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^{[1]} showed that it is unprovable in Peano arithmetic (but it can be proven in stronger systems, such as secondorder arithmetic). This was the third example of a true statement that is unprovable in Peano arithmetic, after Gödel's incompleteness theorem and Gerhard Gentzen's 1943 direct proof of the unprovability of ?_{0}induction in Peano arithmetic. The ParisHarrington theorem was a later example.
Laurence Kirby and Jeff Paris introduced a graphtheoretic hydra game with behavior similar to that of Goodstein sequences: the "Hydra" (named for the mythological multiheaded 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.^{[1]}
Goodstein sequences are defined in terms of a concept called "hereditary basen notation". This notation is very similar to usual basen positional notation, but the usual notation does not suffice for the purposes of Goodstein's theorem.
In ordinary basen 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 a_{i} satisfies , and . For example, in base 2,
Thus the base2 representation of 35 is 100011, which means . Similarly, 100 represented in base 3 is 10201:
Note that the exponents themselves are not written in basen notation. For example, the expressions above include 2^{5} and 3^{4}, and 5>2, 4>3.
To convert a basen representation to hereditary basen notation, first rewrite all of the exponents in basen 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 basen notation.
For example, while 35 in ordinary base2 notation is , it is written in hereditary base2 notation as
using the fact that Similarly, 100 in hereditary base3 notation is
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 base2 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:
Early Goodstein sequences terminate quickly. For example, G(3) terminates at the 6th step:
Base  Hereditary notation  Value  Notes 

2  3  Write 3 in base2 notation  
3  3  Switch the 2 to a 3, then subtract 1  
4  3  Switch the 3 to a 4, then subtract 1. Now there are no more 4s left  
5  2  No 4s left to switch to 5s. Just subtract 1  
6  1  No 5s left to switch to 6s. Just subtract 1  
7  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:
Base  Hereditary notation  Value 

2  4  
3  26  
4  41  
5  60  
6  83  
7  109  
11  253  
12  299  
24  1151  
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:
Hereditary notation  Value 

19  




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.
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, basechanging 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 wellfounded, 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 KirbyParis theorem,^{[1]} 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.
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 b_{1}, b_{2}, b_{3}, … be any sequences of integers. Then let the term 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 and then subtract one. The claim is that this sequence still terminates. The extended proof defines 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 ?. The basechanging 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 , then , hence the ordinal is strictly greater than the ordinal
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 growthrate of can be calibrated by relating it to various standard ordinalindexed hierarchies of functions, such as the functions in the Hardy hierarchy, and the functions in the fastgrowing hierarchy of Löb and Wainer:
Some examples:
n  

1  2  
2  4  
3  6  
4  3·2^{402653211}  2 ? 6.895080803×10^{121210694}  
5  > A(4,4)>  
6  > A(6,6)  
7  > A(8,8)  
8  > A^{3}(3,3) = A(A(61, 61), A(61, 61))  
12  > f_{?+1}(64) > Graham's number  
19 
(For Ackermann function and Graham's number bounds see fastgrowing hierarchy#Functions in fastgrowing hierarchies.)
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.