- Goodstein's theorem
In
mathematical logic , Goodstein's theorem is a statement about thenatural number s made byReuben Goodstein which states that every "Goodstein sequence" eventually terminates at 0. harvtxt|Kirby|Paris|1982 showed that it is unprovable in Peano arithmetic (but it can be proven in stronger systems, such assecond order arithmetic ). This was the first example of a statement unprovable in Peano arithmetic that was not obviously related to mathematical logic. 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. Goodstein's theorem says that the Hydra will eventually be killed, though this may take a very, very long time.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 itin the form , where each is an integer between 0 and "n" − 1; then break up each term into individual powers of "n": becomes . Then write all the exponents "k" in hereditary base "n" notation, and continue recursively until every digit appearing in the expression is "n" or 0 - every non-exponent is "n" and every exponent in a tower of exponents is n or 0 (note that ).
For example, 35 in ordinary base-2 notation is , and in hereditary base-2 notation is
:
The Goodstein sequence on a number "m", notated "G"("m"), is definedas follows: the first element of the sequence is "m". To get the next element, write "m" in hereditary base 2 notation, change all the 2's to 3's, and thensubtract 1 from the result; this is the second element of "G"("m"). To get the third element of "G"("m"), write the previous number in hereditary base 3 notation, change all 3's to 4's, 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 21 + 1 3 The 1 represents 20. 3 31 + 1 − 1 = 3 3 Switch the 2 to a 3, then subtract 1 4 41 − 1 = 1 + 1 + 1 3 Switch the 3 to a 4, and subtract 1. Because the value to be expressed, 3, is less than 4, the representation switches from 41-1 to 40 + 40 + 40, or 1 + 1 + 1 5 1 + 1 + 1 − 1 = 1 + 1 2 Since each of the 1s represents 50, changing the base no longer has an effect. The sequence is now doomed to hit 0. 6 1 + 1 − 1 = 1 1 7 1 − 1 = 0 0 Many later Goodstein sequences increase for a very large number of steps. For example, "G"(4) starts as follows:
Hereditary notation Value 22 4 2·32 + 2·3 + 2 26 2·42 + 2·4 + 1 41 2·52 + 2·5 60 2·62 + 6 + 5 83 2·72 + 7 + 4 109 ... 2·112 + 11 253 2·122 + 11 299 ... Elements of "G"(4) 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 their first and final descent.
The value 0 is reached at base 3 · 2402653211 − 1. However, the example of "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 7625597484990 approximately 1.3 × 10154 approximately 1.8 × 102184 approximately 2.6 × 1036305 approximately 3.8 × 10695974 approximately 6 × 1015151335 approximately 4.3 × 10369693099 ... In spite of this rapid growth, Goodstein's theorem states that every Goodstein sequence eventuallyterminates at 0, no matter what the start value "m" is.
Proof
Goodstein's theorem can be proved (using techniques outside Peanoarithmetic, see below) as follows: Given a Goodstein sequence "G"("m"), we will construct a parallel sequence of
ordinal number s 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, andreplace 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 notchange the element of the parallel sequence: replacing all the 4s in with ω is the same as replacing allthe 4s with 5s and then replacing all the 5s with ω. The'subtracting 1' operation, however, corresponds to decreasing theinfinite ordinal number in the parallel sequence; for example, decreases to 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 aTuring 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.ee also
*
Non-standard arithmetic
*Paris-Harrington theorem References
* Goodstein, R., "On the restricted ordinal theorem", Journal of Symbolic Logic, 9 (1944), 33-41.
* Kirby, L. and Paris, J., "Accessible independence results for Peano arithmetic", Bull. London. Math. Soc., 14 (1982), 285-93.External links
* [http://www.u.arizona.edu/~miller/thesis/node11.html Some elements of a proof that Goodstein's theorem is not a theorem of PA]
* [http://www.cwi.nl/~tromp/pearls.html#goodstein Definitions of Goodstein sequences in the programming languages Ruby and Haskell, as well as a large-scale plot]
* [http://math.andrej.com/2008/02/02/the-hydra-game/ The Hydra game implemented as a Java applet]
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