Goodstein's theorem
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In mathematical logic, Goodstein's theorem is a statement about the natural numbers, made by Reuben Goodstein, which states that every Goodstein sequence eventually terminates at 0. Kirby & Paris (1982) 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 "natural" example of a true statement that is unprovable in Peano arithmetic (after Gerhard Gentzen's 1943 direct proof of the unprovability of ε0- induction in Peano arithmetic and the Paris–Harrington theorem). Earlier statements of this type had either been, except for Gentzen, extremely complicated, ad-hoc constructions (such as the statements generated by the construction given in Gödel's incompleteness theorem) or concerned metamathematics or combinatorial results.Template:Harv
Kirby and Paris gave an interpretation of the Goodstein's theorem as a hydra game: the "Hydra" 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. The Kirby-Paris interpretation of the theorem says 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, very long time.
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Definition of a Goodstein sequence
In order to define a Goodstein sequence, first define hereditary base-n notation. To write a natural number in hereditary base-n notation, first write it in the form <math> a_k n^k + a_{k-1} n^{k-1} + \cdots + a_0 </math>, where each <math> 0 \le a_i < n </math>. Then write all the exponents k in hereditary base n notation, and continue recursively until every digit appearing in the expression is n or less.
For example, 35 in ordinary base-2 notation is <math>2^5 + 2 + 1</math>, and in hereditary base-2 notation is
- <math>2^{2^2+1}+2+1. \, </math>
The Goodstein sequence on a number m, notated G(m), is defined as follows: the first element of the sequence is m. To get the next element, write m in hereditary base 2 notation, change all the 2s to 3s, and then subtract 1 from the result; this is the second element of G(m). To get the third element of G(m), write the second element in hereditary base 3 notation, change all 3s to 4s, and subtract 1 again. Continue until the result is zero, at which point the sequence terminates.
Examples of Goodstein sequences
Early Goodstein sequences terminate quickly; for example G(3):
| Base | Hereditary notation | Value | Notes |
|---|---|---|---|
| 2 | <math> 2^1 + 1 </math> | 3 | Write 3 in base 2 notation |
| 3 | <math> 3^1 + 1 - 1 = 3^1 </math> | 3 | Switch the 2 to a 3, then subtract 1 |
| 4 | <math> 4^1 - 1 = 3 </math> | 3 | Switch the 3 to a 4, then subtract 1. Now there are no more 4s left |
| 5 | <math> 3 - 1 = 2 </math> | 2 | No 4s left to switch to 5s. Just subtract 1 |
| 6 | <math> 2 - 1 = 1 </math> | 1 | |
| 7 | <math> 1 - 1 = 0 </math> | 0 |
Many later Goodstein sequences increase for a very large number of steps. For example, G(4) starts as follows:
| Hereditary notation | Value |
|---|---|
| <math> 2^2 </math> | 4 |
| <math> 3^3 - 1 = 2 \cdot 3^2 + 2 \cdot 3 + 2 </math> | 26 |
| <math> 2 \cdot 4^2 + 2 \cdot 4 + 1 </math> | 41 |
| <math> 2 \cdot 5^2 + 2 \cdot 5 </math> | 60 |
| <math> 2 \cdot 6^2 + 2 \cdot 6 - 1 = 2 \cdot 6^2 + 6 + 5 </math> | 83 |
| <math> 2 \cdot 7^2 + 7 + 4 </math> | 109 |
| <math> \vdots </math> | <math> \vdots </math> |
| <math> 2 \cdot 11^2 + 11 </math> | 253 |
| <math> 2 \cdot 12^2 + 12 - 1 = 2 \cdot 12^2 + 11 </math> | 299 |
| <math> \vdots </math> | <math> \vdots </math> |
Elements of G(4) continue to increase for a while, but at base <math>3 \cdot 2^{402653209}</math>, they reach the maximum of <math>3 \cdot 2^{402653210} - 1</math>, stay there for the next <math>3 \cdot 2^{402653209}</math> steps, and then begin their first and final descent.
The value 0 is reached at base <math>3 \cdot 2^{402653211} - 1</math> (curiously, this is a generalized Woodall number: <math>3 \cdot 2^{402653211} - 1 = 402653184 \cdot 2^{402653184} - 1</math>. This is also the case with all other final bases for starting values greater than 4Template:Citation needed).
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 |
|---|---|
| <math> 2^{2^2} + 2 + 1 </math> | 19 |
| <math> 3^{3^3} + 3 </math> | 7,625,597,484,990 |
| <math> 4^{4^4} + 3 </math> | <math> \approx 1.3 \times 10^{154} </math> |
| <math> 5^{5^5} + 2 </math> | <math> \approx 1.8 \times 10^{2184} </math> |
| <math> 6^{6^6} + 1 </math> | <math> \approx 2.6 \times 10^{36,305} </math> |
| <math> 7^{7^7} </math> | <math> \approx 3.8 \times 10^{695,974} </math> |
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<math> 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)}</math> <math>+ 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</math> <math>+ 7 \cdot 8^{(8+2)} + 7 \cdot 8^{(8+1)} + 7 \cdot 8^8 </math> <math>+ 7 \cdot 8^7 + 7 \cdot 8^6 + 7 \cdot 8^5 + 7 \cdot 8^4 </math> <math>+ 7 \cdot 8^3 + 7 \cdot 8^2 + 7 \cdot 8 + 7</math> | <math> \approx 6 \times 10^{15,151,335} </math> |
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<math>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)}</math> <math>+ 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</math> <math>+ 7 \cdot 9^{(9+2)} + 7 \cdot 9^{(9+1)}+ 7 \cdot 9^9 </math> <math>+ 7 \cdot 9^7 + 7 \cdot 9^6 + 7 \cdot 9^5 + 7 \cdot 9^4 </math> <math>+ 7 \cdot 9^3 + 7 \cdot 9^2 + 7 \cdot 9 + 6</math> | <math> \approx 4.3 \times 10^{369,693,099} </math> |
| <math> \vdots </math> | <math> \vdots </math> |
In spite of this rapid growth, Goodstein's theorem states that every Goodstein sequence eventually terminates at 0, no matter what the start value m is.
Proof
Goodstein's theorem can be proved (using techniques outside Peano arithmetic, see below) as follows: Given a Goodstein sequence G(m), we will construct a parallel sequence of ordinal numbers whose elements are no smaller than those in the given sequence. If the elements of the parallel sequence go to 0, the elements of the Goodstein sequence must also go to 0.
To construct the parallel sequence, take the hereditary base n representation of the (n − 1)-th element of the Goodstein sequence, and replace every instance of n with the first infinite ordinal number ω. Addition, multiplication and exponentiation of ordinal numbers is well defined, and the resulting ordinal number clearly cannot be smaller than the original element.
The 'base-changing' operation of the Goodstein sequence does not change the element of the parallel sequence: replacing all the 4s in <math>4^{4^4} + 4</math> with ω is the same as replacing all the 4s with 5s and then replacing all the 5s with ω. The 'subtracting 1' operation, however, corresponds to decreasing the infinite ordinal number in the parallel sequence; for example, <math>\omega^{\omega^\omega} + \omega</math> decreases to <math>\omega^{\omega^\omega} + 4</math> if the step above is performed. Because the ordinals are well-ordered, there are no infinite strictly decreasing sequences of ordinals. Thus the parallel sequence must terminate at 0 after a finite number of steps. The Goodstein sequence, which is bounded above by the parallel sequence, must terminate at 0 also.
While this proof of Goodstein's theorem is fairly easy, the Kirby-Paris theorem which says 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. What Kirby showed is that Goodstein's theorem leads to Gentzen's theorem, i.e. it can substitute for induction up to ε0.
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 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.
See also
References
- Goodstein, R., On the restricted ordinal theorem, Journal of Symbolic Logic, 9 (1944), pp. 33-41.
- Kirby, L. and Paris, J., Accessible independence results for Peano arithmetic, Bulletin London Mathematical Society, 14 (1982), pp. 285-93.
External links
- Template:Mathworld
- Some elements of a proof that Goodstein's theorem is not a theorem of PA, from an undergraduate thesis by Justin T Miller
- Definitions of Goodstein sequences in the programming languages Ruby and Haskell, as well as a large-scale plot
- The Hydra game implemented as a Java appletcs:Goodsteinova věta
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