Elements of Algebra

490 ADDITIONS. CHAP. I. that are less, and those that are greater than the given fraction, we may also insert different secondary fractions to complete the series; and, for this purpose, we shall follow the same process as before, but taking successively, instead of each number of the upper series, all the integer numbers less than that number, when there are any. So that, considering first the increasing fractions, 1, 1, 1, 15. $4/1, 33/8, 161/39, 2865/694, 86400/20929$ we see that, since unity is above the second, the third, and the fourth, we cannot place any intermediate fraction, either between the first and the second, or between the second and the third, or between the third and the fourth; but as the last fraction stands below the number 15, we may place, between that fraction and the preceding, fourteen intermediate fractions, the numerators* of which will form the arithmetical progression $2865 + 5569$, $2865 + 2 \times 5569$, $2865 + 3 \times 5569$, \ldots their denominators will also form the arithmetical progression $694 + 1349$, $694 + 2 \times 1349$, $694 + 3 \times 1349$, \ldots So that the complete series of increasing fractions will be $4/1, 33/8, 161/39, 2865/694, 8434/2043, 14003/3392, 19572/4741, 25141/6090,$ $30710/7439, 36279/8788, 41848/10137, 47417/11486, 52986/12835, 58555/14184,$ $64124/15533, 69693/16882, 75262/18231, 80831/19580, 86400/20929$. And, as the last fraction is the same as the given fraction, it is evident that this series cannot be carried farther. Hence, if we choose to admit those intercalations only in which the error is too much, the simplest and most exact will be those of one day in four years, or of eight days in thirty-three years, or of thirty-nine in a hundred and sixty-one years, and so on. Let us now consider the decreasing fractions, 7, 3, 16, 1. $39/7, 128/31, 2704/653, 5569/1349$. And first, on account of the number 7, which is above the first fraction, we may place six others before it, the numerators of which will form the arithmetical progression, $4+1, 2 \times 4 + 1, 3 \times 4 + 1, \ldots$ and the denominators of which will form the progression * Because $5569/1349$ is the principal fraction between $2865/694$ and $86400/20929$, as is found in the foregoing series. See page 485. B.