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SUBTRACTION. 217. 1. From 4.156 take .5783.
ANALYSIS. We write the given numbers as in addi.5783
tion, reduce the decimals to a common denominator, 3.5777 and subtract as in integers. Or, we may, in practice,
omit the ciphers necessary to reduce the decimals to a Or,
common denominator, and merely conceive them to be 4.156
nexed, subtracting as otherwise. Hence the fol.5783 lowing 3.5777
RULE. I. Write the numbers so that the decimal points shali stand directly under each other.
II. Subtract as in whole numbers, and place the decimal point in the result directly under the points in the given numbers.
EXAMPLES FOR PRACTICE,
(3.) Minuend, .9876 48.3676 36.5 Subtrahend, .3598
35.875632 Remainder, .6278 24.3876
.624368 4. From 37.456 take 24.367.
Ans. 13.089. 5. From 1.0066 take .15. 6. From 1000 take .001.
Ans. 999.999. 7. From 363 take 2213
Ans. 14.27. 8. From .567 take .55131. 9. From 7j take 5,6
Ans. 1.7708+ 10. From 991 take 119 11. From one take one trillionth. Ans. .999999999999.
12. A speculator having 57436 acres of land, sold at different times 536.74 acres, 1756.19
acres, and 4785.94 acres; how much land has he remaining ?
13. Find the difference between 54321 and 12345, correct to the fifth decimal place.
218. In multiplication of decimals, the location of the decimal point in the product depends upon the following principles :
I. The number of ciphers in the denominator of a decimal is equal to the number of decimal places, (209, VI).
II. If two decimals, in the fractional form, be multiplied to gether, the denominator of the product must contain as many ciphers as there are decimal places in both factors. Therefore,
III. The product of two decimals, expressed in the decimal form, must contain as many decimal places as there are decimals in both factors. 1. Multiply .45 by .7.
ANALYSIS. We first multiply .45
as in whole numbers; then, .7
since the multiplicand has 2 .315
decimal places and the multiplier 1, we point off 2+1=3
decimal places in the product, XL 315 3.315 (III). The reason of this is
further illustrated in the proof, a method applicable to all similar cases.
219. Hence the following
RULE. Multiply as in whole numbers, and from the right hand of the product point off as many figures for decimals as there are decimal places in both factors.
Notes. — 1. If there be not as many figures in the product as there are decimals in both factors, supply the deficiency by prefixing ciphers.
2. To multiply a decimal by 10, 100, 1000, etc., remove the point as many places to the right as there are ciphers on the right of the multiplier.
EXAMPLES FOR PRACTICE.
1. Multiply .75 by .41.
6. Multiply 7.23 by .0156.
Ans. .112788. 7. Multiply .0075 by .005.
Ans. .0000375. 8. Multiply 324 by .324. 9. Multiply 75.64 by .225. 10. Multiply 5.728 by 100.
Ans. 572.8. 11. Multiply .36 by 1000. 12. Multiply :000001 by 1000000. 13. Multiply .576 by 100000. 14. Multiply 73 by 5.1.
Ans. 42.625. 15. Multiply .631 by 24. 16. Multiply 4 by 733
Ans. 31.74. 17. Find the value of 3.425 x 1.265 x 64. Ans. 277.288. 18. Find the value of 32 x .57825 X .25. 19. Find the value of 18.375 X 5.7 X 1.001.
Ans. 104.8422375. 20. If a cubic foot of granite weigh 168.48 pounds, what will be the weight of a granite block that contains 27} cubic feet?
21. When a bushel of corn is worth 2.8 bushels of oats, how many bushels of oats must be given in exchange for 36 bushels of corn and 48 bushels of oats ?
220. To obtain a given number of decimal places in the product.
It is frequently the case in multiplication, that a greater number of decimal figures is obtained in the product, than is necessary for practical accuracy. This may be avoided by contracting each partial product to the required number of decimal places.
To investigate the principles of this method, let us take the two decimals .12345 and .54321, and having reversed the order of the digits in the latter, and written it under the former, multiply each figure of the direct number by the figure below in the reversed number, placing the products with like orders of units in the same column, thus:
.000025 = .00005 X.5
.000001 = 1 x .00001. In this operation we perceive that all the products are of the same order; and this must always be, whether the numbers used be fractional, integral, or mixed. For, as we proceed from right to left in the multiplication, we pass regularly from lower to higher orders in the direct number, and from higher to lower in the reversed number. Hence
221. If one number be written under another with the order of its digits reversed, and each figure of the reversed number be multiplied by the figure above it in the direct number, the products will all be of the same order of units.
1. Multiply 4.78567 by 3.25765, retaining only 3 decimal places in the product.
Analysis. Since the product we obtain 14357 for the first partial product. Then, beginning with the next figure of the multiplier, 2 times 5 are 10, which gives 1 to . be carried to the second partial product; 2 times 8 are 16, and 1 to be carried are 17; writing the 7 under the first figure of the former product, and multiplying the remaining left-hand figures of the multiplicand, we obtain 957 for the second partial product. Then, 5 times 8 are 40, which gives 4 to be carried to the third partial product; 5 times 7 are 35 and 4 are 39 ; writing the 9 in the first column of the products, and proceeding as in the former steps, we obtain 239 for the third partial product. Next, multiplying by 7 in the same manner, we obtain 33 for the fourth partial product. Lastly, beginning 2 places to the right in the multiplicand, 6 times 7 are 42; 6 times 4 are 24, and 4 are 28, which gives 3 to be carried to the fifth partial product; 6 times 0 is 0, and 3 to be carried are 3, which we write for the last partial product. Adding the several partial products, and pointing off 3 decimal places, we have 15.589, the required product.
of any figure by units is of the 4.78567 56752.3
same order as the figure multi
plied, (82, II,) we write 3, the 14357 = 4785 x 3 + 2
units of the multiplier, under 957 478 x 2 + 1
5, the third decimal figure of 239 47 x 5 + 4 4 x 7 + 5
the multiplicand, and the lowest 3
order to be retained in the pro0 x 6+3
duct; and the other figures of 15.589 +, Ans.
the multiplier we write in the inverted order, extending to the left. Then, since the product of 3 and 5 is of the third order, or thousandths, the products of the other corresponding figures at the left, 2 and 8, 5 and 7, 7 and 4, etc., will be thousandths ; and we therefore multiply each figure of the multiplier by the figures above and to the left of it in the multiplicand, carrying from the rejected figures of the multiplicand, as follows: 3 times 6 are 18, and as this is nearer 2 units than one of the next higher order, we must carry 2 to the first contracted product; 3 times 5 are 15, and 2 to be carried are 17; writing the 7 under the 3, and multiplying the other figures at the left in the usual manner,
222. From these principles and illustrations we derive the following
RULE. I. Write the multiplier with the order of its figures reversed, and with the units' place under that figure of the multiplicand which is the lowest decimal to be retained in the product.
II. Find the product of each figure of the multiplier by the figures above and to the left of it in the multiplicand, increasing each partial product by as many units as would have been carried from the rejected part of the multiplicand, and one more when the highest figure in the rejected part of any product is 5 or greater than 5; and write these partial products with the lowest figure of each in the same column.
III. Add the partial products, and from the right hand of the result point off the required number of decimal figures.
Notes.—1. In obtaining the number to be carried to each contracted partial product, it is generally necessary to multiply (mentally) only one figure at the right of the figure above the multiplying figure; but when the figures are large, the multiplication should commence at least two places to the right.
2. Observe, that when the number of units in the highest order of the rejected part of the product is between 5 and 15, carry 1; if between 15 and 25 carry 2; if between 25 and 35 carry 3; and so on.
3. There is always a liability to an error of one or two units in the last place; and as the answer may be either too great or too small by the amount of this