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with laying down the following Rule, by Way of Corollary.

161. The Rule. Begin with the lowest Denomination, fubtracting the Number in that Place of the Minorand from that in the correfponding Place of the Subducend (if Subtraction can be made) and put down the Remainder underneath: But, if the Number of any inferior Species of the Minorand is greater than its Correfpondent in the Subducend, you must increase this Number by a Number which is equal to an Unit in the next fuperior Denomination; (viz. for Example, in the Column of Pence, add 12; and, in Shillings, add 20, &c.) and from this Sum fubtract the Number in the Minorand, and put down the Remainder; but then you must remember to carry or add to the Number in the next fuperior Place of the Subducend. Proceed in this Manner, till you come to the highest Place; which fubtract as whole Numbers.

CHAP. X.

MULTIPLICATION of APPLICATE NUMBERS.

162. S to fimple applicate Numbers, nothing A need be faid; and even Multiplication of

Α

mixed Numbers will be fufficiently explained by two Examples. 1. Multiply 3 Feet 6 Inches by 8.

Solution. Say, 6 x 8 48 Inches (by dividing it by 12) 4 Feet, therefore, under Inches put o, and carry the 4 to the Feet; then 3x8 = 24, +4 you carry 28 Feet, for Anfwer.

Feet Inch.

3: 6

8

28 : 0

153. Example 2. Multiply, 27. 10s. 11d. 3grs. by 57.

Here,

Here, fince it would be troublesome to multiply at once by 57, obferve, that 8 x-7=56, which wants but 1 of 57. the Work may ftand thus:

I

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carry the 7 Shillings,

faying, 10 x 8 = 80+ 7 (you carry)

5: 9:3

87 Shillings (by dividing by 20) 41.75.; . write down 7 under Shillings, and carry 4 to the Pounds, faying, 2× 8=16, +4 20%. which write down. Then multiply this Product by 7, after the fame Manner, which will give 1427. 145. 10d.; thus, we have multiplied by 56, for 8 x 7 = 56; but, fince this wants I of 57, we must add once 27. 10s. 11 d. 3 qrs.; which will give 145l. 5s. 9d. 3qrs., for the Product which was required. Vide Art. 182.

164. Scholium. Since Multiplication is * the repeating a Number, a certain Number of Times, it follows, that, when two Numbers are to be multiplied together, one of them must be an abstract Number; for, fuppofe it was required to multiply 31. by 47., I would ask the Propofer, how many Times he would have 37. taken or repeated? If he fhould answer (as he must according to his Question) 4., it plainly appears to be Nonsense; but, if it had been required to multiply 31. by 4, it would be only to repeat 31. 4 Times, which is very proper, and the Anfwer 3x4 127. Hence it appears, that, when Authors propofe to multiply Money by Money, Weight by Weights, &c. they talk very

51.

improperly and abfurdly. We fhall here only further observe, that as, in Multiplication of abstract Numbers, we may make the Multiplicand the Multiplier, and the Multiplier the Multiplicand, fo we may here alfo, by only confidering the applicate Number as the abstract one, and the abftract as the applicate 95. Number; for Inftance, as 3 x 4 =*4x 3, 34. x41 t 95. must be 41. x 3; or, generally, as a x b = + bxa, al. xbbl.xa..

=

120.

С НАР. XI.

DIVISION of APPLICATE NUMBERS.

HIS admits of two Cafes. 1. To divide

165. T applicate Numbers by applicate Numbers.

Example. Divide 247. by 87. Dividing this, as in abftract Numbers, the Quotient will be 3; fhewing, that 8. is contained in or can be taken out of 24%. three Times. As it would be fomething difficult to divide mixed applicate Numbers by mixed applicate Numbers, without fome Knowledge of Reduction, we fhall omit it here, and refer to Art. 182. And shall only further observe in this Place, that, when an applicate Number is to be divided by an applicate Number, the Quotient must be an abstract Number; for Divifion fhews to find how many Times one Number is contained in another, and, therefore, fuch Authors, as divide an applicate Number by an applicate Number, and make the Quotient an applicate Number, fpeak not only improperly, but abfurdly; for if 47. was to be divided by 21. and we should ask how many Times 21. is contained in 4., and fhould be answered 41.; what would be more ridiculous?.

166. Cafe 2. To divide an applicate by an abftract Number. Here it is proper to-hint, that, be

fides the Definition of Divifion already given, it may, confiftently with that Definition, be defined to be a Rule, which fhews to find a Number, which is contained in a given Number, a given Number of Times; for let x denote a Number in which y is contained z Times, then y, taken z Times, *yxz x;., dividing both Sides of the Equation by z, we

x

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51.

have y That is, any Number, being divided † 108.

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by the Number expreffing the Times another Number is contained in it, will give that other Number.

Example. Divide 1427. 145. 1od. by 7; or, which the fame, divide 142/. 145. 10d. amongft Men, and give each Man an equal Sum.

Solution. Here it is evident we are to find a Number (which, being multiplied by 7, fhall be equal to 142. 145. 10d.; or in other Words) which is contained in 142/. 14s. 10d. feven Times; and therefore, by the above, 142/. 14s. iod, divided by 7, will give the Number required; which, it is plain, must be applicate, and may be found thus: Say, the of 14 is 2, which put down; and the of 2 is o,and 2 remaining; put down the ✪, and, the Remainder being 21. Shillings, carry it to the Shillings; and then we shall have 40+1454 Shillings, the Seventh of which is 7, and 5

40

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d.

142, 14: id

20:

7: 10

remaining; therefore put down the 7 Shillings, and

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Befides this and the former Definition, Divifion will admit of others, viz. 1. To find what Part the Divifor is of the Dividend; for, if the Diviför is contained 3 Times in the Dividend, it must be a third Part of the Dividend; becaufe, being repeated 3 Times, it will be - the Dividend; and for the fame Reason, if it be contained 4 Times in the Dividend, it will be the 4th Part of the Di vid end, E... the Quotient fhews the Part. 24. I.

carry the remaining 5 Shillings, or 5 x 12 60 Pence, to the Pence; and then we have 60 + 1070 Pence; the Seventh of which is 10d. which, being written down compleats the Anfwer, viz. 20l. 7s. 10d. This may be proved by Multiplication of applicate Numbers.

167. Example 2. What is the fourth Part of 2 Yds. 2 Feet. 10 Inches?

Solution. Since 4 cannot be
contained in 2 Yards, under the
Yards put o, and carry the 2
Yards to the Feet, faying, (2
Yards) 6 Feet + 2 Feet 8

Inch.

Yds. Feet
2 2 10

0: 2:

2

Feet; the Fourth of which is 2 Feet, which put down; then the Fourth of 10 Inches is 2, and 2 remaining; put down the 2 Inches, and the 2 remaining, being 2 Parts of 4, is of an Inch; Whence the Answer is 2 Feet 2 Inches.

168. If the Number we are to divide by be fo great, as to be troublesome to divide in one Line, the Operation may be put down as in common Divifion; for Example, divide 145l. 5s. 9d. 3qrs. by 57. See Art. 167.

Solu

2. To find a Number, which is fuch Part of a given Number, as a given Number expreffes. Let x

a

the Number of which it is a Part, b

the required Number, the Number expreffing

the Part; then, by the Nature of the Thing, xXba, ., divid

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