Mixing
What could be better than base 10?
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Most people use base 10; it's obviously the common notation in the modern world.
However, if we could change what became the common notation, would there be a better choice?
I'm aware that it very well may be that there is no intrinsically superior base, but for the purposes of humans, is there a better one?
I've heard from sources such as this and this that base 12 is better, from here that base 8 is better, and, being into computer science, I would say that base 16 is the most handy.
Base 12 does seem to be the most supported non-base 10 number system, mainly due to the following reason pointed out by George Dvorsky:
First and foremost, 12 is a highly composite number — the smallest
number with exactly four divisors: 2, 3, 4, and 6 (six if you count 1
and 12). As noted, 10 has only two. Consequently, 12 is much more
practical when using fractions — it's easier to divide units of
weights and measures into 12 parts, namely halves, thirds, and
quarters.
And, on top of that, previous societies considered very advanced used other systems, such as the Mayans using base 20, and the Babylonians using base 60.
So, summarized, my question is: Is there an intrinsically superior base? If not, is there one that would be best for society's purposes? Or does the best base depend on the context it is being used in?
notation number-systems convention
asked May 6 '13 at 1:25
CisplatinCisplatin
1,96952249
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show 5 more comments
$begingroup$
Most people use base 10; it's obviously the common notation in the modern world.
However, if we could change what became the common notation, would there be a better choice?
I'm aware that it very well may be that there is no intrinsically superior base, but for the purposes of humans, is there a better one?
I've heard from sources such as this and this that base 12 is better, from here that base 8 is better, and, being into computer science, I would say that base 16 is the most handy.
Base 12 does seem to be the most supported non-base 10 number system, mainly due to the following reason pointed out by George Dvorsky:
First and foremost, 12 is a highly composite number — the smallest
number with exactly four divisors: 2, 3, 4, and 6 (six if you count 1
and 12). As noted, 10 has only two. Consequently, 12 is much more
practical when using fractions — it's easier to divide units of
weights and measures into 12 parts, namely halves, thirds, and
quarters.
And, on top of that, previous societies considered very advanced used other systems, such as the Mayans using base 20, and the Babylonians using base 60.
So, summarized, my question is: Is there an intrinsically superior base? If not, is there one that would be best for society's purposes? Or does the best base depend on the context it is being used in?
notation number-systems convention
asked May 6 '13 at 1:25
CisplatinCisplatin
1,96952249
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2
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The usefulness of base-$12$ is certainly being pushed by the Dozenal Society of America. FWIW, if you can get your hands on a copy of Dudley's Mathematical Cranks, there is a delightful discussion on the people who tirelessly push for the more widespread use of other bases.
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– J. M. is not a mathematician
May 6 '13 at 10:57
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But if we used base 12, "The Dirty Dozen" would have been called "The Dirty Ten", which is not as nearly as good title.
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– Asaf Karagila♦
May 6 '13 at 11:04
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@Asaf, I'm sure y'know that $10_{12}$ isn't read as "ten", right? ;)
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– J. M. is not a mathematician
May 6 '13 at 17:39
3
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Probably the biggest qualitative difference in usefulness that I've heard of is that from a practical computational perspective highly composite bases are ideal, whereas prime bases are more suitable for pure mathematical purposes. If there was a base that could reap both benefits simultaneously, there would probably be more of a push to switch to such a system. But of course, there isn't
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– David H
May 12 '13 at 3:27
10
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It's not a XKCD, but I think this comic is mandatory here... cowbirdsinlove.com/43
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– woliveirajr
May 14 '13 at 12:12
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show 5 more comments
$begingroup$
Most people use base 10; it's obviously the common notation in the modern world.
However, if we could change what became the common notation, would there be a better choice?
I'm aware that it very well may be that there is no intrinsically superior base, but for the purposes of humans, is there a better one?
I've heard from sources such as this and this that base 12 is better, from here that base 8 is better, and, being into computer science, I would say that base 16 is the most handy.
Base 12 does seem to be the most supported non-base 10 number system, mainly due to the following reason pointed out by George Dvorsky:
First and foremost, 12 is a highly composite number — the smallest
number with exactly four divisors: 2, 3, 4, and 6 (six if you count 1
and 12). As noted, 10 has only two. Consequently, 12 is much more
practical when using fractions — it's easier to divide units of
weights and measures into 12 parts, namely halves, thirds, and
quarters.
And, on top of that, previous societies considered very advanced used other systems, such as the Mayans using base 20, and the Babylonians using base 60.
So, summarized, my question is: Is there an intrinsically superior base? If not, is there one that would be best for society's purposes? Or does the best base depend on the context it is being used in?
notation number-systems convention
asked May 6 '13 at 1:25
CisplatinCisplatin
1,96952249
$endgroup$
Most people use base 10; it's obviously the common notation in the modern world.
However, if we could change what became the common notation, would there be a better choice?
I'm aware that it very well may be that there is no intrinsically superior base, but for the purposes of humans, is there a better one?
I've heard from sources such as this and this that base 12 is better, from here that base 8 is better, and, being into computer science, I would say that base 16 is the most handy.
Base 12 does seem to be the most supported non-base 10 number system, mainly due to the following reason pointed out by George Dvorsky:
First and foremost, 12 is a highly composite number — the smallest
number with exactly four divisors: 2, 3, 4, and 6 (six if you count 1
and 12). As noted, 10 has only two. Consequently, 12 is much more
practical when using fractions — it's easier to divide units of
weights and measures into 12 parts, namely halves, thirds, and
quarters.
And, on top of that, previous societies considered very advanced used other systems, such as the Mayans using base 20, and the Babylonians using base 60.
So, summarized, my question is: Is there an intrinsically superior base? If not, is there one that would be best for society's purposes? Or does the best base depend on the context it is being used in?
notation number-systems convention
notation number-systems convention
asked May 6 '13 at 1:25
CisplatinCisplatin
1,96952249
asked May 6 '13 at 1:25
CisplatinCisplatin
1,96952249
edited May 6 '13 at 21:36
Cisplatin
asked May 6 '13 at 1:25
CisplatinCisplatin
1,96952249
asked May 6 '13 at 1:25
CisplatinCisplatin
1,96952249
asked May 6 '13 at 1:25
CisplatinCisplatin
1,96952249
1,96952249
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The usefulness of base-$12$ is certainly being pushed by the Dozenal Society of America. FWIW, if you can get your hands on a copy of Dudley's Mathematical Cranks, there is a delightful discussion on the people who tirelessly push for the more widespread use of other bases.
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– J. M. is not a mathematician
May 6 '13 at 10:57
7
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But if we used base 12, "The Dirty Dozen" would have been called "The Dirty Ten", which is not as nearly as good title.
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– Asaf Karagila♦
May 6 '13 at 11:04
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@Asaf, I'm sure y'know that $10_{12}$ isn't read as "ten", right? ;)
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– J. M. is not a mathematician
May 6 '13 at 17:39
3
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Probably the biggest qualitative difference in usefulness that I've heard of is that from a practical computational perspective highly composite bases are ideal, whereas prime bases are more suitable for pure mathematical purposes. If there was a base that could reap both benefits simultaneously, there would probably be more of a push to switch to such a system. But of course, there isn't
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– David H
May 12 '13 at 3:27
10
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It's not a XKCD, but I think this comic is mandatory here... cowbirdsinlove.com/43
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– woliveirajr
May 14 '13 at 12:12
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show 5 more comments
2
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The usefulness of base-$12$ is certainly being pushed by the Dozenal Society of America. FWIW, if you can get your hands on a copy of Dudley's Mathematical Cranks, there is a delightful discussion on the people who tirelessly push for the more widespread use of other bases.
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– J. M. is not a mathematician
May 6 '13 at 10:57
7
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But if we used base 12, "The Dirty Dozen" would have been called "The Dirty Ten", which is not as nearly as good title.
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– Asaf Karagila♦
May 6 '13 at 11:04
4
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@Asaf, I'm sure y'know that $10_{12}$ isn't read as "ten", right? ;)
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– J. M. is not a mathematician
May 6 '13 at 17:39
3
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Probably the biggest qualitative difference in usefulness that I've heard of is that from a practical computational perspective highly composite bases are ideal, whereas prime bases are more suitable for pure mathematical purposes. If there was a base that could reap both benefits simultaneously, there would probably be more of a push to switch to such a system. But of course, there isn't
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– David H
May 12 '13 at 3:27
10
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It's not a XKCD, but I think this comic is mandatory here... cowbirdsinlove.com/43
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– woliveirajr
May 14 '13 at 12:12
2
2
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The usefulness of base-$12$ is certainly being pushed by the Dozenal Society of America. FWIW, if you can get your hands on a copy of Dudley's Mathematical Cranks, there is a delightful discussion on the people who tirelessly push for the more widespread use of other bases.
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– J. M. is not a mathematician
May 6 '13 at 10:57
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The usefulness of base-$12$ is certainly being pushed by the Dozenal Society of America. FWIW, if you can get your hands on a copy of Dudley's Mathematical Cranks, there is a delightful discussion on the people who tirelessly push for the more widespread use of other bases.
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– J. M. is not a mathematician
May 6 '13 at 10:57
7
7
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But if we used base 12, "The Dirty Dozen" would have been called "The Dirty Ten", which is not as nearly as good title.
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– Asaf Karagila♦
May 6 '13 at 11:04
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But if we used base 12, "The Dirty Dozen" would have been called "The Dirty Ten", which is not as nearly as good title.
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– Asaf Karagila♦
May 6 '13 at 11:04
4
4
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@Asaf, I'm sure y'know that $10_{12}$ isn't read as "ten", right? ;)
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– J. M. is not a mathematician
May 6 '13 at 17:39
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@Asaf, I'm sure y'know that $10_{12}$ isn't read as "ten", right? ;)
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– J. M. is not a mathematician
May 6 '13 at 17:39
3
3
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Probably the biggest qualitative difference in usefulness that I've heard of is that from a practical computational perspective highly composite bases are ideal, whereas prime bases are more suitable for pure mathematical purposes. If there was a base that could reap both benefits simultaneously, there would probably be more of a push to switch to such a system. But of course, there isn't
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– David H
May 12 '13 at 3:27
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Probably the biggest qualitative difference in usefulness that I've heard of is that from a practical computational perspective highly composite bases are ideal, whereas prime bases are more suitable for pure mathematical purposes. If there was a base that could reap both benefits simultaneously, there would probably be more of a push to switch to such a system. But of course, there isn't
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– David H
May 12 '13 at 3:27
10
10
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It's not a XKCD, but I think this comic is mandatory here... cowbirdsinlove.com/43
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– woliveirajr
May 14 '13 at 12:12
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It's not a XKCD, but I think this comic is mandatory here... cowbirdsinlove.com/43
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– woliveirajr
May 14 '13 at 12:12
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show 5 more comments
11 Answers
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I like the factorial base,
where the integer part of a real number
is written as
$sum_{i=2}^n a_i i!$
where the $a_i$ are integers such that $0 le a_i < i$
and the fractional part is written as
$sum_{i=2}^{infty} frac{b_i}{i!}$
where the $b_i$ are integers such that$0 le b_i < i$.
The nice thing about this is that
the integer part has a unique representation and
the fractional part terminates if and only if
the number if rational (except for the
case corresponding to
$frac1{n!} = sum_{i=n+1}^{infty} frac{i-1}{i!}$,
the same as 1 = .99999...).
This is a special case of the following result:
If $(B_i)_{i=0}^{infty}$ is an increasing series
of positive integers with $B_0 = 1$,
we can represent all positive integers in the form
$N=sum_{i=1}^m a_i B_i$
where $0 le a_i < B_{i}/B_{i-1}$
and $N < B_m$.
This representation is unique if and only if
$B_{i}/B_{i-1}$ is an integer for all $i$.
The usual decimal, binary, and hexadecimal bases have
$B_i = 2^i, 10^i$, or $16^i$.
The factorial base has
$B_i = (i+1)!$.
I worked this out over 40 years ago
and found it quite interesting.
I am sure the result is several hundered years old.
answered May 6 '13 at 5:35
marty cohenmarty cohen
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Um... call me dumb, but in what base do you represent each $a_i$, considering that no constant finite base will suffice once $i!$ is greater than it?
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– ikdc
Aug 28 '14 at 21:32
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Instead of each $a_i$ satisfying $0 le a_i le B-1$, they satisfy $0 le a_i le i$.
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– marty cohen
Sep 1 '14 at 4:18
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Doesn't that imply that you need a potentially unbounded number of symbols to represent large numbers? If a number requires $n$ factorial digits, then you need $n$ symbols to represent the most significant digit. In a classic base $B$, you only ever need $B$ symbols.
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– ikdc
Sep 1 '14 at 5:29
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Yep. That's why we have the integers.
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– marty cohen
Sep 1 '14 at 14:53
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@Marty Cohen: But how do you represent those integers? Seems you'd still need a "regular" base type system in there as a "sub-digit" system, like how the Babylonian base-60 system used a base-10 "inner" base.
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– The_Sympathizer
Jul 26 '16 at 3:42
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I think base $6$ would make counting on our hands particularly convenient, we would have a $1$'s hand and a $6$'s hand and would be able to count up to $35$.
answered May 12 '13 at 3:23
JimJim
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There is a Korean technique that uses the thumb to represent 5 and so can count up to 99 on two hands. This seems rather more convenient to me than your suggestion.
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– MJD
May 15 '13 at 15:40
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Base 6 also has the advantage of trivial divisibility tests for 2 and 3, and easy tests for 5 (sum of digits) and 7 (sum of digits with alternating sign).
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– Dan
May 16 '13 at 3:40
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@MJD If you want to use your hands and are allowed to distinguish between fingers on each hand, then binary is clearly optimal, allowing you to count from 0 to 1024 (half-open).
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– ikdc
Aug 28 '14 at 21:34
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Multiplication tables would be greatly simplified up to $36$. It would also be handy to use the $36=100_{6}$ characters $0,1,dotsc,9,a,b,dotsc,z$, ready available on any keyboard, to represent numbers. I think this is the reason why $6$ is called perfect. (Beware the Dozenal Society though, they may hijack your cause for theirs...)
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– Oskar Limka
Mar 23 '18 at 7:02
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Brian Hayes in his American Scientist article Third Base argues that "When base 2 is too small and base 10 is too big, base 3 is just right."
Figure 1 has the caption
Most economical radix for a numbering system is $e$ (about $2.718$) when economy is measured as the product of the radix and the width, or number of digits, needed to express a given range of values. Here both the radix and the width are treated as continuous variables.
Figure 2 has the caption
Most economical integer radix is almost always 3, the integer closest to $e$. If the capacity of a numbering system is $r^w$, and the cost of a representation is $rw$, then
$r=3$ is the best integer radix for all but a finite set of capacities. Specifically, ternary is inferior to binary only for 8,487 values of $r^w$; ternary is superior for infinitely many values.
Figure 3 has the caption
Ternary structure may offer the quickest path through a telephone menu system. Putting eight choices (assumed to be equally likely) in a single octonary menu (left) forces the caller to listen to 4.5 menu items on average. A binary structure (middle) has the same
performance, but the ternary tree (right) reduces the average to 3.75.
answered May 6 '13 at 7:52
Joel Reyes NocheJoel Reyes Noche
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Sure, base 3 may minimize the number of beads on an abacus, but would you really want 1/2 to be 0.111111111... recurring?
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– Dan
May 16 '13 at 3:47
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1/3 is 0.3333333 recurring in base 10...
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– Michael
Aug 6 '13 at 17:15
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@michael, that's still tolerable. Dan has a point. Doubling and halving is the most common mental arithmetic people do. I wouldn't want 1/2 to be recurring but 1/3 is more OK.
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– nawfal
May 26 '18 at 7:48
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For computer applications, bases like 2, 8 and 16 are obviously the best. Given that a large percentage of numerical data is stored in and processed by computers, these days, one could argue that what's good for computers is good for society.
Of the three I mentioned, I suppose that 8 or 16 would be better than base 2. Having the price of bananas as a binary number in the supermarket wouldn't work too well. Binary numbers are too long, and they all tend to look alike, so they're hard for people to read.
In the world at large (as opposed to the narrower world of mathematics and computers), reading numbers is probably just as important as doing arithmetic with them. Think of speed limit signs on roads, distances of journeys, prices in stores, or temperatures in weather forecasts. These numbers need to be read and understood quickly (by human beings), and I doubt that this would be possible if they were written in binary. We'd no longer be taking advantage of the wonderful human ability to quickly recognize symbols, and it would be a pity to waste that ability just so that we can make computing easier (in my opinion).
answered May 6 '13 at 3:44
bubbabubba
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Not sure why $4$ was left out of $2, 8, 16$.
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– Lance Pollard
Jan 13 at 16:41
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Because bases 2, 8, 16 are already widely used in computing, whereas 4 is not (as far as I know). But, I suppose it would work OK.
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– bubba
Jan 13 at 23:34
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Okay that makes sense. 4 is not as used b/c computers are 8-bit at least, so 4 is inefficient. But 2 is there because everything is just 1's and 0's.
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– Lance Pollard
Jan 14 at 0:04
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Balanced Nonary (base 9) would probably be really good. The digits go from -4 to 4, so taking the negative of a number would just be taking the negative of each digit, so subtraction is easy. Multiplication and division are particularly easy too if you make the easy conversion to balanced ternary first. Then there's no carrying when multiplying single digits (like in binary), and division is just testing inequalities (if you can divide by 2). Of course, if you want to do things faster, learning a balanced nonary times table would be easier than learning a regular nonary times table since you only really need to know the table for 1,2,3,4 and then handle negatives (and zero) appropriately.
There have even been computers based on balanced ternary.
answered May 8 '13 at 0:07
Mark S.Mark S.
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While bubba raises valid points about base 2 from a practicality standpoint, I myself would defend the choice of base 2 for the following reason: it makes addition and multiplication incredibly easy. This is, in fact, the way computers do these basic operations.
Addition in binary operates under the following rules:
$0 land 0 = 0$
$1 land 0 = 1$
$0 land 1 = 1$
$1 land 1 = 0$ (carry a 1)
Therefore when you do long addition in binary, the algorithm is particularly simple: if there are 2 $0$'s in the column, you put down $0$, if there is one $0$ and one $1$ you put down $1$, if there are 2 $1$'s you put down $0$ and carry $1$ over to the next place value. Imagine the time we could save by teaching kids to add this way. We could start teaching actual mathematics instead!
Long multiplication is just as easy: for every place value you're multiplying by either $0$ or $1$, which makes the computation very simple. I invite you to try out a few simple sums and products in binary to see what I mean.
See http://en.wikipedia.org/wiki/Binary_arithmetic#Addition and http://en.wikipedia.org/wiki/Binary_arithmetic#Multiplication for more on these two operations; the article has details on subtraction, division, and square roots as well.
answered May 6 '13 at 7:09
Gyu Eun LeeGyu Eun Lee
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Binary makes arithmetic easy on paper. But how would it affect mental arithmetic?
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– Jack M
May 24 '13 at 14:09
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I'm not really sure, but mental arithmetic is no easier in any other base - the only difference is the number of digits for larger numbers. I say let people use a calculator. Arithmetic is the job they were born to do, and the job we were born to automate.
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– Gyu Eun Lee
May 24 '13 at 18:00
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It makes numbers a bit longer.
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– PyRulez
Jul 1 '13 at 14:42
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Sure, base 2 is convenient; however, some numbers that have a finite decimal expansion (e.g. $(0.2)_{10}$) but an infinite binary expansion ($(0.00overline{11})_2$) is problematic, as it is a source of roundoff error in floating-point-number systems.
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– jubobs
Sep 30 '13 at 15:23
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@Jubobs Same with $(frac{1}{3})_{10}$: $0.overline{3}$.
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– Cole Johnson
May 25 '14 at 21:49
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In order to answer this question, it is first necessary to ask: What makes one base “better” than other? Some reasonable things to consider are:
Size
There is a tradeoff between the number of distinct digit characters used in a base (Base $b$ has exactly $b$ of these, from $0$ to $b - 1$, inclusive) and the length of the numeral required to represent a given number (which is $O(1/log{b})$).
If the base is too small, then numbers explode into cumbersome long strings of digits. For example, in binary, the current year is 111 1101 1101, and the population of China (according to its 2010 census) was 100 1111 1101 1010 1001 0100 0011 0100. Modern computers can easily work with 32-bit or 64-bit binary numbers, but humans can't, which is why programmers have developed more compact encodings of binary, such as hexadecimal.
On the other hand, if we picked a very large base, like 2520, then you would need only 3 characters to represent the population of China, but typing them would be just as challenging as typing Chinese. And forget about learning the mulitplication table, whose size is $O(b^2)$. The only practical way to use such a large base is to split it into sub-bases, the way base-60 is represented as a mixture of base-6 and base-10.
So, what we want is a happy medium.
Fraction-friendliness
This is the main argument advanced in favor of base-12 or other highly composite bases (2, 4, 6, 12, 24, 36, 48, 60, 120, ...).
If a base has a lot of factors, it makes fractions easier to work with. For example, in base ten, 1/3 is represented as the infinitely repeating 0.333 333 333... (often rounded to 0.33 or 0.333), and this awkwardness crops up in deals like “3 for $5” or +/- grading systems. But in base-12, 1/3 is a nice simple 0.4.
Of course, because there are an infinite number of primes, it's impossible to completely avoid repeating “decimals”. And base-12's simplicity for the fractions 1/3 (0.4), 1/4 (0.3), 1/6 (0.2), 1/8 (0.16) and 1/9 (0.14) comes at the price of making 1/5 (0.24972497...) and 1/10 (0.124972497...) recurring dozenal fractions. But 1/3 is more common than 1/5.
answered May 16 '13 at 3:48
DanDan
4,55511517
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The quater-imaginary base $2i$ is quite amusing, being able to express every complex number using only digits in ${0, 1, 2, 3}$. Being the only base yet proposed here that includes the elements of this extended system, it is clearly the best one for this purpose :)
answered Aug 24 '14 at 2:40
Ryan ReichRyan Reich
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Not all bases are 10 in their own notation. There are a group of alternating bases where the base is not 'ten' but a 'hundred'. The most elegant of these is the long-hundred of the proto-germanics and their decendents. Reckoning in the six-score long-hundred (ie 120), was still common enough in 1350 to pass without comment.
Yes, i have used this base for some thirty years. It's truly elegant, being more efficient than either 10 or 12. It's the first base, for which the (number of proper divisors)/(ln base) is greater than 3.
Also 120 is the smallest multiply perfect number, and has the same features as the perfect numbers. For example, 120 = 1+2+4+8+15+30+60 = 3+5+6+10+12+20+24+40, all of these numbers make the total divisors of 120. The second set corresponds to a set of weights, eg
- 1 oz, 2 oz, 4 oz, 8 oz, 1 lb, 2 lb, 4 lb. 15 oz = 1 lb , 120 oz = 1 clove.
- 1 ct, 2 ct, 4 ct, 8 ct, 1 dr, 2 dr, 4 dr : 15 ct = 1dr , 120 ct = 1 oz
- 1 lb, 2 lb, 4 lb, 8 lb, 1 st, 2 st, 4 st : 15 lb = 1 st, 120 lb = 1 cwt
When one considers not just integers, but also fractions x/y and y/z, arranged by xy, one finds that the expressions of these are very short for the first sixty or so, even in things like 56 (8/7 = 1:17.17.17, vs 7/8 = V5), and this unusual pair at 96 (3/32 = 11:30, 32/3 = 10:80.
answered May 15 '13 at 9:20
Wendy KriegerWendy Krieger
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It is not necessarily true that the multiplication is of the order of $O(b^2)$, since this is a particular implementation of the base, rather than the base itself. The mayans divided their score into four sticks of five dots, and had a true zero in the dot-position (eg fifteen is "3-fives-zero").
One should remember that counting (multiples) and division are separate operations, and that it is possible to use different number-systems for them. Historically, the sixty-wise system is one of divisions: the first column is that of units, and later places are divisions by sixty. Likewise, the romans multiplied by 10's, and divided into 12's.
An alternating base like 60 or 120, supposes two rows on each column of the abacus, where the unit (in the bottom row), is counted to 10 to make one carry up, but divides into 12 to be borrowed into the top row of the lower column. Since one can start in either the top row or bottom row (for counting), the use of tens by twelves or twelves by tens, automatically produces an alternating base.
Using alternating arithmetic then reduces the size of the tables to the order of $O(b)$.
It should be noted that the sumerian system is a division system to avoid division. We see this from recriprocal tables (eg 2 <=> 30 ), and tables of reckoners of multiples of the recriprocals (eg multiples of 4.26.40). Even in their reckoners, multiples are supplied for 1 to 20, and 40. Neugebauer even gives reference to a paper on the seven brothers, ie what is 1/7. It is concluded it lies between 0.8.34.16 and 0.8.34.18.
But i wrangle base 120 for nearly 30 years, and never felt the need to go past 12*12.
answered May 16 '13 at 6:55
wendy.kriegerwendy.krieger
5,83711426
$endgroup$
add a comment |
$begingroup$
Base 10 is a positional fixed base (The value of the digit depends on its position in the number as well as on its value) and it is neither to large nor to small which is best suited for most purposes. This has many advantages and the andvantage this has over factorial base is, that we need only 10 symbols to represent every possible number. Factorial number system is appealing, but it lacks this practical part. As for other fixed number bases: well,we have 10 digits and are used to think in terms of that. It probbably has something to do with the way we learn our first numbers as children. It is similar to measuring an angle in degrees or radians: we are just so much more used to it being expressed in degrees.
It is just the matter of everyday practicality: base 10 and degrees are just so much more practical in everyday life than anything else and mathematics is for everyday live as well as for advanced mathematicians! And the sciences that need other number bases, just make best use of it (like i.e. the case of computer science).
I guess there is NO number base that is wasty superior to others, so fixed base that is neither to small neither to large best suits ALL purposes and if in some particular base we need a better suited base,then we just make use of one.
The ancient Greeks didn't even use either positional number system, nor a decimal one. They didn't even have a standard number system. Their system was similar to that of Romans (not positional) and Babilonians (not decimal but base 60).
answered Dec 10 '14 at 17:32
SashaSasha
212
$endgroup$
add a comment |
protected by Asaf Karagila♦ Dec 10 '14 at 18:43
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$begingroup$
I like the factorial base,
where the integer part of a real number
is written as
$sum_{i=2}^n a_i i!$
where the $a_i$ are integers such that $0 le a_i < i$
and the fractional part is written as
$sum_{i=2}^{infty} frac{b_i}{i!}$
where the $b_i$ are integers such that$0 le b_i < i$.
The nice thing about this is that
the integer part has a unique representation and
the fractional part terminates if and only if
the number if rational (except for the
case corresponding to
$frac1{n!} = sum_{i=n+1}^{infty} frac{i-1}{i!}$,
the same as 1 = .99999...).
This is a special case of the following result:
If $(B_i)_{i=0}^{infty}$ is an increasing series
of positive integers with $B_0 = 1$,
we can represent all positive integers in the form
$N=sum_{i=1}^m a_i B_i$
where $0 le a_i < B_{i}/B_{i-1}$
and $N < B_m$.
This representation is unique if and only if
$B_{i}/B_{i-1}$ is an integer for all $i$.
The usual decimal, binary, and hexadecimal bases have
$B_i = 2^i, 10^i$, or $16^i$.
The factorial base has
$B_i = (i+1)!$.
I worked this out over 40 years ago
and found it quite interesting.
I am sure the result is several hundered years old.
answered May 6 '13 at 5:35
marty cohenmarty cohen
73.6k549128
$endgroup$
1
$begingroup$
Um... call me dumb, but in what base do you represent each $a_i$, considering that no constant finite base will suffice once $i!$ is greater than it?
$endgroup$
– ikdc
Aug 28 '14 at 21:32
$begingroup$
Instead of each $a_i$ satisfying $0 le a_i le B-1$, they satisfy $0 le a_i le i$.
$endgroup$
– marty cohen
Sep 1 '14 at 4:18
2
$begingroup$
Doesn't that imply that you need a potentially unbounded number of symbols to represent large numbers? If a number requires $n$ factorial digits, then you need $n$ symbols to represent the most significant digit. In a classic base $B$, you only ever need $B$ symbols.
$endgroup$
– ikdc
Sep 1 '14 at 5:29
$begingroup$
Yep. That's why we have the integers.
$endgroup$
– marty cohen
Sep 1 '14 at 14:53
$begingroup$
@Marty Cohen: But how do you represent those integers? Seems you'd still need a "regular" base type system in there as a "sub-digit" system, like how the Babylonian base-60 system used a base-10 "inner" base.
$endgroup$
– The_Sympathizer
Jul 26 '16 at 3:42
add a comment |
$begingroup$
I like the factorial base,
where the integer part of a real number
is written as
$sum_{i=2}^n a_i i!$
where the $a_i$ are integers such that $0 le a_i < i$
and the fractional part is written as
$sum_{i=2}^{infty} frac{b_i}{i!}$
where the $b_i$ are integers such that$0 le b_i < i$.
The nice thing about this is that
the integer part has a unique representation and
the fractional part terminates if and only if
the number if rational (except for the
case corresponding to
$frac1{n!} = sum_{i=n+1}^{infty} frac{i-1}{i!}$,
the same as 1 = .99999...).
This is a special case of the following result:
If $(B_i)_{i=0}^{infty}$ is an increasing series
of positive integers with $B_0 = 1$,
we can represent all positive integers in the form
$N=sum_{i=1}^m a_i B_i$
where $0 le a_i < B_{i}/B_{i-1}$
and $N < B_m$.
This representation is unique if and only if
$B_{i}/B_{i-1}$ is an integer for all $i$.
The usual decimal, binary, and hexadecimal bases have
$B_i = 2^i, 10^i$, or $16^i$.
The factorial base has
$B_i = (i+1)!$.
I worked this out over 40 years ago
and found it quite interesting.
I am sure the result is several hundered years old.
answered May 6 '13 at 5:35
marty cohenmarty cohen
73.6k549128
$endgroup$
1
$begingroup$
Um... call me dumb, but in what base do you represent each $a_i$, considering that no constant finite base will suffice once $i!$ is greater than it?
$endgroup$
– ikdc
Aug 28 '14 at 21:32
$begingroup$
Instead of each $a_i$ satisfying $0 le a_i le B-1$, they satisfy $0 le a_i le i$.
$endgroup$
– marty cohen
Sep 1 '14 at 4:18
2
$begingroup$
Doesn't that imply that you need a potentially unbounded number of symbols to represent large numbers? If a number requires $n$ factorial digits, then you need $n$ symbols to represent the most significant digit. In a classic base $B$, you only ever need $B$ symbols.
$endgroup$
– ikdc
Sep 1 '14 at 5:29
$begingroup$
Yep. That's why we have the integers.
$endgroup$
– marty cohen
Sep 1 '14 at 14:53
$begingroup$
@Marty Cohen: But how do you represent those integers? Seems you'd still need a "regular" base type system in there as a "sub-digit" system, like how the Babylonian base-60 system used a base-10 "inner" base.
$endgroup$
– The_Sympathizer
Jul 26 '16 at 3:42
add a comment |
$begingroup$
I like the factorial base,
where the integer part of a real number
is written as
$sum_{i=2}^n a_i i!$
where the $a_i$ are integers such that $0 le a_i < i$
and the fractional part is written as
$sum_{i=2}^{infty} frac{b_i}{i!}$
where the $b_i$ are integers such that$0 le b_i < i$.
The nice thing about this is that
the integer part has a unique representation and
the fractional part terminates if and only if
the number if rational (except for the
case corresponding to
$frac1{n!} = sum_{i=n+1}^{infty} frac{i-1}{i!}$,
the same as 1 = .99999...).
This is a special case of the following result:
If $(B_i)_{i=0}^{infty}$ is an increasing series
of positive integers with $B_0 = 1$,
we can represent all positive integers in the form
$N=sum_{i=1}^m a_i B_i$
where $0 le a_i < B_{i}/B_{i-1}$
and $N < B_m$.
This representation is unique if and only if
$B_{i}/B_{i-1}$ is an integer for all $i$.
The usual decimal, binary, and hexadecimal bases have
$B_i = 2^i, 10^i$, or $16^i$.
The factorial base has
$B_i = (i+1)!$.
I worked this out over 40 years ago
and found it quite interesting.
I am sure the result is several hundered years old.
answered May 6 '13 at 5:35
marty cohenmarty cohen
73.6k549128
$endgroup$
I like the factorial base,
where the integer part of a real number
is written as
$sum_{i=2}^n a_i i!$
where the $a_i$ are integers such that $0 le a_i < i$
and the fractional part is written as
$sum_{i=2}^{infty} frac{b_i}{i!}$
where the $b_i$ are integers such that$0 le b_i < i$.
The nice thing about this is that
the integer part has a unique representation and
the fractional part terminates if and only if
the number if rational (except for the
case corresponding to
$frac1{n!} = sum_{i=n+1}^{infty} frac{i-1}{i!}$,
the same as 1 = .99999...).
This is a special case of the following result:
If $(B_i)_{i=0}^{infty}$ is an increasing series
of positive integers with $B_0 = 1$,
we can represent all positive integers in the form
$N=sum_{i=1}^m a_i B_i$
where $0 le a_i < B_{i}/B_{i-1}$
and $N < B_m$.
This representation is unique if and only if
$B_{i}/B_{i-1}$ is an integer for all $i$.
The usual decimal, binary, and hexadecimal bases have
$B_i = 2^i, 10^i$, or $16^i$.
The factorial base has
$B_i = (i+1)!$.
I worked this out over 40 years ago
and found it quite interesting.
I am sure the result is several hundered years old.
answered May 6 '13 at 5:35
marty cohenmarty cohen
73.6k549128
edited May 6 '13 at 5:59
answered May 6 '13 at 5:35
marty cohenmarty cohen
73.6k549128
answered May 6 '13 at 5:35
marty cohenmarty cohen
73.6k549128
answered May 6 '13 at 5:35
marty cohenmarty cohen
73.6k549128
73.6k549128
1
$begingroup$
Um... call me dumb, but in what base do you represent each $a_i$, considering that no constant finite base will suffice once $i!$ is greater than it?
$endgroup$
– ikdc
Aug 28 '14 at 21:32
$begingroup$
Instead of each $a_i$ satisfying $0 le a_i le B-1$, they satisfy $0 le a_i le i$.
$endgroup$
– marty cohen
Sep 1 '14 at 4:18
2
$begingroup$
Doesn't that imply that you need a potentially unbounded number of symbols to represent large numbers? If a number requires $n$ factorial digits, then you need $n$ symbols to represent the most significant digit. In a classic base $B$, you only ever need $B$ symbols.
$endgroup$
– ikdc
Sep 1 '14 at 5:29
$begingroup$
Yep. That's why we have the integers.
$endgroup$
– marty cohen
Sep 1 '14 at 14:53
$begingroup$
@Marty Cohen: But how do you represent those integers? Seems you'd still need a "regular" base type system in there as a "sub-digit" system, like how the Babylonian base-60 system used a base-10 "inner" base.
$endgroup$
– The_Sympathizer
Jul 26 '16 at 3:42
add a comment |
1
$begingroup$
Um... call me dumb, but in what base do you represent each $a_i$, considering that no constant finite base will suffice once $i!$ is greater than it?
$endgroup$
– ikdc
Aug 28 '14 at 21:32
$begingroup$
Instead of each $a_i$ satisfying $0 le a_i le B-1$, they satisfy $0 le a_i le i$.
$endgroup$
– marty cohen
Sep 1 '14 at 4:18
2
$begingroup$
Doesn't that imply that you need a potentially unbounded number of symbols to represent large numbers? If a number requires $n$ factorial digits, then you need $n$ symbols to represent the most significant digit. In a classic base $B$, you only ever need $B$ symbols.
$endgroup$
– ikdc
Sep 1 '14 at 5:29
$begingroup$
Yep. That's why we have the integers.
$endgroup$
– marty cohen
Sep 1 '14 at 14:53
$begingroup$
@Marty Cohen: But how do you represent those integers? Seems you'd still need a "regular" base type system in there as a "sub-digit" system, like how the Babylonian base-60 system used a base-10 "inner" base.
$endgroup$
– The_Sympathizer
Jul 26 '16 at 3:42
1
1
$begingroup$
Um... call me dumb, but in what base do you represent each $a_i$, considering that no constant finite base will suffice once $i!$ is greater than it?
$endgroup$
– ikdc
Aug 28 '14 at 21:32
$begingroup$
Um... call me dumb, but in what base do you represent each $a_i$, considering that no constant finite base will suffice once $i!$ is greater than it?
$endgroup$
– ikdc
Aug 28 '14 at 21:32
$begingroup$
Instead of each $a_i$ satisfying $0 le a_i le B-1$, they satisfy $0 le a_i le i$.
$endgroup$
– marty cohen
Sep 1 '14 at 4:18
$begingroup$
Instead of each $a_i$ satisfying $0 le a_i le B-1$, they satisfy $0 le a_i le i$.
$endgroup$
– marty cohen
Sep 1 '14 at 4:18
2
2
$begingroup$
Doesn't that imply that you need a potentially unbounded number of symbols to represent large numbers? If a number requires $n$ factorial digits, then you need $n$ symbols to represent the most significant digit. In a classic base $B$, you only ever need $B$ symbols.
$endgroup$
– ikdc
Sep 1 '14 at 5:29
$begingroup$
Doesn't that imply that you need a potentially unbounded number of symbols to represent large numbers? If a number requires $n$ factorial digits, then you need $n$ symbols to represent the most significant digit. In a classic base $B$, you only ever need $B$ symbols.
$endgroup$
– ikdc
Sep 1 '14 at 5:29
$begingroup$
Yep. That's why we have the integers.
$endgroup$
– marty cohen
Sep 1 '14 at 14:53
$begingroup$
Yep. That's why we have the integers.
$endgroup$
– marty cohen
Sep 1 '14 at 14:53
$begingroup$
@Marty Cohen: But how do you represent those integers? Seems you'd still need a "regular" base type system in there as a "sub-digit" system, like how the Babylonian base-60 system used a base-10 "inner" base.
$endgroup$
– The_Sympathizer
Jul 26 '16 at 3:42
$begingroup$
@Marty Cohen: But how do you represent those integers? Seems you'd still need a "regular" base type system in there as a "sub-digit" system, like how the Babylonian base-60 system used a base-10 "inner" base.
$endgroup$
– The_Sympathizer
Jul 26 '16 at 3:42
add a comment |
$begingroup$
I think base $6$ would make counting on our hands particularly convenient, we would have a $1$'s hand and a $6$'s hand and would be able to count up to $35$.
answered May 12 '13 at 3:23
JimJim
24.4k23370
$endgroup$
1
$begingroup$
There is a Korean technique that uses the thumb to represent 5 and so can count up to 99 on two hands. This seems rather more convenient to me than your suggestion.
$endgroup$
– MJD
May 15 '13 at 15:40
3
$begingroup$
Base 6 also has the advantage of trivial divisibility tests for 2 and 3, and easy tests for 5 (sum of digits) and 7 (sum of digits with alternating sign).
$endgroup$
– Dan
May 16 '13 at 3:40
3
$begingroup$
@MJD If you want to use your hands and are allowed to distinguish between fingers on each hand, then binary is clearly optimal, allowing you to count from 0 to 1024 (half-open).
$endgroup$
– ikdc
Aug 28 '14 at 21:34
$begingroup$
Multiplication tables would be greatly simplified up to $36$. It would also be handy to use the $36=100_{6}$ characters $0,1,dotsc,9,a,b,dotsc,z$, ready available on any keyboard, to represent numbers. I think this is the reason why $6$ is called perfect. (Beware the Dozenal Society though, they may hijack your cause for theirs...)
$endgroup$
– Oskar Limka
Mar 23 '18 at 7:02
add a comment |
$begingroup$
I think base $6$ would make counting on our hands particularly convenient, we would have a $1$'s hand and a $6$'s hand and would be able to count up to $35$.
answered May 12 '13 at 3:23
JimJim
24.4k23370
$endgroup$
1
$begingroup$
There is a Korean technique that uses the thumb to represent 5 and so can count up to 99 on two hands. This seems rather more convenient to me than your suggestion.
$endgroup$
– MJD
May 15 '13 at 15:40
3
$begingroup$
Base 6 also has the advantage of trivial divisibility tests for 2 and 3, and easy tests for 5 (sum of digits) and 7 (sum of digits with alternating sign).
$endgroup$
– Dan
May 16 '13 at 3:40
3
$begingroup$
@MJD If you want to use your hands and are allowed to distinguish between fingers on each hand, then binary is clearly optimal, allowing you to count from 0 to 1024 (half-open).
$endgroup$
– ikdc
Aug 28 '14 at 21:34
$begingroup$
Multiplication tables would be greatly simplified up to $36$. It would also be handy to use the $36=100_{6}$ characters $0,1,dotsc,9,a,b,dotsc,z$, ready available on any keyboard, to represent numbers. I think this is the reason why $6$ is called perfect. (Beware the Dozenal Society though, they may hijack your cause for theirs...)
$endgroup$
– Oskar Limka
Mar 23 '18 at 7:02
add a comment |
$begingroup$
I think base $6$ would make counting on our hands particularly convenient, we would have a $1$'s hand and a $6$'s hand and would be able to count up to $35$.
answered May 12 '13 at 3:23
JimJim
24.4k23370
$endgroup$
I think base $6$ would make counting on our hands particularly convenient, we would have a $1$'s hand and a $6$'s hand and would be able to count up to $35$.
answered May 12 '13 at 3:23
JimJim
24.4k23370
answered May 12 '13 at 3:23
JimJim
24.4k23370
answered May 12 '13 at 3:23
JimJim
24.4k23370
answered May 12 '13 at 3:23
JimJim
24.4k23370
24.4k23370
1
$begingroup$
There is a Korean technique that uses the thumb to represent 5 and so can count up to 99 on two hands. This seems rather more convenient to me than your suggestion.
$endgroup$
– MJD
May 15 '13 at 15:40
3
$begingroup$
Base 6 also has the advantage of trivial divisibility tests for 2 and 3, and easy tests for 5 (sum of digits) and 7 (sum of digits with alternating sign).
$endgroup$
– Dan
May 16 '13 at 3:40
3
$begingroup$
@MJD If you want to use your hands and are allowed to distinguish between fingers on each hand, then binary is clearly optimal, allowing you to count from 0 to 1024 (half-open).
$endgroup$
– ikdc
Aug 28 '14 at 21:34
$begingroup$
Multiplication tables would be greatly simplified up to $36$. It would also be handy to use the $36=100_{6}$ characters $0,1,dotsc,9,a,b,dotsc,z$, ready available on any keyboard, to represent numbers. I think this is the reason why $6$ is called perfect. (Beware the Dozenal Society though, they may hijack your cause for theirs...)
$endgroup$
– Oskar Limka
Mar 23 '18 at 7:02
add a comment |
1
$begingroup$
There is a Korean technique that uses the thumb to represent 5 and so can count up to 99 on two hands. This seems rather more convenient to me than your suggestion.
$endgroup$
– MJD
May 15 '13 at 15:40
3
$begingroup$
Base 6 also has the advantage of trivial divisibility tests for 2 and 3, and easy tests for 5 (sum of digits) and 7 (sum of digits with alternating sign).
$endgroup$
– Dan
May 16 '13 at 3:40
3
$begingroup$
@MJD If you want to use your hands and are allowed to distinguish between fingers on each hand, then binary is clearly optimal, allowing you to count from 0 to 1024 (half-open).
$endgroup$
– ikdc
Aug 28 '14 at 21:34
$begingroup$
Multiplication tables would be greatly simplified up to $36$. It would also be handy to use the $36=100_{6}$ characters $0,1,dotsc,9,a,b,dotsc,z$, ready available on any keyboard, to represent numbers. I think this is the reason why $6$ is called perfect. (Beware the Dozenal Society though, they may hijack your cause for theirs...)
$endgroup$
– Oskar Limka
Mar 23 '18 at 7:02
1
1
$begingroup$
There is a Korean technique that uses the thumb to represent 5 and so can count up to 99 on two hands. This seems rather more convenient to me than your suggestion.
$endgroup$
– MJD
May 15 '13 at 15:40
$begingroup$
There is a Korean technique that uses the thumb to represent 5 and so can count up to 99 on two hands. This seems rather more convenient to me than your suggestion.
$endgroup$
– MJD
May 15 '13 at 15:40
3
3
$begingroup$
Base 6 also has the advantage of trivial divisibility tests for 2 and 3, and easy tests for 5 (sum of digits) and 7 (sum of digits with alternating sign).
$endgroup$
– Dan
May 16 '13 at 3:40
$begingroup$
Base 6 also has the advantage of trivial divisibility tests for 2 and 3, and easy tests for 5 (sum of digits) and 7 (sum of digits with alternating sign).
$endgroup$
– Dan
May 16 '13 at 3:40
3
3
$begingroup$
@MJD If you want to use your hands and are allowed to distinguish between fingers on each hand, then binary is clearly optimal, allowing you to count from 0 to 1024 (half-open).
$endgroup$
– ikdc
Aug 28 '14 at 21:34
$begingroup$
@MJD If you want to use your hands and are allowed to distinguish between fingers on each hand, then binary is clearly optimal, allowing you to count from 0 to 1024 (half-open).
$endgroup$
– ikdc
Aug 28 '14 at 21:34
$begingroup$
Multiplication tables would be greatly simplified up to $36$. It would also be handy to use the $36=100_{6}$ characters $0,1,dotsc,9,a,b,dotsc,z$, ready available on any keyboard, to represent numbers. I think this is the reason why $6$ is called perfect. (Beware the Dozenal Society though, they may hijack your cause for theirs...)
$endgroup$
– Oskar Limka
Mar 23 '18 at 7:02
$begingroup$
Multiplication tables would be greatly simplified up to $36$. It would also be handy to use the $36=100_{6}$ characters $0,1,dotsc,9,a,b,dotsc,z$, ready available on any keyboard, to represent numbers. I think this is the reason why $6$ is called perfect. (Beware the Dozenal Society though, they may hijack your cause for theirs...)
$endgroup$
– Oskar Limka
Mar 23 '18 at 7:02
add a comment |
$begingroup$
Brian Hayes in his American Scientist article Third Base argues that "When base 2 is too small and base 10 is too big, base 3 is just right."
Figure 1 has the caption
Most economical radix for a numbering system is $e$ (about $2.718$) when economy is measured as the product of the radix and the width, or number of digits, needed to express a given range of values. Here both the radix and the width are treated as continuous variables.
Figure 2 has the caption
Most economical integer radix is almost always 3, the integer closest to $e$. If the capacity of a numbering system is $r^w$, and the cost of a representation is $rw$, then
$r=3$ is the best integer radix for all but a finite set of capacities. Specifically, ternary is inferior to binary only for 8,487 values of $r^w$; ternary is superior for infinitely many values.
Figure 3 has the caption
Ternary structure may offer the quickest path through a telephone menu system. Putting eight choices (assumed to be equally likely) in a single octonary menu (left) forces the caller to listen to 4.5 menu items on average. A binary structure (middle) has the same
performance, but the ternary tree (right) reduces the average to 3.75.
answered May 6 '13 at 7:52
Joel Reyes NocheJoel Reyes Noche
5,63633149
$endgroup$
9
$begingroup$
Sure, base 3 may minimize the number of beads on an abacus, but would you really want 1/2 to be 0.111111111... recurring?
$endgroup$
– Dan
May 16 '13 at 3:47
9
$begingroup$
1/3 is 0.3333333 recurring in base 10...
$endgroup$
– Michael
Aug 6 '13 at 17:15
$begingroup$
@michael, that's still tolerable. Dan has a point. Doubling and halving is the most common mental arithmetic people do. I wouldn't want 1/2 to be recurring but 1/3 is more OK.
$endgroup$
– nawfal
May 26 '18 at 7:48
add a comment |
$begingroup$
Brian Hayes in his American Scientist article Third Base argues that "When base 2 is too small and base 10 is too big, base 3 is just right."
Figure 1 has the caption
Most economical radix for a numbering system is $e$ (about $2.718$) when economy is measured as the product of the radix and the width, or number of digits, needed to express a given range of values. Here both the radix and the width are treated as continuous variables.
Figure 2 has the caption
Most economical integer radix is almost always 3, the integer closest to $e$. If the capacity of a numbering system is $r^w$, and the cost of a representation is $rw$, then
$r=3$ is the best integer radix for all but a finite set of capacities. Specifically, ternary is inferior to binary only for 8,487 values of $r^w$; ternary is superior for infinitely many values.
Figure 3 has the caption
Ternary structure may offer the quickest path through a telephone menu system. Putting eight choices (assumed to be equally likely) in a single octonary menu (left) forces the caller to listen to 4.5 menu items on average. A binary structure (middle) has the same
performance, but the ternary tree (right) reduces the average to 3.75.
answered May 6 '13 at 7:52
Joel Reyes NocheJoel Reyes Noche
5,63633149
$endgroup$
9
$begingroup$
Sure, base 3 may minimize the number of beads on an abacus, but would you really want 1/2 to be 0.111111111... recurring?
$endgroup$
– Dan
May 16 '13 at 3:47
9
$begingroup$
1/3 is 0.3333333 recurring in base 10...
$endgroup$
– Michael
Aug 6 '13 at 17:15
$begingroup$
@michael, that's still tolerable. Dan has a point. Doubling and halving is the most common mental arithmetic people do. I wouldn't want 1/2 to be recurring but 1/3 is more OK.
$endgroup$
– nawfal
May 26 '18 at 7:48
add a comment |
$begingroup$
Brian Hayes in his American Scientist article Third Base argues that "When base 2 is too small and base 10 is too big, base 3 is just right."
Figure 1 has the caption
Most economical radix for a numbering system is $e$ (about $2.718$) when economy is measured as the product of the radix and the width, or number of digits, needed to express a given range of values. Here both the radix and the width are treated as continuous variables.
Figure 2 has the caption
Most economical integer radix is almost always 3, the integer closest to $e$. If the capacity of a numbering system is $r^w$, and the cost of a representation is $rw$, then
$r=3$ is the best integer radix for all but a finite set of capacities. Specifically, ternary is inferior to binary only for 8,487 values of $r^w$; ternary is superior for infinitely many values.
Figure 3 has the caption
Ternary structure may offer the quickest path through a telephone menu system. Putting eight choices (assumed to be equally likely) in a single octonary menu (left) forces the caller to listen to 4.5 menu items on average. A binary structure (middle) has the same
performance, but the ternary tree (right) reduces the average to 3.75.
answered May 6 '13 at 7:52
Joel Reyes NocheJoel Reyes Noche
5,63633149
$endgroup$
Brian Hayes in his American Scientist article Third Base argues that "When base 2 is too small and base 10 is too big, base 3 is just right."
Figure 1 has the caption
Most economical radix for a numbering system is $e$ (about $2.718$) when economy is measured as the product of the radix and the width, or number of digits, needed to express a given range of values. Here both the radix and the width are treated as continuous variables.
Figure 2 has the caption
Most economical integer radix is almost always 3, the integer closest to $e$. If the capacity of a numbering system is $r^w$, and the cost of a representation is $rw$, then
$r=3$ is the best integer radix for all but a finite set of capacities. Specifically, ternary is inferior to binary only for 8,487 values of $r^w$; ternary is superior for infinitely many values.
Figure 3 has the caption
Ternary structure may offer the quickest path through a telephone menu system. Putting eight choices (assumed to be equally likely) in a single octonary menu (left) forces the caller to listen to 4.5 menu items on average. A binary structure (middle) has the same
performance, but the ternary tree (right) reduces the average to 3.75.
answered May 6 '13 at 7:52
Joel Reyes NocheJoel Reyes Noche
5,63633149
answered May 6 '13 at 7:52
Joel Reyes NocheJoel Reyes Noche
5,63633149
answered May 6 '13 at 7:52
Joel Reyes NocheJoel Reyes Noche
5,63633149
answered May 6 '13 at 7:52
Joel Reyes NocheJoel Reyes Noche
5,63633149
5,63633149
9
$begingroup$
Sure, base 3 may minimize the number of beads on an abacus, but would you really want 1/2 to be 0.111111111... recurring?
$endgroup$
– Dan
May 16 '13 at 3:47
9
$begingroup$
1/3 is 0.3333333 recurring in base 10...
$endgroup$
– Michael
Aug 6 '13 at 17:15
$begingroup$
@michael, that's still tolerable. Dan has a point. Doubling and halving is the most common mental arithmetic people do. I wouldn't want 1/2 to be recurring but 1/3 is more OK.
$endgroup$
– nawfal
May 26 '18 at 7:48
add a comment |
9
$begingroup$
Sure, base 3 may minimize the number of beads on an abacus, but would you really want 1/2 to be 0.111111111... recurring?
$endgroup$
– Dan
May 16 '13 at 3:47
9
$begingroup$
1/3 is 0.3333333 recurring in base 10...
$endgroup$
– Michael
Aug 6 '13 at 17:15
$begingroup$
@michael, that's still tolerable. Dan has a point. Doubling and halving is the most common mental arithmetic people do. I wouldn't want 1/2 to be recurring but 1/3 is more OK.
$endgroup$
– nawfal
May 26 '18 at 7:48
9
9
$begingroup$
Sure, base 3 may minimize the number of beads on an abacus, but would you really want 1/2 to be 0.111111111... recurring?
$endgroup$
– Dan
May 16 '13 at 3:47
$begingroup$
Sure, base 3 may minimize the number of beads on an abacus, but would you really want 1/2 to be 0.111111111... recurring?
$endgroup$
– Dan
May 16 '13 at 3:47
9
9
$begingroup$
1/3 is 0.3333333 recurring in base 10...
$endgroup$
– Michael
Aug 6 '13 at 17:15
$begingroup$
1/3 is 0.3333333 recurring in base 10...
$endgroup$
– Michael
Aug 6 '13 at 17:15
$begingroup$
@michael, that's still tolerable. Dan has a point. Doubling and halving is the most common mental arithmetic people do. I wouldn't want 1/2 to be recurring but 1/3 is more OK.
$endgroup$
– nawfal
May 26 '18 at 7:48
$begingroup$
@michael, that's still tolerable. Dan has a point. Doubling and halving is the most common mental arithmetic people do. I wouldn't want 1/2 to be recurring but 1/3 is more OK.
$endgroup$
– nawfal
May 26 '18 at 7:48
add a comment |
$begingroup$
For computer applications, bases like 2, 8 and 16 are obviously the best. Given that a large percentage of numerical data is stored in and processed by computers, these days, one could argue that what's good for computers is good for society.
Of the three I mentioned, I suppose that 8 or 16 would be better than base 2. Having the price of bananas as a binary number in the supermarket wouldn't work too well. Binary numbers are too long, and they all tend to look alike, so they're hard for people to read.
In the world at large (as opposed to the narrower world of mathematics and computers), reading numbers is probably just as important as doing arithmetic with them. Think of speed limit signs on roads, distances of journeys, prices in stores, or temperatures in weather forecasts. These numbers need to be read and understood quickly (by human beings), and I doubt that this would be possible if they were written in binary. We'd no longer be taking advantage of the wonderful human ability to quickly recognize symbols, and it would be a pity to waste that ability just so that we can make computing easier (in my opinion).
answered May 6 '13 at 3:44
bubbabubba
30.4k33186
$endgroup$
$begingroup$
Not sure why $4$ was left out of $2, 8, 16$.
$endgroup$
– Lance Pollard
Jan 13 at 16:41
$begingroup$
Because bases 2, 8, 16 are already widely used in computing, whereas 4 is not (as far as I know). But, I suppose it would work OK.
$endgroup$
– bubba
Jan 13 at 23:34
$begingroup$
Okay that makes sense. 4 is not as used b/c computers are 8-bit at least, so 4 is inefficient. But 2 is there because everything is just 1's and 0's.
$endgroup$
– Lance Pollard
Jan 14 at 0:04
add a comment |
$begingroup$
For computer applications, bases like 2, 8 and 16 are obviously the best. Given that a large percentage of numerical data is stored in and processed by computers, these days, one could argue that what's good for computers is good for society.
Of the three I mentioned, I suppose that 8 or 16 would be better than base 2. Having the price of bananas as a binary number in the supermarket wouldn't work too well. Binary numbers are too long, and they all tend to look alike, so they're hard for people to read.
In the world at large (as opposed to the narrower world of mathematics and computers), reading numbers is probably just as important as doing arithmetic with them. Think of speed limit signs on roads, distances of journeys, prices in stores, or temperatures in weather forecasts. These numbers need to be read and understood quickly (by human beings), and I doubt that this would be possible if they were written in binary. We'd no longer be taking advantage of the wonderful human ability to quickly recognize symbols, and it would be a pity to waste that ability just so that we can make computing easier (in my opinion).
answered May 6 '13 at 3:44
bubbabubba
30.4k33186
$endgroup$
$begingroup$
Not sure why $4$ was left out of $2, 8, 16$.
$endgroup$
– Lance Pollard
Jan 13 at 16:41
$begingroup$
Because bases 2, 8, 16 are already widely used in computing, whereas 4 is not (as far as I know). But, I suppose it would work OK.
$endgroup$
– bubba
Jan 13 at 23:34
$begingroup$
Okay that makes sense. 4 is not as used b/c computers are 8-bit at least, so 4 is inefficient. But 2 is there because everything is just 1's and 0's.
$endgroup$
– Lance Pollard
Jan 14 at 0:04
add a comment |
$begingroup$
For computer applications, bases like 2, 8 and 16 are obviously the best. Given that a large percentage of numerical data is stored in and processed by computers, these days, one could argue that what's good for computers is good for society.
Of the three I mentioned, I suppose that 8 or 16 would be better than base 2. Having the price of bananas as a binary number in the supermarket wouldn't work too well. Binary numbers are too long, and they all tend to look alike, so they're hard for people to read.
In the world at large (as opposed to the narrower world of mathematics and computers), reading numbers is probably just as important as doing arithmetic with them. Think of speed limit signs on roads, distances of journeys, prices in stores, or temperatures in weather forecasts. These numbers need to be read and understood quickly (by human beings), and I doubt that this would be possible if they were written in binary. We'd no longer be taking advantage of the wonderful human ability to quickly recognize symbols, and it would be a pity to waste that ability just so that we can make computing easier (in my opinion).
answered May 6 '13 at 3:44
bubbabubba
30.4k33186
$endgroup$
For computer applications, bases like 2, 8 and 16 are obviously the best. Given that a large percentage of numerical data is stored in and processed by computers, these days, one could argue that what's good for computers is good for society.
Of the three I mentioned, I suppose that 8 or 16 would be better than base 2. Having the price of bananas as a binary number in the supermarket wouldn't work too well. Binary numbers are too long, and they all tend to look alike, so they're hard for people to read.
In the world at large (as opposed to the narrower world of mathematics and computers), reading numbers is probably just as important as doing arithmetic with them. Think of speed limit signs on roads, distances of journeys, prices in stores, or temperatures in weather forecasts. These numbers need to be read and understood quickly (by human beings), and I doubt that this would be possible if they were written in binary. We'd no longer be taking advantage of the wonderful human ability to quickly recognize symbols, and it would be a pity to waste that ability just so that we can make computing easier (in my opinion).
answered May 6 '13 at 3:44
bubbabubba
30.4k33186
edited May 8 '17 at 0:04
answered May 6 '13 at 3:44
bubbabubba
30.4k33186
answered May 6 '13 at 3:44
bubbabubba
30.4k33186
answered May 6 '13 at 3:44
bubbabubba
30.4k33186
30.4k33186
$begingroup$
Not sure why $4$ was left out of $2, 8, 16$.
$endgroup$
– Lance Pollard
Jan 13 at 16:41
$begingroup$
Because bases 2, 8, 16 are already widely used in computing, whereas 4 is not (as far as I know). But, I suppose it would work OK.
$endgroup$
– bubba
Jan 13 at 23:34
$begingroup$
Okay that makes sense. 4 is not as used b/c computers are 8-bit at least, so 4 is inefficient. But 2 is there because everything is just 1's and 0's.
$endgroup$
– Lance Pollard
Jan 14 at 0:04
add a comment |
$begingroup$
Not sure why $4$ was left out of $2, 8, 16$.
$endgroup$
– Lance Pollard
Jan 13 at 16:41
$begingroup$
Because bases 2, 8, 16 are already widely used in computing, whereas 4 is not (as far as I know). But, I suppose it would work OK.
$endgroup$
– bubba
Jan 13 at 23:34
$begingroup$
Okay that makes sense. 4 is not as used b/c computers are 8-bit at least, so 4 is inefficient. But 2 is there because everything is just 1's and 0's.
$endgroup$
– Lance Pollard
Jan 14 at 0:04
$begingroup$
Not sure why $4$ was left out of $2, 8, 16$.
$endgroup$
– Lance Pollard
Jan 13 at 16:41
$begingroup$
Not sure why $4$ was left out of $2, 8, 16$.
$endgroup$
– Lance Pollard
Jan 13 at 16:41
$begingroup$
Because bases 2, 8, 16 are already widely used in computing, whereas 4 is not (as far as I know). But, I suppose it would work OK.
$endgroup$
– bubba
Jan 13 at 23:34
$begingroup$
Because bases 2, 8, 16 are already widely used in computing, whereas 4 is not (as far as I know). But, I suppose it would work OK.
$endgroup$
– bubba
Jan 13 at 23:34
$begingroup$
Okay that makes sense. 4 is not as used b/c computers are 8-bit at least, so 4 is inefficient. But 2 is there because everything is just 1's and 0's.
$endgroup$
– Lance Pollard
Jan 14 at 0:04
$begingroup$
Okay that makes sense. 4 is not as used b/c computers are 8-bit at least, so 4 is inefficient. But 2 is there because everything is just 1's and 0's.
$endgroup$
– Lance Pollard
Jan 14 at 0:04
add a comment |
$begingroup$
Balanced Nonary (base 9) would probably be really good. The digits go from -4 to 4, so taking the negative of a number would just be taking the negative of each digit, so subtraction is easy. Multiplication and division are particularly easy too if you make the easy conversion to balanced ternary first. Then there's no carrying when multiplying single digits (like in binary), and division is just testing inequalities (if you can divide by 2). Of course, if you want to do things faster, learning a balanced nonary times table would be easier than learning a regular nonary times table since you only really need to know the table for 1,2,3,4 and then handle negatives (and zero) appropriately.
There have even been computers based on balanced ternary.
answered May 8 '13 at 0:07
Mark S.Mark S.
11.8k22670
$endgroup$
add a comment |
$begingroup$
Balanced Nonary (base 9) would probably be really good. The digits go from -4 to 4, so taking the negative of a number would just be taking the negative of each digit, so subtraction is easy. Multiplication and division are particularly easy too if you make the easy conversion to balanced ternary first. Then there's no carrying when multiplying single digits (like in binary), and division is just testing inequalities (if you can divide by 2). Of course, if you want to do things faster, learning a balanced nonary times table would be easier than learning a regular nonary times table since you only really need to know the table for 1,2,3,4 and then handle negatives (and zero) appropriately.
There have even been computers based on balanced ternary.
answered May 8 '13 at 0:07
Mark S.Mark S.
11.8k22670
$endgroup$
add a comment |
$begingroup$
Balanced Nonary (base 9) would probably be really good. The digits go from -4 to 4, so taking the negative of a number would just be taking the negative of each digit, so subtraction is easy. Multiplication and division are particularly easy too if you make the easy conversion to balanced ternary first. Then there's no carrying when multiplying single digits (like in binary), and division is just testing inequalities (if you can divide by 2). Of course, if you want to do things faster, learning a balanced nonary times table would be easier than learning a regular nonary times table since you only really need to know the table for 1,2,3,4 and then handle negatives (and zero) appropriately.
There have even been computers based on balanced ternary.
answered May 8 '13 at 0:07
Mark S.Mark S.
11.8k22670
$endgroup$
Balanced Nonary (base 9) would probably be really good. The digits go from -4 to 4, so taking the negative of a number would just be taking the negative of each digit, so subtraction is easy. Multiplication and division are particularly easy too if you make the easy conversion to balanced ternary first. Then there's no carrying when multiplying single digits (like in binary), and division is just testing inequalities (if you can divide by 2). Of course, if you want to do things faster, learning a balanced nonary times table would be easier than learning a regular nonary times table since you only really need to know the table for 1,2,3,4 and then handle negatives (and zero) appropriately.
There have even been computers based on balanced ternary.
answered May 8 '13 at 0:07
Mark S.Mark S.
11.8k22670
answered May 8 '13 at 0:07
Mark S.Mark S.
11.8k22670
answered May 8 '13 at 0:07
Mark S.Mark S.
11.8k22670
answered May 8 '13 at 0:07
Mark S.Mark S.
11.8k22670
11.8k22670
add a comment |
add a comment |
$begingroup$
While bubba raises valid points about base 2 from a practicality standpoint, I myself would defend the choice of base 2 for the following reason: it makes addition and multiplication incredibly easy. This is, in fact, the way computers do these basic operations.
Addition in binary operates under the following rules:
$0 land 0 = 0$
$1 land 0 = 1$
$0 land 1 = 1$
$1 land 1 = 0$ (carry a 1)
Therefore when you do long addition in binary, the algorithm is particularly simple: if there are 2 $0$'s in the column, you put down $0$, if there is one $0$ and one $1$ you put down $1$, if there are 2 $1$'s you put down $0$ and carry $1$ over to the next place value. Imagine the time we could save by teaching kids to add this way. We could start teaching actual mathematics instead!
Long multiplication is just as easy: for every place value you're multiplying by either $0$ or $1$, which makes the computation very simple. I invite you to try out a few simple sums and products in binary to see what I mean.
See http://en.wikipedia.org/wiki/Binary_arithmetic#Addition and http://en.wikipedia.org/wiki/Binary_arithmetic#Multiplication for more on these two operations; the article has details on subtraction, division, and square roots as well.
answered May 6 '13 at 7:09
Gyu Eun LeeGyu Eun Lee
13.2k2353
$endgroup$
$begingroup$
Binary makes arithmetic easy on paper. But how would it affect mental arithmetic?
$endgroup$
– Jack M
May 24 '13 at 14:09
$begingroup$
I'm not really sure, but mental arithmetic is no easier in any other base - the only difference is the number of digits for larger numbers. I say let people use a calculator. Arithmetic is the job they were born to do, and the job we were born to automate.
$endgroup$
– Gyu Eun Lee
May 24 '13 at 18:00
$begingroup$
It makes numbers a bit longer.
$endgroup$
– PyRulez
Jul 1 '13 at 14:42
1
$begingroup$
Sure, base 2 is convenient; however, some numbers that have a finite decimal expansion (e.g. $(0.2)_{10}$) but an infinite binary expansion ($(0.00overline{11})_2$) is problematic, as it is a source of roundoff error in floating-point-number systems.
$endgroup$
– jubobs
Sep 30 '13 at 15:23
$begingroup$
@Jubobs Same with $(frac{1}{3})_{10}$: $0.overline{3}$.
$endgroup$
– Cole Johnson
May 25 '14 at 21:49
|
show 3 more comments
$begingroup$
While bubba raises valid points about base 2 from a practicality standpoint, I myself would defend the choice of base 2 for the following reason: it makes addition and multiplication incredibly easy. This is, in fact, the way computers do these basic operations.
Addition in binary operates under the following rules:
$0 land 0 = 0$
$1 land 0 = 1$
$0 land 1 = 1$
$1 land 1 = 0$ (carry a 1)
Therefore when you do long addition in binary, the algorithm is particularly simple: if there are 2 $0$'s in the column, you put down $0$, if there is one $0$ and one $1$ you put down $1$, if there are 2 $1$'s you put down $0$ and carry $1$ over to the next place value. Imagine the time we could save by teaching kids to add this way. We could start teaching actual mathematics instead!
Long multiplication is just as easy: for every place value you're multiplying by either $0$ or $1$, which makes the computation very simple. I invite you to try out a few simple sums and products in binary to see what I mean.
See http://en.wikipedia.org/wiki/Binary_arithmetic#Addition and http://en.wikipedia.org/wiki/Binary_arithmetic#Multiplication for more on these two operations; the article has details on subtraction, division, and square roots as well.
answered May 6 '13 at 7:09
Gyu Eun LeeGyu Eun Lee
13.2k2353
$endgroup$
$begingroup$
Binary makes arithmetic easy on paper. But how would it affect mental arithmetic?
$endgroup$
– Jack M
May 24 '13 at 14:09
$begingroup$
I'm not really sure, but mental arithmetic is no easier in any other base - the only difference is the number of digits for larger numbers. I say let people use a calculator. Arithmetic is the job they were born to do, and the job we were born to automate.
$endgroup$
– Gyu Eun Lee
May 24 '13 at 18:00
$begingroup$
It makes numbers a bit longer.
$endgroup$
– PyRulez
Jul 1 '13 at 14:42
1
$begingroup$
Sure, base 2 is convenient; however, some numbers that have a finite decimal expansion (e.g. $(0.2)_{10}$) but an infinite binary expansion ($(0.00overline{11})_2$) is problematic, as it is a source of roundoff error in floating-point-number systems.
$endgroup$
– jubobs
Sep 30 '13 at 15:23
$begingroup$
@Jubobs Same with $(frac{1}{3})_{10}$: $0.overline{3}$.
$endgroup$
– Cole Johnson
May 25 '14 at 21:49
|
show 3 more comments
$begingroup$
While bubba raises valid points about base 2 from a practicality standpoint, I myself would defend the choice of base 2 for the following reason: it makes addition and multiplication incredibly easy. This is, in fact, the way computers do these basic operations.
Addition in binary operates under the following rules:
$0 land 0 = 0$
$1 land 0 = 1$
$0 land 1 = 1$
$1 land 1 = 0$ (carry a 1)
Therefore when you do long addition in binary, the algorithm is particularly simple: if there are 2 $0$'s in the column, you put down $0$, if there is one $0$ and one $1$ you put down $1$, if there are 2 $1$'s you put down $0$ and carry $1$ over to the next place value. Imagine the time we could save by teaching kids to add this way. We could start teaching actual mathematics instead!
Long multiplication is just as easy: for every place value you're multiplying by either $0$ or $1$, which makes the computation very simple. I invite you to try out a few simple sums and products in binary to see what I mean.
See http://en.wikipedia.org/wiki/Binary_arithmetic#Addition and http://en.wikipedia.org/wiki/Binary_arithmetic#Multiplication for more on these two operations; the article has details on subtraction, division, and square roots as well.
answered May 6 '13 at 7:09
Gyu Eun LeeGyu Eun Lee
13.2k2353
$endgroup$
While bubba raises valid points about base 2 from a practicality standpoint, I myself would defend the choice of base 2 for the following reason: it makes addition and multiplication incredibly easy. This is, in fact, the way computers do these basic operations.
Addition in binary operates under the following rules:
$0 land 0 = 0$
$1 land 0 = 1$
$0 land 1 = 1$
$1 land 1 = 0$ (carry a 1)
Therefore when you do long addition in binary, the algorithm is particularly simple: if there are 2 $0$'s in the column, you put down $0$, if there is one $0$ and one $1$ you put down $1$, if there are 2 $1$'s you put down $0$ and carry $1$ over to the next place value. Imagine the time we could save by teaching kids to add this way. We could start teaching actual mathematics instead!
Long multiplication is just as easy: for every place value you're multiplying by either $0$ or $1$, which makes the computation very simple. I invite you to try out a few simple sums and products in binary to see what I mean.
See http://en.wikipedia.org/wiki/Binary_arithmetic#Addition and http://en.wikipedia.org/wiki/Binary_arithmetic#Multiplication for more on these two operations; the article has details on subtraction, division, and square roots as well.
answered May 6 '13 at 7:09
Gyu Eun LeeGyu Eun Lee
13.2k2353
edited May 6 '13 at 7:19
answered May 6 '13 at 7:09
Gyu Eun LeeGyu Eun Lee
13.2k2353
answered May 6 '13 at 7:09
Gyu Eun LeeGyu Eun Lee
13.2k2353
answered May 6 '13 at 7:09
Gyu Eun LeeGyu Eun Lee
13.2k2353
13.2k2353
$begingroup$
Binary makes arithmetic easy on paper. But how would it affect mental arithmetic?
$endgroup$
– Jack M
May 24 '13 at 14:09
$begingroup$
I'm not really sure, but mental arithmetic is no easier in any other base - the only difference is the number of digits for larger numbers. I say let people use a calculator. Arithmetic is the job they were born to do, and the job we were born to automate.
$endgroup$
– Gyu Eun Lee
May 24 '13 at 18:00
$begingroup$
It makes numbers a bit longer.
$endgroup$
– PyRulez
Jul 1 '13 at 14:42
1
$begingroup$
Sure, base 2 is convenient; however, some numbers that have a finite decimal expansion (e.g. $(0.2)_{10}$) but an infinite binary expansion ($(0.00overline{11})_2$) is problematic, as it is a source of roundoff error in floating-point-number systems.
$endgroup$
– jubobs
Sep 30 '13 at 15:23
$begingroup$
@Jubobs Same with $(frac{1}{3})_{10}$: $0.overline{3}$.
$endgroup$
– Cole Johnson
May 25 '14 at 21:49
|
show 3 more comments
$begingroup$
Binary makes arithmetic easy on paper. But how would it affect mental arithmetic?
$endgroup$
– Jack M
May 24 '13 at 14:09
$begingroup$
I'm not really sure, but mental arithmetic is no easier in any other base - the only difference is the number of digits for larger numbers. I say let people use a calculator. Arithmetic is the job they were born to do, and the job we were born to automate.
$endgroup$
– Gyu Eun Lee
May 24 '13 at 18:00
$begingroup$
It makes numbers a bit longer.
$endgroup$
– PyRulez
Jul 1 '13 at 14:42
1
$begingroup$
Sure, base 2 is convenient; however, some numbers that have a finite decimal expansion (e.g. $(0.2)_{10}$) but an infinite binary expansion ($(0.00overline{11})_2$) is problematic, as it is a source of roundoff error in floating-point-number systems.
$endgroup$
– jubobs
Sep 30 '13 at 15:23
$begingroup$
@Jubobs Same with $(frac{1}{3})_{10}$: $0.overline{3}$.
$endgroup$
– Cole Johnson
May 25 '14 at 21:49
$begingroup$
Binary makes arithmetic easy on paper. But how would it affect mental arithmetic?
$endgroup$
– Jack M
May 24 '13 at 14:09
$begingroup$
Binary makes arithmetic easy on paper. But how would it affect mental arithmetic?
$endgroup$
– Jack M
May 24 '13 at 14:09
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I'm not really sure, but mental arithmetic is no easier in any other base - the only difference is the number of digits for larger numbers. I say let people use a calculator. Arithmetic is the job they were born to do, and the job we were born to automate.
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– Gyu Eun Lee
May 24 '13 at 18:00
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I'm not really sure, but mental arithmetic is no easier in any other base - the only difference is the number of digits for larger numbers. I say let people use a calculator. Arithmetic is the job they were born to do, and the job we were born to automate.
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– Gyu Eun Lee
May 24 '13 at 18:00
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It makes numbers a bit longer.
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– PyRulez
Jul 1 '13 at 14:42
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It makes numbers a bit longer.
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– PyRulez
Jul 1 '13 at 14:42
1
1
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Sure, base 2 is convenient; however, some numbers that have a finite decimal expansion (e.g. $(0.2)_{10}$) but an infinite binary expansion ($(0.00overline{11})_2$) is problematic, as it is a source of roundoff error in floating-point-number systems.
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– jubobs
Sep 30 '13 at 15:23
$begingroup$
Sure, base 2 is convenient; however, some numbers that have a finite decimal expansion (e.g. $(0.2)_{10}$) but an infinite binary expansion ($(0.00overline{11})_2$) is problematic, as it is a source of roundoff error in floating-point-number systems.
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– jubobs
Sep 30 '13 at 15:23
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@Jubobs Same with $(frac{1}{3})_{10}$: $0.overline{3}$.
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– Cole Johnson
May 25 '14 at 21:49
$begingroup$
@Jubobs Same with $(frac{1}{3})_{10}$: $0.overline{3}$.
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– Cole Johnson
May 25 '14 at 21:49
|
show 3 more comments
$begingroup$
In order to answer this question, it is first necessary to ask: What makes one base “better” than other? Some reasonable things to consider are:
Size
There is a tradeoff between the number of distinct digit characters used in a base (Base $b$ has exactly $b$ of these, from $0$ to $b - 1$, inclusive) and the length of the numeral required to represent a given number (which is $O(1/log{b})$).
If the base is too small, then numbers explode into cumbersome long strings of digits. For example, in binary, the current year is 111 1101 1101, and the population of China (according to its 2010 census) was 100 1111 1101 1010 1001 0100 0011 0100. Modern computers can easily work with 32-bit or 64-bit binary numbers, but humans can't, which is why programmers have developed more compact encodings of binary, such as hexadecimal.
On the other hand, if we picked a very large base, like 2520, then you would need only 3 characters to represent the population of China, but typing them would be just as challenging as typing Chinese. And forget about learning the mulitplication table, whose size is $O(b^2)$. The only practical way to use such a large base is to split it into sub-bases, the way base-60 is represented as a mixture of base-6 and base-10.
So, what we want is a happy medium.
Fraction-friendliness
This is the main argument advanced in favor of base-12 or other highly composite bases (2, 4, 6, 12, 24, 36, 48, 60, 120, ...).
If a base has a lot of factors, it makes fractions easier to work with. For example, in base ten, 1/3 is represented as the infinitely repeating 0.333 333 333... (often rounded to 0.33 or 0.333), and this awkwardness crops up in deals like “3 for $5” or +/- grading systems. But in base-12, 1/3 is a nice simple 0.4.
Of course, because there are an infinite number of primes, it's impossible to completely avoid repeating “decimals”. And base-12's simplicity for the fractions 1/3 (0.4), 1/4 (0.3), 1/6 (0.2), 1/8 (0.16) and 1/9 (0.14) comes at the price of making 1/5 (0.24972497...) and 1/10 (0.124972497...) recurring dozenal fractions. But 1/3 is more common than 1/5.
answered May 16 '13 at 3:48
DanDan
4,55511517
$endgroup$
add a comment |
$begingroup$
In order to answer this question, it is first necessary to ask: What makes one base “better” than other? Some reasonable things to consider are:
Size
There is a tradeoff between the number of distinct digit characters used in a base (Base $b$ has exactly $b$ of these, from $0$ to $b - 1$, inclusive) and the length of the numeral required to represent a given number (which is $O(1/log{b})$).
If the base is too small, then numbers explode into cumbersome long strings of digits. For example, in binary, the current year is 111 1101 1101, and the population of China (according to its 2010 census) was 100 1111 1101 1010 1001 0100 0011 0100. Modern computers can easily work with 32-bit or 64-bit binary numbers, but humans can't, which is why programmers have developed more compact encodings of binary, such as hexadecimal.
On the other hand, if we picked a very large base, like 2520, then you would need only 3 characters to represent the population of China, but typing them would be just as challenging as typing Chinese. And forget about learning the mulitplication table, whose size is $O(b^2)$. The only practical way to use such a large base is to split it into sub-bases, the way base-60 is represented as a mixture of base-6 and base-10.
So, what we want is a happy medium.
Fraction-friendliness
This is the main argument advanced in favor of base-12 or other highly composite bases (2, 4, 6, 12, 24, 36, 48, 60, 120, ...).
If a base has a lot of factors, it makes fractions easier to work with. For example, in base ten, 1/3 is represented as the infinitely repeating 0.333 333 333... (often rounded to 0.33 or 0.333), and this awkwardness crops up in deals like “3 for $5” or +/- grading systems. But in base-12, 1/3 is a nice simple 0.4.
Of course, because there are an infinite number of primes, it's impossible to completely avoid repeating “decimals”. And base-12's simplicity for the fractions 1/3 (0.4), 1/4 (0.3), 1/6 (0.2), 1/8 (0.16) and 1/9 (0.14) comes at the price of making 1/5 (0.24972497...) and 1/10 (0.124972497...) recurring dozenal fractions. But 1/3 is more common than 1/5.
answered May 16 '13 at 3:48
DanDan
4,55511517
$endgroup$
add a comment |
$begingroup$
In order to answer this question, it is first necessary to ask: What makes one base “better” than other? Some reasonable things to consider are:
Size
There is a tradeoff between the number of distinct digit characters used in a base (Base $b$ has exactly $b$ of these, from $0$ to $b - 1$, inclusive) and the length of the numeral required to represent a given number (which is $O(1/log{b})$).
If the base is too small, then numbers explode into cumbersome long strings of digits. For example, in binary, the current year is 111 1101 1101, and the population of China (according to its 2010 census) was 100 1111 1101 1010 1001 0100 0011 0100. Modern computers can easily work with 32-bit or 64-bit binary numbers, but humans can't, which is why programmers have developed more compact encodings of binary, such as hexadecimal.
On the other hand, if we picked a very large base, like 2520, then you would need only 3 characters to represent the population of China, but typing them would be just as challenging as typing Chinese. And forget about learning the mulitplication table, whose size is $O(b^2)$. The only practical way to use such a large base is to split it into sub-bases, the way base-60 is represented as a mixture of base-6 and base-10.
So, what we want is a happy medium.
Fraction-friendliness
This is the main argument advanced in favor of base-12 or other highly composite bases (2, 4, 6, 12, 24, 36, 48, 60, 120, ...).
If a base has a lot of factors, it makes fractions easier to work with. For example, in base ten, 1/3 is represented as the infinitely repeating 0.333 333 333... (often rounded to 0.33 or 0.333), and this awkwardness crops up in deals like “3 for $5” or +/- grading systems. But in base-12, 1/3 is a nice simple 0.4.
Of course, because there are an infinite number of primes, it's impossible to completely avoid repeating “decimals”. And base-12's simplicity for the fractions 1/3 (0.4), 1/4 (0.3), 1/6 (0.2), 1/8 (0.16) and 1/9 (0.14) comes at the price of making 1/5 (0.24972497...) and 1/10 (0.124972497...) recurring dozenal fractions. But 1/3 is more common than 1/5.
answered May 16 '13 at 3:48
DanDan
4,55511517
$endgroup$
In order to answer this question, it is first necessary to ask: What makes one base “better” than other? Some reasonable things to consider are:
Size
There is a tradeoff between the number of distinct digit characters used in a base (Base $b$ has exactly $b$ of these, from $0$ to $b - 1$, inclusive) and the length of the numeral required to represent a given number (which is $O(1/log{b})$).
If the base is too small, then numbers explode into cumbersome long strings of digits. For example, in binary, the current year is 111 1101 1101, and the population of China (according to its 2010 census) was 100 1111 1101 1010 1001 0100 0011 0100. Modern computers can easily work with 32-bit or 64-bit binary numbers, but humans can't, which is why programmers have developed more compact encodings of binary, such as hexadecimal.
On the other hand, if we picked a very large base, like 2520, then you would need only 3 characters to represent the population of China, but typing them would be just as challenging as typing Chinese. And forget about learning the mulitplication table, whose size is $O(b^2)$. The only practical way to use such a large base is to split it into sub-bases, the way base-60 is represented as a mixture of base-6 and base-10.
So, what we want is a happy medium.
Fraction-friendliness
This is the main argument advanced in favor of base-12 or other highly composite bases (2, 4, 6, 12, 24, 36, 48, 60, 120, ...).
If a base has a lot of factors, it makes fractions easier to work with. For example, in base ten, 1/3 is represented as the infinitely repeating 0.333 333 333... (often rounded to 0.33 or 0.333), and this awkwardness crops up in deals like “3 for $5” or +/- grading systems. But in base-12, 1/3 is a nice simple 0.4.
Of course, because there are an infinite number of primes, it's impossible to completely avoid repeating “decimals”. And base-12's simplicity for the fractions 1/3 (0.4), 1/4 (0.3), 1/6 (0.2), 1/8 (0.16) and 1/9 (0.14) comes at the price of making 1/5 (0.24972497...) and 1/10 (0.124972497...) recurring dozenal fractions. But 1/3 is more common than 1/5.
answered May 16 '13 at 3:48
DanDan
4,55511517
answered May 16 '13 at 3:48
DanDan
4,55511517
answered May 16 '13 at 3:48
DanDan
4,55511517
answered May 16 '13 at 3:48
DanDan
4,55511517
4,55511517
add a comment |
add a comment |
$begingroup$
The quater-imaginary base $2i$ is quite amusing, being able to express every complex number using only digits in ${0, 1, 2, 3}$. Being the only base yet proposed here that includes the elements of this extended system, it is clearly the best one for this purpose :)
answered Aug 24 '14 at 2:40
Ryan ReichRyan Reich
5,3911627
$endgroup$
add a comment |
$begingroup$
The quater-imaginary base $2i$ is quite amusing, being able to express every complex number using only digits in ${0, 1, 2, 3}$. Being the only base yet proposed here that includes the elements of this extended system, it is clearly the best one for this purpose :)
answered Aug 24 '14 at 2:40
Ryan ReichRyan Reich
5,3911627
$endgroup$
add a comment |
$begingroup$
The quater-imaginary base $2i$ is quite amusing, being able to express every complex number using only digits in ${0, 1, 2, 3}$. Being the only base yet proposed here that includes the elements of this extended system, it is clearly the best one for this purpose :)
answered Aug 24 '14 at 2:40
Ryan ReichRyan Reich
5,3911627
$endgroup$
The quater-imaginary base $2i$ is quite amusing, being able to express every complex number using only digits in ${0, 1, 2, 3}$. Being the only base yet proposed here that includes the elements of this extended system, it is clearly the best one for this purpose :)
answered Aug 24 '14 at 2:40
Ryan ReichRyan Reich
5,3911627
answered Aug 24 '14 at 2:40
Ryan ReichRyan Reich
5,3911627
answered Aug 24 '14 at 2:40
Ryan ReichRyan Reich
5,3911627
answered Aug 24 '14 at 2:40
Ryan ReichRyan Reich
5,3911627
5,3911627
add a comment |
add a comment |
$begingroup$
Not all bases are 10 in their own notation. There are a group of alternating bases where the base is not 'ten' but a 'hundred'. The most elegant of these is the long-hundred of the proto-germanics and their decendents. Reckoning in the six-score long-hundred (ie 120), was still common enough in 1350 to pass without comment.
Yes, i have used this base for some thirty years. It's truly elegant, being more efficient than either 10 or 12. It's the first base, for which the (number of proper divisors)/(ln base) is greater than 3.
Also 120 is the smallest multiply perfect number, and has the same features as the perfect numbers. For example, 120 = 1+2+4+8+15+30+60 = 3+5+6+10+12+20+24+40, all of these numbers make the total divisors of 120. The second set corresponds to a set of weights, eg
- 1 oz, 2 oz, 4 oz, 8 oz, 1 lb, 2 lb, 4 lb. 15 oz = 1 lb , 120 oz = 1 clove.
- 1 ct, 2 ct, 4 ct, 8 ct, 1 dr, 2 dr, 4 dr : 15 ct = 1dr , 120 ct = 1 oz
- 1 lb, 2 lb, 4 lb, 8 lb, 1 st, 2 st, 4 st : 15 lb = 1 st, 120 lb = 1 cwt
When one considers not just integers, but also fractions x/y and y/z, arranged by xy, one finds that the expressions of these are very short for the first sixty or so, even in things like 56 (8/7 = 1:17.17.17, vs 7/8 = V5), and this unusual pair at 96 (3/32 = 11:30, 32/3 = 10:80.
answered May 15 '13 at 9:20
Wendy KriegerWendy Krieger
311
$endgroup$
add a comment |
$begingroup$
Not all bases are 10 in their own notation. There are a group of alternating bases where the base is not 'ten' but a 'hundred'. The most elegant of these is the long-hundred of the proto-germanics and their decendents. Reckoning in the six-score long-hundred (ie 120), was still common enough in 1350 to pass without comment.
Yes, i have used this base for some thirty years. It's truly elegant, being more efficient than either 10 or 12. It's the first base, for which the (number of proper divisors)/(ln base) is greater than 3.
Also 120 is the smallest multiply perfect number, and has the same features as the perfect numbers. For example, 120 = 1+2+4+8+15+30+60 = 3+5+6+10+12+20+24+40, all of these numbers make the total divisors of 120. The second set corresponds to a set of weights, eg
- 1 oz, 2 oz, 4 oz, 8 oz, 1 lb, 2 lb, 4 lb. 15 oz = 1 lb , 120 oz = 1 clove.
- 1 ct, 2 ct, 4 ct, 8 ct, 1 dr, 2 dr, 4 dr : 15 ct = 1dr , 120 ct = 1 oz
- 1 lb, 2 lb, 4 lb, 8 lb, 1 st, 2 st, 4 st : 15 lb = 1 st, 120 lb = 1 cwt
When one considers not just integers, but also fractions x/y and y/z, arranged by xy, one finds that the expressions of these are very short for the first sixty or so, even in things like 56 (8/7 = 1:17.17.17, vs 7/8 = V5), and this unusual pair at 96 (3/32 = 11:30, 32/3 = 10:80.
answered May 15 '13 at 9:20
Wendy KriegerWendy Krieger
311
$endgroup$
add a comment |
$begingroup$
Not all bases are 10 in their own notation. There are a group of alternating bases where the base is not 'ten' but a 'hundred'. The most elegant of these is the long-hundred of the proto-germanics and their decendents. Reckoning in the six-score long-hundred (ie 120), was still common enough in 1350 to pass without comment.
Yes, i have used this base for some thirty years. It's truly elegant, being more efficient than either 10 or 12. It's the first base, for which the (number of proper divisors)/(ln base) is greater than 3.
Also 120 is the smallest multiply perfect number, and has the same features as the perfect numbers. For example, 120 = 1+2+4+8+15+30+60 = 3+5+6+10+12+20+24+40, all of these numbers make the total divisors of 120. The second set corresponds to a set of weights, eg
- 1 oz, 2 oz, 4 oz, 8 oz, 1 lb, 2 lb, 4 lb. 15 oz = 1 lb , 120 oz = 1 clove.
- 1 ct, 2 ct, 4 ct, 8 ct, 1 dr, 2 dr, 4 dr : 15 ct = 1dr , 120 ct = 1 oz
- 1 lb, 2 lb, 4 lb, 8 lb, 1 st, 2 st, 4 st : 15 lb = 1 st, 120 lb = 1 cwt
When one considers not just integers, but also fractions x/y and y/z, arranged by xy, one finds that the expressions of these are very short for the first sixty or so, even in things like 56 (8/7 = 1:17.17.17, vs 7/8 = V5), and this unusual pair at 96 (3/32 = 11:30, 32/3 = 10:80.
answered May 15 '13 at 9:20
Wendy KriegerWendy Krieger
311
$endgroup$
Not all bases are 10 in their own notation. There are a group of alternating bases where the base is not 'ten' but a 'hundred'. The most elegant of these is the long-hundred of the proto-germanics and their decendents. Reckoning in the six-score long-hundred (ie 120), was still common enough in 1350 to pass without comment.
Yes, i have used this base for some thirty years. It's truly elegant, being more efficient than either 10 or 12. It's the first base, for which the (number of proper divisors)/(ln base) is greater than 3.
Also 120 is the smallest multiply perfect number, and has the same features as the perfect numbers. For example, 120 = 1+2+4+8+15+30+60 = 3+5+6+10+12+20+24+40, all of these numbers make the total divisors of 120. The second set corresponds to a set of weights, eg
- 1 oz, 2 oz, 4 oz, 8 oz, 1 lb, 2 lb, 4 lb. 15 oz = 1 lb , 120 oz = 1 clove.
- 1 ct, 2 ct, 4 ct, 8 ct, 1 dr, 2 dr, 4 dr : 15 ct = 1dr , 120 ct = 1 oz
- 1 lb, 2 lb, 4 lb, 8 lb, 1 st, 2 st, 4 st : 15 lb = 1 st, 120 lb = 1 cwt
When one considers not just integers, but also fractions x/y and y/z, arranged by xy, one finds that the expressions of these are very short for the first sixty or so, even in things like 56 (8/7 = 1:17.17.17, vs 7/8 = V5), and this unusual pair at 96 (3/32 = 11:30, 32/3 = 10:80.
answered May 15 '13 at 9:20
Wendy KriegerWendy Krieger
311
answered May 15 '13 at 9:20
Wendy KriegerWendy Krieger
311
answered May 15 '13 at 9:20
Wendy KriegerWendy Krieger
311
answered May 15 '13 at 9:20
Wendy KriegerWendy Krieger
311
311
add a comment |
add a comment |
$begingroup$
It is not necessarily true that the multiplication is of the order of $O(b^2)$, since this is a particular implementation of the base, rather than the base itself. The mayans divided their score into four sticks of five dots, and had a true zero in the dot-position (eg fifteen is "3-fives-zero").
One should remember that counting (multiples) and division are separate operations, and that it is possible to use different number-systems for them. Historically, the sixty-wise system is one of divisions: the first column is that of units, and later places are divisions by sixty. Likewise, the romans multiplied by 10's, and divided into 12's.
An alternating base like 60 or 120, supposes two rows on each column of the abacus, where the unit (in the bottom row), is counted to 10 to make one carry up, but divides into 12 to be borrowed into the top row of the lower column. Since one can start in either the top row or bottom row (for counting), the use of tens by twelves or twelves by tens, automatically produces an alternating base.
Using alternating arithmetic then reduces the size of the tables to the order of $O(b)$.
It should be noted that the sumerian system is a division system to avoid division. We see this from recriprocal tables (eg 2 <=> 30 ), and tables of reckoners of multiples of the recriprocals (eg multiples of 4.26.40). Even in their reckoners, multiples are supplied for 1 to 20, and 40. Neugebauer even gives reference to a paper on the seven brothers, ie what is 1/7. It is concluded it lies between 0.8.34.16 and 0.8.34.18.
But i wrangle base 120 for nearly 30 years, and never felt the need to go past 12*12.
answered May 16 '13 at 6:55
wendy.kriegerwendy.krieger
5,83711426
$endgroup$
add a comment |
$begingroup$
It is not necessarily true that the multiplication is of the order of $O(b^2)$, since this is a particular implementation of the base, rather than the base itself. The mayans divided their score into four sticks of five dots, and had a true zero in the dot-position (eg fifteen is "3-fives-zero").
One should remember that counting (multiples) and division are separate operations, and that it is possible to use different number-systems for them. Historically, the sixty-wise system is one of divisions: the first column is that of units, and later places are divisions by sixty. Likewise, the romans multiplied by 10's, and divided into 12's.
An alternating base like 60 or 120, supposes two rows on each column of the abacus, where the unit (in the bottom row), is counted to 10 to make one carry up, but divides into 12 to be borrowed into the top row of the lower column. Since one can start in either the top row or bottom row (for counting), the use of tens by twelves or twelves by tens, automatically produces an alternating base.
Using alternating arithmetic then reduces the size of the tables to the order of $O(b)$.
It should be noted that the sumerian system is a division system to avoid division. We see this from recriprocal tables (eg 2 <=> 30 ), and tables of reckoners of multiples of the recriprocals (eg multiples of 4.26.40). Even in their reckoners, multiples are supplied for 1 to 20, and 40. Neugebauer even gives reference to a paper on the seven brothers, ie what is 1/7. It is concluded it lies between 0.8.34.16 and 0.8.34.18.
But i wrangle base 120 for nearly 30 years, and never felt the need to go past 12*12.
answered May 16 '13 at 6:55
wendy.kriegerwendy.krieger
5,83711426
$endgroup$
add a comment |
$begingroup$
It is not necessarily true that the multiplication is of the order of $O(b^2)$, since this is a particular implementation of the base, rather than the base itself. The mayans divided their score into four sticks of five dots, and had a true zero in the dot-position (eg fifteen is "3-fives-zero").
One should remember that counting (multiples) and division are separate operations, and that it is possible to use different number-systems for them. Historically, the sixty-wise system is one of divisions: the first column is that of units, and later places are divisions by sixty. Likewise, the romans multiplied by 10's, and divided into 12's.
An alternating base like 60 or 120, supposes two rows on each column of the abacus, where the unit (in the bottom row), is counted to 10 to make one carry up, but divides into 12 to be borrowed into the top row of the lower column. Since one can start in either the top row or bottom row (for counting), the use of tens by twelves or twelves by tens, automatically produces an alternating base.
Using alternating arithmetic then reduces the size of the tables to the order of $O(b)$.
It should be noted that the sumerian system is a division system to avoid division. We see this from recriprocal tables (eg 2 <=> 30 ), and tables of reckoners of multiples of the recriprocals (eg multiples of 4.26.40). Even in their reckoners, multiples are supplied for 1 to 20, and 40. Neugebauer even gives reference to a paper on the seven brothers, ie what is 1/7. It is concluded it lies between 0.8.34.16 and 0.8.34.18.
But i wrangle base 120 for nearly 30 years, and never felt the need to go past 12*12.
answered May 16 '13 at 6:55
wendy.kriegerwendy.krieger
5,83711426
$endgroup$
It is not necessarily true that the multiplication is of the order of $O(b^2)$, since this is a particular implementation of the base, rather than the base itself. The mayans divided their score into four sticks of five dots, and had a true zero in the dot-position (eg fifteen is "3-fives-zero").
One should remember that counting (multiples) and division are separate operations, and that it is possible to use different number-systems for them. Historically, the sixty-wise system is one of divisions: the first column is that of units, and later places are divisions by sixty. Likewise, the romans multiplied by 10's, and divided into 12's.
An alternating base like 60 or 120, supposes two rows on each column of the abacus, where the unit (in the bottom row), is counted to 10 to make one carry up, but divides into 12 to be borrowed into the top row of the lower column. Since one can start in either the top row or bottom row (for counting), the use of tens by twelves or twelves by tens, automatically produces an alternating base.
Using alternating arithmetic then reduces the size of the tables to the order of $O(b)$.
It should be noted that the sumerian system is a division system to avoid division. We see this from recriprocal tables (eg 2 <=> 30 ), and tables of reckoners of multiples of the recriprocals (eg multiples of 4.26.40). Even in their reckoners, multiples are supplied for 1 to 20, and 40. Neugebauer even gives reference to a paper on the seven brothers, ie what is 1/7. It is concluded it lies between 0.8.34.16 and 0.8.34.18.
But i wrangle base 120 for nearly 30 years, and never felt the need to go past 12*12.
answered May 16 '13 at 6:55
wendy.kriegerwendy.krieger
5,83711426
answered May 16 '13 at 6:55
wendy.kriegerwendy.krieger
5,83711426
answered May 16 '13 at 6:55
wendy.kriegerwendy.krieger
5,83711426
answered May 16 '13 at 6:55
wendy.kriegerwendy.krieger
5,83711426
5,83711426
add a comment |
add a comment |
$begingroup$
Base 10 is a positional fixed base (The value of the digit depends on its position in the number as well as on its value) and it is neither to large nor to small which is best suited for most purposes. This has many advantages and the andvantage this has over factorial base is, that we need only 10 symbols to represent every possible number. Factorial number system is appealing, but it lacks this practical part. As for other fixed number bases: well,we have 10 digits and are used to think in terms of that. It probbably has something to do with the way we learn our first numbers as children. It is similar to measuring an angle in degrees or radians: we are just so much more used to it being expressed in degrees.
It is just the matter of everyday practicality: base 10 and degrees are just so much more practical in everyday life than anything else and mathematics is for everyday live as well as for advanced mathematicians! And the sciences that need other number bases, just make best use of it (like i.e. the case of computer science).
I guess there is NO number base that is wasty superior to others, so fixed base that is neither to small neither to large best suits ALL purposes and if in some particular base we need a better suited base,then we just make use of one.
The ancient Greeks didn't even use either positional number system, nor a decimal one. They didn't even have a standard number system. Their system was similar to that of Romans (not positional) and Babilonians (not decimal but base 60).
answered Dec 10 '14 at 17:32
SashaSasha
212
$endgroup$
add a comment |
$begingroup$
Base 10 is a positional fixed base (The value of the digit depends on its position in the number as well as on its value) and it is neither to large nor to small which is best suited for most purposes. This has many advantages and the andvantage this has over factorial base is, that we need only 10 symbols to represent every possible number. Factorial number system is appealing, but it lacks this practical part. As for other fixed number bases: well,we have 10 digits and are used to think in terms of that. It probbably has something to do with the way we learn our first numbers as children. It is similar to measuring an angle in degrees or radians: we are just so much more used to it being expressed in degrees.
It is just the matter of everyday practicality: base 10 and degrees are just so much more practical in everyday life than anything else and mathematics is for everyday live as well as for advanced mathematicians! And the sciences that need other number bases, just make best use of it (like i.e. the case of computer science).
I guess there is NO number base that is wasty superior to others, so fixed base that is neither to small neither to large best suits ALL purposes and if in some particular base we need a better suited base,then we just make use of one.
The ancient Greeks didn't even use either positional number system, nor a decimal one. They didn't even have a standard number system. Their system was similar to that of Romans (not positional) and Babilonians (not decimal but base 60).
answered Dec 10 '14 at 17:32
SashaSasha
212
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Base 10 is a positional fixed base (The value of the digit depends on its position in the number as well as on its value) and it is neither to large nor to small which is best suited for most purposes. This has many advantages and the andvantage this has over factorial base is, that we need only 10 symbols to represent every possible number. Factorial number system is appealing, but it lacks this practical part. As for other fixed number bases: well,we have 10 digits and are used to think in terms of that. It probbably has something to do with the way we learn our first numbers as children. It is similar to measuring an angle in degrees or radians: we are just so much more used to it being expressed in degrees.
It is just the matter of everyday practicality: base 10 and degrees are just so much more practical in everyday life than anything else and mathematics is for everyday live as well as for advanced mathematicians! And the sciences that need other number bases, just make best use of it (like i.e. the case of computer science).
I guess there is NO number base that is wasty superior to others, so fixed base that is neither to small neither to large best suits ALL purposes and if in some particular base we need a better suited base,then we just make use of one.
The ancient Greeks didn't even use either positional number system, nor a decimal one. They didn't even have a standard number system. Their system was similar to that of Romans (not positional) and Babilonians (not decimal but base 60).
answered Dec 10 '14 at 17:32
SashaSasha
212
$endgroup$
Base 10 is a positional fixed base (The value of the digit depends on its position in the number as well as on its value) and it is neither to large nor to small which is best suited for most purposes. This has many advantages and the andvantage this has over factorial base is, that we need only 10 symbols to represent every possible number. Factorial number system is appealing, but it lacks this practical part. As for other fixed number bases: well,we have 10 digits and are used to think in terms of that. It probbably has something to do with the way we learn our first numbers as children. It is similar to measuring an angle in degrees or radians: we are just so much more used to it being expressed in degrees.
It is just the matter of everyday practicality: base 10 and degrees are just so much more practical in everyday life than anything else and mathematics is for everyday live as well as for advanced mathematicians! And the sciences that need other number bases, just make best use of it (like i.e. the case of computer science).
I guess there is NO number base that is wasty superior to others, so fixed base that is neither to small neither to large best suits ALL purposes and if in some particular base we need a better suited base,then we just make use of one.
The ancient Greeks didn't even use either positional number system, nor a decimal one. They didn't even have a standard number system. Their system was similar to that of Romans (not positional) and Babilonians (not decimal but base 60).
answered Dec 10 '14 at 17:32
SashaSasha
212
answered Dec 10 '14 at 17:32
SashaSasha
212
answered Dec 10 '14 at 17:32
SashaSasha
212
answered Dec 10 '14 at 17:32
SashaSasha
212
212
add a comment |
add a comment |
protected by Asaf Karagila♦ Dec 10 '14 at 18:43
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The usefulness of base-$12$ is certainly being pushed by the Dozenal Society of America. FWIW, if you can get your hands on a copy of Dudley's Mathematical Cranks, there is a delightful discussion on the people who tirelessly push for the more widespread use of other bases.
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– J. M. is not a mathematician
May 6 '13 at 10:57
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But if we used base 12, "The Dirty Dozen" would have been called "The Dirty Ten", which is not as nearly as good title.
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– Asaf Karagila♦
May 6 '13 at 11:04
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@Asaf, I'm sure y'know that $10_{12}$ isn't read as "ten", right? ;)
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– J. M. is not a mathematician
May 6 '13 at 17:39
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Probably the biggest qualitative difference in usefulness that I've heard of is that from a practical computational perspective highly composite bases are ideal, whereas prime bases are more suitable for pure mathematical purposes. If there was a base that could reap both benefits simultaneously, there would probably be more of a push to switch to such a system. But of course, there isn't
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– David H
May 12 '13 at 3:27
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It's not a XKCD, but I think this comic is mandatory here... cowbirdsinlove.com/43
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– woliveirajr
May 14 '13 at 12:12