Page:Elementary algebra (1896).djvu/421

403 Rh and we see that the numerator of the third convergent may be formed by multiplying the numerator of the second convergent by the third quotient, and adding the numerator of the first convergent; also that the denominator may be formed in a similar manner.

Suppose that the successive convergents are formed in a similar way; let the numerators be denoted by p_1, p_2, p_3, and the denominators by q_1, q_2, q_3,.

Assume that the law of formation holds for the nth convergent; that is, suppose

p_n = a_np_{n-1} + p_{n-2}, q_n = a_nq_{n-1} + q_{n-2}

The (n+ 1)th convergent differs from the nth only in having the quotient a_n+ {1}{a_{n+1}} in the place of a_n; hence the (n + 1)th convergent

$$=\frac{(a_n+ \frac{1}{a_{n+1}})p_{n-1} + p_{n-2}}{(a_n+ \frac{1}{a_{n+1}})q_{n-1} + q_{n-2}} = \frac{a_{n+1}(a_n p_{n-1} + p_{n-2}) +p_{n-1}}{a_{n+1}(a_n q_{n-1} + q_{n-2}) +q_{n-1}}$$

$$=\frac{a_{n+1}p_n+p_{n-1}} {a_{n+1}q_n+q_{n-1}}$$ by supposition.

If therefore we put

p_{n+1} = a_{n+1}p_n+p_{n-1}, q_{n+1} = a_{n+1}q_n+q_{n-1}

we see that the numerator and denominator of the (n +1)th convergent follow the law which was supposed to hold in the case of the nth. But the law does hold in the case of the third convergent, hence it holds for the fourth, and so on; therefore it holds universally.

Ex. Reduce 674 313 to acontinued fraction and calculate the successive convergents.

By Art 499, 674 313= 2 + {1}{6 +} {1}{1 +} {1}{1 +} {1}{11 +2}.

The successive quotients are 2, 6, 1, 1, 11, 2. The successive convergents are 2 1, 13 6, 15 7, 28 13, 323 150, 674 313.