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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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    @JoelCohen: The same way you do for any other base. See http://en.wikipedia.org/wiki/Non-integer_representation for example.2012-05-06

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Not for an integer base, but it just happens that $\pi=0.1111111...=0.\bar{1}$ in base $\kappa=\frac{1}{\pi}+1 \approx 1.31831$. :)

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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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    @MAR, that's already answered in some of the comments.2015-11-02
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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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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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If $\pi$ were periodic in any base, then it would be rational, and therefore periodic in every base. This does not happen.