math

cross-posted from: https://sh.itjust.works/post/48633930

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 = 1234

More 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 PVD

This 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 PVD

However 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

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So there you have it. Is it useful? Probably not. Is it interesting? Of course!