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### Converting Repeating Decimals to Fractions

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Date: 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
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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/
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Associated Topics:
Middle School Fractions