# If a rational number has a finite decimal representation, then it has a finite representation in base $b$ for any $b>1?$

Is it true that if a rational number has a finite decimal representation, then it has a finite representation in base $b$ for any $b>1?$

I would like to know if there is a book where this subject is fully detailed.

EDIT: Improve the question!

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Not true. 0.1 base 10 = 0.00011001100110011.... in base 2. –  Apprentice Queue Mar 30 '12 at 16:25
The answer to your question is fairly easy as written: a number with finite representation in base $10$ has finite representation in base $100$. –  Arturo Magidin Mar 30 '12 at 16:26
@ArturoMagidin: I think I messed up my question! Unfortunately, I do not know how to improve it. –  spohreis Mar 30 '12 at 16:30
If you meant "finite representation in base $b$ for any $b\gt 1$", then write i like that. It currently reads "...has finite representation in relation to at least one base $b$, $b\neq 10$." –  Arturo Magidin Mar 30 '12 at 16:31

Theorem. A rational number $\frac{a}{b}$, $\gcd(a,b)=1$, has finite representation in base $k$ if and only if there exists $n\gt 0$ such that $b|k^n$.

Proof. Suppose $b|k^n$. Then $k^n = bq$, so we can write $$\frac{a}{b} = \frac{aq}{bq} = \frac{aq}{k^n}$$ which has finite representation.

Conversely, if $\frac{a}{b}$ has finite representation, then we can write it as $\frac{r}{k^n}$ for some $n\gt 0$, hence $br=ak^n$; since $\gcd(a,b)=1$, then $b|k^n$. $\Box$

Corollary. A rational number $\frac{a}{b}$, $\gcd(a,b)=1$, has finite representation in base $k$ if and only if the only primes that divide $b$ also divide $k$.

If we interpret your question as: "if $\frac{a}{b}$ has finite decimal representation, then it has finite representation in base $k$ for any $k\gt 1$", then the answer is "no". $\frac{1}{2}$ does not have finite representation in base $3$. If we interpret it as "if $\frac{a}{b}$ has finite decimal representation, then it has finite representation in base $k$ for some $k$," then the answer is "yes" (though it trivially does in base $10^n$ for any $n$).

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Thank you very much! –  spohreis Mar 30 '12 at 16:32

An $\rm x\in\mathbb{Q}$ has a finite digital representation in base $\rm b$ if and only if $\rm x= n/b^k$ for some integers $\rm n,k$, with $\rm k\ge 0$. Since $\rm x= nm^k /(mb)^k$, $\rm x$ is also has finite representation in base $\rm mb$ for any $\rm m\in\mathbb{N}$.

Wait, do you mean some other base, or any other base? These are distinct questions.

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$\frac{1}{13}$ in base $10 : 0.0\overline{769230}$

$\frac{1}{13}$ in base $13 : 0.1$

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Every number with a finite expansion in base $a$ has a finite expansion in base $b$ if and only if there exists a positive integer $n$ such that $a\mid b^n$.

The expansion in base $b$ of every decimal is finite if and only if $10\mid b$.

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Hardy and Wright has a good analysis of the theory of "decimals" in different bases.

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