# What is the most efficient numerical base system?

I remember reading somewhere that base $e$ is the most "efficient" base system because of its ratio of possible characters to number length. For example, binary is "inefficient" because each represented number is very long when represented with only two digits (0 and 1). Base 10 is also considered "inefficient" because there are so many numbers to remember (0-9) even though each represented number is shorter than binary. Apparently the maximum "efficiency" occurs at base $e$. How is this possible? Can irrational bases exist? How is "efficiency" of a base numerically calculated?

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– Foo Barrigno Jul 18 '13 at 15:52
Relevant – MJD Jul 18 '13 at 15:55
Another source of references: mathworld.wolfram.com/Base.html – dfeuer Jul 18 '13 at 15:56
@FooBarrigno: I would say, more this in general, and that for the optimality of $\mathrm e$ – Ilya Jul 18 '13 at 15:57
Does that sense of optimality have any practical value? – dfeuer Jul 18 '13 at 15:59

First, let's specify the sources. From Wikipedia:

The base $e$ is the most economical choice of radix $\beta > 1$ (Hayes 2001), where the radix economy is measured as the product of the radix and the length of the string of symbols needed to express a given range of values.

Then we have a longish article by Hayes in American Scientist, which I don't feel like reading. The matter boils down to: the representation of number $n$ in base $\beta$ (integer or not) takes $\approx \log n/\log \beta$ digits. If your idea of "economy" is the product of this length with $\beta$ then of course, you are going to minimize $\beta /\log \beta$ and find that the minimum is at $\beta=e$. For example, with Wolfram Alpha, which can plot this function and compute its derivative.

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It seems a bit counter-intuitive that an irrational number is the most "efficient choice", since it is irrational. For instance how does on "economically" represent the number 1. It seems that base three would be a better choice in practice since it is the integer closest to three. – Baby Dragon Jul 20 '13 at 17:51
@BabyDragon: In base $e$, the number "one" is represented as "$1$", and the number "two" as "$2$". The number "three" is represented as $1.0200112\dots$, denoting $e + 0e^{-1} + 2e^{-2} + 0e^{-3} + 0e^{-4} + 1e^{-5} + 1e^{-6} + 2e^{-7} + \dots$. See en.wikipedia.org/wiki/Non-integer_representation – ShreevatsaR Jul 21 '13 at 6:49
@ShreevatsaR Wouldn't "three" be $2.21211\ldots=2e^0+2e^{-1}+e^{-2}+2e^{-3}+e^{-4}+e^{-5}+\cdots$? It looks like you skipped over the ones place, and are actually expanding $10.0200112\ldots$, which would be an alternative. – alex.jordan Jul 15 '15 at 18:13
@alex.jordan: Indeed, looks like I made exactly that mistake, thanks for pointing it out. – ShreevatsaR Jul 15 '15 at 19:12

Assume there are V independent states of information, then we can present V/N digits in base N.

The amount of information we can represent: $I=N^{V/N}$

The value of N that makes I the maximum is the most "efficient" base

Natural log both sides: lnI=(V/N)lnN

Take the derivative: (lnI)'=V(1-lnN)/$N^2$

When lnI gets to an extreme, (lnI)'=0, i.e. N=e

Take the second derivative: (lnI)''=V(2lnN-3)/$N^3$

When N=e, (lnI)''=$-1/N^3$, which is negative

Therefore, I reaches its maximum when N=e.

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