# Re: MATH/COMP/PHIL: "Omega Man"

From: Lee Corbin (lcorbin@ricochet.net)
Date: Tue Apr 03 2001 - 23:33:19 MDT

Hal wrote

>Lee writes:
>
>> Here is a puzzle that I formulated on this subject that's fun.
>> A great mathematician is at a party, and two students come
>> up to him. One says, "I have found a proof that Goldbach's
>> Conjecture is unprovable!". The mathematician snorts, "Go
>> away, you crackpot". Then the other student says, "I have
>> found a proof that whether or not there exist infinitely
>> many prime pairs is unprovable!", and the mathematician's
>> eyes light up and he says "Oh, really!? Please tell me
>> more!". Why was the mathematician eager to hear out the
>> second student, but not the first?
>
>Goldbach's Conjecture is that every even number is the sum of two primes.
>If it is false, there must be an even number which is not the sum of
>two primes. If such a number exists, then by showing it we could show
>that GC is false and GC would not be unprovable (that is, it would be
>provably false). For GC to be unprovable, therefore, would mean that
>no such number exists. But this is simply the statement of GC itself,
>so to show it to be unprovable (undecidable) is to show it to be true.
>Hence you can't show GC to be unprovable.

This is what I had in mind. In other words, suppose that
there was a proof that GC could not be proved or disproved.
Then since it could not be disproved, it means that you
could not find any such even number. But since one might
pick any even number, that in turn would imply that there
is no even number that is not the sum of two primes. But
that would prove the theorem.

>However we can imagine a weaker set of axioms, one which does not allow
>us to make all the proofs we have now. I don't know enough mathematics
>to give a specific example, but by analogy think of Euclidean geometry
>without the parallel postulate. You can still make a lot of proofs,
>just not all of them.
>
>It's conceivable that in this weaker axiom system, we could show that
>GC is unprovable.

But I would suppose that in the weaker system, you would still
have even numbers, and so might possibly disprove the conjecture
by picking one and exhaustively (but finitely) trying all prime
sums below it. So I don't see the analogy.

>Now, how does this fit with the original argument above, which seemed
>sound? Again, think of the analogy to Euclidean geometry minus the
>parallel postulate. For a long time it was thought that this set of
>axioms might fully characterize the plane, and people spent years trying

>to derive the parallel postulate from the others. However it turned out
>that this was wrong; this truncated geometry does not fully characterize
>the plane. There are non-planar systems which satisfy the remaining
>axioms but which behave differently with respect to parallel lines.
>The truncated axiom system leaves a certain ambiguity with respect to
>certain kinds of questions.

The first four postulates together with the fifth do characterize
the "plane" and nowadays this is what we mean by "plane". On a
plane, it's true that there exists only one parallel line through
a point parallel to a given line.

>After all, intelligent men for centuries still thought that the parallel
>postulate might follow from the others. We could be making a similar
>mistake about the completeness of our arithemetical axiom systems.

Indeed we do or are, in the following sense. No one has yet
written down axioms that capture the integers exactly as we
know them. So I submit that integers are unlike the plane
in this sense. (The Peano postulates, for example, do not
exclude the existence of super big integers that are like
ordinary integers except that if you start counting from 1,
you can never get to them.)

We seem to have been happy to conclude that we had nailed
down our prior conception of "a plane", but we (platonists)
realize that the integers---already out there waiting for
our observations (I am heartened by Chaitin being a platonist,
or so Jim F.'s article would seem to indicate)--- the integers,
have not been captured (yet?).

Lee

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