Perhaps I'm old-fashioned, or gullible, but I prefer to believe (the fiction) that I can insist that a bank -- the national bank, no less -- do my bidding:
I PROMISE TO PAY THE BEARER ON DEMAND THE SUM OF TWENTY POUNDS
Economics with different numerical base systems.
- Thread starter erisiamk
- Start date
Darth Angelus
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- Apr 3, 2012
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- 477
There would need to be a recalculation as a part of the translation.
When we are talking about small numbers, it would be fairly simple, but for larger numbers, with multiple digits, it would be most convenient to use a conversion program, as human working memory would probably be overloaded pretty fast.
The ones digit is worth the same amount in both systems (1), but the further you get from that digit, the difference would grow exponentially.
That is, the next digit from there in the increasing direction (left, in our way of writing, but that could also differ), which would be the tens digit in our 10-based system, would be the eights digit instead. Therefore, that digit is worth 0.8 times the amount in our system. The next one (hundreds, in the 10-based system) is worth 64 instead of 100, which is 0.64 (or 0.8^2) times the amount. The next one after that is worth 512 instead of 1000, which is 0.512 times the amount.
In the general case, a digit is worth 0.8^n times what it would be in our system, where n is the number of steps in increasing direction (left) from the ones digit. With the ones digit, n would be 0, and anything to the power of 0 = 1. With decimals (digits in decreasing direction (right) from the ones digit), n would be negative, and therefore they would actually have higher value than in our system.
The point is, each (nonzero) digit in an 8-based number has to be multiplied by a power of 8 (instead of 10, like we are used to), and then the results will have to be summed up.
This can be a bit of a large operation, when multiple digits are involved, and it is easy to mix up numbers in your head if you work with several digits. Computers are better suited for this task, as they will do it quickly and without mistakes.
As for money, they would be likely to have notes and coins that are powers of 2, in their currency.
We tend to have money starting with 2 or 5, which is doubling and halving a power of 10 respectively. 8 (unlike 10) is a power of 2, so any power of 8 is also a power of two (with a three times higher exponent). Doubling or halving a power of two would just be increasing or decreasing the exponent by 1.
Here is a little table of likely note or coin values, in the following format...
Power of 2, 10-based number, 8-based number
2^-6, 0.015625, 0.01
2^-5, 0.03125, 0.02
2^-4, 0.0625, 0.04
2^-3, 0.125, 0.1
2^-2, 0.25, 0.2
2^-1, 0.5, 0.4
2^0, 1, 1
2^1, 2, 2
2^2, 4, 4
2^3, 8, 10
2^4, 16, 20
2^5, 32, 40
2^6, 64, 100
2^7, 128, 200
2^8, 256, 400
2^9, 512, 1000
...
Etc...
When we are talking about small numbers, it would be fairly simple, but for larger numbers, with multiple digits, it would be most convenient to use a conversion program, as human working memory would probably be overloaded pretty fast.
The ones digit is worth the same amount in both systems (1), but the further you get from that digit, the difference would grow exponentially.
That is, the next digit from there in the increasing direction (left, in our way of writing, but that could also differ), which would be the tens digit in our 10-based system, would be the eights digit instead. Therefore, that digit is worth 0.8 times the amount in our system. The next one (hundreds, in the 10-based system) is worth 64 instead of 100, which is 0.64 (or 0.8^2) times the amount. The next one after that is worth 512 instead of 1000, which is 0.512 times the amount.
In the general case, a digit is worth 0.8^n times what it would be in our system, where n is the number of steps in increasing direction (left) from the ones digit. With the ones digit, n would be 0, and anything to the power of 0 = 1. With decimals (digits in decreasing direction (right) from the ones digit), n would be negative, and therefore they would actually have higher value than in our system.
The point is, each (nonzero) digit in an 8-based number has to be multiplied by a power of 8 (instead of 10, like we are used to), and then the results will have to be summed up.
This can be a bit of a large operation, when multiple digits are involved, and it is easy to mix up numbers in your head if you work with several digits. Computers are better suited for this task, as they will do it quickly and without mistakes.
As for money, they would be likely to have notes and coins that are powers of 2, in their currency.
We tend to have money starting with 2 or 5, which is doubling and halving a power of 10 respectively. 8 (unlike 10) is a power of 2, so any power of 8 is also a power of two (with a three times higher exponent). Doubling or halving a power of two would just be increasing or decreasing the exponent by 1.
Here is a little table of likely note or coin values, in the following format...
Power of 2, 10-based number, 8-based number
2^-6, 0.015625, 0.01
2^-5, 0.03125, 0.02
2^-4, 0.0625, 0.04
2^-3, 0.125, 0.1
2^-2, 0.25, 0.2
2^-1, 0.5, 0.4
2^0, 1, 1
2^1, 2, 2
2^2, 4, 4
2^3, 8, 10
2^4, 16, 20
2^5, 32, 40
2^6, 64, 100
2^7, 128, 200
2^8, 256, 400
2^9, 512, 1000
...
Etc...
psychotick
Dangerously confused
Hi,
I was actually thinking of using binary notational concepts. So you start with a bit, and then eight bits make a byte. Only problem was what came after that - eight bytes to a butt? Eight butts to a bust? When does a bust become bent? And of course just how big is a bint? Plus there was always the question of value of the currency - perhaps best expressed by the common question - just how much beer can I get for a bent bint?
There it goes again - my failure of imagination!
Cheers, Greg.
I was actually thinking of using binary notational concepts. So you start with a bit, and then eight bits make a byte. Only problem was what came after that - eight bytes to a butt? Eight butts to a bust? When does a bust become bent? And of course just how big is a bint? Plus there was always the question of value of the currency - perhaps best expressed by the common question - just how much beer can I get for a bent bint?
There it goes again - my failure of imagination!
Cheers, Greg.
Darth Angelus
Well-Known Member
- Joined
- Apr 3, 2012
- Messages
- 477
That's handy.
(Would it sound odd if one ran a tenner through a base convertor....
)
(Would it sound odd if one ran a tenner through a base convertor....
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