or exponent denotes the number of factors employed. This method of notation furnishes some important conclusions; for, since 52 = 5 × 5 and 5* = 5 × 5 × 5 × 5; therefore we have, 5' x 5o (5 x 5 x 5 x 5) X (5 × 5) = 5° = 5*+2; consequently powers of the same number are multiplied by adding their indices and divided by subtractiny the index of the divisor from that of the dividend. Thus, since 11 5"=5*X 5' x 53. = 5' X 5' x 5' ÷ 5 = 625 × 625 X 125 The 12th power of 5 is 5' x 5' X 5' is 5a × 53 × 53 × 53 = = = the cube of 5 or (5*)3, or it the fourth power of 5 or (53); hence (53)* (5)3 = 512; and, therefore, the power of a power of a number is expressed by multiplying the index or exponent by the degree of the power. Hence, 9° 91X2X3 9 × 92 × 93 9 x 81 x 729 531441, and the same result will be obtained if either of the following indicated operations be performed : = = = = = = = 25 = The operation of evolution is in 106. A root of a number is such a number as, being multiplied by itself one, two, three, or (n-1) times, produces the given number; and the operation by which the root is obtained is termed Evolution. Thus the second or square root of 25 is 5, since 5 × 5 5o; the third, or cube root of 27 is 3, for 33 3 x 3 x 3 27, and the square root of 0625 is 25, since 25 × 25 = ・0625. Also the cube root of 8 2 2 2 2 8 is because X X 27 3' 3 3 3 27 dicated by the sign, accompanied with a small figure in the opening. The sign is called the radical sign, and the figure in its opening indicates the particular root to be determined. In the case of the square root the figure 2 in the radical sign is always omitted; hence the square root of 25 is expressed by 25, the cube, or third root of 27, by 3/27, and so on. Again, by involution (82) 3 = 82X3 86; therefore, conversely, the cube, or third root of 8° is 82, where the index 2 is found by dividing 6, the index of the power, by 3 the index of the root. generally the n root of a power of a number is that power of the number whose index is the nth part of the index of the proposed power. Thus the square root of 2° is 23; the cube root of 26 is 2 and the nth root of 2" is 2. = Hence Suppose, now, we take a number with a fractional index, as 8, and treat the fraction exactly as it were an integer; then since 8a × 83 83+3 =3 8°, we have 8 × 8 – 81+1 must necessarily produce 8; hence / 8 = will represent the second or square root. = S1 = 83 = 83++++ 83, and the Hence 8, we have in like manner 8; but 38 × 38 × 38 = 8 hence 38 = ; index characterizes the third or cube root; and so on. (11)* = 11*×+ = (11*)* or (11)*; hence ; and also 11 = √11 =√√11; (11)* = 111×+×+ = 111×3 = 11°×+ = 1331* = √ 1331. EXTRACTION OF THE SQUARE ROOT. = 107. The square root of a number is that number which multiplied by itself produces the proposed number. There are many numbers whose square roots cannot be determined exactly, as 5, 7, 10, etc., but they may always be found to any degree of accuracy by means of decimals. And since 102 100, 100 = 10000, etc., it follows that the squares of all numbers between 1 and 10 must consist of 1 or 2 figures; the squares of all numbers between 10 and 100 must consist of more than 2, but not more than 4 figures; and the squares of all numbers between 100 and 1000 must consist of more than 4, but not more than 6 figures, etc.; hence, if a number be proposed to find its square root, we must place a point over every second figure, beginning at the unit's place; thus dividing the number into periods of two figures each, excepting the last period, which may consist of either one or two figures as the case may be. There will be just as many figures in the root as there are periods. With respect to a number composed of integers and decimals, the points will necessarily fall above the second, fourth, sixth, etc. decimals, counting from the unit's place, which we must do in all cases, and when there is no unit, as in the case of decimals, its place must be supplied with a cipher, and a point placed over it; hence the number of decimals must be one of the even numbers, 2, 4, 6, 8, etc. A cipher must be added to the right of an odd number of decimals to complete the period. = 108. Let the several parts of a number be denoted by a, b, c; then (a+b+c) is the same as a + b + c repeated a times, b times, and c times. But a+b+c repeated a times is a2 + ab + ac, b times = ab+b2 + bc, and c times = ac + bc + c2; hence by adding all together we get the square of a+b+c expressed in either of the following forms : (a+b+c)2=a2+(2 a+b)b, or (a+b+c)2=c2+2c (a+b) +b2+2ab. + a2 109. From the latter of these forms the rule for squaring will be, square each part, and multiply all that precede by twice that part; and a reverse rule for extracting the square root immediately presents itself. Let n denote the given number, and take a number a whose square does not exceed n. Find the remainder; take a second number b, such that the remainder will bear the subtraction of the square of b and twice b multiplied by the preceding part a. If there be a remainder, take a third number c and find whether the second remainder will allow of the subtraction of the square of c and twice e multiplied by a+b. Proceed in this manner till the process terminate, or until the root be obtained to the nearest unit. The first form, however, affords an easy process for forming the numbers to be subtracted. For we have only to double the sum of all the parts which have been obtained, add the new part, and multiply the sum by the new part. Thus after a is subtracted; (2a + b) × b is the next subtrahend; {2 (a + b) + c} x c is the third, and so on. root of 273529. Begin 273529 (500+20 + 3 250000 23529 3129 3129 or 523 Ex. Let it be required to extract the square at the unit's place and point off the periods as already directed; then since there are 3 periods, the first figure of the root will be in the place of hundreds. This first figure will be 5, because 500 is less and 6002 is greater than 270000. Subtracting 250000 from the number leaves the remainder 23529. Let 2 tens or 20 be the next part; then by the processes given above, the number to be subtracted will be 20o +2× 20×500, or (2×500+20) × 20, viz., 20400. Lastly, take 3 as the next part of the root; then, as before, 32+2 × 3 × 520, or (2 × 520 + 3) × 3, gives 3129, which, being exactly equal to the second remainder, shows that the root is It will be covenient to arrange the operation of forming the numbers to be subtracted on the left of the successive remainders, as in the following example: Ex. Find the square root of 293764. 293764 (500+ 40 + 2 = 523. In the first of these, the numbers are written at length, but in the second the ciphers on the right are omitted, and each period is annexed to the remainder as it is wanted. The parts of the root a, b, c may be any whatever; but when they are confined to the nearest number of hundreds, tens, and units, the parts 2a, and 2a + 26 which form so large a portion of the factors 2a + b, 2a + 2b + c soon become available for the determination of the succeeding parts. Thus 1000 is contained in 43764 more than 40 but less than 50 times. In this way the successive figures of the root can be written in the places which they occupy in the decimal scale, as in the second mode of operation. 110. When the proposed number has not an exact square root, it may be obtained to any degree of accuracy by means of decimals, and when one more than half the number of figures in the root have been obtained, the remaining figures may be found by dividing the last remainder by its corresponding divisor as in contracted division of decimals. Ex. Find the square root of 10, and also of 17.108 each to 6 places of decimals. 111. The square root of a fraction is equal to the square root of the numerator divided by the square root of the denominator. If the terms of the fraction have not exact square roots, the fraction may be reduced to a decimal, and its root extracted, or if the denominator has no exact root, the terms of the fractions may be multiplied by such a number as will render the denominator a complete square. Thus : = 112. The cube root of a number is that number which multiplied twice by itself produces the proposed number. Since 10 1000, 100% = 1000000, etc., the cubes of all numbers between 1 and 10 must consist of 1, 2, or 3 figures; the cubes of all numbers between 10 and 100 must consist of more than 3 but not more than 6 figures, and so on; hence the number whose cube root is to be found is to be divided into periods of three figures each, beginning at the unit's place, and putting a point over it, and a point over every third figure towards the left in integers and the right in decimals. In decimals supply one or two ciphers, if necessary, to complete the period of 3 figures. 113. Let a and b represent the tens and units of the root; then (a+b) is the same as a + b repeated a times and 6 times. But a + b repeated a times is a2+ab, and repeated 6 times is ab+b2; hence a+b repeated a+b times, or (a + b)2 = a2 +2ab+b2. Again VOL. I. E (a+b) is the same as a2 + 2 ab + b2 repeated a times and b times; that is (a+b)3 = (a3 + 2a2 b + ab2) + (a2 b + 2 a b2 + b3) = a3 + 3 ab + 3 a b2 + b3. This may be put in either of the forms a3 + (3a2 + 3 ab + b2) b or a3 + 3 ab (a + b) + 63. From this result the following simple rule for finding the cube root of a number is deduced. Let n be the number, and take a number a whose cube does not exceed n. Find the remainder, take a second number b, such that the remainder may bear the subtraction of the cube of b, and the continued product of thrice a, the second number b, and the sum of a and b. If there be a remainder, consider a + b as the first number, and proceed as before. 3 a2 3a + b 3 ab+b2) 3 a +3 ab + B ba 3a2+6 ab + 3b2 The following mode of forming the successive numbers to be subtracted is the most convenient in practice. Write down 3 times the first number, and three times its square separately, the former one line lower than the other, and to the left of it as in the margin. To 3 a add the second number b, and multiply the sum by b, placing the product below 3 a2 and adding it thereto. The sum 3 a2 + 3 ab+b being multiplied by b, produces the entire number to be subtracted. As the part 3 a2 forms a large part of the factor 3 a2 + 3 a b +b2, it soon becomes available for the determination of the next figure, by using it as a trial divisor. To show how the process may be continued, change a into a + b, and b into c in the arrangement in the margin; then 3 a becomes 3 a +36 and 3 a2 becomes 3 a2 + 6 ab+ 362; now to obtain these, we have only to add 2b to the one column, and b2 to the sum of the last two lines of the other column. Then to 3 a + 3b add the next number c; multiply the sum by c, placing the product below 3 a2 + 6 a b + 3 b2; then adding and multiplying the sum by c, the next entire number to be subtracted will be obtained. In this way the cube root may be extracted with the greatest facility, and when the root cannot be accurately obtained, it may be approximated to, and the work contracted as in the following example. 90 2700 46656 (30+6 = 36 6 576 19656 96 3276 19656 Ex. 1. Find the cube root of 46656. Here the number of tens is evidently 3, for 303 is less, and 403 greater than 46656, or 33 is ess and 43 greater than the second period 46. Writing 3 times 30, and 3 times the square of 30 on the left, we ask how many times is 2700 contained in the remainder 19656? The quotient is 7, which will be found a unit too much; taking 6 then it is added to 90, and the sum 96 is multiplied by 6. The product 576 is written below 2700 and added thereto; then the sum 3276 is multiplied by 6, and the product being equal to the remainder, the process terminates. In the next example we shall omit the ciphers, and place the figures as they arise in their proper places. |
