the tens can have no figure less than hundreds, and the product of the tens by the units, none less than tens. If the number contains more places of figures, the same reasoning will apply as in the square root. The number must be divided into periods of three figures each; for we have (10) 1000, (100)31.000.000, (1000) 1.000.000.000, from which it will appear, that the cube of tens can have no significant figure in the first period, the cube of hundreds none in the first two, and the cube of thousands none in the first three. The number of periods will, therefore, denote the number of figures in the root. Let it be required to extract the cube root of 199176704. 199.176.704(584 125 Div. (5)2 x3 = 25 x 375.00 | 74.176 1st dividend. 8764 70.112 Complete divisor. 2d div. (58)3×3=3364×3=10092.00 | 4064704 2d dividend. 58x4x3= 696.0 2d complete divisor. (4)2 = 16 1016176 4064704 In this example, I first seek the root of the two left hand periods, which I find to be 58. I then call this 58 tens, and proceed to find the last figure of the root by taking three times the square of 58 tens, for a divisor, and bringing down the last period to the remainder, for a dividend. The divisor is completed in the same manner as in the last example. If the given number contained four periods, the same method would be followed, always considering the figures of the root, already found, so many tens, and the figure sought, so many units. For the rule for extracting the cube root of numbers, see Arith. art. (57), p. 70. 3 EXAMPLES. 1. √7645373 = 197. 2. √52734375 — 375. 3. V334255384 = 694. GENERAL METHOD OF EXTRACTING THE ROOTS OF NUM BERS OF ANY POWERS WHATEVER. (52.) The roots of any powers whatever may be extracted by a method analogous to that given (48) for extracting the roots of algebraic quantities. The first two terms of the developement of any power of a binomial, as (a+b)" is always a"+na"-1b (43). Hence it will appear, that if we make a the tens of any number, and b the units, we can approximate to the units by raising the tens to a power one less, or a"-1, and multiplying this power by n, the index of the root to be extracted, for a divisor. To find in what part of the number we are to look for the power of the tens, we are to point off the number in periods of n figures each; for, in raising 10 to any power, we add a cipher for every multiplication; thus, (10) 100, (10)3 = 1000, (10)* 10000; and since in raising 10 to the nth power, it is to be multiplied by itself n-1 times, we shall have for the nth power of 10, 1 and n ciphers. Hence, the nth power of the tens of any number, can have no significant figure in the first n figures, on the right. = Let it now be required to extract the 4th root of 331776. In this example, n= 4; I therefore point off four figures from the right, as containing no part of the fourth power of the tens. The greatest fourth power in the left hand period is 16, the fourth power of 2. I therefore conclude that 2 is the tens of the root: to find the units, I raise 2 to the third power, and multiply it by 4, the index of the root, for a divi sor. It then corresponds to na" of the developement of (a+b)". For a dividend, I bring down to the remainder the last period. I have now, for a divisor, 32, and for a dividend, 179.376; and since 32 is four times the cube of the tens, it can occupy no lower place than that of thousands. I therefore reject the last three places of figures in the dividend, and see how many times the divisor is contained in 179. I find the quotient to be 5. But the dividend being what remains after the fourth power of the tens is taken out, corresponds to 4a3b+6a2b2+4ab3 + b2, the first term of which, divided by 4a3, which in this case is 32, will give, for a quotient, b the units of the root. It is evident, therefore, that the whole of this dividend divided by 32, will give a quotient somewhat too large, and on this account I take 4, instead of 5, for the units of the root. Having raised 24 to the fourth power, I find it exactly equal to the given number, which shows that it is the root required. The greater the index of the root to be extracted, the more will the quotient exceed the units of the root, for the higher the power of a binomial, the greater will be the number of terms remaining after the first term is taken away. If there are more periods than one, the root will consist of more than two places of figures; but the same principles will apply; for having obtained two figures of the root, we may consider them as expressing so many tens, and proceed as before. RULE. 1. Divide the number into periods of as many figures each, as there are units in the index of the root. 2. Extract the root of the left hand period, and subtract its power from that period, and to the remainder bring down the first figure of the next period for a dividend. 3. Involve the root, already found, to a power one less than the index of the required root, and multiply this power by the index of the root, for a divisor. 4. See how many times this divisor is contained in the dividend, and place the result in the root. 5. Involve the whole root to the given power, and subtract it from the two left hand periods, and to the remainder bring down the first figure of the next period for a new dividend, and form a divisor from the whole root already found, in the same manner as before; proceed in like manner, till all the periods are brought down, always observing to subtract the root when involved to the given power, from as many of the left hand periods of the given number, as there are figures in the root. 2d Div. 3(56)2=9408)9424 2d dividend. NOTE. If the index of the root to be extracted, be such a number as can be decomposed into factors, we can obtain the root by successively extracting the roots which have these factors for indices, upon the principles laid down in Art. (49). EXTRACTION OF ROOTS BY APPROXIMATION. (53.) There are but few numbers, comparatively, that will admit of exact roots; these are called perfect powers. 1, 4, 9, 16, 25, 36, 49, &c. 1, 8, 27, 64, 125, 216, 343, &c. are perfect squares and cubes; but all the intermediate numbers will admit of no exact root, it being impossible to find any number, which, multiplied by itself once or twice, will produce any one of those numbers. Indeed, it can be shown, that the root of a number which is not a perfect power, can be neither a whole number nor a fraction. In the first place, the root cannot be a whole number, for then the number itself would be a perfect power of that root, which is contrary to the supposition. C To show that the root of an imperfect power cannot be a fraction, it is necessary to prove that the powers of any irreducible fraction, as are irreducible. To this purpose, we remark, that the terms of such a fraction being prime to each other, they can have no factor in common; and hence, their prime factors must be different. Let the prime factors of c, be, de....f, and the prime factors of b, hk.... l; then, dxex....f hxkx....l' C b |