Examples of the subtraction of compound numbers.

1st. From 35 lbs. 8 ozs. troy weight, take 19 lbs. 10 ozs. 13 dwts. 81 grs. The reduction of fractional expressions to compound numbers and to common denominators, with other preparations requisite for the addition of compound numbers, are also required for the subtraction of one compound number from another.

In the present example it is necessary to find the value of oz. which by Article 304=2 dwts. 204 grs.

It is also necessary to reduce #gr. and gr. to the same denominator. The fractions thus reduced are gr. and 28 gr.

35 8

lbs. Ozs. dwts. grs.
2 2019

Detail of the process.

The subtraction is commenced with the lowest denomination, grains. 1st. (2014-87) grs.=12 grs. (Art. 184).

The remainder, 12, is written under the column of grs.

2d. (2-13) dwts., without borrowing the subtraction is not possible; 1 of the next higher denomination (ounces) is borrowed :

1 oz. 20 dwts. (20+2) dwts.=22 dwts. and (22-13) dwts.=9 dwts. The remainder, 9, is written under the column of dwts., and 1, for that borrowed, is carried to the ounces of the subtrahend.

3d. 10 ozs.+1 oz. borrowed=11 ozs.

(8-11) ozs., without borrowing the subtraction is not possible; 1 of the next higher denomination (lbs.) is borrowed:

1 lb. 12 ozs. and (12+8) ozs. 20 ozs. and (20-11) oz. 9 ozs.

9 is written under the column of ozs. and 1 carried to the simple units of the lbs. for the pound borrowed.

4th. (1+19) lbs.=20 lbs. and (35—20) lbs.=15 lbs.

15 is written under the column of lbs.; and the difference of the given compound numbers is found to be equal to 15 lbs. 9 ozs. 9 dwts. 12 grs.

In subtraction of whole numbers, when the partial subtrahend exceeds the minuend, 1 of the next higher order of units (and equal to 10 units of the partial subtrahend or minuend) is borrowed, and added to the partial minuend, to render the subtraction possible.

In subtraction of compound numbers, when the partial subtrahend is greater than the minuend, 1 is borrowed from the next higher denomination, reduced to the denomination of the partial minuend, and added to it. The number so added is not 10 or any constant number, but depending on the number of units of the partial minuend which make 1 unit of the next higher denomination of that weight or measure by which the given compound numbers have been valued.

In this consists the difference between subtraction of whole and subtraction of compound numbers.

2d Example. From 579 £. 12 sh. subtract 289 £. 17 sh. 6 d. 23 qrs. Writing zeros to fill the places of the deficient pence and farthings of the minuend,

£ sh. d. qrs.
579 £. 12 sh.
. 579 12 0 0
From which subtract 289 17 6 23

O qrs.-23 qrs., the subtraction is not possible; one of the next higher denomination (pence) is borrowed:

1 d. 4 qrs. and 4 qrs.-23 qrs.=14 grs.

14 is written under the column of farthings, and 1 is carried to the pence of the subtrahend for the penny borrowed.

2d. Subtraction of the pence:

1d.+6d. 7 d., partial subtrahend.

Od.-7 d., to render the subtraction possible, 1sh. 12 d. is borrowed from the next higher denomination.

12d.-7 d. 5 d.

5 is written under the column of pence, and 1 for that borrowed is carried to the shillings of the subtrahend.

3d. Subtraction of the shillings:

1 sh.+17 sh.=18 sh., partial subtrahend.

12 sh.-18 sh.; to render the subtraction possible, 1 £.=20 sh. is borrowed from the next higher denomination.

20 sh.+12 sh.=32 sh. and 32 sh.-18 sh.=14 sh.

14 is written under the column of shillings, and 1, for 1 £. borrowed, is carried to the pounds of the subtrahend.

4th. Subtraction of the pounds:

1 £.+289 £.=290 £., and 579 £.-290 £.=289 £.

289 is written under the column of pounds, and the result is found to be 289 £. 14 sh. 5 d. 13 qrs.

These examples and remarks seem sufficient to direct the process of subtracting one compound number from another.

The rule for such subtraction is,

Prepare the quantities in the same manner as for addition of compound numbers.

Write the minuend in a horizontal line, and under its several denominations the corresponding denominations of the subtrahend.

Subtract the number in the lowest denomination of the subtrahend from the number in the lowest denomination of the minuend, if the subtraction can be effected; but if it cannot, borrow one of the next higher denomination, reduce this to units of the lowest denomination, and add the result to the number in the lowest denomination of the minuend. Subtract the number in the subtrahend from the sum; carry one for that borrowed to the next denomination of the subtrahend; subtract the number thus augmented from the corresponding denomination of the minuend, if the subtraction is possible; if not, augment the number in the minuend by as many units of the denomination to which it belongs as make one unit of the next higher denomination; subtract the number in the subtrahend from the

sum.

And proceed in this manner till the calculation is completed.
Exercises in subtraction of compound numbers.

1st. From 73 £. 15 sh. 10 d. subtract 48 £. 12 sh. 6d. ?

Ans. 25 £. 3 sh. 4 d.

2d. From 708 £. 14 sh. 64 d. subtract 278 £. 17 sh. 103 d.?
Ans. 429 £. 16 sh. 7 d.

3d. From 95 £. 8 sh. 7 d. subtract 68 £. 13 sh. 6 d.?

Ans. 26 £. 15 sh. Old.

4th. From 17 cwts. 1 qr. 21 lbs. subtract 13 cwt. 2 qrs. 24 lbs? Ans. 3 cwt. 2 qrs. 25 lbs.

5th. From 13 lb. 5 oz. 13 dwt. 17 gr. subtract 9 lb. 7 oz. 14 dwt. Ans. 3 lbs. 9 oz. 18 dwt. 20 gr.

21 grs.?

6th. From 18 tuns 2 hhds. 48 gals. subtract 13 tuns 2 hhds. 54 gals. ...Ans. 4 tuns 13 hhds. 56 gals. 2 qts.

2 qts ?.....

7th. Required the difference between 17.92 £. and 26 £. 12 sh.? Ans. 8 £. 13 sh. 7 d.

8th. Required the difference between 12 acres and 37693 sq. yds.? Ans. 2 ro. 1 po. 274 yds. 9th. Required the difference between 20 mi. 3 fur. 25 po. 24 yds. and 13 mi. 7 fur. 37 po. 37 yds.?.

10th. Required the difference

crowns?........

Ans. 6 mi. 3 fur. 27 po. 333 yds.

between 4 of 65-2 £. and 456-625 .Ans. 104 £. 16 sh. 10 d.

11th. From 28 miles take 1275 yds.?

Ans. 1 mi. 6 fur. 25 po. 417 yds.

12th. Required the difference between 46 ac. 3 ro. 17 po. 173 yds.
and 230 square chains?....Ans. 23 ac. 3 ro. 6 po. 4 yds.
13th. Required the difference between 10075 links long
measure and 20 miles?
...Ans. 19 mi. 8398 po.
.......Ans. 14 sh. 3 d.

14th. From £. subtract & sh.?
15th. Find, in troy weight, the difference between lb. troy
and lb. avoirdupois?..............Ans. 19 dwts. 9 gr. troy.
16th. Required the difference between 26.5 po. and 703 yds.
sq. measure?.....
Ans. 24 po. 5 yds.
17th. Required the difference between 271 £. and 3·125 £. ?

80

18th. Two pieces whose lengths are 25 feet 7

1

Ans. 4 sh. 9 d. inches and 19 feet 10 inches are cut from a plank whose length is 17 yds. 2.7 feet. How much is left?

Ans. 2 yds. 2 feet 2.3 in.

MULTIPLICATION OF COMPOUND NUMBERS.

320. In multiplication of whole numbers, if the multiplicand is repeated as often as the multiplier contains unity, the sum of these repetitions is the product of the numbers.

The repetitions of the multiplicand are made by repeating the units of each order contained in the multiplicand.

Consequently, if each of the denominations of a compound number is repeated as often as a multiplier contains unity, the result is the product of the compound number by that multiplier.

Examples of the multiplication of compound numbers.

1st. Multiply 3 miles 6 fur. 32 po. by 9.

The multiplicand and multiplier may be arranged as in whole numbers, thus,

The product is equal to 32 po. x9+6 fur.×9+3 mi. × 9.

32 po. x9=288 po.=7 fur. 8 po.

8 is written under the poles of the multiplicand, and 7 reserved for addition to the product of the furlongs by 9.

6 fur. x9=54 fur. and 54+7 carried=61 fur. 7 miles 5 fur.

5 is written under the furlongs of the multiplicand, and 7 carried to the product of the miles by 9.

3 mi. x9=27 mi. and 27+7 carried=34 miles.

34 is written under the miles of the multiplicand; and the required product is 34 miles 5 fur. 8 po.

The partial products of the several denominations of the multiplicand by the multiplier may be formed separately, and afterwards combined into one sum by the rules for the reduction and addition of compound numbers.

But the method taken in the preceding example, namely, that of making the reduction and the combination of the partial products to proceed simultaneously, is less laborious and more concise.

a. This method is, however, advantageous so long only as the multiplier is a number so small that the multiplications and reductions can be made mentally.

To extend the limits of mental calculation, a multiplication table to 12x12 is generally employed in the multiplication of compound numbers. In this treatise a table of products from 1x1 to 25 × 25 is given. The labour of committing to memory the whole of this table would be considerable; but the part of it to 12 × 12 may be easily learnt.

2d Example. Multiply 8 £. 15 sh. 94 d. by 132. The product required is equal to

d. x 132+9 d. × 132+15 sh. × 132+8 £. × 132.

But the multiplications and reductions of these partial products cannot be made mentally; therefore, unless some other process is devised, the partial products must be formed separately, and the complete product composed by addition of these partial products into one compound number.

What is required is to find a compound number equal to 132 repetitions of 8 £. 15 sh. 94 d; and this without the necessity of multiplying by a number too great to allow the partial multiplications and reductions to be made mentally.

Now, it is proved in Articles 69 and 71 that, whether a number is multiplied by any multiplier, or by the factors of that multiplier taken in any order, the product is the same,

As often, therefore, as a multiplier can be resolved into factors, each not greater than the greatest factor contained in the multiplication table which is used (generally 12), the product of a multiplicand which is a compound number by that multiplier can be obtained by repetitions of the method followed in Example 1.

132, the multiplier of Example 2, is composed of the prime factors 22, 3, and 11; but as any factor not greater than 12 can be used, it will be sufficient to decompose 132 into the factors 11 and 12, and to multiply the multiplicand by 11, and this product by 12.

The calculation is subjoined. A lengthened detail, after the explanation of the first example, seems unnecessary.

£ sh. d.

8 15 94

11

96 13 53 11 repetitions of multiplicand.

12

1160 1 9=11x12 or 132 repetitions of the multiplicand.

The product required is 1160 £. 1 sh. 9 d. It could have been obtained by multiplying 8 £. 15 sh. 94 d. by 2, this product by 2, this by 3, and this again by 11; 2, 2, 3, 11, being the prime factors of 132. But it is more concise to multiply at once by 12 than successively by 2, 2, 3, the prime factors of 12.

b. Any multiplier which contains among its simple factors a prime number greater than 12 cannot be decomposed into factors of which none shall exceed 12, and whose continual product shall be equal to the given multiplier.

The method of Article 132 consequently fails for such examples.

But although the product cannot, in these cases, be obtained by continual multiplication of the multiplicand by the factors of the multiplier, it can be formed by breaking down the multiplier into parts which are either composed of factors not greater than 12, or are themselves not greater than 12, forming the separate products of the multiplicand by these parts of the multiplier, and taking the sum of the partial products (Art. 70).

To illustrate these remarks, let it be required.

3d Example. To multiply 12 lb. 3 oz. 16 dwt. troy by 59, the multiplier is a prine number (Art. 103); it cannot therefore be decomposed into any factors whose product shall be 59; but it can be broken into the parts 3 and 56, of which 3 is less than 12, and 56 is the product of the factors 7 and 8, both of which are also less than 12. And (denoting the multiplicand by m for the sake of brevity)

mx7x8+mx3 or m× (56+3)=m×59 (Art. 70).

That is, if to the continual product of the multiplicand by the factors 7 and 8, the product of the multiplicand by 3 is added, the result is equal to 59 repetitions of the multiplicand, which is the number required. Calculation.

lb. oz. dwt.

12 3 16x3

7

86 2 12=7 repetitions of the multiplicand.

689

8

816=7x8 or 56 repetitions of multiplicand.

36 11 8 3 repetitions of multiplicand.

726 8 4=56+3 or 59 repetitions of multiplicand.

Since 5960-1=6x10-1,

m x 59 =mx6x10-m,

that is, if m is multiplied by 6, this product by 10 and once m is subtracted from the result, the remainder contains 60-1 or 59 repetitions of m.

Whence, in general, if a product is found in excess, and the number of repetitions of the multiplicand in excess are subtracted, the remainder contains the number of repetitions required.

The product of 12 lb. 3 oz. 16 dwt. by 59 is found by this method as follows,

c. Any whole number can be decomposed into as many parts as it contains different orders of units. Of these parts (regard being had to the relative values of the different units), the first is less than 10, the second is a multiple of 10, the third a multiple of 10%, the fourth a multiple of 103, &c.

Decomposing a multiplier in this manner, the sum of the partial products of the multiplicand by the several orders of units of the multiplier is equal to the complete product of the multiplicand and multiplier.

The number which expresses the relative value of the units of any order, n, higher than the first, being one of the series,

it is evident that the partial products can be formed without the necessity of multiplying by any factor greater than 10; and these products are, in practice, easily derived from each other.