Elements of Algebra

CHAP. I. ADDITIONS. 491 $1, 2, 3, \ldots$*; also, an account of the number 3, we may place two intermediate fractions between the first and the second; and between the second and the third we may place fifteen, on account of the number 16 which is above the third; but between this and the last we cannot insert any, because the number above it is unity. Farther, we must remark, that, as the preceding series is not terminated by the given fraction, we may continue it as far as we please, as we have shewn, Art. 18. So that we shall have this series of decreasing fractions, $\frac{5}{1}$ $\frac{9}{2}$ $\frac{13}{3}$ $\frac{17}{4}$ $\frac{21}{5}$ $\frac{25}{6}$ $\frac{29}{7}$ $\frac{62}{15}$ $\frac{95}{23}$ $\frac{128}{31}$ $\frac{289}{70}$ $\frac{450}{109}$ $\frac{611}{148}$ $\frac{772}{187}$ $\frac{933}{226}$ $\frac{1094}{265}$ $\frac{1255}{304}$ $\frac{1416}{343}$ $\frac{1577}{382}$ $\frac{1738}{421}$ $\frac{1899}{460}$ $\frac{2060}{499}$ $\frac{2221}{538}$ $\frac{2382}{577}$ $\frac{2543}{616}$ $\frac{2704}{655}$ $\frac{5569}{1349}$ $\frac{91969}{22278}$ $\frac{178369}{43207}$ $\frac{264769}{64136}$ $\frac{351169}{85065}$ $\frac{437569}{105994}, \ldots$ which are all less than the fraction proposed, and approach nearer to it than any other fractions expressed in simpler terms. Hence we may conclude, that if we only attend to the intercalations, in which the error is too small, the simplest and most exact are those of one day in five years, or of two days in nine years, or of three days in thirteen years, $\ldots$ In the Gregorian calendar, only ninety-seven days are intercalated in four hundred years; but it is evident, from the preceding series, that it would be much more exact, to intercalate a hundred and nine days in four hundred and fifty years. But it must be observed, that in the Gregorian reformation, the determination of the year given by Copernicus was made use of, which is $365^d. 5^h. 49'. 20''$; and substituting this, instead of the fraction $\frac{86400}{20929}$, we shall have $\frac{86400}{20960}$, or rather $\frac{540}{131}$; whence we may find, by the preceding method, the quotients $4, 8, 5, 3$, and from them the principal fractions, $4, 8, 5, 3.$ $\frac{1}{4}$ $\frac{33}{8}$ $\frac{169}{41}$ $\frac{540}{131}$ which, except the first two, are quite different from the fractions found before. However, we do not perceive among them the fraction $\frac{97}{400}$ adopted in the Gregorian calendar; and this fraction cannot even be found among the intermediate fractions, which may be inserted in * See page 485.