Number bases done differently (just something i thought of)
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Normally, we use a place-value system. This uses exponentials and multiplication.
1234 ^^^^ |||| |||└ 4 * 10^0 = 4 ||└ 3 * 10^1 = 30 |└ 2 * 10^2 = 200 1 * 10^3 = 1000 1000 + 200 + 30 + 4 = 1234More generally, let d be the value of the digit, and n be the digit's position. So the value of the digit is d * 10^n^ if you're using base 10; or d * B^n^ where B is the base.
1234 ^^^^ |||| |||└ d = 4, n = 0 ||└ d = 3, n = 1 |└ d = 2, n = 2 d = 1, n = 3
What I came up with was a base system that was polynomial, and a system that was purely exponential, no multiplication.
In the polynomial system, each digit is d^n^. We will start n at 1.
polynomial: 1234 ^^^^ |||| |||└ 4^1 = 4 in Place-Value Decimal (PVD) ||└ 3^2 = 9 PVD |└ 2^3 = 8 PVD 1^4 = 1 PVD 1234 poly = 1 + 8 + 9 + 4 PVD = 22 PVDThis runs into some weird stuff, for example:
- Small digits in high positions can have a lower magnitude than large digits in low positions
- 1 in any place will always equal 1
- Numbers with differing digits being equal!
202 poly = 31 poly PVD: 2^3 + 2^1 = 3^2 + 1^1 8 + 2 = 9 + 1 = 10
In the purely exponential system, each digit is n^d^. This is a bit more similar to place value, and it is kind of like a mixed-base system.
1234 ^^^^ |||| |||└ 1^4 = 1 ||└ 2^3 = 8 |└ 3^2 = 9 4^1 = 4 1234 exp = 4 + 8 + 9 + 1 PVD = 22 PVDHowever it still runs into some of the same problems as the polynomial one.
- Small digits in high positions can have a lower magnitude than large digits in low positions (especially if the digit is 1)
- The digit in the ones place will always equal 1
- Numbers with differing digits can still be equal
200 exp = 31 exp PVD: 3^2 = 2^3 + 1^1 9 = 8 + 1
So there you have it. Is it useful? Probably not. Is it interesting? Of course!
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Normally, we use a place-value system. This uses exponentials and multiplication.
1234 ^^^^ |||| |||└ 4 * 10^0 = 4 ||└ 3 * 10^1 = 30 |└ 2 * 10^2 = 200 1 * 10^3 = 1000 1000 + 200 + 30 + 4 = 1234More generally, let d be the value of the digit, and n be the digit's position. So the value of the digit is d * 10^n^ if you're using base 10; or d * B^n^ where B is the base.
1234 ^^^^ |||| |||└ d = 4, n = 0 ||└ d = 3, n = 1 |└ d = 2, n = 2 d = 1, n = 3
What I came up with was a base system that was polynomial, and a system that was purely exponential, no multiplication.
In the polynomial system, each digit is d^n^. We will start n at 1.
polynomial: 1234 ^^^^ |||| |||└ 4^1 = 4 in Place-Value Decimal (PVD) ||└ 3^2 = 9 PVD |└ 2^3 = 8 PVD 1^4 = 1 PVD 1234 poly = 1 + 8 + 9 + 4 PVD = 22 PVDThis runs into some weird stuff, for example:
- Small digits in high positions can have a lower magnitude than large digits in low positions
- 1 in any place will always equal 1
- Numbers with differing digits being equal!
202 poly = 31 poly PVD: 2^3 + 2^1 = 3^2 + 1^1 8 + 2 = 9 + 1 = 10
In the purely exponential system, each digit is n^d^. This is a bit more similar to place value, and it is kind of like a mixed-base system.
1234 ^^^^ |||| |||└ 1^4 = 1 ||└ 2^3 = 8 |└ 3^2 = 9 4^1 = 4 1234 exp = 4 + 8 + 9 + 1 PVD = 22 PVDHowever it still runs into some of the same problems as the polynomial one.
- Small digits in high positions can have a lower magnitude than large digits in low positions (especially if the digit is 1)
- The digit in the ones place will always equal 1
- Numbers with differing digits can still be equal
200 exp = 31 exp PVD: 3^2 = 2^3 + 1^1 9 = 8 + 1
So there you have it. Is it useful? Probably not. Is it interesting? Of course!
Did you just recreate binary or am I reading this wrong?
