Mathematical logic and foundations Arithmetic and incompleteness Theorem
Goodstein's theorem
Builds on Peano axiomsOrdinal number
Statement
Let be an integer. The hereditary base- notation of writes as a sum with and . Then it writes each exponent in hereditary base- notation again. Let be the number obtained by replacing every in this notation by , with . The Goodstein sequence of is and while .
- For every the Goodstein sequence of reaches .
- Statement 1, written as a sentence of first-order arithmetic, is not provable in Peano arithmetic .
Remarks
Proof idea for 1: replace each base in the hereditary notation of by the ordinal . This gives an ordinal , and the change of base does not change it while subtracting makes it strictly smaller. A strictly decreasing sequence of ordinals is finite, so the sequence reaches . Goodstein proved part 1 in 1944; Kirby and Paris proved part 2 in 1982, by showing that plus statement 1 proves the consistency of .
Example: for , the sequence is , then , then in base , becomes , then , , . For the sequence also reaches , but only after more than steps.
Going deeper
For a student
The theorem shows how ordinals below measure the strength of : transfinite induction up to epsilon nought proves it, while does not. This ties it to Gentzen's consistency proof and to ordinal analysis. Like the Paris–Harrington theorem, it is a natural true statement independent of .
Reading path
To understand this page from the ground up, read these 3 pages in order. The foundations come first.
Show the path as a graph
- no evidence level
- Ordinal number concept
- Peano axioms concept
- Goodstein's theorem statement
Further reading
About this page
| Tag | XQMNJF241J |
|---|---|
| URI | m2://XQMNJF241J |
| Kind | statement, theorem |
| Writing | W0 stub |
| MSC | 03F30, 03F15 |
| Status | This copy of the site has no build data. Run m2 status XQMNJF241J. |
| About | Peano axioms |
| Uses | Peano axioms, Ordinal number |
| Same as | |
| Changed | 2 November 2026 |
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