Exploring Two-Digit Number Counting with Different Bases

[Digit]

A general two digit counting number is:
d1*b + d0 = C ; where C is the count
Let b = 2 as an example

d
1 0 C
------
0 0 0
0 1 1
1 0 2
1 1 3

b is the count of symbols in the digit, but it can start at 0

d1*0 + d0 = C ; works fine
d0 = C ; and can be any base.

d
1 0 C
-------
0 0 0
0 1 1
1 0 0
1 1 1

Does this mean anything?

 

[chroot]

A number system with zero numerals is rather useless.

- Warren

 

[Digit]

You still have d0 left and it can have any base you want.
Why does anything have to be useful to be interesting?

 

[chroot]

A number system with zero numerals is rather uninteresting, too. You might as well stare at the bottom of an empty bucket.

- Warren

 

[Digit]

When b = 0 d0 = C
C is a counting number. You can count any other number system.
Any multi digit number can be turned into a single digit by choosing the proper b. What this amounts to is that only digits are real.
That is, all numbers cycle on some b set.

d4*A^4 + d3*B^3 + d2*C^2 + d1*D + d0 = N is a number of 4 digits, all with different bases.
The electron shells around atoms have these bases:
2 8 18 18 32 32 are the bases of electron shells.

What caused this post was working with multiple digits. On my number wheel, base 0 means the wheel cannot turn. However with multiple digits you still have d0 = C. C cannot have a 0 base or we cannot count. (setting a base = 0 does not set all all number bases to 0
Warren.)
Number base is a property of numbers, but most of the time, by making multiple digits or a large b, it can be ignored.
Exploring the guts of arithmetic is all I have left to do.

 

[HallsofIvy]

Digit: You seem to have a misconception of what a "base n" number system is-

If we write a number is base n, we might well have "an+ b" so that it has "digits" a and b but a and b themselves must be less that n! That is why "base 0" is both useless and uninteresting.

 

[Digit]

Yes, I know that.
What I was not sure of is where the base counting number starts.
I have put digit on a wheel. With base 0 or 1 you cannot turn the wheel.
However with a*n + b = c you can make n = 0 for digit 1 and you still have digit 0 where b = c. b = c has a base also, but it does not have to be 0.
I think base is also the count of the symbols of the digit, including 0. However I cannot prove that. Base = 0 violates that since there are no symbols. Base starting at 1 has one symbol but no states.
So maybe we can say that base 0 numbers are always 0 and base 1 numbers are all 1. Can't prove that either.
d1*b1^2 + d2*b2 + d3 = c is a number evern when b1 <> b2.
What kind of number?
Also, do negative bases make a mirror image number?

 

[HallsofIvy]

Base 0 has no "numbers" but base 1 certainly does. It has the one digit "0" (except that normally we replace it with the symbol 1) and

1: 1
2: 11
3: 111
4: 1111
5: 11111
6: 111111
7: 1111111

Look familiar?

 

[Digit]

Yes.
Base 1 numbers have one symbol. Matters not what the symbol is.
As you have shown, it is always that symbol.

Thanks for the comments. Nothing much more to say on this subject.