Rule 2. For an unlimited number of places.

Take half the number of shillings for the number of tenths, and reserve 5 hundredths, or 50 thousandths, if there is an odd shilling.

Reduce the pence into farthings, consider them as so many thousandths, with the addition of I thousandth if the pence are 6 or above, and 50 thousandths if there is an odd shilling.

Then to continue the decimal parts, divide the given pence, or the pence above 6, by 6, annexing as many ciphers as may be necessary to the number of the pence.*

N. B. If there are farthings apply their decimal values to the pence, before applying the ciphers; that is, annex 25 for 1 farthing, 5 for d., or 75 for 3 farthings.

EXAMPLE 1.

To value 18 s. 4 d. and 17 s. 11 d. as decimals of a £.

18 s. being 9-10 ths and 4 d. being 16 farthings, we obtain 916 for the first 3 figures in the first calculation; then the operation is continued by dividing 4000, &c. by 6 for the remaining figures 666, &c.

17 s. in a like manner give 850 thousandths, and 11 d. or 44 farthings give 45-1000 ths, making together 895-1000 ths; then as the pence above 6 are 5, we divide 5000, &c. by 6, for the remaining figures 833, &c.

EXAMPLE 2.

To value 14 s. 4 d. and 17 s. 73 d. as decimals of a £.

14 s. 4 d. produce as before 0.718 thousandths; and the 44, that is 4.50, divided by 6 produce the exact additional parts 75. 17 s. 7 d. produce 0.882 thousandths, and the 12 d. above 6 d., that is 1.75, divided by 6, produces the infinite decima! 29166, &c.

The reason of this part of the rule evidently is from 1-1000 th being the exact addition to the number of the farthings when the pence are 6, and consequently for any other number of pence the additional parts are the 6 th part of so many pence.

EXAMPLE 3.

To value 18 s. 04 d. and 17 s. 6 d. as decimals of a £

18 s. 01 d. produce 901 thousandths; then 4 d., or 0.25 pence, divided by 6 produces 0416, &c.

17 s. 6 d. produce 877 thousandths, and d., or 0.5 pence. divided by 6 produces 0833.

N. B. We see from this example, that when there is only a. ord. above 1 s. or 6 d., there will be a cipher in the fourth place of decimals: also from the preceding calculations we observe, that when the excess above 1 s. or 6 d. is 3 farthings, or any multiple of 3 farthings, as 1 d., 24 d., 3 d., 32 d., 41⁄2 d., or 5 d., the decimal will terminate at the 5 th place with halfpence, or at the 6 th place with farthings; but in other cases, the decimal 3 or 6 repeats, at or before the 7 th place of decimals.

EXERCISE.

Find the decimal value of the following sums to 7 places of decimals, when the number of places is unlimited:*

N. B. It forms a good proof to this Exercise to add together the decimal values of these sums, and see that the amount is the decimal value of the amount of these sums. To perform this addition with accuracy, we must divide the amount of the column of the 7 th place of decimals by 9, and putting down the remainder carry the quotient to the 6 th column, which with the rest of the columns, is added in the common manner. See Addition of Circulating Decimals.

DECIMAL ADDITION.

Rule. Arrange the given numbers so that the similar parts may be in the same columns, and find the amount as in the Addition of whole numbers.

Example. To find the amount of £ 36.047, £ 4.67, £ 83.4045, and £ 516.2.

£

36.047

4.67

83.4045

516.2

£ 640.3215

When the numbers are properly arranged, the decimal points will stand one under another; and as 10 in each lower denomination make 1 of the next higher, the addition is necessarily performed as in the addition of whole numbers.

For the practice of beginners the columns of figures may be headed as in the Numeration Table, page 147, or the fractional value of each may be expressed thus:

Ex. 1. Find the amount of £ 48.6, £ 27.3, and £ 101.7. 2. Add together s. 11.28, s. 7.145, s. 8.036, and s. 14.0007. 3. Add together lb. 6.07, lb. 100.082, and lb. 1.600766. 4. Add together £ 0.366, £ 0.58, and £ 0.00455.

Ex. 1. £ 177 12 0.

AMOUNTS.

Ex. 2. £ 2 05 Ex. 3. lb. 107.12.
Ex. 4. 0 19 0.

In the addition and subtraction of decimal quantities, if they are not in the same denomination they must be reduced into either the higher or the lower denominations according to the Rule, page 151; thus in Ex. 2. of Subtraction, if the 3.72 was £ instead of s., then either the £ 3.72 should be multiplied by 20 to bring them into s., making 74.4 s., or the 2.30706 should be divided by 20 to bring them into the decimal of a £, making £ 0.115353; the difference would then be either s. 72.09294, or £ 3.604647.

DECIMAL SUBTRACTION.

Rule. After having arranged the numbers as in Decimal Addition, find the difference as in the Subtraction of whole numbers.

EXAMPLE 1.

To subtract £ 60.437 from £ 82.141.

£ 82.141

60.437

£ 21.704

EXAMPLE 2.

To subtract s. 2.30706 from s. 3.72, and also from s. 5.0.

In subtractions like these, as many ciphers may be annexed to the greater quantity, as are required to make the number of the figures equal to those in the lower quantity; and in the performance of such calculations it will be seen, that the right hand figure of the remainder and the one above it are equal to ten, and that the sum of the other parts or figures, until we come to the figure in the upper line, is each equal to 9.

DECIMAL MULTIPLICATION.

Rule. Multiply as in the Multiplication of integral numbers, without regarding the difference between whole numbers and decimals.

Then, if the multiplier is a whole number, there will be as many places of decimals in the product as there are decimal places in the multiplicand; and if the multiplier is also a decimal number, there will be as many more places of decimals in the product as there are decimal places in the multiplier.*

EXAMPLE 1.

To multiply £37.31 by 10, by 100, and by 1000. £ 373.11 st product

£ 3731.0 2 nd

£ 37310.0

3 rd

In multiplications of decimals by 10, 100, 1000, &c. all that there is to be done, is to remove the decimal point one, two, three, &c. places to the right.

These are worked upon the same principles as common multiplication.

To multiply £4263 by 0.25.

EXAMPLE 3.

£ 4263
.25

£ 1065.75

As there are not any places of decimals in this multiplicand, but two places in the multiplier, we cut off two places from the right of the product.

The second part of this rule may be expressed thus - then from the right of the product cut off as many places for decimals, as there are decimal places in both the multiplier and the multiplicand.