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. OPERATION. 4.78567 56752.3 143574785 × 3 + 2 239 = 33 = = 4 x 7 + 5 15.589, Ans. ANALYSIS. Since the product of any figure by units is of the same order as the figure multiplied, (82, II,) we write 3, the units of the multiplier, under 5, the third decimal figure of the multiplicand, and the lowest order to be retained in the product; and the other figures of 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, 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 mul tiplicand, 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. 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 error, the uncertainty may be indicated by the double sign, ±, read, plus, or minus, and placed after the product. 4. When the number of decimal places in the multiplicand is less than the number to be retained in the product, supply the deficiency by annexing ciphers. EXAMPLES FOR PRACTICE. 1. Multiply 236.45 by 32.46357, retaining 2 decimal places, and 2.563789 by .0347263, retaining 6 decimal places in the product. 2. Multiply 36.275 by 4.3678, retaining 1 decimal place in the product. Ans. 158.4 ±. 3. Multiply .24367 by 36.75, retaining 2 decimal places in the product. 4. Multiply 4256.785 by .00564, rejecting all beyond the third decimal place in the product. Ans. 24.008 ±. 5. Multiply 357.84327 by 1.007806, retaining 4 decimal places in the product. 6. Multiply 400.756 by 1.367583, retaining 2 decimal places in the product. Ans. 548.07 ±. 7. Multiply 432.5672 by 1.0666666, retaining 3 decimal places in the product. 8. Multiply 48.4367 by 23, extending the product to three decimal places. Ans. 103.418. 9. Multiply 73 by 3318, extending the product to three decimal places. 376 4391 10. The first satellite of Uranus moves in its orbit 142.8373 + degrees in 1 day; find how many degrees it will move in 2.52035 days, carrying the answer to two decimal places. Ans. 360.00 degrees. 11. A gallon of distilled water weighs 8.33888 pounds; how many pounds in 35.8756 gallons? 12. One French metre is equal to how many yards in 478.7862 metres. Ans. 299.16 pounds. 1.09356959 English yards; Ans. 523.58 yards. 13. The polar radius of the earth is 6356078.96 metres, and the equatorial radius, 6377397.6 metres; find the two radii, and their difference, to the nearest hundredth of a mile, 1 metre being equal to 0.000621346 of a mile. DIVISION. 223. In division of decimals the location of the decimal point in the quotient depends upon the following principles: I. If one decimal number in the fractional form be divided by another also in the fractional form, the denominator of the quotient must contain as many ciphers as the number of ciphers in the denominator of the dividend exceeds the number in the denominator of the divisor. Therefore, II. The quotient of one number divided by another in the deci mal form must contain as many decimal places as the number of decimal places in the dividend exceed the number in the divisor. 1. Divide 34.368 by 5.37. OPERATION. 5.37) 34.368 (6.4 2 148 2 148 PROOF. 34368 1000 539 10 100 X = = 6.4 = ANALYSIS. We first divide as in whole numbers; then, since the dividend has 3 decimal places and the divisor 2, we point off 3 2 1 decimal place in the quotient, (II). The correctness of the work is shown in the proof, where the dividend and divisor are written as common fractions. For, when we have canceled the denominator of the divisor from the denominator of the dividend, the denominator of the quotient must contain as many ciphers as the number in the dividend exceeds those in the divisor. 224. Hence the following RULE. Divide as in whole numbers, and from the right hand of the quotient point off as many places for decimals as the decimal places in the dividend exceed those in the divisor. NOTES.-1. If the number of figures in the quotient be less than the excess of the decimal places in the dividend over those in the divisor, the deficiency must be supplied by prefixing ciphers. 2. If there be a remainder after dividing the dividend, annex ciphers, and continue the division: the ciphers annexed are decimals of the dividend. 3. The dividend should always contain at least as many decimal places as the divisor, before commencing the division; the quotient figures will then be integers till all the decimals of the dividend have been used in the partial dividends. 4. To divide a decimal by 10, 100, 1000, etc., remove the point as many places to the left as there are ciphers on the right of the divisor. 14. If 25 men build 154.125 rods of fence in a day, how much does each man build? 15. How many coats can be made from 16.2 yards of cloth, allowing 2.7 yards for each coat? 16. If a man- travel 36.34 miles a day, how long will it take him to travel 674 miles? Ans. 18.547+days. 17. How many revolutions will a wheel 14.25 feet in circumference make in going a distance of 1 mile or 5280 feet? |