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### Fraction Equivalents of Decimals that Repeat ... Mostly

```Date: 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]

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]

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]

I hope this helps. Please write back if you have any further questions

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Algebra
Middle School Fractions

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