be the convergent formed from the quotients a, b, ... down to k inclusive; and let be the convergent immediately preceding; then, as in Art. 607, P' Let be the convergent formed from the quotients r, s,. کہا ... From (1) and (2) by eliminating y we obtain a quadratic equation in x with rational coefficients. To find x under an irrational form we should take the positive value of y found from (2), that is, from Q'y2 + (Q-P) y - P=0, and substitute it in (1). EXAMPLES OF CONTINUED FRACTIONS FROM QUADRATIC SURDS. Express the following fourteen surds as continued fractions, and find the first four convergents to each: 17. Shew that the 9th convergent to (33) will give the true value to at least 6 places of decimals. 18. Find limits of the error when 211 is taken for √(23). 916 19. Shew that differs from (23) by a quantity less 1151 20. Find limits of the error when is taken for (23). 21. Find limits of the error when the 8th convergent is taken shew that the second convergent differs from the true value by a quantity less than 1÷a (4a2 + 1); and thence by making a=7, 99 shew that 70 differs from (2) by a quantity less than 1 13790* 25. Shew that the 3rd convergent to √(a2+a+1) is (2a +1). √3 13 26. Find convergents to ; shew that exceeds the true value by a quantity less than 4 1 2910° 27. Find the 6th convergent to √(3). 30 28. Find the 6th convergent to the positive root of 2x-3x-6= 0. 29. Find the 6th convergent to each root of 30. Find the 7th convergent to the greater root of XLVI. INDETERMINATE EQUATIONS OF THE FIRST DEGREE. 623. When only one equation is given involving more than one variable, we can generally solve the equation in an infinite number of ways; for example, if ax + by = c, we may ascribe any value we please to x, and then determine the corresponding value of y. Similarly, if there be any number of equations involving more than the same number of variables, there will be an infinite number of systems of solutions. Such equations are called indeterminate equations. 624. In some cases, however, the nature of the problem may be such, that we only want those solutions in which the variables have positive integral values. In this case the number of solutions may be limited, as we shall see. We shall proceed then to some propositions respecting the solution of indeterminate equations in positive integers. The coefficients and constant terms in these equations will be assumed to be integers. 625. Neither of the equations ax + by = c, ax-by=c can be solved in integers if a and b have a divisor which does not divide c. For, if possible, suppose that either of the equations has such a solution; then divide both sides of the equation by the common divisor; thus the left-hand member is integral and the right-hand member fractional, which is impossible. If a, b, c have any common divisor, it may be removed by division, so that we shall in future suppose that a and b have no common divisor. 626. Given one solution of ax-by=c in positive integers, to find the general solution. Suppose xa, y = ẞ is one solution of ax-by=c, so that aa-bẞ=c. By subtraction a (x − a) − b (y – ẞ) = 0; therefore α α y-B b Ꮳ - a = Since is in its lowest terms, and x and y are to have b integral values, we must have (as will be shewn in the chapter on the Theory of Numbers), x — a = bt, y-B = at, where t is an integer; therefore x = a + bt, y = B+ at. Hence if one solution is known, we may by ascribing to t different positive integral values, obtain as many solutions as we please. We may also give to t such negative integral values as make bt and at numerically less than a and ẞ respectively. We shall now shew that one solution can always be found. 627. A solution of the equation ax-by=c in positive integers can always be found. α Let be converted into a continued fraction, and the succesb P sive convergents formed; let ? be the convergent immediately α preceding; then aq-bp = ± 1. First suppose aq-bp = 1, therefore aqc-bpc = c. Hence x=qc, y=pc is a solution of ax — by = c. Next suppose aq-bp-1, then a (b-q)-b (a-p)= 1; therefore a (b-q)c- b (a − p) c = c. Hence (b-q)c, y=(a-p)c is a solution of ax- -by= = C. = If a=1, the preceding method is inapplicable; in this case the equation becomes x-by= c; we can obtain solutions obviously by giving to y any positive integral value, and then making x=c+ by. Similarly if b = 1. 628. Given one solution of the equation ax + by =c in positive integers, to find the general solution. Suppose that x=a, y =ß is one solution of ax + by=c, so that aa+bẞ= c. By subtraction, a (x − a) + b ( y − ẞ) = 0; therefore α α b = B-y Ꮳ α Since is in its lowest terms and x and y are to have inte b 629. It may happen that there is no such solution of the equation ax + by = c. For example, if c is less than a + b, it is impossible that c = ax + by for positive integral values of x and y, excluding zero values. By the following method we can find a solution when one exists. Let be converted into a continued fraction, and let 2 α Ъ α be the convergent immediately preceding; then aq − bp = ± 1. |