# Goodstein's theorem

{{#invoke:Hatnote|hatnote}}Template:Main other In mathematical logic, Goodstein's theorem is a statement about the natural numbers, proved by Reuben Goodstein in 1944, which states that every Goodstein sequence eventually terminates at 0. Kirby and Paris showed that it is unprovable in Peano arithmetic (but it can be proven in stronger systems, such as second order arithmetic). This was the third "natural" example of a true statement that is unprovable in Peano arithmetic (after Gerhard Gentzen's 1943 direct proof of the unprovability of ε0-induction in Peano arithmetic and the Paris–Harrington theorem). Earlier statements of this type had either been, except for Gentzen, extremely complicated, ad-hoc constructions (such as the statements generated by the construction given in Gödel's incompleteness theorem) or concerned metamathematics or combinatorial results.

Laurence Kirby and Jeff Paris introduced a graph theoretic hydra game with behavior similar to that of Goodstein sequences: the "Hydra" is a rooted tree, and a move consists of cutting off one of its "heads" (a branch of the tree), to which the hydra responds by growing a finite number of new heads according to certain rules. Kirby and Paris proved that the Hydra will eventually be killed, regardless of the strategy that Hercules uses to chop off its heads, though this may take a very, very, very long time.

## Hereditary base-n notation

Goodstein sequences are defined in terms of a concept called "hereditary base-n notation". This notation is very similar to usual base-n positional notation, but the usual notation does not suffice for the purposes of Goodstein's theorem.

In ordinary base-n notation, where n is a natural number greater than 1, an arbitrary natural number m is written as a sum of multiples of powers of n:

$m=a_{k}n^{k}+a_{k-1}n^{k-1}+\cdots +a_{0},$ $35=32+2+1=2^{5}+2^{1}+2^{0}.$ Thus the base 2 representation of 35 is $2^{5}+2^{1}+2^{0}$ . (This expression could be written in binary notation as 100011.) Similarly, one can write 100 in base 3:

$100=81+18+1=3^{4}+2\cdot 3^{2}+1.$ Note that the exponents themselves are not written in base-n notation. For example, the expressions above include $2^{5}$ and $3^{4}$ .

To convert a base-n representation to hereditary base n notation, first rewrite all of the exponents in base-n notation. Then rewrite any exponents inside the exponents, and continue in this way until every digit appearing in the expression is n or less.

For example, while 35 in ordinary base-2 notation is $2^{5}+2+1$ , it is written in hereditary base-2 notation as

$35=2^{2^{2}+1}+2+1,$ using the fact that $5=2^{2}+1.$ Similarly, 100 in hereditary base 3 notation is

$100=3^{3+1}+2\cdot 3^{2}+1.$ ## Goodstein sequences

The Goodstein sequence G(m) of a number m is a sequence of natural numbers. The first element in the sequence G(m) is m itself. To get the second, G(m)(2), write m in hereditary base 2 notation, change all the 2s to 3s, and then subtract 1 from the result. In general, the $n+1^{st}$ term G(m)(n+1) of the Goodstein sequence of m is as follows: take the hereditary base n+1 representation of G(m)(n), and replace each occurrence of the base n+1 with n+2 and then subtract one. Note that the next term depends both on the previous term and on the index n. Continue until the result is zero, at which point the sequence terminates.

Early Goodstein sequences terminate quickly. For example, G(3) terminates at the sixth step:

Base Hereditary notation Value Notes
2 $2^{1}+1$ 3 Write 3 in base 2 notation
3 $3^{1}+1-1=3^{1}$ 3 Switch the 2 to a 3, then subtract 1
4 $4^{1}-1=3$ 3 Switch the 3 to a 4, then subtract 1. Now there are no more 4s left
5 $3-1=2$ 2 No 4s left to switch to 5s. Just subtract 1
6 $2-1=1$ 1 No 5s left to switch to 6s. Just subtract 1
7 $1-1=0$ 0 No 6s left to switch to 7s. Just subtract 1

Later Goodstein sequences increase for a very large number of steps. For example, G(4) starts as follows:

Elements of G(4) continue to increase for a while, but at base $3\cdot 2^{402653209}$ , they reach the maximum of $3\cdot 2^{402653210}-1$ , stay there for the next $3\cdot 2^{402653209}$ steps, and then begin their first and final descent.

The value 0 is reached at base $3\cdot 2^{402653211}-1$ (curiously, this is a Woodall number: $3\cdot 2^{402653211}-1=402653184\cdot 2^{402653184}-1$ . This is also the case with all other final bases for starting values greater than 4{{ safesubst:#invoke:Unsubst||date=__DATE__ |\$B= {{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }}).

However, even G(4) doesn't give a good idea of just how quickly the elements of a Goodstein sequence can increase. G(19) increases much more rapidly, and starts as follows:

In spite of this rapid growth, Goodstein's theorem states that every Goodstein sequence eventually terminates at 0, no matter what the starting value is.

## Proof of Goodstein's theorem

Goodstein's theorem can be proved (using techniques outside Peano arithmetic, see below) using results in ordinal numbers. Actually, the whole power of ordinal numbers is not needed. Define $P$ to be the set containing the positive integers and the symbol $\omega$ and closed under addition, multiplication, and exponentiation base $\omega$ with the usual interpretations. The includes things like $4\omega ^{7\omega ^{3}+2\omega +8}+5$ . These could be thought of as ordinal numbers in Cantor normal form where $\omega$ is the first infinite ordinal number or even more simply as functions from the positive integers $\omega$ to the positive integers. (The later ignores the complexities of ordinal numbers like whether these operation are non-Commutative and strange things like $\varepsilon _{0}=\omega ^{\varepsilon _{0}}$ .) There is a total ordering of these functions $p_{1},p_{2}\in P$ as done with Big O notation. Specifically, $p_{1} if and only if there exists an integer $\omega _{0}$ such that $\forall \omega \geq \omega _{0},p_{1}(\omega ) . The only property used of these objects in $P$ is that there are no infinitely decreasing sequences of them. Specifically, any decreasing sequence that can only terminate at zero, does in fact terminate at zero. The intuition is as follows. The first obvious way to form an infinitely long decreasing sequence is $\omega ,\omega -1,\omega -2,\omega -3,...$ , however this is not allowed because subtraction is not allowed. In any decreasing sequence, the next object after $\omega$ has to be some positive integer like $1000000000$ . Though this integer could arbitrarily large, it must be finite and hence when it is continually decreased, it must eventually terminate at zero. The function $5\omega ^{3}$ does not need to decrease in one step all the way to $4\omega ^{3}$ , because the function $4\omega ^{3}+1000000000\omega ^{2}+1000000000\omega +1000000000$ is also strictly smaller. However, inductively these smaller order terms when decreased must eventually terminate at zero, giving us $4\omega ^{3}$ . In this way, this 4 must continue to decrease until we get to $\omega ^{3}$ at which point it must decrease to a lower order function like $1000000000\omega ^{2}$ . Just as this exponent decreased from 3 to 2, inductively we know that the exponent in something like $\omega ^{\omega ^{3}}$ must also eventually decrease to zero.

Given a Goodstein sequence G(m), we prove that it eventually terminates at zero as follows. We construct a parallel sequence P(m) of functions $p\in P$ (or of ordinal numbers) which is strictly decreasing and hence must terminate. Then G(m) must terminate too, and it can terminate only when it goes to 0. A common misunderstanding of this proof is to believe that G(m) goes to 0 because it is dominated by P(m). In fact that P(m) dominates G(m) plays no role at all. The important points is: G(m)(k) exists if and only if P(m)(k) exists (parallelism). Then if P(m) terminates, so does G(m). And G(m) can terminate only when it comes to 0.

Each term $P(m)(n)=f(G(m)(n),n+1)$ in the parallel sequence P(m) of functions $p\in P$ is obtained by applying a function f on the term G(m)(n) of the Goodstein sequence of m as follows: take the hereditary base n+1 representation of G(m)(n), and replace each occurrence of the base n+1 with $\omega$ . For example $G(3)(1)=3=2^{1}+2^{0}$ and $P(3)(1)=f(G(3)(1))=\omega ^{1}+\omega ^{0}=\omega +1$ .

The base-changing operation of the Goodstein sequence when going from G(m)(n) to G(m)(n+1) does not change the value of f (that's the main point of the construction), thus after the minus 1 operation, P(m)(n+1) will be strictly smaller than P(m)(n). For example, $f(3\cdot 4^{4^{4}}+4,4)=3\omega ^{\omega ^{\omega }}+\omega =f(3\cdot 5^{5^{5}}+5,5)$ , hence the ordinal $f(3\cdot 4^{4^{4}}+4,4)$ is strictly greater than the ordinal $f((3\cdot 5^{5^{5}}+5)-1,5).$ If the sequence G(m) did not go to 0, it would not terminate and would be infinite (since G(m)(k+1) would always exist). Consequently, P(m) also would be infinite (since in its turn P(m)(k+1) would always exist too). But P(m) is strictly decreasing and the standard order < on ordinals is well-founded, therefore an infinite strictly decreasing sequence cannot exist, or equivalently, every strictly decreasing sequence of ordinals do terminate (and cannot be infinite). This contradiction shows that G(m) terminates, and since it terminates, goes to 0 (by the way, since there exists a natural number k such that G(m)(k)=0, by construction of P(m) we have that P(m)(k)=0).

While this proof of Goodstein's theorem is fairly easy, the KirbyâParis theorem which says that Goodstein's theorem is not a theorem of Peano arithmetic, is technical and considerably more difficult. It makes use of countable nonstandard models of Peano arithmetic. What Kirby showed is that Goodstein's theorem leads to Gentzen's theorem, i.e. it can substitute for induction up to Îµ0.

## Extended Goodstein's theorem

Suppose the definition Goodstein sequence is changed so that instead of replacing each occurrence of the base b with b+1 it was replaces it with b+2. Would the sequence still terminate? More generally, let $b_{1},b_{2},b_{2},...$ be any sequences of integers. Then let the $n+1^{st}$ term G(m)(n+1) of the extended Goodstein sequence of m be as follows: take the hereditary base $b_{n}$ representation of G(m)(n), and replace each occurrence of the base $b_{n}$ with $b_{n+1}$ and then subtract one. The claim is that this sequence still terminates. The extended proof defines $P(m)(n)=f(G(m)(n),n)$ as follows: take the hereditary base $b_{n}$ representation of $G(m)(n)$ , and replace each occurrence of the base $b_{n}$ with the first infinite ordinal number Ï. The base-changing operation of the Goodstein sequence when going from G(m)(n) to G(m)(n+1) still does not change the value of f. For example, if $b_{n}=4$ and if $b_{n+1}=9$ , then $f(3\cdot 4^{4^{4}}+4,4)=3\omega ^{\omega ^{\omega }}+\omega =f(3\cdot 9^{9^{9}}+9,9)$ , hence the ordinal $f(3\cdot 4^{4^{4}}+4,4)$ is strictly greater than the ordinal $f((3\cdot 9^{9^{9}}+9)-1,9).$ ## Sequence length as a function of the starting value

The Goodstein function, ${\mathcal {G}}:\mathbb {N} \to \mathbb {N} \,\!$ , is defined such that ${\mathcal {G}}(n)$ is the length of the Goodstein sequence that starts with n. (This is a total function since every Goodstein sequence terminates.) The extreme growth-rate of ${\mathcal {G}}\,\!$ can be calibrated by relating it to various standard ordinal-indexed hierarchies of functions, such as the functions $H_{\alpha }\,\!$ in the Hardy hierarchy, and the functions $f_{\alpha }\,\!$ in the fast-growing hierarchy of Löb and Wainer:

• Kirby and Paris (1982) proved that
${\mathcal {G}}\,\!$ has approximately the same growth-rate as $H_{\epsilon _{0}}\,\!$ (which is the same as that of $f_{\epsilon _{0}}\,\!$ ); more precisely, ${\mathcal {G}}\,\!$ dominates $H_{\alpha }\,\!$ for every $\alpha <\epsilon _{0}\,\!$ , and $H_{\epsilon _{0}}\,\!$ dominates ${\mathcal {G}}\,\!.$ (For any two functions $f,g:\mathbb {N} \to \mathbb {N} \,\!$ , $f\,\!$ is said to dominate $g\,\!$ if $f(n)>g(n)\,\!$ for all sufficiently large $n\,\!$ .)
• Cichon (1983) showed that
${\mathcal {G}}(n)=H_{R_{2}^{\omega }(n+1)}(1)-1,$ where $R_{2}^{\omega }(n)$ is the result of putting n in hereditary base-2 notation and then replacing all 2s with ω (as was done in the proof of Goodstein's theorem).
${\mathcal {G}}(n)=f_{R_{2}^{\omega }(m_{1})}(f_{R_{2}^{\omega }(m_{2})}(\cdots (f_{R_{2}^{\omega }(m_{k})}(3))\cdots ))-2$ .

Some examples:

## Application to computable functions

Goodstein's theorem can be used to construct a total computable function that Peano arithmetic cannot prove to be total. The Goodstein sequence of a number can be effectively enumerated by a Turing machine; thus the function which maps n to the number of steps required for the Goodstein sequence of n to terminate is computable by a particular Turing machine. This machine merely enumerates the Goodstein sequence of n and, when the sequence reaches 0, returns the length of the sequence. Because every Goodstein sequence eventually terminates, this function is total. But because Peano arithmetic does not prove that every Goodstein sequence terminates, Peano arithmetic does not prove that this Turing machine computes a total function.