From: "Alberto Monteiro" <[EMAIL PROTECTED]>
> Robert Shaw wrote:
> > If you start with 3 you can only generate a countable infinity
> > of numbers. Take the factorial n times, then the square root
> > m times. Because of the sparsity of the factorials I'd suspect
> > this set isn't dense.
> >
> My intuition says that this set *is* dense - of course, it's
> hard to prove stuff with numbers in the range of 3!!!!!!!
True.
> > Generally, composing just two functions will give an uncountable
> >
> Countable :-)
>
Uncountable, if you allow infinite sequences.
If you've got functions f and g you can do the
diagonal argument on any list of sequences.
> > infinity of numbers which I think means the set of numbers
> > must be dense over a nonzero interval.
> >
> This is not true in general; if you take off an interval
> from any dense set, the resulting set isn't dense.
>
I'm not clear what you mean here.
Can you give a uncountable set of numbers whose
closure doesn't include any connected sets?
> > Sometimes.
> > Galois proved that from x+1, x^(1/2), x^(1/3) and x^(1/5)
> > you can't get the solution of x^5+x=1.
> >
> It was stronger than that
Much stronger, the solutions of almost all polynomials
aren't constructable even with all nth roots available.
> - and x^5+x = 1 *has*
> a simple solution, because x^5 + x - 1 = x^5 + x^2 - x^2 + x - 1 =
> x^2 (x + 1) (x^2 - x + 1) - (x^2 - x + 1) etc.
>
Special cases, x^5+x=pi doesn't.
> >
> > We can add one because of the trigometric identities
> > but multiplying would seem to require an infinite
> > set of such identities. I'd suspect there isn't any finite
> > set of functions whose compositions include n*x for
> > all integer n.
> >
> Hmmm.... I'm not so sure. Those trigonometric functions
> always find funny ways of cancelling each other :-)
>
Which is why I'd like to see a proof.
> Ok, non-cheating. Let's try to construct 7^(1/7)
>
