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"

Question

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.

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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 $a∈Z$.

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Thank you!

Answer 1

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

Answer 2

$$\frac{p}{2^{a}5^b}$$

Let's say $a \ge b$ and $a-b=r$ then we can write

$$\frac{p}{2^{a}5^b}=\frac{5^rp}{2^{a}5^b5^r}=\frac{5^rp}{2^a5^{a}}=\frac{5^rp}{10^{a}}$$

If $b \ge a$ and $b-a=r$, we get:

$$\frac{p}{2^{a}5^b}=\frac{2^rp}{10^{b}}$$