Proof of "'$x \in R$ being rational and the denominator of its min expression being $2^a5^b$' implying that it has a finite decimal expansion" Please could someone give me feedback on my proof for the proposition in the title, I am not sure if it is infallible, or if I could be more concise. * * * If we are given that $x∈Q$, and that it has a minimal expression involving $2^a 5^b$ with $a,b∈N$, we can assume that it can be written in the form $p/(2^a 5^b )$, where $p∈Z$ Now, we need to make $a=b$ so that $2^a 5^b=10^a=10^b$, we do this by taking the smaller one of $a$ and $b$, setting it as $x$, and the bigger one as $y$. Then we append either $2^{y-a}$ or $5^{y-b}$ to the top of our fraction (depending on whether $a$ or $b$ is the smaller power. Now we have $p$ multiplied by some positive integer; divided by $10^x$, s.t there is clearly some finite decimal expansion of $x=p/10^a$ as it is just $p$ shifted $a$ places past the decimal point where $a∈Z$. * * * Thank you!

Your answer is correct. An easier way to get it is by noting that $$\frac{p}{2^a5^b}=\frac{2^b5^ap}{2^{a+b}5^{a+b}}=\frac{2^b5^ap}{10^{a+b}}.$$