Expand Messages
• ... I m not sure that you can clearly define irreducible in the logical order . Or what you mean by using 1 theorem only once. After all, I m sure that most
Message 1 of 12 , May 1, 2001
• 0 Attachment
> Yes, I see it now,
> (p + 1)^p == p + 1 mod p
> (p^p + 1)^p == p^p + 1 mod p etc
>
> So we have a strange object, an open `closed' form. The closure
> here being only in the logical order, and semi-open in the algebraic
> order (extended via the theorem). I did say the theory starts.
>
>
> Of course you can only use 1 theorem once in any form, no repeated
> applications etc. Forms must be irreducible in the logical as well
> as algebraic order. Once again the point of this is that we can

I'm not sure that you can clearly define "irreducible in the logical
order". Or what you mean by using 1 theorem only once. After all,
I'm sure that most of the methods we use actually use certain
theorems meny times over: Example

Theorem: for any integers x and y, x*y = y*x.

Does that mean they are discounted?

To my mind, every theorem is reducible in a logical sense, except
of course, the axioms of the logical system...

Yours, Mike H...

> now produce ever larger numbers of which we know the factor.
> There is an error in my previous posting, q = 3*p + 1 is difficult
> for q and p odd prime. If q = k*p +m in a Generalised Sophie
> Germain then k and m have opposite parity
>
> regards,
>
> Paul Mills
>
>
>
>
>
> Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
> The Prime Pages : http://www.primepages.org
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>

Michael Hartley : Michael.Hartley@...