2 48 11. √17, √1.5, √5 + 1' √3 − 1 Give the values correct to four places of decimals of— of 31416 12. √93 √5 + √2 √10 - 2 3.1416 of 40 32.16 Find the cube roots of— 13. 8 a3by12, 125 x12y3, a3 + 6a2b+ 12 ab2 + 863. 2 -1 16. x3 + x−3 + 3 (x + x−1), x3y−3 + 3 x2y−2 + 3 xy−1 + 1. Find the cube roots of 17. 5849513501832, 1371-330631. 18. 20-346417; 037, L08 Give the value of the following correct to four places of decimals: 21. ( √7 + 2) ( √7 − 1), (5 + √3) (4 + √12). 24. a3 (b − c) – b3 (a − c) + c3 (a - b), where a = √1.2, CHAPTER IV. GREATEST COMMON MEASURE AND LEAST COMMON MULTIPLE. Greatest Common Measure. 44. In Arithmetic (page 24) we defined the G.C.M. of two or more numbers as their highest common factor. In Algebra the same definition will suffice, provided we understand by the term highest common factor, the factor of highest dimen sions (Art. 18). This, it need hardly be remarked, does not necessarily correspond to the factor of highest numerical value. 45. To find the G.C.M. of two quantities. RULE.-Let A and B be the quantities, of which A is not of lower dimensions than B. Divide A by B, until a remainder is obtained of lower dimensions than B. Take this remainder as a new divisor, and the preceding divisor A as a new dividend, and divide till a remainder is again obtained of lower dimensions than the divisor; and so on. The last divisor is the G.C.M. Before giving the general theory of the G.C.M. we shall work out a few examples. Ex. 1. Find the G. C.M. of x2 6x27 and 2x2-11x-63. According to the above rule, the operation is as follows: Ex. 2. Find the G.C.M. of 10 x3 + 31 x2 63 x and 14x3 + 51 x2 54 x. We may tell by inspection that x is a common factor, which we therefore strike out of both, only taking care to reserve it. The quantities then become 10 x2 + 31 x 63, and 14x2 + 51x 54. We may now proceed according to rule, taking the former as divisor. We see, however, that the coefficient of the first term of the dividend is not exactly divisible by the coefficient of the first term of the divisor. Multiply therefore (to avoid fractions) the dividend by such a number as will make it so divisible, viz., by 5. This will not affect the G.C.M., as 5 is not a factor of the first expression, viz., 10 x2 + 31 x 63. It may as well be here mentioned that the G.C.M. of two quantities cannot be affected by the multiplication or division of one of the quantities by any quantity which is not a measure of the other. We shall, for a similar reason, reject certain factors or introduce them into any of the remainders or dividends during the operation. (See Art. 47). Rejecting the factor 19 of this remainder, we have— Hence, 2x + 9 is the last divisor, and multiplying this by x, the common measure struck out at the commencement, we find the G.C.M to be x (2 x + 9) or 2 x2 + 9 x. Ex. 3. Find the G. C.M. of x-7 x3-3x2-5x3 + 42x2-34x-21, and x3-11x+25x3 +19x2-49x-21. x5 – 11 x1 + 25 x3 + 19 x2 - 49 x - 21) x6 52 + 42x219 x3 49 x2 4 25 28 x1 4 205 24x3+91 x2 100 x3 + 76 x2 44 x 16x 124 x 34x-21(x+4 34x-21 (x+4 15 x2 + 183 x + 63 Multiplying the preceding divisor by 16, and taking the result for a dividend, we have16 x1 124 x3 + 15 x2 + 183x + 63)16 x (Multiplying this remainder by 4) 16x5 176x+400 x3 + 304x2 124 x 15 x3 + 183x2 + 336(x taking the result for a dividend, we have— 135 x2 + 1647 x + 567( − 2 x 2018 x2 + 1050 x Dividing this remainder by 14371, and taking the quotient It will be seen that we have introduced and rejected factors during the operation in order to avoid fractional coefficients. This, as will be seen from the general theory, will not affect the result, provided that no factor thus introduced or rejected is a measure of the corresponding divisor or dividend, as the case may be. Theory of the Greatest Common Measure. 46. Let A and B be the two algebraical quantities, and the operation as indicated by the rule (Art. 45) be performed. Thus, let A be divided by B, with quotient p and remainder C. Then let B be divided by C, with quotient q, and remainder D. Lastly, let °C be divided by D, with quotient r, and remainder zero. Then we are required to show that D is the G.C.M. of A and B. B)A (p C) B(q D)C(r rD (1.) D is a common measure of A and B. 0 Now, we have C = rD, B = qC + D, A = pB + C. Hence, D is a measure of C, and therefore of qC. It is therefore a measure of qC + D or B. Hence, also, D is a measure of pB, and since it is also a measure of C, it must be a measure of pB + C or A. But we have shown it to be a measure of B. Hence, D is a common measure of A and B. (2.) D is the G.C.M. of A and B. For every measure of A and B will divide A - pB or C; and hence every measure of A and B will divide B - qC or D. Now, D cannot be divided by any quantity higher than D, and, therefore, there cannot exist a measure of A and B higher than D. Hence, D is the G.C.M. of A and B. |