What is the most efficient numerical base system?

Question

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?

Answer 1

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.

Answer 2

The numeral for the integer $n$ in base $b$ requires approximately $\frac{\log n}{\log b}$ digits. (Except for the case of $n=1$, where it requires $n$ digits.)

Let $c(b)$ denote the “cost” of representing each base-$b$ digit. Then the total cost of representing $n$ is $\frac{c(b)\log n}{\log b}$. Independent of the specific choice of $n$, the “most efficient” base is the one that minimizes the quantity $f(b) = \frac{c(b)}{\log b}$.

One possible cost function is $c(b) = b$, where the cost is directly proportional to the base. This applies to things like abacuses with $b$ beads in each position, or Egyptian numerals (where each base-$b$ “digit” is notated in a unary format). Then $f(b) = \frac{b}{\log b}$. This value is optimized when $f'(b) = \frac{\log b - 1}{(\log b)^2} = 0$, or $\log b = 1$, or $b = e$.

However, different cost functions will produce different optimal bases.

For example, if $c(b) = \log b$, then $f(b) = 1$, and all bases are equally efficient.

If $c(b) = \sqrt{b}$, then optimality is achieved by $b \approx 7.39$.

If $c(b) = b^{0.4343}$, then $b \approx 10$.

I'm not sure what cost function works best with human psychology. But it's probably not a simple power function like the above. I'd expect the “cost” function to be much lower for integer bases than for fractional ones. And account for whether commonly-encountered fractions (1/2, 1/3, 1/4, etc.) have simple representations. For more discussion, see What could be better than base 10?