If the principle holds up to and including any convergent, it holds for the next convergent. Suppose it holds up to and including the kth convergent. We then as a denominator where the preceding has dr. Therefore, if we substitute without assuming that the principle holds beyond the kth convergent. = 1 dk+1 = dk+1(d2Nk−1 + Nk−2) + Nk−1 − dk+1Nk + Nk−1 ̧ a result in accordance with the principle. Therefore, since the principle holds to and including the third convergent, it holds for the fourth; then, since it holds for the fourth, it holds for the fifth; and so on. This method of proof is called Proof by Mathematical Induction. Properties of Convergents. 8. (i.) The successive convergents, beginning with the zeroth, are alternately less and greater than the continued fraction. The symbol, read difference between, is placed between two numbers to indicate that the less is to be subtracted from the greater. E.g., 3~4 4~3=4−3=1. (ii.) The difference between any two consecutive convergents is 1 divided by the product of their denominators. (iii.) Each convergent is nearer in value to the continued fraction than any preceding convergent. (iv.) The convergents of even order continually increase, but are always less than the continued fraction; while the convergents of odd order continually decrease, but are always greater than the continued fraction. Ey., 3 10 1 3 45 151 7 47 In a terminating continued fraction, the last convergent will, of course, be the continued fraction, and therefore neither greater nor less than itself. In the first convergent, the partial quotient d1 is too small by 1 hence di In the second convergent, the second partial quotient d is too small is also too great; 1 by ; hence 1 is too small. 1 1 di + da And so on. (ii.) Since da dit da we have only to prove If it holds for any two consecutive convergents, it holds for the second of these two and the next convergent. We have N ̧D ̧+1 ~ D ̧Ñ ̧+1 = Na(dr+1Dx + Dr−1)~ Dɛ(Ngdk+1 + Na−1) Therefore, if the principle holds for (iii.) Then, But N+1 differs from V only in having d+1 where V has Dk+1 k-1 ~ = KD+ Dk-1 NDK-1~ NK-1 Di Dk (KDk + Dk-1) N1−1 ~ у – Nê−1 、 KN1⁄2 + Nk−1 − K (Nk−1Dk ~ NkDk−1) and DR-1 Hence, any convergent is nearer in value to the continued fraction than the immediately preceding convergent, and consequently than any preceding convergent. (iv.) The proof follows at once from (iii.) and (i.). Limit to Error of Any Convergent. 9. Since, by Art. 8 (i.), the value of a continued fraction is between the values of any two consecutive convergents, it must differ from either of them by less than they differ from each other. Therefore, an error of taking N for the continued fraction Dx Hence, to find a convergent which differs from the continued 1 fraction by less than we have only to compute successive m convergents up to, wherein d. D <m. D Ex. Find an approximation to 3.14159, correct to five decimal places. To Reduce a Quadratic Surd to a Continued Fraction. 10. The general method may be illustrated by particular examples. Ex. Reduce √14 to a continued fraction. Since the greatest integer contained in √14 is 3, we assume Since the greatest integer in this value of d, is 1, we assume |