# C_{0} - C_{0} = [-1, 1]

### Febraruary 12, 2011

Stuart Anderson

We want to solve A = B - C with A in [0, 1] and B, C in the Cantor set. As in Proof #1, A, B, and C are written as ternary strings, with B and C free of the digit 1. Let's think of the problem as

Now the basic limitation of Proof #1 is that the carry value propagates from right to left, so that one must have a starting point for the calculation at the right end of the digit sequence. Hence only finite digit sequences can be constructed, and topological arguments are needed to obtain the solution for the rest.

This can be avoided by the following trick borrowed from computer science: in designing an algorithm for a recursive calculation one tries to reformulate the problem to make as much of it as possible non-recursive, leaving only a minimal core of recursion. Let us then try to split the calculation into a recursive step which computes ONLY THE CARRY, and a non-recursive step which then specifies the solution.

At each digit location, one of the following 6 possibilities holds, where

- i = the carry into this digit,
- o = the carry out to the left neighbor digit, and
- x is a free choice of 0 or 2:

o i | | 0 0 | 1 1 | 1 0 | 0 1 | 0 0 | 1 1 | ||||||||

a | | 0 | 0 | 1 | 1 | 2 | 2 | ||||||||

+ c | | x | 2 | 2 | 0 | 0 | x | ||||||||

--- | | --- | --- | --- | --- | --- | --- | ||||||||

= b | | x | 0 | 0 | 2 | 2 | x |

When a = 0 or 2, i = o, and when a = 1, i = ˜o (the binary complement of o), and both of these relations are symmetrical in i and o. It is therefore clear that the carry can be propagated from left to right instead of right to left. Simply start with

With the carry string available, the computation of each digit of B and C is independent and can be carried out non-recursively. Specifically, let

B and C are also the SMALLEST pair that works, since we have chosen zero for every value of x. If we further define E by

Further, the arbitrary choices in E can be described by choosing a binary string X in

For every A in ^{z}, where z is the number of occurances of the digit 0 in D. This is true even if z is infinite, since ^{card(N)} = card([0, 1]).

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