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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 eventually terminates at 0. Kirby and Paris[1] 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.[1]

Laurie Kirby and Jeff 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.

## 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, but the usual notation does not suffice for the purposes of Goodstein's theorem.

In 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:

$m = a_k n^k + a_{k-1} n^{k-1} + \cdots + a_0,$

where each coefficient $a_i$ satisfies $0 \le a_i < n$, and $a_k\not=0$. For example, in base 2,

$35 = 32 + 2 + 1 = 2^5 + 2^1 + 2^0.$

Thus the base 2 representation of 35 is $2^5 + 2^1 + 2^0$. (This expression could be written in binary notation as 100011.) Similarly, one can write 100 in base 3:

$100 = 81 + 18 + 1 = 3^4 + 2\cdot 3^2 + 1.$

Note that the exponents themselves are not written in base-n notation. For example, the expressions above include $2^5$ and $3^4$.

To convert a base-n representation to hereditary base n notation, first rewrite all of the exponents in base-n notation. Then rewrite any exponents inside the exponents, and continue in this way until every digit appearing in the expression is n or less.

For example, while 35 in ordinary base-2 notation is $2^5 + 2 + 1$, it is written in hereditary base-2 notation as

$35 = 2^{2^2+1}+2+1,$

using the fact that $5 = 2^2 + 1.$ Similarly, 100 in hereditary base 3 notation is

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

## Goodstein sequences

The Goodstein sequence G(m) of a number m is a sequence of natural numbers. The first element in the sequence G(m) is m itself. 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.

Early Goodstein sequences terminate quickly. For example, G(3) terminates at the sixth step:

Base Hereditary notation Value Notes
2 $2^1 + 1$ 3 Write 3 in base 2 notation
3 $3^1 + 1 - 1 = 3^1$ 3 Switch the 2 to a 3, then subtract 1
4 $4^1 - 1 = 3$ 3 Switch the 3 to a 4, then subtract 1. Now there are no more 4s left
5 $3 - 1 = 2$ 2 No 4s left to switch to 5s. Just subtract 1
6 $2 - 1 = 1$ 1
7 $1 - 1 = 0$ 0

Later Goodstein sequences increase for a very large number of steps. For example, G(4) starts as follows:

Hereditary notation Value
$2^2$ 4
$3^3 - 1 = 2 \cdot 3^2 + 2 \cdot 3 + 2$ 26
$2 \cdot 4^2 + 2 \cdot 4 + 1$ 41
$2 \cdot 5^2 + 2 \cdot 5$ 60
$2 \cdot 6^2 + 2 \cdot 6 - 1 = 2 \cdot 6^2 + 6 + 5$ 83
$2 \cdot 7^2 + 7 + 4$ 109
$\vdots$ $\vdots$
$2 \cdot 11^2 + 11$ 253
$2 \cdot 12^2 + 12 - 1 = 2 \cdot 12^2 + 11$ 299
$\vdots$ $\vdots$

Elements of G(4) continue to increase for a while, but at base $3 \cdot 2^{402653209}$, they reach the maximum of $3 \cdot 2^{402653210} - 1$, stay there for the next $3 \cdot 2^{402653209}$ steps, and then begin their first and final descent.

The value 0 is reached at base $3 \cdot 2^{402653211} - 1$ (curiously, this is a generalized Woodall number: $3 \cdot 2^{402653211} - 1 = 402653184 \cdot 2^{402653184} - 1$. This is also the case with all other final bases for starting values greater than 4[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
$2^{2^2} + 2 + 1$ 19
$3^{3^3} + 3$ 7,625,597,484,990
$4^{4^4} + 3$ $\approx 1.3 \times 10^{154}$
$5^{5^5} + 2$ $\approx 1.8 \times 10^{2184}$
$6^{6^6} + 1$ $\approx 2.6 \times 10^{36,305}$
$7^{7^7}$ $\approx 3.8 \times 10^{695,974}$

$8^{8^8} - 1 = 7 \cdot 8^{(7 \cdot 8^7 + 7 \cdot 8^6 + 7 \cdot 8^5 + 7 \cdot 8^4 + 7 \cdot 8^3 + 7 \cdot 8^2 + 7 \cdot 8 + 7)}$ $+ 7 \cdot 8^{(7 \cdot 8^7 + 7 \cdot 8^6 + 7 \cdot 8^5 + 7 \cdot 8^4 + 7 \cdot 8^3 + 7 \cdot 8^2 + 7 \cdot 8 + 6)} + \cdots$ $+ 7 \cdot 8^{(8+2)} + 7 \cdot 8^{(8+1)} + 7 \cdot 8^8$ $+ 7 \cdot 8^7 + 7 \cdot 8^6 + 7 \cdot 8^5 + 7 \cdot 8^4$ $+ 7 \cdot 8^3 + 7 \cdot 8^2 + 7 \cdot 8 + 7$

$\approx 6 \times 10^{15,151,335}$

$7 \cdot 9^{(7 \cdot 9^7 + 7 \cdot 9^6 + 7 \cdot 9^5 + 7 \cdot 9^4 + 7 \cdot 9^3 + 7 \cdot 9^2 + 7 \cdot 9 + 7)}$ $+ 7 \cdot 9^{(7 \cdot 9^7 + 7 \cdot 9^6 + 7 \cdot 9^5 + 7 \cdot 9^4 + 7 \cdot 9^3 + 7 \cdot 9^2 + 7 \cdot 9 + 6)} + \cdots$ $+ 7 \cdot 9^{(9+2)} + 7 \cdot 9^{(9+1)}+ 7 \cdot 9^9$ $+ 7 \cdot 9^7 + 7 \cdot 9^6 + 7 \cdot 9^5 + 7 \cdot 9^4$ $+ 7 \cdot 9^3 + 7 \cdot 9^2 + 7 \cdot 9 + 6$

$\approx 4.3 \times 10^{369,693,099}$
$\vdots$ $\vdots$

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

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 $4^{4^4} + 4$ 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, $\omega^{\omega^\omega} + \omega$ decreases to $\omega^{\omega^\omega} + 4$ 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.

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.

## Sequence length as a function of the starting value

The Goodstein function, $\mathcal{G}: \mathbb{N} \to \mathbb{N} \,\!$, is defined such that $\mathcal{G}(n)$ is the length of the Goodstein sequence that starts with n. (This is a total function since every Goodstein sequence terminates.) The extreme growth-rate of $\mathcal{G}\,\!$ can be calibrated by relating it to various standard ordinal-indexed hierarchies of functions, such as the functions $H_\alpha\,\!$ in the Hardy hierarchy, and the functions $f_\alpha\,\!$ in the fast-growing hierarchy of Löb and Wainer:

• Kirby and Paris (1982) proved that
$\mathcal{G}\,\!$ has approximately the same growth-rate as $H_{\epsilon_0}\,\!$ (which is the same as that of $f_{\epsilon_0}\,\!$); more precisely, $\mathcal{G}\,\!$ dominates $H_\alpha\,\!$ for every $\alpha < \epsilon_0\,\!$, and $H_{\epsilon_0}\,\!$ dominates $\mathcal{G}\,\!.$
(For any two functions $f, g: \mathbb{N} \to \mathbb{N} \,\!$, $f\,\!$ is said to dominate $g\,\!$ if $f(n) > g(n)\,\!$ for all sufficiently large $n\,\!$.)
• Cichon (1983) showed that
$\mathcal{G}(n) = H_{R_2^\omega(n+1)}(1) - 1,$
where $R_2^\omega(n)$ 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).
• Caicedo (2007) showed that if $n = 2^{m_1} + 2^{m_2} + \cdots + 2^{m_k}$ with $m_1 > m_2 > \cdots > m_k,$ then
$\mathcal{G}(n) = f_{R_2^\omega(m_1)}(f_{R_2^\omega(m_2)}(\cdots(f_{R_2^\omega(m_k)}(3))\cdots)) - 2$.

Some examples:

n $\mathcal{G}(n)$
1 $2^0$ $2 - 1$ $H_\omega(1) - 1$ $f_0(3) - 2$ 2
2 $2^1$ $2^1 + 1 - 1$ $H_{\omega + 1}(1) - 1$ $f_1(3) - 2$ 4
3 $2^1 + 2^0$ $2^2 - 1$ $H_{\omega^\omega}(1) - 1$ $f_1(f_0(3)) - 2$ 6
4 $2^2$ $2^2 + 1 - 1$ $H_{\omega^\omega + 1}(1) - 1$ $f_\omega(3) - 2$ 3·2402653211 − 2
5 $2^2 + 2^0$ $2^2 + 2 - 1$ $H_{\omega^\omega + \omega}(1) - 1$ $f_\omega(f_0(3)) - 2$ > A(4,4)
6 $2^2 + 2^1$ $2^2 + 2 + 1 - 1$ $H_{\omega^\omega + \omega + 1}(1) - 1$ $f_\omega(f_1(3)) - 2$ > A(6,6)
7 $2^2 + 2^1 + 2^0$ $2^{2 + 1} - 1$ $H_{\omega^{\omega + 1}}(1) - 1$ $f_\omega(f_1(f_0(3))) - 2$ > A(8,8)
8 $2^{2 + 1}$ $2^{2 + 1} + 1 - 1$ $H_{\omega^{\omega + 1} + 1}(1) - 1$ $f_{\omega + 1}(3) - 2$ > A3(3,3) = A(A(61, 61), A(61, 61))
$\vdots$
12 $2^{2 + 1} + 2^2$ $2^{2 + 1} + 2^2 + 1 - 1$ $H_{\omega^{\omega + 1} + \omega^\omega + 1}(1) - 1$ $f_{\omega + 1}(f_\omega(3)) - 2$ > fω+1(64) > Graham's number
$\vdots$
19 $2^{2^2} + 2^1 + 2^0$ $2^{2^2} + 2^2 - 1$ $H_{\omega^{\omega^\omega} + \omega^\omega}(1) - 1$ $f_{\omega^\omega}(f_1(f_0(3))) - 2$

## 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.

## References

1. ^ a b Kirby, L.; Paris, J. (1982). "Accessible Independence Results for Peano Arithmetic". Bulletin of the London Mathematical Society 14 (4): 285. doi:10.1112/blms/14.4.285.

## Bibliography

 22 videos foundNext >
 Ridiculously huge numbers (part 11)Continuation of part 10, in which I introduce the full verson of the Goodstein sequence and Goodstein function. (I lean heavily on the very good Wikipedia ar... Goodstein Function in Terms of Fast-Growing Function Hierarchieshttp://demonstrations.wolfram.com/GoodsteinFunctionInTermsOfFastGrowingFunctionHierarchies/ The Wolfram Demonstrations Project contains thousands of free int... Beaker Breaking Demo by Dr. Goodstein, Cal TechCopyright of The Annenberg/CPB project The short clip is from The Mechanical Universe and Beyond by Dr. David L Goodstein at Cal Tech Demo shows the power of... Ridiculously huge numbers (part 10)An introduction to weak goodstein sequences and the weak Goodstein function, which is a toy model of an important example from proof theory. This also serves... Mechanical Universe...and BeyondDavid Goodstein demonstrates making an electric current with a bar magnet. The history of making electricity is traced from windmills to steam engines to Far... Gomory's Theoremhttp://demonstrations.wolfram.com/GomorysTheorem/ The Wolfram Demonstrations Project contains thousands of free interactive visualizations, with new entries ... Godelizationhttp://demonstrations.wolfram.com/Goedelization/ The Wolfram Demonstrations Project contains thousands of free interactive visualizations, with new entries a... Conscious Understanding: What is its Physical Basis?Google Tech Talk March 10, 2010 ABSTRACT Presented by Sir Roger Penrose. Powerful arguments can be given, to support the case that the quality of human under... Ridiculously huge numbers (part 5)Continuation of part 4, introducing (part of) the fast-growing hierarchy. GoedelzahlGrundlage für das Gödel-Paradox ist die Verschlüsselung der Zeichensprache von Zahlentheorie und Aussagenlogik.
 22 videos foundNext >

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