Fraction Equivalents of Decimals that Repeat ... MostlyDate: 05/04/2010 at 18:05:52 From: john Subject: Reccuring decimals into fractions I understand how to find the fraction equivalent of recurring decimals such as 0.727272, where all the digits to the right of the decimal occur in the part that repeats: x = 0.727272 100x = 72.727272 99x = 72 So 0.727272 = 72/99, which may be simplified. =) But I do not understand how to work out recurring decimals that contain a number that is not recurring. For example, x =? 0.0727272 1000x =? 72.0727272 I'm not sure what to do next.... If I take away x from that 1000x, the answer will be incorrect, as the zero in the tenths place does not recur. That's why I need your help, Dr. Maths! Date: 05/04/2010 at 20:20:55 From: Doctor Greenie Subject: Re: Reccuring decimals into fractions Hi, John -- At the risk of confusing you, let me show you several options you have for finding the common fraction equivalent of a repeating decimal. If you want to stick with your formal algebraic approach, you still have at least a couple of options. The basic idea, taken from the process you demonstrate, is that there are two digits in the repeating part of the decimal. That means we need two numbers, one of which is 100 times the other, so that when we subtract, the like digits "line up" for easy subtraction. We could still do "x" and "100x".... 100x = 7.272727... x = 0.072727... --------------- 99x = 7.2 990x = 72 x = 72/990 (which can be simplified) With that calculation, after we did the subtraction we still had a digit to the right of the decimal point in our equation that we had to deal with. To avoid that, we could instead use "10x" and "1000x": 1000x = 72.727272... 10x = 0.727272... ----------------- 990x = 72 x = 72/990 With your particular example, where the "extra" digit to the right of the decimal point is a leading 0, we don't really have to use the algebraic method. The leading zero in the tenths place just means we have divided something by 10: 0.0727272... = (0.727272...)/10 = (72/99)/10 = 72/990 If the "extra" leading digit is not 0, we can't use that simple last method; but the algebraic method still works. So let's look at some options for converting 0.3727272... to its common fraction equivalent. 1000x = 372.727272.... 10x = 3.727272.... ------------------ 990x = 369 x = 369/990 Personally, if I don't have pencil and paper handy, I convert that example like this: 0.3727272... = (3.727272...)/10 = (3 72/99)/10 = ((3*99 + 72)/99)/10 = (297 + 72)/990 = 369/990 Next, here is a method used by some high school math team members I have worked with: 0.3727272... = 0.3 + 0.0727272... = 3/10 + 72/990 = (3*99 + 72)/990 = 369/990 You can see the arithmetic involved is basically the same as with the method I use. Finally, below is a shortcut I have seen used by students on high school math teams, who are called upon to perform conversions like this more often than just once in a while. If you need to find these kinds of equivalences often, this is a nice method to learn.... numerator: 372 - 3 [the number consisting of the "extra" digit(s) plus one cycle of the repeating digits; minus the repeating digit(s)] denominator: 990 [two 9's because there are two repeating digits; and one 0 because there is one "extra" digit] answer: 369/990 If you look closely at the algebraic method for performing the conversion, you can see how this shortcut works. Here are two more random examples to demonstrate the formal algebraic conversion process and the shortcut: (1) 0.3455555... (one repeating digit; two "extra" digits) algebraic conversion: 1000x = 345.5555... 100x = 34.5555... ------------------ 900x = 311 x = 311/900 shortcut: numerator: 345 - 34 = 311 denominator: 900 [one repeating digit; two "extra" digits] answer: 311/900 And (2) 0.8345345345... (three repeating digits; one "extra" digit) algebraic conversion: 10000x = 8345.345345... 10x = 8.345345... ---------------------- 9990x = 8337 x = 8337/9990 Shortcut: numerator: 8345 - 8 = 8337 denominator: 9990 [three repeating digits; one "extra" digit] answer: 8337/9990 I hope this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/