Converting Repeating Decimals to FractionsDate: 11/19/97 at 23:19:17 From: Oscar Duarte Subject: Repeating decimals to fractions Hello, I was asked this question the other day and I could not remember how to convert a repeating decimal into a fraction. For example, I know .333333333333 is 1/3 but what is the trick to it? Thanks, Oscar Date: 11/25/97 at 16:23:24 From: Doctor Mark Subject: Re: Repeating decimals to fractions Hello, Oscar As you may already know, every fraction (technically, a fraction is called a rational number) either terminates - ends with a string of 0's - or is a repeating decimal (of course, you could think of the string of zeros as just a special kind of repeating decimal). The decimals that neither terminate nor repeat cannot be repreented by fractions (with integers on the top and bottom), and are called "irrational" numbers. So the question that occurs, as it did to you, is how to find out what fraction that repeating decimal is equal to. And you're right, there is a trick to it. Here's the trick. Suppose that you have a repeating decimal, and it looks like .(a)(a)(a)... where (a) is some sequence of repeating digits (technically, (a) is called the "repetend," i.e., "the thing which is repeated"). For instance, for 1/3 = .3333333..., (a) is 3 for 1/11 = .09090909..., (a) is 09 for 1/7 = .142857142857..., (a) is 142857 Still following me? Oh yes, the trick... First, you have to count the number of digits in the repetend. When (a) is 3, the number of digits is 1, when (a) is 09, the number of digits is 2, and when (a) is 142857, the number of digits is 6. Now, multiply your repeating decimal by a power of 10, namely, the power of 10 which is a 1 followed by a number of zeros equal to the number of digits in the repetend. That's a mouthful, so let's see how it works in the examples above: For 0.33333..., the repetend is 3, and that has *one* digit, so multiply by a 1 followed by *one* zero, i.e., by 10 For 0.090909..., the repetend is 09, which has *two* digits, so multiply by a 1 followed by *two* zeros, i.e., by 100 For 0.142857142857..., the repetend is 142857, which has *six* digits, so multiply by a 1 followed by *six* zeros, i.e., by 1,000,000 (one million). If we multiply the repeating decimal by a power of 10 in this way, we end up with a decimal which has the repetend to the LEFT of the decimal point, and the same repeating decimal we started out with to the RIGHT of the decimal point: Multipy 0.33333... by 10, and we get 3.33333... Multiply 0.090909... by 100, and we get 9.090909... Multiply 0.142857142857... by 1000000, and we get 142857.142857... But now note that after we multiply by this appropriate power of 10, we get the sum of an integer (which is numerically equal to whatever the repetend was) and the repeating decimal we started out with. If we let x be the repeating decimal we started out with, we find: If x = .333.., then 10x = 3.333... = 3 + .333... = 3 + x. That is, we get 10x = 3 + x. If you remember your algebra, subtracting x from both sides of this equation gives us 9x = 3, so that x = 3/9 = 1/3, after we reduce the fraction to lowest terms. If x = .0909..., I hope you can see that we get 100x = 9 + x, or 99x = 9, or x = 9/99 = 1/11. And if x = .142857142857..., do you see that 1000000x = 142857 + x? Solve that for x, and you get x = 142857/999999 = 1/7 (though the reduction to lowest terms takes a little longer here if you forget that we know that this better be the reduction, since that's how we got the repeating decimal in the first place). If you have followed this argument, then maybe you can see that in general, if the repeating decimal has (a) as the repetend, then the fraction that is represented by that repeating decimal is just (a)/R where R is a number with the same number of digits as (a), but all the digits are 9's. Thus, 0.567567... = 567/999 (= 21/37 after reduction) 0.42014201... = 4201/9999 (is already reduced to lowest terms) and so on. So now that you feel you understand this, let me throw a monkeywrench into the whole thing: If we looked at 0.333333..., who said the repetend was 3? Couldn't it just as well be 33, or 333, or even 333333333333333333333? The answer is that it could, but (and this takes a little more work) it ends up giving you the same fraction after reduction to lowest terms. The study of repeating decimals is an interesting one, and there are many deep and interesting questions one can ask about them. Let me leave you with one of these. If you look at the repeating decimals formed by taking 1/n, where n is a prime number that is not 2 or 5, you might notice that the length of the (smallest choice of) repetend [i.e., 3, not 333, for 0.3333...] is ** always less than or equal to n-1, ** always a divisor of n-1, and ** equal to n-1 only for some of these prime numbers. For instance, 1/3 = 0.333... has a one digit repetend, and one is a divisor of 2 (= 3-1). 1/11 = 0.0909... has a two digit repetend, and two is a divisor of 10 (= 11-1). And 1/7 = 0.142857142857... has a six-digit repetend, and six certainly divides 6 (= 7-1). [That is, 1/7 is one of the numbers 1/n which has a repetend of "maximum" length n-1.] The reasons for this are more complicated, and I can't explain them here. But if you go into the library and look up any book on Number Theory, it's explained there. -Doctor Mark, 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- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/