(navigation image)
Home American Libraries | Canadian Libraries | Universal Library | Community Texts | Project Gutenberg | Biodiversity Heritage Library | Children's Library | Additional Collections
Search: Advanced Search
Anonymous User (login or join us)
Upload
See other formats

Full text of "History Of The Theory Of Numbers - I"

178                         HlSTOKY  OF THE TfiEOBY OF NUMBERS.                   [CHAP. VI
T. Ghezzi139 considered a proper irreducible fraction m/p with p prime to the base b of numeration.    Let b belong to the exponent n modulo p.   In
mb**pqi+rlt         r1b^pq2+r2t. - .,        (Xr^p, 0<r2<p,. . .,
r1}. . ., rn are distinct and rn=ra.   Multiply the respective equations by b11-1, 6n~2, . . . and add; we see that
A similar proof shows that m/p equals a fraction with the denominator b'(6n 1) when 6 = aia2a3, p = pia1ra28a3^ the a's being primes and P! relatively prime to b, while 6' is the least power of b having the divisor a1ra2*a3', and n is the exponent to which b belongs modulo plt
F. Stasi140 gave a long proof showing that the length of the period for I/a does not exceed that for I/a. If the period A for l/p has m digits and n = pq is prune to 10, the length of the period for l/n is m if A is divisible by q] is mi if A is prime to q and if the least A(10wAs~1)+ . . . +1) divisible by # has m=i; and is wy if A=A'a, q = a.q', with A', q' relatively prime, while the least A! (10m(*:~1)+ ... +1) divisible by q' has k=j. For a prime 5, let
= ph    10w-l'
and let Ah be the first of the periods of successive powers of l/p not divisible by p; then the period for l/ph+k has mpk digits. If pt is a prime ?^2, 5, and r\ is the length of the period for 1/p^ and if l/p/* is the highest power of I/pi with a period of r{ digits, the length of the period for l/p/1* is r/ = rt-pl-a<~/?< and that for l/Tlpf* is a multiple of the 1. c. m. of the r/.
If n is prime to 10 and if r1} . . . , rm = 1 are the successive remainders on reducing l/n to a decimal, then r?^r2i (mod n). Hence if l/n has a period of 2i digits, r?^l (mod n) and conversely. But if it has a period of 2^+1 digits, r?+i==10 and conversely.
*K. W. Lichtenecker141 gave the length of the period for l/p, when p is a prime ^307, and the factors of 10n  1, n^lO.
L. Pasternak142 noted that, after multiplying the terms of a fraction by 9, 3 or 7, we may assume the denominator N  10m  1. To convert Ro/N into a decimal, we have 10Rk,i  Nyk-\-Rk (k  1, 2,. . .) Set Rk= 10zk-{-ek, ek^ 9. Since yk^ 9, ek = yk and Rk_i = mek+zk. Hence the successive digits of the period are the unit digits of the successive remainders.
E. Maillet143 defined a unique development a0+ ai/n+a2/n2+ ... of an arbitrary number, where the a,- are integers satisfying certain conditions. He studied the conditions that the development be limited or periodic.
13I1 Boll. Matematica Gior. Sc.-Didat., 9, 1910, 263-9. mlbid., 11, 1912, 226-246.
141Zeitschr. fur das Realschulwesen, 37, 1912, 338-349. 142L'enseignement math., 14, 1912, 285-9.
M*T,'iTt.prmdin.iw HPB mn+.>i     OH    1 01 ^