Goodstein's theorem

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Template:Short description 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 ParisTemplate:Sfn showed in 1982 that Goodstein's theorem 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 also 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 by "Hercules" 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.Template:Sfn

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 for natural numbers, 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=aknk+ak−1nk−1+⋯+a0,

where each coefficient ai satisfies Template:Nowrap, and Template:Nowrap.

For example, the base-3 notation of 100:

100=81+18+1=34+2⋅32+30.

Here, the exponents of n themselves are not written in base-n notation, as is seen in the case 34, above.

To convert a base-n notation 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 Template:Nowrap). Then rewrite any exponent inside the exponents again 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, 100 in hereditary base-3 notation is

100=331+1+2⋅32+1.

Goodstein sequences

The Goodstein sequence Gm of a number Template:Mvar is a sequence of natural numbers. The first element in the sequence, written as Gm(1), is Template:Mvar itself. The second element, Gm(2), is obtained by writing Template:Mvar in hereditary base-2 notation, changing all the 2s to 3s, then subtracting 1 from the result. In general, the term Gm(n+1) of the Goodstein sequence of Template:Mvar is computed by:

Gm(n+1) depends both on Gm(n) and on the index Template:Mvar. Sometimes Gm(n) is written as Gn−1(m). [1]

A Goodstein sequence terminates when its element reaches 0. Early Goodstein sequences terminate quickly. For example, G3 terminates at the sixth step (the column labeled "Hereditary notation" shows how the value is calculated):

Base Hereditary notation Value Computation
2 21+1 3
3 31+1−1=31 3 Obtained by writing the last element (3) in hereditary base-2 notation, replacing all 2s with 3s, then subtracting 1.
4 41−1=3 3 Obtained by writing the last element (3) in hereditary base-3 notation, replacing all 3s with 4s, then subtracting 1.
5 3−1=2 2 Obtained by subtracting the last element (3) by 1, since there are no 4s to replace.
6 2−1=1 1 Obtained by subtracting the last element (2) by 1, since there are no 5s to replace.
7 1−1=0 0 Obtained by subtracting the last element (1) by 1, since there are no 6s to replace.

Later Goodstein sequences increase for a very large number of steps. For example, G4 starts as follows (Template:OEIS2C):

Base Hereditary notation Value
2 221 4
3 331−1=2⋅32+2⋅31+2 26
4 2⋅42+2⋅41+1 41
5 2⋅52+2⋅51 60
6 2⋅62+2⋅6−1=2⋅62+61+5 83
7 2⋅72+71+4 109
⋮ ⋮ ⋮
11 2⋅112+111 253
12 2⋅122+121−1=2⋅122+11 299
⋮ ⋮ ⋮
24 2⋅242−1=242+23⋅241+23 1151
⋮ ⋮ ⋮
B=3⋅2402653209−1 2⋅B1 3⋅2402653210−2
B=3⋅2402653209 2⋅B1−1=B1+(B−1) 3⋅2402653210−1
⋮ ⋮ ⋮

Elements of G4 continue to increase for a while, but at base 3⋅2402653209, they reach the maximum of 3⋅2402653210−1, stay there for the next 3⋅2402653209 steps, and then begin descending by 1, reaching 0 when the base reaches 3⋅2402653211−1. The exponent here is equal to 3⋅227+27,[2] so the base is equal to n⋅2n−1, a Woodall number with n=3⋅227.

For another example, G19 increases much more rapidly and starts as follows:

Hereditary notation Value
222+21+1 19
333+31 Template:Val
444+3 ≈1.3×10154
555+2 ≈1.8×102184
666+1 ≈2.6×1036305
777 ≈3.8×10695974

888−1=7⋅87⋅87+7⋅86+7⋅85+7⋅84+7⋅83+7⋅82+7⋅8+7 +7⋅87⋅87+7⋅86+7⋅85+7⋅84+7⋅83+7⋅82+7⋅8+6+⋯ +7⋅88+2+7⋅88+1+7⋅88 +7⋅87+7⋅86+7⋅85+7⋅84 +7⋅83+7⋅82+7⋅81+7

≈6.0×1015151335

7⋅97⋅97+7⋅96+7⋅95+7⋅94+7⋅93+7⋅92+7⋅9+7 +7⋅97⋅97+7⋅96+7⋅95+7⋅94+7⋅93+7⋅92+7⋅9+6+⋯ +7⋅99+2+7⋅99+1+7⋅99 +7⋅97+7⋅96+7⋅95+7⋅94 +7⋅93+7⋅92+7⋅91+6

≈5.6×1035942384
⋮ ⋮

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 Gm, we construct a parallel sequence Pm of ordinal numbers in Cantor normal form that is strictly decreasing and terminates. A common misunderstanding of this proof is to believe that Gm goes to 0 because it is dominated by Pm. Actually, the fact that Pm dominates Gm plays no role at all. The important point is: Gm(k) exists if and only if Pm(k) exists (parallelism), and comparison between two members of Gm is preserved when comparing corresponding entries of Pm.Template:Sfn Then if Pm terminates, so does Gm. By infinite regress, Gm must reach 0, which guarantees termination.

We define a function f=f(u,k) that 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(331+1+2⋅32+1,3)=ωω1+1+ω2⋅2+1=ωω+1+ω2⋅2+1.

Each term Pm(n) of the sequence Pm is then defined as f(Gm(n),n+1). For example, G3(1)=3=21+20 and P3(1)=f(21+20,2)=ω1+ω0=ω+1. Addition, multiplication and exponentiation of ordinal numbers are well defined.

We claim that f(Gm(n),n+1)>f(Gm(n+1),n+2):

Let G'm(n) be Gm(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 Gm(n+1)=G'm(n)−1.

Then f(Gm(n),n+1)=f(G'm(n),n+2).[Note 1] Now we apply the minus 1 operation, and f(G'm(n),n+2)>f(Gm(n+1),n+2), as G'm(n)=Gm(n+1)+1.[Note 2]

For example, G4(1)=22 and G4(2)=2⋅32+2⋅3+2, so f(22,2)=ωω and f(2⋅32+2⋅3+2,3)=ω2⋅2+ω⋅2+2, which is strictly smaller. Note that in order to calculate f(Gm(n),n+1), we first need to write Gm(n) in hereditary base n+1 notation, as for instance the expression ωω−1 is not an ordinal.

Thus the sequence Pm 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 Pm(n) is calculated directly from Gm(n). Hence the sequence Gm must terminate as well, meaning that it must reach 0.

While this proof of Goodstein's theorem is fairly easy, the Kirby–Paris theorem,Template:Sfn 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

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 b1, b2, b3,… be any non-decreasing sequence of integers with b1≥2. Then let the (n+1)st term Gm(n+1) of the extended Goodstein sequence of m be as follows:

  • Take the hereditary base bn representation of Gm(n).
  • Replace each occurrence of the base bn with bn+1.
  • Subtract one.

A simple modification of the above proof shows that this sequence still terminates. For example, if bn=4 and if bn+1=9, then f(3⋅444+4,4)=3ωωω+ω=f(3⋅999+9,9), hence the ordinal f(3⋅444+4,4) is strictly greater than the ordinal f((3⋅999+9)−1,9).

The extended version is in fact the one considered in Goodstein's original paper,Template:Sfn 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≤b1b1b1 (equivalent to transfinite induction up to ωωω).

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.Template:Sfn 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.Template:Sfn 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 bn=n+1, thus giving an alternative proof to the same result Kirby and Paris proved.Template:Sfn

Sequence length as a function of the starting value

The Goodstein function, 𝒢:ℕ→ℕ, is defined such that 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 can be calibrated by relating it to various standard ordinal-indexed hierarchies of functions, such as the functions Hα in the Hardy hierarchy, and the functions fα in the fast-growing hierarchy of Löb and Wainer:

  • Kirby and Paris (1982) proved that
𝒢 has approximately the same growth-rate as Hϵ0 (which is the same as that of fϵ0); more precisely, 𝒢 dominates Hα for every α<ϵ0, and Hϵ0 dominates 𝒢.
(For any two functions f,g:ℕ→ℕ, f is said to dominate g if f(n)>g(n) for all sufficiently large n.)
  • Cichon (1983) showed that
𝒢(n)=HR2ω(n+1)(1)−1,
where R2ω(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 n=2m1+2m2+⋯+2mk with m1>m2>⋯>mk, then
𝒢(n)=fR2ω(m1)(fR2ω(m2)(⋯(fR2ω(mk)(3))⋯))−2.

Some examples:

n 𝒢(n)
1 20 2−1 Hω(1)−1 f0(3)−2 2
2 21 21+1−1 Hω+1(1)−1 f1(3)−2 4
3 21+20 22−1 Hωω(1)−1 f1(f0(3))−2 6
4 22 22+1−1 Hωω+1(1)−1 fω(3)−2 3·2402653211 − 2 ≈ 6.895080803×10121210694
5 22+20 22+2−1 Hωω+ω(1)−1 fω(f0(3))−2 > A(4,4) > 10101019727
6 22+21 22+2+1−1 Hωω+ω+1(1)−1 fω(f1(3))−2 > A(6,6)
7 22+21+20 22+1−1 Hωω+1(1)−1 fω(f1(f0(3)))−2 > A(8,8)
8 22+1 22+1+1−1 Hωω+1+1(1)−1 fω+1(3)−2 > A3(3,3) = A(A(61, 61), A(61, 61))
⋮
12 22+1+22 22+1+22+1−1 Hωω+1+ωω+1(1)−1 fω+1(fω(3))−2 > fω+1(64) > Graham's number
⋮
19 222+21+20 222+22−1 Hωωω+ωω(1)−1 fωω(f1(f0(3)))−2

(For Ackermann function and Graham's number bounds see Template:Section link.)

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 that 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.

See also

Notes

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References

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Bibliography


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