Ideal Base for a Number System

[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, calendar) 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, calendar) 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.