Fractions with Terminating Decimals

Late last month, we have talked about fractions with terminating decimals as well fractions with non-terminating decimals. We ended up with a conjecture that a fraction is a terminating decimal if its denominator has only the following factors: 2 (or its powers), 5 (or its powers) or both.  In this post,we refine this conjecture. This conjecture is the same as saying

A rational fraction \frac{a}{b} in the lowest terms has a terminating decimal if and only if the integer b has no prime factor other than 2 and 5.

Note that we have already explained the only if part in the preceding post. It remains to show that if part which is

if \frac{a}{b} is in lowest terms and b contains at most 2 and 5 as factors, then the fraction is a terminating decimal.

For a specific example, we have

\displaystyle \frac{9471}{3200} = \frac{9741}{(2^7)(5^2)}= \frac{(9741)(5^5)}{(2^7)(5^7)}

which is equal to

 

\displaystyle \frac{30440625}{10^7} = 3.0440645.

Now let us prove this statement.

Suppose the denominator b = 2^m5^n where m and n are non-negative integers. There are two possible cases: n \leq m or n > m.

If n \leq m, then we multiply the denominator by \frac{5^{m-n}}{5^{m-n}}. This means that

\displaystyle \frac{a}{b} = \frac{a}{2^m5^n} = \frac{a(5^{m-n})}{2^m5^n(5^{m-n})} = \frac{a5^{m-n}}{2^m5^m}.

This result can be simplified further to

 

\frac{a(5^{m-n})}{10^m}.

Now, since a is an integer and m and n are non-negative integers, the numerator is an integer and the denominator is a power of 10. Therefore, \frac{a}{b} is a terminating decimal.

Now for the second case, let n > m.

\displaystyle \frac{a}{b} = \frac{a}{2^m5^n} = \frac{a(2^{n-m})}{2^m5^n(2^{n-m})} = \frac{a2^{n-m}}{2^n5^n}

 

which is equal to

\displaystyle \frac{a2^{n-m}}{10^n}.

 

Again, a is an integer, n and m are integers, and the denominator is a power of 10. Therefore, the fraction \frac{a}{b} is a terminating decimal.

Reference: Rational and Irrational Numbers by Ivan Niven

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