Goodstein s theorem

In mathematical logic, Goodstein's theorem is a statement about the natural numbers that is undecidable in Peano arithmetic but can be proven to be true using the stronger axiom system of set theory, in particular using the axiom of infinity. The theorem states that every Goodstein sequence eventually terminates at 0. It stands as an example that not all undecidable theorems are peculiar or contrived, as those constructed by Gödel's incompleteness theorem are sometimes considered.

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, we first write it in the form a_kn^k + a_{k - 1}n^{k - 1} + ... + a₀, where each a_i is an integer between 0 and n-1; we then recursively write all the exponents in hereditary base n notation. For example, 35 in ordinary base-2 notation is 2⁵ + 2 + 1, and in hereditary base-2 notation is

$2^{2^2+1}+2+1$ .

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 2's to 3's, 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 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):

Hereditary notation	Value
2+1	3
3+1 − 1 = 3	3
4 − 1 = 1+1+1	3
1+1+1 − 1 = 1+1	2
1+1 − 1 = 1	1
1 − 1 = 0	0

Elements of later Goodstein sequences appear to increase for a substantial time. 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 * 2^402653209, they reach the maximum of 3 * 2^402653210-1, stay there for the next 3 * 2^402653209 steps, and then begin their first and final descent.

The value 0 is reached at base 3 * 2^402653211 - 1, which, curiously, is a Woodall number, just as all other final bases for starting values greater than 4. 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
$2^{2^2}+2+1$	19
$3^{3^3}+3$	7625597484990
$4^{4^4}+3$	approximately 1.3 × 10¹⁵⁴
$5^{5^5}+2$	approximately 1.8 × 10²¹⁸⁴
$6^{6^6}+1$	approximately 2.6 × 10³⁶³⁰⁵
$7^{7^7}$	approximately 3.8 × 10⁶⁹⁵⁹⁷⁴
$7 \times 8^{(7 \times 8^7 + 7 \times 8^6 + 7 \times 8^5 + 7 \times 8^4 + 7 \times 8^3 + 7 \times 8^2 + 7 \times 8 + 7)}$ $+ 7 \times 8^{(7 \times 8^7 + 7 \times 8^6 + 7 \times 8^5 + 7 \times 8^4 + 7 \times 8^3 + 7 \times 8^2 + 7 \times 8 + 6)}$ $+ ... \,\;$ $+ 7 \times 8^{(8+2)} + 7 \times 8^{(8+1)} + 7 \times 8^8 + 7 \times 8^7 + 7 \times 8^6$ $+ 7 \times 8^5 + 7 \times 8^4 + 7 \times 8^3 + 7 \times 8^2 + 7 \times 8 + 7$	approximately 6 × 10^15151335
$7 \times 9^{(7 \times 9^7 + 7 \times 9^6 + 7 \times 9^5 + 7 \times 9^4 + 7 \times 9^3 + 7 \times 9^2 + 7 \times 9 + 7)}$ $+ 7 \times 9^{(7 \times 9^7 + 7 \times 9^6 + 7 \times 9^5 + 7 \times 9^4 + 7 \times 9^3 + 7 \times 9^2 + 7 \times 9 + 6)}$ $+ ... \,\;$ $+ 7 \times 9^{(9+2)} + 7 \times 9^{(9+1)} + 7 \times 9^9 + 7 \times 9^7 + 7 \times 9^6$ $+ 7 \times 9^5 + 7 \times 9^4 + 7 \times 9^3 + 7 \times 9^2 + 7 \times 9 + 6$	approximately 4.3 × 10^369693099
...

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 4's in 4^(4^4)+4 with ω is the same as replacing all the 4's with 5's and then replacing all the 5's with ω. The 'subtracting 1' operation, however, corresponds to decreasing the infinite ordinal number in the parallel sequence; for example, ω^(ω^ω) + ω decreases to ω^(ω^ω) + 4 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.

Need a proof sketch that the theorem cannot be proved in PA

(a link to a detailed proof) http://www.u.arizona.edu/~miller/thesis/node11.html

Here is an excellent online seminar about this theorem and its significance. http://www2.maths.bris.ac.uk/~maadb/research/seminars/online/fgfut/index.html

Categories: Mathematical logic | Set theory | Theorems

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