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and having multipled and proved as in whole numbers, as there are three decimal places in the two factors, we point off the three right-hand places of the product.

Again, we multiply, as in whole numbers, ,1853 by ,013, thus :

,1853

,013

,0024089 and, as there are 7 decimals in the two factors, and but 5 figures in the product 24089, we place two ciphers on the left, and prefix the comma.

The student may prove the following multiplications, by multiplying both ways, and by casting out the nines :

Examples
1. ,153 X ,82 X ,017=,00213282.

,153
,82

,017
306

,01394 1224

,153 ,12546

4182 ,017

20910
,00213282

,00213282
2. 7,813 x 3,69 X 2,17 =
3. 11,917 x 1261,4 X 2,84 =
4. ,05418 X 4,816 X 75,15 =
5. 1926,48 x 17,34 x 1,632

6. ,00625 X ,039 X ,05 X ,8. 7. Required the difference between the sum of the products of the six preceding examples, and one million.

Answer. Nine hundred and two thousand, seven hundred and nine units; and four hundred and twenty-six millions, seven hundred and thirty-two thousand, four hundred and ninety-eight billionths.

235. When the product is only required to a proposed degree of exactness, the work may be abridged by the following method, given by Bezout: We reverse the order of the figures of the multiplier and write it under the multiplicand, so that its unit figure may stand under the second place on the right of that to which the product is required. We then multiply by each figure of the multiplier, beginning with the figure under which it stands, and place the first figure of each new product under that of the preceding one. Having added the products, we suppress the two right-hand figures, observing however, to increase by a unit the last of those which remain when the suppressed figures exceed 50. Lastly, we point off the places of the proposed limit.

Required, the product of ,2345678 X 85,276 within a thousandth.

,2345678
67258

1876536
117280

4690
1638
138

20,00289 As the figures suppressed exceed 50, we increase the next figure 2 by a unit, and have 20,003 for the product within a thousandth. The true product is 20,0030037128.

In placing the unit figure of the multiplier under the second grade below the proposed limit, the product is of the order of that grade. Also, by the inverted order of the figures, the next lower order of the multiplicand is multiplied by a tenfold higher order of the multiplier; and the next higher order of the multiplicand by a tenfold lower order of the multiplier, and so on in succession. Therefore, all the products are of the same order, which accounts for placing their first figures under each other.

Again, because the part rejected cannot equal a unit of the order multiplied, the product of this part by

any figure cannot equal a unit of the next higher order: therefore, each product is within a unit of the next order on the right of the proposed limit. Hence, the rule, as an approximation, may in all ordinary cases be relied on.

When there is no unit figure in the multiplier, place 0 in its stead. Also, when there are not enough of decimal places in the multiplicand, supply the deficiency with ciphers.

/

Find the product of ,227538917 X ,5664178 to the seventh decimal inclusive.

,227538917

87146650 113769455 13652334 1365228 91012 2275 1589

176 128882080

Product ,1288821. The student may prove the following examples by multiplying in the ordinary way:

Examples. 1. Required, the product of 376,273495 multiplied by 2,73486, correct to one-thousandth. Answer, 1029,055 +.

2. Required, the product of 83,4679215 X 67,89341, correct to a hundred-thousandth. Answer, 5666,92182.

3. Required, the product of 75,82344 X ,196497, true to five decimals.

Answer, 14,89908. 4. Required, the product of ,7358462199 X ,324162549, correct to nine decimals. Answer, ,238533786 +.

Division of Decimals. 236. When the divisor is a whole number, the quotient, (150) being of the same order as the dividend, must have the same number of decimals.

237. When the divisor is not a whole number, we place as many ciphers on the right of the dividend, when integral, or remove its comma, when decimal, (annexing ciphers if necessary,) as many places to the right, as there are decimal places in the divisor, in which we then suppress the comma. Both numbers being thus (228) multiplied by the same number, the quotient (165) is not altered. The divisor being now a whole number, we divide as usual, and point off for decimals, in the quotient as many places as there are decimal places in the dividend.

When the divisor is rendered integral, ciphers on the left of its highest grade are effaced. But ciphers between the comma and the highest order of the dividend, though neglected in the operation, are retained to ascertain the place of the comma in the quotient. Also, we place any number of ciphers at pleasure, as decimal places, on the right of the dividend, when it does not contain the divisor:

Examples 1. 5236 ,128=5236000 - 128 = 40906,25 We here first multiply both numbers by 1000, which ren. ders the divisor integral, and (165) does not affect the quotient. Then, to avoid confusion in distinguishing the annexed ciphers in integral, from those in decimal places, we at once determine the place of the comma, thus: As the part 523, which is tens of thousands, contains the divisor, the quotient figure 4 is (150) tens of thousands; that is, there will be 5 integral figures in the quotient: therefore, having found these, we place the comma on the right.

2. 6,375 = 85=,075 Having divided, as in whole numbers, and found 75 for the quotient, as there must be (236) as many decimal figures in the quotient as in the dividend, we place a cipher on the left of 75, and, prefixing the comma, we have ,075 for the true quotient.

238. When the divisor and dividend have each the same number of decimal places, the comma in both may be suppressed : because the equimultiples of any two numbers (165) give the same quotient as the numbers do. Thus :

6,375 – ,075 = 6375 - 75=85, (237.) 239. If, after we have rendered the divisor integral, the dividend does not contain it, we place a comma in the quotient, and also on the right of the dividend, if integral. We then place just as many ciphers on the right of the dividend as will make it contain the divisor ; and, on the right of the comma, in the quotient, as many ciphers, less one, as there are now decimal places in the dividend; because the quotient figure itself will occupy the place of the last cipher in the dividend, which (150) is its true order.

Examples. 1. ,128 = 81,92 = 12,8 + 8192= 12,800 8192 Having thus prepared the numbers, as the dividend now contains three decimals, we write ,001 in the quotient, which, when complete, is ,0015625.

2. ,128 = ,0015625 =,1280000 + ,0015625=1280000 • 15625=81,92

Having rendered the numbers integral, as the first partial dividend is tens, we know that there will be two integral figures: as soon, therefore, as we obtain these, we place a comma on the right and continue the operation.

240. Ciphers on the right of an integral divisor may always be suppressed by removing the comma in the dividend one place to the left for every such cipher.

Escamples. 1. 2248318,477 - 970000 =

224,8318477 +97=2,3178541 2. 27445863,6 = 36000 = 27445,8636 = 36=27445,8636 + (6 X 6) =

4574,3106 = 6=762,3851 241. When both numbers terminate with the decimal form of some well-known fraction, having, in its denominator, no factor prime to 10, such as ,25 and ,75, which are 4 and į ; ,125; 375; ,625, and ,875; which are i, ,0625; ,1875; ,3125; ,4375; ,5625; ,6875; „8125, and ,9375; which are to, in f& is, 16, 15, 18, and 1&; the work may often be greatly facilitated by multiplying both numbers by the denominator of the known fraction, in doing which we recognize, without calculation, the numerator of the fraction, for the product of the terminating figures.

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Examples. 1. 536,25 + ,75 = 2145 3=715 We here consider ,25 and ,75 as the numerators of the fractions which they represent. Wherefore, in multiplying the dividend by 4, we begin on the left of ,25, and say, 4 times 6 is 24, and 1 is 25, &c.

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