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Number bases done differently (just something i thought of) (sh.itjust.works)

submitted 8 months ago* (last edited 8 months ago) by DeltaWingDragon@sh.itjust.works to c/nowhereelsetoshare@sh.itjust.works

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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 = 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

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

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[–] slazer2au@lemmy.world 1 points 8 months ago (1 children)

Did you just recreate binary or am I reading this wrong?

[–] DeltaWingDragon@sh.itjust.works 3 points 8 months ago

This isn't binary. Binary is just an ordinary place-value system with base 2 instead of base 10.

Here's a TLDR of my post:

Let d be any digit in a number, and n be its position.
Ordinary number systems have a base B, and each digit is equal to d * B^n^. Common values for B are 10, 16, 8, 2 and 12.
In my invented polynomial number system, each digit is equal to d^n^.
In my invented exponential number system, each digit is equal to n^d^.

[–] flyinryan@lemmy.zip 1 points 1 week ago (1 children)

For a system like this, what defines what can be used as a digit? It's not base ten but there aren't really examples beyond 1-4. Could you use hexadecimals digits to make large numbers with something like

BAFD
^^^^
||||
|||└ D^1 = 13 in Place-Value Decimal (PVD)
||└ F^2 = 256 PVD
|└ A^3 = 1000 PVD
B^4 = 14641 PVD

1A3D poly = 13 + 256 + 1000 + 14641 = 15910 PVD

I guess I'm asking how or if a polynomial system interacts with a base system since it's still involved with the digits of a poly number.

[–] DeltaWingDragon@sh.itjust.works 1 points 1 week ago

Only limit to the digits is what's used as a common base.

If I was making it a programming library, it would only support base 10, and common programming bases like base 16, 8, and 2.

And base 12 (for clocks and old civilizations), base 20 (Kaktovik and Old French), base 60 (Babylonian).

And maybe also base 64 (for encoding purposes), and base 40 (someone might use this somewhere).

Actually it can theoretically support any natural number of digits, even approaching infinity. (Except for 1 which is just tally marks)