Which Fractions Repeat?Date: 07/21/98 at 00:51:14 From: Melissa Subject: Fractions Is there any way you can determine whether a decimal is going to be terminating or repeating without doing the conversion? Also, is there any way you can tell how many decimal places it will have if it is indeed repeating? Finally, is there a way to determine what the length of the decimal places will be if a fraction in lowest terms is repeating? I don't have a clue as to how one can determine these things without doing the conversion. Thanks for your help. Melissa Date: 07/21/98 at 15:19:40 From: Doctor Rob Subject: Re: Fractions Thanks for asking, Melissa. The way to answer the first question is to factor the denominator of the fraction (after it is reduced to lowest terms) into a product of powers of prime numbers. If the prime factors are all 2 or 5, then the decimal will terminate. The answer to the second question is that if any other prime number appears as a factor, then the decimal will be nonterminating but repeating. Example: If the denominator is 51200, then the decimal will terminate, because 51200 = 2^11 * 5^2, and the only prime numbers dividing are 2 and 5. If the denominator is 288, the decimal will repeat, since 288 = 2^5 * 3^2, and the prime number 3 appears in the factorization. The answer to the third question is more difficult. It involves some pretty complicated mathematics known as Number Theory. Probably the simplest way to figure this out is the following. Write the denominator in the form N = 2^a * 5^b * M, where neither 2 nor 5 divides into M evenly, and a and b are some whole number exponents. Then find the smallest number all of whose digits are 9's which is a multiple of M. The number of 9's required is the length of the repeating part of the decimal expansion of your fraction. Example: N = 26 = 2*13, so a = 1, b = 0, and M = 13. Then divide: ---------- 13 ) 999.... 13 doesn't go into 9 even once. It goes into 99 seven times, and 7*13 = 91, so: 7 ---------- 13 ) 999.... 91 ---- 89 Continue this, bringing down one nine at a time from the dividend, and following the rules for long division, until a remainder of 0 is obtained: 76923 --------- 13 ) 999999 91 ---- 89 78 ---- 119 117 ----- 29 26 ---- 39 39 --- 0 This tells us that the repeating part of the decimal expansion of some number of 13ths contains 6 digits. This implies that the repeating part of the decimal expansion of some number of 26ths also contains six digits. You try one with denominator 85. If a technically correct answer to the third question is needed, here it is. In terms of number theory, you need to find the order of 10 modulo M, that is, the smallest power of 10 which leaves a remainder of 1 when divided by M. It is known to be a divisor of lambda(M), but which divisor can only be determined by testing. Lambda(M) can be found as the least common multiple of the numbers (p-1)*p^(e-1), where p is any prime dividing M, and p^e is the highest power of p dividing M. As an example, if M = 3^4*7 = 567, then: lamdba(567) = LCM([3-1]*3^3, [7-1]*7^0) = LCM(2*3^3, 2*3) = 54 The actual repeating part has length 18, a divisor of 54, since 3^4 is a divisor of 10^9 - 1, and 7 is a divisor of 10^6 - 1, and LCM(9, 6) = 18, so both 3^4 and 7 divide 10^18 - 1. - Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/