117 INVOLUTION. INVOLUTION is the term applied to the multiplication of a number one or more times by itself; thus 2 x 2 = 4; the result is called a POWER of that number. The first power of a number is the number itself before being multiplied; thus The second power, termed the sQUARE, is the number multiplied by itself; thus The third power, termed the CUBE, is the number multiplied by itself, and the product again multiplied by it; thus First Power, 2 2 x 2 = 4 3 x 3 x 3 = 27 And so on, with the higher powers. The power of a number is indicated by writing the number with a small figure above it; thus, 32 means 3 x 3 = 9; 53 means 5x5x5 = 125. " Exercises. cubes of 16, 37, 46? Ans. 2809, 25921, 75076 4096, 50653, 97336 1. What are the squares of 53, 161, 274? 2. "1 3. If the side of a square table measure 62 inches, how many square inches are contained in the surface of the table? Ans. 3844 sq. inches. 4. The side of a cubic block of granite measures 6 feet; how many solid feet in the block?-and what is its weight, if 1 cubic foot weigh 2654 oz. ? Ans. 216 Solid feet; Weight, 319 cwts. 3 qrs. 17 lbs. EVOLUTION, OR THE EXTRACTION OF ROOTS. EVOLUTION is the process of finding or extracting the roots of numbers. The Root of any number, is that number which, on being multiplied one or more times by itself, produces the given number. The Square root is that number which, on being multiplied by itself, produces the given number; thus, 4 is the square root of 16, because 4 multiplied by 4 (4 × 4) produces 16. The Cube root is that number which, on being multiplied by itself, and the product again multiplied by it, produces the given number; thus, 3 is the cube root of 27, because 3 multiplied by 3, and the product again by 3 ( 3 × 3 × 3) produces 27. The sign (termed the radical sign) placed before a number, indicates that the square root of that number is to be extracted; thus / 25 = 5, The sign / placed before a number, indicates that the cube root of that number is to be extracted; thus 3/729 = 9. The extraction of the square and cube roots, serves various useful purposes in connection with the measurement of fields, walls, solid bodies, &c. as will appear from the exercises under the rules for extracting the square and cube roots of numbers. EXTRACTION OF THE SQUARE ROOT. RULE.-1. Point off the given number, by means of commas, into periods of two figures each; beginning at the right in integers, and counting towards the left; and beginning at the left in decimals, and counting towards the right: thus-7,38,69 37,62,1. If, in pointing off the periods, one figure remain over, either at the left in integers, or right in decimals, it is considered as a period, although consisting of only one figure. 2 )5,92,92,25 ( 2435 Ans. 2 4 44 )192 3 1449 3. Form a new DIVISOR as follows:Add the previous divisor to itself (in other words, double it), to form part of the new divisor; then find how often this is contained in the dividend *—exclusive of its last figure-and annex the quotient to the partial divisor, which is thereby completed: also place the quotient as the next figure of the root. 4. Multiply the completed divisor by this quotient, and subtract the product from the dividend; then annex to any remainder, the next period of two figures, to form another dividend. 5. Form another DIVISOR as follows: -Add to the previous divisor its last figure, to form part of the new divisor; and, as already described in paragraph 3, find how often this is contained in the dividend-exclusive of its last figure-and annex the quotient to the partial divisor, which is thereby completed: also place the quotient as the next figure of the root. 6. Multiply the completed divisor; and so on, as already described in paragraph 4. 7. Proceed to form new divisors, according to paragraph 5; and go on with the rest of the process according to paragraph 4; till all the periods have been brought down and operated upon; when, if there is no remainder, the extraction of the root is completed. See Note 2, page 119. Here the first period, 5, is divided by 2, the greatest square root contained in it; and to the remainder, 1, is annexed the next period, 92, forming the new dividend, 192. Here 2, the previous divisor, is doubled, making 4 for part of a new divisor; and 4 being contained 4 times in 19 (19,2, the divi. dend, with the last figure left out), 4 is annexed to the partial divisor, thereby forming the complete divisor, 44: the quotient, 4, is also placed as the next figure of the root. Here the divisor, 44, is multiplied by the quotient, 4, and the product, 176, is subtracted from the dividend, 192: to the remainder, 16, is then annexed the next period, 92, to form another dividend. Here, to the previous divisor, 44, is added 4, its last figure, making 48 for part of a new divisor; and 48 being contained 3 times in 169 (169,2, the dividend, with its last figure left out), 3 is annexed to the divisor to complete it, making 483 the 3 is also placed as the next figure of the root. The divisor, 483, is then multiplied by the quotient, 3, and the product, 1449, is subtracted from the dividend, 1692; to the remainder, 243, is annexed the next period, 25, forming a new dividend, 24325; and the process of extracting the next figure of the root is then proceeded with as before. There being now no more periods to bring down, and no remainder, the process is completed. A less figure sometimes requires to be taken, see Note 3, page 119. NOTE 1.-When there is a remainder after all the periods have been brought down, annex a period of two nothings to form a new dividend; and then proceed with the further extraction of the root: the figure of the root thus obtained is a decimal. The process may be carried to any degree of minuteness by annexing more nothings. There are always as many figures in the root as there are periods in the given number; and those figures are decimals in the root, which are extracted from the decimals in the given number. NOTE 2.-If at any time, on bringing down a new period to form a dividend, the partial divisor is found to be greater than the dividend, a nothing [0] must be placed in the root to express this: the next period is then brought down, and annexed to the dividend, and the extraction of a new figure of the root proceeded with as before. NOTE 3.-It sometimes happens that the completed divisor obtained in the manner described above, proves to be too large, as on multiplying it, it is found to be greater than the dividend: when this is the case, the quotient placed in the root and annexed to the divisor must be reduced as much as is found necessary; thus, if the figure is 9, it must be made 8. NOTE 4.-The square root of a fraction, is found by extracting the roots of the numerator and denominator. Exercises. Find the square root of the following numbers : THE MEAN PROPORTIONAL between two numbers-that is, a number that is as many times greater than the one given number, as it is less than the other-is found by multiplying the two given numbers together, and extracting the square root of the product. 15. Find the mean proportional between 16 and 144, Ans. 48 16. 11 19 and 41, 27.91057 17. A cheese, when put into one scale of a false balance, was found to weigh 43 lbs., but when put into the other, it weighed 89 lbs.; what was the true weight of the cheese? Ans. 61-86275 lbs. THE LENGTH OF THE SIDE OF A SQUARE whose area is given, is found by extracting the square root of the area. 18. The area of a circle is 7085 square inches; what is the side of a square whose area is equal to that of the circle? Ans. 84.1724 inches. 19. A gentleman has three fields: the first measures 2 ac. 1 ro. 30 po.; the second, 3 ac. 2 ro. 15 po.; and the third, 1 ac. 3 ro. 27 po. He wishes another of a square form equal in area to all the three; how many poles must its side measure? 20. A person has a field measuring 3 ac. 1 ro. 35 po., to exchange for a square one of inferior quality, but 3 how many poles in the length of its side? Ans. 35-665109 which he wishes times as large; Ans. 44.0738 FOR ADDITIONAL EXERCISES, SEE PAGE 122. EXTRACTION OF THE CUBE ROOT. RULE.-1. Point off the given number, by means of commas, into periods of three figures each; beginning at the right in integers, and counting towards the left; and beginning at the left in decimals, and counting towards the right: thus, 1,784,453·547,46. If, in pointing off the periods, one or two figures remain over, either at the left in integers, or right in decimals, this remainder is considered as a period, although not consisting of three figures. 2. Find the greatest cube root contained in the first period, and place it at the right of the given number, as the first figure of the root then subtract the cube of the root from the first period, and to the remainder, if any, annex the next period of three figures to form a dividend. Example.-What is the cube root of 178453547 ? 3. Form a DIVISOR for this dividend as follows: Multiply the square of the root found by 300, for a partial divisor. Find how often this partial divisor is contained in the dividend,* and place the quotient as the next figure of the required root: then multiply the previous figure of the root by this quotient, and the product by 30, placing the result below the partial divisor: also place the square of the same quotient (that is, the square of the figure just annexed to the root) below the last product; then add the three sums together, and their amount is the complete divisor. Here, the greatest cube root contained in the first period, 178, is 5, which is placed as the first figure of the required root: the cube of 5*namely, 125-is then subtracted from the first period, and to the remainder, 53, is annexed the next period, 453, forming the dividend, 53453. The partial divisor is contained 6 times in the dividend; the quotient, 6, is therefore placed as the next figure of the required root: the root already found, 5, is then multiplied by the 6, and the product by 30, making 900, which is placed below the partial divisor, 7500. The square of the quotient, 6-namely, 36-is placed below the last product, 900. The three sums, 7500, 900, and 36 are then added together, forming the complete divisor, 8436. It is here contained 7 times in the dividend, but as it is found, on carrying the process further, that 7 is too many, G the next lower figure, is taken. 4. Multiply the complete divisor by the last figure of the root; subtract the product from the dividend; and to any remainder, annex the next period of three figures, to form a new dividend. M 5. Form a new DIVISOR as follows:Place the square of the last figure of the root under the previous divisor; add it to the three lines of figures above it; and to the amount annex two nothings, to form a partial divisor. Find how often the partial divisor is contained in the dividend; and, as already described in paragraph 3, place the quotient as the next figure of the required root: then multiply the previous figures of the root by this quotient, and the product by 30, placing the result below the partial divisor: also place the square of the quotient (that is, the square of the figure just annexed to the root) below the last product; then add the three sums together, and their amount is the complete divisor. 6. Multiply the complete divisor, and so on, as already described in paragraph 4. 7. Proceed to form new divisors according to paragraph 5; and go on with the rest of the process according to paragraph 4, till all the periods have been brought down and operated upon; when, if there is no remainder, the extraction of the root is completed. Here, the divisor, 8436, is multiplied by 6, the last figure of the root, and the product is subtracted from the dividend, 53453: to the remainder, 2837, is brought down the next period of three figures, forming the new dividend, 2837547. Here, 36, the square of 6, the last figure of the root, is placed below the complete divisor, 8436; it is then added to the three lines above it—namely, 8436, 36, and 900-and their amount, with two nothings annexed, forms the partial divisor, 940800. The partial divisor is contained 3 times in the dividend; the quotient, 3, is therefore placed as the next figure of the root: the root already found, 56, is then multiplied by the 3, and the product by 30, making 5040, which is placed under the partial divisor. The square of the quotient, 3-namely, 9-is placed below the last product, 5040. The three sums, 940800, added 5040, and 9, are then together, forming the complete divisor, 945849. Here, the divisor, 945849, is multiplied by 3, the last figure of the root, and the product is subtracted from the dividend, 2837547: there being no remainder, and no more periods to bring down, the extraction of the root is completed. NOTE 1.-When there is a remainder after all the periods have been brought down, annex a period of three nothings, to form a new dividend; and then proceed with the further extraction of the root: the figure of the root thus obtained is a decimal. The process may be carried to any degree of minuteness by annexing more nothings. There are always as many figures in the root as there are periods in the given number; and those figures are decimals in the root, which are extracted from the decimals in the given number. NOTE 2.-If at any time, on bringing down a new period to form a dividend, the partial divisor is found to be greater than the dividend, a nothing must be placed in the root, and two nothings annexed to the partial divisor: the next period is then brought down, and annexed to the dividend, and the extraction of a new figure of the root proceeded with as before. NOTE 3.-The cube root of a fraction is found by extracting the roots of the numerator and denominator. |