Undecidability (was reasoning)

[ER <ression@hotmail.com>]

--- In HPFGU-OTChatter at yahoogroups.com, "David <dfrankiswork at n...>" 
<dfrankiswork at n...> wrote:

>For suppose we add the axiom that it is false. [snip] Which means 
>that adding the axiom that the GC is false means that the GC can't 
>be undecidable. Thus we have proved, by contradiction, that if GC is 
>undecidable it is true.

You've lost me here - I don't understand where proof by contradiction 
(Reductio ad absurdum) comes into it? Clearly, if GC is false it is 
decidable. Therefore it cannot be false and undecidable. Thus, if it 
is undecidable, it must be true. Which is the conclusion you gave, 
it's just that I don't understand where RAD (RAA?) comes into it?

I'm also intrigued that if is undecidable, then we know it's true - 
doesn't that make it decidable! Er ...

I also think that one can't add an axiom such as "GC is false". It 
wouldn't be considered cricket. Axioms have to be self-evident and 
internally-consistent, and "GC is false" doesn't fit the bill.

So, back to my original (hello Heidi!) question (which you may have 
answered, but if so it went over my head) ... take the Continuum 
Hypothesis (CH) as an example (it's Sunday night, I've had a glass or 
two of red, so why not). It is, as you said, undecidable. I think it 
was Cohen who put the final nail in the coffin, but Godel had a good 
go as well - he showed that it's impossible to prove that the size is 
not Aleph-1. Anyway, under Zermelo-Fraenkel (ZF) we can never know 
its size for sure. The thrust of my question was (and I probably put 
it badly, certainly "dream up" was a poor choice of words, one must 
select very carefully indeed) - can one introduce a new, self-
evident, axiom to ZF that allows CH to be decidable? Can this process 
be extended to any axiom-based regime such that one can eventually 
construct systems to allow any theorem to be decidable? Or will some 
things remain undecidable no matter what we do?

Am I making sense? Probably not. Now my head does hurt :)

Casual readers, who have not being paying attention, should note that 
if -

(a) GC is true and decidable, HH is a sure-bet. I think the smart 
money is on this :)

(b) GC is true, but undecidable, RH is coming up fast on the outside.

(c) GC is false (and thus decidable), Hermione and the Trolley Lady 
are linked in some mysterious way.

Isn't mathematics wonderful!


ER