B Ideal Base for a Number System

[Isaac0427]

That's of course different from saying it's impossible!

To this we must distinguish the difference between theoretically impossible and practically impossible. Roman numeral integration with trig functions is practically impossible.

 

[TheCanadian]

$5C91_{16}$, or as usually written in C,C++, etc., 0x5C91.

Just curious, what exactly does each term in this notation represent?

 

[fresh_42]

Just curious, what exactly does each term in this notation represent?

23697 (decimal) or \[ (ASCII) or *j (EBCDIC)

 

[SammyS]

There's a straightforward conversion from binary to hexadecimal (base-16), or $5\text C91_{16}$, or as usually written in C,C++, etc., 0x5C91.

Just curious, what exactly does each term in this notation represent?

Do you know how numbers are represented in bases other than base ten? (Yes, I prefer to write out the base in words.)
The 1 is in the 1's place, the 9 in the 16's place, the C is in the 16² place, etc.

It's base sixteen so we need 6 extra digits over those used for decimal. C, the third letter (in the alphabet), represents 9 + 3 = 12 .

 

[ProfuselyQuarky]

Did you try the famous egg-drop experiment yet?

Yeah, but I wasted a lot of eggs before anything good happened. I especially like playing with pH indicators. I used to have a lovely bottle of universal indicator solution that was fun to play with until one of my younger siblings learned to walk and tried to drink the bright pink vinegar solution on the table one day.

I have to side with micromass on this. To a mathemetician, if you have proven conclusively that you can do something then actually doing it is just a detail.

In engineering on the other hand knowing that something CAN be done is a far cry from actually doing it and the difference matters.

True, however, considering that the problem in this thread would only be solved for recreation, the satisfying part IS finding the answer.

To this we must distinguish the difference between theoretically impossible and practically impossible. Roman numeral integration with trig functions is practically impossible.

Micromass was just referring to the fact that you said that you said that it would "take too long". Just because something is extremely tedious, it doesn't mean that it can't be done. "Taking too long" and "is impossible" are different things.

 

[bob012345]

Not even technically, unless you're talking about tally marks.
In base-2, the digits are 0 and 1. In base-3, the digits are 0, 1, and 2. In base-n (n > 1), the digits are 0, 1, ..., n - 1. In "base-1" the only possible digit is 0.

In any base-n system, and arbitrary number is the sum of multiples of powers of the base. For example, the decimal number 123 means $1 \times 10^2 + 2 \times 10^1 + 3 \times 10^0$. In "base 1" the only multiplier available is 0, and every power of 1 is also 1.

I had a long discussion on Compuserve about this more than 20 years ago.

I think you can still use base 1. For example, 3 is 1x1^2 + 1X1^1 + 1X1^0 written as 111 since even in binary, the powers are not restricted to 0 or 1.

 

[golfrmyx]

In dozenal, only 3/10, 6/10 and 9/10 can be written in the form of a decimal with a finite number of didgets. Something about that bothers me, but that's just me.

Seems there is something wrong with your dozenal system thinking. You have twelve digits and thus 1/12, 2/12, 3/12 etc. are all the decimal forms. Except you would not call them decimal. You would need a term involving dozen such as dozenal or some form of the term twelve.

 

[Mark44]

I think you can still use base 1. For example, 3 is 1x1^2 + 1X1^1 + 1X1^0 written as 111 since even in binary, the powers are not restricted to 0 or 1.

Your example is way overcomplicated. Since 1^n = 1 for any integer value of 1, you can discard all of the factors 1^2, 1^2, and 1^0 and write 3 as 1 + 1 + 1, or as ||| using tally marks. In this scheme 5 would be |||||. Even Roman numerals would be an improvement over this.

 

[bob012345]

Your example is way overcomplicated. Since 1^n = 1 for any integer value of 1, you can discard all of the factors 1^2, 1^2, and 1^0 and write 3 as 1 + 1 + 1, or as ||| using tally marks. In this scheme 5 would be |||||. Even Roman numerals would be an improvement over this.

I was just showing how it fits into the usual base schemes. Of course tally marks are base one from time immemorial.

 

[golfrmyx]

This is just a fun question I thought of:
If you take away all knowledge of base-10 being the most natural number system, something we were just taught to think, and you could decide what number system we use, what would you pick? What do you think would make the most sense? Personally, I think base-8, but I'm curious to see what others on PF would choose.

In any digit system, you need symbols for those digits with a name, such as one or two. The more digits or symbols, the more memory difficulty and words to identify the total. Such as using letters of the alphabet for a 24 digit system. What would you call "cz"? The beauty of a 24 digit system is that ones needs to use less digits to number anything in very high quantities such as in the millions but how hard would it be to understand? It requires memorizing the entire system. A ten digit system is easier to memorize that one of twenty four. An eight digit system is thus easier than a ten digit system but requires more digits in a very high number answer such as in the millions, instead of seven digits you would need about twenty percent more digits, thus about nine instead of seven. This requires more writing or typing time and space.

 

[micromass]

I was just showily how it fits into the usual base schemes. Of course tally marks are base one from time immemorial.

Does it though? A decimal number would be of the form $\sum a_i 10^i$ where $a_i \in \{0,...,9\}$. So the correct generalization is a number of the form $\sum a_i p^i$ where $a_i\in \{0,...,p-1\}$. If $p=1$, then we would have $\sum a_i 1^i$ where $a_i \in \{0\}$. So only $0$ can be written as such.

 

[Mark44]

Your example is way overcomplicated. Since 1^n = 1 for any integer value of 1, you can discard all of the factors 1^2, 1^2, and 1^0 and write 3 as 1 + 1 + 1, or as ||| using tally marks. In this scheme 5 would be |||||. Even Roman numerals would be an improvement over this.

I was just showing how it fits into the usual base schemes. Of course tally marks are base one from time immemorial.

I disagree -- there is no "base-1" in the usual meaning of "base-n" counting. For a given base, n, the digits are 0, 1, 2, ... , n - 1. For base-1 (and not including SammyS's tweak in post #19), the only possible digit is 0.

For base-2 (binary), a number is made up of a sum of terms of the form $d \times 2^r$, where d can be 0 or 1. In base-3 (ternary or trinary), a number is made up of a sum of terms of the form $d \times 3^r$, where d here can be 0, 1, or 2. In all of the actual base-n counting systems, the notation is positional -- the value of a particular digit depends on its location in the number. In tally counting, that's not the case -- you merely add up the digits.

 

[bob012345]

I disagree -- there is no "base-1" in the usual meaning of "base-n" counting. For a given base, n, the digits are 0, 1, 2, ... , n - 1. For base-1 (and not including SammyS's tweak in post #19), the only possible digit is 0.

For base-2 (binary), a number is made up of a sum of terms of the form $d \times 2^r$, where d can be 0 or 1. In base-3 (ternary or trinary), a number is made up of a sum of terms of the form $d \times 3^r$, where d here can be 0, 1, or 2. In all of the actual base-n counting systems, the notation is positional -- the value of a particular digit depends on its location in the number. In tally counting, that's not the case -- you merely add up the digits.

Thanks. In my defense I point out that the usual system is not completely consistent, one uses powers beyond the allowed digits. For example in binary we always say only 0 or 1 are allowed but then we say 8 is 1 X 2^3 + 0 X 2^2 + 0 X 2^1 + 0 X 2^0. We used 2 and 3 so for base 1 we can stretch things a bit too. Basically, just change the definition a bit for base 1.

 

[bob012345]

Does it though? A decimal number would be of the form $\sum a_i 10^i$ where $a_i \in \{0,...,9\}$. So the correct generalization is a number of the form $\sum a_i p^i$ where $a_i\in \{0,...,p-1\}$. If $p=1$, then we would have $\sum a_i 1^i$ where $a_i \in \{0\}$. So only $0$ can be written as such.

Change the definition a bit to fit.

 

[jbriggs444]

In all of the actual base-n counting systems, the notation is positional -- the value of a particular digit depends on its location in the number. In tally counting, that's not the case -- you merely add up the digits.

One can regard tally counting as a degenerate place value system -- in which the value of all the places is the same. The bit that feels "wrong" about tally counting is that that the unfilled places have a value of zero. But zero is not a valid digit. It is as if you are using binary with codes 1 and " " in place of 1 and 0, but doing it stupidly with powers of 1 for place value instead of powers of 2.

 

[Mark44]

Thanks. In my defense I point out that the usual system is not completely consistent, one uses powers beyond the allowed digits.

That's not inconsistent. It's only the "digits" (the multipliers of the power of the base) that are among the set {0, 1, ..., n - 1}. The exponents don't have to be represented by the digits.

For example in binary we always say only 0 or 1 are allowed

as digits

but then we say 8 is 1 X 2^3 + 0 X 2^2 + 0 X 2^1 + 0 X 2^0. We used 2 and 3 so for base 1 we can stretch things a bit too. Basically, just change the definition a bit for base 1.

 

[MarkJW]

So, I'd like to live in a world where we use base-16. At the very least, it wouldn't hurt to make computers easier to deal with, and it's not like it's any less efficient than base-10. The only drawback is that we maybe would have to come up with 6 more symbols, because if we were starting from scratch, I doubt we'd want to borrow letters from the alphabet for our numbers. Also, binary and octal are more inefficient than base-10, so I wouldn't want those.

I would propose binary, but use hexadecimal for communication numbers in writing.

 

[ToddSformo]

Technically, yes. Practically, no.

How is the dozenal system not a sweet spot also? :)

I was wondering something a bit different: how does one determine whether a "wrong" answer in base 10 could be a correct answer in another base? I always liked the Ma Pa Kettle math problem where 25/5 = 14 or where 14+14+14+14+14 = 25. Is there a base in which this is valid? If so, how does one figure it out? Thanks
-Todd

 

[Mark44]

I always liked the Ma Pa Kettle math problem where 25/5 = 14 or where 14+14+14+14+14 = 25. Is there a base in which this is valid? If so, how does one figure it out?

One could check the addition in a few bases, with the base being 6 or larger. For example, in base-6, $14_6 + 14_6 + 14_6 + 14_6 + 14_6 = 122_6$. (In decimal, this is 10 + 10 + 10 + 10 + 10 = 50.)

In base-7, $14_7 + 14_7 + 14_7 + 14_7 + 14_7 = 106_7$, so we're at least going in the right direction.

As an aside, there was a similar routine that Abbott & Costello did, where Costello "proved" that 7 X 13 = 28, and equivalently, that 28/7 = 13. See http://www.bing.com/videos/search?q...552BF221ED7288337D80552BF221ED72883&FORM=VIRE

 

[fresh_42]

One could check the addition in a few bases, with the base being 6 or larger. For example, in base-6, $14_6 + 14_6 + 14_6 + 14_6 + 14_6 = 122_6$. (In decimal, this is 10 + 10 + 10 + 10 + 10 = 50.)

In base-7, $14_7 + 14_7 + 14_7 + 14_7 + 14_7 = 106_7$, so we're at least going in the right direction.

As an aside, there was a similar routine that Abbott & Costello did, where Costello "proved" that 7 X 13 = 28, and equivalently, that 28/7 = 13. See http://www.bing.com/videos/search?q...552BF221ED7288337D80552BF221ED72883&FORM=VIRE

I like to counter $2+2=1$ when people post the famous question whether $2+2$ equals $4$ or $5$.

 

[Mark Harder]

Binary, definitely. Aesthetic and subjective notions aside, binary is an irreducible system in the sense that it is the smallest algebraic field, which contains only positive integers (notations for floating point numbers exist, too). The 1 and 0 system represents decisions, the states of switches, anything digital, the information content of a signal or of anything else, I think (As in the game of '20 Questions', where you discover the identity of a secret object by asking enough yes/no questions.) As long as I can use computers, I don't care how long a binary number is, as long as it's not infinite, in which case every number representation is impossible to contain in a computer's finite memory.

Barely relevant: I'm binge watching House, MD on Netflix and in one episode House and team were trying to determine which medication out of a handful they were giving to a patient was damaging him in a specific way that could be observed. They decided to remove the drugs from the patient one at a time until the patient's negative symptoms went away. Subtract one med, wait for specific negative symptom to vanish, repeat if there's no change. Given that the problem drug would kill the patient in time (for the episode to end), and barring critical medical reasons for keeping more than one med in the patient at a time, wouldn't a better algorithm be to remove a randomly chosen half of the drugs from the regimen, wait and watch, keep giving the patient the half-batch on which he thrived and throw out the other half? There'd be a 50% chance that the first trial would improve the patient's health; endgame. Otherwise, remove a random half of the remainder of the regimen and repeat. Randomly removing one out of N drugs at a time, and that's only a 1/N probability. You got to go with binary, sez me.

 

[SammyS]

...
I always liked the Ma Pa Kettle math problem where 25/5 = 14 or where 14+14+14+14+14 = 25. Is there a base in which this is valid? If so, how does one figure it out? Thanks
-Todd

In base B :

14 = 1B + 4 and 25 = 2B + 5

So 14+14+14+14+14 is 5B + 5×4 → 5B + 20 .

Solve 5B + 20 = 2B + 5 .

It gives B = -5 which doesn't make sense.

 

[golfrmyx]

In dozenal, only 3/10, 6/10 and 9/10 can be written in the form of a decimal with a finite number of didgets. Something about that bothers me, but that's just me.

Correction. any smaller number divided by a larger number gives you the decimal form for an answer as a percentage. This would still be the case with any base system because anything you call a dozen is still a dozen as well as 96 out of one hundred is still that many no matter what words you use to call that quantity. Thia is probably the best reason for a base 10 digit system.

 

[golfrmyx]

Binary, definitely. Aesthetic and subjective notions aside, binary is an irreducible system in the sense that it is the smallest algebraic field, which contains only positive integers (notations for floating point numbers exist, too). The 1 and 0 system represents decisions, the states of switches, anything digital, the information content of a signal or of anything else, I think (As in the game of '20 Questions', where you discover the identity of a secret object by asking enough yes/no questions.) As long as I can use computers, I don't care how long a binary number is, as long as it's not infinite, in which case every number representation is impossible to contain in a computer's finite memory.

Barely relevant: I'm binge watching House, MD on Netflix and in one episode House and team were trying to determine which medication out of a handful they were giving to a patient was damaging him in a specific way that could be observed. They decided to remove the drugs from the patient one at a time until the patient's negative symptoms went away. Subtract one med, wait for specific negative symptom to vanish, repeat if there's no change. Given that the problem drug would kill the patient in time (for the episode to end), and barring critical medical reasons for keeping more than one med in the patient at a time, wouldn't a better algorithm be to remove a randomly chosen half of the drugs from the regimen, wait and watch, keep giving the patient the half-batch on which he thrived and throw out the other half? There'd be a 50% chance that the first trial would improve the patient's health; endgame. Otherwise, remove a random half of the remainder of the regimen and repeat. Randomly removing one out of N drugs at a time, and that's only a 1/N probability. You got to go with binary, sez me.

In base B :

14 = 1B + 4 and 25 = 2B + 5

So 14+14+14+14+14 is 5B + 5×4 → 5B + 20 .

Solve 5B + 20 = 2B + 5 .

It gives B = -5 which doesn't make sense.

More simply, look at the figures and see the total to find a simple answer. It is a five digit system with 1 and 4 used the same as in our two digit system. 0, 1, 2, 3 and 4 would be the digits. 4 "a" plus 1 "a" is "aaaaa". or five with our normal ten digit system. take them five times and you get 25. But showing how to express that total in the five digit system would take a lot of space. 4 +1 = 10 or "aaaaa". Thus 10 + 10 = 20 or "aaaaa + aaaaa" . 20 x 2 = 40 or "aaaaa + aaaaa + aaaaa + aaaaa" which is the same as 10 x 4 = 40. thus 10 x 10 = 100 or "aaaaa + aaaaa + aaaaa + aaaaa + aaaaa".
Check the math now. Since 10 is equal to five in a ten digit system and five times five is twenty five, the system is correct so far. I think. Notice in the five digit system 100 is the five digit form of 25 in the ten digit form.

This problem was in the first few pages of my sons fifth grade math book in 1977 in Houston, Texas and was assigned as home work. The book had iot messed up because it was using the ten digit system for answers to the five digit system without clarifying which system digits were being used. I called the teacher and she said it was for extra credit and really should have been in the back of the book. It should never have been in any school book at any level.

 

[jbriggs444]

It gives B = -5 which doesn't make sense.

Sense can be made of such a base... https://en.wikipedia.org/wiki/Negative_base

 

[Isaac0427]

Correction. any smaller number divided by a larger number gives you the decimal form for an answer as a percentage.

I said with a finite number of digits.

 

[fresh_42]

It should never have been in any school book at any level.

It is somehow relieving to hear that your math schoolbooks are as stupid as ours. I've read definitions and stuff in them I never met again. Not to talk about inconsistencies. Did your son's teacher explain why they confront 5th grades with base 5 systems? I may understand it as a part of a history lesson about counting systems of early tribes rather than in regular maths. 2 (computing), 12 (dozen, gross) or 60 (time, calender) occur in real life, but 5?

 

[jbriggs444]

It is somehow relieving to hear that your math schoolbooks are as stupid as ours. I've read definitions and stuff in them I never met again. Not to talk about inconsistencies. Did your son's teacher explain why they confront 5th grades with base 5 systems? I may understand it as a part of a history lesson about counting systems of early tribes rather than in regular maths. 2 (computing), 12 (dozen, gross) or 60 (time, calender) occur in real life, but 5?

In the U.S., I believe that it dates to the "new math" in the 1960's. Rather than emphasizing calculation, there was an attempt to inject some more pure or abstract math content into the curriculum. As I recall (having been in elementary school at the time), we would begin each new school year with some content involving Venn diagrams and non-decimal arithmetic before reverting to more calculation-oriented material for the rest of the year.

A debate on the merits of the "new math" is, of course, not on topic here, so I'll not try to justify or criticize the program.

 

[fresh_42]

In the U.S., I believe that it dates to the "new math" in the 1960's. Rather than emphasizing calculation, there was an attempt to inject some more pure or abstract math content into the curriculum. As I recall (having been in elementary school at the time), we would begin each new school year with some content involving Venn diagrams and non-decimal arithmetic before reverting to more calculation-oriented material for the rest of the year.

A debate on the merits of the "new math" is, of course, not on topic here, so I'll not try to justify or criticize the program.

Yep, but interesting anyway. "Set theory" has been a cuss word here for some time. I once tutored a grandmother in Venn diagrams because she want's to help her granddaughter with her homework!

 

[gjonesy]

The example above is just counting sticks which already has been too poor 5,000 years ago.

Definitely fingers and toes system...its convenient and I carry that calculator with me everywhere I go. I suppose wearing sandals would make things even easier...lol j/k

 

[bob012345]

As an aside, consider dividing $1000 into envelopes so you can make any amount in combination. What is the minimum number of envelopes needed and how much money in each one? Now, you all will get the answer quickly but then, is that the best way to do it? What is the minimum number for the simplest system as far as counting quickly, say in just a few seconds?

 

[Mark44]

As an aside, consider dividing $1000 into envelopes so you can make any amount in combination.

Including coins?

Also, is there a connection to the subject of this thread?

 

[cosmik debris]

I am not so keen on integer bases, Imaginary bases are cool.

 

[ProfuselyQuarky]

I am not so keen on integer bases, Imaginary bases are cool.

But they're not exactly practical for daily usage :)

 

[fresh_42]

But they're not exactly practical for daily usage :)

Yep, got in deep trouble last time I tried to pay my burger with imaginary based money ...

 

[ProfuselyQuarky]

Yep, got in deep trouble last time I tried to pay my burger with imaginary based money ...

Taking out the pink bills from your Monopoly board game, are you?

 

[cosmik debris]

Yep, got in deep trouble last time I tried to pay my burger with imaginary based money ...

As long as it's an even amount of money I think it is real.

 

[fresh_42]

As long as it's an even amount of money I think it is real.

And what do you call paying with $6000000_i$? A stick-up?

But I admit it would be great fun to rewrite est. 1,000,000,000,000 lines of code of banking software.

 

[jbriggs444]

But I admit it would be great fun to rewrite est. 1,000,000,000,000 lines of code of banking software.

The Y2i problem? -- long term gainful employment solving an imaginary crisis?

Edit: But if you imagine spending long years at an imaginary pay rate, the result will be a real negative.

 

[bob012345]

Including coins?

Also, is there a connection to the subject of this thread?

No, integer dollar amounts. It's a simple riddle and when you solve it you see the connection which is about counting in bases.

 

[jbriggs444]

No, integer dollar amounts. It's a simple riddle and when you solve it you see the connection which is about counting in bases.

So you are going for a notion of radix economy based on the sum of the digits in the set of strings that encode values from 0 to 1000. This comes from a metric of goodness that is the maximum (over all values) of the minimum (over all conforming envelope selections) number of envelopes that produce the value. This under the constraint that all values must be encodable with at least one selection of envelopes. This amounts to a place value system where the place values are the denominations of the envelopes and the allowable digits are zero through the multiplicity of each denomination.

It seems clear that with this notion of radix economy that binary wins and you want place values of 1, 2, 4, 8, 16, 32, 64, 128, 256, 489 or some minor variant.

Of course, the standard notion of radix economy uses a different metric of goodness and results in a different optimal base.

 

[bob012345]

So you are going for a notion of radix economy based on the sum of the digits in the set of strings that encode values from 0 to 1000. This comes from a metric of goodness that is the maximum (over all values) of the minimum (over all conforming envelope selections) number of envelopes that produce the value. This under the constraint that all values must be encodable with at least one selection of envelopes. This amounts to a place value system where the place values are the denominations of the envelopes and the allowable digits are zero through the multiplicity of each denomination.

It seems clear that with this notion of radix economy that binary wins and you want place values of 1, 2, 4, 8, 16, 32, 64, 128, 256, 489 or some minor variant.

Of course, the standard notion of radix economy uses a different metric of goodness and results in a different optimal base.

Correct, you have ten envelopes and can make any sum to 1000 whereas a strictly decimal system would have 28 envelopes (9 hundreds, 9 tens, 10 ones). But if I give you a random number can you dish out the right envelopes in say about 2 or 3 seconds? What's the easiest system with the minimum envelopes?

 

[jbriggs444]

Correct, you have ten envelopes and can make any sum to 1000 whereas a strictly decimal system would have 28 envelopes (9 hundreds, 9 tens, 10 ones). But if I give you a random number can you dish out the right envelopes in say about 2 or 3 seconds? What's the easiest system with the minimum envelopes?

Optimizing for two criteria is not, in general, feasible. You are, perhaps, asking for the easiest system consistent with having the minimum number of envelopes total. In the case at hand, that's 10 envelopes. I do not see a quick (2 or 3 second) algorithm for a human teller given a decimal input in the range from 0 to 1000 to select an appropriate collection from an array of ten envelopes.

With 12 envelopes and a 3 (or 4 in the edge case) digit decimal request, it's easy, of course.

 

[bob012345]

Optimizing for two criteria is not, in general, feasible. You are, perhaps, asking for the easiest system consistent with having the minimum number of envelopes total. In the case at hand, that's 10 envelopes. I do not see a quick (2 or 3 second) algorithm for a human teller given a decimal input in the range from 0 to 1000 to select an appropriate collection from an array of ten envelopes.

With 12 envelopes and a 3 (or 4 in the edge case) digit decimal request, it's easy, of course.

Exactly. One possibility might be 1,2,3,4, 10,20,30,30 and 100,200,300,300. I think three seconds may be possible.

 

[Mark Harder]

Maybe it's obvious to all, but just consider polynomials in x given by p(x) = a₀ x⁰ + a₁ x¹ + a₂ x² + ... ,
and expressions for numbers (Integrals, at least) in bases. They can be expressed as polynomials. In base B:
p(B) = a₀ B⁰ + a₁ B¹ + a₂ B² + ...
For example take 123₁₀: 123(B=10) = 3₁₀ 10⁰ + 2₁₀ 10¹ + 1 10²

I don't know if the congruence between numerical systems and polynomials is good for anything, but it's interesting to think that when we write numbers in any base we desire, we're dealing with polynomials in that base with coefficients drawn from the base.

 

[SammyS]

Of course in base seven we write seven as 10 .

 

[jbriggs444]

I don't know if the congruence between numerical systems and polynomials is good for anything, but it's interesting to think that when we write numbers in any base we desire, we're dealing with polynomials in that base with coefficients drawn from the base.

Yes, for any [finite] digit string there is a corresponding polynomial using those digits as its coefficients. For a particular digit-string, one can find its associated value in base b by evaluating the corresponding polynomial function p at b.

The idea of "polynomials in that base" does not ring true. More typically one would talk of polynomial functions over a particular dummy variable (e.g. "a polynomial in x"). Or of formal polynomials over a particular field or ring (e.g. "a polynomial over the reals").

 

[Ralph Dratman]

Base 60 is one of the oldest used systems. (3300 B.C.)
https://en.wikipedia.org/wiki/Sexagesimal
A few decades ago as Assembler was still often in use 8 and 16 were useful to know.

But base 60 has fifty new "digits" beyond 0-9 for the K-12 math committee to invent, along with 1725 new multiplication facts that every child has to learn. I think in that case very few people would ever pass 4th grade. Certainly I never would have.

 

[Ralph Dratman]

I'm aware of that, thanks

It's just one of those odd things I forget, I guess. I'd edit my posts to make them accurate, but I no longer can edit them.

Why can you not edit your posts?

 

[Ralph Dratman]

Hm, base-8 is an interesting choice! Why do you choose that one? I think that base-12 would be the most reasonable number system to use. The transition from the decimal system to dozenal would be really easy. Base-12 is also convenient because 12 has a lot of factors--1, 2, 3, 4,and 6. Quick arithmetic while you're at the store or trying to divide something in your head during daily life would be simpler, I think.

I quite like binary, but the thought of living in a world that runs on the impractical base-1 is just ugly . . .

I think you hit it with twelve. Shall we use A and B for the extra digits? I like this already. About the only sad part is that we have 21 new multiplication facts to learn. I fear that would have broken me in 4th grade. I can remember crying over the multiplication table even in the old ten system we used in those days.