<p>Let you want to represent integer values from 0 to 999. You need 3 digits, each has 10 states, 30 states overall. When using binary digits, you need 10 digits with 2 states, 20 states overall, So binary representation is more "dense". In general case, p*ln(N)/ln(p) states is required. This formula has minimum at p=e. However, this is pure theoretical speculation. Numeral system with the base 3 is slightly more dense than that of 2, and 3-based computers were really built, but appeared more complex than binary computers.</p>

CC BY-SA 3.0
2013-11-04T12:44:38.830Z
Thank you Alexey! Actually, I´ve read it on the Wiki http://en.wikipedia.org/wiki/Radix_economy, topic called Radix Economy. What I wanted to figure out is how they construct numeral system for non-integer base, say e. And was disappointed with the answer :)
Roman
CC BY-SA 3.0
2013-10-31T18:39:37.620Z
2013-10-31T18:39:37.620Z
Alexei Kaigorodov
<p>What is an "informational density" and why numeral system with the base of e (2,71828...) has the maximum informational density? How do you calculate "informational density" of a given numeral system?</p>

CC BY-SA 3.0
2013-11-01T16:17:34.850Z
[kolmogorov complexity applications in number theory](http://cstheory.stackexchange.com/questions/18816/kolmogorov-complexity-applications-in-number-theory)
vzn
<it.information-theory>
Numeral system information density
Roman
# Numeral system information density

(Asked by: username_0 on 2013-10-20)

What is an "informational density" and why numeral system with the base of e (2,71828...) has the maximum informational density? How do you calculate "informational density" of a given numeral system?

**Tags:** it.information-theory
> **Comments:**
> **username_1**: [kolmogorov complexity applications in number theory](http://cstheory.stackexchange.com/questions/18816/kolmogorov-complexity-applications-in-number-theory)  

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## 1 answers

### Answer by username_2 on 2013-10-31. Score: 1 | Accepted answer

Let you want to represent integer values from 0 to 999. You need 3 digits, each has 10 states, 30 states overall. When using binary digits, you need 10 digits with 2 states, 20 states overall, So binary representation is more "dense". In general case, p\*ln(N)/ln(p) states is required. This formula has minimum at p=e. However, this is pure theoretical speculation. Numeral system with the base 3 is slightly more dense than that of 2, and 3-based computers were really built, but appeared more complex than binary computers.

> **Comments:**
> **username_0**: Thank you Alexey! Actually, I´ve read it on the Wiki http://en.wikipedia.org/wiki/Radix_economy, topic called Radix Economy. What I wanted to figure out is how they construct numeral system for non-integer base, say e. And was disappointed with the answer :)  

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