# Big Numbers

Take a number, say, 264, and write it in binary:

$$264 = 2^8 + 2^3.$$

Because we like binary so much, write the exponents in binary too:

\begin{align*} 264 & = 2^{2^3} + 2^{2 + 1} \\ {} & = 2^{2^{2 + 1}} + 2^{2 + 1} \end{align*}

and so on, until every number is 2 or less. This is the hereditary base-2 representation of 264. Now, transform that number by increasing the base from 2 to 3, and subtracting one:

\begin{align*} G_2(264) & = 3^{3^{3 + 1}} + 3^{3 + 1} - 1 \\ & = 3^{3^{3 + 1}} + 2 \cdot 3^3 + 2 \cdot 3^2 + 2 \cdot 3 + 2 \\ & \approx 4.4 \cdot 10^{38} \end{align*}

And again:

\begin{align*} G_3(G_2(264)) & = 4^{4^{4 + 1}} + 2 \cdot 4^4 + 2 \cdot 4^2 + 2 \cdot 4 + 1 \\ & \approx 3.2 \cdot 10^{616} \end{align*}

Goodstein sequences are defined by these repeated applications of $$G_n$$. The Goodstein sequence starting at $$m$$ is $$m_1 = m$$, $$m_2 = G_2(m_1)$$, $$m_3 = G_3(m_2)$$, etc.

## Goodstein's theorem

Goodstein's theorem states that every Goodstein sequence eventually reaches zero. Despite the deceivingly explosive initial increase (the next value for $$m = 265$$ above is $$m_4 \approx 2.5 \cdot 10^{10921}$$), the surprising truth is that $$m_k = 0$$ for some finite $$k$$, for all initial values of $$m$$.

This counterintuitive theorem is actually fairly easy to prove. We simply find an upper bound on the base $$b$$ used in the hereditary base-$$b$$ representations in the sequence. Finding the upper bound is actually the easiest part:

$$b < \infty.$$ "But wait," you say, "that's obvious and unhelpful!" Well, by using a particular kind of infinity, this upper bound becomes useful. Precisely, we use $\omega$, the first transfinite ordinal number. For $$m = 264$$ as before, the bounds look like this:

$$\begin{array}{cclcl} m_1 & = & 2^{2^{2 + 1}} + 2^{2 + 1} & \le & \omega^{\omega^{\omega + 1}} + \omega^{\omega + 1} \\ m_2 & = & 3^{3^{3 + 1}} + 2 \cdot 3^3 + 2 \cdot 3^2 + 2 \cdot 3 + 2 & \le & \omega^{\omega^{\omega + 1}} + \omega^\omega \cdot 2 + \omega^2 \cdot 2 + \omega \cdot 2 + 2 \\ m_3 & = & 4^{4^{4 + 1}} + 2 \cdot 4^4 + 2 \cdot 4^2 + 2 \cdot 4 + 1 & \le & \omega^{\omega^{\omega + 1}} + \omega^\omega \cdot 2 + \omega^2 \cdot 2 + \omega \cdot 2 + 1 \\ m_4 & = & 5^{5^{5 + 1}} + 2 \cdot 5^5 + 2 \cdot 5^2 + 2 \cdot 5 & \le & \omega^{\omega^{\omega + 1}} + \omega^\omega \cdot 2 + \omega^2 \cdot 2 + \omega \cdot 2 \\ m_5 & = & 6^{6^{6 + 1}} + 2 \cdot 6^6 + 2 \cdot 6^2 + 6 + 5 & \le & \omega^{\omega^{\omega + 1}} + \omega^\omega \cdot 2 + \omega^2 \cdot 2 + \omega + 5 \\ m_6 & = & 7^{7^{7 + 1}} + 2 \cdot 7^7 + 2 \cdot 7^2 + 7 + 4 & \le & \omega^{\omega^{\omega + 1}} + \omega^\omega \cdot 2 + \omega^2 \cdot 2 + \omega + 4 \\ \vdots \end{array}$$

These upper bounds are well-defined using ordinal arithmetic. They are also strictly decreasing, and since the ordinals are well-ordered, every such strictly-decreasing sequence must reach zero in a finite number of steps. Note that we write $$\omega \cdot 2$$ rather than $$2 \cdot \omega$$ above because ordinal arithmetic is not commutative; in fact, $$2 \cdot \omega = \omega < \omega \cdot 2[/latex]. ## Independence Despite being a theorem about natural numbers, Goodstein's theorem is independent of the Peano axioms. It's a natural, concrete example of Gödel's incompleteness theorem for those axioms. Kirby and Paris proved its independency by essentially showing that any proof of Goodstein's theorem requires induction over sets too large for Peano's axioms to handle. They showed that Goodstein's theorem implies Gentzen's theorem. ## The Goodstein function How long are Goodstein sequences? It turns out they are extremely long. Let the Goodstein function [latex]\mathcal{G}(m)$$ be the length of the Goodstein sequence starting at $$m_1 = m$$.

$$\mathcal{G}(12)$$ is larger than Graham's number. $$\mathcal{G}(m)$$ grows faster than the Ackermann function $$A(m, m)$$. $$\mathcal{G}(m)$$ grows faster than the iterated Ackermann function $$A^{A(m, m)}(m, m) = A(A(\cdots), A(\cdots))$$. In fact, $$\mathcal{G}(m)$$ grows faster than any function that can be proven to be total in Peano arithmetic. That means $$\mathcal{G}(m)$$ grows faster than any function that can be shown to even exist by induction over the natural numbers.

## 2 thoughts on “Big Numbers”

1. bubbe says:

I could never understand Greek either.