# 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 \(d > 1\).

# 2.2 Complex

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.