Elements of Algebra

CHAP. I. ADDITIONS. 483 $\frac{B+A}{B'+A'}$, $\frac{2B+A}{2B'+A'}$, $\frac{3B+A}{3B'+A'}$ Now, it is evident, that the denominators of these fractions form an increasing arithmetical series from $A'$ to $C'$; and we shall see that the fractions themselves also increase continually from $\frac{A}{A'}$ to $\frac{C}{C'}$; so that it would now be impossible to insert in the series $\frac{A}{A'}$, $\frac{B+A}{B'+A'}$, $\frac{2B+A}{2B'+A'}$, $\frac{3B+A}{3B'+A'}$, $\frac{4B+A}{4B'+A'}$, or $\frac{C}{C'}$, any fraction, whose value would fall between the values of two consecutive fractions, and whose denominator also would be found between the denominators of the same fractions: for, if we take the differences of the above fractions, since $BA' - AB' = 1$, we have, $\frac{B+A}{B'+A'} - \frac{A}{A'} = \frac{1}{A'(B'+A')}$ $\frac{2B+A}{2B'+A'} - \frac{B+A}{B'+A'} = \frac{1}{(B'+A') \times (2B'+A')}$ $\frac{3B+A}{3B'+A'} - \frac{2B+A}{2B'+A'} = \frac{1}{(2B'+A') \times (3B'+A')}$ $\frac{C}{C'} - \frac{3B+A}{3B'+A'} = \frac{1}{(3B'+A')C'}$ whence we immediately perceive, that the fractions $\frac{A}{A'}$, $\frac{B+A}{B'+A'}$, &c. continually increase, since their differences are all positive; then, as those differences are equal to unity, if divided by the product of the two denominators, we may prove, by a reasoning analogous to that which we employed (Art. 12), that it is impossible for any fraction, $\frac{m}{n}$, to fall between two consecutive fractions of the preceding series, if the denominator $n$ fall between the denominators of those fractions; or, in general, if it be less than the greater of the two denominators. Farther, as the fractions of which we speak are all greater than the real value of $a$, and the fraction $\frac{B}{B'}$ is less than it, it is evident that each of those fractions will approximate towards the value of the quantity $a$, so that the difference I I 2