The integer next below 8+2 is 4; and we have Thus 8 is equal to a periodic continued fraction with one nonperiodic element, which is half the last quotient of the recurring portion; and it will be proved later on that this law holds good for every quadratic surd. 363. We now proceed to shew how to convert any quadratic surd into a continued fraction. Let N be any quadratic surd, and let a be the integer next below N; then Since Na is positive and less than 1, it follows that The process can be continued in this way to any extent may be desired. Thus N = a + 1 1 1 364. To shew that any quadratic surd is equal to a recurring continued fraction. r It is first necessary to prove that the quantities which, in the preceding Article, are called ɑ, ɑ„, ɑ., ......., 71, 72, 73, ••• are all positive integers. It is known that N is a positive integer, and that a, b, c, d,... are all positive integers. We have the following relations : Now it is obvious from (i) that r, is an integer. Thus a, br, -a, and r,=1+2ab-b'r,; whence it follows that a and r, are integers, since r, is an integer. whence it follows that a, and r, are integers, since a, and r, are integers. Then again, from (iv) we have a=dra, and r1 = 1 + 2a,d − d3r ̧ ; and r3 whence it follows that a, and r, are integers, since are integers. And so on; so that a, and r, are integral for all values of n. We have now to prove that a, and r„ are positive for all values of n. We know that a, b, c, &c. are the positive integers √N+ a √N+α &c. Hence √N — a, next below √N, 1 √N-α2, √N-a,, &c., and therefore also N- a2, √N+a ̧ it Again, since b is the integer next below follows that N + a < br,+r,. Now, a cannot be equal or greater than br,, for then N< r,, and therefore a <r1; therefore a <br1, since r, is positive and b a positive integer. Hence a <br1, so that a, is positive. √N +αg, it Again, since c is the integer next below follows that √N+a2<cr2+~1⁄2• = r And we cannot have acr,, for then √N<r,, and therefore a, <r,<cr„, since r, is positive and c a positive integer. Thus a,< cr2, so that a, is positive. And so on; so that a, is positive for all values of n. Having shewn that the quantities r1, r, r,, &c. and a, aga, &c. are all positive integers, it follows from the n n-1 = 2 relation rr, N-a, that a, is less than N, so that ana; hence the only possible values of a„ are 1, 2,..., a. Then, from the relation a+an+1=k.r, where k is a positive integer, it follows that cannot be greater than 2a. n n cannot have more than 2ax a different values; and therefore after 2a quotients, at most, there must be a recurrence. 365. Theorem. Any quadratic surd can be reduced to a periodic continued fraction with one non-recurring element, the last recurring quotient being twice the quotient which does not recur; also the quotients of the recurring period, exclusive of the last, are the same when read backwards or forwards. Let N be the quadratic surd. Then, from the preceding Article, we know that √N is equal to a periodic continued fraction. We also know that any periodic continued fraction is equal to one of the roots of a quadratic equation with rational coefficients; and the only quadratic equation in a with rational coefficients of which N is one root is the equation x2 - N = 0. Now the roots of x2 N = 0 are both greater than unity in absolute magnitude, and the roots are of different signs: it therefore follows from Articles 360 and 361 that the continued fraction which is equal to N must be a mixed recurring continued fraction with one non-recurring element. root of a quadratic equation with rational coefficients; and as this positive root is Na, the negative root must be — √N-a. Hence, Art. 360, we have 366. To shew that any series can be expressed as a continued fraction. Then the sum of n terms of the series (i) is equal to the nth convergent of the continued fraction This can be proved by induction, as follows. Assume that the sum of the first n terms of (i) is equal to the nth convergent of (ii). Another term of the series is taken into account by changing u into u,+U+1; and, by changing u into un+un+ un S. A. n 1' n-1 + u will n become 30 |