342 
Proceedings of Royal Society of Edinburgh. [sess. 
B = 1.C + 0.D + E, leaving E = . 2599210. Here, however, on 
comparing C, D, E, we find that D may be taken thrice from C, 
and we get C = 3. D + 0.E + F, leaving F = 175597 ; and thus the 
work proceeds, as shown in the accompanying scheme : — 
1.5874011 =A= 1.B+0.C+D 
1.2599210 = B = 1.C+0.D+E 
1.0000000- C = 3.D + 0.E + F 
.3274801 = D = 1.E + 3.F+G 
.2599210 = E =14.F +0.G +H 
175597 = F = 
148800 = G = 
140852 = H = 
26797 = 1 = 
7948 = K = 
6867 = L = 
2953 = M = 
1081 = N = 
961 = 0 = 
791 = P = 
120= Q- 
50 = K 
1.G + 0.H + I 
1.H + 0.I + K 
5.1 +0.K + L 
3.K+0.L + M 
1. L 4- O.M + N 
2. M + 0.N + O 
2.H+0.O + P 
1.0 + O.P+Q 
1. P +1.Q +K 
6.Q +1.R+S 
2. E +1.S 
21 = S . 
In the work so carried on, there is no appearance of recurrence 
among the quotients, such as is seen when two quantities are compared 
for the purpose of finding the square-root. But here we have 
followed the purely arbitrary rule of making as many subtractions 
as possible of the second from the first of the three quantities. We 
might have written, 
C-2.D + 1.E + F 
or C = l.D + 2.E + F, 
or even C = O.D + 3.E + F . 
In such case we should have a change in the sequence of the 
quotients, but any of these, if continued to exhaust the absolute 
numbers proposed, would necessarily result in reproducing those 
very numbers. 
