. Carnegie Institution of Washington publication. CHAP. II] SOLUTION OF ax+by = c. 49 J. L. Lagrange22 used the method of Saunderson13 and noted that the process is equivalent to the usual one of converting b/a into a continued fraction. He23 gave a more popular account [results as in Lagrange17]. C. F. Gauss24 employed the notations £ = O,0] = 0a + l, C=[a,0, y^ = yB+a, [or, 0, 7, 5]= &C+B, Apply the process to a and 6 which are relatively prime and positive, with a ^ b; let a = ab + c, b = (3c + d, c = yd + e, •—, m = pn + 1, so that a = O, M, ' • ' , 7, 0, «1 6 = [n, M, • • ' , 7
. Carnegie Institution of Washington publication. CHAP. II] SOLUTION OF ax+by = c. 49 J. L. Lagrange22 used the method of Saunderson13 and noted that the process is equivalent to the usual one of converting b/a into a continued fraction. He23 gave a more popular account [results as in Lagrange17]. C. F. Gauss24 employed the notations £ = O,0] = 0a + l, C=[a,0, y^ = yB+a, [or, 0, 7, 5]= &C+B, Apply the process to a and 6 which are relatively prime and positive, with a ^ b; let a = ab + c, b = (3c + d, c = yd + e, •—, m = pn + 1, so that a = O, M, ' • ' , 7, 0, «1 6 = [n, M, • • ' , 7, 0]- Take z = [>, • • •, 7, 0], y = [M, • • -, 7, 0, «]• Then ax = by + (- 1)* if fc is the number of the terms a, 0, • • • , p, n. Cf . Pilatte25 solved aix + axi = 6, where fli and a are relatively prime, a > ai, by applying the greatest common divisor process: a = a\q\ 02, #1 Replacing a by its value, we get x = x2 — q&i, where xz = (b — must be integral. Thus a2x{ + a^x2 = b. Proceeding similarly with the latter equation, we get a3xz + azx3 = b, • • • , zn_i + an-ixn = b. Eliminat- ing Xn-ij xn-z, • • • , we get x = d= ab =F axw, where a is an integer deter- mined by the process. P. Nicholson26 gave a method best explained by his example 500 - llx llx - r y = —^~ = 14 - -^5— , r = 10. Divide 35z by llx — r to get the remainder 2x + 3r. Then divide llx — r by 2x + 3r to get the remainder x — 16r, in which the coefficient of x is unity. The remainder 20 from the division of 16r = 160 by 35 is the least positive x. But in the example 200 - 5x 5x-r 2/ = — ir- = 18--n-, r = 2, we reach the remainder x + 2r in which the sign is plus; thus 11 — 2r = 7 is the least x. G. Libri27 gave as the least positive integral solution x of ax + b = cy, where a and c are relatively prime,. sin 22 Jour, de l'6cole polyt., cah. 5, 1798, 93-114; Oeuvres, VII, 291-313. 23 Ibid., cahs. 7, 8, 1812, 174-9, 208-9; Reprint
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