In reducing a common fraction to a decimal, when the number of decimal places equals the number of units in the denominator less one, all the possible different remainders will have been used, and hence the dividends, and therefore the quotients which constitute the circulate, will begin to repeat at this point, if not before. 7. When the reciprocal of any prime number is reduced to a repetend, the remainder which occurs at the close of the period is 1. For, since the reduction of the fraction to a circulate commenced with a dividend of 1 with ciphers annexed, that the quotients may repeat we must begin at the close of the period with the same dividend, and therefore the remainder at the close of the period must be 1. 8. The number of places in a repetend is always equal to the prime denominator of the common fraction producing it, less one, or to some factor of this number. For, the repetend must end when it reaches the point where it has as many places less 1 as there are units in the denominator of the producing fraction; hence if it ends before this, the number of places must be an exact part of the denominator less 1, that it may end when it has as many places as the denominator less 1. 9. When the reciprocal of any prime number is reduced to a repetend, the remainder which occurs when the number of decimal places is one less than the prime is 1. For, since the number of decimal places in the period equals the denominator less 1, or is a factor of the denominator less 1, at the close of a period consisting of as many places as the denominator less 1, there will be an exact number of repeating periods, and therefore the remainder will be 1. 10. A number consisting of as many 9's as there are units in any prime except 2 and 5, less 1, is divisible by that prime. For, if we divide 1 with ciphers annexed by a prime, after a number of places 1 less than the prime the remainder is 1; hence 1 with the same number of ciphers annexed minus 1, would be exactly divisible by the prime, but this number will be a series of 9's; therefore, etc. Thus, 999999 is divisible by 7. 11. A number consisting of as many l's as there are units in any prime (except 3), less 1, is divisible by that prime. For, since the prime is a divisor of a series of 9's, Prin. 10, which is equal to 9 times a series of l's, and 9 and the prime are relatively prime, it must be a divisor of a series of 1's. Thus, 111111 is divisible by 7; 1111111111 is divisible by 11. 12. A number consisting of any digit used as many times as there are units in a prime (except 3), less 1, is divisible by that prime. For, since such a series of l's is divisible by the prime, any number of times such a series will be divisible by the prime. Hence 222222, 333333, 444444, etc., are divisible by 7. 13. The same perfect repetend will express the values of all proper fractions having the same prime denominator, by starting at different places. Thus, _=,14285714285, etc. But =.1%, hence the part that follows 1 in the repetend of 4 is the repetend of ļ, that is, š=.42857i. Again, =.14%, hence the part that follows .14 in the repetend of is the repetend of ?, that is, =285714. In a similar manner we find & .857142, 15.571428; and the same thing we see is generally true. 14. In reducing the reciprocal of a prime to a decimal, if we obtain a remainder 1 less than the prime, we have onehalf of the period, and the remaining half can be found by subtracting the terms of the first half respectively from 9. Take }, and let us suppose, in decimating, we have reached a remainder of 6; now what follows will be the repetend of 4, and the repetend of , added to the repetend of ļ must equal 1, since % +=1; hence the sum of these two repetends must equal .999999 etc. (since .999 etc.= 1). Now in adding the terms of these two repetends together, that the sum may be a series of 9's, there must be just as many places before the point where 6 occurred as a remainder as after, hence 6 occurred as a remainder when we were half through the series. Again, since the sum of the terms of the latter and the former half of the repetend equals a series of 9's, each term of the first half of the repetend subtracted from 9 will give the corresponding term of the latter half of the series. Notes.—1. All perfect repetends possess this property, and a large number of those which are not perfect. Repetends possessing this property are called complementary repetends. 2. The last two properties are of great practical value in reducing common fractions to repetends. COMPLEMENTARY REPETENDS. 330. Complementary Repetends are those in which the terms of the first half of the period are respectively equal to 9 minus the corresponding terms of the second half of the period. 331. Complementary Repetends include all perfect repetends, and many that are not perfect. The following curious properties illustrate the principles presented : 1. If the last half of the terms of a perfect repetend are written in order under the first half, and added to the terms in the first half, the sum will be a succession of 9's. Thus, the fraction b=.052631578947368421, and this repetend written and added as suggested, will give 052631578 999999999 2. If the remainders obtained in reducing the common fraction to a repetend are written in the same way, and added, each sum will be the denominator of the common fraction. Thus, the remainders in reducing i, are 10, 5, 12, 6, 3, 11, 15, 17, 18 9, 14, 7, 13, 16, 8, 4, 2, 1 which, added, give 19, 19, 19, 19, 19, 19, 19, 19, 19 3. If we subtract the unit term of the denominator of the common fraction from 10 and multiply any term of the repetend by the remainder, the unit term of the product will be the unit term of the corresponding remainder. Thus, in ty=.0, 5, 2, 6, 3, 1, 5, etc., terms of repetend. 10-9=1 0, 5, 2, 6, 3, 1, 5, etc., unit terms of product and remainders. 4. A complementary repetend, by beginning at different points, will be the repetend of all proper fractions having the same denominator as the fraction which produced it. Thus, i;=.52631 etc, which begins with the 2d figure of the circulate equal to g. Again, =,263157 etc., which begins with the 3d figure of the circulate equal to 1o, etc. 5. The numerator of the fraction equal to any one of the several repetends beginning with the successive terms of a complementary repetend, is the remainder left when the preceding term of the repetend was obtained. Thus, in reducing {to a circulate, when the first 5 of the circulates was obtained, 5 was the remainder, and 5 is the numerator of the fraction equal to the circulate .26315 etc. Which of the following give complementary repetends? CONTINUED FRACTIONS. 332. A Continued Fraction is a fraction whose numerator is 1, and denominator an integer plus a fraction whose numerator is also 1 and denominator an integer plus a similar fraction, and so on. Thus, 1425 = } li; 1; 1; 1; ibi NOTE.-For a fuller discussion of circulates see the author's Philosophy of Arithmetic. +. 333. Several recent autbors, for convenience, write a continued fraction with the sign of addition between the de 1 1 1 1 nominators; thus, 2+3+4+5 334. There are Two Cases: 1st. To reduce a common fraction to a continued fraction ; 2d. To reduce a continued fraction to a common fraction. NOTE.—Continued fractions were proposed about the year 1670, by Lord Brouncker, President of the Royal Society. CASE I. OPERATION. 68 157 335. To reduce a common fraction to a continued fraction. 1. Reduce in to a continued fraction. SOLUTION.—Dividing both numerator and denominator by 68, we have 1 divided by 2+ 1. Dividing the terms of aš by 21, we have i divided by 3+1 Dividing again by 5, we have 1 divided by 4+}, which completes the reduction, as the numerator of the last fraction is unity. The terms }, , }, etc., are called the ti first, second, third, etc., partial fractions. It + will be seen that the same result may be obtained by dividing as in finding the greatest common divisor, and taking the reciprocals of the several quotients for the partial fractions. Hence we derive the following Rule.—Find the greatest common divisor of the terms of the given fraclion; the reciprocals of the successive quotients will be the partial fractions which constitute the continued fraction required. 2. Reduce 21 to a continued fraction. Ans. It 288 3. Reduce to a continued fraction. Ans. 4. Reduce 24 375 to a continued fraction. Ans. 17 OPERATION. CASE II. 336. To reduce a continued fraction to a common fraction. 1. Reduce $7% TIF to a common fraction. SOLUTION 1st.-Reducing the complex fraction formed by the last two partial fractions to a simple fraction, we have . Taking this result, and the preceding partial fraction together, and reduc +1=1 ing, we have ing. Writing this with the preceding IP = 15, Ans. partial fraction, and reducing, we have 13, the value of the fraction. SOLUTION 2D.-Approximate values may be obtained by beginning at the first partial fraction and reducing respectively two, three, or more, of the partial fractions to simple fractions. Thus, the first approximate value is }; the second is 3 the true ; 于是 value is 16. By exhibiting the second method in an analytic form, a law may be discovered which gives us a simple and practical rule for the reduction. 2. Find the approximate values of the continued fraction + the third is FIFT = 4, 2d SOLUTION.—The work may be written as follows: } = 3, 1st approx. val. 1 3x5+1 3x5+1 3x5+1 (3x2+1)x5+27x5+2 1 =2+5+1 16 X 4+3 1933, true value. 3x (5+1)+1 37 X 4+7 -34, 3d “ *FIE + EF=2+5+1 |