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When I typed this question in google I found this link: http://octomatics.org/ Just from the graphic point of view: this system seems to be easier (when he explains that you can overlap the line). He also talks about how base 8 is useful for computers. Is base 8 really the best,what makes one base better than the other. Is there some mathematical argument that we can use to decide which is the best base to use? Thank you for your insight, I think the answer has to be entangled with other areas of knowledge as cognitive sciences or simply how large are the numbers that are used more often but: Arithmetic is something that almost everybody does so it is probably worth a while to try to see if we are making it as easy as possible. Thank you very much in advance.

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Depending on the base one needs a certain number of symbols, to represent each place. For example base 2 only requires the symbols 0,1, while base 10 requires 0,1,2,3,4,5,6,7,8,9. –  Ethan Dec 14 '12 at 23:56
"Useful for computers" means electronics would be more pertinent than cognitive science. –  anon Dec 14 '12 at 23:59
Base $\omega$, in which every natural number is representable by a unique digit! :-) –  Asaf Karagila Dec 15 '12 at 0:02
Wait, sorry I didnt mean usefull for computers like to make computers go faster, I meant It is easier to go from binary to base 2 like when you are seeing code or whatnot. –  user4140 Dec 15 '12 at 0:04
It's worth making the historical note that octal was very commonly used in connection with computers up until the 1980s (its chief advantage being that it didn't require any unfamiliar numerals), but it's been almost completely supplanted by hexadecimal (base 16) these days. The reason being that computers tend to work in chunks of 8 bits, or 16, or 32, etc. all of which are divided evenly by 4-bit hexadecimal digits, but not divided evenly by 3-bit octal digits. –  hobbs Dec 15 '12 at 20:10
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4 Answers

up vote 5 down vote accepted

We could say that there are two desirable goals for a base system:

  1. A minimized number of symbols (e.g. there are two symbols in binary, 0 and 1).
  2. Minimal digital lengths for each number (e.g. the number 10101 has a length of 5 digits in binary).

Let us say our base is $b$, and hence we have exactly $b$ symbols to represent numbers with. The digital length of a number is then represented as $\log_bN$ for a positive number $N.$

There are many ways to attempt to minimize both of these values simultaneously. One way is to minimize $||\langle b,\space \log_bN \rangle||_p$, where $||\small \vec x||_p$ is the p-norm of $\vec x$, and where $N$ can be any arbitrary number (hence $N$ becomes a weighting parameter of sorts).

The results you obtain with the above algorithm are completely dependent on what values you choose for $N$ and $p$. So while there is no absolute answer, you can obtain one by assigning fixed values to these two parameters.

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Quick LaTeX tip: You can use "\log" rather than "log" to get $\log$ rather than $log$. Also, you can use \rangle and \langle to get $\langle$ and $\rangle$ rather than $<$ and $>$. –  JavaMan Dec 15 '12 at 1:39
@JavaMan Hmm, I knew about the \log one, but not \rangle and \langle. That'll come in handy - thanks! –  Ryan Dec 16 '12 at 5:43
Glad I could help! –  JavaMan Dec 17 '12 at 1:42
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A while back I tried doing some pencil-and-paper arithmetic in base 16 (extraction of square roots) to see what it was like, and found that it was hard mainly because I don't have the base-16 multiplication table memorized, and there is a lot to memorize. My conclusion at the end was that the easiest way to do base-16 arithmetic is to do it in base 4 and then convert at the end; the conversion is trivial. All the numerals have twice as many digits, but it takes much less than half as long to calculate each digit, so it is much faster. Converting to base 2 goes too far and loses the tradeoff; now the numerals are twice as long again, but you do not gain enough speed to make it worth while.

For base $n$, you have to memorize $\frac12\bigl(n^2-3n+2\bigr)$ multiplication table entries, since $0\times n$ and $1\times n$ are trivial, and $m\times n = n\times m$. This increases rapidly with $n$: in base 10 you have 36 products to memorize; in base 16 there are 105. For base 4, you only have to remember $2\times 2 = 10, 2\times 3 = 12, $ and $3\times 3 = 21$.

I did not try any of the usual methods for speeding up paper-and-pencil multiplication, such as Napier's bones, or Genaille-Lucas rulers. But I don't think these would help enough. If you don't have the multiplication tables down pat, you lose a lot of number sense that you probably take for granted. For example, you can't estimate the answer to division problems. The square root algorithm requires that you guess the answer to a lot of sub-problems of the type "How many times does 2D,409 go into 1C0,000?" and whereas this sort of thing is easy in base 10 because you have years of practice (What's $1,835,000\div 185,353$ to one decimal place?) it is hard in base 16. So knowing the multiplication tables is important, and in base 16 there is a lot to know.

I found the exercise very enlightening, and I recommend it. If nothing else, it may be interesting to re-experience feelings that you have not had since third grade.

Converting to and from base 8 is not so felicitous, but the multiplication table is not too hard to memorize. Still I think the advantage over bases 4 or 16 are minimal, if they exist at all. I think the worst choice for pencil calculation would be a large prime number, since there's no easy conversion to a smaller base as there is from bases 8 or 16 to base 4.

The desiderata for computer calculation are of course completely different, and for abacus calculation are different again. For computers I have seen claims that base 3 is in some sense optimal, but I think 70 years of engineering practice refutes that completely. There is a certain sense in which modern computers operate not in base 2 but in base 256, but it seems to be something of a philosophical question which it is; it depends on what level of operation you are looking at.

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I think the claims that base 3 is optimal stem from an assumption that a trit is exactly 3/2 times as hard to work with as a bit, but numbers become log(2)/log(3) ~= 0.631 times as long, the product of those two factors being 0.946, so ternary is notionally 5% more efficient than binary. Going from base 3 to 4 you get instead 4 log 3 / 3 log 4, which is greater than 1. The ideal base by this notion is base e, but 3 is the nearest integer. Of course, it's that first assumption that's hard to justify. –  hobbs Dec 15 '12 at 20:03
@hobbs Taking the product of those two is a bit arbitrary though. I'd prefer to add a free parameter $s$. So rather than minimizing $\frac {n+1}{n} \times \frac {\log {n}}{\log {n+1}}$, you would minimize $(\frac {n+1}{n})^s \times \frac {\log {n}}{\log {n+1}}$. –  Ryan Dec 18 '12 at 6:57
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Base 8 has direct conversion back and forth binary and would work almost equally well.

Having base-2 rather than anything else is a matter of electronics/hardware concerns before even getting into the values-in-any-base zone. Each separate bit is converted from an analog stream to binary through a measure it is judged-- it is cast as 1 if it is closer to 1 than it is to 0 and vice verse (An electronics expert would comment further on this). Splitting a signal among more choices would decrease its accuracy and is heavier on the hardware costs. The hardware was more expansive than the software on it back in the early days while these were all being set.

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Slight note: there's nothing particularly special about base 8. Any base $m$ that is of the form $m = 2^n$ has a direct conversion to and from binary: simply group and substitute. –  Nik Bougalis Dec 15 '12 at 4:50
@NikBougalis and which is clear enough. –  ashley Dec 15 '12 at 5:04
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To have the largest readership possible, along with the greatest probability of your readers reading your work until the end, and the probability of your readers also finding your work easy to read, at present, it seems you want base ten.

There doesn't exist any mathematical argument as to which base system comes as the best until we have criteria to determine what "the best" would mean. Who do we want to talk to? For what purpose do we talk to them? What sort of background do they have? What sort of tools do they have? What can those tools do?

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