Favorite numeral system?

[°Leprechaun]

Another question: what is your favorite number?

17 choices because 17 is a cool number
16 day poll because 16 is a cool number, although 17 is cooler

https://en.wikipedia.org/wiki/List_of_numeral_systems

https://en.wikipedia.org/wiki/List_of_numbers_in_various_languages

[°Leprechaun]

Can you think of a number that has an "a" in it? (spelled out in words, not in the number itself)

[Why]

ninty two quadrillion dollars

[°Leprechaun]

Well, quadrillion was the only number that I could think of. Of course, dollars as well.
and quarters.

[dead0man]

hexadecimal (for some reason called base 16 in the poll) is my favorite


I could do basic math in binary, octal, decimal (obviously) and hexadecimal at one point in my life....probably still could I guess, but I'd have to remember/relearn the tricks.  Oddly, it almost NEVER comes up in real life that one needs to divide two octal numbers.

[SUSAN CRUSHBONE]

dozenal (normal)

[DavidB.]

I don't like math (normal)

[°Leprechaun]

hexadecimal (for some reason called base 16 in the poll) is my favorite


I could do basic math in binary, octal, decimal (obviously) and hexadecimal at one point in my life....probably still could I guess, but I'd have to remember/relearn the tricks.  Oddly, it almost NEVER comes up in real life that one needs to divide two octal numbers.

The only reason is that I was being lazy, I guess.

[angus]

other:  Maya

Also, 8 has the letter a in German.  40 has the letter a in Italian, Spanish, French, and Portuguese.  In English, one thousand has the letter a, as does one thousand one, one thousand two, one thousand three, as do an infinite number of other integers. 

[°Leprechaun]

other:  Maya

Also, 8 has the letter a in German.  40 has the letter a in Italian, Spanish, French, and Portuguese.  In English, one thousand has the letter a, as does one thousand one, one thousand two, one thousand three, as do an infinite number of other integers. 

I knew that about other languages. I wasn't thinking when I was thinking about English, however and you are right, I don't know why I forgot about one thousand. Come to think of it you could say a hundred instead of one hundred, as well. You could also say one thousand and one instead of one thousand one, so there are a lot of "A"s. Not before one hundred, however unless you say twenty and one, which most people don't unless they really like the letter a. The point is that a isn't as common as e,i, and o. U is the same as A, I guess. I also don't know if any of this is really important, so sorry for wasting  time. I think math is very important, however and that's the point.

[True Federalist (진정한 연방 주의자)]

Base 60, like the old time Babylonians used, and we still use in keeping time.

[muon2]

What's with the votes for base 11-15? I've worked a lot with alternate bases, and this isn't a range that comes up much. At best there is base 12, but that only includes the concepts of dozen and gross (a dozen dozen), but nothing beyond that so it's limited to smaller counts.

[dead0man]

What's with the votes for base 11-15? I've worked a lot with alternate bases, and this isn't a range that comes up much. At best there is base 12, but that only includes the concepts of dozen and gross (a dozen dozen), but nothing beyond that so it's limited to smaller counts.

It is indeed odd that nearly a quarter of us have voted for that range.

[muon2]

Base 3 doesn't come up much, but there is an argument that it is the most efficient numeral system.

Suppose that the cost of a number is related to the product of the number of digits in a number times the number of choices for a digit (the base). For example the number 134₁₀ has a cost of 3*10=30, 10000110₂ has a cost of 8*2=16, so it's more efficient to store that number in binary than decimal. This model of cost is actually not unreasonable given our knowledge of information storage.

The number of digits in an arbitrary number k is equal to the logarithm of a number in the base n it is represented by (rounded down).  Eliminating the rounding factor I can write the cost in terms of the natural logarithm (ln) as

C(k,n) = (ln k/ln n) * n.

The cost as a function of the base is minimized when the derivative of the cost function is zero.

dC(k,n)/dn = 0

Finding the derivative reduces this equation to

ln n - 1 = 0

The solution is n = e, the base of the natural logarithm. e is an irrational number approximately equal to 2.718 and can't be used as a base for discrete counting numbers. However, the closest integer to e is 3, so that is the most efficient integer base. To use my earlier example 134₁₀ = 11222₃ (81+27+2*9+2*3+2*1). The cost of 11222₃ is 5*3 = 15, so it comes up slightly better than binary.

[dead0man]

and my brain just locked up.  Do you know how long it takes to reboot this thing?

[°Leprechaun]

and my brain just locked up.  Do you know how long it takes to reboot this thing?

I think most people have a problem with math, so you're not alone there.

[angus]

Base 3 doesn't come up much, but there is an argument that it is the most efficient numeral system.

Suppose that the cost of a number is related to the product of the number of digits in a number times the number of choices for a digit (the base). For example the number 134₁₀ has a cost of 3*10=30, 10000110₂ has a cost of 8*2=16, so it's more efficient to store that number in binary than decimal. This model of cost is actually not unreasonable given our knowledge of information storage.

The number of digits in an arbitrary number k is equal to the logarithm of a number in the base n it is represented by (rounded down).  Eliminating the rounding factor I can write the cost in terms of the natural logarithm (ln) as

C(k,n) = (ln k/ln n) * n.

The cost as a function of the base is minimized when the derivative of the cost function is zero.

dC(k,n)/dn = 0

Finding the derivative reduces this equation to

ln n - 1 = 0

The solution is n = e, the base of the natural logarithm. e is an irrational number approximately equal to 2.718 and can't be used as a base for discrete counting numbers. However, the closest integer to e is 3, so that is the most efficient integer base. To use my earlier example 134₁₀ = 11222₃ (81+27+2*9+2*3+2*1). The cost of 11222₃ is 5*3 = 15, so it comes up slightly better than binary.

I follow, except that in the third paragraph you might need the phrase "the ratio of the natural log of the number to the natural log of the base" after the word equal.  What you have in prose is not quite correct, I think.

The math looks okay.  I started with dC/dn=0 and applied both the product and chain rule and got ln(n)=1.  (fun exercise, by the way!)

This yields n=e.  

Also, I'm wondering about the next bit.  (I actually started to post "e" when I initially posted, to be a smart-aleck, but then I remembered that we need an integer for the base.  Then I googled Klingon thinking that they might have a cool numbering system; alas they use base 10 as well.  Then I remembered the Maya have a nifty system based on 1, 5, and multiples of 20 and 40.)  Anyway, you then say that since e won't work, we must choose 3, which is the nearest integer.  But does it follow that the if the problem is minimized for n=e, and since we need an integer, then n=3 is necessarily the integer?

[muon2]

It's a reasonable question. I haven't seen a proof that 3 is better than 2 but consider this table of costs. As I look at the beginning base 3 is better than base 2 more often than the reverse. I suspect the trend continues as base 3 pulls away as the more economical choice.

range	base 2 cost	base 3 cost
81-127	14	15
128-242	16	15
243-255	16	18
256-511	18	18
512-728	20	18
729-1023	20	21
1024-2047	22	21
2048-2186	24	21
2187-4095	24	24
4096-6580	26	24
6581-8191	26	27
8192-16383	28	27
16384-19682	30	27
19683-32767	30	30
32768-59048	32	30
59049-65535	32	33
65536-131071	34	33
131072-177146	36	33
177147-262143	36	36
262144-524287	38	36
524288-531440	40	36
531441-1048575	40	39

edit: I checked 2³⁰ - 1 = 1,073,741,823 takes 30 digits in base 2 (cost = 60) and 18 digits in base 3 (cost = 54) so my supposition seems to hold.

[°Leprechaun]

by the way my vote was for base 4, because of what 222 equals

[angus]

...my supposition seems to hold.

seems so, thus far.  I guess what we want to demonstrate is something like:

 lim  C_n=3  <  lim  C_n=2
n→∞               n→∞

Or, more broadly,

 lim  C_n=3  <  lim  C_{n∈{N ∀ n≠3}}
n→∞               n→∞

Proof by the Second Principle of Mathematical Induction, perhaps, assuming that you can demonstrate conclusively for 2 and 4? 

Well, it's an interesting subject.  I hadn't encountered the idea of the cost of expressing numbers before.  The next time I'm at a cocktail party during which the subject of favorite bases comes up, I'll be able to suggest 3 in a way that makes me look really smart.

[Antonio the Sixth]

Either base 8 or base 16. I love powers of 2.

Having more than 20 symbols would start making it difficult to remember things.

[Figueira]

Normally I'd say base 12 (although I like bases 6, 8, and 14 as well) but I had to choose the one named after me.

[muon2]

...my supposition seems to hold.

seems so, thus far.  I guess what we want to demonstrate is something like:

 lim  C_n=3  <  lim  C_n=2
n→∞               n→∞

Or, more broadly,

 lim  C_n=3  <  lim  C_{n∈{N ∀ n≠3}}
n→∞               n→∞

Proof by the Second Principle of Mathematical Induction, perhaps, assuming that you can demonstrate conclusively for 2 and 4? 

Well, it's an interesting subject.  I hadn't encountered the idea of the cost of expressing numbers before.  The next time I'm at a cocktail party during which the subject of favorite bases comes up, I'll be able to suggest 3 in a way that makes me look really smart.

The cost grows without limit because it grows as ln(k). If I look at C/ln(k) = n/ln(n) that gives me a measure of the generalized cost independent of the specific number. For base 2 it is 2.89 (same for base 4) and for base 3 it is 2.73. That adds more credence to 3 as the most efficient base.

[muon2]

...my supposition seems to hold.

seems so, thus far.  I guess what we want to demonstrate is something like:

 lim  C_n=3  <  lim  C_n=2
n→∞               n→∞

Or, more broadly,

 lim  C_n=3  <  lim  C_{n∈{N ∀ n≠3}}
n→∞               n→∞

Proof by the Second Principle of Mathematical Induction, perhaps, assuming that you can demonstrate conclusively for 2 and 4? 

Well, it's an interesting subject.  I hadn't encountered the idea of the cost of expressing numbers before.  The next time I'm at a cocktail party during which the subject of favorite bases comes up, I'll be able to suggest 3 in a way that makes me look really smart.

The cost grows without limit because it grows as ln(k). If I look at C/ln(k) = n/ln(n) that gives me a measure of the generalized cost independent of the specific number. For base 2 it is 2.89 (same for base 4) and for base 3 it is 2.73. That adds more credence to 3 as the most efficient base.

[angus]

yes, I meant as k approaches infinity, not n.  

It's clear that n/ln(n) increases without bound for n≥3.  This is independent of k.  I made smoothed curve Microsoft Excel plot of ln(n)/n for the first 15 integers, where I used 1.2 for the first independent value to avoid overflow.  



I guess I was thinking of an elegant syllogistic proof.  It has been a long time since I have thought about those things.  The proofs and demonstrations we do in thermodynamics are algebraic derivations involving simple substitutions with differential and integral calculus manipulations.  Nothing too complicated.   

It turns out that Wikipedia has a fairly accessible article on Radix Economy.  It shows, in tabular form, that e minimizes C, and that 3 gives the lowest C among the integers under 60.  It also offers a direct proof that e is most efficient, as you did.  Then it simply states that "it follows that 3 is the integer base with the lowest average radix economy."  I'm not sure that isn't a non-sequitur, but I'd have to give it more thought.