Goodstein's theorem (original) (raw)

Theorem about natural numbers

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 (as defined below) eventually terminates at 0. Laurence Kirby and Jeff Paris[1] showed that it is unprovable in Peano arithmetic (but it can be proven in stronger systems, such as second-order arithmetic or Zermelo-Fraenkel set theory). 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.[1]

Hereditary base-n notation

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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:

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

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

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

100 = 81 + 18 + 1 = 3 4 + 2 ⋅ 3 2 + 3 0 . {\displaystyle 100=81+18+1=3^{4}+2\cdot 3^{2}+3^{0}.} {\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 ≤ ai < 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

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

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

Goodstein sequences

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The Goodstein sequence G m {\displaystyle G_{m}} {\displaystyle G_{m}} of a number m is a sequence of natural numbers. The first element in the sequence G m {\displaystyle G_{m}} {\displaystyle G_{m}} is m itself. To get the second, G m ( 2 ) {\displaystyle G_{m}(2)} {\displaystyle 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 1 + _n_th term, G m ( n + 1 ) {\displaystyle G_{m}(n+1)} {\displaystyle G_{m}(n+1)}, of the Goodstein sequence of m is as follows:

Early Goodstein sequences terminate quickly. For example, G 3 {\displaystyle G_{3}} {\displaystyle G_{3}} terminates at the 6th step:

Base Hereditary notation Value Notes
2 2 1 + 1 {\displaystyle 2^{1}+1} {\displaystyle 2^{1}+1} 3 Write 3 in base-2 notation
3 3 1 + 1 − 1 = 3 1 {\displaystyle 3^{1}+1-1=3^{1}} {\displaystyle 3^{1}+1-1=3^{1}} 3 Switch the 2 to a 3, then subtract 1
4 4 1 − 1 = 3 {\displaystyle 4^{1}-1=3} {\displaystyle 4^{1}-1=3} 3 Switch the 3 to a 4, then subtract 1. Now there are no more 4's left
5 3 − 1 = 2 {\displaystyle 3-1=2} {\displaystyle 3-1=2} 2 No 4's left to switch to 5's. Just subtract 1
6 2 − 1 = 1 {\displaystyle 2-1=1} {\displaystyle 2-1=1} 1 No 5's left to switch to 6's. Just subtract 1
7 1 − 1 = 0 {\displaystyle 1-1=0} {\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 {\displaystyle G_{4}} {\displaystyle G_{4}} OEIS: A056193 starts as follows:

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

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

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

Hereditary notation Value
2 2 2 + 2 + 1 {\displaystyle 2^{2^{2}}+2+1} {\displaystyle 2^{2^{2}}+2+1} 19
3 3 3 + 3 {\displaystyle 3^{3^{3}}+3} {\displaystyle 3^{3^{3}}+3} 7625597484990
4 4 4 + 3 {\displaystyle 4^{4^{4}}+3} {\displaystyle 4^{4^{4}}+3} ≈ 1.3 × 10 154 {\displaystyle \approx 1.3\times 10^{154}} {\displaystyle \approx 1.3\times 10^{154}}
5 5 5 + 2 {\displaystyle 5^{5^{5}}+2} {\displaystyle 5^{5^{5}}+2} ≈ 1.8 × 10 2 184 {\displaystyle \approx 1.8\times 10^{2\,184}} {\displaystyle \approx 1.8\times 10^{2\,184}}
6 6 6 + 1 {\displaystyle 6^{6^{6}}+1} {\displaystyle 6^{6^{6}}+1} ≈ 2.6 × 10 36 305 {\displaystyle \approx 2.6\times 10^{36\,305}} {\displaystyle \approx 2.6\times 10^{36\,305}}
7 7 7 {\displaystyle 7^{7^{7}}} {\displaystyle 7^{7^{7}}} ≈ 3.8 × 10 695 974 {\displaystyle \approx 3.8\times 10^{695\,974}} {\displaystyle \approx 3.8\times 10^{695\,974}}
8 8 8 − 1 = 7 ⋅ 8 7 ⋅ 8 7 + 7 ⋅ 8 6 + 7 ⋅ 8 5 + 7 ⋅ 8 4 + 7 ⋅ 8 3 + 7 ⋅ 8 2 + 7 ⋅ 8 + 7 {\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 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}} + 7 ⋅ 8 7 ⋅ 8 7 + 7 ⋅ 8 6 + 7 ⋅ 8 5 + 7 ⋅ 8 4 + 7 ⋅ 8 3 + 7 ⋅ 8 2 + 7 ⋅ 8 + 6 + ⋯ {\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^{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 } + 7 ⋅ 8 8 + 2 + 7 ⋅ 8 8 + 1 + 7 ⋅ 8 8 {\displaystyle {}+7\cdot 8^{8+2}+7\cdot 8^{8+1}+7\cdot 8^{8}} {\displaystyle {}+7\cdot 8^{8+2}+7\cdot 8^{8+1}+7\cdot 8^{8}} + 7 ⋅ 8 7 + 7 ⋅ 8 6 + 7 ⋅ 8 5 + 7 ⋅ 8 4 {\displaystyle {}+7\cdot 8^{7}+7\cdot 8^{6}+7\cdot 8^{5}+7\cdot 8^{4}} {\displaystyle {}+7\cdot 8^{7}+7\cdot 8^{6}+7\cdot 8^{5}+7\cdot 8^{4}} + 7 ⋅ 8 3 + 7 ⋅ 8 2 + 7 ⋅ 8 + 7 {\displaystyle {}+7\cdot 8^{3}+7\cdot 8^{2}+7\cdot 8+7} {\displaystyle {}+7\cdot 8^{3}+7\cdot 8^{2}+7\cdot 8+7} ≈ 6.0 × 10 15 151 335 {\displaystyle \approx 6.0\times 10^{15\,151\,335}} {\displaystyle \approx 6.0\times 10^{15\,151\,335}}
7 ⋅ 9 7 ⋅ 9 7 + 7 ⋅ 9 6 + 7 ⋅ 9 5 + 7 ⋅ 9 4 + 7 ⋅ 9 3 + 7 ⋅ 9 2 + 7 ⋅ 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+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+7}} + 7 ⋅ 9 7 ⋅ 9 7 + 7 ⋅ 9 6 + 7 ⋅ 9 5 + 7 ⋅ 9 4 + 7 ⋅ 9 3 + 7 ⋅ 9 2 + 7 ⋅ 9 + 6 + ⋯ {\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^{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 } + 7 ⋅ 9 9 + 2 + 7 ⋅ 9 9 + 1 + 7 ⋅ 9 9 {\displaystyle {}+7\cdot 9^{9+2}+7\cdot 9^{9+1}+7\cdot 9^{9}} {\displaystyle {}+7\cdot 9^{9+2}+7\cdot 9^{9+1}+7\cdot 9^{9}} + 7 ⋅ 9 7 + 7 ⋅ 9 6 + 7 ⋅ 9 5 + 7 ⋅ 9 4 {\displaystyle {}+7\cdot 9^{7}+7\cdot 9^{6}+7\cdot 9^{5}+7\cdot 9^{4}} {\displaystyle {}+7\cdot 9^{7}+7\cdot 9^{6}+7\cdot 9^{5}+7\cdot 9^{4}} + 7 ⋅ 9 3 + 7 ⋅ 9 2 + 7 ⋅ 9 + 6 {\displaystyle {}+7\cdot 9^{3}+7\cdot 9^{2}+7\cdot 9+6} {\displaystyle {}+7\cdot 9^{3}+7\cdot 9^{2}+7\cdot 9+6} ≈ 5.6 × 10 35 942 384 {\displaystyle \approx 5.6\times 10^{35\,942\,384}} {\displaystyle \approx 5.6\times 10^{35\,942\,384}}
⋮ {\displaystyle \vdots } {\displaystyle \vdots } ⋮ {\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.

Proof of Goodstein's theorem

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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).[2] Then if P(m) terminates, so does G(m). By infinite regress, G(m) must reach 0, which guarantees termination.

We define a function f = f ( u , k ) {\displaystyle f=f(u,k)} {\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, f ( 100 , 3 ) = f ( 3 3 1 + 1 + 2 ⋅ 3 2 + 1 , 3 ) = ω ω 1 + 1 + ω 2 ⋅ 2 + 1 = ω ω + 1 + ω 2 ⋅ 2 + 1 {\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} {\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 f ( G ( m ) ( n ) , n + 1 ) > f ( G ( m ) ( n + 1 ) , n + 2 ) {\displaystyle f(G(m)(n),n+1)>f(G(m)(n+1),n+2)} {\displaystyle f(G(m)(n),n+1)>f(G(m)(n+1),n+2)}:

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

Then f ( G ( m ) ( n ) , n + 1 ) = f ( G ′ ( m ) ( n ) , n + 2 ) {\displaystyle f(G(m)(n),n+1)=f(G'(m)(n),n+2)} {\displaystyle f(G(m)(n),n+1)=f(G'(m)(n),n+2)}. Now we apply the minus 1 operation, and f ( G ′ ( m ) ( n ) , n + 2 ) > f ( G ( m ) ( n + 1 ) , n + 2 ) {\displaystyle f(G'(m)(n),n+2)>f(G(m)(n+1),n+2)} {\displaystyle f(G'(m)(n),n+2)>f(G(m)(n+1),n+2)}, as G ′ ( m ) ( n ) = G ( m ) ( n + 1 ) + 1 {\displaystyle G'(m)(n)=G(m)(n+1)+1} {\displaystyle G'(m)(n)=G(m)(n+1)+1}. For example, G ( 4 ) ( 1 ) = 2 2 {\displaystyle G(4)(1)=2^{2}} {\displaystyle G(4)(1)=2^{2}} and G ( 4 ) ( 2 ) = 2 ⋅ 3 2 + 2 ⋅ 3 + 2 {\displaystyle G(4)(2)=2\cdot 3^{2}+2\cdot 3+2} {\displaystyle G(4)(2)=2\cdot 3^{2}+2\cdot 3+2}, so f ( 2 2 , 2 ) = ω ω {\displaystyle f(2^{2},2)=\omega ^{\omega }} {\displaystyle f(2^{2},2)=\omega ^{\omega }} and f ( 2 ⋅ 3 2 + 2 ⋅ 3 + 2 , 3 ) = ω 2 ⋅ 2 + ω ⋅ 2 + 2 {\displaystyle f(2\cdot 3^{2}+2\cdot 3+2,3)=\omega ^{2}\cdot 2+\omega \cdot 2+2} {\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 ω ω − 1 {\displaystyle \omega ^{\omega }-1} {\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,[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.

Extended Goodstein's theorem

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The above proof still works if the definition of the Goodstein sequence is changed so that the base-changing operation replaces each occurrence of the base b with b + 2 instead of b + 1. More generally, let _b_1, _b_2, _b_3, ... be any non-decreasing sequence of integers with _b_1 ≥ 2. Then let the (n + 1)-stterm G(m)(n + 1) of the extended Goodstein sequence of m be as follows:

An simple modification of the above proof shows that this sequence still terminates. For example, if bn = 4 and if b n+1 = 9, then f ( 3 ⋅ 4 4 4 + 4 , 4 ) = 3 ω ω ω + ω = f ( 3 ⋅ 9 9 9 + 9 , 9 ) {\displaystyle f(3\cdot 4^{4^{4}}+4,4)=3\omega ^{\omega ^{\omega }}+\omega =f(3\cdot 9^{9^{9}}+9,9)} {\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 f ( 3 ⋅ 4 4 4 + 4 , 4 ) {\displaystyle f(3\cdot 4^{4^{4}}+4,4)} {\displaystyle f(3\cdot 4^{4^{4}}+4,4)} is strictly greater than the ordinal f ( ( 3 ⋅ 9 9 9 + 9 ) − 1 , 9 ) . {\displaystyle f{\big (}(3\cdot 9^{9^{9}}+9)-1,9{\big )}.} {\displaystyle f{\big (}(3\cdot 9^{9^{9}}+9)-1,9{\big )}.}

The extended version is in fact the one considered in Goodstein's original paper,[3] where Goodstein proved that it is equivalent to the restricted ordinal theorem (i.e. the claim that transfinite induction below ε0 is valid), and gave a finitist proof for the case where m ≤ b 1 b 1 b 1 {\displaystyle m\leq b_{1}^{b_{1}^{b_{1}}}} {\displaystyle m\leq b_{1}^{b_{1}^{b_{1}}}} (equivalent to transfinite induction up to ω ω ω {\displaystyle \omega ^{\omega ^{\omega }}} {\displaystyle \omega ^{\omega ^{\omega }}}).

The extended Goodstein's theorem without any restriction on the sequence bn is not formalizable in Peano arithmetic (PA), since such an arbitrary infinite sequence cannot be represented in PA. This seems to be what kept Goodstein from claiming back in 1944 that the extended Goodstein's theorem is unprovable in PA due to Gödel's second incompleteness theorem and Gentzen's proof of the consistency of PA using ε0-induction.[4] However, inspection of Gentzen's proof shows that it only needs the fact that there is no primitive recursive strictly decreasing infinite sequence of ordinals, so limiting bn to primitive recursive sequences would have allowed Goodstein to prove an unprovability result.[4] Furthermore, with the relatively elementary technique of the Grzegorczyk hierarchy, it can be shown that every primitive recursive strictly decreasing infinite sequence of ordinals can be "slowed down" so that it can be transformed to a Goodstein sequence where b n = n + 1, thus giving an alternative proof to the same result Kirby and Paris proved.[4]

Sequence length as a function of the starting value

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The Goodstein function, G : N → N {\displaystyle {\mathcal {G}}:\mathbb {N} \to \mathbb {N} } {\displaystyle {\mathcal {G}}:\mathbb {N} \to \mathbb {N} }, is defined such that G ( n ) {\displaystyle {\mathcal {G}}(n)} {\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 G {\displaystyle {\mathcal {G}}} {\displaystyle {\mathcal {G}}} can be calibrated by relating it to various standard ordinal-indexed hierarchies of functions, such as the functions H α {\displaystyle H_{\alpha }} {\displaystyle H_{\alpha }} in the Hardy hierarchy, and the functions f α {\displaystyle f_{\alpha }} {\displaystyle f_{\alpha }} in the fast-growing hierarchy of Löb and Wainer:

G {\displaystyle {\mathcal {G}}} {\displaystyle {\mathcal {G}}} has approximately the same growth-rate as H ϵ 0 {\displaystyle H_{\epsilon _{0}}} {\displaystyle H_{\epsilon _{0}}} (which is the same as that of f ϵ 0 {\displaystyle f_{\epsilon _{0}}} {\displaystyle f_{\epsilon _{0}}}); more precisely, G {\displaystyle {\mathcal {G}}} {\displaystyle {\mathcal {G}}} dominates H α {\displaystyle H_{\alpha }} {\displaystyle H_{\alpha }} for every α < ϵ 0 {\displaystyle \alpha <\epsilon _{0}} {\displaystyle \alpha <\epsilon _{0}}, and H ϵ 0 {\displaystyle H_{\epsilon _{0}}} {\displaystyle H_{\epsilon _{0}}} dominates G . {\displaystyle {\mathcal {G}}\,\!.} {\displaystyle {\mathcal {G}}\,\!.}

(For any two functions f , g : N → N {\displaystyle f,g:\mathbb {N} \to \mathbb {N} } {\displaystyle f,g:\mathbb {N} \to \mathbb {N} }, f {\displaystyle f} {\displaystyle f} is said to dominate g {\displaystyle g} {\displaystyle g} if f ( n ) > g ( n ) {\displaystyle f(n)>g(n)} {\displaystyle f(n)>g(n)} for all sufficiently large n {\displaystyle n} {\displaystyle n}.)

G ( n ) = H R 2 ω ( n + 1 ) ( 1 ) − 1 , {\displaystyle {\mathcal {G}}(n)=H_{R_{2}^{\omega }(n+1)}(1)-1,} {\displaystyle {\mathcal {G}}(n)=H_{R_{2}^{\omega }(n+1)}(1)-1,}

where R 2 ω ( n ) {\displaystyle R_{2}^{\omega }(n)} {\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).

G ( n ) = f R 2 ω ( m 1 ) ( f R 2 ω ( m 2 ) ( ⋯ ( f R 2 ω ( m k ) ( 3 ) ) ⋯ ) ) − 2 {\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} {\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 G ( n ) {\displaystyle {\mathcal {G}}(n)} {\displaystyle {\mathcal {G}}(n)}
1 2 0 {\displaystyle 2^{0}} {\displaystyle 2^{0}} 2 − 1 {\displaystyle 2-1} {\displaystyle 2-1} H ω ( 1 ) − 1 {\displaystyle H_{\omega }(1)-1} {\displaystyle H_{\omega }(1)-1} f 0 ( 3 ) − 2 {\displaystyle f_{0}(3)-2} {\displaystyle f_{0}(3)-2} 2
2 2 1 {\displaystyle 2^{1}} {\displaystyle 2^{1}} 2 1 + 1 − 1 {\displaystyle 2^{1}+1-1} {\displaystyle 2^{1}+1-1} H ω + 1 ( 1 ) − 1 {\displaystyle H_{\omega +1}(1)-1} {\displaystyle H_{\omega +1}(1)-1} f 1 ( 3 ) − 2 {\displaystyle f_{1}(3)-2} {\displaystyle f_{1}(3)-2} 4
3 2 1 + 2 0 {\displaystyle 2^{1}+2^{0}} {\displaystyle 2^{1}+2^{0}} 2 2 − 1 {\displaystyle 2^{2}-1} {\displaystyle 2^{2}-1} H ω ω ( 1 ) − 1 {\displaystyle H_{\omega ^{\omega }}(1)-1} {\displaystyle H_{\omega ^{\omega }}(1)-1} f 1 ( f 0 ( 3 ) ) − 2 {\displaystyle f_{1}(f_{0}(3))-2} {\displaystyle f_{1}(f_{0}(3))-2} 6
4 2 2 {\displaystyle 2^{2}} {\displaystyle 2^{2}} 2 2 + 1 − 1 {\displaystyle 2^{2}+1-1} {\displaystyle 2^{2}+1-1} H ω ω + 1 ( 1 ) − 1 {\displaystyle H_{\omega ^{\omega }+1}(1)-1} {\displaystyle H_{\omega ^{\omega }+1}(1)-1} f ω ( 3 ) − 2 {\displaystyle f_{\omega }(3)-2} {\displaystyle f_{\omega }(3)-2} 3·2402653211 − 2 ≈ 6.895080803×10121210694
5 2 2 + 2 0 {\displaystyle 2^{2}+2^{0}} {\displaystyle 2^{2}+2^{0}} 2 2 + 2 − 1 {\displaystyle 2^{2}+2-1} {\displaystyle 2^{2}+2-1} H ω ω + ω ( 1 ) − 1 {\displaystyle H_{\omega ^{\omega }+\omega }(1)-1} {\displaystyle H_{\omega ^{\omega }+\omega }(1)-1} f ω ( f 0 ( 3 ) ) − 2 {\displaystyle f_{\omega }(f_{0}(3))-2} {\displaystyle f_{\omega }(f_{0}(3))-2} > A(4,4) > 10101019727
6 2 2 + 2 1 {\displaystyle 2^{2}+2^{1}} {\displaystyle 2^{2}+2^{1}} 2 2 + 2 + 1 − 1 {\displaystyle 2^{2}+2+1-1} {\displaystyle 2^{2}+2+1-1} H ω ω + ω + 1 ( 1 ) − 1 {\displaystyle H_{\omega ^{\omega }+\omega +1}(1)-1} {\displaystyle H_{\omega ^{\omega }+\omega +1}(1)-1} f ω ( f 1 ( 3 ) ) − 2 {\displaystyle f_{\omega }(f_{1}(3))-2} {\displaystyle f_{\omega }(f_{1}(3))-2} > A(6,6)
7 2 2 + 2 1 + 2 0 {\displaystyle 2^{2}+2^{1}+2^{0}} {\displaystyle 2^{2}+2^{1}+2^{0}} 2 2 + 1 − 1 {\displaystyle 2^{2+1}-1} {\displaystyle 2^{2+1}-1} H ω ω + 1 ( 1 ) − 1 {\displaystyle H_{\omega ^{\omega +1}}(1)-1} {\displaystyle H_{\omega ^{\omega +1}}(1)-1} f ω ( f 1 ( f 0 ( 3 ) ) ) − 2 {\displaystyle f_{\omega }(f_{1}(f_{0}(3)))-2} {\displaystyle f_{\omega }(f_{1}(f_{0}(3)))-2} > A(8,8)
8 2 2 + 1 {\displaystyle 2^{2+1}} {\displaystyle 2^{2+1}} 2 2 + 1 + 1 − 1 {\displaystyle 2^{2+1}+1-1} {\displaystyle 2^{2+1}+1-1} H ω ω + 1 + 1 ( 1 ) − 1 {\displaystyle H_{\omega ^{\omega +1}+1}(1)-1} {\displaystyle H_{\omega ^{\omega +1}+1}(1)-1} f ω + 1 ( 3 ) − 2 {\displaystyle f_{\omega +1}(3)-2} {\displaystyle f_{\omega +1}(3)-2} > _A_3(3,3) = A(A(61, 61), A(61, 61))
⋮ {\displaystyle \vdots } {\displaystyle \vdots }
12 2 2 + 1 + 2 2 {\displaystyle 2^{2+1}+2^{2}} {\displaystyle 2^{2+1}+2^{2}} 2 2 + 1 + 2 2 + 1 − 1 {\displaystyle 2^{2+1}+2^{2}+1-1} {\displaystyle 2^{2+1}+2^{2}+1-1} H ω ω + 1 + ω ω + 1 ( 1 ) − 1 {\displaystyle H_{\omega ^{\omega +1}+\omega ^{\omega }+1}(1)-1} {\displaystyle H_{\omega ^{\omega +1}+\omega ^{\omega }+1}(1)-1} f ω + 1 ( f ω ( 3 ) ) − 2 {\displaystyle f_{\omega +1}(f_{\omega }(3))-2} {\displaystyle f_{\omega +1}(f_{\omega }(3))-2} > _f_ω+1(64) > Graham's number
⋮ {\displaystyle \vdots } {\displaystyle \vdots }
19 2 2 2 + 2 1 + 2 0 {\displaystyle 2^{2^{2}}+2^{1}+2^{0}} {\displaystyle 2^{2^{2}}+2^{1}+2^{0}} 2 2 2 + 2 2 − 1 {\displaystyle 2^{2^{2}}+2^{2}-1} {\displaystyle 2^{2^{2}}+2^{2}-1} H ω ω ω + ω ω ( 1 ) − 1 {\displaystyle H_{\omega ^{\omega ^{\omega }}+\omega ^{\omega }}(1)-1} {\displaystyle H_{\omega ^{\omega ^{\omega }}+\omega ^{\omega }}(1)-1} f ω ω ( f 1 ( f 0 ( 3 ) ) ) − 2 {\displaystyle f_{\omega ^{\omega }}(f_{1}(f_{0}(3)))-2} {\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.)

Application to computable functions

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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.

  1. ^ a b c Kirby & Paris 1982.
  2. ^ Rathjen 2014, lemma 2.2.
  3. ^ Goodstein 1944.
  4. ^ a b c Rathjen 2014.