Is $\pi$ periodic in any base-k numeral system, where k is integer ? And what is the status of this problem?
Asked
Active
Viewed 1,976 times
8
-
19Yes. For example, in base $\pi$, $\pi$ is written $10$. – Chris Eagle May 06 '12 at 11:41
-
1No, periodic numbers are algebraic (or even rational), $\pi$ is neither. – Asaf Karagila May 06 '12 at 11:41
-
5Evidently, it depends on what you mean by "numeral system." – Gerry Myerson May 06 '12 at 11:44
-
@Chris Eagle: how do you write numbers in base $\pi$? – Joel Cohen May 06 '12 at 11:56
-
6@JoelCohen: The same way you do for any other base. See http://en.wikipedia.org/wiki/Non-integer_representation for example. – Chris Eagle May 06 '12 at 11:58
-
@ChrisEagle: This might bake your noodle, but: is pi rational, then, in base pi? Is it really a member of R, the reals (or is that an artifact of base 10 bias)? – Marcos Jan 22 '24 at 08:44
-
@ChrisEagle: When you write in octal, you use the numbers 0-7. What numbers do you use in base pi? – Marcos Jan 22 '24 at 08:53
-
@AsafKaragila: If the base system was an irrational number, the pi could potentially be periodic, yes? – Marcos Jan 23 '24 at 04:53
5 Answers
13
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 $k$.
Martin Sleziak
- 53,687
Adam Rubinson
- 20,052
12
$\pi$ 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.
Gerry Myerson
- 179,216
0
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 $\pi$.
Jay
- 3,822
0
If $\pi$ were periodic in any base, then it would be rational, and therefore periodic in every base. This does not happen.
wendy.krieger
- 6,856
-1
Not for an integer base, but it just happens that $\pi=0.1111111...=0.\bar{1}$ in base $\kappa=\frac{1}{\pi}+1 \approx 1.31831$. :)
e.tadeu
- 99