find another figure. The rest of the work proceeds in the same manner, and the root true to four places of decimals, is found to be 19.1049. The truth of this result is proved by multiplying 19.1049 by itself, and adding the remainder to the product; as the result will be exactly 365.* 2. Required the square root of . Here, the square root of 35 (=5×7), which is found to be 5.9160798, divided by 7, the result will be .84515425, the root required. The same result would be obtained by extracting the root of .71'42′85/71, the decimal equivalent to the given fraction. The reason of this is evident, since 5 5X7 35 and the square root of 25 is equivalent to the 7 7X7 49' square root of 35 divided by the square root of 49, or 7. 3. Required the square root of 21. Here, 24 is equivalent to 2, and the square root of is equal to, or 12. This result might also be obtained by extracting the root of 2.25. This would be found to be 1.5. or 14, as before. Exercises.-1. Required the square root of 106929. Ans. 327. Ans. 12345. Ans. 2345. Ans. 8.8600225. 2. Required the square root of 152399025. 36 Ans. .02149. Ans. 3.3, or 31. Ans. 1.16', or 1. *The principle on which the preceding rule depends, is, that the square of the sum of two numbers is equal to the squares of the numbers with twice their product. Thus, the square of 34 is equal to the squares of 30 and of 4 with twice the product of 30 and 4; that is, to 900-+-2x30×4+ 16 1156. Here, in extracting the second root of 1156, we separate it into two parts, 1100 and 56. Thus, 1100 contains 900, the square of 30, with the remainder 200; the first part of the root is therefore 30, and the remainder 200+56, or 256. Now, according to the principle above mentioned, this remainder must be twice the product of 30 and the part of the root still to be found, together with the square of that part. Now, dividing 256 by 60, the double of 30, we find for quotient 4; then this part being added to 60, the sum is 64, which being multiplied by 4, the product 256, is evidently twice the product of 30 and 4, together with the square of 4. In the same manner the operation may be illustrated in every case. The rule, however, is best demonstrated by Algebra; see my treatise on the subject, page 231, second edition. PROBLEM II. To extract the third or cube root of a given number. 180. RULE. Commencing at the unit figure, cut off periods of three figures each, till all the figures of the given number are exhausted. Then find the greatest cube number contained in the first period, and place the cube root of it in the quotient, Subtract its cube from the first period, and bring down the next three figures; divide the number thus brought down by 300 times the square of the first figure of the root, and it will give the second figure; add 300 times the square of the first figure, 30 times the product of the first and second figures, and the square of the second figure together, for a divisor; then multiply this divisor by the second figure, and subtract the result from the dividend, and then bring down the next period, and so proceed till all the periods are brought down. To extract the cube root of a fraction, reduce it to a decimal, and then extract the root; or multiply the numerator by the square of the denominator, find the cube root of the product, and divide it by the denomi nator. The cube root of a mixed number is generally best found by reducing the fractional part to a decimal, if it be not so already, and then extracting the root. It may also be found by reducing the given number to an improper fraction, and then working according to the preceding directions. Example.-Required the cube root of 48228544. 48'228'544(364 Here, by dividing the given number into periods, we find for the first period 48: then 3, the root of the greatest integral cube contained in 48, is put in the quotient, and its cube taken from the first period. To the remainder the next period is brought down, which makes for dividend 21228. The first part of the divisor is found by multiplying the square of 3 by 300; in finding, in the next place, what figure must be annexed to the part of the root already found, though 2700 would be contained 8 times in 21228, yet we can readily find by trial, that 6 is the figure which must be put in the quotient, since 2700 is not the complete divisor. Now, to find the complete divisor, we must add 30 times the product of 3 and 6, or 540, and also the square of 6, or 36, to 2700; then, multiplying the sum thus found 3276 by 6, and subtracting the product from 21228, there remains 1572. To this remainder the next period 544 is brought down. The rest of the work proceeds in the same manner, and the root is found to be 364. The truth of this is proved by raising 364 to the third power, as the result will be exactly 48228544.* * The reason of the preceding rule will appear evident from the following illustration. The cube of 25, for instance, is equivalent to the cube of 20 added to the cube of 5, together with the sum of 300×4×5 +30×2×5×5; or, which is the same thing, 25 is equal to 20+5, and, therefore, 25 cubed is equal to 20+5 cubed; but 20+5 cubed is equivalent to 8000+300×4×5+30×2×5×5+125, or to 203+(300x4 +30x2x5+5×5)x5=48228544. Thus, 20+5 20+5 Multiplied 20×20+5×20 Multiplied {20x20+2×5×20+25=2d power. 20×20×20+2×5×20×20+20×25 5×20×20+2×20×25+125 8000+3×5×20×20+3x20x25+125=3d power. or 8000+300×4×5+30×2×25+125. Exercises.-1. Required the cube root of 34567. Ans. 32.57521043, &c. 2. Required the cube root of 782140. Ans. 92.1357479, &c. 3. Required the cube root of 123456789. Ans. 497.933859, &c. 4. Required the cube root of 389017. Ans. 73. Ans. 103. Ans. 52.74, &c. 7. What is the cube root of .0001357? Ans. .05138, &c. 8. What is the cube root of 13? 9. What is the cube root of 150? Ans. 2.3908. 10. What is the cube root of? Ans. .6436595897, &c. Repeat the rule for finding any power of a given number. What is evolution? How is the root of a given number denoted ? Repeat the rule for extracting the square root of a given number. Repeat the rule for extracting the cube root of a given number. CHAPTER XII. POSITION.* 181. 'Position is a rule by which, from the assumption of one or more false answers to a problem, the true one is obtained. Hence, the rule is evident. In the same manner, the operation may be illustrated in every case. For a demonstration of this rule in gene-" ral terms, see my Treatise on Algebra, Theoretical and Practical. *This rule is sometimes called the rule of false, or the rule of false position, or the rule of trial and error. It might properly be called the rule f supposition. It admits of two varieties, single position and double position. In single position the answer is obtained by one assumption in double position it is obtained by two.* Single Position.† 182. RULE.-Assume any number, and perform on it the operations mentioned in the question as being performed on the required number. Then, as the result thus obtained is to the assumed number, so is the result given in the question to the number required. Example.-Required a number to which if one-half, onethird, one-fourth, and one-fifth of itself be added, the sum -may be 1664, Suppose the number to be 60: then, if to 60 one half, one-third, one-fourth, and one fifth of itself be added, the sum is 137. Hence, according to the rule, as 137: 60 :: 1644 720, the number required. The truth of the result is proved by adding to 720 one-half, one-third, &c. of itself, and the sum is found to be 1644.‡ Exercises.-1. Divide $2000 between A, B, and C, giving A as much as B, and a fifth part more, and C as much as both together. Ans. A's part $545.45.4, B's $454.54.55, C's $1000. 2. One-third of a ship belongs to A, and one-fifth to B, and A's part is worth $1000 more than B's required the value of the ship. Ans. $7500. 3. A father bequeaths to his three sons $7000, in such a manner, that if the share of the eldest be multiplied by 5, * Single position may be employed in resolving problems in which the required number is any how increased or diminished in a given ratio; such as when it is increased or diminished by any part of itself, or when it is multiplied or divided by any number. † Every question that can be resolved by this rule, may also be resolved by the rule for double position, or without position, by some of the preceding rules; and hence this rule is of little importance. The number 60 was here assumed, not as being near the truth, but as being a multiple of 2, 3, 4, and 5; and in this way the operation was kept free from fractions. By the assumption of any other number, however, the answer would have been found correctly, but often not so easily. The reason of the operation is obvious from the principles of proportion. |