Prove that $\frac{p}{q}$ is a rational number with a finite decimal expression if $p$ is an integer and $q=(2^n)(5^m)$

Question

Let $p,q$ be two integers and $q=(2^n)(5^m)$. Then $\frac pq$ is a rational number with a finite decimal expression.

Any ideas how to do this? I've been thinking about it all day but I have no idea how to use the hypothesis that $q=(2^n)(5^m)$

Answer 1

Assuming $m$ and $n$ are positive integers[*]. Then consider

$$\frac p q = \frac p {2^n5^m} = \frac {p \cdot 2^{\max(m,n) -2}\cdot 5^{\max(m,n) -m}}{2^n5^m2^{\max(m,n)-n}5^{\max(m,n)-m}} = \frac { {p \cdot 2^{\max(m,n) -2}\cdot5^{\max(m,n) -m}}}{2^{\max(m,n)}5^{\max(m,n)}} = \frac { {p \cdot 2^{\max(m,n) -2}\cdot 5^{\max(m,n) -m}}}{10^{\max(m,n)}}= \frac P{10^k}$$

where $P = p \cdot 2^{\max(m,n) -2}\cdot 5^{\max(m,n) -m}$ is an integer and $k = \max(m,n)$ so $10^k$ is a power of 10.

An integer divided by a power of 10 is a terminating decimal clearly.

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[*] If $m, n$ aren't necessarily positive integers then $1/3 = 1/2^{\log_3 2}5^0$ is a counter example.

Answer 2

Rather than just give you the answer, do you see how to multiply both the numerator and denominator of $\frac{p}{q}$ by appropriate powers of 2 and 5 so that you obtain a denominator which is an integer power of 10?

Answer 3

Consider $ \frac{d}{2} $ where d is between 1 and 9. All these cases have finite expansions. Likewise for $ \frac{d}{5} $. Any finite expansion will end with a digit between 1 and 9 so any finite expansion will remain finite.

We can use this to create an inductive proof. $ \frac{p}{2^05^0} $ is finite because all integers are finite.

If $ \frac{p}{2^n5^m} $ is finite then $ \frac{p}{2^{n+1}5^m} $ and $ \frac{p}{2^n5^{m+1}} $ are finite due to the argument given above.

So it is true for natural numbers n and m.

Answer 4

$$ \frac{19}{2^7 \cdot 5^4} = \frac{19\cdot 5^3}{2^7 \cdot 5^7} = \frac{19\cdot 5^3}{10^7}. $$

Do something similar with exponents other than the ones here, i.e. other than $7$ and $4$, and get a power of $10$ in the denominator. Then you're done.