Goodstein's theorem

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 Paris 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.

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:

:<math>m = a_k n^k + a_{k-1} n^{k-1} + \cdots + a_0,</math>

where each coefficient ai satisfies , and .

For example, the base-3 notation of 100:

:<math>100 = 81 + 18 + 1 = 3^4 + 2 \cdot 3^2 + 3^0.</math>

Note that 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 ). 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

:<math>100 = 3^{3^1+1} + 2 \cdot 3^2 + 1.</math>

Goodstein sequences

The Goodstein sequence <math>G_m</math> of a number m is a sequence of natural numbers. The first element in the sequence, written as <math>G_m(1)</math>, is m itself. To get the second, <math>G_m (2)</math>, write m in hereditary base-2 notation, change all the 2s to 3s, and then subtract 1 from the result. In general, the term <math>G_m (n+1)</math> of the Goodstein sequence of m is computed as follows:

Take the hereditary base-() representation of <math>G_m (n)</math>.
Replace each occurrence of the base () with .
Subtract one.

Note that <math>G_m (n+1)</math> depends both on <math>G_m (n)</math> and on the index n. Note also that sometimes what we have called <math>G_m(n)</math> is denoted <math>G_{n-1}(m).</math>

Now continue computing further values in the Goodstein sequence, until you reach 0, at which point the sequence terminates.

Early Goodstein sequences terminate quickly. For example, <math>G_3</math> terminates at the sixth step (the column labeled "Hereditary notation" first shows how the value is calculated):

{| class="wikitable" border="1"

! Base !! Hereditary notation !! Value !! Notes

| 2 || <math> 2^1 + 1 </math>|| 3 || <math>G_3(1).</math> Write 3 in hereditary base-2 notation

| 3 || <math> 3^1 + 1 - 1 = 3^1 </math>|| 3 || <math>G_3(2).</math> Switch the 2s to 3s, then subtract 1. Write in hereditary base-3 notation.

| 4 || <math> 4^1 - 1 = 3 </math>|| 3 || Switch the 3s to 4s, then subtract 1. Now there are no more 4s left

| 5 || <math> 3^1 - 1 = 2 </math>|| 2 || No 4s left to switch to 5s. Just subtract 1

| 6 || <math> 2^1 - 1 = 1 </math>|| 1 || No 5s left to switch to 6s. Just subtract 1

| 7 || <math> 1^1 - 1 = 0 </math>|| 0 || No 6s left to switch to 7s. Just subtract 1

Later Goodstein sequences increase for a very large number of steps. For example, <math>G_4</math> starts as follows:

{| class="wikitable" border="1"

! Base !! Hereditary notation !! Value

| 2 || <math> 2^{2^1} </math> || 4

| 3 || <math> 3^{3^1} - 1 = 2 \cdot 3^2 + 2 \cdot 3^1 + 2 </math>|| 26

| 4 || <math> 2 \cdot 4^2 + 2 \cdot 4^1 + 1 </math>|| 41

| 5 || <math> 2 \cdot 5^2 + 2 \cdot 5^1 </math>|| 60

| 6 || <math> 2 \cdot 6^2 + 2 \cdot 6 - 1 = 2 \cdot 6^2 + 6^1 + 5 </math>|| 83

| 7 || <math> 2 \cdot 7^2 + 7^1 + 4 </math>|| 109

|- align="center"

| <math> \vdots </math> || <math> \vdots </math> || <math> \vdots </math>

| 11 || <math> 2 \cdot 11^2 + 11^1 </math>|| 253

| 12 || <math> 2 \cdot 12^2 + 12^1 - 1 = 2 \cdot 12^2 + 11 </math>|| 299

|- align="center"

| <math> \vdots </math> || <math> \vdots </math> || <math> \vdots </math>

| 24 || <math> 2 \cdot 24^2 - 1 = 24^2 + 23 \cdot 24^1 + 23 </math>|| 1151

|- align="center"

| <math> \vdots </math> || <math> \vdots </math> || <math> \vdots </math>

| <math> B = 3 \cdot 2^{402\,653\,209} - 1 </math> || <math> 2 \cdot B^1 </math> || <math> 3 \cdot 2^{402\,653\,210} - 2 </math>

| <math> B = 3 \cdot 2^{402\,653\,209} </math> || <math> 2 \cdot B^1 - 1 = B^1 + (B-1) </math> || <math> 3 \cdot 2^{402\,653\,210} - 1 </math>

|- align="center"

| <math> \vdots </math> || <math> \vdots </math> || <math> \vdots </math>

Elements of <math>G_4</math> continue to increase for a while, but at base <math>3 \cdot 2^{402\,653\,209}</math>,

they reach the maximum of <math>3 \cdot 2^{402\,653\,210} - 1</math>, stay there for the next <math>3 \cdot 2^{402\,653\,209}</math> steps, and then begin descending by 1, reaching 0 when the base reaches <math>3\cdot 2^{402\,653\,211}-1.</math> The exponent here is equal to <math>3\cdot 2^{27}+27,</math> so the base is equal to <math>n\cdot 2^n-1,</math> a Woodall number with <math>n=3\cdot 2^{27}.</math>

However, even <math>G_4</math> does not give a good idea of just how quickly the elements of a Goodstein sequence can increase.

<math>G_{19}</math> increases much more rapidly and starts as follows:

{| class="wikitable" border="1"

! Hereditary notation !! Value

| <math> 2^{2^2} + 2^1 + 1 </math>|| 19

| <math> 3^{3^3} + 3^1 </math>||

| <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^{2\,184} </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>

<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> \approx 6.0 \times 10^{15\,151\,335} </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> \approx 5.6 \times 10^{35\,942\,384} </math>

|- align="center"

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

We define a function <math>f=f(u,k)</math> that computes the hereditary base <math>k</math> representation of <math>u</math> and then replaces each occurrence of the base <math>k</math> with the first infinite ordinal number <math>\omega</math>. For example, <math>f(100,3)=f(3^{3^1+1}+2\cdot3^2+1,3)=\omega^{\omega^1+1} + \omega^2\cdot2 + 1 = \omega^{\omega+1} + \omega^2\cdot2 + 1</math>.

Each term <math>P_m(n)</math> of the sequence <math>P_m</math> is then defined as <math>f(G_m(n),n+1)</math>. For example, <math>G_3(1) = 3 = 2^1 + 2^0</math> and <math>P_3(1) = f(2^1 + 2^0,2) = \omega^1 + \omega^0 = \omega + 1</math>. Addition, multiplication and exponentiation of ordinal numbers are well defined.

We claim that <math>f(G_m(n),n+1) > f(G_m(n+1),n+2)</math>:

Let <math>G'_m(n)</math> be <math>G_m(n)</math> 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 <math>G_m(n+1)= G'_m(n)-1</math>.

Then <math>f(G_m(n),n+1) = f(G'_m(n),n+2)</math>.

References

Bibliography

External links

Some elements of a proof that Goodstein's theorem is not a theorem of PA, from an undergraduate thesis by Justin T Miller
A Classification of non standard models of Peano Arithmetic by Goodstein's theorem - Thesis by Dan Kaplan, Franklan and Marshall College Library
Definition of Goodstein sequences in Haskell and the lambda calculus
Goodstein Sequences: The Power of a Detour via Infinity - good exposition with illustrations of Goodstein Sequences and the hydra game.
Goodstein Calculator