toward the right for the decimal. We must be careful to have the last group on the right of the decimal point contain three figures, annexing ciphers when necessary. 240. Notice that if a denotes the first term, and b the second term of the root, the first complete divisor is 3 a2 + 3 ab + b2, and the second trial divisor is 3 (a + b)3, that is, 3a2+6 ab +362, which may be obtained by adding to the preceding complete divisor its second term and twice its third term. Extract the cube root of 5 to five places of decimals. After the first two figures of the root are found, the next trial divisor is obtained by bringing down the sum of the 210 and 49 obtained in completing the preceding divisor; then adding the three numbers connected by the brace, and annexing two ciphers to the result. The last two figures of the root are found by division. The rule in such cases is that two less than the number of figures already obtained may be found by division without error, the divisor being three times the square of the part of the root already found. Find to four decimal places the cube root of: 241. Since the fourth power is the square of the square, and the sixth power the square of the cube, the fourth root is the square root of the square root, and the sixth root is the cube root of the square root. In like manner, the eighth, ninth, twelfth, root may be found. ..... EXERCISE 89. Find the fourth root of: 1. 81 x + 108 x3 + 54 x2 + 12 x + 1. 2. 16 x1 — 32 ax3 + 24 a2x2 — 8 a3x + a1. 3. 1+4x+x3 + 4x2+10x + 16x3 +10x2+19x+16x5. Find the sixth root of: 4. 1+6 d + d + 6 d3 + 15 d + 20 d° + 15 ď2. 6. 118y+135y2 - 540 y3 +1215 y* - 1458 y5+ 729 yo. Find the eighth root of: 7. 1-8y+28 y2 — 56 y3 +70y* — 56 y5 + 28 yo — 8 y1 + y3. CHAPTER XVI. THEORY OF EXPONENTS. 242. If n is a positive integer, we have defined an to mean the product obtained by taking a as a factor n times, and called a" the nth power of a; we have also defined Va as a number which taken n times as a factor gives the product a, and called a the nth root of a. 243. By this definition of a" the exponent n denotes simply repetitions of a as a factor; and such expressions a2, -8 as a a have no meaning. It is found convenient, however, to extend the meaning of a" to include fractional and negative values of n. 244. If we do not define the meaning of a" when n is fractional or negative, but require that the meaning of a" must in all cases be such that the fundamental index law shall always hold true, namely, we shall find that this condition alone will be sufficient to define the meaning of a" for all cases. 245. Meaning of a Zero Exponent. By the index law, Therefore, the zero power of any number is equal to unity. 246. Meaning of a Fractional Exponent. By the index law, The meaning, therefore, of a", where m and n are positive integers, is the nth root of the mth power of a. Hence, The numerator of a fractional exponent indicates a power and the denominator a root. 247. Meaning of a Negative Exponent. By the index law, if n is a positive integer, That is, a2 and a-" are reciprocals of each other (§ 169), so that a-" = 1 ans and an = 1 αι 248. Hence, we can change any factor from the numerator of a fraction to the denominator, or from the denominator to the numerator, provided we change the sign of its exponent. ab2 c3d3 Thus, may be written ab2c-3d-3, or 1 249. We have now assigned definite meanings to fractional exponents and negative exponents, by assuming that the index law for multiplication, am × a" = am+n, is true for all values of the exponents m and n. It remains to show that the index laws established for division, involution, and evolution apply to fractional and negative exponents. 250. Index Law of Division for all Values of m and n. To divide by a number is to multiply the dividend by the reciprocal of the divisor. Therefore, for all values of m and n, 251. Index Law of Involution and Evolution for all Values of m and n. To prove (am)n = amn for all values of m and n. CASE 1. Let m have any value, and let n be a positive Ρ CASE 2. Let m have any value, and n = p and q being positive integers. Then, (am)1 = √(am)p = √amp mp =aq. CASE 3. Let m have any value, and n = positive integer or a positive fraction. r, r being a |
