Arrangement of Successive Convergents in Order of Accuracy. 657 
Write down 0/1 and 1/0 
0 + 10 + 2x1. 
then Y^^' 1 _^ 2 X 0 ' ^^^^ 
then;-^-. .;-^^-.-.3/land5/2 
and so on, the number of terms in each set being determined by the corre- 
sponding partial quotient. 
The terms to be removed are now the earlier members of a set. 
When these are removed we obtain a series of fractions of least com- 
plexity nearest x in absolute value by reading along the successive lines 
completely from left to right, e. g. the fractions nearest 888/365 with denomi- 
nators less than 10, 20, 40, 100 are 17/7, 17/7, 90/37, 163/67, all being in 
defect. 
The particular fraction 1/0 is exceptional. 
The actual differences are tabulated below in the order of the previous 
table : 
2-43 00 
(1-43), -43 (-57), -067 
(•10), -033, -0042 (-0115), (-0046), -0019, -00054 
•00045, -000041 (-00011), (-000050), -000023, -000009, 0. 
A further example is added to show the special points mentioned in § 2 : 
2 + i + i +i + i 
The table of convergents is 
0/1 1/0 
1/1, 2/1 (3/1), (5/2), 7/3, 9/4 
11/3, 20/9 (29/13), 49/22, 69/31, 89/40 
(109/49), 198/89 (287/129), 485/215, 683/307, 881/396. 
The convergents 3/1, 29/13, and 287/129 can be removed at once. 
In settling the doubtful ones remember that we use go instead of the 
first partial quotient, and oo for any partial quotient after the last or before 
the first. 
For exclusion of 1/1 4<oo. .-.1/1 is excluded. 
For 5/2 2 < 00 . 5/2 is excluded. 
For 11/3 4 = 4, 2 >oo; this not true. .-. 11/3 is included. 
For 49/22 2 = 2, 4 = 4, oo = oo. .-. f|- and are equidistant. 
For 109/49 4 = 4, oo > 2. .-. 109/49 is excluded. 
485/218 00 < 2 ; this is not true. .-. 485/218 is included. 
