Ideal Base for a Number System

[Isaac0427]

In our system of numbers, integrate by parts 4 times starting with u=x4u=x4u = x^4, then eventually you'll just be left with one trig function. It's a bit trickier with Roman numerals.

That's what I did. I got a huge expression that required me to multiply Roman numerals at least 5 times. The only way to do that is to distribute and simplify, which would take way too long.

 

[micromass]

That's what I did. I got a huge expression that required me to multiply Roman numerals at least 5 times. The only way to do that is to distribute and simplify, which would take way too long.

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

 

[ProfuselyQuarky]

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

So let's see you do it, micromass

 

[micromass]

So let's see you do it, micromass

I have proven the answer exists and is unique. That is enough for a mathematician.

 

[ProfuselyQuarky]

I have proven the answer exists and is unique. That is enough for a mathematician.

How is that enough? That's a very unsatisfactory way to do things.

I know that I can bake a cake, but who cares until I actually do it?

 

[micromass]

How is that enough? That's a very unsatisfactory way to do things.

I know that I can bake a cake, but who cares until I actually do it?

You must be very interested in engineering and experimental physics

 

[ProfuselyQuarky]

You must be very interested in engineering and experimental physics

Not so much engineering, but I like biochemistry and, yes, experimental physics. Of course, there's not so much "experimenting" I can do considering that I'm limited to a garage that doesn't meet the standards of "lab"

 

[micromass]

Not so much engineering, but I like biochemistry and, yes, experimental physics. Of course, there's not so much "experimenting" I can do considering that I'm limited to a garage that doesn't meet the standards of "lab"

Did you try the famous egg-drop experiment yet?

 

[jim mcnamara]

On a different note: Athabascan languages use base 4. Everything relates to the cyclic nature of things; 4 seasons. If Navajos had invented baseball it would be four strikes and you're out. The Navajo speakers I knew used decimal numbers. Why? Not because they view them as better, rather just to be able to function in business transactions since everyone else uses base 10. Langauge is meant for communication.
Numbers are an important part of communication in Western culture.

Answer to the OP;
So, if you want to see what numbering would be like if we 'started over', look at other language families. You will find lots of differences.
Humans are disgustingly inventive from that point of view.

As @Mark44 points out base 60 was also used and is still part of everyday life.

I would like to suggest a concept: there is not always a best, rather a set of choices of varying subjective usefulness. For a number base, consider the Lingua Franca, decimal, as the first choice, then go from there if changes are really needed - other than as an intellectual fun exercise.

 

[phinds]

How is that enough? That's a very unsatisfactory way to do things.

I know that I can bake a cake, but who cares until I actually do it?

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.

 

[Dr_Zinj]

The title of the thread is actually a deception.
You use different based number systems for different purposes. Depending on the purpose, some systems work better than others.
For instance, I like base 2 for counting to 32 on a single hand. (I only have the normal 5 fingered hand.)

 

[fresh_42]

The title of the thread is actually a deception.
You use different based number systems for different purposes. Depending on the purpose, some systems work better than others.
For instance, I like base 2 for counting to 32 on a single hand. (I only have the normal 5 fingered hand.)

31, I guess. And there are some very tough numbers in between ...

 

[phinds]

For instance, I like base 2 for counting to 32 on a single hand. (I only have the normal 5 fingered hand.)

Uh ... you probably shouldn't do that in public or the 4 might get you in trouble

 

[fresh_42]

Uh ... you probably shouldn't do that in public or the 4 might get you in trouble

18 can be even more dangerous!

 

[phinds]

18 can be even more dangerous!

That one depends more on the culture

 

[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 $5$.