Goodstein's Theorem

Laurence Kirby and Jeff 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 example of a true statement about natural numbers that is unprovable in Peano arithmetic, after the examples provided by Gödel's incompleteness theorem and Gerhard Gentzen's 1943 direct proof of the unprovability of ε₀-induction in Peano arithmetic. The Paris–Harrington theorem gave another example.

Kirby and 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.

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.

To achieve the 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:

${\displaystyle m=a_{k}n^{k}+a_{k-1}n^{k-1}+\cdots +a_{0},}$ 

where each coefficient a_i satisfies 0 ≤ a_i < n, and a_k ≠ 0. For example, to achieve the base 2 notation, one writes

${\displaystyle 35=32+2+1=2^{5}+2^{1}+2^{0}.}$ 

Thus the base-2 representation of 35 is 100011, which means 2⁵ + 2 + 1. Similarly, 100 represented in base-3 is 10201:

${\displaystyle 100=81+18+1=3^{4}+2\cdot 3^{2}+3^{0}.}$ 

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 notation (which is a step in achieving base-n representation) to a hereditary base-n notation, first rewrite all of the exponents as a sum of powers of n (with the limitation on the coefficients 0 ≤ a_i < n). Then rewrite any exponent inside the exponents in base-n notation (with the same limitation on the coefficients), and continue in this way until every number appearing in the expression (except the bases themselves) is written in base-n notation.

For example, while 35 in ordinary base-2 notation is 2⁵ + 2 + 1, it is written in hereditary base-2 notation as

${\displaystyle 35=2^{2^{2^{1}}+1}+2^{1}+1,}$ 

using the fact that 5 = 2²¹ + 1. Similarly, 100 in hereditary base-3 notation is

${\displaystyle 100=3^{3^{1}+1}+2\cdot 3^{2}+1.}$ 

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).
• Replace each occurrence of the base-n + 1 with n + 2.
• 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:

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

Later Goodstein sequences increase for a very large number of steps. For example, G(4) OEIS: A056193 starts as follows:

Base	Hereditary notation	Value
2	${\displaystyle 2^{2^{1}}}$	4
3	${\displaystyle 3^{3^{1}}-1=2\cdot 3^{2}+2\cdot 3+2}$	26
4	${\displaystyle 2\cdot 4^{2}+2\cdot 4+1}$	41
5	${\displaystyle 2\cdot 5^{2}+2\cdot 5}$	60
6	${\displaystyle 2\cdot 6^{2}+2\cdot 6-1=2\cdot 6^{2}+6+5}$	83
7	${\displaystyle 2\cdot 7^{2}+7+4}$	109
${\displaystyle \vdots }$	${\displaystyle \vdots }$	${\displaystyle \vdots }$
11	${\displaystyle 2\cdot 11^{2}+11}$	253
12	${\displaystyle 2\cdot 12^{2}+12-1=2\cdot 12^{2}+11}$	299
${\displaystyle \vdots }$	${\displaystyle \vdots }$	${\displaystyle \vdots }$
24	${\displaystyle 2\cdot 24^{2}-1=24^{2}+23\cdot 24+23}$	1151
${\displaystyle \vdots }$	${\displaystyle \vdots }$	${\displaystyle \vdots }$
${\displaystyle B=3\cdot 2^{402\,653\,209}-1}$	${\displaystyle 2\cdot B^{1}}$	${\displaystyle 3\cdot 2^{402\,653\,210}-2}$
${\displaystyle B=3\cdot 2^{402\,653\,209}}$	${\displaystyle 2\cdot B^{1}-1=B^{1}+(B-1)}$	${\displaystyle 3\cdot 2^{402\,653\,210}-1}$
${\displaystyle \vdots }$	${\displaystyle \vdots }$	${\displaystyle \vdots }$

Elements of G(4) continue to increase for a while, but at base ${\displaystyle 3\cdot 2^{402\,653\,209}}$ , they reach the maximum of ${\displaystyle 3\cdot 2^{402\,653\,210}-1}$ , stay there for the next ${\displaystyle 3\cdot 2^{402\,653\,209}}$  steps, and then begin their descent.

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
${\displaystyle 2^{2^{2}}+2+1}$	19
${\displaystyle 3^{3^{3}}+3}$	7625597484990
${\displaystyle 4^{4^{4}}+3}$	${\displaystyle \approx 1.3\times 10^{154}}$
${\displaystyle 5^{5^{5}}+2}$	${\displaystyle \approx 1.8\times 10^{2\,184}}$
${\displaystyle 6^{6^{6}}+1}$	${\displaystyle \approx 2.6\times 10^{36\,305}}$
${\displaystyle 7^{7^{7}}}$	${\displaystyle \approx 3.8\times 10^{695\,974}}$
${\displaystyle 8^{8^{8}}-1=7\cdot 8^{7\cdot 8^{7}+7\cdot 8^{6}+7\cdot 8^{5}+7\cdot 8^{4}+7\cdot 8^{3}+7\cdot 8^{2}+7\cdot 8+7}}$  ${\displaystyle {}+7\cdot 8^{7\cdot 8^{7}+7\cdot 8^{6}+7\cdot 8^{5}+7\cdot 8^{4}+7\cdot 8^{3}+7\cdot 8^{2}+7\cdot 8+6}+\cdots }$  ${\displaystyle {}+7\cdot 8^{8+2}+7\cdot 8^{8+1}+7\cdot 8^{8}}$  ${\displaystyle {}+7\cdot 8^{7}+7\cdot 8^{6}+7\cdot 8^{5}+7\cdot 8^{4}}$  ${\displaystyle {}+7\cdot 8^{3}+7\cdot 8^{2}+7\cdot 8+7}$	${\displaystyle \approx 6.0\times 10^{15\,151\,335}}$
${\displaystyle 7\cdot 9^{7\cdot 9^{7}+7\cdot 9^{6}+7\cdot 9^{5}+7\cdot 9^{4}+7\cdot 9^{3}+7\cdot 9^{2}+7\cdot 9+7}}$  ${\displaystyle {}+7\cdot 9^{7\cdot 9^{7}+7\cdot 9^{6}+7\cdot 9^{5}+7\cdot 9^{4}+7\cdot 9^{3}+7\cdot 9^{2}+7\cdot 9+6}+\cdots }$  ${\displaystyle {}+7\cdot 9^{9+2}+7\cdot 9^{9+1}+7\cdot 9^{9}}$  ${\displaystyle {}+7\cdot 9^{7}+7\cdot 9^{6}+7\cdot 9^{5}+7\cdot 9^{4}}$  ${\displaystyle {}+7\cdot 9^{3}+7\cdot 9^{2}+7\cdot 9+6}$	${\displaystyle \approx 5.6\times 10^{35\,942\,384}}$
${\displaystyle \vdots }$	${\displaystyle \vdots }$

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 in Cantor normal form which is strictly decreasing and terminates. 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), and comparison between two members of G(m) is preserved when comparing corresponding entries of P(m). Then if P(m) terminates, so does G(m). By infinite regress, G(m) must reach 0, which guarantees termination.

We define a function ${\displaystyle f=f(u,k)}$  which computes the hereditary base k representation of u and then replaces each occurrence of the base k with the first infinite ordinal number ω. For example, ${\displaystyle f(100,3)=f(3^{3^{1}+1}+2\cdot 3^{2}+1,3)=\omega ^{\omega ^{1}+1}+\omega ^{2}\cdot 2+1=\omega ^{\omega +1}+\omega ^{2}\cdot 2+1}$ .

Each term P(m)(n) of the sequence P(m) is then defined as f(G(m)(n),n+1). For example, G(3)(1) = 3 = 2¹ + 2⁰ and P(3)(1) = f(2¹ + 2⁰,2) = ω¹ + ω⁰ = ω + 1. Addition, multiplication and exponentiation of ordinal numbers are well defined.

We claim that ${\displaystyle f(G(m)(n),n+1)>f(G(m)(n+1),n+2)}$ :

Let ${\displaystyle G'(m)(n)}$  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 ${\displaystyle G(m)(n+1)=G'(m)(n)-1}$ .

Then ${\displaystyle f(G(m)(n),n+1)=f(G'(m)(n),n+2)}$ . Now we apply the minus 1 operation, and ${\displaystyle f(G'(m)(n),n+2)>f(G(m)(n+1),n+2)}$ , as ${\displaystyle G'(m)(n)=G(m)(n+1)+1}$ . For example, ${\displaystyle G(4)(1)=2^{2}}$  and ${\displaystyle G(4)(2)=2\cdot 3^{2}+2\cdot 3+2}$ , so ${\displaystyle f(2^{2},2)=\omega ^{\omega }}$  and ${\displaystyle f(2\cdot 3^{2}+2\cdot 3+2,3)=\omega ^{2}\cdot 2+\omega \cdot 2+2}$ , 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 n+1 notation, as for instance the expression ${\displaystyle \omega ^{\omega }-1}$  is not an ordinal.

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.

Suppose the definition of the Goodstein sequence is changed so that instead of replacing each occurrence of the base b with b + 1 it replaces it with b + 2. Would the sequence still terminate? More generally, let b₁, b₂, b₃, ... 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 ω. 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 ${\displaystyle f(3\cdot 4^{4^{4}}+4,4)=3\omega ^{\omega ^{\omega }}+\omega =f(3\cdot 9^{9^{9}}+9,9)}$ , hence the ordinal ${\displaystyle f(3\cdot 4^{4^{4}}+4,4)}$  is strictly greater than the ordinal ${\displaystyle f{\big (}(3\cdot 9^{9^{9}}+9)-1,9{\big )}.}$ 

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

• Kirby and Paris (1982) proved that

${\displaystyle {\mathcal {G}}}$  has approximately the same growth-rate as ${\displaystyle H_{\epsilon _{0}}}$  (which is the same as that of ${\displaystyle f_{\epsilon _{0}}}$ ); more precisely, ${\displaystyle {\mathcal {G}}}$  dominates ${\displaystyle H_{\alpha }}$  for every ${\displaystyle \alpha <\epsilon _{0}}$ , and ${\displaystyle H_{\epsilon _{0}}}$  dominates ${\displaystyle {\mathcal {G}}\,\!.}$ 
(For any two functions ${\displaystyle f,g:\mathbb {N} \to \mathbb {N} }$ , ${\displaystyle f}$  is said to dominate ${\displaystyle g}$  if ${\displaystyle f(n)>g(n)}$  for all sufficiently large ${\displaystyle n}$ .)

• Cichon (1983) showed that

${\displaystyle {\mathcal {G}}(n)=H_{R_{2}^{\omega }(n+1)}(1)-1,}$ 
where ${\displaystyle R_{2}^{\omega }(n)}$  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).

• Caicedo (2007) showed that if ${\displaystyle n=2^{m_{1}}+2^{m_{2}}+\cdots +2^{m_{k}}}$  with ${\displaystyle m_{1}>m_{2}>\cdots >m_{k},}$  then

${\displaystyle {\mathcal {G}}(n)=f_{R_{2}^{\omega }(m_{1})}(f_{R_{2}^{\omega }(m_{2})}(\cdots (f_{R_{2}^{\omega }(m_{k})}(3))\cdots ))-2}$ .

Some examples:

n			${\displaystyle {\mathcal {G}}(n)}$
1	${\displaystyle 2^{0}}$	${\displaystyle 2-1}$	${\displaystyle H_{\omega }(1)-1}$	${\displaystyle f_{0}(3)-2}$	2
2	${\displaystyle 2^{1}}$	${\displaystyle 2^{1}+1-1}$	${\displaystyle H_{\omega +1}(1)-1}$	${\displaystyle f_{1}(3)-2}$	4
3	${\displaystyle 2^{1}+2^{0}}$	${\displaystyle 2^{2}-1}$	${\displaystyle H_{\omega ^{\omega }}(1)-1}$	${\displaystyle f_{1}(f_{0}(3))-2}$	6
4	${\displaystyle 2^{2}}$	${\displaystyle 2^{2}+1-1}$	${\displaystyle H_{\omega ^{\omega }+1}(1)-1}$	${\displaystyle f_{\omega }(3)-2}$	3·2402653211 − 2 ≈ 6.895080803×10121210694
5	${\displaystyle 2^{2}+2^{0}}$	${\displaystyle 2^{2}+2-1}$	${\displaystyle H_{\omega ^{\omega }+\omega }(1)-1}$	${\displaystyle f_{\omega }(f_{0}(3))-2}$	> A(4,4) > 10101019727
6	${\displaystyle 2^{2}+2^{1}}$	${\displaystyle 2^{2}+2+1-1}$	${\displaystyle H_{\omega ^{\omega }+\omega +1}(1)-1}$	${\displaystyle f_{\omega }(f_{1}(3))-2}$	> A(6,6)
7	${\displaystyle 2^{2}+2^{1}+2^{0}}$	${\displaystyle 2^{2+1}-1}$	${\displaystyle H_{\omega ^{\omega +1}}(1)-1}$	${\displaystyle f_{\omega }(f_{1}(f_{0}(3)))-2}$	> A(8,8)
8	${\displaystyle 2^{2+1}}$	${\displaystyle 2^{2+1}+1-1}$	${\displaystyle H_{\omega ^{\omega +1}+1}(1)-1}$	${\displaystyle f_{\omega +1}(3)-2}$	> A3(3,3) = A(A(61, 61), A(61, 61))
${\displaystyle \vdots }$
12	${\displaystyle 2^{2+1}+2^{2}}$	${\displaystyle 2^{2+1}+2^{2}+1-1}$	${\displaystyle H_{\omega ^{\omega +1}+\omega ^{\omega }+1}(1)-1}$	${\displaystyle f_{\omega +1}(f_{\omega }(3))-2}$	> fω+1(64) > Graham's number
${\displaystyle \vdots }$
19	${\displaystyle 2^{2^{2}}+2^{1}+2^{0}}$	${\displaystyle 2^{2^{2}}+2^{2}-1}$	${\displaystyle H_{\omega ^{\omega ^{\omega }}+\omega ^{\omega }}(1)-1}$	${\displaystyle f_{\omega ^{\omega }}(f_{1}(f_{0}(3)))-2}$

(For Ackermann function and Graham's number bounds see fast-growing hierarchy#Functions in fast-growing 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.

• Non-standard model of arithmetic
• Fast-growing hierarchy
• Paris–Harrington theorem
• Kanamori–McAloon theorem
• Kruskal's tree theorem