# # Exact Mandelbrot

## # 1 Claim

The Mandelbrot set cannot be calculated with exact rational arithmetic, because the precision required goes up by a factor of two each iteration, so by as few as $$100$$ iterations, the storage space of the computer has long been exhausted.

## # 2 Proof

### # 2.1 Real

Let $$c = \frac{m}{2^n}$$ where $$m$$ is odd and $$n > 0$$.

Induction step:

Let $$z = \frac{M}{2^N}$$ where $$M$$ is odd and $$2N > n > 0$$.

Then $z^2 + c = \frac{M^2 + 2^{2N - n} m}{2^{2N}}$ in lowest terms because $$M^2$$ is odd and $$2^{2N - n} m$$ is even, so their sum is odd.

Base case:

$$c^2 + c = \frac{m^2 + 2^n m}{2^{2n}}$$

It is given that $$n > 0$$, then $$2n \ge 2 > n \ge 1$$.

The argument works the same way with $$2$$ multipliers replaced by $$3$$ (and other prime bases) and “odd” replaced by $$\not= 0\pmod{3}$$, “even” by $$= 0\pmod{3}$$.

For composite bases $$b$$ I think (but I’m not sure) that you need $$m \not = 0 \pmod{p}$$ for each prime factor $$p$$ of $$b$$.

The argument also works changing the power from $$2$$ to $$3$$ or any integer $$d > 1$$.

To be continued…

## # 3 Alternatives

Continued fractions and continued logarithms allow exact computation with rational numbers with different space/time tradeoffs, by using (implicit) rational intervals that are computed as narrowly as needed to resolve inequalities (like $$|z|^2 > 2^2$$) to a definite answer.