Why decimal form of $\frac{A}{2^m\times5^n\times3^p\times7^q\cdots}$ has both repeating and non repeating numbers?

Question

I read in my book that, decimal form of the fraction $\cfrac{A}{\underbrace{2^m\times5^n}_{\text{at least one}}\times\underbrace{3^p\times7^q}_{\text{at least one}}\times\cdots}\quad(A\in\mathbb{N})$, has both repeating numbers and non repeating numbers. I don't understand the reason of that. I know the decimal form of $\cfrac{A}{2^m\times5^n}$ is determinable because we can rewrite it as $A\times(0.5)^m\times(0.2)^n$. hence with other prime factors in denominator like $3^p$ and $7^q$ decimal is not determinable so it has repeating numbers. but for $\cfrac{A}{{2^m\times5^n}\times{3^p\times7^q}\times\cdots}$ which is a combination of two forms I don't know how to show it mathematically that contain both repeating and non repeating numbers in decimal form. I can rewrite it as : $$A\times(0.5)^m\times(0.2)^n\times\frac{1}{3^p\times 7^q\cdots}$$ It is product of determinable and repeating decimals but it doesn't help me.

Answer 1

I assume you're talking about the fact that every rational number, in decimal form, has an initial segment followed by an infinitely repeating segment. For instance, $\frac{11}{90}=0.1\bar{2}=0.1222\cdots$.

Consider a rational number $a/b$. We can factor it as $10^{-k}c/d$ where $d$ is coprime to $10$, so without loss of generality $b$ is coprime to $10$. For simplicity, I will also assume $a/b>0$ (a little bit of modification is necessary for negatives). Furthermore, we can subtract out the integer part to assume $a<b$.

The powers of ten must eventually repeat mod $b$, so there is an $n$ for which $10^n-1=b\bar{b}$. If we multiply and divide $a/b$ by $\bar{b}$ it is amenable to the geometric sum formula:

$$ \frac{a}{b}=\frac{a\bar{b}}{b\bar{b}}=\frac{a\bar{b}}{10^n-1}=\frac{1}{10^n}\frac{a\bar{b}}{1-\frac{1}{10^n}}=\frac{a\bar{b}}{10^n}+\frac{a\bar{b}}{10^{2n}}+\cdots $$

Since $a\bar{b}<10^n$, if we substitute $a\bar{b}$s decimal representation into the above we see that we get the decimal digits of $a\bar{b}$ (interpreted as having $n$ digits, with leading $0$s if necessary) repeat over and over again.

Adding back in the integer part we subtracted off can only affect the first so many digits, and similarly multiplying the whole shebang by $10^{-k}$ only shifts everything.

Answer 2

I just came up with this answer:

Consider fraction $\cfrac{A}{2^m\times5^n\times3^p\times7^q}$ , Without Loss Of Generality assume $2^m\times5^n<3^p\times7^q$. Let's decompose it:

$$A\times\cfrac{1}{2^m.5^n.3^p.7^q}=A\times\cfrac{1}{3^p.7^q-2^m.5^n}\left(\cfrac{3^p.7^q-2^m.5^n}{2^m.5^n.3^p.7^q}\right)$$

$$=\cfrac{A}{3^p.7^q-2^m.5^n}\times\left(\cfrac{1}{2^m.5^n}-\cfrac{1}{3^p.7^q}\right)$$

If we write $\cfrac{1}{2^m.5^n}-\cfrac{1}{3^p.7^q}$ in decimals, we have subtraction of determinable decimal and repeating decimal, it is not hard to see the result is a decimal number which has initial segments followed by repeating segments. multiplying it by $\cfrac{A}{3^p.7^q-2^m.5^n}$ doesn't influence on the conclusion.