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Is $\pi$ periodic in any base-k numeral system, where k is integer ? And what is the status of this problem?

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Yes. For example, in base $\pi$, $\pi$ is written $10$. –  Chris Eagle May 6 '12 at 11:41
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No, periodic numbers are algebraic (or even rational), $\pi$ is neither. –  Asaf Karagila May 6 '12 at 11:41
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Evidently, it depends on what you mean by "numeral system." –  Gerry Myerson May 6 '12 at 11:44
    
@Chris Eagle: how do you write numbers in base $\pi$? –  Joel Cohen May 6 '12 at 11:56
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@JoelCohen: The same way you do for any other base. See en.wikipedia.org/wiki/Non-integer_representation for example. –  Chris Eagle May 6 '12 at 11:58

4 Answers 4

up vote 7 down vote accepted

No. In order for $\pi$ to be periodic in base $k$, it must be true that $\pi \equiv m(\pi) \pmod{k}$ for some integer $m$.

By definition of mod, this means that $m(\pi) = \pi + nk$ $\Rightarrow$ $\pi = nk/(m-1)$, which is rational. Since we know that $\pi$ is irrational, we get a contradiction.

In fact you can apply the same argument for all irrational numbers. You can conclude that any irrational number is non-periodic in $k$.

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If $\pi$ were periodic in any base, then it would be rational, and therefore periodic in every base. This does not happen.

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According to the wikipedia article on non-integer representation in base $\pi$ the circumference of a circle of diameter $1$ is $\pi$, which is represented by $10_{\pi}$. This is the base $\pi$ representation of $\pi$.

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$\pi$ is irrational - that was settled hundreds of years ago. That implies that the expression of $\pi$ to any integer base $b$ will be aperiodic. If you have some other kind of numeral system in mind, please edit your question accordingly.

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