349. As a test of the correctness of the converging fractions, it may be observed, that the difference between any two consecutive fractions is always a fraction having +1 or -1 for its numerator, according as that which is subtracted from the other is in an odd or even place. For A &c. correctly obtained according to the C Rule, is in its lowest terms; for if and be any two consecutive ones, B D have a common measure greater than 1, it will measure AD and BC, and, therefore, AD ~ BC or 1, which is impossible, since it is greater than 1. 350. Hence also we may determine the limit of the error in taking a b the other, (Art. 345); therefore the difference between and either of them 22 Or, since D> B, à fortiori, Thus in Ex. Art. 345, differs from the value of 3.14159 by a quantity less than 7 by a quantity less than 351. To find the value of a continued fraction, when the quotients q, r, s, &c. recur in any certain order. by the solution of which quadratic equation the value of ≈ may obtained. be a x − b = 0; whence the value of x is to be found. 352. The roots of quadratic equations may sometimes be approximated to by means of continued fractions. Let x2 - px - q = 0; then x = Thus, so that x = 오 P+ p+ P+ &c. sometimes be made use of to which, though not a contiuued fraction of such a form as those to which the preceding theory applies, may yet calculate the value of x. 353. To find in the form of a continued fraction the value of x which satisfies the equation a* b. Substitute for x the numbers 0, 1, 2, 3, &c. until two consecutive numbers are found, n, and n + 1, such that then it appears that x < n + 1, and > n; so that x = n + and <a, by substituting the numbers between 1 and a for y in the last equation, two consecutive numbers, p, and p + 1, tinuing the process in the same manner, the fraction, expressing the value of x, may be continued. Ex. Required the value of x in 10" 2. = By substituting 0 and 1 for x, it appears that x > 0, and <1; let Again, it appears that > 3, and < 4; let z = 3 + INDETERMINATE EQUATIONS AND UNLIMITED 354. WHEN there are more unknown quantities than independent equations, the number of corresponding values which those quantities admit is indefinite (Art. 198). This number may be lessened by rejecting all the values which are not integers; it may be farther lessened by rejecting all the negative values; and still farther, by rejecting all values which are not square or cube numbers; &c. By restrictions of this kind, the number of answers may be confined within definite limits; and problems are not wanting, in which such restrictions must be made. 355. If a simple equation express the relation of two unknown quantities, and their corresponding integral values be required, divide the whole equation by the coefficient which is the less of the two, and suppose that part of the result, which is in a fractional form, equal to some whole number; thus a new simple equation is obtained, with which we may proceed as before; let the operation be repeated, till the coefficient of one of the unknown quantities is 1, and the coefficient of the other a whole number; then an integral value of the former may be obtained by substituting 0, or any whole number, for the other; and from the preceding equations integral values of the original unknown quantities may be found. Ex. 1. Let 5x + 7y=29; to find the corresponding integral values of a and y. Dividing the whole equation by 5, the less coefficient, If 80, then a = 3 and y = 2, the only positive whole numbers which answer the conditions of the equation; for, if x and y are positive integers, 5s cannot be greater than 2, that is, s cannot be greater than and s cannot be negative, for then x would be negative. 2 5' If negative values of x and y are not excluded, then an indefinite number of such solutions may be found by putting 1, 2, 3, &c. — 1, − 2 -3, &c. for s. Ex. 2. To find a number which being divided by 3, 4, 5, gives the remainders 2, 3, 4, respectively. Let be the number, x = 5−y + 2(2-y) 2-y =s, or y=2-5s, |