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Laurence Kirby and Jeff Paris introduced a graph-theoretic hydra game with behavior similar to that of Goodstein sequences: the "Hydra" (named for the mythological multi-headed Hydra of Lerna) 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 long time. Just like for Goodstein sequences, Kirby and Paris showed that it cannot be proven in Peano arithmetic alone.
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:
where each coefficient ai satisfies , and . For example, in base 2,
Thus the base-2 representation of 35 is 100011, which means . Similarly, 100 represented in base 3 is 10201:
Note that the exponents themselves are not written in base-n notation. For example, the expressions above include 25 and 34, and 5>2, 4>3.
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 number appearing in the expression (except the bases themselves) has been converted to base-n notation.
For example, while 35 in ordinary base-2 notation is , it is written in hereditary base-2 notation as
using the fact that Similarly, 100 in hereditary base-3 notation is
The Goodstein sequenceG(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 term of the Goodstein sequence of m is as follows:
Take the hereditary base- representation of G(m)(n).
Replace each occurrence of the base- with .
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 6th step:
Write 3 in base-2 notation
Switch the 2 to a 3, then subtract 1
Switch the 3 to a 4, then subtract 1. Now there are no more 4s left
No 4s left to switch to 5s. Just subtract 1
No 5s left to switch to 6s. Just subtract 1
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 ,
they reach the maximum of , stay there for the next steps, and then begin their first descent.
The value 0 is reached at base . (Curiously, this is a Woodall number: . This is also the case with all other final bases for starting values greater than 4.)
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) as follows: Given a Goodstein sequence G(m), we construct a parallel sequence P(m) of ordinal numbers which is strictly decreasing and terminates. 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). Actually, the fact that P(m) dominates G(m) plays no role at all. The important point 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.
We define a function which computes the hereditary base representation of and then replaces each occurrence of the base with the first infinite ordinal number ?. For example, .
Each term P(m)(n) of the sequence P(m) is then defined as f(G(m)(n),n+1). For example, and . Addition, multiplication and exponentiation of ordinal numbers are well defined.
We claim that :
Let be G(m)(n) after applying the first,
base-changing operation in generating the next element of the Goodstein sequence,
but before the second minus 1 operation in this generation.
Observe that .
Then clearly, . Now we apply the minus 1 operation, and , as .
For example, and , so and , which is strictly smaller. Note that in order to calculate f(G(m)(n),n+1), we first need to write G(m)(n) in hereditary base notation, as for instance the expression either makes no sense or is equal to .
Thus the sequence P(m) is strictly decreasing. As the standard order < on ordinals is well-founded, an infinite strictly decreasing sequence cannot exist, or equivalently, every strictly decreasing sequence of ordinals terminates (and cannot be infinite). But P(m)(n) is calculated directly from G(m)(n). Hence the sequence G(m) must terminate as well, meaning that it must reach 0.
While this proof of Goodstein's theorem is fairly easy, the Kirby-Paris theorem, which shows 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.
Extended Goodstein's theorem
Suppose the definition of the Goodstein sequence is changed so that instead of
replacing each occurrence of the base b with
it replaces it with . Would the sequence still terminate?
More generally, let b1, b2, b3, … be any sequences of integers.
Then let the
term of the extended Goodstein sequence of m be as
follows: take the hereditary base bn representation of
G(m)(n), and replace each occurrence of the base bn
with and then subtract one.
The claim is that this sequence still terminates.
The extended proof defines as
follows: take the hereditary base bn representation of
G(m)(n), and replace each occurrence of the base
bn with the first infinite ordinal number ?.
The base-changing operation of the Goodstein sequence when going
from G(m)(n) to still does not change the value of f.
For example, if and if ,
, hence the ordinal is strictly greater than the ordinal
Sequence length as a function of the starting value
The Goodstein function, , is defined such that 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 can be calibrated by relating it to various standard ordinal-indexed hierarchies of functions, such as the functions in the Hardy hierarchy, and the functions in the fast-growing hierarchy of Löb and Wainer:
Kirby and Paris (1982) proved that
has approximately the same growth-rate as (which is the same as that of ); more precisely, dominates for every , and dominates
(For any two functions , is said to dominate if for all sufficiently large .)
Cichon (1983) showed that
where is the result of putting n in hereditary base-2 notation and then replacing all 2s with ? (as was done in the proof of Goodstein's theorem).
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.
Cichon, E. (1983), "A Short Proof of Two Recently Discovered Independence Results Using Recursive Theoretic Methods", Proceedings of the American Mathematical Society, 87 (4): 704-706, doi:10.2307/2043364, JSTOR2043364.