| Ex. | 7392 | 42785 | 37846 | 93956 | |-----|------|-------|-------|-------| | | 365 | 91 | 235 | 8704 | | | 36645| 42785 | 189230| 375824| | | 43974| 385065| 118588| 657692| | | 21987| 3893435| 75692 | 751648| | | 2675085| 8893810| 817793024|

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

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

**Contractions and Varieties in Multiplication.**

1st. If the multiplier be a composite number, we may multiply successively by the component parts.

| Ex. | 7638 by 45, or 5 times 9 | |-----|--------------------------| | | 45 | | | 38190 | | | 36552 | | | 343710 |

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

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

3dly, To multiply by 9, which is one less than 10, we may annex a cipher, and subtract the multiplicand from the number it composes. To multiply by 99999, or any number of 9's, annex as many ciphers, and subtract the multiplicand. The reason is obvious; and a like rule may be found though the unit place be different from 9.

4thly, Sometimes a line of the product is more easily obtained from a former line of the same than from the multiplicand.

| Ex. 1st. | 1372 | 2d. | 1348 | |----------|------|-----|------| | | 84 | | 36 | | | 5488 | | 8088 | | | 10976| | 4044 | | | 115248| | 48528|

In the first example, instead of multiplying by 8, we may multiply 5488 by 2; and, in the second, instead of multiplying by 3, we may divide 8088 by 2.

5thly, Sometimes the product of two or more figures may be obtained at once, from the product of a figure already found.

| Ex. 1st. | 14356 | 2d. | 3462321 | |----------|-------|-----|---------| | | 648 | | 96484 | | | 114848| | 13849284| | | 918754| | 166191408| | | 9302088| | 332382816|

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

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

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

| Ex. 1st. | 3125 | or | 3125 | 2d. | 32452 | |----------|------|----|------|-----|-------| | | 642 | | 642 | | 52575 | | | 18750| | 18750| | 162260| | | 12500| | 131250| | 811300| | | 6250 | | 2006250| | 2433900| | | 2006250| | | | 1706163900|

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

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

**Compound Multiplication.**

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

| Ex. 1st. | £15 | 3 | 8 by 32 = 8×4 | |----------|-----|---|---------------| | | | | | | | £121| 9 | 4 = 32 times. | | | | | | | | £485| 17| 4 = 32 times. | | | | | | | | £17 | 3 | 8 by 75 = 5×5×3 | | | | | | | | £51 | 11| 3 times. | | | | | | | | £257| 15| 15 times. | | | | | | | | £1288| 15| 75 times. |

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

Note 2. The operation may be proved by taking the

Reduction component parts in a different order, or dividing the multiplier in a different manner.

RULE III.—If the multiplier be a prime number, multiply first by the composite number next lower, then by the difference, and add the products.

| £35 17 | 9 by 67 = 64 + 3 | | --- | --- | | 8 | 64 = 8 × 8 is 64, we multiply twice by 8, which gives £2296. |

Here, because 8 times 8 16s. equal to 64 times the multiplicand; then we find the amount of 3 times the multiplicand, which is £107. 13s. 8d.; and it is evident that these added amount to 67, the multiplicand.

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

| £17 4 | 5 by 106 = 108 — 2 | | --- | --- | | 12 | 108 = 9 × 12 by 12 and 9, the component parts of 108, and obtain a product of £1859. 17s. equal to 108 times the multiplicand; and, as this is twice oftener than was required, we subtract the multiplicand doubled, and the remainder is the number sought.

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

| £34 8 | 2 by 5 = £172 1 0 4 = 5 times. | | --- | --- | | 10 times | £344 2 1 by 6 = 2064 12 6 = 60 times. | | 100 times | £3441 10 by 4 = 13764 3 4 = 400 times. | | 1000 times | £34410 8 by 3 = 103231 5 = 3000 times. | | | £119232 1 10 4 = 3465 times. |

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

REDUCTION.

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

Ex. To reduce £46. 13s. 8½d. to farthings.

| £46 | 20 | | --- | --- | | 920 shillings in £46 | £46 13 8½ | | 13 | 20 | | 933 shillings in £46 13 | 933 | | 12 | 12 | | 11196 pence in £46 13 | 11204 | | 8 | 4 | | 11204 pence in £46 13 8 | 44819 | | 4 | 3 | | 44816 farthings in £46 13 8 | 44819 farthings in £46 13 8½ |

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

CHAP. V.—DIVISION.

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

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

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

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

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

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

To divide 5936 by 8.

From 5936 Take 5600 = 700 times 8. Rem. 336 From which take 320 = 40 times 8. Rem. 16 From which take 16 = 2 times 8. 0 742 times 8 in all.

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

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

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

3. Subtract the product from the figures above it.

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

Examples.] 1st, 8)5936(742 2d, 63)30114(478 56 ···· 252 ···· 33 491 32 441 16 504 16 504

3d, 365)974932(2671 730 ···· 2449 2190 2593 2555 382 365

Remainder 17

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

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

If there be a remainder, the division is completed by a vulgar fraction, whose numerator is the remainder, and its denominator the divisor. Thus, in Ex. 3 the quotient is 2671, and the remainder 17; and the quotient completed is 2671 17/365.

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

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

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

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

Contractions and Varieties in Division.

1st. When the divisor does not exceed 12, the whole computation may be performed without setting down any figures except the quotient.

Ex. 7)35868(5124 or 7)35868

5124

2dly. When the divisor is a composite number, and one of the component parts also measures the dividend, we may divide successively by the component parts.

Ex. 1st, 30114 by 63 2d, 975 by 105=5×7×3 9)30114 5975 7)3346 3195 Quotient 478 7)65

This method might be also used although the component parts of the divisor do not measure the dividend; but the management of the remainder requires the doctrine of vulgar fractions.

3dly. When there are ciphers annexed to the divisor, cut them off, and cut off an equal number of figures from the dividend; annex those figures to the remainder.

Ex. To divide 378643 by 5200.

5200)378643(7213

364

146 104 4243

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

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

Note.—Since 4 times 25 make 100, instead of dividing by 25, we may multiply the dividend by 4, and cut off the two last figures from the product. The figures left will be the quotient, and those cut off the remainder. In like manner, to divide by 125, which multiplied by 8 produces 1000, we may multiply the dividend by 8, and cut off three figures, which will be the remainder, and those left the quotient.

4thly. When the divisor consists of several figures, we may try them separately, by inquiring how often the first figure of the divisor is contained in the first figure of the dividend, and then considering whether the second and following figures of the divisor be contained as often in the corresponding ones of the dividend with the remainder (if any) prefixed. If not, we must begin again, and make trial of a lower number. When the remainder is nine upwards, we may be sure the division will hold through the lower places; and it is unnecessary to continue the trial farther.

5thly. We may make a table of the products of the divisor, multiplied by the nine digits, in order to discover more readily how often it is contained in each dividend. This is convenient when the dividend is very long, or when it is required to divide frequently by the same divisor. To explain the reason of this, we must recollect, that whatever number of hundreds any dividend contains, it contains an equal number of 9's, together with an equal number of units. In Ex. 1, the dividend contains 3241 hundreds, and a remainder of 23. It therefore contains 3241 times 99, and also 3241, besides the remainder already mentioned.—Again, 3241 contains 32 hundreds and a remainder of 41. It therefore contains 32 99's, and also 32, besides the remainder of 41. Consequently the dividend contains 99, altogether, 3241 times and 32 times, that is, 3273 times, and the remainder consists of 23, 41, and 32, added, which make 96.

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

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

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

**Compound Division.**

**Rule I.**—When the dividend only consists of different denominations, divide the higher denomination, and reduce the remainder to the next lower, taking in (p. 547, Reduction) the given number of that denomination, and continue the division.

**Examples.**

| Divide £465. 12s. 8d. | Divide 345 cwt. 1 qr. 8 lb. | |----------------------|--------------------------| | by 72 | by 22 | | £ 4 6 5 | Cwt. q. lb. | | 12 8 | 3 4 5 | | 6 9 4 | 1 8 | | 432 | 22 | | 33 | | | 20 | | | Quotient | | | 3273 and rem. 96 | | | Quotient 5533-58 rem.| | | 3d, 999)4765923 | | | 476 | | | 476999 | | | Quotient 477 | |

Or we might divide by the component parts of 72 as explained under (2dy, p. 548).

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

| To divide £38. 13s. by | To divide 96 cwt. 1 qr. 20 lb. by 3 cwt. 2 qr. 8 lb. | |------------------------|-----------------------------------------------------| | £3 4 5 | Cwt. q. lb. | | £38 13 | 3 2 8 | | 20 | )96 1 20 | | 64 | 773 | | 12 | 12 | | 773 | )9276(12 quot. | | | 130 | | | 773 | | | 28 | | | 1546 | | | 400 | | | )10800(27 quot. |

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

The use of division is to find either of the factors by whose multiplication a given number is produced when the other factor is given, and therefore is of two kinds, since either the multiplier or the multiplicand may be Division, given. If the former be given, it discovers what number is which is contained so many times in another. If the latter be given, it discovers how many times one number is contained in another. Thus, it answers the questions of an opposite kind to those mentioned under Rule IV, p. 547, as, to find the quantity of a single parcel or share; to find the value, weight, or measure, of a single article; to find how much work is done, provisions consumed, interest incurred, or the like, in a single day, &c.

The last use of division is a kind of reduction exactly opposite to that described under Rule V, p. 547. The manner of conducting and arranging it, when there are several denominations in the question, will appear from the following examples.

1 To reduce 15783 pence to pounds, shillings, and pence. 20 20 12 12)15783(1315(65 24)174865(7286(364(30 12 120 37 115 68 128 04 36 100 48 120 18 15 206 86 12 192 80 63 145 6 60 144 3 1

Answer, £65. 15s. 3d.

2. To reduce 174865 grs. to lbs. oz. and dwt. Troy. 20 20 12 168 60 36 12)174865(7286(364(30 12 120 37 115 68 128 04 36 100 48 120 18 15 206 86 12 192 80 63 145 6 60 144 3 1

Answer, 30 lb. 4 oz. 6 dwt. 1 gr.

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

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

Rule for Mixed Reduction.—Reduce the given denomination by multiplication to some lower one which is an aliquot part of both; then reduce that by division to the denomination required.

Ex.—Reduce £31742 to guineas.

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

Note 1. Guineas may be reduced to pounds by adding one twentieth part of the number. 2. Pounds may be reduced to merks by adding one half. 3. Merks may be reduced to pounds by subtracting one third.

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

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

3) 15 9 price of 3 yards. 5 3 price of 1 yard, by Rule IV, p. 547.

£1 16 9 price of 7 yards (by par. ult. p. 549, col. 2.)

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

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

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

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

1st, If several numbers be divided by any divisor (the remainders being always added to the next number), the sum of the quotients, and the last remainder, will be the same as those obtained when the sum of the number is divided by the same divisor. Thus, 19, 15, and 23, contain together as many 5's, as many 7's, &c. as their sum 57 does, and the remainders are the same; and, in this way, addition may be proven by division. It is from the correspondence of the remainders that the proof by casting out the 9's is deduced.

2dy, If any figure with ciphers annexed be divided by 9, the quotient consists entirely of that figure; and the remainder is also the same. Thus, 40 divided by 9 quotes 4, remainder 4; and 400 divided by 9 quotes 44, remainder 4. The same holds with all the digits, and the reason will be easily understood; every digit, with a ci Arithmetic.

3000 divided by 9 quotes 333, remainder 3 700 9 quotes 77, remainder 7 60 9 quotes 6, remainder 6 5 9 quotes 0, remainder 5

3765 416, sum of rem. 21

Again: 21 divided by 9 quotes 2, remainder 3; wherefore, 3765 divided by 9 quotes 418, remainder 3; for the reason given. Hence we may collect the following rules for practice:

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

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

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

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

Ex. 1st, 9276 res. 6 × 8 = 48 res. 3.

8 74208 res. 3.

2d, 7898 res. 5 × 3 = 15 res. 6.

48 res. 3.

63184 31592

379104 res. 6.

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

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

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

res. 6. 210 390 378

Rem. 12 - - - res. 3.

And the result of the dividend is 6.

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

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

1st, If an error of 9, or any of its multiples, be committed, the results will nevertheless agree; and so the error will remain undiscovered. And this will always be the case when a figure is placed or reckoned in a wrong column, which is one of the most frequent causes of error.

2dly, When it appears by the disagreement of the results that an error has been committed, the particular figure or figures in which the error lies are not pointed out, and consequently it is not easily corrected.

Chap. VI.—Rule of Proportion.

Sect. I.—Simple Proportion.

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

Proportionals are marked down in the following manner:

6 : 3 :: 8 : 4 12 : 36 :: 9 : 27 9 : 6 :: 24 : 16 16 : 24 :: 10 : 15

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

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

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

18 : 27 :: 34 : 51

34 108 81 18918(51) 90 18 18

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

\[ \frac{18}{27} \]

Multiplied by 34, \( \frac{612}{918} \)

Divided by 18, \( \frac{34}{51} \)

The products 612, 918, and the quotients 34, 51, have therefore the same proportion to each other that 18 has to 27. In the course of this operation, the products of the first and third terms are divided by the first; therefore the quotient is equal to the third.

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

When any of the terms is in more denominations than one, we may reduce them all to the lowest. But this is not always necessary. The first and second should not be reduced lower than directed, p. 549, col. 2, par. penult.; and when either the second or third is a simple number, the other, though in different denominations, may be multiplied without reduction.

\[ \begin{align*} \text{Ex. } & 5 : 7 :: 25 : 11 \\ & 7 \\ & 5) 178 \quad 18 \quad 9 \quad (35 \quad 15 \quad 9 \end{align*} \]

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

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

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

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

\[ \begin{align*} \text{H. H. A.} \\ 30 :: 42 :: 12 : \text{answer.} \end{align*} \]

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

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

\[ \begin{align*} \text{d. d. H.} \\ 8 : 5 :: 40 : \text{answer.} \end{align*} \]

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

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

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

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

\[ \begin{align*} \text{Y. Y. s.} \\ 36 : 27 :: 42 \\ 4 : 3 :: 42 \\ 3 \\ 4) 126(31 6, \text{the answer.} \end{align*} \]

**Sect. II.—Compound Proportion.**

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

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

\[ \begin{align*} \text{M. M. B. B.} \\ 18 : 24 :: 6 : 8 \\ 6 \\ 18) 144(8 \end{align*} \]

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

\[ \begin{align*} \text{D. D. B. B.} \\ 28 : 56 :: 8 : 16 \\ 8 \\ 28) 448(16 \end{align*} \]

In the course of this operation, the original number of bolls, 6, is first multiplied into 24, then divided by 18, then multiplied into 56, then divided by 28. It would answer the same purpose to collect the multipliers into one product and the divisors into another, and then to multiply the given number of bolls by the former, and divide the product by the latter, p. 550, col. 2.

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

Men 18 : 24 :: 6 bolls

Days 28 : 56

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

\[ \begin{align*} 144 & 144 \\ 36 & 120 \\ 504 & 1344 \\ 6 \\ 504) 8064(16, \text{answer.} \end{align*} \]

In general, state the several particulars on which the question depends, as so many simple proportions, attending to the sense of the question to discover whether the proportions should be stated directly or inversely; then multiply all the terms in the first rank together, and all those in the second rank together, and work with the product as directed in the simple rule (Sect. i. p. 551.)

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

Men 150 : 100 :: 27 days.

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

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

Ex. If 18 men consume £30 value of corn in 9 months when the price is 16s. per boll, how many will consume £54 value in 6 months when the price is 12s. per boll?

In this question the proportion depends upon three particulars—the value of corn, the time, and price; the first of which is direct, because the greater the value of provisions is, the more time is required to consume them; but the second and third are inverse, for the greater the time and price are, fewer men will consume an equal value.

Value 30 : 54 :: 18 men.

Months 6 : 9.

Here we observe 6 in the first rank measures 54 in the second; so we dash them out, and place the quotient 9 in the second rank. Next, because 30 and 9 are both measured by 3, we dash them out, and place down the quotients 10 and 3; then, because 12 and 16 are both measured by 4, we dash them out and place down the quotients 3 and 4. Lastly, because there is now 3 in both columns, we dash them out, and work with the remaining terms, according to the rule.

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

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

112 : 104 :: 100

100 : 89

110 : 117

If it be required how many lbs. avoirdupois will make 100 lb. of Seville, then the terms must be placed in the different columns thus—

104 : 112 :: 100

89 : 100

117 : 110

SECT. III.—DISTRIBUTIVE PROPORTION.

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

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

A's stock £150

B's 320

C's 350

D's 500

Whole stock £1320

Proof £730

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

Ex. 2. A bankrupt owes A £146, B £170, C £45, D £480, and E £72; his whole effects are only £342 7s. 6d. How much should each have?

A's debt £146 913 : 146 :: £342 7 6 : £54 15

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

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

D's 480 913 : 480 :: 342 7 6 : 160

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

£193 £342 7 6

This might also be calculated by finding what composition the bankrupt was able to pay per pound, which is obtained by dividing the amount of his effects by the amount of his debts, and comes to 7s. 6d., and then finding by the rules of practice how much each debt came to at that rate.

CHAP. VII.—RULES FOR PRACTICE.

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

SECT. I.—COMPUTATION OF PRICES.

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

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

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

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

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

CASE I. When the rate is an aliquot part of a pound, decide the quantity by the number which may be bought for a pound. Table of the aliquot parts of £1.

| Shillings | Pence | |-----------|-------| | 10 | 6s. | | 5 | 3s. | | 4 | 2s. | | 3 | 1s. | | 2 | 1s. | | 1 | 1s. |

Ex. 1st.] What is the value of 7463 yards, at 4s.? 57463 £1492. 12s.

Ex. 2d.] What is the value of 1773 yards, at 3d.? 8(0)773 £222. 8s. 3d.

In the first example we divide by 5 because 4s. is \( \frac{1}{5} \) of a pound; the quotient 1492 shows how many pounds they amount to; besides which there remain three yards at 4s., and these come to 12s. In the second example we divide by 80, as directed, and the quotient gives £22, and the remainder 13 yards, which at 3d. come to 8s. 3d.

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

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

Ex. 1st.] 4573 at 13s. 13 13719 4573 20159449 £2972. 9s.

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

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

**Case III.** When the rate consists of pence only.

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

1 penny is \( \frac{1}{12} \) of a shilling, 2d. of ditto. 3d. of ditto. 4d. of ditto. 5d. is the sum of 4d. and 1d. or of 2d. and 3d. 7d. is the sum of 4d. and 3d. or of 6d. and 1d. 8d. is the sum of 6d. and 2d. or the double of 4d. 9d. is the sum of 6d. and 3d. 10d. is the sum of 6d. and 4d. 11d. is the sum of 6d. 3d. and 2d.

Ex. 1st.] 7423 at 4d. 3)7423 20)2474 £123 14 4

Here, because 4d. is one third of a shilling, we divide by 3, which gives the price in shillings, and reduce these by division to pounds.

Ex. 2d.] At 6d. = \( \frac{1}{2} \) of 1s. At 3d. = \( \frac{1}{3} \) of 6d. At 9d. = \( \frac{1}{3} \) of 1s. At 6d. = \( \frac{1}{2} \) of 1s. At 3d. = \( \frac{1}{3} \) of 6d. At 9d. = \( \frac{1}{3} \) of 1s.

Ex. 3d.] 4856 at 11d. At 6d. = \( \frac{1}{2} \) of 1s. At 3d. = \( \frac{1}{3} \) of 6d. At 2d. = \( \frac{1}{2} \) of 6d. At 11d. = \( \frac{1}{11} \) of 6d.

Ex. 4th.] 3572 at 7½ At 6d. = \( \frac{1}{2} \) of 1s. At 1½d. = \( \frac{1}{2} \) of 6d. At 7½ = \( \frac{1}{2} \) of 1s.

Here we suppose first 6d., Pragie and then 3d. to be paid for each article; half the quantity is the number of shillings they would cost at 6d. each, half of that is the cost at 3d., and these added and reduced give the answer.

Here we calculate what the articles would cost at 6d., at 3d., and at 2d., and add the values.

It is sometimes easier to calculate at two rates whose difference is the rate required, and subtract the lesser value from the greater. Thus, the last example may be wrought by subtracting the value at a penny from the value at a shilling. The remainder must be the value at 11d.

At 1s. At ld. = \( \frac{1}{10} \) of 1s. At 11d. = \( \frac{1}{11} \) of 6d.

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

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

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

Ex. 1st.] 37843 at 1 farthing. 2d.] 23754 at 4d. 4)37843 farthing. 2)23754 halfpence 12) 94603 pence 12) 11877 pence 788 43 899 9 39 8 43 249 9 9 72564 Or, 72564 3 3 At 3d. 36283 d. At 2d. 18141 d. 4)217692 farthing. 12) 54423 pence 12) 54423 4535 3 4535 3 £226 15 3 £226 15 3

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

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

Ex. 1st.] 3287 at 5¼d. At 4d. = \( \frac{1}{4} \) of 1s. At 1d. = \( \frac{1}{4} \) of 4d. At 1f. = \( \frac{1}{4} \) of ld. At 5¼ At 1438 0¾ £71 18 0¾

Ex. 2d.] 4573 at 2½ At 2d. = \( \frac{1}{2} \) of 1s. At 1½d. = \( \frac{1}{2} \) of 2d. At 1½d. = \( \frac{1}{2} \) of ld. At 2½ 1047 11¾ £52 7 11¾

Ex. 3d.] 2842 at 3½ At 3d. = \( \frac{1}{3} \) of 1s. At 3f. = \( \frac{1}{3} \) of 3d. At 3½ 888 1¾ 244 8 1½

Ex. 4th.] 3572 at 7½ At 6d. = \( \frac{1}{2} \) of 1s. At 1½d. = \( \frac{1}{2} \) of 6d. At 7½ 2232 6 £111 12 6 It is sometimes best to join some of the pence with the farthings in the calculation. Thus, in Ex. 4, we reckon the value at 6d. and at 3 halfpence, which makes 7½.

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

**Case VI. When the rate consists of shillings and lower denominations.**

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

**Ex. 1st,** 4258 at 17s. 3d.

| 17 | 29906 | |----|-------| | | 4258 |

17s.

| 3d. = ¼ of 1s. | 1064 6 | |----------------|--------| | 17s. 3d. | 73450 6 |

£3672 10 6

**Ex. 2d,** 5482 at 12s. 4½d.

| 12 | 65784 | |----|-------| | | 1370 6 | | | 685 3 |

12s. 4½d.

| 3d. = ¼ of 1s. | 1370 6 | |----------------|--------| | 11d. = ½ of 3d. | 685 3 |

| 67839 9 |

£3391 19 9

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

**Ex. 1st,** 3894 at 17s. 6d.

| 10s. = £1947 | | 5s. = 973 10 | | 2s. 6d. = 486 15 |

17s. 6d.

| 3d. = ¼ of 1s. | 1370 6 | |----------------|--------| | 11d. = ½ of 3d. | 685 3 |

| 3407 5 |

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

**Ex. 3d,** 63790 at 5s. 4d.

| 4s. = £12758 | | 1s. 4d. = ¾ of 4s. | 4252 13 4 | | 5s. = ½ of 10s. | 916 |

5s. 4d.

| 17010 13 4 15s. |

| 3d. = ¼ of 5s. 45 16 | | 14s. 9d. = £2702 4 |

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

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

| 12½628 | | 20469 | | £23 9 | | 5 | | £117 5 | | 7 |

| £820 15 |

**Case VII. When the rate consists of pounds and lower denominations.**

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

**Ex. 1st,** 3592 at £3. 12s. 8d.

| £3 | | 12s. = ¼ of £3 | | 8d. = ½ of 12s. |

| 10776 | | 2155 4 | | 119 14 8 |

| £13050 18 8 |

**Ex. 2d,** 543 at £2. 5s. 10½d. Practice.

| £2 | | 5s. = ¼ of £1 | | 10d. = ½ of 5s. |

| 1086 | | 135 15 | | 22 12 6 |

| 1 2 7 ½ |

| £2 5 10½ | | £1245 10 1½ |

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

**Ex. 1st,** 3592 at £3. 12s. 8d.

| 72 | | 7184 | | 25144 |

| 20 | | 72 | | 14732 |

| 45 | | 18415 | | 14732 |

| 258624 | | 1197 4 | | 1197 4 |

At 45s.

| 165735 | | 307 11 | | 165427 0 1 |

| 261018 8 | | 8d. = ¼ of 1s. |

| £13050 18 8 |

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

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

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

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

**Note 1.** 7 lb. 14 lb. and 21 lb. are aliquot parts of 1 qr.; and 16 lb. is ¼ of 1 cwt.; and are therefore easily calculated.

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

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

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

**Sect. II.—Deductions on Weights, &c.**

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

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

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

Cloff or Draught is a further allowance granted on some goods in London of 2 lb. on every 3 cwt. to turn the scale in favour of the purchaser. The method of calculating these, and the like, will appear from the following examples. Ex. 1st. What is the nett weight of 17 cwt. 2 q. 14 lb., tare 18 lb. per cwt.

| Cwt. g. lb. | Cwt. g. lb. | |-------------|-------------| | 17 2 14 | 17 2 14 | | 6 | |

16 lb. = ¼ cwt. 2 2 2

2 lb. = ½ of 16 lb. 1 7½

18 lb. 2 3 9½ tare

317 1

14 3 4½ nett. (28) 317½ lb. Cwt. g. lb.

4) 11 9½ (2 3 9½ tare.

In the first method we add the tare at 16 lb., which is ¼ of the gross weight, to the tare at 2 lb., which is ½ of the former. In the second we multiply the gross weight by 18; the tare is 1 lb. for each cwt. of the product, and is reduced by division to higher denominations.

Ex. 2d. What is the tret of 158 cwt. 3 q. 4 lb.

| Cwt. g. lb. | Cwt. g. lb. | |-------------|-------------| | 26) 158 | 3 4 | | 156 | |

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

This allowance being 2 lb. on every 3 cwt., might be found by taking ⅓ of the number of cwt. and multiplying it by 2. It is better to begin with multiplication, for the reason given, p. 550, col. 2, par. 2.

Ex. 3d. What is the cloff on 28 cwt. 2 q.?

| Cwt. g. lb. | Cwt. g. lb. | |-------------|-------------| | 3) 57 | (19 lb. |

This allowance being 2 lb. on every 3 cwt., might be found by taking ⅓ of the number of cwt. and multiplying it by 2. It is better to begin with multiplication, for the reason given, p. 550, col. 2, par. 2.

SECT. III.—COMMISSION, &c.

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

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

Ex. What is the commission on £728 at 2¾ per cent.?

| 728 | 2 | |-----|---| | 2 per cent. | 1456 | | ¾ | 364 | | 182 | 100 | | 20 | 40 | | 12 | 480 | | 4 | Answ. £20 0 4¾ |

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

When the given sum is an exact number of 10 pounds, the calculation may be done without setting down any figures. Every £10 at ½ per cent. is a shilling, and at other rates in proportion. Thus, £170 at ½ per cent. is 17s., and at ¼ per cent. 8s. 6d.

SECT. IV.—INTEREST.

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

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

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

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

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

It would have served the same purpose, and been easier, to multiply at first by 17, the number of days; and instead of dividing separately by 365, and by 20, to divide at once by 7300, the product of 365 multiplied by 20; and this division may be facilitated by the table inserted p. 549, col. 1.

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

I. To calculate interest at 5 per cent.

Multiply the principal by the number of days, and divide the product by 7300.

II. To calculate interest at any other rate.

Find what it comes to at 5 per cent. and take a proper proportion of the same for the rate required.

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

| 34675 | 17 | |-------|----| | 242725 | 2 | | 34675 | 2 | | 73(00)589475(80 15 | | 584 | | 5475 | | 20 | | 109500 | | 73 | | 365 | | 365 |

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

| £304 | 3 | 4 | |------|---|---| | 73(00)2433 | 6 | 8(6) 8 |

| 20 | |-----| | 48666 | | 438 | | 4866 | | 12 | | 58400 | | 584 | | 0 |

Interest at 5 per cent. £0 6 8 Deduct ½ 0 1 4 Interest at 4 per cent. £0 5 4

CHAP. VIII.—VULGAR FRACTIONS.

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

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

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

**Problem I.—To reduce mixed numbers to improper fractions.**

Multiply the integer by the denominator of the fraction, and to the product add the numerator. The sum is the numerator of the improper fraction sought, and is placed above the given denominator.

Ex. \( \frac{5}{3} = \frac{17}{3} \).

- 5 integer. - 3 denominator. - 15 product. - 2 numerator given. - 17 numerator sought.

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

**Problem II.—To reduce improper fractions to whole or mixed numbers.**

Divide the numerator by the denominator.

Ex. \( \frac{112}{17} = 6\frac{12}{17} \).

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

**Problem III.—To reduce fractions to lower terms.**

Divide both numerator and denominator by any number which measures both, and place the quotients in the form of a fraction.

Example. \( \frac{135}{360} = \frac{9}{24} = \frac{3}{8} \).

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

It is generally sufficient, in practice, to divide by such measures as are found to answer on inspection, or by the rules given p. 547, col. 2. But if it be required to reduce a fraction to the lowest possible terms, we must divide the numerator and denominator by the greatest number which measures both. What number this is may not be obvious, but will always be found by the following rule.

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

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

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

To explain the reason of this, we must observe, that any number which measures two others will also measure their sum and their difference, and will measure any multiple of either. In the foregoing example, any number which measures 589 and 475 will measure their difference, 114, and will measure 456, which is a multiple of 114; and any number which measures 475 and 456 will also measure their difference, 19. Consequently, no number greater than 19 can measure 589 and 475. Again, 19 will measure them both, for it measures 114, and therefore measures 456, which is a multiple of 114 and 475, which is just 19 more than 456; and because it measures 475 and 114, it will measure their sum, 589. To reduce \( \frac{475}{589} \) to the lowest possible terms, we divide both numbers by 19, and it comes to \( \frac{25}{31} \).

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

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

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

1st, Multiply the numerator of each fraction by all the deno- minators except its own; the products are numerators to the respective fractions sought. 2d, Multiply all the denomi- nators into each other; the product is the common denominator.

Ex. \( \frac{3}{4} \) and \( \frac{5}{6} \) and \( \frac{7}{8} \) and \( \frac{9}{10} \) and \( \frac{11}{12} \).

\( 4 \times 9 \times 8 = 288 \) first numerator. \( 7 \times 5 \times 8 = 280 \) second numerator. \( 3 \times 5 \times 9 = 135 \) third numerator. \( 5 \times 9 \times 8 = 360 \) common denominator.

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

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

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

SECT. I.—ADDITION OF VULGAR FRACTIONS.

Rule.—Reduce them, if necessary, to a common denom- inator; add the numerators, and place the sum above the denominator.

Ex. 1st, \( \frac{3}{4} + \frac{5}{6} = \frac{11}{12} \) by Problem IV.

2d, \( \frac{7}{8} + \frac{9}{10} = \frac{16}{20} \) by Problem II.

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

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

SECT. II.—SUBTRACTION OF VULGAR FRACTIONS.

Rule.—Reduce the fractions to a common denominator; subtract the numerator of the subtrahend from the numer- ator of the minuend, and place the remainder above the de- nominator.

Ex. Subtract \( \frac{3}{4} \) from \( \frac{5}{6} \) remainder \( \frac{11}{12} \)

\( \frac{5}{6} - \frac{3}{4} = \frac{1}{12} \) by Prob. IV.

To subtract a fraction from an integer, subtract the numerator from the denominator, and place the remain-

der above the denominator; prefix to this the integer di- minished by unity.

Ex. Subtract \( \frac{3}{4} \) from 12. remainder \( \frac{11}{12} \).

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

Ex. Subtract \( \frac{145}{2} \) from \( \frac{248}{3} \)

\( \frac{248}{3} - \frac{145}{2} = \frac{103}{6} \)

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

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

SECT. III.—MULTIPLICATION OF VULGAR FRACTIONS.

Rule.—Multiply the numerators of the factors together for the numerator of the product, and the denominators to- gether for the denominator of the product.

Ex. 1st, \( \frac{3}{4} \times \frac{5}{6} = \frac{15}{24} \) \( \frac{15}{24} = \frac{5}{8} \) by Prob. L

\( \frac{5}{8} \times \frac{7}{9} = \frac{35}{72} \) by ditto

\( \frac{35}{72} \times \frac{1}{2} = \frac{35}{144} \)

To multiply \( \frac{3}{4} \) by \( \frac{5}{6} \) is the same as to find what two third parts of \( \frac{3}{4} \) comes to. If one third part only had been re- quired, it would have been obtained by multiplying the denominator 7 by 3, because the value of fractions is le- sened when their denominators are increased; and this comes to \( \frac{3}{7} \); and, because two thirds were required, we must double that fraction, which is done by multiplying the numerator by 2, and comes to \( \frac{6}{7} \). Hence we infer that fractions of fractions, or compound fractions, such as \( \frac{3}{4} \) of \( \frac{5}{6} \), are reduced to simple ones by multiplication. The same method is followed when the compound fraction is expressed in three parts or more.

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

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

Ex. \( 3756 \times \frac{3}{5} = 2253 \)

\( 5 \times 11268 = 2253 \)

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

Ex. \( 138 \times \frac{5}{3} = 793 \)

\( 138 \times \frac{3}{5} = 83 \)

\( 4 \times 141 = 108 \)

\( 793 + 83 = 793 \) 3d. To multiply a mixed number by a fraction, we may multiply the integer by the fraction, and the two fractions together, and add the products.

Ex. \( \frac{15}{7} \times \frac{3}{4} = \frac{3}{4} \)

\( \frac{15}{7} \times \frac{3}{4} = \frac{3}{4} \)

\( \frac{3}{4} \times \frac{3}{4} = \frac{3}{4} \)

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

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

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

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

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

product \( \frac{56}{12} \) as before.

SECT. IV.—DIVISION OF VULGAR FRACTIONS.

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

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

Ex. Divide \( \frac{2}{3} \) by \( \frac{5}{7} \). quotient \( \frac{14}{15} \).

To explain the reason of this operation, let us suppose it required to divide \( \frac{2}{3} \) by 7, or to take one seventh part of that fraction. This is obtained by multiplying the denominator by 7; for the value of fractions is diminished by increasing their denominators, and comes to \( \frac{2}{3} \). Again, because \( \frac{2}{3} \) is nine times less than seven, the quotient of any number divided by \( \frac{2}{3} \) will be nine times greater than the quotient of the same number divided by 7. Therefore we multiply \( \frac{2}{3} \) by 9, and obtain \( \frac{14}{15} \).

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

Ex. \( \frac{14}{15} \) divided by \( \frac{2}{3} \) quotes 4.

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

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

Ex. Divide 368 by \( \frac{7}{5} \) quotient \( \frac{2576}{315} \).

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

Ex. Divide \( \frac{8}{5} \) by 5 quotient \( \frac{3}{10} \).

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

Ex. To divide \( \frac{576}{11} \) by 7 quotient \( \frac{82}{7} \).

Here we divide 576 by 7, the quotient is 82, and the remainder \( \frac{2}{7} \), to which we annex the fraction \( \frac{1}{7} \), and reduce \( \frac{2}{7} \) to an improper fraction \( \frac{1}{7} \), and multiply its denominator by 7, which gives \( \frac{1}{7} \).

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

Problem V.—To reduce lower denominations to fractions of higher.

Place the given number for the numerator, and the value of the higher for the denominator.

Examples.

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

Problem VI.—To value fractions of higher denominations.

Multiply the numerator by the value of the given denomination, and divide the product by the denominator; if there be a remainder, multiply it by the value of the next denomination, and continue the division.

Ex. 1st. Required the value of \( \frac{17}{12} \) of Ll.

| 17 | 12 | |----|----| | 20 | 8 |

Ex. 2nd. Required the value of \( \frac{8}{15} \) of \( \frac{1}{cwt} \).

| 8 | 15 | |---|---| | 27 | 5 |

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

Sometimes the value of the fraction does not amount to an unit of the lowest denomination; but it may be reduced to a fraction of that or any other denomination by multiplying the numerator according to the value of the places. Thus \( \frac{1}{289} \) of a pound is equal to \( \frac{1}{289} \) of a shilling, or \( \frac{1}{289} \) of a penny, \( \frac{1}{289} \) of a farthing.

CHAP. IX.—DECIMAL FRACTIONS.

SECT. I.—NOTATION AND REDUCTION.

Decimal fractions are such as have 10, or some power of 10 (that is 100, 1000, &c.), for a denominator; such are these,

\[ \begin{array}{cccc} \frac{1}{10} & \frac{1}{100} & \frac{1}{1000} & \frac{1}{10000} \\ \end{array} \]

They are more simply written thus:

\[ \begin{array}{cccc} .1 & .01 & .001 & .0001 \\ \end{array} \]

the number of figures after the point being always the same as the number of ciphers in the denominators.

In decimal fractions, as thus written, the figure next Decimal Fractions.

The point, to the right, indicates so many tenths; the next so many hundredths, and so on. Thus, in the fraction .346, the figure 3 expresses 3 tenths, 4 denotes 4 hundredths, and 6, 6 thousandths.

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

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

- 47 signifies four, and seven tenth parts. - 47 four tenth parts, and seven hundredth parts, or 47 hundredth parts. - 047 four hundredth parts, and seven thousandth parts, or 47 thousandth parts. - 407 four tenth parts, and seven thousandth parts, or 407 thousandth parts. - 407 four, and seven hundredth parts. - 4007 four, and seven thousandth parts.

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

Ex. 1st. \( \frac{1}{2} = 0.5 \) \( \frac{1}{2} = 0.5 \) \( \frac{1}{2} = 0.5 \)

Ex. 2nd. \( \frac{1}{2} = 0.5 \) \( \frac{1}{2} = 0.5 \) \( \frac{1}{2} = 0.5 \)

Ex. 3rd. \( \frac{1}{2} = 0.5 \) \( \frac{1}{2} = 0.5 \) \( \frac{1}{2} = 0.5 \)

Ex. 4th. \( \frac{1}{2} = 0.5 \) \( \frac{1}{2} = 0.5 \) \( \frac{1}{2} = 0.5 \)

Ex. 5th. \( \frac{1}{2} = 0.5 \) \( \frac{1}{2} = 0.5 \) \( \frac{1}{2} = 0.5 \)

Ex. 6th. \( \frac{1}{2} = 0.5 \) \( \frac{1}{2} = 0.5 \) \( \frac{1}{2} = 0.5 \)

Here, in order to reduce 21 lb. to a decimal of 1 gr., we annex a cipher, and divide by 28, the value of 1 gr. This gives 75. Then we reduce 275 qrs. to a decimal of 1 cwt. by dividing by 4, the value of 1 cwt., and it comes to 6875. Lastly, 56875 cwt. is reduced to a decimal of a ton by dividing by 20, and comes to 284375.

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

Example. What is the value of .425 of £1?

\[ \begin{array}{c} \text{sh. } 8:500 \\ \text{d. } 3:000 \\ \end{array} \]

Sect. II.—Arithmetic of Terminate Decimals.

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

In addition and subtraction, Arrange units under units, tenth parts under tenth parts, and proceed as in integers. In multiplication, allow as many decimal places in the product as there are in both factors. If the product has not so many places, supply them by prefixing ciphers on the left hand.

Ex. 1st. 1:37 2d. 43:75 3d. -1572

1:8 .48 .12

1096 35000 -018864

137 17500

2:466 21:0000

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

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

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

Therefore, if there be the same number of decimal places in the divisor and dividend, there will be none in the quotient.

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

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

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

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

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

Multiplication by fractions corresponds with division by integers, and division by fractions with multiplication by integers; when we multiply by $\frac{1}{2}$ or $\frac{1}{3}$, we obtain the same answer as when we divide by 2, and every integer has a correspondent decimal, which may be called its reciprocal. Multiplication by that decimal supplies the place of division by the integer, and division supplies the place of multiplication.

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

Ex. Required the reciprocal of 625.

$$\begin{array}{c} 625 \div 1:000:0016 \\ \hline 625 \\ 3750 \\ 3750 \\ \hline 0 \end{array}$$

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

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

Ex. .0016(5)16-0000(322500)

$$\begin{array}{c} 516 \\ 625 \\ 36 \\ 32 \\ 40 \\ 32 \\ 80 \\ 80 \\ \hline 0 \end{array}$$

Sect. III.—Approximate Decimals.

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

Decimals thus restricted are called approximate.

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

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

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

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

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

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

**Addition.**

| Cut. qrs. lb. | Cut. qrs. lb. | |---------------|---------------| | 3 2 14 = 3-625 | 3 2 2 = 3-51785 | | 2 3 22 = 2-94642 | 1 1 19 = 1-41964 | | 3 3 19 = 3-01964 | 2 - 11 = 2-09821 | | 4 1 25 = 4-47321 | |

14 3 24 14-96427

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

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

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

**Multiplication.**

| Meth. 1st, 8278+ | Meth. 2d, 8278 | Meth. 3d, 8278 | |------------------|---------------|---------------| | 2153+ | 2153 | 3512 | | 24834 | 16556 | 16556 | | 41390 | 8278 | 827 | | 8278 | 413 90 | 413 | | 16556 | 24834 | 24 | | 17822534 | 17822534 | 1782 |

Here the last four places are quite uncertain. The right-hand figure of each particular product is obtained by multiplying 8 into the figures of the multiplier; but if the multiplicand had been extended, the carriage from the right-hand place would have been taken in; consequently the right-hand place of each particular product, and the four places of the total product, which depend on these, are quite uncertain. Since part of the operation therefore is useless, we may omit it; and for this purpose it will be convenient to begin (as in p. 547, col. 1, fifth variety) at the highest place of the multiplier. We may perceive that all the figures on the right hand of the line in Meth. 2 serve no purpose, and may be left out if we only multiply the figures on the multiplicand, whose products are placed on the left hand of the line. This is readily done by inverting the multiplier in Meth. 3, and beginning each product with the multiplication of that figure which stands above the figure of the multiplier that produces it, and including the carriage from the right-hand place.

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

**Division.**

\[ \begin{array}{c} 3724 \\ -798(64237 + (2144 or 3724))79864237(2144) \\ 7448 \\ 5384 \\ 3724 \\ 16602 \\ 14896 \\ 17063 \\ 14896 \\ 2167 \\ \end{array} \]

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

**Chap. X.—Interterminate Decimals.**

**Sect. I.—Reduction of Interterminate Decimals.**

We have seen that some vulgar fractions admit of being converted into exact decimal fractions, while others have not that property, but proceed interminably, the numerator being either the same figure, or else a combination of figures always repeated. The fraction \( \frac{1}{3} \) is of the first mentioned kind, its decimal value being \( \frac{1}{3} \); that is, .375; again, \( \frac{1}{3} = .333 \), &c. and \( \frac{1}{3} = .714285, .714285, \ldots \), &c. are examples of the second kind. Let us suppose a fraction reduced to its lowest terms, \( \frac{1}{5} \), for example. Now, to convert this into a decimal, we annex ciphers to the numerator (that is, we multiply it by some power of 10), and divide by the denominator, the decimal denominator being always that power of ten by which the numerator was multiplied. In the case of \( \frac{1}{5} \), the numerator is multiplied by 1000, which is exactly divisible by 5; for when the numbers are expressed by the product of their simple factors we have \( 1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 \) and \( 5 = 2 \times 2 \times 5 \), and \( \frac{1000}{5} = 5 \times 5 \times 5 = 125 \), therefore the decimal value of \( \frac{1}{5} \) is .2, that is, .375. Here it appears that the decimal fraction has a finite form, because the divisors of the denominator of the vulgar fraction are also divisors of a power of 10.

A like remark may be made concerning the fraction \( \frac{1}{5} = \frac{1}{5} \), the two equal divisors 5 being divisors of 10, and therefore their product a divisor of 100, and the fraction \( \frac{1}{5} = .2 \).

In general, it appears that if the simple factors of the denominators of a vulgar fraction be all either 2 or 5, its decimal value will be finite, otherwise not.

The fraction \( \frac{1}{9} \) is therefore, from its nature, incapable of being converted into a finite decimal; but this fraction, when converted into a decimal, being .1111, &c. ad infinitum, therefore \( \frac{1}{9} \) will be .2222, &c., and \( \frac{1}{9} = .333 \), &c., and so on to \( \frac{1}{9} \), which will be .888, &c. Hence it is manifest that every repeating decimal is equivalent to a vulgar fraction whose denominator is 9.

Again, since \( \frac{1}{9} = .010101 \), &c. ad infinitum, therefore \( \frac{1}{9} = .020202 \), &c.; and in like manner \( \frac{1}{9} = .1717 \), &c. and so on, from \( \frac{1}{9} \) to \( \frac{1}{9} \), which is equal to .898989, &c. Therefore every circulating decimal of two figures is convertible into a vulgar fraction whose denominator is 99.

And since \( \frac{1}{99} = .001001001 \), &c., therefore any fraction whose numerator consists of three figures, and deno-

Decimal minator 999, must produce a circulate of three figures; for example, \( \frac{365}{999} = 0.365365365 \), &c.; and in like manner, since \( \frac{1}{9999} = 0.00010001 \), &c., therefore every fraction whose numerator consists of not more than four figures, and which has 9999 for a denominator, must produce a circulate of four figures, and so on.

It is now sufficiently manifest that all circulating decimals may be generated from vulgar fractions whose denominators are formed by a repetition of the figure 9; and hence it follows that every circulate may be converted into a vulgar fraction of that form.

RULES for reducing intermediate decimals to vulgar fractions.

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

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

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

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

Thus, \( \frac{235}{62} = \frac{3837}{9999} \).

From the whole decimal 23562

We subtract the finite part 235

and the remainder 23327 is the numerator.

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

\( \frac{235}{62} \) may be divided into \( \frac{235}{100} \) by Rule I.

and \( \frac{000}{62} = \frac{62}{9999} \) by Rules II. III.

In order to add these vulgar fractions, we reduce them to a common denominator; and for that purpose we multiply both terms of the former by 99, which gives \( \frac{23265}{9999} \); then we add the numerators.

235, or, by method explained p. 546, col. 1, par. 3,

\[ \begin{array}{cccc} 99 & 23500 & 23265 & 23562 \\ 2115 & 235 & 62 & 235 \\ 23265 & 23265 & 23327 & 23327 \\ \end{array} \]

Sum of numerators.

The value of circulating decimals is not altered though one or more places be separated from the circle and considered as a finite part, providing the circle be completed.

For example, .27 may be written \( \frac{27}{100} \), or \( \frac{27}{9999} \), which is also the value of .27; and if two or more circles be joined, the value of the decimal is still the same. Thus, \( \frac{2727}{9999} \), which is reduced by dividing the terms by 101 to \( \frac{27}{99} \).

All circulating decimals may be reduced to a similar form, having a like number both of finite and circulating places.

For this purpose, we extend the finite part of each as far as the longest, and then extend all the circles to so many places as may be a multiple of the number of places in each.

Ex. .34725, extended, \( \frac{34725725725725}{99999999999999} \)

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

SECT. II.—ADDITION AND SUBTRACTION OF INTERMEDIATE DECIMALS.

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

\[ \begin{array}{ccc} \text{Ex. } & \text{37524} & \text{or } \frac{37524}{9999} \\ & -8 & -88888 \\ & -643 & -643 \\ & -73 & -73333 \\ \end{array} \]

\[ \begin{array}{ccc} \text{Ex. } & \text{37524} & \text{or } \frac{37524}{9999} \\ & -8 & -88888 \\ & -643 & -643 \\ & -73 & -73333 \\ \end{array} \]

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

\[ \begin{array}{ccc} \text{Ex. } & \text{1st, } & \text{93566} \\ & -84738 & -58427 \\ & -08827 & -11172 \\ \end{array} \]

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

To add circulating decimals. Extend them till they become similar (p. 563, col. 1, par. tle &c.); and when you add the right-hand column, include the figure which would have been carried if the circle had been extended farther.

\[ \begin{array}{ccc} \text{Ex. } & \text{Extended.} & \text{Ex. } & \text{Extended.} \\ 574, & 574,574, & 874, & 874,87474, \\ 2698, & 269,869, & 1468, & 146,333333, \\ 428, & 428, & 158, & 158,585858, \\ 37,983, & 379,839, & 32, & 323,232323, \\ 1652,284, & 1503,026390, \\ \end{array} \]

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

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

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

To subtract circulating decimals. Extend them till they become similar; and when you subtract the right-hand figure, consider whether 1 would have been borrowed if the circles had been extended farther, and make allowance accordingly.

\[ \begin{array}{ccc} -573, & -974, & -974974, \\ -486, & -86, & -868686, \\ -085, & -106288, & -3606060, \\ \end{array} \]

or -360

SECT. III.—MULTIPLICATION OF INTERMEDIATE DECIMALS.

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

Ex. \( \frac{13494}{9999} \)

\[ \begin{array}{ccc} -367 & -94464 & -809666 \\ & 404833 & -04952461 \\ \end{array} \]

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

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

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

Ex. \(3746 \times 235\)

\[ \begin{array}{c} -235 \\ 18732 \\ 112393 \\ 749292 \\ -0880419, \end{array} \]

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

Ex. 1st,

\[ \begin{array}{ccc} -958 \times 8 & 2d, & -784 \times 36, \\ -8 & -36, \\ 7664 & 4704 \\ 7664 & 2352 \\ 7664 & 28224 \\ 7664 & 28224 \\ -8515 & 28509, \end{array} \]

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

The multiplication of interminate decimals may be often facilitated by reducing the multiplier to a vulgar fraction, and proceeding as directed p. 558, col. 2, par 8. Thus,

\[ \begin{array}{ccc} 3d, & -384 \times \frac{7}{9} = \frac{2}{3}, \\ 7 & 23 \\ 9)26768 & 1152 \\ -29742 & 768 \\ -90)8832 & -09813, \end{array} \]

Therefore, in order to multiply by \(g\), we take one third part of the multiplier; and, to multiply by \(g\), we take two thirds of the same. Thus,

\[ \begin{array}{ccc} 5th, & -784 \times \frac{3}{8} = \frac{1}{3}, \\ 3)784 & 261 \\ -261 & 261 \\ 3)17522 & 58406 \end{array} \]

As the denominator of the vulgar fractions always consists of 8's or of 9's with ciphers annexed, we may use the contraction explained p. 549, col. 1, par 1, &c.; and this will lead us exactly to the same operation which was explained p. 564, col. I, par. I, &c. on the principles of decimal arithmetic.

\[ \begin{array}{ccc} 7th, & -735 \times 326 = \frac{3}{5}, \\ 323 & 3 \\ 2205 & 323 \\ 1470 & 1668 \\ 2205 & 834 \\ 99(0)297405 & 999)101470, \\ 2374.05 & 101, \\ 23.74 & 101,571, \\ -239803, \end{array} \]

When the multiplier is a mixed repeater or circulate, we may proceed as in Ex. 4th and 7th; or we may divide the multiplier into two parts, of which the first is finite, and the second a pure repeater or circulate, with ciphers prefixed, and multiply separately by these, and add the products. Thus,

\[ \begin{array}{ccc} -384 \times 25 = 0768 \\ \text{and by } 05 = 02136 \\ -09813 \end{array} \]

In the following examples the multiplicand is a repeater, and therefore the multiplication by the numerator of the vulgar fraction is performed as directed p. 564, col. 1.

\[ \begin{array}{ccc} 9th, & -683 \times 5 = \frac{5}{9}, \\ 5 & 237 \\ 9)3416(-37,962, \\ -27 & 1896 \\ & 12666 \\ & 99)15010(-15,16, \\ & 99 \end{array} \]

In the following examples the multiplicand is a circulate, and therefore the multiplication by the numerator is performed as directed, p. 563, col. 2, par ult.

\[ \begin{array}{ccc} 11th, & -381 \times 5 = \frac{5}{9}, \\ 48 & 48 \\ 3054 & 48 \\ 15272 \\ 9(0)18327,(203,63, \\ 18 \\ *032 \\ 27 \\ 57 \\ 54 \\ *32 \end{array} \]

In Ex. 12 we have omitted the products of the divisor, and only marked down the remainders. These are found

In these accounts the quotient is never finite. It may repeat if the dividend repeat; or, if the dividend circulate, it may circulate an equal number of places, often more, and never fewer. The greatest possible extent of the circle is found by multiplying the divisor into the number of places in the circle of the dividend. Thus, a circulate of 3 places, divided by 3, quotes a circulate of 3 times 3 or 9 places.

CASE II.—When the divisor is interminate, the multiplications and subtractions must be performed according to the directions given for repeating and circulating decimals.

Ex. 1st. Divide .37845 by 5.

\[ \begin{array}{c} 5)37845(-68121 \\ 33333 \\ \hline 45116 \\ 44444 \\ \hline 672 \\ 555 \\ \hline 116 \\ 111 \\ \hline 5 \\ 5 \\ \hline 0 \\ \end{array} \]

The number of places in the circle of the product is sometimes very great, though there be few places in the factors; but it never exceeds the product of the denominator of the multiplier, multiplied by the number of places in the circle of the multiplicand. Therefore, if the multiplier be \( \frac{1}{2} \) or \( \frac{1}{6} \), the product may circulate three times as many places as the multiplicand; if the multiplier be any other repeater, nine times as many; if the multiplier be a circulate of two places, ninety-nine times as many: thus, in the last example, .01, a circulate of two places, multiplied by .01, a circulate of two places, produces a circulate of twice 99, or 198 places. And the reason of this limit may be inferred from the nature of the operation; for the greatest possible number of remainders, including 0, is equal to the divisor 99; and each remainder may afford two dividuals, if both the circulating figures, 3 and 6, occur to be annexed to it. If the multiplier circulate three places, the circle of the product, for a like reason, may extend 999 times as far as that of the multiplicand. But the number of places is often much less.

SECT. IV.—DIVISION OF INTERMINATE DECIMALS.

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

Ex. 1st. Divide .5326 by 7

\[ \begin{array}{c} 7)5326(-76095238 \\ 49 \\ \hline 42 \\ 42 \\ \hline 36 \\ 35 \\ \hline 16 \\ 14 \\ \hline 26 \\ 21 \\ \hline 56 \\ 56 \\ \hline \end{array} \]

Ex. 2d. Divide .843 by 5.

\[ \begin{array}{c} 5)843(-1686 \\ 49 \\ \hline 34 \\ 30 \\ \hline 43 \\ 40 \\ \hline 33 \\ 30 \\ \hline 33 \\ \hline \end{array} \]

Ex. 3d. Divide .65328 by 8.

\[ \begin{array}{c} 8)65328(-81661 \\ 56 \\ \hline 56 \\ \hline \end{array} \]

Note 1. Division by \( \frac{1}{2} \) triples the dividend, and division by \( \frac{1}{6} \) increases the dividend one half.

Note 2. When the divisor circulates, the denominator of the vulgar fraction consists of 9's, and the multiplication is sooner performed by the contraction explained p.546, col.1. It may be wrought in the same way when the divisor repeats, and the denominator of consequence is 9.

Note 3. If a repeating dividend be divided by a repeating or circulating divisor; or, if a circulating dividend be divided by a similar circulating divisor; or, if the number of places in the circle of the divisor be a multiple of the number in the dividend; then the product of the dividend multiplied by the denominator of the divisor will be terminate, since like figures are subtracted from like in the contracted multiplication, and consequently no remainder left.

Note 4. In other cases the original and multiplied dividend are similar, and the form of the quotient is the same as in the case of a finite divisor. Note 5. If the terms be similar, or extended till they become so, the quotient is the same as if they were finite, and the operation may be conducted accordingly; for the quotient of vulgar fractions that have the same denominator is equal to the quotient of their numerators.

CHAP. XI.—OF THE EXTRACTION OF ROOTS.

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

If the root of any power not exceeding the seventh power be a single digit, it may be obtained by inspection from the following Table of powers.

| 1st power | 2d power | 3d power | 4th power | 5th power | 6th power | 7th power | |-----------|----------|----------|-----------|-----------|-----------|-----------| | 1 | 1 | 1 | 1 | 1 | 1 | 1 | | 2 | 4 | 8 | 16 | 32 | 64 | 128 | | 3 | 9 | 27 | 81 | 243 | 729 | 2187 | | 4 | 16 | 64 | 256 | 1024 | 4096 | 16384 | | 5 | 25 | 125 | 625 | 3125 | 15625 | 78125 | | 6 | 36 | 216 | 1296 | 7776 | 46656 | 279936 | | 7 | 49 | 343 | 2401 | 16807 | 117649 | 893543 | | 8 | 64 | 512 | 4096 | 32768 | 262144 | 2097152 | | 9 | 81 | 729 | 6561 | 59049 | 531441 | 4782969 |

SECT. I.—EXTRACTION OF THE SQUARE ROOT.

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

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

III.—Double the quot for the first part of the divisor; inquire how often this first part is contained in the whole resolvent, excluding the unit's place; and place the figure denoting the answer both in the quot and on the right of the first part; and you have the divisor complete.

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

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

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

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

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

Square number 133225(365 root. 365 9 365 1 div. 66)432 resolvent. 1825 396 product. 2190 1095 2 div. 725)3625 resolvent. 3625 product. 133225 proof.

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

72-00000000(8.48528137 root.