# In what base would $\pi$ have the fewest decimals? [closed]

What is the numeral system in which $\pi$ would have the lowest number of decimals possible?

## closed as off-topic by Najib Idrissi, Peter Franek, Claude Leibovici, Daniel W. Farlow, Lord_FarinSep 6 '15 at 9:37

This question appears to be off-topic. The users who voted to close gave these specific reasons:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Najib Idrissi, Lord_Farin
• "This question is not about mathematics, within the scope defined in the help center." – Peter Franek, Claude Leibovici, Daniel W. Farlow
If this question can be reworded to fit the rules in the help center, please edit the question.

• Could that random down vote expose reasons? Thanks. – Nathan Parker Sep 4 '15 at 9:22
• What's with the close votes for the 'not about mathematics' reason? This question certainly seems about mathematics. If the question had replaced $\pi$ with an arbitrary number $a$, or with some rational number like $\frac{12}{167}$, I doubt people would vote to close for this reason. – Peter Woolfitt Sep 4 '15 at 10:27
• It's really lame that people downvoted this. How DARE someone not know that this question misunderstands what an irrational number is???. This is a question based upon a faulty premise, sure, but it's fully answerable and obviously asked in good faith. Surely the purpose of this place is to learn and share knowledge, not to scoff at those who know less than yourself. – Some_Guy Apr 28 '17 at 20:02

$\pi$ is an irrational number meaning it can never be expressed as a ratio of integers. In particular this means $\pi$ has infinitely many digits after the decimal point in any integer base $b>1$, for if not, we would have

$$\pi=c_nb^n+c_{n-1}b^{n-1}+\dots +c_kb^k$$

where each of the coefficients $c_i$ are integers between $0$ and $b-1$. If $k\ge0$ this directly implies $\pi$ is an integer, while if $k<0$, we have

$$\pi=\frac{c_nb^{n-k}+c_{n-1}b^{n-1-k}+\dots +c_k}{b^{-k}}$$

so $\pi$ would be rational.

Since $\pi$ is irrational, it has no finite decimal representation in any integer base $b>1$.

However, we can use non integer bases. In fact, we can represent numbers using base $\pi$ itself! $$\pi=10_\pi$$

• Thank you very much, really helped. – Nathan Parker Sep 4 '15 at 10:46