# 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