[SCIENCE.]
A NEW RULE FOR DIVISION IN ARITHMETIC.
The ordinary process of long division is rather difficult, owing to the necessity of guessing at the successive figures which form the divisor. In case the repeating decimal expressing the exact quotient is required, the following method will be found convenient:
Rule for division.
First. Treat the divisor as follows:
If its last figure is a 0, strike this off, and treat what is left as the divisor.
If its last figure is a 5, multiply the whole by 2, and treat the product as the divisor.
If its last figure is an even number, multiply the whole by 5, and treat the product as a divisor.
Repeat this treatment until these precepts cease to be applicable. Call the result the prepared divisor.
Second. From the prepared divisor cut off the last figure: and, if this be a 9, change it to a 1, or if it be a 1, change it to a 9; otherwise keep it unchanged. Call this figure the extraneous multiplier.
Multiply the extraneous multiplier into the divisor thus truncated, and increase the product by 1, unless the extraneous multiplier be 7, when increase the product by 5. Call the result the current multiplier.
Third. Multiply together the extraneous multiplier and all the multipliers used in the process of obtaining the prepared divisor. Use the product to multiply the dividend, calling the result the prepared dividend.
Fourth. From the prepared dividend cut off the last figure, multiply this by the current multiplier, and add the product to the truncated dividend. Call the sum the modified dividend, and treat this in the same way. Continue this process until a modified dividend is reached which equals the original prepared dividend or some previous modified dividend; so that, were the process continued, the same figures would recur.
Fifth. Consider the series of last figures which have been successively cut off from the prepared dividend and from the modified dividends as constituting a number, the figure first cut off being in the units' place, the next in the tens' place, and so on. Call this the first infinite number, because its left-hand portion consists of a series of figures repeating itself indefinitely toward the left. Imagine another infinite number, identical with the first in the repeating part of the latter, but differing from this in that the same series is repeated uninterruptedly and indefinitely toward the right into the decimal places.
Subtract the first infinite number from the second, and shift the decimal point as many places to the left as there were zeros dropped in the process of obtaining the prepared divisor.
The result is the quotient sought.
Examples.
1. The following is taken at random. Divide 1883 by 365.
First. The divisor, since it ends in 5, must be multiplied by 2, giving 730. Dropping the O, we have 73 for the prepared divisor.
Second. The last figure of the prepared divisor being 3, this is the extraneous multiplier. Multiplying the truncated divisor, 7, by the extraneous multiplier, 3, and adding 1, we have 22 for the current multiplier.
Third. The dividend, 1883, has now to be multiplied by the product of 3, the extraneous multiplier, and 2, the multiplier used in preparing the divisor. The product, 11298, is the prepared dividend.
Fourth. From the prepared dividend, 11298, we cut off the last figure 8, and multiply this by the current multiplier, 22. The product, 176, is added to the truncated dividend, 1129, and gives 1305 for the first modified divisor. The whole operation is shown thus:
1 8 8 3
6
-------
1 1 2 9|8
1 7 6 -
-----
1 3 0|5
1 1 0 -
-----
2|4 0
8 8 ---
---
|9 0
-----
1 9|8
1 7 6 -
-----
1 9|5
1 1 0 -
-----
1 2|9
1 9 8 -
-----
2|1 0
2 2 ---
2 4
We stop at this point because 24 was a previous modified dividend, written under the form 240 above. Our two infinite numbers (which need not in practice be written down) are, with their difference:
. .
10,958,904,058 . .
10,958,904,109.5890410958904
----------------------------
. .
51.5890410958904
. .
Here the quotient sought is 5.158904109.
Example 2. Find the reciprocal of 333667.
The whole work is here given:
3 3 3 6 6|7 |7
2 3 3 5 6 7 - 1 6 3 4 9 6|9
2 1 0 2 1 0 3 -
-------------
2 2 6 5 5 9|9
2 1 0 2 1 0 3 -
-------------
2 3 2 8 6 6|2
4 6 7 1 3 4 -
-----------
7 0 0 0 0 0
. .
Answer, 0.000002997.
Example 3. Find the reciprocal of 41.
Solution.—
4|1 |9
----- -----
3 7|9 3 3|3
- 1 1 1 -
-----
1 4|4
1 4 8 -
-----
1 6|2
7 4 -
---
9 0
. .
Answer, 0.02439.
C.S. PEIRCE.