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I recently stumbled upon an observation: the fraction $\frac{x}{y}$ terminates if and only if $y$ only has prime factors $2$ and $5$.

For example:

$$\frac{1}{20} = \frac{1}{2\cdot2\cdot5} = 0.05$$ $$\frac{1}{6} = \frac{1}{2\cdot3} = 0.1\bar6$$

I think this is true because fractions are in the form:

$$\frac{a}{10} + \frac{b}{100} + \frac{c}{1000} + \ldots$$ $$\frac{a}{2\cdot5} + \frac{b}{2\cdot2\cdot5\cdot5} + \frac{c}{2\cdot2\cdot2\cdot5\cdot5\cdot5} + \ldots$$

How can I rigorously prove this?

user01101001
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    If the fraction terminates, that means that $x/y \times 10^N$ becomes an integer for a sufficiently large positive integer $N$. If $x/y$ is in lowest terms (otherwise the proposition is not even true!) what can you say about the factors of $y$, using this observation? –  Apr 27 '16 at 20:28

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HINT: You basically have it. To say that you have a terminating decimal means that $$\frac xy = \frac m{10^s}$$ for some integer $m$ and positive integer $s$. Since the only factors of $10^s$ are ..., the only possible factors of $q$ are ... . (Why? What are you tacitly assuming about $x$ and $y$?)

Ted Shifrin
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