The theorem states that every Goodstein sequence (see below) eventually terminates at 0.
Table of contents |
2 Examples of Goodstein sequences 3 Proof |
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 after approximately 2.6 × 1060605351 steps, the elements begin to decrease again, and eventually get to 0. 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 |
---|---|
222+2+1 | 19 |
333+3 | 7625597484990 |
444+3 | approximately 1.3 × 10154 |
555+2 | approximately 1.8 × 102184 |
666+1 | approximately 2.6 × 1036305 |
777 | approximately 3.8 × 10695974 |
7·87·87 +
7·86 +
7·85 +
7·84 +
7·83 +
7·82 +
7·8 + 7
+ 7·87·87 + 7·86 + 7·85 + 7·84 + 7·83 + 7·82 + 7·8 + 6 +
... + 7·82 + 7·8 + 7
|
approximately 6 × 1015151335 |
7·97·97 +
7·96 +
7·95 +
7·94 +
7·93 +
7·92 +
7·9 + 7
+ 7·97·97 + 7·96 + 7·95 + 7·94 + 7·93 + 7·92 + 7·9 + 6 +
... + 7·92 + 7·9 + 6
|
approximately 4.3 × 10369693099 |
... |
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.
Goodstein's theorem can be proved (using techniques outside Peano
arithmetic, see below) as follows: Given a Goodstein sequence G(m), we will contruct 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.
Proof