Infinite decimal expression of rational numbers I'm asked to prove that if a non-integer rational number is written as a non reductible fraction $a/b$ and $a,b $are co-primes and $b$ is not multiple of 2 or 5, then the decimal expression of the number is infinite.
I can get as far as proving that if $r$ is a rational number with finite decimal expression,then the decimal expression must be of the form $\frac{a_110^{n-1}+...+a_n}{2^n5^n}$ but altought this proves when the decimal expression is finite, I think is not a good proof, since I'm not using any of the hypothestis.
Could you help me and give me some advice?
 A: Hint: Suppose we have that $x$ has a finite decimal expansion, allowing us to write
$$x=\sum_{i=m}^{n}a_i\cdot 10^i$$
for some $n,m$ and  $a_m,a_{m-1},\dots, a_n$. We then have that
$$10^{|m|}x\in\mathbb{Z}$$
Conversely, if $10^{p}x\in \mathbb{Z}$ for some $p$, then we have that $10^{p}x$ has a terminating decimal expansion, i.e.
$$10^{p}x=\sum_{i=m}^{n}a_i\cdot 10^{i}$$
and so
$$x=\sum_{i=m}^{n}a_i\cdot 10^{i-p}$$
has a terminating decimal expansion. Thus

$x$ has terminating decimal expansion if and only if $10^{m}x\in\mathbb{Z}$ for some $m\in\mathbb{Z}$.

A: All you need is a bit of logic.  The given statement is equivalent to

if $a/b$ is not reducible (that is, $a$ and $b$ are relatively prime) and is not an integer and the decimal of $a/b$ is finite, then $b$ is a multiple of $2$ or $5$.

You have almost proved this.  If the decimal is finite then you have shown
$$\frac ab=\frac{a_110^{n-1}+...+a_n}{2^n5^n}\ .$$
If the numerator and denominator have any common factors, cancel them to get $b$ in the denominator: then $b=2^l5^m$ for some $l,m$ and since $b>1$, it is a multiple of $2$ or $5$.
A: If every terminating decimal, written as a fraction $a/b$, has $b$ divisible by $2$ or $5$, then a fraction $a/b$ where $b>1$ is not divisible by $2$ or $5$ must not correspond to a terminating decimal.  But it has some decimal expansion, so that must be non-terminating.
